Insert a rational number between 2 9 \dfrac{2}{9} 9 2 3 8 \dfrac{3}{8} 8 3
Answer
The L.C.M of 9 and 8 is 72.
2 9 = 2 × 8 9 × 8 = 16 72 3 8 = 3 × 9 8 × 9 = 27 72 Since 16 < 27 , 2 9 < 3 8 \dfrac{2}{9} = \dfrac{2 \times 8}{9 \times 8} = \dfrac{16}{72} \\[0.5em] \dfrac{3}{8} = \dfrac{3 \times 9}{8 \times 9} = \dfrac{27}{72} \\[0.5em] \text{Since } 16 \lt 27, \dfrac{2}{9} \lt \dfrac{3}{8} 9 2 = 9 × 8 2 × 8 = 72 16 8 3 = 8 × 9 3 × 9 = 72 27 Since 16 < 27 , 9 2 < 8 3
A rational number between 2 9 \dfrac{2}{9} 9 2 3 8 \dfrac{3}{8} 8 3
= 2 9 + 3 8 2 = 16 + 27 72 2 = 43 144 = \dfrac{\dfrac{2}{9} + \dfrac{3}{8}}{2} \\[0.5em] = \dfrac{\dfrac{16 + 27}{72}}{2} \\[0.5em] = \bold{\dfrac{43}{144}} \\[0.5em] = 2 9 2 + 8 3 = 2 72 16 + 27 = 144 43
Numbers in descending order are:
3 8 , 43 144 , 2 9 \bold{\dfrac{3}{8}}, \bold{\dfrac{43}{144}}, \bold{\dfrac{2}{9}} 8 3 , 144 43 , 9 2
Insert two rational numbers between 1 3 \dfrac{1}{3} 3 1 1 4 \dfrac{1}{4} 4 1
Answer
The L.C.M of 3 and 4 is 12.
1 3 = 1 × 4 3 × 4 = 4 12 1 4 = 1 × 3 4 × 3 = 3 12 Since 3 < 4 , 1 4 < 1 3 \dfrac{1}{3} = \dfrac{1 \times 4}{3 \times 4} = \dfrac{4}{12} \\[0.5em] \dfrac{1}{4} = \dfrac{1 \times 3}{4 \times 3} = \dfrac{3}{12} \\[0.5em] \text{Since } 3 \lt 4, \dfrac{1}{4} \lt \dfrac{1}{3} 3 1 = 3 × 4 1 × 4 = 12 4 4 1 = 4 × 3 1 × 3 = 12 3 Since 3 < 4 , 4 1 < 3 1
A rational number between 1 4 \dfrac{1}{4} 4 1 1 3 \dfrac{1}{3} 3 1
= 1 4 + 1 3 2 = 4 + 3 12 2 = 7 24 = \dfrac{\dfrac{1}{4} + \dfrac{1}{3}}{2} \\[0.5em] = \dfrac{\dfrac{4 + 3}{12}}{2} \\[0.5em] = \bold{\dfrac{7}{24}} \\[0.5em] = 2 4 1 + 3 1 = 2 12 4 + 3 = 24 7
A rational number between 1 4 \dfrac{1}{4} 4 1 7 24 \dfrac{7}{24} 24 7
= 1 4 + 7 24 2 = 6 + 7 24 2 = 13 48 = \dfrac{\dfrac{1}{4} + \dfrac{7}{24}}{2} \\[0.5em] = \dfrac{\dfrac{6 + 7}{24}}{2} \\[0.5em] = \bold{\dfrac{13}{48}} \\[0.5em] = 2 4 1 + 24 7 = 2 24 6 + 7 = 48 13
Numbers in ascending order are:
1 4 , 13 48 , 7 24 , 1 3 \bold{\dfrac{1}{4}}, \bold{\dfrac{13}{48}}, \bold{\dfrac{7}{24}}, \bold{\dfrac{1}{3}} 4 1 , 48 13 , 24 7 , 3 1
Insert two rational numbers between − 1 3 -\dfrac{1}{3} − 3 1 − 1 2 -\dfrac{1}{2} − 2 1
Answer
The L.C.M of 2 and 3 is 6.
− 1 3 = − 1 × 2 3 × 2 = − 2 6 − 1 2 = − 1 × 3 2 × 3 = − 3 6 Since − 3 < − 2 , − 1 2 < − 1 3 -\dfrac{1}{3} = -\dfrac{1 \times 2}{3 \times 2} = -\dfrac{2}{6} \\[0.5em] -\dfrac{1}{2} = -\dfrac{1 \times 3}{2 \times 3} = -\dfrac{3}{6} \\[0.5em] \text{Since } -3 \lt -2, -\dfrac{1}{2} \lt -\dfrac{1}{3} − 3 1 = − 3 × 2 1 × 2 = − 6 2 − 2 1 = − 2 × 3 1 × 3 = − 6 3 Since − 3 < − 2 , − 2 1 < − 3 1
A rational number between -1 2 \dfrac{1}{2} 2 1 1 3 \dfrac{1}{3} 3 1
= − 1 2 + ( − 1 3 ) 2 = − 3 + ( − 2 ) 6 2 = − 3 − 2 6 2 = − 5 12 = \dfrac{-\dfrac{1}{2} + \Big(-\dfrac{1}{3}\Big)}{2} \\[0.5em] = \dfrac{\dfrac{-3 + (-2)}{6}}{2} \\[0.5em] = \dfrac{\dfrac{-3 -2}{6}}{2} \\[0.5em] = \bold{-\dfrac{5}{12}} \\[0.5em] = 2 − 2 1 + ( − 3 1 ) = 2 6 − 3 + ( − 2 ) = 2 6 − 3 − 2 = − 12 5
A rational number between -1 2 \dfrac{1}{2} 2 1 5 12 \dfrac{5}{12} 12 5
= − 1 2 + ( − 5 12 ) 2 = − 6 + ( − 5 ) 12 2 = − 6 − 5 12 2 = − 11 24 = \dfrac{-\dfrac{1}{2} + \Big(-\dfrac{5}{12}\Big)}{2} \\[0.5em] = \dfrac{\dfrac{-6 + (-5)}{12}}{2} \\[0.5em] = \dfrac{\dfrac{-6 -5}{12}}{2} \\[0.5em] = \bold{-\dfrac{11}{24}} \\[0.5em] = 2 − 2 1 + ( − 12 5 ) = 2 12 − 6 + ( − 5 ) = 2 12 − 6 − 5 = − 24 11
Numbers in ascending order are:
− 1 2 , − 11 24 , − 5 12 , − 1 3 \bold{-\dfrac{1}{2}}, \bold{-\dfrac{11}{24}}, \bold{-\dfrac{5}{12}}, \bold{-\dfrac{1}{3}} − 2 1 , − 24 11 , − 12 5 , − 3 1
Insert three rational numbers between 1 3 \dfrac{1}{3} 3 1 4 5 \dfrac{4}{5} 5 4
Answer
The L.C.M of 3 and 5 is 15.
1 3 = 1 × 5 3 × 5 = 5 15 4 5 = 4 × 3 5 × 3 = 12 15 Since 5 < 12 , 1 3 < 4 5 \dfrac{1}{3} = \dfrac{1 \times 5}{3 \times 5} = \dfrac{5}{15} \\[0.5em] \dfrac{4}{5} = \dfrac{4 \times 3}{5 \times 3} = \dfrac{12}{15} \\[0.5em] \text{Since } 5 \lt 12, \dfrac{1}{3} \lt \dfrac{4}{5} 3 1 = 3 × 5 1 × 5 = 15 5 5 4 = 5 × 3 4 × 3 = 15 12 Since 5 < 12 , 3 1 < 5 4
A rational number between 1 3 \dfrac{1}{3} 3 1 4 5 \dfrac{4}{5} 5 4
= 1 3 + 4 5 2 = 5 + 12 15 2 = 17 30 = \dfrac{\dfrac{1}{3} + \dfrac{4}{5}}{2} \\[0.5em] = \dfrac{\dfrac{5 + 12}{15}}{2} \\[0.5em] = \bold{\dfrac{17}{30}} \\[0.5em] = 2 3 1 + 5 4 = 2 15 5 + 12 = 30 17
A rational number between 1 3 \dfrac{1}{3} 3 1 17 30 \dfrac{17}{30} 30 17
= 1 3 + 17 30 2 = 10 + 17 30 2 = 27 60 = \dfrac{\dfrac{1}{3} + \dfrac{17}{30}}{2} \\[0.5em] = \dfrac{\dfrac{10 + 17}{30}}{2} \\[0.5em] = \bold{\dfrac{27}{60}} \\[0.5em] = 2 3 1 + 30 17 = 2 30 10 + 17 = 60 27
A rational number between 17 30 \dfrac{17}{30} 30 17 4 5 \dfrac{4}{5} 5 4
= 17 30 + 4 5 2 = 17 + 24 30 2 = 41 60 = \dfrac{\dfrac{17}{30} + \dfrac{4}{5}}{2} \\[0.5em] = \dfrac{\dfrac{17 + 24}{30}}{2} \\[0.5em] = \bold{\dfrac{41}{60}} \\[0.5em] = 2 30 17 + 5 4 = 2 30 17 + 24 = 60 41
Numbers in descending order are:
4 5 , 41 60 , 17 30 , 27 60 , 1 3 \bold{\dfrac{4}{5}}, \bold{\dfrac{41}{60}}, \bold{\dfrac{17}{30}}, \bold{\dfrac{27}{60}}, \bold{\dfrac{1}{3}} 5 4 , 60 41 , 30 17 , 60 27 , 3 1
Using concept of decimals, insert
(i) three rational numbers between 3 and 3.5
(ii) four rational numbers between 1 4 \dfrac{1}{4} 4 1 2 5 \dfrac{2}{5} 5 2
(iii) five rational numbers between 1 1 2 1\dfrac{1}{2} 1 2 1 1 3 4 1\dfrac{3}{4} 1 4 3
Answer
(i) We want three rational numbers between 3 and 3.5.
We know that,
Terminating decimals are rational numbers.
Let us take 3.1, 3.2, 3.3
Thus, we have :
⇒ 3 < 3.1 < 3.2 < 3.3 <3.5
Hence, three rational numbers between 3 and 3.5 are 3.1, 3.2 and 3.3.
(ii) Converting fraction in decimal form,
1 4 \dfrac{1}{4} 4 1
2 5 \dfrac{2}{5} 5 2
We know that,
Terminating decimals are rational numbers.
Let us take 0.28, 0.30, 0.32, 0.34
Thus, we have :
⇒ 0.25 < 0.28 < 0.30 < 0.32 < 0.34 < 0.40
Hence, four rational numbers between 1 4 and 2 5 \dfrac{1}{4} \text{ and } \dfrac{2}{5} 4 1 and 5 2
(iii) We want five rational numbers between 1 1 2 1\dfrac{1}{2} 1 2 1 1 3 4 1\dfrac{3}{4} 1 4 3
Numbers : 1 1 2 = 3 2 = 1.50 1\dfrac{1}{2} = \dfrac{3}{2} = 1.50 1 2 1 = 2 3 = 1.50 1 3 4 = 7 4 = 1.75 1\dfrac{3}{4} = \dfrac{7}{4} = 1.75 1 4 3 = 4 7 = 1.75
We know that,
Terminating decimals are rational numbers.
Let us take 1.61, 1.62, 1.63, 1.64, 1.65
Thus, we have :
⇒ 1.50 < 1.61 < 1.62 < 1.63 <1.64 < 1.65 <1.75
Hence, five rational numbers between 1 1 2 and 1 3 4 1\dfrac{1}{2} \text{ and } 1\dfrac{3}{4} 1 2 1 and 1 4 3
Find six rational numbers between 3 and 4.
Answer
The numbers 3 and 4 can be written as 3 1 \dfrac{3}{1} 1 3 4 1 \dfrac{4}{1} 1 4
Since we want to find six rational numbers between the given numbers, multiplying the numerator and denominator of the above numbers by 6 + 1 i.e. by 7, we get 21 7 \dfrac{21}{7} 7 21 28 7 \dfrac{28}{7} 7 28
As 21 < 22 < 23 < 24 < 25 < 26 < 27 < 28 , 21 7 < 22 7 < 23 7 < 24 7 < 25 7 < 26 7 < 27 7 < 28 7 \text{As } 21 \lt 22 \lt 23 \lt 24 \lt 25 \lt 26 \lt 27 \lt 28, \\[0.7em] \dfrac{21}{7} \lt \dfrac{22}{7} \lt \dfrac{23}{7} \lt \dfrac{24}{7} \lt \dfrac{25}{7} \lt \dfrac{26}{7} \lt \dfrac{27}{7} \lt \dfrac{28}{7} As 21 < 22 < 23 < 24 < 25 < 26 < 27 < 28 , 7 21 < 7 22 < 7 23 < 7 24 < 7 25 < 7 26 < 7 27 < 7 28
Therefore, six rational numbers between 3 and 4 are:
22 7 , 23 7 , 24 7 , 25 7 , 26 7 , 27 7 \bold{\dfrac{22}{7}}, \bold{\dfrac{23}{7}}, \bold{\dfrac{24}{7}}, \bold{\dfrac{25}{7}}, \bold{\dfrac{26}{7}}, \bold{\dfrac{27}{7}} 7 22 , 7 23 , 7 24 , 7 25 , 7 26 , 7 27
Find five rational numbers between 3 5 \dfrac{3}{5} 5 3 4 5 \dfrac{4}{5} 5 4
Answer
Since we want to find five rational numbers between the given numbers, multiplying the numerator and denominator of the above numbers by 5 + 1 i.e. by 6, we get 18 30 \dfrac{18}{30} 30 18 24 30 \dfrac{24}{30} 30 24
As 18 < 19 < 20 < 21 < 22 < 23 < 24 , 18 30 < 19 30 < 20 30 < 21 30 < 22 30 < 23 30 < 24 30 ⇒ 3 5 < 19 30 < 2 3 < 7 10 < 11 15 < 23 30 < 4 5 \text{As } 18 \lt 19 \lt 20 \lt 21 \lt 22 \lt 23 \lt 24, \\[0.7em] \dfrac{18}{30} \lt \dfrac{19}{30} \lt \dfrac{20}{30} \lt \dfrac{21}{30} \lt \dfrac{22}{30} \lt \dfrac{23}{30} \lt \dfrac{24}{30} \\[0.7em] \Rightarrow \dfrac{3}{5} \lt \dfrac{19}{30} \lt \dfrac{2}{3} \lt \dfrac{7}{10} \lt \dfrac{11}{15} \lt \dfrac{23}{30} \lt \dfrac{4}{5} As 18 < 19 < 20 < 21 < 22 < 23 < 24 , 30 18 < 30 19 < 30 20 < 30 21 < 30 22 < 30 23 < 30 24 ⇒ 5 3 < 30 19 < 3 2 < 10 7 < 15 11 < 30 23 < 5 4
Therefore, six rational numbers between 3 5 \dfrac{3}{5} 5 3 4 5 \dfrac{4}{5} 5 4
19 30 , 2 3 , 7 10 , 11 15 , 23 30 \bold{\dfrac{19}{30}}, \bold{\dfrac{2}{3}}, \bold{\dfrac{7}{10}}, \bold{\dfrac{11}{15}}, \bold{\dfrac{23}{30}} 30 19 , 3 2 , 10 7 , 15 11 , 30 23
Find ten rational numbers between − 2 5 -\dfrac{2}{5} − 5 2 1 7 \dfrac{1}{7} 7 1
Answer
Writing the given numbers with same denominator 35 (L.C.M. of 5 and 7), we get:
− 2 5 = − 14 35 1 7 = 5 35 -\dfrac{2}{5} = -\dfrac{14}{35} \\[0.5em] \dfrac{1}{7} = \dfrac{5}{35} − 5 2 = − 35 14 7 1 = 35 5
As − 14 < − 13 < − 12 < − 11 < − 10 < − 9 < − 8 < − 7 < 0 < 1 < 2 < 5 , − 14 35 < − 13 35 < − 12 35 < − 11 35 < − 10 35 < − 9 35 < − 8 35 < − 7 35 < 0 < 1 35 < 2 35 < 5 35 ⇒ − 2 5 < − 13 35 < − 12 35 < − 11 35 < − 2 7 < − 9 35 < − 8 35 < − 1 5 < 0 < 1 35 < 2 35 < 5 35 \text{As } -14 \lt -13 \lt -12 \lt -11 \lt -10 \lt -9 \lt -8 \lt -7 \lt 0 \lt 1 \lt 2 \lt 5, \\[0.7em] -\dfrac{14}{35} \lt -\dfrac{13}{35} \lt -\dfrac{12}{35} \lt -\dfrac{11}{35} \lt -\dfrac{10}{35} \lt -\dfrac{9}{35} \lt -\dfrac{8}{35} \lt -\dfrac{7}{35} \lt 0 \lt \dfrac{1}{35} \lt \dfrac{2}{35} \lt \dfrac{5}{35} \\[0.7em] \Rightarrow -\dfrac{2}{5} \lt -\dfrac{13}{35} \lt -\dfrac{12}{35} \lt -\dfrac{11}{35} \lt -\dfrac{2}{7} \lt -\dfrac{9}{35} \lt -\dfrac{8}{35} \lt -\dfrac{1}{5} \lt 0 \lt \dfrac{1}{35} \lt \dfrac{2}{35} \lt \dfrac{5}{35} As − 14 < − 13 < − 12 < − 11 < − 10 < − 9 < − 8 < − 7 < 0 < 1 < 2 < 5 , − 35 14 < − 35 13 < − 35 12 < − 35 11 < − 35 10 < − 35 9 < − 35 8 < − 35 7 < 0 < 35 1 < 35 2 < 35 5 ⇒ − 5 2 < − 35 13 < − 35 12 < − 35 11 < − 7 2 < − 35 9 < − 35 8 < − 5 1 < 0 < 35 1 < 35 2 < 35 5
Therefore, ten rational numbers between − 2 5 -\dfrac{2}{5} − 5 2 1 7 \dfrac{1}{7} 7 1
− 13 35 , − 12 35 , − 11 35 , − 2 7 , − 9 35 , − 8 35 , − 1 5 , 0 , 1 35 , 2 35 \bold{-\dfrac{13}{35}}, \bold{-\dfrac{12}{35}}, \bold{-\dfrac{11}{35}}, \bold{-\dfrac{2}{7}}, \bold{-\dfrac{9}{35}}, \\[0.5em] \bold{-\dfrac{8}{35}}, \bold{-\dfrac{1}{5}}, 0, \bold{\dfrac{1}{35}}, \bold{\dfrac{2}{35}} − 35 13 , − 35 12 , − 35 11 , − 7 2 , − 35 9 , − 35 8 , − 5 1 , 0 , 35 1 , 35 2
Find six rational numbers between 1 2 \dfrac{1}{2} 2 1 2 3 \dfrac{2}{3} 3 2
Answer
Writing the given numbers with same denominator 6 (L.C.M. of 2 and 3), we get:
1 2 = 3 6 2 3 = 4 6 \dfrac{1}{2} = \dfrac{3}{6} \\[0.5em] \dfrac{2}{3} = \dfrac{4}{6} 2 1 = 6 3 3 2 = 6 4
Since we want to find six rational numbers between the given numbers, multiplying the numerator and denominator of the above numbers by 6 + 1 i.e. by 7, we get 21 42 \dfrac{21}{42} 42 21 28 42 \dfrac{28}{42} 42 28
As 21 < 22 < 23 < 24 < 25 < 26 < 27 < 28 , 21 42 < 22 42 < 23 42 < 24 42 < 25 42 < 26 42 < 27 42 < 28 42 ⇒ 1 2 < 11 21 < 23 42 < 4 7 < 25 42 < 13 21 < 9 14 < 2 3 \text{As } 21 \lt 22 \lt 23 \lt 24 \lt 25 \lt 26 \lt 27 \lt 28, \\[0.7em] \dfrac{21}{42} \lt \dfrac{22}{42} \lt \dfrac{23}{42} \lt \dfrac{24}{42} \lt \dfrac{25}{42} \lt \dfrac{26}{42} \lt \dfrac{27}{42} \lt \dfrac{28}{42} \\[0.7em] \Rightarrow \dfrac{1}{2} \lt \dfrac{11}{21} \lt \dfrac{23}{42} \lt \dfrac{4}{7} \lt \dfrac{25}{42} \lt \dfrac{13}{21} \lt \dfrac{9}{14} \lt \dfrac{2}{3} As 21 < 22 < 23 < 24 < 25 < 26 < 27 < 28 , 42 21 < 42 22 < 42 23 < 42 24 < 42 25 < 42 26 < 42 27 < 42 28 ⇒ 2 1 < 21 11 < 42 23 < 7 4 < 42 25 < 21 13 < 14 9 < 3 2
Therefore, six rational numbers between 1 2 \dfrac{1}{2} 2 1 2 3 \dfrac{2}{3} 3 2
11 21 , 23 42 , 4 7 , 25 42 , 13 21 , 9 14 \bold{\dfrac{11}{21}}, \bold{\dfrac{23}{42}}, \bold{\dfrac{4}{7}}, \bold{\dfrac{25}{42}}, \bold{\dfrac{13}{21}}, \bold{\dfrac{9}{14}} 21 11 , 42 23 , 7 4 , 42 25 , 21 13 , 14 9
Prove that 5 \sqrt{5} 5
Answer
Let 5 \sqrt{5} 5
5 = p q , \sqrt{5} = \dfrac{p}{q}, 5 = q p ,
where p, q are integers, q ≠ 0 and p, q have no common factors (except 1)
⇒ 5 = p 2 q 2 ⇒ p 2 = 5 q 2 ....(i) \Rightarrow 5 = \dfrac{p^2}{q^2} \\[0.5em] \Rightarrow p^2 = 5q^2 \qquad \text{....(i)} ⇒ 5 = q 2 p 2 ⇒ p 2 = 5 q 2 ....(i)
As 5 divides 5q2 , so 5 divides p2 but 5 is prime
⇒ 5 divides p (Theorem 1) \Rightarrow 5 \text{ divides } p \qquad \text{(Theorem 1)} ⇒ 5 divides p (Theorem 1)
Let p = 5m, where m is an integer.
Substituting this value of p in (i), we get
( 5 m ) 2 = 5 q 2 ⇒ 25 m 2 = 5 q 2 ⇒ 5 m 2 = q 2 (5m)^2 = 5q^2 \\[0.5em] \Rightarrow 25m^2 = 5q^2 \\[0.5em] \Rightarrow 5m^2 = q^2 \\[0.5em] ( 5 m ) 2 = 5 q 2 ⇒ 25 m 2 = 5 q 2 ⇒ 5 m 2 = q 2
As 5 divides 5m2 , so 5 divides q2 but 5 is prime
⇒ 5 divides q (Theorem 1) \Rightarrow 5 \text{ divides } q \qquad \text{(Theorem 1)} ⇒ 5 divides q (Theorem 1)
Thus, p and q have a common factor 5. This contradicts that p and q have no common factors (except 1).
Hence, 5 \sqrt{5} 5 5 \sqrt{5} 5
Prove that 7 \sqrt{7} 7
Answer
Let 7 \sqrt{7} 7
7 = p q , \sqrt{7} = \dfrac{p}{q}, 7 = q p ,
where p, q are integers, q ≠ 0 and p, q have no common factors (except 1)
⇒ 7 = p 2 q 2 ⇒ p 2 = 7 q 2 ....(i) \Rightarrow 7 = \dfrac{p^2}{q^2} \\[0.5em] \Rightarrow p^2 = 7q^2 \qquad \text{....(i)} ⇒ 7 = q 2 p 2 ⇒ p 2 = 7 q 2 ....(i)
As 7 divides 7q2 , so 7 divides p2 but 7 is prime
⇒ 7 divides p (Theorem 1) \Rightarrow 7 \text{ divides } p \qquad \text{(Theorem 1)} ⇒ 7 divides p (Theorem 1)
Let p = 7m, where m is an integer.
Substituting this value of p in (i), we get
( 7 m ) 2 = 7 q 2 ⇒ 49 m 2 = 7 q 2 ⇒ 7 m 2 = q 2 (7m)^2 = 7q^2 \\[0.5em] \Rightarrow 49m^2 = 7q^2 \\[0.5em] \Rightarrow 7m^2 = q^2 \\[0.5em] ( 7 m ) 2 = 7 q 2 ⇒ 49 m 2 = 7 q 2 ⇒ 7 m 2 = q 2
As 7 divides 7m2 , so 7 divides q2 but 7 is prime
⇒ 7 divides q (Theorem 1) \Rightarrow 7 \text{ divides } q \qquad \text{(Theorem 1)} ⇒ 7 divides q (Theorem 1)
Thus, p and q have a common factor 7. This contradicts that p and q have no common factors (except 1).
Hence, 7 \sqrt{7} 7 7 \sqrt{7} 7
Prove that 6 \sqrt{6} 6
Answer
Suppose that 6 \sqrt{6} 6
6 = p q , \sqrt{6} = \dfrac{p}{q}, 6 = q p ,
where p, q are integers, q ≠ 0 and p, q have no common factors (except 1)
⇒ 6 = p 2 q 2 ⇒ p 2 = 6 q 2 ....(i) \Rightarrow 6 = \dfrac{p^2}{q^2} \\[0.5em] \Rightarrow p^2 = 6q^2 \qquad \text{....(i)} ⇒ 6 = q 2 p 2 ⇒ p 2 = 6 q 2 ....(i)
As 2 divides 6q2 , so 2 divides p2 but 2 is prime
⇒ 2 divides p (Theorem 1) \Rightarrow 2 \text{ divides } p \qquad \text{(Theorem 1)} ⇒ 2 divides p (Theorem 1)
Let p = 2k, where k is some integer.
Substituting this value of p in (i), we get
( 2 k ) 2 = 6 q 2 ⇒ 4 k 2 = 6 q 2 ⇒ 2 k 2 = 3 q 2 (2k)^2 = 6q^2 \\[0.5em] \Rightarrow 4k^2 = 6q^2 \\[0.5em] \Rightarrow 2k^2 = 3q^2 \\[0.5em] ( 2 k ) 2 = 6 q 2 ⇒ 4 k 2 = 6 q 2 ⇒ 2 k 2 = 3 q 2
As 2 divides 2k2 , so 2 divides 3q2
⇒ \Rightarrow ⇒ 2
But 2 does not divide 3, therefore, 2 divides q2
⇒ \Rightarrow ⇒
Thus, p and q have a common factor 2. This contradicts that p and q have no common factors (except 1).
Hence, our supposition is wrong. Therefore, 6 \sqrt{6} 6 6 \sqrt{6} 6
Prove that 1 11 \dfrac{1}{\sqrt{11}} 11 1
Answer
Let 1 11 \dfrac{1}{\sqrt{11}} 11 1
1 11 = p q , \dfrac{1}{\sqrt{11}} = \dfrac{p}{q}, 11 1 = q p ,
where p, q are integers, q ≠ 0 and p, q have no common factors (except 1)
⇒ 1 11 = p 2 q 2 ⇒ q 2 = 11 p 2 ....(i) \Rightarrow \dfrac{1}{11} = \dfrac{p^2}{q^2} \\[0.5em] \Rightarrow q^2 = 11p^2 \qquad \text{....(i)} ⇒ 11 1 = q 2 p 2 ⇒ q 2 = 11 p 2 ....(i)
As 11 divides 11p2 , so 11 divides q2 but 11 is prime
⇒ 11 divides q (Theorem 1) \Rightarrow 11 \text{ divides } q \qquad \text{(Theorem 1)} ⇒ 11 divides q (Theorem 1)
Let q = 11m, where m is an integer.
Substituting this value of q in (i), we get
( 11 m ) 2 = 11 p 2 ⇒ 121 m 2 = 11 p 2 ⇒ 11 m 2 = p 2 (11m)^2 = 11p^2 \\[0.5em] \Rightarrow 121m^2 = 11p^2 \\[0.5em] \Rightarrow 11m^2 = p^2 \\[0.5em] ( 11 m ) 2 = 11 p 2 ⇒ 121 m 2 = 11 p 2 ⇒ 11 m 2 = p 2
As 11 divides 11m2 , so 11 divides p2 but 11 is prime
⇒ 11 divides p (Theorem 1) \Rightarrow 11 \text{ divides } p \qquad \text{(Theorem 1)} ⇒ 11 divides p (Theorem 1)
Thus, p and q have a common factor 11. This contradicts that p and q have no common factors (except 1).
Hence, 1 11 \dfrac{1}{\sqrt{11}} 11 1 1 11 \dfrac{1}{\sqrt{11}} 11 1
Prove that 2 \sqrt{2} 2 3 − 2 3 - \sqrt{2} 3 − 2
Answer
Let 2 \sqrt{2} 2
2 = p q , \sqrt{2} = \dfrac{p}{q}, 2 = q p ,
where p, q are integers, q ≠ 0 and p, q have no common factors (except 1)
⇒ 2 = p 2 q 2 ⇒ p 2 = 2 q 2 ....(i) \Rightarrow 2 = \dfrac{p^2}{q^2} \\[0.5em] \Rightarrow p^2 = 2q^2 \qquad \text{....(i)} ⇒ 2 = q 2 p 2 ⇒ p 2 = 2 q 2 ....(i)
As 2 divides 2q2 , so 2 divides p2 but 2 is prime
⇒ 2 divides p (Theorem 1) \Rightarrow 2 \text{ divides } p \qquad \text{(Theorem 1)} ⇒ 2 divides p (Theorem 1)
Let p = 2m, where m is an integer.
Substituting this value of p in (i), we get
( 2 m ) 2 = 2 q 2 ⇒ 4 m 2 = 2 q 2 ⇒ 2 m 2 = q 2 (2m)^2 = 2q^2 \\[0.5em] \Rightarrow 4m^2 = 2q^2 \\[0.5em] \Rightarrow 2m^2 = q^2 \\[0.5em] ( 2 m ) 2 = 2 q 2 ⇒ 4 m 2 = 2 q 2 ⇒ 2 m 2 = q 2
As 2 divides 2m2 , so 2 divides q2 but 2 is prime
⇒ 2 divides q (Theorem 1) \Rightarrow 2 \text{ divides } q \qquad \text{(Theorem 1)} ⇒ 2 divides q (Theorem 1)
Thus, p and q have a common factor 2. This contradicts that p and q have no common factors (except 1).
Hence, 2 \sqrt{2} 2 2 \sqrt{2} 2
Suppose that 3 − 2 3 - \sqrt{2} 3 − 2
Then, 3 − 2 3 - \sqrt{2} 3 − 2
⇒ − 2 = r − 3 ⇒ 2 = 3 − r \Rightarrow - \sqrt{2} = r - 3 \\[0.5em] \Rightarrow \sqrt{2} = 3 - r \\[0.5em] ⇒ − 2 = r − 3 ⇒ 2 = 3 − r
As r is rational and r ≠ 0, so 3 - r is rational
⇒ 2 \Rightarrow \sqrt{2} ⇒ 2
But this contradicts that 2 \sqrt{2} 2
∴ 3 − 2 3 - \sqrt{2} 3 − 2
Prove that 3 \sqrt{3} 3 2 5 3 \dfrac{2}{5}\sqrt{3} 5 2 3
Answer
Let 3 \sqrt{3} 3
3 = p q , \sqrt{3} = \dfrac{p}{q}, 3 = q p ,
where p, q are integers, q ≠ 0 and p, q have no common factors (except 1)
⇒ 3 = p 2 q 2 ⇒ p 2 = 3 q 2 ....(i) \Rightarrow 3 = \dfrac{p^2}{q^2} \\[0.5em] \Rightarrow p^2 = 3q^2 \qquad \text{....(i)} ⇒ 3 = q 2 p 2 ⇒ p 2 = 3 q 2 ....(i)
As 3 divides 3q2 , so 3 divides p2 but 3 is prime
⇒ 3 divides p (Theorem 1) \Rightarrow 3 \text{ divides } p \qquad \text{(Theorem 1)} ⇒ 3 divides p (Theorem 1)
Let p = 3m, where m is an integer.
Substituting this value of p in (i), we get
( 3 m ) 2 = 3 q 2 ⇒ 9 m 2 = 3 q 2 ⇒ 3 m 2 = q 2 (3m)^2 = 3q^2 \\[0.5em] \Rightarrow 9m^2 = 3q^2 \\[0.5em] \Rightarrow 3m^2 = q^2 \\[0.5em] ( 3 m ) 2 = 3 q 2 ⇒ 9 m 2 = 3 q 2 ⇒ 3 m 2 = q 2
As 3 divides 3m2 , so 3 divides q2 but 3 is prime
⇒ 3 divides q (Theorem 1) \Rightarrow 3 \text{ divides } q \qquad \text{(Theorem 1)} ⇒ 3 divides q (Theorem 1)
Thus, p and q have a common factor 3. This contradicts that p and q have no common factors (except 1).
Hence, 3 \sqrt{3} 3 3 \sqrt{3} 3
Suppose that 2 5 3 \dfrac{2}{5}\sqrt{3} 5 2 3
Then, 2 5 3 \dfrac{2}{5}\sqrt{3} 5 2 3
⇒ 3 = 5 2 r \Rightarrow \sqrt{3} = \dfrac{5}{2}r \\[0.5em] ⇒ 3 = 2 5 r
As r is rational and r ≠ 0, so 5 2 r \dfrac{5}{2}r 2 5 r
⇒ 3 \Rightarrow \sqrt{3} ⇒ 3
But this contradicts that 3 \sqrt{3} 3
∴ 2 5 3 \dfrac{2}{5}\sqrt{3} 5 2 3
Prove that 5 \sqrt{5} 5 − 3 + 2 5 -3 + 2\sqrt{5} − 3 + 2 5
Answer
Let 5 \sqrt{5} 5
5 = p q , \sqrt{5} = \dfrac{p}{q}, 5 = q p ,
where p, q are integers, q ≠ 0 and p, q have no common factors (except 1)
⇒ 5 = p 2 q 2 ⇒ p 2 = 5 q 2 \Rightarrow 5 = \dfrac{p^2}{q^2} \\[0.5em] \Rightarrow p^2 = 5q^2 ⇒ 5 = q 2 p 2 ⇒ p 2 = 5 q 2
As 5 divides 5q2 , so 5 divides p2 but 5 is prime
⇒ 5 divides p (Theorem 1) \Rightarrow 5 \text{ divides } p \qquad \text{(Theorem 1)} ⇒ 5 divides p (Theorem 1)
Let p = 5m, where m is an integer.
Substituting this value of p in (i), we get
( 5 m ) 2 = 5 q 2 ⇒ 25 m 2 = 5 q 2 ⇒ 5 m 2 = q 2 (5m)^2 = 5q^2 \\[0.5em] \Rightarrow 25m^2 = 5q^2 \\[0.5em] \Rightarrow 5m^2 = q^2 \\[0.5em] ( 5 m ) 2 = 5 q 2 ⇒ 25 m 2 = 5 q 2 ⇒ 5 m 2 = q 2
As 5 divides 5m2 , so 5 divides q2 but 5 is prime
⇒ 5 divides q (Theorem 1) \Rightarrow 5 \text{ divides } q \qquad \text{(Theorem 1)} ⇒ 5 divides q (Theorem 1)
Thus, p and q have a common factor 5. This contradicts that p and q have no common factors (except 1).
Hence, 5 \sqrt{5} 5 5 \sqrt{5} 5
Suppose that − 3 + 2 5 -3 + 2\sqrt{5} − 3 + 2 5
Then, − 3 + 2 5 -3 + 2\sqrt{5} − 3 + 2 5
⇒ 2 5 = r + 3 ⇒ 5 = r + 3 2 \Rightarrow 2\sqrt{5} = r + 3 \\[0.5em] \Rightarrow \sqrt{5} = \dfrac{r + 3}{2} \\[0.5em] ⇒ 2 5 = r + 3 ⇒ 5 = 2 r + 3
As r is rational and r ≠ 0, so r + 3 2 \dfrac{r + 3}{2} 2 r + 3
⇒ 5 \Rightarrow \sqrt{5} ⇒ 5
But this contradicts that 5 \sqrt{5} 5
∴ − 3 + 2 5 -3 + 2\sqrt{5} − 3 + 2 5
Prove that the following numbers are irrational:
(i) 5 + 2 (ii) 3 − 5 3 (iii) 2 3 − 7 (iv) 2 + 5 \begin{matrix} \text{(i)} & 5 + \sqrt{2} \\[0.5em] \text{(ii)} & 3 - 5\sqrt{3} \\[0.5em] \text{(iii)} & 2\sqrt{3} - 7 \\[0.5em] \text{(iv)} & \sqrt{2} + \sqrt{5} \end{matrix} (i) (ii) (iii) (iv) 5 + 2 3 − 5 3 2 3 − 7 2 + 5
Answer
(i) 5 + 2 \text{(i) } 5 + \sqrt{2} (i) 5 + 2
Let us assume that 5 + 2 5 + \sqrt{2} 5 + 2
Then,
5 + 2 = r ⇒ 2 = r − 5 5 + \sqrt{2} = r \\[0.5em] \Rightarrow \sqrt{2} = r - 5 5 + 2 = r ⇒ 2 = r − 5
As r is rational, r - 5 is rational
⇒ 2 \Rightarrow \sqrt{2} ⇒ 2
But this contradicts the fact that 2 \sqrt{2} 2
Hence, our assumption is wrong.
∴ 5 + 2 5 + \sqrt{2} 5 + 2
(ii) 3 − 5 3 \text{(ii) } 3 - 5\sqrt{3} (ii) 3 − 5 3
Let us assume that 3 − 5 3 3 - 5\sqrt{3} 3 − 5 3
Then,
3 − 5 3 = r ⇒ 5 3 = 3 − r ⇒ 3 = 3 − r 5 3 - 5\sqrt{3} = r \\[0.5em] \Rightarrow 5\sqrt{3} = 3 - r \\[0.5em] \Rightarrow \sqrt{3} = \dfrac{3 - r}{5} \\[0.5em] 3 − 5 3 = r ⇒ 5 3 = 3 − r ⇒ 3 = 5 3 − r
As r is rational, 3 - r is rational
⇒ 3 − r 5 \Rightarrow \dfrac{3 - r}{5} ⇒ 5 3 − r
⇒ 3 \Rightarrow \sqrt{3} ⇒ 3
But this contradicts the fact that 3 \sqrt{3} 3
Hence, our assumption is wrong.
∴ 3 − 5 3 3 - 5\sqrt{3} 3 − 5 3
(iii) 2 3 − 7 \text{(iii) } 2\sqrt{3} - 7 (iii) 2 3 − 7
Let us assume that 2 3 − 7 2\sqrt{3} - 7 2 3 − 7
Then,
2 3 − 7 = r ⇒ 2 3 = r + 7 ⇒ 3 = r + 7 2 2\sqrt{3} - 7 = r \\[0.5em] \Rightarrow 2\sqrt{3} = r + 7 \\[0.5em] \Rightarrow \sqrt{3} = \dfrac{r + 7}{2} \\[0.5em] 2 3 − 7 = r ⇒ 2 3 = r + 7 ⇒ 3 = 2 r + 7
As r is rational, r + 7 is rational
⇒ r + 7 2 \Rightarrow \dfrac{r + 7}{2} ⇒ 2 r + 7
⇒ 3 \Rightarrow \sqrt{3} ⇒ 3
But this contradicts the fact that 3 \sqrt{3} 3
Hence, our assumption is wrong.
∴ 2 3 − 7 2\sqrt{3} - 7 2 3 − 7
(iv) 2 + 5 \text{(iv) } \sqrt{2} + \sqrt{5} (iv) 2 + 5
Let us assume that 2 + 5 \sqrt{2} + \sqrt{5} 2 + 5
Then,
2 + 5 = r ⇒ 5 = r − 2 ⇒ ( 5 ) 2 = ( r − 2 ) 2 ⇒ 5 = r 2 + 2 − 2 2 r ⇒ 2 2 r = r 2 + 2 − 5 ⇒ 2 2 r = r 2 − 3 ⇒ 2 = r 2 − 3 2 r \sqrt{2} + \sqrt{5} = r \\[0.5em] \Rightarrow \sqrt{5} = r - \sqrt{2} \\[0.5em] \Rightarrow (\sqrt{5})^2 = (r - \sqrt{2})^2 \\[0.5em] \Rightarrow 5 = r^2 + 2 - 2\sqrt{2}r \\[0.5em] \Rightarrow 2\sqrt{2}r = r^2 + 2 - 5 \\[0.5em] \Rightarrow 2\sqrt{2}r = r^2 - 3 \\[0.5em] \Rightarrow \sqrt{2} = \dfrac{r^2 - 3}{2r} \\[0.5em] 2 + 5 = r ⇒ 5 = r − 2 ⇒ ( 5 ) 2 = ( r − 2 ) 2 ⇒ 5 = r 2 + 2 − 2 2 r ⇒ 2 2 r = r 2 + 2 − 5 ⇒ 2 2 r = r 2 − 3 ⇒ 2 = 2 r r 2 − 3
As r is rational,
⇒ r 2 − 3 2 r \Rightarrow \dfrac{r^2 - 3}{2r} ⇒ 2 r r 2 − 3
⇒ 2 \Rightarrow \sqrt{2} ⇒ 2
But this contradicts the fact that 2 \sqrt{2} 2
Hence, our assumption is wrong.
∴ 2 + 5 \sqrt{2} + \sqrt{5} 2 + 5
Locate 10 \sqrt{10} 10 17 \sqrt{17} 17
Answer
Locating 10 \sqrt{10} 10
Representing 10 as the sum of squares of two natural numbers:
10 = 9 + 1 = 32 + 12
Let l be the number line. If point O represents number 0 and point A represents number 3, then draw a line segment OA = 3 units.
At A, draw AC ⟂ OA. From AC, cut off AB = 1 unit.
We observe that OAB is a right angled triangle at A. By Pythagoras theorem, we get:
O B 2 = O A 2 + A B 2 ⇒ O B 2 = 3 2 + 1 2 ⇒ O B 2 = 9 + 1 ⇒ O B 2 = 10 ⇒ O B = 10 units OB^2 = OA^2 + AB^2 \\[0.5em] \Rightarrow OB^2 = 3^2 + 1^2 \\[0.5em] \Rightarrow OB^2 = 9 + 1 \\[0.5em] \Rightarrow OB^2 = 10 \\[0.5em] \Rightarrow OB = \sqrt{10} \text{ units} \\[0.5em] O B 2 = O A 2 + A B 2 ⇒ O B 2 = 3 2 + 1 2 ⇒ O B 2 = 9 + 1 ⇒ O B 2 = 10 ⇒ OB = 10 units
With O as centre and radius = OB, we draw an arc of a circle to meet the number line l at point P.
As OP = OB = 10 \sqrt{10} 10 10 \sqrt{10} 10
Locating 17 \sqrt{17} 17
Representing 17 as the sum of squares of two natural numbers:
17 = 16 + 1 = 42 + 12
Let l be the number line. If point O represents number 0 and point A represents number 4, then draw a line segment OA = 4 units.
At A, draw AC ⟂ OA. From AC, cut off AB = 1 unit.
We observe that OAB is a right angled triangle at A. By Pythagoras theorem, we get:
O B 2 = O A 2 + A B 2 ⇒ O B 2 = 4 2 + 1 2 ⇒ O B 2 = 16 + 1 ⇒ O B 2 = 17 ⇒ O B = 17 units OB^2 = OA^2 + AB^2 \\[0.5em] \Rightarrow OB^2 = 4^2 + 1^2 \\[0.5em] \Rightarrow OB^2 = 16 + 1 \\[0.5em] \Rightarrow OB^2 = 17 \\[0.5em] \Rightarrow OB = \sqrt{17} \text{units} \\[0.5em] O B 2 = O A 2 + A B 2 ⇒ O B 2 = 4 2 + 1 2 ⇒ O B 2 = 16 + 1 ⇒ O B 2 = 17 ⇒ OB = 17 units
With O as centre and radius = OB, we draw an arc of a circle to meet the number line l at point P.
As OP = OB = 17 \sqrt{17} 17 17 \sqrt{17} 17
Write the decimal expansion of each of the following numbers and say what kind of decimal expansion each has:
(i) 36 100 (ii) 4 1 8 (iii) 2 9 (iv) 2 11 (v) 3 13 (vi) 329 400 \begin{matrix} \text{(i)} & \dfrac{36}{100} \\[1.5em] \text{(ii)} & 4\dfrac{1}{8} \\[1.5em] \text{(iii)} & \dfrac{2}{9} \\[1.5em] \text{(iv)} & \dfrac{2}{11} \\[1.5em] \text{(v)} & \dfrac{3}{13} \\[1.5em] \text{(vi)} & \dfrac{329}{400} \\[1.5em] \end{matrix} (i) (ii) (iii) (iv) (v) (vi) 100 36 4 8 1 9 2 11 2 13 3 400 329
Answer
(i) 36 100 \text{(i) } \dfrac{36}{100} (i) 100 36
∴ 36 100 = 0.36 \therefore \dfrac{36}{100} = 0.36 ∴ 100 36 = 0.36
Remainder becomes zero.
Decimal expansion of 36 100 \bold{\dfrac{36}{100}} 100 36
(ii) 4 1 8 \text{(ii) } 4\dfrac{1}{8} (ii) 4 8 1
∴ 4 1 8 = 4.125 \therefore 4\dfrac{1}{8} = 4.125 ∴ 4 8 1 = 4.125
Remainder becomes zero.
Decimal expansion of 4 1 8 \bold{4\dfrac{1}{8}} 4 8 1
(iii) 2 9 \text{(iii) } \dfrac{2}{9} (iii) 9 2
∴ 2 9 = 0.2222..... = 0. 2 ‾ \therefore \dfrac{2}{9} = 0.2222..... = 0.\overline{2} ∴ 9 2 = 0.2222..... = 0. 2
Remainder is repeating.
Decimal expansion of 2 9 \bold{\dfrac{2}{9}} 9 2
(iv) 2 11 \text{(iv) } \dfrac{2}{11} (iv) 11 2
∴ 2 11 = 0.1818..... = 0. 18 ‾ \therefore \dfrac{2}{11} = 0.1818..... = 0.\overline{18} ∴ 11 2 = 0.1818..... = 0. 18
Remainder is repeating.
Decimal expansion of 2 11 \bold{\dfrac{2}{11}} 11 2
(v) 3 13 \text{(v) } \dfrac{3}{13} (v) 13 3
∴ 3 13 = 0.2307692307..... = 0. 230769 ‾ \therefore \dfrac{3}{13} = 0.2307692307..... = 0.\overline{230769} ∴ 13 3 = 0.2307692307..... = 0. 230769
Remainder is repeating.
Decimal expansion of 3 13 \bold{\dfrac{3}{13}} 13 3
(vi) 329 400 \text{(vi) } \dfrac{329}{400} (vi) 400 329
∴ 329 400 = 0.8225 \therefore \dfrac{329}{400} = 0.8225 ∴ 400 329 = 0.8225
Remainder becomes zero.
Decimal expansion of 329 400 \bold{\dfrac{329}{400}} 400 329
Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion:
(i) 13 3125 (ii) 17 8 (iii) 23 75 (iv) 6 15 (v) 1258 625 (vi) 77 210 \begin{matrix} \text{(i)} & \dfrac{13}{3125} \\[1.5em] \text{(ii)} & \dfrac{17}{8} \\[1.5em] \text{(iii)} & \dfrac{23}{75} \\[1.5em] \text{(iv)} & \dfrac{6}{15} \\[1.5em] \text{(v)} & \dfrac{1258}{625} \\[1.5em] \text{(vi)} & \dfrac{77}{210} \\[1.5em] \end{matrix} (i) (ii) (iii) (iv) (v) (vi) 3125 13 8 17 75 23 15 6 625 1258 210 77
Answer
(i) 13 3125 \text{(i) } \dfrac{13}{3125} (i) 3125 13
The given number 13 3125 \dfrac{13}{3125} 3125 13
Prime factorization of denominator 3125:
5 3125 5 625 5 125 5 25 5 5 1 \begin{array}{l|l} 5 & 3125 \\ \hline 5 & 625 \\ \hline 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array} 5 5 5 5 5 3125 625 125 25 5 1
3125 = 5 x 5 x 5 x 5 x 5 x 15 x 15 0 x 55 [∵ 20 = 1]
Denominator is of the form 2m x 5n , where m, n are non-negative integers.
∴ The given number 13 3125 \dfrac{13}{3125} 3125 13
(ii) 17 8 \text{(ii) } \dfrac{17}{8} (ii) 8 17
The given number 17 8 \dfrac{17}{8} 8 17
Prime factorization of denominator 8:
2 8 2 4 2 2 1 \begin{array}{l|l} 2 & 8 \\ \hline 2 & 4 \\ \hline 2 & 2 \\ \hline & 1 \end{array} 2 2 2 8 4 2 1
8 = 2 x 2 x 2 x 13 x 13 x 50 [∵ 50 = 1]
Denominator is of the form 2m x 5n , where m, n are non-negative integers.
∴ The given number 17 8 \dfrac{17}{8} 8 17
(iii) 23 75 \text{(iii) } \dfrac{23}{75} (iii) 75 23
The given number 23 75 \dfrac{23}{75} 75 23
Prime factorization of denominator 75:
3 75 5 25 5 5 1 \begin{array}{l|l} 3 & 75 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array} 3 5 5 75 25 5 1
75 = 3 x 5 x 5 x 12 x 12 x 20 [∵ 20 = 1]
Denominator has a prime factor 3 other than 2 or 5.
∴ The given number 23 75 \dfrac{23}{75} 75 23
(iv) 6 15 \text{(iv) } \dfrac{6}{15} (iv) 15 6
Both numerator and denominator contain common factor 3. Reducing the number to its lowest form:
6 15 = 3 × 2 3 × 5 = 2 5 \dfrac{6}{15} = \dfrac{\cancel{3} \times 2}{\cancel{3} \times 5} \\[0.5em] = \dfrac{2}{5} 15 6 = 3 × 5 3 × 2 = 5 2
The Denominator 5 = 20 x 51
Denominator is of the form 2m x 5n , where m, n are non-negative integers.
∴ The given number 6 15 \dfrac{6}{15} 15 6
(v) 1258 625 \text{(v) } \dfrac{1258}{625} (v) 625 1258
The given number 1258 625 \dfrac{1258}{625} 625 1258
Prime factorization of denominator 625:
5 625 5 125 5 25 5 5 1 \begin{array}{l|l} 5 & 625 \\ \hline 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array} 5 5 5 5 625 125 25 5 1
625 = 5 x 5 x 5 x 5 x 14 x 14 0 x 54 [∵ 20 = 1]
Denominator is of the form 2m x 5n , where m, n are non-negative integers.
∴ The given number 1258 625 \dfrac{1258}{625} 625 1258
(vi) 77 210 \text{(vi) } \dfrac{77}{210} (vi) 210 77
Both numerator and denominator contain common factor 7. Reducing the number to its lowest form:
77 210 = 7 × 11 7 × 30 = 11 30 \dfrac{77}{210} = \dfrac{\cancel{7} \times 11}{\cancel{7} \times 30} \\[0.5em] = \dfrac{11}{30} 210 77 = 7 × 30 7 × 11 = 30 11
Prime factorization of denominator 30:
2 30 3 15 5 5 1 \begin{array}{l|l} 2 & 30 \\ \hline 3 & 15 \\ \hline 5 & 5 \\ \hline & 1 \end{array} 2 3 5 30 15 5 1
30 = 2 x 3 x 5
Denominator has a prime factor 3 other than 2 or 5.
∴ The given number 77 210 \dfrac{77}{210} 210 77
Without actually performing the long division, find if 987 10500 \dfrac{987}{10500} 10500 987
Answer
GCD of numerator and denominator is 21. Reducing the number to its lowest form:
987 10500 = 21 × 47 21 × 500 = 47 500 \dfrac{987}{10500} = \dfrac{\cancel{21} \times 47}{\cancel{21} \times 500} \\[0.5em] = \dfrac{47}{500} 10500 987 = 21 × 500 21 × 47 = 500 47
Prime factorization of denominator 500:
2 500 2 250 5 125 5 25 5 5 1 \begin{array}{l|l} 2 & 500 \\ \hline 2 & 250 \\ \hline 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array} 2 2 5 5 5 500 250 125 25 5 1
500 = 2 x 2 x 5 x 5 x 5 = 22 x 53
Denominator is of the form 2m x 5n , where m, n are non-negative integers.
∴ The given number 987 10500 \dfrac{987}{10500} 10500 987
Write the decimal expansions of the following numbers which have terminating decimal expansions:
(i) 17 8 (ii) 13 3125 (iii) 7 80 (iv) 6 15 (v) 2 2 × 7 5 4 (vi) 237 1500 \begin{matrix} \text{(i)} & \dfrac{17}{8} \\[1.5em] \text{(ii)} & \dfrac{13}{3125} \\[1.5em] \text{(iii)} & \dfrac{7}{80} \\[1.5em] \text{(iv)} & \dfrac{6}{15} \\[1.5em] \text{(v)} & \dfrac{2^2 \times 7}{5^4} \\[1.5em] \text{(vi)} & \dfrac{237}{1500} \\[1.5em] \end{matrix} (i) (ii) (iii) (iv) (v) (vi) 8 17 3125 13 80 7 15 6 5 4 2 2 × 7 1500 237
Answer
(i) 17 8 \text{(i) } \dfrac{17}{8} (i) 8 17
The given number 17 8 \dfrac{17}{8} 8 17
Prime factorization of denominator 8:
2 8 2 4 2 2 1 \begin{array}{l|l} 2 & 8 \\ \hline 2 & 4 \\ \hline 2 & 2 \\ \hline & 1 \end{array} 2 2 2 8 4 2 1
8 = 2 x 2 x 2 x 13 x 50 [∵ 50 = 1]
17 8 = 17 2 3 = 17 × 5 3 2 3 × 5 3 = 17 × 125 ( 2 × 5 ) 3 = 2125 10 3 = 2.125 ∴ 17 8 = 2.125 \dfrac{17}{8} = \dfrac{17}{2^3} \\[0.5em] = \dfrac{17 \times 5^3}{2^3 \times 5^3} \\[0.5em] = \dfrac{17 \times 125}{(2 \times 5)^3} \\[0.5em] = \dfrac{2125}{10^3} \\[0.5em] = 2.125 \\[0.5em] \bold{\therefore \dfrac{17}{8} = 2.125} 8 17 = 2 3 17 = 2 3 × 5 3 17 × 5 3 = ( 2 × 5 ) 3 17 × 125 = 1 0 3 2125 = 2.125 ∴ 8 17 = 2.125
(ii) 13 3125 \text{(ii) } \dfrac{13}{3125} (ii) 3125 13
The given number 13 3125 \dfrac{13}{3125} 3125 13
Prime factorization of denominator 3125:
5 3125 5 625 5 125 5 25 5 5 1 \begin{array}{l|l} 5 & 3125 \\ \hline 5 & 625 \\ \hline 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array} 5 5 5 5 5 3125 625 125 25 5 1
3125 = 5 x 5 x 5 x 5 x 5 x 15 x 15 0 x 55 [∵ 20 = 1]
13 3125 = 13 5 5 = 13 × 2 5 2 5 × 5 5 = 13 × 32 ( 2 × 5 ) 5 = 416 10 5 = 0.00416 ∴ 13 3125 = 0.00416 \dfrac{13}{3125} = \dfrac{13}{5^5} \\[0.5em] = \dfrac{13 \times 2^5}{2^5 \times 5^5} \\[0.5em] = \dfrac{13 \times 32}{(2 \times 5)^5} \\[0.5em] = \dfrac{416}{10^5} \\[0.5em] = 0.00416 \\[0.5em] \bold{\therefore \dfrac{13}{3125} = 0.00416} 3125 13 = 5 5 13 = 2 5 × 5 5 13 × 2 5 = ( 2 × 5 ) 5 13 × 32 = 1 0 5 416 = 0.00416 ∴ 3125 13 = 0.00416
(iii) 7 80 \text{(iii) } \dfrac{7}{80} (iii) 80 7
The given number 7 80 \dfrac{7}{80} 80 7
Prime factorization of denominator 80:
2 80 2 40 2 20 2 10 5 5 1 \begin{array}{l|l} 2 & 80 \\ \hline 2 & 40 \\ \hline 2 & 20 \\ \hline 2 & 10 \\ \hline 5 & 5 \\ \hline & 1 \end{array} 2 2 2 2 5 80 40 20 10 5 1
80 = 2 x 2 x 2 x 2 x 54 x 54 x 51
7 80 = 7 2 4 × 5 1 = 7 × 5 3 2 4 × 5 1 × 5 3 = 7 × 125 2 4 × 5 4 = 7 × 125 ( 2 × 5 ) 4 = 875 10 4 = 0.0875 ∴ 7 80 = 0.0875 \dfrac{7}{80} = \dfrac{7}{2^4 \times 5^1} \\[0.5em] = \dfrac{7 \times 5^3}{2^4 \times 5^1 \times 5^3} \\[0.5em] = \dfrac{7 \times 125}{2^4 \times 5^4} \\[0.5em] = \dfrac{7 \times 125}{(2 \times 5)^4} \\[0.5em] = \dfrac{875}{10^4} \\[0.5em] = 0.0875 \\[0.5em] \bold{\therefore \dfrac{7}{80} = 0.0875} 80 7 = 2 4 × 5 1 7 = 2 4 × 5 1 × 5 3 7 × 5 3 = 2 4 × 5 4 7 × 125 = ( 2 × 5 ) 4 7 × 125 = 1 0 4 875 = 0.0875 ∴ 80 7 = 0.0875
(iv) 6 15 \text{(iv) } \dfrac{6}{15} (iv) 15 6
GCD of numerator and denominator is 3. Reducing the number to its lowest form:
6 15 = 3 × 2 3 × 5 = 2 5 \dfrac{6}{15} = \dfrac{\cancel{3} \times 2}{\cancel{3} \times 5} \\[0.5em] = \dfrac{2}{5} 15 6 = 3 × 5 3 × 2 = 5 2
6 15 = 2 5 = 2 × 2 2 × 5 = 4 10 = 0.4 ∴ 6 15 = 0.4 \dfrac{6}{15} = \dfrac{2}{5} \\[0.5em] = \dfrac{2 \times 2}{2 \times 5} \\[0.5em] = \dfrac{4}{10} \\[0.5em] = 0.4 \\[0.5em] \bold{\therefore \dfrac{6}{15} = 0.4} 15 6 = 5 2 = 2 × 5 2 × 2 = 10 4 = 0.4 ∴ 15 6 = 0.4
(v) 2 2 × 7 5 4 \text{(v) } \dfrac{2^2 \times 7}{5^4} (v) 5 4 2 2 × 7
2 2 × 7 5 4 = 2 2 × 7 × 2 4 5 4 × 2 4 = 4 × 7 × 16 ( 2 × 5 ) 4 = 448 10 4 = 0.0448 ∴ 2 2 × 7 5 4 = 0.0448 \dfrac{2^2 \times 7}{5^4} = \dfrac{2^2 \times 7 \times 2^4}{5^4 \times 2^4} \\[0.5em] = \dfrac{4 \times 7 \times 16}{(2 \times 5)^4} \\[0.5em] = \dfrac{448}{10^4} \\[0.5em] = 0.0448 \\[0.5em] \bold{\therefore \dfrac{2^2 \times 7}{5^4} = 0.0448} 5 4 2 2 × 7 = 5 4 × 2 4 2 2 × 7 × 2 4 = ( 2 × 5 ) 4 4 × 7 × 16 = 1 0 4 448 = 0.0448 ∴ 5 4 2 2 × 7 = 0.0448
(vi) 237 1500 \text{(vi) } \dfrac{237}{1500} (vi) 1500 237
GCD of numerator and denominator is 3. Reducing the number to its lowest form:
237 1500 = 3 × 79 3 × 500 = 79 500 \dfrac{237}{1500} = \dfrac{\cancel{3} \times 79}{\cancel{3} \times 500} \\[0.5em] = \dfrac{79}{500} 1500 237 = 3 × 500 3 × 79 = 500 79
237 1500 = 79 500 = 79 × 2 500 × 2 = 158 10 3 = 0.158 ∴ 237 1500 = 0.158 \dfrac{237}{1500} = \dfrac{79}{500} \\[0.5em] = \dfrac{79 \times 2}{500 \times 2} \\[0.5em] = \dfrac{158}{10^3} \\[0.5em] = 0.158 \\[0.5em] \bold{\therefore \dfrac{237}{1500} = 0.158} 1500 237 = 500 79 = 500 × 2 79 × 2 = 1 0 3 158 = 0.158 ∴ 1500 237 = 0.158
Write the denominator of the rational number 257 5000 \dfrac{257}{5000} 5000 257 m × 5n where m, n are non-negative integers. Hence, write its decimal expansion without actual division.
Answer
The given number 257 5000 \dfrac{257}{5000} 5000 257
Prime factorization of denominator 5000:
2 5000 2 2500 2 1250 5 625 5 125 5 25 5 5 1 \begin{array}{l|l} 2 & 5000 \\ \hline 2 & 2500 \\ \hline 2 & 1250 \\ \hline 5 & 625 \\ \hline 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array} 2 2 2 5 5 5 5 5000 2500 1250 625 125 25 5 1
5000 = 2 x 2 x 2 x 5 x 5 x 5 x 53 x 54
Hence, denominator of rational number 257 5000 \dfrac{257}{5000} 5000 257 m × 5n is 23 x 54 where m = 3 and n = 4.
257 5000 = 257 2 3 × 5 4 = 257 × 2 2 3 × 5 4 × 2 = 514 2 4 × 5 4 = 514 ( 2 × 5 ) 4 = 514 10 4 = 0.0514 ∴ 257 5000 = 0.0514 \dfrac{257}{5000} = \dfrac{257}{2^3 \times 5^4} \\[0.5em] = \dfrac{257 \times 2}{2^3 \times 5^4 \times 2} \\[0.5em] = \dfrac{514}{2^4 \times 5^4} \\[0.5em] = \dfrac{514}{(2 \times 5)^4} \\[0.5em] = \dfrac{514}{10^4} \\[0.5em] = 0.0514 \\[0.5em] \bold{\therefore \dfrac{257}{5000} = 0.0514} 5000 257 = 2 3 × 5 4 257 = 2 3 × 5 4 × 2 257 × 2 = 2 4 × 5 4 514 = ( 2 × 5 ) 4 514 = 1 0 4 514 = 0.0514 ∴ 5000 257 = 0.0514
Write the decimal expansion of 1 7 \dfrac{1}{7} 7 1 2 7 \dfrac{2}{7} 7 2 3 7 \dfrac{3}{7} 7 3 4 7 \dfrac{4}{7} 7 4 5 7 \dfrac{5}{7} 7 5 6 7 \dfrac{6}{7} 7 6
Answer
The fraction : 1 7 \dfrac{1}{7} 7 1
Decimal expansion of 1 7 \dfrac{1}{7} 7 1 0. 142857 ‾ \bold{0.\overline{142857}} 0. 142857
Since, it is recurring
2 7 = 2 × 1 7 = 2 × 0. 142857 ‾ = 0. 285714 ‾ \dfrac{2}{7}=2\times\dfrac{1}{7}=2\times0.\overline{142857}=\bold{0.\overline{285714}} 7 2 = 2 × 7 1 = 2 × 0. 142857 = 0. 285714
3 7 = 3 × 1 7 = 3 × 0. 142857 ‾ = 0. 428571 ‾ \dfrac{3}{7}=3\times\dfrac{1}{7}=3\times0.\overline{142857}=\bold{0.\overline{428571}} 7 3 = 3 × 7 1 = 3 × 0. 142857 = 0. 428571
4 7 = 4 × 1 7 = 4 × 0. 142857 ‾ = 0. 571428 ‾ \dfrac{4}{7}=4\times\dfrac{1}{7}=4\times0.\overline{142857}=\bold{0.\overline{571428}} 7 4 = 4 × 7 1 = 4 × 0. 142857 = 0. 571428
5 7 = 5 × 1 7 = 5 × 0. 142857 ‾ = 0. 714285 ‾ \dfrac{5}{7}=5\times\dfrac{1}{7}=5\times0.\overline{142857}=\bold{0.\overline{714285}} 7 5 = 5 × 7 1 = 5 × 0. 142857 = 0. 714285
6 7 = 6 × 1 7 = 6 × 0. 142857 ‾ = 0. 857142 ‾ \dfrac{6}{7}=6\times\dfrac{1}{7}=6\times0.\overline{142857}=\bold{0.\overline{857142}} 7 6 = 6 × 7 1 = 6 × 0. 142857 = 0. 857142
Express the following numbers in the form p q \dfrac{p}{q} q p
(i) 0. 3 ‾ (ii) 5. 2 ‾ (iii) 0.404040... (iv) 0.4 7 ‾ (v) 0.1 34 ‾ (vi) 0. 001 ‾ \begin{matrix} \text{(i)} & 0.\overline{3} \\[1.5em] \text{(ii)} & 5.\overline{2} \\[1.5em] \text{(iii)} & 0.404040... \\[1.5em] \text{(iv)} & 0.4\overline{7} \\[1.5em] \text{(v)} & 0.1\overline{34} \\[1.5em] \text{(vi)} & 0.\overline{001} \\[1.5em] \end{matrix} (i) (ii) (iii) (iv) (v) (vi) 0. 3 5. 2 0.404040... 0.4 7 0.1 34 0. 001
Answer
(i) Let x = 0. 3 ‾ 0.\overline{3} 0. 3 ....(i) \qquad \text{....(i)} ....(i)
As there is one repeating digit after decimal point,
So multiplying both sides of (i) by 10
we get,
10x = 3.3333.......(ii) \qquad \text{....(ii)} ....(ii)
Subtracting (i) from (ii), we get
9x = 3
x = 3 9 \dfrac{3}{9} 9 3 1 3 \bold{\dfrac{1}{3}} 3 1
which is in the form of p q \dfrac{p}{q} q p
(ii) Let x = 5. 2 ‾ 5.\overline{2} 5. 2 ....(i) \qquad \text{....(i)} ....(i)
As there is one repeating digit after decimal point,
So multiplying both sides of (i) by 10
we get,
10x = 52.2222.......(ii) \qquad \text{....(ii)} ....(ii)
Subtracting (i) from (ii), we get
9x = 47
x = 47 9 \bold{\dfrac{47}{9}} 9 47
Which is in the form of p q \dfrac{p}{q} q p
(iii) Let x = 0. 40 ‾ 0.\overline{40} 0. 40 ....(i) \qquad \text{....(i)} ....(i)
As there are two repeating digit after decimal point,
So multiplying both sides of (i) by 100
we get,
100x = 40.4040.......(ii) \qquad \text{....(ii)} ....(ii)
Subtracting (i) from (ii), we get
99x = 40
x = 40 99 \bold{\dfrac{40}{99}} 99 40
Which is in the form of p q \dfrac{p}{q} q p
(iv) Let x = 0.4 7 ‾ 0.4\overline{7} 0.4 7 ....(i) \qquad \text{....(i)} ....(i)
As there is one repeating digit after decimal point ,
So multiplying both sides of (i) by 10
we get,
10x=4.7777.......(ii) \qquad \text{....(ii)} ....(ii)
Multiply by 100 on both sides
100x=47.77777.........(iii) \qquad \text{....(iii)} ....(iii)
subtracting (ii) from (iii), we get
100x-10x=47.7777.. -4.777...
90x= 43
x = 43 90 \bold{\dfrac{43}{90}} 90 43
Which is in the form of p q \dfrac{p}{q} q p
(v) Let x = 0.1 34 ‾ 0.1\overline{34} 0.1 34 ....(i) \qquad \text{....(i)} ....(i)
So multiplying both sides of (i) by 10
we get,
10x=1.343434.......(ii) \qquad \text{....(ii)} ....(ii)
Again multiply by 100 on both sides ,
1000x =134.3434.........(iii) \qquad \text{....(iii)} ....(iii)
Subtracting (ii) from (iii), we get
1000x - 10x = 134.3434... - 1.3434...
990x = 133
x = 133 990 \bold{\dfrac{133}{990}} 990 133
which is in the form of p q \dfrac{p}{q} q p
(vi) Let x = 0. 001 ‾ 0.\overline{001} 0. 001 ....(i) \qquad \text{....(i)} ....(i)
So multiplying both sides of (i) by 1000,
we get,
1000x = 1.001001.......(ii) \qquad \text{....(ii)} ....(ii)
Subtracting (i) from (ii), we get
1000x - x = 1.001001... - 0.001001...
999x = 1
x = 1 999 \bold{\dfrac{1}{999}} 999 1
which is in the form of p q \dfrac{p}{q} q p
Classify the following numbers as rational or irrational:
(i) 23 (ii) 225 (iii) 0.3796 (iv) 7.478478... (v) 1.101001000100001... (vi) 345.0 456 ‾ \begin{matrix} \text{(i)} & \sqrt{23} \\[1.5em] \text{(ii)} & \sqrt{225} \\[1.5em] \text{(iii)} & 0.3796 \\[1.5em] \text{(iv)} & 7.478478... \\[1.5em] \text{(v)} & 1.101001000100001... \\[1.5em] \text{(vi)} & 345.0\overline{456} \\[1.5em] \end{matrix} (i) (ii) (iii) (iv) (v) (vi) 23 225 0.3796 7.478478... 1.101001000100001... 345.0 456
Answer
Rational numbers are in the form c, q ≠ 0, and p and q are integers.
(i) 23 \bold{\sqrt{23}} 23 p q \dfrac{p}{q} q p
(ii) 225 \sqrt{225} 225 15 × 15 \sqrt{15×15} 15 × 15 15 1 \dfrac{15}{1} 1 15
As it can be written in the form p q \dfrac{p}{q} q p
∴ 225 \bold{\sqrt{225}} 225 .
(iii) 0.3796 = 3796 1000 \dfrac{3796}{1000} 1000 3796
Since, the decimal expansion is terminating decimal.
∴ 0.3796 is a rational number .
(iv) 7.478478
Let x = 7.478478 ....(i) \text{....(i)} ....(i)
Since there is three repeating digit after decimal point,
Multiplying both sides by 1000, we get
1000x = 7478.478478... ....(ii) \text{....(ii)} ....(ii)
Subtracting (i) from (ii) we get,
999x = 7471
x = 7471 999 \bold{\dfrac{7471}{999}} 999 7471
∴ It is non terminating , repeating rational number .
(v) 1.101001000100001...
Since number of 0's are increasing between two consecutive terms as we move further , So it is non terminating, non repeating decimal.
∴ 1.101001000100001... is an irrational number .
(vi) 345.0 456 ‾ 345.0\overline{456} 345.0 456
345.0 456 ‾ 345.0\overline{456} 345.0 456
Let x = 345.0456456...
Multiply both sides by 10, we get
10x = 3450.456456.. ....(i) \text{....(i)} ....(i)
Since, after decimal there are three repeating digit:
Multiply both sides by 1000, we get
10000x = 3450456.456456... ....(ii) \text{....(ii)} ....(ii)
Subtracting (i) from (ii) ,
9990x = 3447006
x = 3447006 9990 \bold{\dfrac{3447006}{9990}} 9990 3447006
since, it is non-terminating , repeating decimal.
∴ 345.0 456 ‾ \bold{345.0\overline{456}} 345.0 456
The following real numbers have decimal expansions as given below. In each case, state whether they are rational or not. If they are rational and expressed in the form p q \dfrac{p}{q} q p
(i) 37.09158 (ii) 423. 04567 ‾ (iii) 8.9010010001... (iv) 2.3476817681... \begin{matrix} \text{(i)} & 37.09158 \\[1.5em] \text{(ii)} & 423.\overline{04567}\\[1.5em] \text{(iii)} & 8.9010010001... \\[1.5em] \text{(iv)} & 2.3476817681... \\[1.5em] \end{matrix} (i) (ii) (iii) (iv) 37.09158 423. 04567 8.9010010001... 2.3476817681...
Answer
(i) 37.07158
This can be written as 37.09158 = 3709158 100000 \dfrac{3709158}{100000} 100000 3709158
Since , it is terminating decimal
It is Rational number and the prime factors of its denominator q will be 2 or 5 or both .
(ii) 423. 04567 ‾ 423.\overline{04567} 423. 04567
since it has non-terminating recurring decimal,
423. 04567 ‾ 423.\overline{04567} 423. 04567
It is a rational number which is non-terminating and repeating. Its denominator q will have prime factors other than 2 or 5 .
(iii) 8.9010010001...
Since, it is non-terminating non-repeating decimal number
∴ It is not a Rational number .
(iv) 2.3476817681... = 2.34 7681 ‾ 2.34\overline{7681} 2.34 7681
Since, it is a non-terminating repeating decimal number,
∴ It is a Rational number and its denominator q will have prime factors other than 2 or 5.
Insert an irrational number between the following :
(i) 1 3 \dfrac{1}{3} 3 1 1 2 \dfrac{1}{2} 2 1
(ii) − 2 5 -\dfrac{2}{5} − 5 2 1 2 \dfrac{1}{2} 2 1
(iii) 0 and 0.1
Answer
(i) One irrational number between 1 3 \dfrac{1}{3} 3 1 1 2 \dfrac{1}{2} 2 1
1 3 \dfrac{1}{3} 3 1
1 2 \dfrac{1}{2} 2 1
So there are infinite irrational number between 1 3 \dfrac{1}{3} 3 1 1 2 \dfrac{1}{2} 2 1
One irrational number among them can be 0.4040040004... \bold{0.4040040004...} 0.4040040004...
(ii) One irrational number between − 2 5 -\dfrac{2}{5} − 5 2 1 2 \dfrac{1}{2} 2 1
− 2 5 -\dfrac{2}{5} − 5 2
1 2 \dfrac{1}{2} 2 1
So, there are infinite irrational numbers between − 2 5 -\dfrac{2}{5} − 5 2 1 2 \dfrac{1}{2} 2 1
One irrational number among them can be 0.2020020002... \bold{0.2020020002...} 0.2020020002...
(iii) One irrational number among 0 and 0.1 , can be 0.050050005... \bold{0.050050005...} 0.050050005...
Insert two irrational number between 2 and 3.
Answer
Consider, the squares ( 2 ) 2 (2)^2 ( 2 ) 2 ( 3 ) 2 (3)^2 ( 3 ) 2
Two irrational numbers can be te squares root of any natural number between 4 and 9.
As, As, 4 < 5 < 6 < 9 4 \lt 5 \lt 6\lt 9 4 < 5 < 6 < 9
4 < 5 < 6 < 9 \sqrt{4} \lt \sqrt{5} \lt \sqrt{6} \lt \sqrt{9} 4 < 5 < 6 < 9
therefore , 5 \sqrt{5} 5 6 \sqrt{6} 6
2 < 5 < 6 < 3 2 \lt \sqrt{5} \lt \sqrt{6} \lt \\ 3 2 < 5 < 6 < 3
Hence, two irrational number between 2 and 3 or 4 \sqrt{4} 4 9 \sqrt{9} 9 5 \sqrt{5} 5 6 \sqrt{6} 6 .
Write two irrational numbers between 4 9 \dfrac{4}{9} 9 4 7 11 \dfrac{7}{11} 11 7
Answer
4 9 \dfrac{4}{9} 9 4
7 11 \dfrac{7}{11} 11 7
So, two irrational number between 4 9 \dfrac{4}{9} 9 4 7 11 \dfrac{7}{11} 11 7
Find a rational number between 2 \sqrt{2} 2 3 \sqrt{3} 3
Answer
Consider the squares of 2 \sqrt{2} 2 3 \sqrt{3} 3
( 2 ) 2 {(\sqrt2)^2} ( 2 ) 2 ( 3 ) 2 {(\sqrt3)^2} ( 3 ) 2
Take any rational number between 2 and 3 which is a perfect squares of a rational number,
One such number is 2.25 and
2.25 = ( 1.5 ) 2 (1.5)^2 ( 1.5 ) 2
2.25 \sqrt{2.25} 2.25
As, 2 < 2.25 < 3 2 \lt 2.25 \lt 3 2 < 2.25 < 3
2 < 2.25 < 3 \sqrt{2} \lt \sqrt{2.25} \lt \sqrt{3} 2 < 2.25 < 3
2 < 1.5 < 3 \sqrt{2} \lt 1.5 \lt \sqrt{3} 2 < 1.5 < 3
Hence , one rational number between 2 \sqrt{2} 2 3 \sqrt{3} 3 .
Find two rational numbers between 2 3 2\sqrt{3} 2 3 15 \sqrt{15} 15
Answer
2 3 2\sqrt{3} 2 3 4 × 3 \sqrt{4 × 3} 4 × 3 12 \sqrt{12} 12
So, we need to find two irrational number between 12 \sqrt{12} 12 15 \sqrt{15} 15
Since , 12 < 12.25 < 12.96 < 15 ⇒ 12 < 12.25 < 12.96 < 15 \text{Since}, 12 \lt 12.25 \lt 12.96 \lt 15 \\[0.5em] \Rightarrow \sqrt{12} \lt \sqrt{12.25} \lt \sqrt{12.96} \lt \sqrt{15} Since , 12 < 12.25 < 12.96 < 15 ⇒ 12 < 12.25 < 12.96 < 15
Hence, two rational number between 12 \sqrt{12} 12 15 \sqrt{15} 15 12.25 \bold{\sqrt{12.25}} 12.25 12.96 \bold{\sqrt{12.96}} 12.96 .
Insert an irrational number between 5 \sqrt{5} 5 7 \sqrt{7} 7
Answer
consider the squares of 5 \sqrt{5} 5 7 \sqrt{7} 7
( 5 ) 2 {(\sqrt5)^2} ( 5 ) 2 ( 7 ) 2 {(\sqrt7)^2} ( 7 ) 2
As, 5 < 6 < 7 5 \lt 6\lt 7 5 < 6 < 7
5 < 6 < 7 \sqrt{5} \lt \sqrt{6} \lt \sqrt{7} 5 < 6 < 7 6 \sqrt{6} 6 5 \sqrt{5} 5 7 \sqrt{7} 7
Hence, irrational number between 5 \sqrt{5} 5 7 \sqrt{7} 7 6 \bold{\sqrt{6}} 6 .
Insert two irrational numbers between 3 \sqrt{3} 3 7 \sqrt{7} 7
Answer
consider the squares of 3 \sqrt{3} 3 7 \sqrt{7} 7
( 3 ) 2 {(\sqrt3)^2} ( 3 ) 2 ( 7 ) 2 {(\sqrt7)^2} ( 7 ) 2
As, 3 < 5 < 6 < 7 3 \lt 5 \lt 6\lt 7 3 < 5 < 6 < 7
3 < 5 < 6 < 7 \sqrt{3} \lt \sqrt{5} \lt \sqrt{6} \lt \sqrt{7} 3 < 5 < 6 < 7 5 \sqrt{5} 5 6 \sqrt{6} 6 3 \sqrt{3} 3 7 \sqrt{7} 7
Hence, two irrational number between 3 \sqrt{3} 3 7 \sqrt{7} 7 5 \bold{\sqrt{5}} 5 6 \bold{\sqrt{6}} 6 .
Simplify the following:
(i) 45 − 3 20 + 4 5 (ii) 3 3 + 2 27 + 7 3 (iii) 6 5 × 2 5 (iv) 8 15 ÷ 2 3 (v) 24 8 + 54 9 (vi) 3 8 + 1 2 \begin{matrix} \text{(i)} & \sqrt{45} - 3\sqrt{20} + 4\sqrt{5} \\[1.5em] \text{(ii)} & 3\sqrt{3} + 2\sqrt{27} + \dfrac{7}{\sqrt{3}} \\[1.5em] \text{(iii)} & 6\sqrt{5} × 2\sqrt{5} \\[1.5em] \text{(iv)} & 8\sqrt{15} ÷ 2\sqrt{3} \\[1.5em] \text{(v)} & \dfrac{\sqrt{24}}{8} + \dfrac{\sqrt{54}}{9} \\[1.5em] \text{(vi)} & \dfrac{3}{\sqrt{8}} + \dfrac{1}{\sqrt{2}} \\[1.5em] \end{matrix} (i) (ii) (iii) (iv) (v) (vi) 45 − 3 20 + 4 5 3 3 + 2 27 + 3 7 6 5 × 2 5 8 15 ÷ 2 3 8 24 + 9 54 8 3 + 2 1
Answer
(i) 45 − 3 20 + 4 5 = 3 × 3 × 5 − 3 5 × 4 + 4 5 = 3 5 − 3 × 2 5 + 4 5 = 3 5 − 6 5 + 4 5 = 5 ( 3 − 6 + 4 ) = 5 \text{(i) } \sqrt{45} - 3\sqrt{20} + 4\sqrt{5} \\[1.5em] =\sqrt{3 × 3 × 5} - 3\sqrt{5 × 4} + 4\sqrt{5} \\[1.5em] = 3\sqrt{5} - 3 × 2\sqrt{5} + 4\sqrt5 \\[1.5em] = 3\sqrt{5} - 6\sqrt{5} + 4\sqrt5 \\[1.5em] = \sqrt{5}(3 - 6 + 4) \\[1.5em] = \bold{\sqrt{5}} (i) 45 − 3 20 + 4 5 = 3 × 3 × 5 − 3 5 × 4 + 4 5 = 3 5 − 3 × 2 5 + 4 5 = 3 5 − 6 5 + 4 5 = 5 ( 3 − 6 + 4 ) = 5
(ii) 3 3 + 2 27 + 7 3 = 3 3 + 2 3 × 3 × 3 + 7 3 = 3 3 + 2 × 3 3 + 7 3 = 3 3 + 6 3 + 7 3 × 3 3 = 3 × ( 3 + 6 + 7 3 ) = 34 3 3 \text{(ii) } 3\sqrt{3} + 2\sqrt{27} + \dfrac{7}{\sqrt{3}} \\[1.5em] = 3\sqrt{3} + 2\sqrt{3 × 3 × 3} + \dfrac{7}{\sqrt{3}} \\[1.5em] = 3\sqrt{3} + 2 × 3\sqrt{3} + \dfrac{7}{\sqrt{3}} \\[1.5em] = 3\sqrt{3} + 6\sqrt{3} + \dfrac{7}{\sqrt{3}} × \dfrac{\sqrt3}{\sqrt3} \\[1.5em] = \sqrt{3} × (3 + 6 + \dfrac{7}{3}) = \bold{\dfrac{34}{3}{\sqrt{3}}} \\[1.5em] (ii) 3 3 + 2 27 + 3 7 = 3 3 + 2 3 × 3 × 3 + 3 7 = 3 3 + 2 × 3 3 + 3 7 = 3 3 + 6 3 + 3 7 × 3 3 = 3 × ( 3 + 6 + 3 7 ) = 3 34 3
(iii) 6 5 × 2 5 = 12 × ( 5 × 5 ) = 12 × ( 5 ) 2 = 12 × 5 = 60 \text{(iii) } 6\sqrt{5} × 2\sqrt{5} \\[1.5em] = 12 × (\sqrt{5} × \sqrt{5}) \\[1.5em] = 12 × (\sqrt{5})^2 \\[1.5em] = 12 × 5 \\[1.5em] = \bold{60} \\[1.5em] (iii) 6 5 × 2 5 = 12 × ( 5 × 5 ) = 12 × ( 5 ) 2 = 12 × 5 = 60
(iv) 8 15 ÷ 2 3 = 8 15 2 3 = 8 3 × 5 2 3 = 8 3 5 2 3 = 8 5 2 = 4 5 \text{(iv) } 8\sqrt{15} ÷ 2\sqrt{3} \\[1.5em] = \dfrac{8\sqrt{15}}{2\sqrt{3}} \\[1.5em] = \dfrac{8\sqrt{3×5}}{2\sqrt{3}} \\[1.5em] = \dfrac{8\sqrt{3}\sqrt{5}}{2\sqrt{3}} \\[1.5em] = \dfrac{8\sqrt{5}}{2} \\[1.5em] = \bold{4\sqrt{5}} \\[1.5em] (iv) 8 15 ÷ 2 3 = 2 3 8 15 = 2 3 8 3 × 5 = 2 3 8 3 5 = 2 8 5 = 4 5
(v) 24 8 + 54 9 = 2 × 2 × 6 8 + 3 × 3 × 6 9 = 2 6 8 + 3 6 9 = 6 4 + 6 3 = 6 × ( 1 4 + 1 3 ) = 6 × ( 3 + 4 12 ) = 7 12 6 \text{(v) } \dfrac{\sqrt{24}}{8} + \dfrac{\sqrt{54}}{9} \\[1.5em] = \dfrac{\sqrt{2 × 2 × 6}}{8} + \dfrac{\sqrt{3 × 3 × 6}}{9} \\[1.5em] = \dfrac{2\sqrt{6}}{8} + \dfrac{3\sqrt6}{9}\\[1.5em] = \dfrac{\sqrt{6}}{4} + \dfrac{\sqrt6}{3} \\[1.5em] = \sqrt{6} × {(\dfrac{1}{4} + \dfrac{1}{3})} \\[1.5em] = \sqrt{6} × (\dfrac{3 + 4}{12}) \\[1.5em] = \bold{\dfrac{7}{12}{\sqrt{6}}} (v) 8 24 + 9 54 = 8 2 × 2 × 6 + 9 3 × 3 × 6 = 8 2 6 + 9 3 6 = 4 6 + 3 6 = 6 × ( 4 1 + 3 1 ) = 6 × ( 12 3 + 4 ) = 12 7 6
(vi) 3 8 + 1 2 = 3 2 × 2 × 2 + 1 2 = 3 2 2 + 1 2 = 1 2 × ( 3 2 + 1 ) = 1 2 × ( 3 + 2 2 ) = 1 2 × 5 2 = 1 2 × 2 2 × 5 2 = 5 2 4 \text{(vi) } \dfrac{3}{\sqrt{8}} + \dfrac{1}{\sqrt{2}} \\[1.5em] = \dfrac{3}{\sqrt{2 × 2 × 2}} + \dfrac{1}{\sqrt{2}} \\[1.5em] = \dfrac{3}{2\sqrt{2}} + \dfrac{1}{\sqrt2} \\[1.5em] = \dfrac{1}{\sqrt2} × (\dfrac{3}{2} + 1) \\[1.5em] = \dfrac{1}{\sqrt2} × (\dfrac{3 + 2}{2}) \\[1.5em] = \dfrac{1}{\sqrt2} × \dfrac{5}{2} \\[1.5em] = \dfrac{1}{\sqrt2} × \dfrac{\sqrt{2}}{\sqrt{2}} × \dfrac{5}{2} \\[1.5em] = \bold{\dfrac{5\sqrt{2}}{4}} \\[1.5em] (vi) 8 3 + 2 1 = 2 × 2 × 2 3 + 2 1 = 2 2 3 + 2 1 = 2 1 × ( 2 3 + 1 ) = 2 1 × ( 2 3 + 2 ) = 2 1 × 2 5 = 2 1 × 2 2 × 2 5 = 4 5 2
Simplify the following:
(i) ( 5 + 7 ) ( 2 + 5 ) (ii) ( 5 + 5 ) ( 5 − 5 ) (iii) ( 5 + 2 ) 2 (iv) ( 3 − 7 ) 2 (v) ( 2 + 3 ) ( 5 + 7 ) (vi) ( 4 + 5 ) ( 3 − 7 ) \begin{matrix} \text{(i)} & (5 + \sqrt{7})(2 + \sqrt{5}) \\[1.5em] \text{(ii)} & (5 + \sqrt{5})(5 - \sqrt{5}) \\[1.5em] \text{(iii)} & (\sqrt{5} + \sqrt{2})^2 \\[1.5em] \text{(iv)} & (\sqrt{3} - \sqrt{7})^2 \\[1.5em] \text{(v)} & (\sqrt{2} + \sqrt{3})(\sqrt{5} + \sqrt{7}) \\[1.5em] \text{(vi)} & (4 + \sqrt{5})(\sqrt{3} - \sqrt{7}) \\[1.5em] \end{matrix} (i) (ii) (iii) (iv) (v) (vi) ( 5 + 7 ) ( 2 + 5 ) ( 5 + 5 ) ( 5 − 5 ) ( 5 + 2 ) 2 ( 3 − 7 ) 2 ( 2 + 3 ) ( 5 + 7 ) ( 4 + 5 ) ( 3 − 7 )
Answer
(i) ( 5 + 7 ) ( 2 + 5 ) = 10 + 5 5 + 2 7 + 7 5 = 10 + 5 5 + 2 7 + 35 \text{(i) } (5 + \sqrt{7})(2 + \sqrt{5}) \\[1.5em] = 10 + 5\sqrt{5} + 2\sqrt{7} + \sqrt{7}\sqrt{5} \\[1.5em] = \bold{10 + 5\sqrt{5} + 2\sqrt{7} + \sqrt{35}} \\[1.5em] (i) ( 5 + 7 ) ( 2 + 5 ) = 10 + 5 5 + 2 7 + 7 5 = 10 + 5 5 + 2 7 + 35
(ii) ( 5 + 5 ) ( 5 − 5 ) Using identity : ( a + b ) ( a − b ) = a 2 − b 2 = 5 2 − ( 5 ) 2 = 25 − 5 = 20 \text{(ii) } (5 + \sqrt{5})(5 - \sqrt{5}) \\[1.5em] \text{Using identity} : (a + b)(a - b) = a^2 - b^2 \\[1.5em] = 5^2 - (\sqrt{5})^2 \\[1.5em] = \bold{25 - 5 = 20} \\[1.5em] (ii) ( 5 + 5 ) ( 5 − 5 ) Using identity : ( a + b ) ( a − b ) = a 2 − b 2 = 5 2 − ( 5 ) 2 = 25 − 5 = 20
(iii) ( 5 + 2 ) 2 Using identity : ( a + b ) 2 = a 2 + 2 a b + b 2 = ( 5 + 2 ) 2 = ( 5 ) 2 + 2 × 5 × 2 + ( 2 ) 2 = 5 + 2 10 + 2 = 7 + 2 10 \text{(iii) } (\sqrt{5} + \sqrt{2})^2 \\[1.5em] \text{Using identity} : (a + b)^2 = a^2 + 2ab + b^2 \\[1.5em] = (\sqrt{5} + \sqrt{2})^2 = {(\sqrt{5})}^2 + 2 × \sqrt{5} ×\sqrt{2} + {(\sqrt{2})}^2 \\[1.5em] = 5 + 2\sqrt{10} + 2 \\[1.5em] = \bold{7 + 2\sqrt{10}} \\[1.5em] (iii) ( 5 + 2 ) 2 Using identity : ( a + b ) 2 = a 2 + 2 ab + b 2 = ( 5 + 2 ) 2 = ( 5 ) 2 + 2 × 5 × 2 + ( 2 ) 2 = 5 + 2 10 + 2 = 7 + 2 10
(iv) ( 3 − 7 ) 2 Using identity : ( a − b ) 2 = a 2 − 2 a b + b 2 = ( 3 − 7 ) 2 = ( 3 ) 2 − 2 × 3 × 7 + ( 7 ) 2 = 3 − 2 21 + 7 = 10 − 2 21 \text{(iv) } (\sqrt{3} - \sqrt{7})^2 \\[1.5em] \text{Using identity} : (a - b)^2 = a^2 - 2ab + b^2 \\[1.5em] = (\sqrt{3} - \sqrt{7})^2 = {(\sqrt{3})}^2 - 2 × \sqrt{3} ×\sqrt{7} + {(\sqrt{7})}^2 \\[1.5em] = 3 - 2\sqrt{21} + 7 \\[1.5em] = \bold{10 - 2\sqrt{21}} \\[1.5em] (iv) ( 3 − 7 ) 2 Using identity : ( a − b ) 2 = a 2 − 2 ab + b 2 = ( 3 − 7 ) 2 = ( 3 ) 2 − 2 × 3 × 7 + ( 7 ) 2 = 3 − 2 21 + 7 = 10 − 2 21
(v) ( 2 + 3 ) ( 5 + 7 ) = 2 × 5 + 2 × 7 + 3 × 5 + 3 × 7 = 10 + 14 + 15 + 21 \text{(v) } (\sqrt{2} + \sqrt{3})(\sqrt{5} + \sqrt{7}) \\[1.5em] = \sqrt{2} × \sqrt{5} + \sqrt{2} × \sqrt{7} + \sqrt{3} × \sqrt{5} + \sqrt{3} × \sqrt{7} \\[1.5em] = \bold{\sqrt{10} + \sqrt{14} + \sqrt{15} + \sqrt{21}} \\[1.5em] (v) ( 2 + 3 ) ( 5 + 7 ) = 2 × 5 + 2 × 7 + 3 × 5 + 3 × 7 = 10 + 14 + 15 + 21
(vi) ( 4 + 5 ) ( 3 − 7 ) = 4 3 − 4 7 + 5 × 3 − 5 × 7 = 4 3 − 4 7 + 15 − 35 \text{(vi) } (4 + \sqrt{5})(\sqrt{3} - \sqrt{7}) \\[1.5em] = 4\sqrt{3} - 4\sqrt{7} + \sqrt{5} × \sqrt{3} - \sqrt{5} × \sqrt{7} \\[1.5em] = \bold{4\sqrt{3} - 4\sqrt{7} + \sqrt{15} - \sqrt{35}} \\[1.5em] (vi) ( 4 + 5 ) ( 3 − 7 ) = 4 3 − 4 7 + 5 × 3 − 5 × 7 = 4 3 − 4 7 + 15 − 35
If 2 \sqrt{2} 2
(i) 8 + 50 + 72 + 98 (ii) 3 32 − 2 50 + 4 128 − 20 18 \begin{matrix} \text{(i)} & \sqrt{8} + \sqrt{50} + \sqrt{72} + \sqrt{98} \\[1.5em] \text{(ii)} & 3\sqrt{32} - 2\sqrt{50} + 4\sqrt{128} - 20\sqrt{18} \\[1.5em] \end{matrix} (i) (ii) 8 + 50 + 72 + 98 3 32 − 2 50 + 4 128 − 20 18
Answer
(i) 8 + 50 + 72 + 98 = 2 × 2 × 2 + 5 × 5 × 2 + 6 × 6 × 2 + 2 × 7 × 7 = 2 2 + 5 2 + 6 2 + 7 2 = ( 2 + 5 + 6 + 7 ) × 2 = ( 20 ) × 2 = 20 × 1.414 = 28.28 \text{(i) } \sqrt{8} + \sqrt{50} + \sqrt{72} + \sqrt{98} \\[1.5em] = \sqrt{2 × 2 × 2} + \sqrt{5 × 5 × 2} + \sqrt{6 × 6 × 2} + \sqrt{2 × 7 × 7} \\[1.5em] = 2\sqrt{2} + 5\sqrt{2} + 6\sqrt{2} + 7\sqrt{2} \\[1.5em] = (2 + 5 + 6 + 7) × \sqrt{2} \\[1.5em] = (20) × \sqrt{2} \\[1.5em] = \bold{20 × 1.414 = 28.28 } \\[1.5em] (i) 8 + 50 + 72 + 98 = 2 × 2 × 2 + 5 × 5 × 2 + 6 × 6 × 2 + 2 × 7 × 7 = 2 2 + 5 2 + 6 2 + 7 2 = ( 2 + 5 + 6 + 7 ) × 2 = ( 20 ) × 2 = 20 × 1.414 = 28.28
(ii) 3 32 − 2 50 + 4 128 − 20 18 = 3 2 × 4 × 4 − 2 5 × 5 × 2 + 4 8 × 8 × 2 − 20 2 × 3 × 3 = 12 2 − 10 2 + 32 2 − 60 2 = ( 12 − 10 + 32 − 60 ) × 2 = ( 44 − 70 ) × 2 = ( − 26 ) × 2 = − 26 × 1.414 = − 36.764 \text{(ii) } 3\sqrt{32} - 2\sqrt{50} + 4\sqrt{128} - 20\sqrt{18} \\[1.5em] = 3\sqrt{2 × 4 × 4 } - 2\sqrt{5 × 5 × 2} + 4\sqrt{8 × 8 × 2} - 20\sqrt{2 × 3 × 3} \\[1.5em] = 12\sqrt{2} - 10\sqrt{2} + 32\sqrt{2} - 60\sqrt{2} \\[1.5em] = (12 - 10 + 32 - 60) × \sqrt{2} \\[1.5em] = (44 - 70) × \sqrt{2} \\[1.5em] = (-26) × \sqrt{2} \\[1.5em] = \bold{-26 × 1.414 = -36.764 } \\[1.5em] (ii) 3 32 − 2 50 + 4 128 − 20 18 = 3 2 × 4 × 4 − 2 5 × 5 × 2 + 4 8 × 8 × 2 − 20 2 × 3 × 3 = 12 2 − 10 2 + 32 2 − 60 2 = ( 12 − 10 + 32 − 60 ) × 2 = ( 44 − 70 ) × 2 = ( − 26 ) × 2 = − 26 × 1.414 = − 36.764
If 3 \sqrt{3} 3
(i) 27 + 75 + 108 − 243 (ii) 5 12 − 3 48 + 6 75 + 7 108 \begin{matrix} \text{(i)} & \sqrt{27} + \sqrt{75} + \sqrt{108} - \sqrt{243} \\[1.5em] \text{(ii)} & 5\sqrt{12} - 3\sqrt{48} + 6\sqrt{75} + 7\sqrt{108} \\[1.5em] \end{matrix} (i) (ii) 27 + 75 + 108 − 243 5 12 − 3 48 + 6 75 + 7 108
Answer
(i) 27 + 75 + 108 − 243 = 3 × 3 × 3 + 3 × 5 × 5 + 2 × 2 × 3 × 3 × 3 − 3 × 3 × 3 × 3 × 3 = 3 3 + 5 3 + 6 3 − 9 3 = ( 3 + 5 + 6 − 9 ) × 3 = ( 14 − 9 ) × 3 = 5 × 1.732 = 8.660 \text{(i) } \sqrt{27} + \sqrt{75} + \sqrt{108} - \sqrt{243} \\[1.5em] = \sqrt{3 × 3 × 3 } + \sqrt{3 × 5 × 5} + \sqrt{2 × 2 × 3 × 3 ×3 } - \sqrt{3 × 3 × 3 × 3 × 3} \\[1.5em] = 3\sqrt{3} + 5\sqrt{3} + 6\sqrt{3} - 9\sqrt{3} \\[1.5em] = (3 + 5 + 6 - 9) × \sqrt{3} \\[1.5em] = (14 - 9) × \sqrt{3} \\[1.5em] = \bold{5 × 1.732 = 8.660} \\[1.5em] (i) 27 + 75 + 108 − 243 = 3 × 3 × 3 + 3 × 5 × 5 + 2 × 2 × 3 × 3 × 3 − 3 × 3 × 3 × 3 × 3 = 3 3 + 5 3 + 6 3 − 9 3 = ( 3 + 5 + 6 − 9 ) × 3 = ( 14 − 9 ) × 3 = 5 × 1.732 = 8.660
(ii) 5 12 − 3 48 + 6 75 + 7 108 = 5 2 × 2 × 3 − 3 2 × 2 × 2 × 2 × 3 + 6 5 × 5 × 3 + 7 2 × 2 × 3 × 3 × 3 = 5 × 2 3 − 4 × 3 3 + 6 × 5 3 + 7 × 6 3 = 10 3 − 12 3 + 30 3 + 42 3 = ( 10 − 12 + 30 + 42 ) × 3 = ( 82 − 12 ) × 3 = 70 × 1.732 = 121.24 \text{(ii) } 5\sqrt{12} - 3\sqrt{48} + 6\sqrt{75} + 7\sqrt{108} \\[1.5em] = 5\sqrt{2 × 2 × 3 } - 3\sqrt{2 × 2 × 2 × 2 × 3} + 6\sqrt{5 × 5 × 3 } + 7\sqrt{2 × 2 × 3 × 3 × 3} \\[1.5em] = 5 × 2\sqrt{3} - 4 × 3\sqrt{3} + 6 × 5\sqrt{3} + 7 × 6\sqrt{3} \\[1.5em] = 10\sqrt{3} - 12\sqrt{3} + 30\sqrt{3} + 42\sqrt{3} \\[1.5em] = (10 - 12 + 30 +42) × \sqrt{3} \\[1.5em] = (82 - 12) × \sqrt{3} \\[1.5em] = \bold{70 × 1.732 = 121.24} \\[1.5em] (ii) 5 12 − 3 48 + 6 75 + 7 108 = 5 2 × 2 × 3 − 3 2 × 2 × 2 × 2 × 3 + 6 5 × 5 × 3 + 7 2 × 2 × 3 × 3 × 3 = 5 × 2 3 − 4 × 3 3 + 6 × 5 3 + 7 × 6 3 = 10 3 − 12 3 + 30 3 + 42 3 = ( 10 − 12 + 30 + 42 ) × 3 = ( 82 − 12 ) × 3 = 70 × 1.732 = 121.24
State which of the following numbers are irrational :
(i) 4 9 , − 3 70 , 7 25 , 16 5 (ii) − 2 49 , 3 200 , 25 3 , − 49 16 \begin{matrix} \text{(i)} & \sqrt{\dfrac{4}{9}}, -{\dfrac{3}{70}},\sqrt{\dfrac{7}{25}},\sqrt{\dfrac{16}{5}} \\[1.5em] \text{(ii)} & -{\sqrt{\dfrac{2}{49}}}, {\dfrac{3}{200}},\sqrt{\dfrac{25}{3}},-{\sqrt{\dfrac{49}{16}}} \\[1.5em] \end{matrix} (i) (ii) 9 4 , − 70 3 , 25 7 , 5 16 − 49 2 , 200 3 , 3 25 , − 16 49
Answer
(i) 4 9 = 2 3 − 3 70 = − 3 70 7 25 = 7 5 16 5 = 4 5 \text{(i) } \sqrt{\dfrac{4}{9}} = \dfrac{2}{3} \\[1.5em] -\dfrac{3}{70} = -\dfrac{3}{70} \\[1.5em] \sqrt{\dfrac{7}{25}} = \dfrac{\sqrt{7}}{5} \\[1.5em] \sqrt{\dfrac{16}{5}} = \dfrac{4}{\sqrt{5}} \\[1.5em] (i) 9 4 = 3 2 − 70 3 = − 70 3 25 7 = 5 7 5 16 = 5 4
∴ 7 25 \bold{\sqrt{\dfrac{7}{25}}} 25 7 16 5 \bold{\sqrt{\dfrac{16}{5}}} 5 16 p q \dfrac{{p}}{q} q p
4 9 \bold{\sqrt{\dfrac{4}{9}}} 9 4 − 3 70 \bold{- \dfrac{3}{70}} − 70 3 p q \dfrac{{p}}{q} q p
(ii) − 2 49 = − 2 7 3 200 = 3 200 25 3 = 5 3 − 49 16 = − 7 4 \text{(ii) } - \sqrt{\dfrac{2}{49}} = - \dfrac{\sqrt{2}}{7} \\[1.5em] \dfrac{3}{200} = \dfrac{3}{200} \\[1.5em] \sqrt{\dfrac{25}{3}} = \dfrac{5}{\sqrt{3}} \\[1.5em] -\sqrt{\dfrac{49}{16}} = -\dfrac{7}{{4}} \\[1.5em] (ii) − 49 2 = − 7 2 200 3 = 200 3 3 25 = 3 5 − 16 49 = − 4 7
∴ − 2 49 \bold{- \sqrt{\dfrac{2}{49}}} − 49 2 25 3 \bold{\sqrt{\dfrac{25}{3}}} 3 25 p q \dfrac{{p}}{q} q p
− 49 16 -\bold{\sqrt{\dfrac{49}{16}}} − 16 49 3 200 \bold{\dfrac{3}{200}} 200 3 p q \dfrac{{p}}{q} q p
State which of the following numbers will change into non-terminating, non-recurring decimals :
(i) − 3 2 (ii) 256 81 (iii) 27 × 16 (iv) 5 36 \begin{matrix} \text{(i)} & - 3\sqrt{2} \\[1.5em] \text{(ii)} & \sqrt{\dfrac{256}{81}} \\[1.5em] \text{(iii)} & \sqrt{27 × 16} \\[1.5em] \text{(iv)} & \sqrt{\dfrac{5}{36}} \\[1.5em] \end{matrix} (i) (ii) (iii) (iv) − 3 2 81 256 27 × 16 36 5
Answer (i) − 3 2 \text{(i) } -3\sqrt{2} (i) − 3 2
it is an irrational number.
We know that 2 \sqrt2 2 − 3 2 \bold{-3\sqrt{2}} − 3 2
(ii) 256 81 \text{(ii) } \sqrt{\dfrac{256}{81}} (ii) 81 256
Here, 256 81 = 16 9 \sqrt{\dfrac{256}{81}} = \dfrac{16}{9} 81 256 = 9 16
(iii) ( 27 × 16 ) = 27 × 16 = 3 3 × 4 = 12 3 \text{(iii) } \sqrt{(27 × 16)} = \sqrt{27} × \sqrt{16} = 3\sqrt{3} × 4 = 12\sqrt{3} (iii) ( 27 × 16 ) = 27 × 16 = 3 3 × 4 = 12 3
It is an irrational number.
We know that 3 \sqrt{3} 3
So , 12 3 12\sqrt{3} 12 3
Hence, 27 × 16 \bold{\sqrt{27 × 16}} 27 × 16
(iv) 5 36 \text{(iv) }\sqrt{\dfrac{5}{36}} (iv) 36 5
Here, 5 36 \sqrt{\dfrac{5}{36}} 36 5 5 6 \dfrac{\sqrt{5}}{6} 6 5
As, 5 6 \dfrac{\sqrt{5}}{6} 6 5 5 36 \sqrt{\dfrac{5}{36}} 36 5
State which of the following numbers are irrational:
(i) 3 − 7 25 (ii) − 2 3 + 2 3 (iii) 3 3 (iv) − 2 7 5 3 (v) ( 2 − 3 ) ( 2 + 3 ) (vi) ( 3 + 5 ) 2 (vii) ( 2 5 7 ) 2 (viii) ( 3 − 6 ) 2 \begin{matrix} \text{(i)} & 3-\sqrt{\dfrac{7}{25}}\\[1.5em] \text{(ii)} & -\dfrac{2}{3}+\sqrt[3]{2} \\[1.5em] \text{(iii)} & \dfrac{3}{\sqrt{3}} \\[1.5em] \text{(iv)} & -\dfrac{2}{7}\sqrt[3]{5} \\[1.5em] \text{(v)} & (2-\sqrt{3})(2+\sqrt{3}) \\[1.5em] \text{(vi)} & (3+\sqrt{5})^2 \\[1.5em] \text{(vii)} &(\dfrac{2}{5}\sqrt{7})^2 \\[1.5em] \text{(viii)} & (3-\sqrt{6})^2 \\[1.5em] \end{matrix} (i) (ii) (iii) (iv) (v) (vi) (vii) (viii) 3 − 25 7 − 3 2 + 3 2 3 3 − 7 2 3 5 ( 2 − 3 ) ( 2 + 3 ) ( 3 + 5 ) 2 ( 5 2 7 ) 2 ( 3 − 6 ) 2
Answer
(i) 3 − 7 25 = 3 − 7 ( 5 × 5 ) = 3 − 7 5 \text{(i) } 3 - \sqrt{\dfrac{7}{25}} = 3 - \dfrac{\sqrt7}{(\sqrt{5 × 5})} = 3 - \dfrac{\sqrt7}{5} (i) 3 − 25 7 = 3 − ( 5 × 5 ) 7 = 3 − 5 7
As , 3 − 7 5 3 - \dfrac{\sqrt{7}}{5} 3 − 5 7
∴ 3 − 7 25 \bold{3 - \sqrt{\dfrac{7}{25}}} 3 − 25 7 .
(ii) − 2 3 + 2 3 \text{(ii) } -\dfrac{2}{3} + \sqrt[3]{2} (ii) − 3 2 + 3 2
Here, 2 is not perfect cube
∴ − 2 3 + 2 3 -\dfrac{2}{3} + \sqrt[3]{2} − 3 2 + 3 2 .
(iii) 3 3 = 3 3 × 3 3 = 3 3 3 = 3 \text{(iii) } \dfrac{3}{\sqrt{3}} = \dfrac{3}{\sqrt{3}} × \dfrac{\sqrt{3}}{\sqrt{3}} = \dfrac{3\sqrt{3}}{3} = \sqrt{3} (iii) 3 3 = 3 3 × 3 3 = 3 3 3 = 3
As, 3 \sqrt{3} 3
∴ 3 3 \dfrac{3}{\sqrt{3}} 3 3 .
(iv) − 2 7 5 3 \text{(iv) } -\dfrac{2}{7}\sqrt[3]{5} (iv) − 7 2 3 5
Here, 5 is not perfect cube
∴ − 2 7 5 3 -\dfrac{2}{7}\sqrt[3]{5} − 7 2 3 5 .
(v) ( 2 − 3 ) ( 2 + 3 ) \text{(v) } (2-\sqrt{3})(2+\sqrt{3}) (v) ( 2 − 3 ) ( 2 + 3 )
Using identity : ( a + b ) ( a − b ) = a 2 − b 2 (a + b)(a - b) = a^2 - b^2 ( a + b ) ( a − b ) = a 2 − b 2
( 2 − 3 ) ( 2 + 3 ) = 2 2 − ( 3 ) 2 = 4 − 3 = 1 (2-\sqrt{3})(2+\sqrt{3}) = 2^2 - (\sqrt3)^2 = 4 - 3 = 1 ( 2 − 3 ) ( 2 + 3 ) = 2 2 − ( 3 ) 2 = 4 − 3 = 1
Hence , ( 2 − 3 ) ( 2 + 3 ) (2-\sqrt{3})(2+\sqrt{3}) ( 2 − 3 ) ( 2 + 3 ) .
(vi) ( 3 + 5 ) 2 \text{(vi) }(3 + \sqrt{5})^2 (vi) ( 3 + 5 ) 2
Using identity : (a + b)2 = a2 + 2ab + b2
( 3 + 5 ) 2 = 3 2 + 2 × 3 × 5 + ( 5 ) 2 = 9 + 6 5 + 5 = 9 + 5 + 6 5 = 14 + 6 5 (3 + \sqrt{5})^2 = 3^2 + 2 × 3 × \sqrt{5} + (\sqrt{5})^2 \\[1.5em] = 9 + 6\sqrt{5} + 5 \\[1.5em] = 9 + 5 + 6\sqrt{5} \\[1.5em] = 14 + 6\sqrt{5} ( 3 + 5 ) 2 = 3 2 + 2 × 3 × 5 + ( 5 ) 2 = 9 + 6 5 + 5 = 9 + 5 + 6 5 = 14 + 6 5
As, 14 + 6 5 6\sqrt{5} 6 5
∴ ( 3 + 5 ) 2 (3+\sqrt{5})^2 ( 3 + 5 ) 2
(vii) ( 2 5 7 ) 2 \text{(vii) } \Big(\dfrac{2}{5}\sqrt{7}\Big)^2 (vii) ( 5 2 7 ) 2
( 2 5 7 ) 2 = 2 5 7 × 2 5 7 = 4 25 × ( 7 ) 2 = 4 25 × 7 = 28 25 \Big(\dfrac{2}{5}\sqrt{7}\Big)^2 = \dfrac{2}{5}\sqrt{7} × \dfrac{2}{5}\sqrt{7} \\[1.5em] = \dfrac{4}{25} × (\sqrt7)^2 \\[1.5em] = \dfrac{4}{25} × 7 \\[1.5em] = \dfrac{28}{25} ( 5 2 7 ) 2 = 5 2 7 × 5 2 7 = 25 4 × ( 7 ) 2 = 25 4 × 7 = 25 28
As, 28 25 \dfrac{28}{25} 25 28
∴ ( 2 5 7 ) 2 (\dfrac{2}{5}\sqrt{7})^2 ( 5 2 7 ) 2
(viii) ( 3 − 6 ) 2 \text{(viii) } (3 - \sqrt{6})^2 (viii) ( 3 − 6 ) 2
Using identity : (a + b)2 = a2 + 2ab + b2
( 3 − 6 ) 2 = 3 2 − 2 × 3 × 6 + ( 6 ) 2 = 9 − 6 6 + 6 = 9 + 6 − 6 6 = 15 − 6 6 (3 - \sqrt{6})^2 = 3^2 - 2 × 3 × \sqrt{6} + (\sqrt{6})^2 \\[1.5em] = 9 - 6\sqrt{6} + 6 \\[1.5em] = 9 + 6 - 6\sqrt{6} \\[1.5em] = 15 - 6\sqrt{6} ( 3 − 6 ) 2 = 3 2 − 2 × 3 × 6 + ( 6 ) 2 = 9 − 6 6 + 6 = 9 + 6 − 6 6 = 15 − 6 6
As, 15 - 6 6 6\sqrt{6} 6 6
∴ ( 3 − 6 ) 2 (3 - \sqrt{6})^2 ( 3 − 6 ) 2 .
Prove that the following numbers are irrational:
(i) 2 3 (ii) 3 3 (iii) 5 4 \begin{matrix} \text{(i)} & \sqrt[3]{2} \\[1.5em] \text{(ii)} & \sqrt[3]{3} \\[1.5em] \text{(iii)} & \sqrt[4]{5} \\[1.5em] \end{matrix} (i) (ii) (iii) 3 2 3 3 4 5
Answer
(i) Suppose that 2 3 \sqrt[3]{2} 3 2 p q \dfrac{p}{q} q p
⇒ 2 = ( p q ) 3 ⇒ p 3 = 2 q 3 ....(i) \Rightarrow 2 = \Big(\dfrac{p}{q}\Big)^3 \\[1.5em] \Rightarrow p^3 = 2q^3 \qquad \text{....(i)} ⇒ 2 = ( q p ) 3 ⇒ p 3 = 2 q 3 ....(i)
As 2 divides 2q3 ⇒ \Rightarrow ⇒ 3 ⇒ \Rightarrow ⇒
Let p = 2k , where k is an integer.
Substituting this value of p in (i), we get
⇒ \phantom{\Rightarrow} ⇒ 3 = 2q3 ⇒ \Rightarrow ⇒ 3 = 2q3 ⇒ \Rightarrow ⇒ 3 = q3
As 2 divides 4k3 ⇒ \Rightarrow ⇒ 3
⇒ \Rightarrow ⇒
Thus, p and q have a common factor 2. This contradicts that p and q have no common factors (except 1).
Hence, our supposition is wrong. It follows that 2 3 \sqrt[3]{2} 3 2 p q \dfrac{p}{q} q p
∴ 2 3 \bold{\sqrt[3]{2}} 3 2
(ii) Suppose that 3 3 \sqrt[3]{3} 3 3 p q \dfrac{p}{q} q p
⇒ 3 = ( p q ) 3 ⇒ p 3 = 3 q 3 ....(i) \Rightarrow 3 = \Big(\dfrac{p}{q}\Big)^3 \\[1.5em] \Rightarrow p^3 = 3q^3 \qquad \text{....(i)} ⇒ 3 = ( q p ) 3 ⇒ p 3 = 3 q 3 ....(i)
As 3 divides 3q3 ⇒ \Rightarrow ⇒ 3 ⇒ \Rightarrow ⇒
Let p = 3k, where k is an integer.
Substituting this value of p in (i), we get
⇒ \phantom{\Rightarrow} ⇒ 3 = 3q3 ⇒ \Rightarrow ⇒ 3 = 3q3 ⇒ \Rightarrow ⇒ 3 = q3
As 3 divides 9k3 ⇒ \Rightarrow ⇒ 3 ⇒ \Rightarrow ⇒
Thus, p and q have a common factor 3. This contradicts that p and q have no common factors (except 1).
Hence, our supposition is wrong . It follows that 3 3 \sqrt[3]{3} 3 3 p q \dfrac{p}{q} q p
Therefore, 3 3 \bold{\sqrt[3]{3}} 3 3
(iii) Suppose that 5 4 \sqrt[4]{5} 4 5 p q \dfrac{p}{q} q p
⇒ 5 = ( p q ) 4 ⇒ p 4 = 5 q 4 ....(i) \Rightarrow 5 = \Big(\dfrac{p}{q}\Big)^4 \\[1.5em] \Rightarrow p^4 = 5q^4 \qquad \text{....(i)} ⇒ 5 = ( q p ) 4 ⇒ p 4 = 5 q 4 ....(i)
As 5 divides 5q4 ⇒ \Rightarrow ⇒ 4 ⇒ \Rightarrow ⇒
Let p= 5k, where k is an integer.
Substituting this value of p in (i), we get
⇒ \phantom{\Rightarrow} ⇒ 4 = 5q4 ⇒ \Rightarrow ⇒ 4 = 5q4 ⇒ \Rightarrow ⇒ 4 = q4
As 5 divides 125k4 ⇒ \Rightarrow ⇒ 4 ⇒ \Rightarrow ⇒
Thus, p and q have a common factor 5. This contradicts that p and q have no common factors (except 1).
Hence, our supposition is wrong . It follows that 5 4 \sqrt[4]{5} 4 5 p q \dfrac{p}{q} q p
Therefore, 5 4 \bold{\sqrt[4]{5}} 4 5
Find the greatest and the smallest real numbers among the following real numbers :
(i) 2 3 , 3 2 , − 7 , 15 (ii) − 3 2 , 9 5 , − 4 , 4 3 5 , 3 2 3 \begin{matrix} \text{(i)} & 2\sqrt{3},\dfrac{3}{\sqrt{2}},-\sqrt{7},\sqrt{15} \\[1.5em] \text{(ii)} & -3\sqrt{2},\dfrac{9}{\sqrt{5}},-4,{\dfrac{4}{3}}{\sqrt{5}},{\dfrac{3}{2}}{\sqrt{3}} \\[1.5em] \end{matrix} (i) (ii) 2 3 , 2 3 , − 7 , 15 − 3 2 , 5 9 , − 4 , 3 4 5 , 2 3 3
Answer
(i) We will write all the numbers as square roots under one radical.
2 3 = 4 × 3 = 12 3 2 = 9 2 = 4.5 − 7 15 2\sqrt{3}=\sqrt{4 × 3} = \sqrt{12} \\[1.5em] \dfrac{3}{\sqrt{2}} = \sqrt{\dfrac{9}{2}} = \sqrt{4.5} \\[1.5em] -\sqrt{7} \\[1.5em] \sqrt{15} \\[1.5em] 2 3 = 4 × 3 = 12 2 3 = 2 9 = 4.5 − 7 15
∴ The greatest real number is 15 \bold{\sqrt{15}} 15 − 7 \bold{-\sqrt{7}} − 7 .
(ii) We will write all the numbers as square roots under one radical.
− 3 2 = − ( 9 × 2 ) = − 18 9 5 = 81 5 = 16.2 − 4 = − 16 4 3 5 = 16 × 5 9 = 80 9 = 8.88 3 2 3 = 9 × 3 4 = 27 4 = 6.75 -3\sqrt{2}=-\sqrt{(9×2)} = -\sqrt{18} \\[1.5em] \dfrac{9}{\sqrt{5}} = \sqrt{\dfrac{81}{5}} = \sqrt{16.2} \\[1.5em] -4 = -\sqrt{16} \\[1.5em] {\dfrac{4}{3}}\sqrt{5} = \dfrac{\sqrt{16}×\sqrt{5}}{\sqrt{9}} = \sqrt{\dfrac{80}{9}} = \sqrt{8.88} \\[1.5em] {\dfrac{3}{2}}\sqrt{3} = \dfrac{\sqrt{9} × \sqrt{3}}{\sqrt{4}} = \sqrt{\dfrac{27}{4}} = \sqrt{6.75} \\[1.5em] − 3 2 = − ( 9 × 2 ) = − 18 5 9 = 5 81 = 16.2 − 4 = − 16 3 4 5 = 9 16 × 5 = 9 80 = 8.88 2 3 3 = 4 9 × 3 = 4 27 = 6.75
Here, 16.2 \sqrt{16.2} 16.2 − 18 -\sqrt{18} − 18
∴ The greatest real number is 9 5 \bold{\dfrac{9}{\sqrt{5}}} 5 9 − 3 2 . \bold{-3\sqrt{2}}. − 3 2 .
Write the following numbers in ascending order:
(i) 3 2 , 2 3 , 15 , 4 (ii) 3 2 , 2 8 , 4 , 50 , 4 3 \begin{matrix} \text{(i)} & 3\sqrt{2} , 2\sqrt{3} , \sqrt{15} , 4 \\[1.5em] \text{(ii)} & 3\sqrt{2} , 2\sqrt{8} , 4, \sqrt{50} ,4\sqrt{3} \\[1.5em] \end{matrix} (i) (ii) 3 2 , 2 3 , 15 , 4 3 2 , 2 8 , 4 , 50 , 4 3
Answer
(i) Write all the numbers as square root under one radical :
3 2 = 9 × 2 = 9 × 2 = 18 2 3 = 4 × 3 = 4 × 3 = 12 15 = 15 4 = 16 Since , 12 < 15 < 16 < 18 ⇒ 12 < 15 < 16 < 18 ⇒ 2 3 < 15 < 4 < 3 2 3\sqrt{2} = \sqrt{9} × \sqrt{2} = \sqrt{9 × 2} = \sqrt{18} \\[1.5em] 2\sqrt{3} = \sqrt{4} × \sqrt{3} = \sqrt{4 × 3} = \sqrt{12} \\[1.5em] \sqrt{15} = \sqrt{15} \\[1.5em] 4 = \sqrt{16} \\[1.5em] \text{Since} , 12 \lt 15 \lt 16 \lt 18 \\[1.5em] \Rightarrow \sqrt{12} \lt \sqrt{15} \lt \sqrt{16} \lt \sqrt{18} \\[1.5em] \Rightarrow 2\sqrt3 \lt \sqrt{15} \lt 4 \lt 3\sqrt{2} 3 2 = 9 × 2 = 9 × 2 = 18 2 3 = 4 × 3 = 4 × 3 = 12 15 = 15 4 = 16 Since , 12 < 15 < 16 < 18 ⇒ 12 < 15 < 16 < 18 ⇒ 2 3 < 15 < 4 < 3 2
Hence, the given numbers in ascending order are, 2 3 , 15 , 4 , 3 2 2\bold{\sqrt{3}} , \bold{\sqrt{15}} , \bold{4} , \bold{3\sqrt{2}} 2 3 , 15 , 4 , 3 2
(ii) Write all the numbers as square root under one radical :
3 2 = 9 × 2 = 9 × 2 = 18 2 8 = 4 × 8 = 4 × 8 = 32 4 = 16 50 = 50 4 3 = 16 × 3 = 16 × 3 = 48 Since , 16 < 18 < 32 < 48 < 50 16 < 18 < 32 < 48 < 50 ⇒ 4 < 3 2 < 2 8 < 4 3 < 50 3\sqrt{2} = \sqrt{9} × \sqrt{2} = \sqrt{9 × 2} = \sqrt{18} \\[1.5em] 2\sqrt{8} = \sqrt{4} × \sqrt{8} = \sqrt{4 × 8} = \sqrt{32} \\[1.5em] 4 = \sqrt{16} \\[1.5em] \sqrt{50} = \sqrt{50} \\[1.5em] 4\sqrt{3} = \sqrt{16} × \sqrt{3} = \sqrt{16 × 3} = \sqrt{48} \\[1.5em] \text{Since} , 16 \lt 18 \lt 32 \lt 48 \lt 50 \\[1.5em] \sqrt{16} \lt \sqrt{18} \lt \sqrt{32} \lt \sqrt{48} \lt \sqrt{50} \\[1.5em] \Rightarrow 4 \lt 3\sqrt2 \lt 2\sqrt{8} \lt 4\sqrt{3} \lt \sqrt{50} 3 2 = 9 × 2 = 9 × 2 = 18 2 8 = 4 × 8 = 4 × 8 = 32 4 = 16 50 = 50 4 3 = 16 × 3 = 16 × 3 = 48 Since , 16 < 18 < 32 < 48 < 50 16 < 18 < 32 < 48 < 50 ⇒ 4 < 3 2 < 2 8 < 4 3 < 50
Hence , 3 2 , 2 8 , 4 3 , 4 50 3\bold{\sqrt{2}} , 2\bold{\sqrt{8}} , 4\bold{\sqrt{3}} , 4\bold{\sqrt{50}} 3 2 , 2 8 , 4 3 , 4 50 .
Write the following real numbers in descending order:
(i) 9 2 , 3 2 5 , 4 3 , 3 6 5 (ii) 5 3 , 7 3 2 , − 3 , 3 5 , 2 7 . \begin{matrix} \text{(i)} & \dfrac{9}{\sqrt{2}} , {\dfrac{3}{2}}\sqrt{5} , 4\sqrt{3} , 3\sqrt{\dfrac{6}{5}} \\[1.5em] \text{(ii)} & \dfrac{5}{\sqrt{3}} , {\dfrac{7}{3}}{\sqrt{2}} , -\sqrt{3} , 3\sqrt{5} , 2\sqrt{7}. \\[1.5em] \end{matrix} (i) (ii) 2 9 , 2 3 5 , 4 3 , 3 5 6 3 5 , 3 7 2 , − 3 , 3 5 , 2 7 .
Answer
(i) Write all the numbers as square root under one radical :
9 2 = 81 2 = 81 2 = 40.5 3 2 5 = 9 4 × 5 = 9 × 5 4 = 45 4 = 11.25 4 3 = 16 × 3 = 16 × 3 = 48 3 6 5 = 9 × 6 5 = 9 × 6 5 = 54 5 = 10.8 Since , 48 > 40.5 > 11.25 > 10.8 ⇒ 48 > 40.5 > 11.25 > 10.8 ⇒ 4 3 > 9 2 > 3 2 5 > 3 6 5 \dfrac{9}{\sqrt{2}} = \dfrac{\sqrt{81}}{\sqrt{2}} = \sqrt{\dfrac{81}{2}} = \sqrt{40.5} \\[1.5em] {\dfrac{3}{2}}{\sqrt{5}} = \sqrt{\dfrac{9}{4}}×{\sqrt{5}} = \sqrt{\dfrac{9 × 5}{4}} = \sqrt{\dfrac{45}{4}} = \sqrt{11.25} \\[1.5em] 4\sqrt{3} = \sqrt{16} × \sqrt{3} = \sqrt{16 × 3} = \sqrt{48} \\[1.5em] 3\sqrt{\dfrac{6}{5}} = \sqrt{9} × \sqrt{\dfrac{6}{5}} = \sqrt{\dfrac{9 × 6}{5}} = \sqrt{\dfrac{54}{5}} = \sqrt{10.8} \\[1.5em] \text{Since} , 48 \gt 40.5 \gt 11.25 \gt 10.8 \\[1.5em] \Rightarrow \sqrt{48} \gt \sqrt{40.5} \gt \sqrt{11.25} \gt \sqrt{10.8} \\[1.5em] \Rightarrow 4\sqrt3 \gt \dfrac{9}{\sqrt{2}} \gt {\dfrac{3}{2}}{\sqrt{5}} \gt 3\sqrt{\dfrac{6}{5}} 2 9 = 2 81 = 2 81 = 40.5 2 3 5 = 4 9 × 5 = 4 9 × 5 = 4 45 = 11.25 4 3 = 16 × 3 = 16 × 3 = 48 3 5 6 = 9 × 5 6 = 5 9 × 6 = 5 54 = 10.8 Since , 48 > 40.5 > 11.25 > 10.8 ⇒ 48 > 40.5 > 11.25 > 10.8 ⇒ 4 3 > 2 9 > 2 3 5 > 3 5 6
Hence, the given numbers in descending order are 4 3 , 9 2 , 3 2 5 , 3 6 5 \bold{4\sqrt{3}} , \bold{\dfrac{9}{\sqrt{2}}} , \bold{{\dfrac{3}{2}}{\sqrt{5}}} , \bold{3\sqrt{\dfrac{6}{5}}} 4 3 , 2 9 , 2 3 5 , 3 5 6
(ii) Write all the numbers as square root under one radical :
5 3 = 25 3 = 25 3 = 8.33 7 3 2 = 49 9 × 2 = 49 × 2 9 = 98 9 = 10.88 − 3 = − 3 3 5 = 9 × 5 = 45 2 7 = 4 × 7 = 28 Since , 45 > 28 > 10.88 > 8.33 > − 3 45 > 28 > 10.88 > 8.33 > − 3 ⇒ 3 5 > 2 7 > 7 3 2 > 5 3 > − 3 \dfrac{5}{\sqrt{3}} = \dfrac{\sqrt{25}}{\sqrt{3}} = \sqrt{\dfrac{25}{3}} = \sqrt{8.33} \\[1.5em] {\dfrac{7}{3}}{\sqrt{2}} = \sqrt{\dfrac{49}{9}} ×{\sqrt{2}} = \sqrt{\dfrac{49 × 2}{9}} = \sqrt{\dfrac{98}{9}} = \sqrt{10.88} \\[1.5em] -\sqrt{3} = -\sqrt{3} \\[1.5em] 3\sqrt{5} = \sqrt{9 × 5} = \sqrt{45} \\[1.5em] 2\sqrt{7} = \sqrt{4 × 7} = \sqrt{28} \\[1.5em] \text{Since} , 45 \gt 28 \gt 10.88 \gt 8.33 \gt -3 \\[1.5em] \sqrt{45} \gt \sqrt{28} \gt \sqrt{10.88} \gt \sqrt{8.33} \gt -\sqrt{3} \\[1.5em] \Rightarrow 3\sqrt{5} \gt 2\sqrt{7} \gt {\dfrac{7}{3}}{\sqrt{2}} \gt \dfrac{5}{\sqrt{3}} \gt -\sqrt{3} 3 5 = 3 25 = 3 25 = 8.33 3 7 2 = 9 49 × 2 = 9 49 × 2 = 9 98 = 10.88 − 3 = − 3 3 5 = 9 × 5 = 45 2 7 = 4 × 7 = 28 Since , 45 > 28 > 10.88 > 8.33 > − 3 45 > 28 > 10.88 > 8.33 > − 3 ⇒ 3 5 > 2 7 > 3 7 2 > 3 5 > − 3
Hence, 3 5 , 2 7 , 7 3 2 , 5 3 , − 3 \bold{3\sqrt{5}}, \bold{2\sqrt{7}}, \bold{{{\dfrac{7}{3}}{\sqrt{2}}}}, \bold{\dfrac{5}{\sqrt{3}}}, -\bold{\sqrt{3}} 3 5 , 2 7 , 3 7 2 , 3 5 , − 3
Arrange the following numbers in ascending order : 2 3 , 3 , 5 6 \sqrt[3]{2} , \sqrt3 , \sqrt[6]{5} 3 2 , 3 , 6 5
Answer
L.C.M of 3, 2, 6 is 6 :
2 3 = 2 1 3 = ( 2 2 ) 1 6 = ( 4 ) 1 6 3 = 3 1 2 = ( 3 3 ) 1 6 = ( 27 ) 1 6 5 6 = ( 5 ) 1 6 As, 4 < 5 < 27 ⇒ ( 4 ) 1 6 < ( 5 ) 1 6 < ( 27 ) 1 6 ⇒ 2 3 < 5 6 < 3 \sqrt[3]{2} = 2^\dfrac{1}{3} = (2^2)^\dfrac{1}{6} = (4)^\dfrac{1}{6} \\[1.5em] \sqrt{3} = 3^\dfrac{1}{2} = (3^3)^\dfrac{1}{6} = (27)^\dfrac{1}{6} \\[1.5em] \sqrt[6]{5} = (5)^\dfrac{1}{6} \\[1.5em] \text{As, } 4 \lt 5 \lt 27 \\[1.5em] \Rightarrow (4)^\dfrac{1}{6} \lt (5)^\dfrac{1}{6} \lt (27)^\dfrac{1}{6} \\[1.5em] \Rightarrow \sqrt[3]{2} \lt \sqrt[6]{5} \lt \sqrt3 \\[1.5em] 3 2 = 2 3 1 = ( 2 2 ) 6 1 = ( 4 ) 6 1 3 = 3 2 1 = ( 3 3 ) 6 1 = ( 27 ) 6 1 6 5 = ( 5 ) 6 1 As, 4 < 5 < 27 ⇒ ( 4 ) 6 1 < ( 5 ) 6 1 < ( 27 ) 6 1 ⇒ 3 2 < 6 5 < 3
Hence , the given number in ascending order are 2 3 , 5 6 , 3 \sqrt[3]{2} , \sqrt[6]{5} , \sqrt{3} 3 2 , 6 5 , 3
Rationalise the denominator of the following :
(i) 3 4 5 (ii) 5 7 3 (iii) 3 4 − 7 (iv) 17 3 2 + 1 (v) 16 41 − 5 (vi) 1 7 − 6 (vii) 1 5 + 2 (viii) 2 + 3 2 − 3 \begin{matrix} \text{(i)} & \dfrac{3}{4\sqrt{5}} \\[1.5em] \text{(ii)} & \dfrac{5\sqrt{7}}{\sqrt{3}} \\[1.5em] \text{(iii)} & \dfrac{3}{4 - \sqrt{7}} \\[1.5em] \text{(iv)} & \dfrac{17}{3\sqrt{2} + 1} \\[1.5em] \text{(v)} & \dfrac{16}{\sqrt{41}-5} \\[1.5em] \text{(vi)} & \dfrac{1}{\sqrt{7} - \sqrt{6}} \\[1.5em] \text{(vii)} & \dfrac{1}{\sqrt{5} + \sqrt{2}} \\[1.5em] \text{(viii)} & \dfrac{\sqrt{2} + \sqrt{3}}{\sqrt{2} - \sqrt{3}} \\[1.5em] \end{matrix} (i) (ii) (iii) (iv) (v) (vi) (vii) (viii) 4 5 3 3 5 7 4 − 7 3 3 2 + 1 17 41 − 5 16 7 − 6 1 5 + 2 1 2 − 3 2 + 3
Answer
(i) \text{(i)} (i)
3 4 5 \dfrac{3}{4\sqrt{5}} 4 5 3
Let us rationalise the denominator,
Then,
3 4 5 = 3 × 5 4 5 × 5 ⇒ 3 5 4 × 5 ⇒ 3 5 20 \dfrac{3}{4\sqrt{5}} = \dfrac{3×\sqrt{5}}{4\sqrt{5} × \sqrt{5}} \\[1.5em] \Rightarrow\dfrac{3\sqrt{5}}{4 × 5} \\[1.5em] \bold{\Rightarrow\dfrac{3\sqrt{5}}{20}} 4 5 3 = 4 5 × 5 3 × 5 ⇒ 4 × 5 3 5 ⇒ 20 3 5
(ii) \text{(ii)} (ii)
5 7 3 \dfrac{5\sqrt{7}}{\sqrt{3}} 3 5 7
Let us rationalise the denominator,
Then,
5 7 3 = 5 7 × 3 3 × 3 ⇒ 5 21 3 \dfrac{5\sqrt{7}}{\sqrt{3}} = \dfrac{5\sqrt{7}×\sqrt{3}}{\sqrt{3} × \sqrt{3}} \\[1.5em] \bold{\Rightarrow\dfrac{5\sqrt{21}}{3}} 3 5 7 = 3 × 3 5 7 × 3 ⇒ 3 5 21
(iii) \text{(iii)} (iii)
3 4 − 7 \dfrac{3}{4 - \sqrt{7}} 4 − 7 3
Let us rationalise the denominator,
Then,
3 4 − 7 = 3 4 − 7 × 4 + 7 4 + 7 ⇒ 3 ( 4 + 7 ) ( 4 ) 2 − ( 7 ) 2 ⇒ 3 ( 4 + 7 ) ( 16 ) − ( 7 ) ⇒ 3 ( 4 + 7 ) 9 ⇒ ( 4 + 7 ) 3 \dfrac{3}{4 - \sqrt{7}} = \dfrac{3}{4 - \sqrt{7}} × \dfrac{4 + \sqrt{7}}{4 + \sqrt{7}} \\[1.5em] \Rightarrow\dfrac{3(4 + \sqrt{7})}{(4)^2 - (\sqrt{7})^2} \\[1.5em] \Rightarrow\dfrac{3(4 + \sqrt{7})}{(16) - (7)} \\[1.5em] \Rightarrow\dfrac{3(4 + \sqrt{7})}{9} \\[1.5em] \bold{\Rightarrow\dfrac{(4 + \sqrt{7})}{3}} \\[1.5em] 4 − 7 3 = 4 − 7 3 × 4 + 7 4 + 7 ⇒ ( 4 ) 2 − ( 7 ) 2 3 ( 4 + 7 ) ⇒ ( 16 ) − ( 7 ) 3 ( 4 + 7 ) ⇒ 9 3 ( 4 + 7 ) ⇒ 3 ( 4 + 7 )
(iv) \text{(iv)} (iv)
17 3 2 + 1 \dfrac{17}{3\sqrt{2}+1} 3 2 + 1 17
Let us rationalise the denominator,
Then,
17 3 2 + 1 = 17 3 2 + 1 × 3 2 − 1 3 2 − 1 ⇒ 17 ( 3 2 − 1 ) ( 3 2 ) 2 − 1 ⇒ 17 ( 3 2 − 1 ) 17 ⇒ ( 3 2 − 1 ) \dfrac{17}{3\sqrt{2} + 1} = \dfrac{17}{3\sqrt{2} + 1}×\dfrac{3\sqrt{2} - 1}{3\sqrt{2} - 1} \\[1.5em] \Rightarrow\dfrac{17(3\sqrt{2} - 1)}{(3\sqrt{2})^2 - 1} \\[1.5em] \Rightarrow\dfrac{17(3\sqrt{2} - 1)}{17} \\[1.5em] \bold{\Rightarrow(3\sqrt{2} - 1)} \\[1.5em] 3 2 + 1 17 = 3 2 + 1 17 × 3 2 − 1 3 2 − 1 ⇒ ( 3 2 ) 2 − 1 17 ( 3 2 − 1 ) ⇒ 17 17 ( 3 2 − 1 ) ⇒ ( 3 2 − 1 )
(v) \text{(v)} (v)
16 41 − 5 \dfrac{16}{\sqrt{41}-5} 41 − 5 16
Let us rationalise the denominator,
Then,
16 41 − 5 = 16 41 − 5 × 41 + 5 41 + 5 ⇒ 16 ( 41 + 5 ) ( 41 ) 2 − 5 2 ⇒ 16 ( 41 + 5 ) 41 − 25 ⇒ 16 ( 41 + 5 ) 16 41 + 5 \dfrac{16}{\sqrt{41} - 5} = \dfrac{16}{\sqrt{41} - 5}×\dfrac{\sqrt{41} + 5}{\sqrt{41} + 5} \\[1.5em] \Rightarrow\dfrac{16({\sqrt{41} + 5})}{(\sqrt{41})^2 - 5^2} \\[1.5em] \Rightarrow\dfrac{16({\sqrt{41} + 5})}{41 - 25} \\[1.5em] {\Rightarrow\dfrac{16({\sqrt{41} + 5})}{16}} \\[1.5em] \bold{\sqrt{41} + 5} 41 − 5 16 = 41 − 5 16 × 41 + 5 41 + 5 ⇒ ( 41 ) 2 − 5 2 16 ( 41 + 5 ) ⇒ 41 − 25 16 ( 41 + 5 ) ⇒ 16 16 ( 41 + 5 ) 41 + 5
(vi) \text{(vi)} (vi)
1 7 − 6 \dfrac{1}{\sqrt{7} - \sqrt{6}} 7 − 6 1
Let us rationalise the denominator,
Then,
1 7 − 6 = 1 7 − 6 × 7 + 6 7 + 6 ⇒ ( 7 + 6 ) ( 7 ) 2 − ( 6 ) 2 ⇒ 7 + 6 7 − 6 ⇒ 7 + 6 \dfrac{1}{\sqrt{7} - \sqrt{6}} = \dfrac{1}{\sqrt{7} - \sqrt{6}} × \dfrac{\sqrt{7} + \sqrt{6}}{\sqrt{7} + \sqrt{6}} \\[1.5em] \Rightarrow\dfrac{({\sqrt{7} + \sqrt{6}})}{(\sqrt{7})^2 - (\sqrt{6})^2} \\[1.5em] \Rightarrow\dfrac{\sqrt{7} + \sqrt{6}}{7 - 6 } \\[1.5em] \bold{{\Rightarrow\sqrt{7} + \sqrt{6}}} \\[1.5em] 7 − 6 1 = 7 − 6 1 × 7 + 6 7 + 6 ⇒ ( 7 ) 2 − ( 6 ) 2 ( 7 + 6 ) ⇒ 7 − 6 7 + 6 ⇒ 7 + 6
(vii) \text{(vii)} (vii)
1 5 + 2 \dfrac{1}{\sqrt{5}+\sqrt{2}} 5 + 2 1
Let us rationalise the denominator,
Then,
1 5 + 2 = 1 5 + 2 × 5 − 2 5 − 2 ⇒ 5 − 2 ( 5 ) 2 − ( 2 ) 2 ⇒ 5 − 2 3 \dfrac{1}{\sqrt{5} + \sqrt{2}} = \dfrac{1}{\sqrt{5} +\sqrt{2}} × \dfrac{\sqrt{5} - \sqrt{2}}{\sqrt{5} - \sqrt{2}} \\[1.5em] \Rightarrow\dfrac{{\sqrt{5} - \sqrt{2}}}{(\sqrt{5})^2 - (\sqrt{2})^2} \\[1.5em] \bold{\Rightarrow\dfrac{{\sqrt{5} - \sqrt{2}}}{3}} \\[1.5em] 5 + 2 1 = 5 + 2 1 × 5 − 2 5 − 2 ⇒ ( 5 ) 2 − ( 2 ) 2 5 − 2 ⇒ 3 5 − 2
(viii) \text{(viii)} (viii)
2 + 3 2 − 3 \dfrac{\sqrt{2} + \sqrt{3}}{\sqrt{2} - \sqrt{3}} 2 − 3 2 + 3
Let us rationalise the denominator,
Then,
2 + 3 2 − 3 = 2 + 3 2 − 3 × 2 + 3 2 + 3 = ( 2 + 3 ) 2 ( 2 ) 2 − ( 3 ) 2 = ( 2 ) 2 + ( 3 ) 2 + 2 × 2 × 3 ( 2 ) 2 − ( 3 ) 2 = 2 + 3 + 2 2 3 2 − 3 = − ( 5 + 2 6 ) \dfrac{\sqrt{2} + \sqrt{3}}{\sqrt{2} - \sqrt{3}} = \dfrac{\sqrt{2} + \sqrt{3}}{\sqrt{2} - \sqrt{3}} × \dfrac{\sqrt{2} + \sqrt{3}}{\sqrt{2} + \sqrt{3}} \\[1.5em] = \dfrac{{(\sqrt{2} + \sqrt{3})^2}}{(\sqrt{2})^2 - (\sqrt{3})^2} \\[1.5em] = \dfrac{{(\sqrt{2})^2 + (\sqrt{3})^2} + 2 × \sqrt{2} × \sqrt{3}}{(\sqrt{2})^2 - (\sqrt{3})^2} \\[1.5em] = \dfrac{2 + 3 + 2\sqrt{2}\sqrt{3}}{2 - 3} \\[1.5em] = \bold{-(5 + 2\sqrt{6})} \\[1.5em] 2 − 3 2 + 3 = 2 − 3 2 + 3 × 2 + 3 2 + 3 = ( 2 ) 2 − ( 3 ) 2 ( 2 + 3 ) 2 = ( 2 ) 2 − ( 3 ) 2 ( 2 ) 2 + ( 3 ) 2 + 2 × 2 × 3 = 2 − 3 2 + 3 + 2 2 3 = − ( 5 + 2 6 )
Simplify each of the following by rationalising the denominator:
(i) 7 + 3 5 7 − 3 5 (ii) 3 − 2 2 3 + 2 2 (iii) 5 − 3 14 7 + 2 14 \begin{matrix} \text{(i)} & \dfrac{7 + 3\sqrt{5}}{7 - 3\sqrt{5}} \\[1.5em] \text{(ii)} & \dfrac{3 - 2\sqrt{2}}{3 + 2\sqrt{2}} \\[1.5em] \text{(iii)} & \dfrac{5 - 3\sqrt{14}}{7 + 2\sqrt{14}} \\[1.5em] \end{matrix} (i) (ii) (iii) 7 − 3 5 7 + 3 5 3 + 2 2 3 − 2 2 7 + 2 14 5 − 3 14
Answer
( i ) \text(i) ( i )
7 + 3 5 7 − 3 5 \dfrac{7 + 3\sqrt{5}}{7 - 3\sqrt{5}} 7 − 3 5 7 + 3 5
Let us rationalise the denominator,
Then,
7 + 3 5 7 − 3 5 = 7 + 3 5 7 − 3 5 × 7 + 3 5 7 + 3 5 ⇒ ( 7 + 3 5 ) 2 7 2 − ( 3 5 ) 2 ⇒ 7 2 + ( 3 5 ) 2 + 2 × 7 × 3 5 49 − 45 ⇒ 49 + 45 + 42 5 49 − 45 ⇒ 94 + 42 5 49 − 45 ⇒ 2 × ( 47 + 21 5 ) 4 ⇒ ( 47 + 21 5 ) 2 \dfrac{7 + 3\sqrt{5}}{7 - 3\sqrt{5}} = \dfrac{7 + 3\sqrt{5}}{7 - 3\sqrt{5}} × \dfrac{7 + 3\sqrt{5}}{7 + 3\sqrt{5}} \\[1.5em] \Rightarrow\dfrac{(7 + 3\sqrt{5})^2}{7^2 - (3\sqrt{5})^2} \\[1.5em] \Rightarrow\dfrac{7^2 + (3\sqrt{5})^2 + 2 × 7 × 3\sqrt{5}}{49 - 45} \\[1.5em] \Rightarrow\dfrac{49 + 45 + 42\sqrt{5}}{49 - 45} \\[1.5em] \Rightarrow\dfrac{94 + 42\sqrt{5}}{49 - 45} \\[1.5em] \Rightarrow\dfrac{2 × (47 + 21\sqrt{5})}{4} \\[1.5em] \bold{\Rightarrow\dfrac{(47 + 21\sqrt{5})}{2}} \\[1.5em] 7 − 3 5 7 + 3 5 = 7 − 3 5 7 + 3 5 × 7 + 3 5 7 + 3 5 ⇒ 7 2 − ( 3 5 ) 2 ( 7 + 3 5 ) 2 ⇒ 49 − 45 7 2 + ( 3 5 ) 2 + 2 × 7 × 3 5 ⇒ 49 − 45 49 + 45 + 42 5 ⇒ 49 − 45 94 + 42 5 ⇒ 4 2 × ( 47 + 21 5 ) ⇒ 2 ( 47 + 21 5 )
( i i ) \text(ii) ( ii )
3 − 2 2 3 + 2 2 \dfrac{3 - 2\sqrt{2}}{3 + 2\sqrt{2}} 3 + 2 2 3 − 2 2
3 − 2 2 3 + 2 2 = 3 − 2 2 3 + 2 2 × 3 − 2 2 3 − 2 2 ⇒ ( 3 − 2 2 ) 2 3 2 − ( 2 2 ) 2 ⇒ 3 2 + ( 2 2 ) 2 − 2 × 3 × 2 2 9 − 8 ⇒ 9 + 8 − 12 2 1 ⇒ ( 17 − 12 2 ) 1 ⇒ 17 − 12 2 \dfrac{3 - 2\sqrt{2}}{3 + 2\sqrt{2}} = \dfrac{3 - 2\sqrt{2}}{3 + 2\sqrt{2}} × \dfrac{3 - 2\sqrt{2}}{3 - 2\sqrt{2}} \\[1.5em] \Rightarrow\dfrac{(3 - 2\sqrt{2})^2}{3^2 - (2\sqrt{2})^2} \\[1.5em] \Rightarrow\dfrac{3^2 + (2\sqrt{2})^2 - 2 × 3 × 2\sqrt{2}}{9-8} \\[1.5em] \Rightarrow\dfrac{9 + 8 - 12\sqrt{2}}{1} \\[1.5em] \Rightarrow\dfrac{(17 - 12\sqrt{2})}{1} \\[1.5em] \bold{\Rightarrow{17 - 12\sqrt{2}}} \\[1.5em] 3 + 2 2 3 − 2 2 = 3 + 2 2 3 − 2 2 × 3 − 2 2 3 − 2 2 ⇒ 3 2 − ( 2 2 ) 2 ( 3 − 2 2 ) 2 ⇒ 9 − 8 3 2 + ( 2 2 ) 2 − 2 × 3 × 2 2 ⇒ 1 9 + 8 − 12 2 ⇒ 1 ( 17 − 12 2 ) ⇒ 17 − 12 2
( i i i ) \text(iii) ( iii )
5 − 3 14 7 + 2 14 \dfrac{5 - 3\sqrt{14}}{7 + 2\sqrt{14}} 7 + 2 14 5 − 3 14
5 − 3 14 7 + 2 14 = 5 − 3 14 7 + 2 14 × 7 − 2 14 7 − 2 14 ⇒ ( 5 − 3 14 ) ( 7 − 2 14 ) 7 2 − ( 2 14 ) 2 ⇒ 35 − 10 14 − 21 14 + ( 6 × 14 ) 49 − 56 ⇒ 119 − 31 14 − 7 ⇒ − 119 + 31 14 7 \dfrac{5 - 3\sqrt{14}}{7 + 2\sqrt{14}} = \dfrac{5 - 3\sqrt{14}}{7 + 2\sqrt{14}} × \dfrac{7 - 2\sqrt{14}}{7 - 2\sqrt{14}} \\[1.5em] \Rightarrow\dfrac{(5 - 3\sqrt{14})(7 - 2\sqrt{14})}{7^2 - (2\sqrt{14})^2} \\[1.5em] \Rightarrow\dfrac{35 - 10\sqrt{14} - 21\sqrt{14} + (6 × 14)}{49 - 56} \\[1.5em] \Rightarrow\dfrac{119 - 31\sqrt{14}}{-7} \\[1.5em] \bold{\Rightarrow\dfrac{-119+31\sqrt{14}}{7}} \\[1.5em] 7 + 2 14 5 − 3 14 = 7 + 2 14 5 − 3 14 × 7 − 2 14 7 − 2 14 ⇒ 7 2 − ( 2 14 ) 2 ( 5 − 3 14 ) ( 7 − 2 14 ) ⇒ 49 − 56 35 − 10 14 − 21 14 + ( 6 × 14 ) ⇒ − 7 119 − 31 14 ⇒ 7 − 119 + 31 14
Simplify : 7 3 10 + 3 − 2 5 6 + 5 − 3 2 15 + 3 2 \dfrac{7\sqrt{3}}{\sqrt{10} + \sqrt{3}} - \dfrac{2\sqrt{5}}{\sqrt{6} + \sqrt{5}} - \dfrac{3\sqrt{2}}{\sqrt{15} + 3\sqrt{2}} 10 + 3 7 3 − 6 + 5 2 5 − 15 + 3 2 3 2
Answer
7 3 10 + 3 − 2 5 6 + 5 − 3 2 15 + 3 2 \dfrac{7\sqrt{3}}{\sqrt{10} + \sqrt{3}} - \dfrac{2\sqrt{5}}{\sqrt{6} + \sqrt{5}} - \dfrac{3\sqrt{2}}{\sqrt{15} + 3\sqrt{2}} 10 + 3 7 3 − 6 + 5 2 5 − 15 + 3 2 3 2 ....(i) \qquad \text{....(i)} ....(i)
Simplifying each term individually,
7 3 10 + 3 \dfrac{7\sqrt{3}}{\sqrt{10} + \sqrt{3}} 10 + 3 7 3
Let us rationalize its denominator,
7 3 10 + 3 = 7 3 10 + 3 × 10 − 3 10 − 3 ⇒ ( 7 3 ) ( 10 − 3 ) ( 10 ) 2 − ( 3 ) 2 ⇒ 7 3 × 10 − 7 3 × 3 ( 10 ) 2 − ( 3 ) 2 ⇒ 7 3 × 10 − 7 3 × 3 ( 10 ) 2 − ( 3 ) 2 ⇒ 7 30 − ( 7 × 3 ) 10 − 3 ⇒ ( 7 30 − ( 7 × 3 ) 7 ) ⇒ 7 × ( 30 − 3 7 ) ⇒ 30 − 3 ....(ii) \dfrac{7\sqrt{3}}{\sqrt{10} + \sqrt{3}} = \dfrac{7\sqrt{3}}{\sqrt{10} + \sqrt{3}} × \dfrac{\sqrt{10} - \sqrt{3}}{\sqrt{10} - \sqrt{3}} \\[1.5em] \Rightarrow\dfrac{(7\sqrt{3})(\sqrt{10} - \sqrt{3}) }{(\sqrt{10})^2 - (\sqrt{3})^2} \\[1.5em]\Rightarrow\dfrac{7\sqrt{3} × \sqrt{10} - 7\sqrt{3} × \sqrt{3} }{(\sqrt{10})^2 - (\sqrt{3})^2} \\[1.5em] \Rightarrow\dfrac{7\sqrt{3 × 10} - 7\sqrt{3 × 3} }{(\sqrt{10})^2 - (\sqrt{3})^2} \\[1.5em] \Rightarrow\dfrac{7\sqrt{30}-(7×3)}{10-3} \\[1.5em] \Rightarrow\Big(\dfrac{7\sqrt{30}-(7×3)}{7}\Big) \\[1.5em] \Rightarrow 7 ×\Big(\dfrac{\sqrt{30} - 3}{7}\Big) \\[1.5em] \bold{\Rightarrow{\sqrt{30}-3}} \qquad \text{....(ii)} \\[1.5em] 10 + 3 7 3 = 10 + 3 7 3 × 10 − 3 10 − 3 ⇒ ( 10 ) 2 − ( 3 ) 2 ( 7 3 ) ( 10 − 3 ) ⇒ ( 10 ) 2 − ( 3 ) 2 7 3 × 10 − 7 3 × 3 ⇒ ( 10 ) 2 − ( 3 ) 2 7 3 × 10 − 7 3 × 3 ⇒ 10 − 3 7 30 − ( 7 × 3 ) ⇒ ( 7 7 30 − ( 7 × 3 ) ) ⇒ 7 × ( 7 30 − 3 ) ⇒ 30 − 3 ....(ii)
2 5 6 + 5 \dfrac{2\sqrt{5}}{\sqrt{6} + \sqrt{5}} 6 + 5 2 5
Let us rationalize its denominator,
2 5 6 + 5 = 2 5 6 + 5 × 6 − 5 6 − 5 ⇒ ( 2 5 ) ( 6 − 5 ) ( 6 ) 2 − ( 5 ) 2 ⇒ 2 5 × 6 − 2 5 × 5 ( 6 ) 2 − ( 5 ) 2 ⇒ 2 5 × 6 − 2 5 × 5 6 − 5 ⇒ 2 30 − 10 1 ⇒ 2 30 − 10 ....(iii) \dfrac{2\sqrt{5}}{\sqrt{6} + \sqrt{5}} = \dfrac{2\sqrt{5}}{\sqrt{6} + \sqrt{5}} × \dfrac{\sqrt6 - \sqrt{5}}{\sqrt{6} - \sqrt{5}} \\[1.5em] \Rightarrow\dfrac{(2\sqrt{5})(\sqrt{6} - \sqrt{5}) }{(\sqrt{6})^2 - (\sqrt{5})^2} \\[1.5em] \Rightarrow\dfrac{2\sqrt{5} × \sqrt{6} - 2\sqrt{5} ×\sqrt{5} }{(\sqrt{6})^2 - (\sqrt{5})^2} \\[1.5em] \Rightarrow\dfrac{2\sqrt{5 × 6} - 2\sqrt{5 × 5} }{6 - 5} \\[1.5em] \Rightarrow\dfrac{2\sqrt{30} - 10}{1} \\[1.5em] \bold{\Rightarrow{2\sqrt{30} - 10}} \qquad \text{....(iii)} \\[1.5em] 6 + 5 2 5 = 6 + 5 2 5 × 6 − 5 6 − 5 ⇒ ( 6 ) 2 − ( 5 ) 2 ( 2 5 ) ( 6 − 5 ) ⇒ ( 6 ) 2 − ( 5 ) 2 2 5 × 6 − 2 5 × 5 ⇒ 6 − 5 2 5 × 6 − 2 5 × 5 ⇒ 1 2 30 − 10 ⇒ 2 30 − 10 ....(iii)
3 2 15 + 3 2 \dfrac{3\sqrt{2}}{\sqrt{15} + 3\sqrt{2}} 15 + 3 2 3 2
Let us rationalize its denominator,
3 2 15 + 3 2 = 3 2 15 + 3 2 × 15 − 3 2 15 − 3 2 ⇒ ( 3 2 ) ( 15 − 3 2 ) ( 15 ) 2 − ( 3 2 ) 2 ⇒ 3 2 × 15 − 3 2 × 3 2 ( 15 ) 2 − ( 3 2 ) 2 ⇒ 3 30 − 18 15 − 18 ⇒ 3 30 − 18 − 3 ⇒ [ − 3 ( − 30 + 6 ) − 3 ] ⇒ − ( − 30 + 6 1 ) ⇒ 6 − 30 ....(iv) \dfrac{3\sqrt{2}}{\sqrt{15} + 3\sqrt{2}} = \dfrac{3\sqrt{2}}{\sqrt{15} + 3\sqrt{2}} × \dfrac{\sqrt{15} - 3\sqrt{2}}{\sqrt{15} - 3\sqrt{2}} \\[1.5em] \Rightarrow\dfrac{(3\sqrt{2})(\sqrt{15} - 3\sqrt{2}) }{(\sqrt{15})^2 - (3\sqrt{2})^2} \\[1.5em] \Rightarrow\dfrac{3\sqrt{2} × \sqrt{15} - 3\sqrt{2} × 3\sqrt{2} }{(\sqrt{15})^2 - (3\sqrt{2})^2} \\[1.5em] \Rightarrow\dfrac{3\sqrt{30} - 18}{15 - 18} \\[1.5em] \Rightarrow\dfrac{3\sqrt{30} - 18}{-3} \\[1.5em] \Rightarrow\Big[\dfrac{-3(-\sqrt{30} + 6)}{-3}\Big] \\[1.5em] \Rightarrow-\Big(\dfrac{-\sqrt{30} + 6}{1}\Big) \\[1.5em] \bold{\Rightarrow{6 - \sqrt{30}}} \qquad \text{....(iv)} \\[1.5em] 15 + 3 2 3 2 = 15 + 3 2 3 2 × 15 − 3 2 15 − 3 2 ⇒ ( 15 ) 2 − ( 3 2 ) 2 ( 3 2 ) ( 15 − 3 2 ) ⇒ ( 15 ) 2 − ( 3 2 ) 2 3 2 × 15 − 3 2 × 3 2 ⇒ 15 − 18 3 30 − 18 ⇒ − 3 3 30 − 18 ⇒ [ − 3 − 3 ( − 30 + 6 ) ] ⇒ − ( 1 − 30 + 6 ) ⇒ 6 − 30 ....(iv)
Using (ii) , (iii) , (iv) in equation (i):
7 3 10 + 3 − 2 5 6 + 5 − 3 2 15 + 3 2 = ( 30 − 3 ) − ( 2 30 − 10 ) − ( 6 − 30 ) \dfrac{7\sqrt{3}}{\sqrt{10} + \sqrt{3}} - \dfrac{2\sqrt{5}}{\sqrt{6} + \sqrt{5}} - \dfrac{3\sqrt{2}}{\sqrt{15} + 3\sqrt{2}} =(\sqrt{30} - 3)-(2\sqrt{30} - 10) - (6 -\sqrt{30}) \\[1.5em] 10 + 3 7 3 − 6 + 5 2 5 − 15 + 3 2 3 2 = ( 30 − 3 ) − ( 2 30 − 10 ) − ( 6 − 30 ) ⇒ 30 − 3 − 2 30 + 10 − 6 + 30 \Rightarrow \sqrt{30} - 3 - 2\sqrt{30} + 10 - 6 + \sqrt{30} \\[1.5em] ⇒ 30 − 3 − 2 30 + 10 − 6 + 30 ⇒ 2 30 − 2 30 − 3 + 10 − 6 \Rightarrow 2\sqrt{30} - 2\sqrt{30} -3 + 10 - 6 \\[1.5em] ⇒ 2 30 − 2 30 − 3 + 10 − 6 ⇒ 1 \bold{\Rightarrow 1} ⇒ 1
Simplify : 1 4 + 5 + 1 5 + 6 + 1 6 + 7 + 1 7 + 8 + 1 8 + 9 \dfrac{1}{\sqrt{4} + \sqrt{5}} + \dfrac{1}{\sqrt{5} + \sqrt{6}} + \dfrac{1}{\sqrt{6} + \sqrt{7}} + \dfrac{1}{\sqrt{7} + \sqrt{8}} +\dfrac{1}{\sqrt{8} + \sqrt{9}} 4 + 5 1 + 5 + 6 1 + 6 + 7 1 + 7 + 8 1 + 8 + 9 1
Answer
1 4 + 5 + 1 5 + 6 + 1 6 + 7 + 1 7 + 8 + 1 8 + 9 ....(i) \dfrac{1}{\sqrt{4} + \sqrt{5}} + \dfrac{1}{\sqrt{5} + \sqrt{6}} + \dfrac{1}{\sqrt{6} + \sqrt{7}} + \dfrac{1}{\sqrt{7} + \sqrt{8}} +\dfrac{1}{\sqrt{8} + \sqrt{9}} \qquad \text{....(i)} 4 + 5 1 + 5 + 6 1 + 6 + 7 1 + 7 + 8 1 + 8 + 9 1 ....(i)
Simplifying each term individually,
1 4 + 5 \dfrac{1}{\sqrt{4} + \sqrt{5}} 4 + 5 1
Let us rationalise its denominator,
Then,
1 4 + 5 = 1 4 + 5 × 4 − 5 4 − 5 ⇒ 4 − 5 ( 4 ) 2 − ( 5 ) 2 ⇒ 4 − 5 4 − 5 ⇒ − ( 4 − 5 ) ⇒ ( 5 − 4 ) ....(ii) \dfrac{1}{\sqrt{4} + \sqrt{5}} = \dfrac{1}{\sqrt{4} + \sqrt{5}} × \dfrac{\sqrt{4} - \sqrt{5}}{\sqrt{4} - \sqrt{5}} \\[1.5em] \Rightarrow\dfrac{{\sqrt{4} - \sqrt{5}}}{(\sqrt{4})^2 - (\sqrt{5})^2} \\[1.5em] \Rightarrow\dfrac{{\sqrt{4} - \sqrt{5}}}{4 - 5} \\[1.5em] \Rightarrow{-(\sqrt{4} - \sqrt{5})} \\[1.5em] \Rightarrow{(\sqrt{5} - \sqrt{4})} \qquad \text{....(ii)} \\[1.5em] 4 + 5 1 = 4 + 5 1 × 4 − 5 4 − 5 ⇒ ( 4 ) 2 − ( 5 ) 2 4 − 5 ⇒ 4 − 5 4 − 5 ⇒ − ( 4 − 5 ) ⇒ ( 5 − 4 ) ....(ii)
1 5 + 6 \dfrac{1}{\sqrt{5} + \sqrt{6}} 5 + 6 1
Let us rationalise its denominator,
Then,
1 5 + 6 = 1 5 + 6 × 5 − 6 5 − 6 ⇒ 5 − 6 ( 5 ) 2 − ( 6 ) 2 ⇒ 5 − 6 5 − 6 ⇒ − ( 5 − 6 ) ⇒ ( 6 − 5 ) ....(iii) \dfrac{1}{\sqrt{5} + \sqrt{6}} = \dfrac{1}{\sqrt{5}+\sqrt{6}}×\dfrac{\sqrt{5} - \sqrt{6}}{\sqrt{5} - \sqrt{6}} \\[1.5em] \Rightarrow\dfrac{{\sqrt{5} - \sqrt{6}}}{(\sqrt{5})^2 - (\sqrt{6})^2} \\[1.5em] \Rightarrow\dfrac{{\sqrt{5} - \sqrt{6}}}{5 - 6} \\[1.5em] \Rightarrow{-(\sqrt{5} - \sqrt{6})} \\[1.5em] \Rightarrow{(\sqrt{6} - \sqrt{5})} \qquad \text{....(iii)} \\[1.5em] 5 + 6 1 = 5 + 6 1 × 5 − 6 5 − 6 ⇒ ( 5 ) 2 − ( 6 ) 2 5 − 6 ⇒ 5 − 6 5 − 6 ⇒ − ( 5 − 6 ) ⇒ ( 6 − 5 ) ....(iii)
1 6 + 7 \dfrac{1}{\sqrt{6} + \sqrt{7}} 6 + 7 1
Let us rationalise its denominator,
Then,
1 6 + 7 = 1 6 + 7 × 6 − 7 6 − 7 ⇒ 6 − 7 ( 6 ) 2 − ( 7 ) 2 ⇒ 6 − 7 6 − 7 ⇒ − ( 6 − 7 ) ⇒ ( 7 − 6 ) ....(iv) \dfrac{1}{\sqrt{6} + \sqrt{7}} = \dfrac{1}{\sqrt{6}+ \sqrt{7}} × \dfrac{\sqrt{6} - \sqrt{7}}{\sqrt{6} - \sqrt{7}} \\[1.5em] \Rightarrow\dfrac{{\sqrt{6} - \sqrt{7}}}{(\sqrt{6})^2 - (\sqrt{7})^2} \\[1.5em] \Rightarrow\dfrac{{\sqrt{6} - \sqrt{7}}}{6-7} \\[1.5em] \Rightarrow{-(\sqrt{6} - \sqrt{7})} \\[1.5em] \Rightarrow{(\sqrt{7} - \sqrt{6})} \qquad \text{....(iv)} \\[1.5em] 6 + 7 1 = 6 + 7 1 × 6 − 7 6 − 7 ⇒ ( 6 ) 2 − ( 7 ) 2 6 − 7 ⇒ 6 − 7 6 − 7 ⇒ − ( 6 − 7 ) ⇒ ( 7 − 6 ) ....(iv)
1 7 + 8 \dfrac{1}{\sqrt7+\sqrt8} 7 + 8 1
Let us rationalise its denominator,
Then,
1 7 + 8 = 1 7 + 8 × 7 − 8 7 − 8 ⇒ 7 − 8 ( 7 ) 2 − ( 8 ) 2 ⇒ 7 − 8 7 − 8 ⇒ − ( 7 − 8 ) ⇒ ( 8 − 7 ) ....(v) \dfrac{1}{\sqrt{7} + \sqrt{8}} = \dfrac{1}{\sqrt{7}+ \sqrt{8}} × \dfrac{\sqrt{7} - \sqrt{8}}{\sqrt{7} - \sqrt{8}} \\[1.5em] \Rightarrow\dfrac{{\sqrt{7} - \sqrt{8}}}{(\sqrt{7})^2 - (\sqrt{8})^2} \\[1.5em] \Rightarrow\dfrac{{\sqrt{7} - \sqrt{8}}}{7-8} \\[1.5em] \Rightarrow{-(\sqrt{7} - \sqrt{8})} \\[1.5em] \Rightarrow{(\sqrt{8} - \sqrt{7})} \qquad \text{....(v)} \\[1.5em] 7 + 8 1 = 7 + 8 1 × 7 − 8 7 − 8 ⇒ ( 7 ) 2 − ( 8 ) 2 7 − 8 ⇒ 7 − 8 7 − 8 ⇒ − ( 7 − 8 ) ⇒ ( 8 − 7 ) ....(v)
1 4 + 5 \dfrac{1}{\sqrt{4} + \sqrt{5}} 4 + 5 1
Let us rationalise its denominator,
Then,
1 8 + 9 = 1 8 + 9 × 8 − 9 8 − 9 ⇒ 8 − 9 ( 8 ) 2 − ( 9 ) 2 ⇒ 8 − 9 8 − 9 ⇒ − ( 8 − 9 ) ⇒ ( 9 − 8 ) ....(vi) \dfrac{1}{\sqrt{8} + \sqrt{9}} = \dfrac{1}{\sqrt{8}+ \sqrt{9}} × \dfrac{\sqrt{8} - \sqrt{9}}{\sqrt{8} - \sqrt{9}} \\[1.5em] \Rightarrow\dfrac{{\sqrt {8} - \sqrt{9}}}{(\sqrt{8})^2 - (\sqrt{9})^2} \\[1.5em] \Rightarrow\dfrac{{\sqrt{8} - \sqrt{9}}}{8 - 9} \\[1.5em] \Rightarrow{-(\sqrt{8} - \sqrt{9})} \\[1.5em] \Rightarrow{(\sqrt{9}-\sqrt{8})} \qquad \text{....(vi)} \\[1.5em] 8 + 9 1 = 8 + 9 1 × 8 − 9 8 − 9 ⇒ ( 8 ) 2 − ( 9 ) 2 8 − 9 ⇒ 8 − 9 8 − 9 ⇒ − ( 8 − 9 ) ⇒ ( 9 − 8 ) ....(vi)
Using (ii) , (iii) , (iv) , (v) , (vi) in equation (i):
1 4 + 5 + 1 5 + 6 + 1 6 + 7 + 1 7 + 8 + 1 8 + 9 = 5 − 4 + 6 − 5 + 7 − 6 + 8 − 7 + 9 − 8 = 9 − 4 = 3 − 2 = 1 \dfrac{1}{\sqrt4+\sqrt5} + \dfrac{1}{\sqrt5+\sqrt6} + \dfrac{1}{\sqrt6+\sqrt7} + \dfrac{1}{\sqrt7+\sqrt8} +\dfrac{1}{\sqrt8+\sqrt9} \\[1.5em] = \sqrt{5} - \sqrt{4} + \sqrt{6} - \sqrt{5} + \sqrt{7} - \sqrt{6} + \sqrt{8} - \sqrt{7} + \sqrt{9} - \sqrt{8} \\[1.5em] = \sqrt{9} - \sqrt{4} \\[1.5em] = \bold{3 - 2 = 1 } \\[1.5em] 4 + 5 1 + 5 + 6 1 + 6 + 7 1 + 7 + 8 1 + 8 + 9 1 = 5 − 4 + 6 − 5 + 7 − 6 + 8 − 7 + 9 − 8 = 9 − 4 = 3 − 2 = 1
Given a and b are rational numbers. Find a and b if :
(i) 3 − 5 3 + 2 5 = − 19 11 + a 5 (ii) 2 + 3 3 2 − 2 3 = a − b 6 (iii) 7 + 5 7 − 5 − 7 − 5 7 + 5 = a + 7 11 b 5 (iv) 2 2 − 3 2 2 + 2 3 = a + b 24 \begin{matrix} \text{(i)} & \dfrac{3 - \sqrt{5}}{3 + 2\sqrt{5}} = -\dfrac{19}{11} + a\sqrt{5} \\[1.5em] \text{(ii)} & \dfrac{\sqrt{2} + \sqrt{3}}{3\sqrt{2}-2\sqrt{3}} = a - b\sqrt{6} \\[1.5em] \text{(iii)} & \dfrac{7 + \sqrt{5}}{7 - \sqrt{5}} - \dfrac{7 - \sqrt{5}}{7 + \sqrt{5}} = a + \dfrac{7}{11}b\sqrt{5} \\[1.5em] \text{(iv)} & \dfrac{2\sqrt{2} - \sqrt{3}}{2\sqrt{2} + 2\sqrt{3}} = a + b\sqrt{24} \end{matrix} (i) (ii) (iii) (iv) 3 + 2 5 3 − 5 = − 11 19 + a 5 3 2 − 2 3 2 + 3 = a − b 6 7 − 5 7 + 5 − 7 + 5 7 − 5 = a + 11 7 b 5 2 2 + 2 3 2 2 − 3 = a + b 24
Answer
(i) Since, it is given that
3 − 5 3 + 2 5 \dfrac{3 - \sqrt{5}}{3 + 2\sqrt{5}} 3 + 2 5 3 − 5 − 19 11 + a 5 -\dfrac{19}{11} + a\sqrt{5} − 11 19 + a 5
On solving,
( 3 − 5 ) ( 3 + 2 5 ) × ( 3 − 2 5 ) ( 3 − 2 5 ) ⇒ 9 − 6 5 − 3 5 + 10 3 2 − ( 2 5 ) 2 ⇒ 19 − 9 5 9 − 20 ⇒ − 9 5 + 19 − 11 ⇒ − 19 11 + 9 5 11 ∴ − 19 11 + 9 5 11 = − 19 11 + a 5 \dfrac{(3 - \sqrt{5})}{(3 + 2\sqrt{5})} × \dfrac{(3-2\sqrt{5})}{(3-2\sqrt{5})} \\[1.5em] \Rightarrow \dfrac{9 - 6\sqrt{5} - 3\sqrt{5} + 10 }{3^2 - (2\sqrt{5})^2} \\[1.5em] \Rightarrow \dfrac{19 - 9{\sqrt{5}}}{9 - 20} \\[1.5em] \Rightarrow \dfrac{-9{\sqrt{5}} + 19}{-11} \\[1.5em] \Rightarrow -\dfrac{19}{11} + \dfrac{9\sqrt{5}}{11} \\[1.5em] \therefore -\dfrac{19}{11} + \dfrac{9\sqrt5}{11} = -\dfrac{19}{11} + a\sqrt{5} ( 3 + 2 5 ) ( 3 − 5 ) × ( 3 − 2 5 ) ( 3 − 2 5 ) ⇒ 3 2 − ( 2 5 ) 2 9 − 6 5 − 3 5 + 10 ⇒ 9 − 20 19 − 9 5 ⇒ − 11 − 9 5 + 19 ⇒ − 11 19 + 11 9 5 ∴ − 11 19 + 11 9 5 = − 11 19 + a 5
Hence, a = 9 11 \dfrac{9}{11} 11 9
(ii) Since, it is given that
2 + 3 3 2 − 2 3 = a − b 6 \dfrac{\sqrt{2} + \sqrt{3}}{3\sqrt{2} - 2\sqrt{3}} = a - b\sqrt6 3 2 − 2 3 2 + 3 = a − b 6
On solving,
2 + 3 3 2 − 2 3 × ( 3 2 + 2 3 ) ( 3 2 + 2 3 ) ⇒ 6 + 2 6 + 3 6 + 6 ( 3 2 ) 2 − ( 2 3 ) 2 ⇒ 12 + 5 6 18 − 12 ⇒ 12 + 5 6 6 ⇒ 12 6 + 5 6 6 ⇒ 2 + 5 6 6 ∴ 2 − ( − 5 6 ) 6 = a − b 6 \dfrac{\sqrt{2} + \sqrt3}{3\sqrt{2} - 2\sqrt{3}} × \dfrac{(3\sqrt{2} + 2\sqrt{3})}{(3\sqrt{2} + 2\sqrt{3})} \\[1.5em] \Rightarrow \dfrac{6 + 2\sqrt{6} + 3\sqrt6 + 6 }{(3\sqrt{2})^2 -(2\sqrt{3})^2} \\[1.5em] \Rightarrow \dfrac{12 + 5{\sqrt6}}{18-12} \\[1.5em] \Rightarrow \dfrac{12 + 5{\sqrt{6}}}{6} \\[1.5em] \Rightarrow \dfrac{12}{6} + \dfrac{5\sqrt{6}}{6} \\[1.5em] \Rightarrow 2 + \dfrac{5\sqrt{6}}{6} \\[1.5em] \therefore 2 - \dfrac{(-5\sqrt{6})}{6} = a - b\sqrt{6} 3 2 − 2 3 2 + 3 × ( 3 2 + 2 3 ) ( 3 2 + 2 3 ) ⇒ ( 3 2 ) 2 − ( 2 3 ) 2 6 + 2 6 + 3 6 + 6 ⇒ 18 − 12 12 + 5 6 ⇒ 6 12 + 5 6 ⇒ 6 12 + 6 5 6 ⇒ 2 + 6 5 6 ∴ 2 − 6 ( − 5 6 ) = a − b 6
Hence, a = 2 and b = − 5 6 \dfrac{-5}{6} 6 − 5
(iii) Since, it is given that
( 7 + 5 ) ( 7 − 5 ) − ( 7 − 5 ) ( 7 + 5 ) = a + 7 11 b 5 \dfrac{(7 + \sqrt5)}{(7 - \sqrt{5})} - \dfrac{(7 - \sqrt5)}{(7 + \sqrt{5})} = a+\dfrac{7}{11}b\sqrt{5} ( 7 − 5 ) ( 7 + 5 ) − ( 7 + 5 ) ( 7 − 5 ) = a + 11 7 b 5
On solving,
( 7 + 5 ) ( 7 + 5 ) − ( 7 − 5 ) ( 7 − 5 ) ( 7 − 5 ) ( 7 + 5 ) ( 7 + 5 ) 2 − ( 7 − 5 ) 2 ( 7 − 5 ) ( 7 + 5 ) ⇒ ( 7 2 + 2 × 7 × 5 + ( 5 ) 2 ) − ( 7 2 − 2 × 7 × 5 + ( 5 ) 2 ) 7 2 − ( 5 ) 2 ⇒ ( 49 + 2 × 7 5 + 5 ) − ( 49 − 2 × 7 5 + 5 ) 7 2 − ( 5 ) 2 ⇒ 49 + 14 5 + 5 − 49 + 14 5 − 5 7 2 − ( 5 ) 2 ⇒ 28 5 49 − 5 ⇒ 28 5 44 ⇒ 7 5 11 ∴ a + 7 11 b 5 = 7 5 11 ⇒ a + 7 11 b 5 = 0 + 7 11 × 1 × 5 \dfrac{(7 + \sqrt{5})(7 + \sqrt{5}) - (7-\sqrt{5})(7 - \sqrt{5})}{(7 - \sqrt{5})(7 + \sqrt{5})} \\[1.5em] \dfrac{(7 + \sqrt{5})^2 - (7 - \sqrt{5})^2}{(7 - \sqrt{5})(7+\sqrt{5})} \\[1.5em] \Rightarrow \dfrac{(7^2 + 2 × 7 × \sqrt5 + (\sqrt{5})^2)-(7^2 - 2 × 7 × \sqrt5 + (\sqrt{5})^2)}{7^2-(\sqrt5)^2} \\[1.5em] \Rightarrow \dfrac{(49 + 2 × 7\sqrt{5} + 5)-(49 - 2 × 7\sqrt{5}+ 5)}{7^2 - (\sqrt{5})^2} \\[1.5em] \Rightarrow \dfrac{49+14\sqrt{5} + 5 - 49 + 14\sqrt{5} - 5}{7^2 - (\sqrt5)^2} \\[1.5em] \Rightarrow \dfrac{28{\sqrt{5}}}{49 - 5} \\[1.5em] \Rightarrow \dfrac{28{\sqrt{5}}}{44} \\[1.5em] \Rightarrow \dfrac{7{\sqrt{5}}}{11} \\[1.5em] \therefore a+\dfrac{7}{11}b\sqrt{5} = \dfrac{7{\sqrt5}}{11} \\[1.5em] \Rightarrow a + \dfrac{7}{11}b\sqrt{5} = 0 + \dfrac{7}{11} × 1 × \sqrt{5} ( 7 − 5 ) ( 7 + 5 ) ( 7 + 5 ) ( 7 + 5 ) − ( 7 − 5 ) ( 7 − 5 ) ( 7 − 5 ) ( 7 + 5 ) ( 7 + 5 ) 2 − ( 7 − 5 ) 2 ⇒ 7 2 − ( 5 ) 2 ( 7 2 + 2 × 7 × 5 + ( 5 ) 2 ) − ( 7 2 − 2 × 7 × 5 + ( 5 ) 2 ) ⇒ 7 2 − ( 5 ) 2 ( 49 + 2 × 7 5 + 5 ) − ( 49 − 2 × 7 5 + 5 ) ⇒ 7 2 − ( 5 ) 2 49 + 14 5 + 5 − 49 + 14 5 − 5 ⇒ 49 − 5 28 5 ⇒ 44 28 5 ⇒ 11 7 5 ∴ a + 11 7 b 5 = 11 7 5 ⇒ a + 11 7 b 5 = 0 + 11 7 × 1 × 5
Hence, value of a = 0 and b = 1.
(iv) Since, it is given that
2 2 − 3 2 2 + 2 3 = a + b 24 \dfrac{2\sqrt{2} - \sqrt{3}}{2\sqrt{2} + 2\sqrt{3}} = a + b\sqrt{24} 2 2 + 2 3 2 2 − 3 = a + b 24
Rationalizing, 2 2 − 3 2 2 + 2 3 \dfrac{2\sqrt{2} - \sqrt{3}}{2\sqrt{2} + 2\sqrt{3}} 2 2 + 2 3 2 2 − 3
⇒ 2 2 − 3 2 2 + 2 3 × 2 2 − 2 3 2 2 − 2 3 ⇒ ( 2 2 − 3 ) ( 2 2 − 2 3 ) ( 2 2 ) 2 − ( 2 3 ) 2 ⇒ 8 − 4 6 − 2 6 + 6 8 − 12 ⇒ 14 − 6 6 − 4 ⇒ − 14 4 + 6 6 4 ⇒ − 7 2 + 3 2 6 ⇒ − 7 2 + 3 6 2 × 2 2 ⇒ − 7 2 + 3 6 × 2 2 4 ⇒ − 7 2 + 3 24 4 . \Rightarrow \dfrac{2\sqrt{2} - \sqrt{3}}{2\sqrt{2} + 2\sqrt{3}} \times \dfrac{2\sqrt{2} - 2\sqrt{3}}{2\sqrt{2} - 2\sqrt{3}} \\[1em] \Rightarrow \dfrac{(2\sqrt{2} - \sqrt{3})(2\sqrt{2} - 2\sqrt{3})}{(2\sqrt{2})^2 - (2\sqrt{3})^2} \\[1em] \Rightarrow \dfrac{8 - 4\sqrt{6} - 2\sqrt{6} + 6}{8 - 12} \\[1em] \Rightarrow \dfrac{14 - 6\sqrt{6}}{-4} \\[1em] \Rightarrow -\dfrac{14}{4} + \dfrac{6\sqrt{6}}{4} \\[1em] \Rightarrow -\dfrac{7}{2} + \dfrac{3}{2}\sqrt{6} \\[1em] \Rightarrow -\dfrac{7}{2} + \dfrac{3\sqrt{6}}{2} \times \dfrac{2}{2} \\[1em] \Rightarrow -\dfrac{7}{2} + \dfrac{3\sqrt{6 \times 2^2}}{4} \\[1em] \Rightarrow -\dfrac{7}{2} + \dfrac{3\sqrt{24}}{4}. ⇒ 2 2 + 2 3 2 2 − 3 × 2 2 − 2 3 2 2 − 2 3 ⇒ ( 2 2 ) 2 − ( 2 3 ) 2 ( 2 2 − 3 ) ( 2 2 − 2 3 ) ⇒ 8 − 12 8 − 4 6 − 2 6 + 6 ⇒ − 4 14 − 6 6 ⇒ − 4 14 + 4 6 6 ⇒ − 2 7 + 2 3 6 ⇒ − 2 7 + 2 3 6 × 2 2 ⇒ − 2 7 + 4 3 6 × 2 2 ⇒ − 2 7 + 4 3 24 .
Comparing − 7 2 + 3 24 4 -\dfrac{7}{2} + \dfrac{3\sqrt{24}}{4} − 2 7 + 4 3 24 b 24 b\sqrt{24} b 24
a = − 7 2 , b = 3 4 -\dfrac{7}{2}, b = \dfrac{3}{4} − 2 7 , b = 4 3
Hence, a = − 7 2 , b = 3 4 -\dfrac{7}{2}, b = \dfrac{3}{4} − 2 7 , b = 4 3
If 7 + 3 5 3 + 5 − 7 − 3 5 3 − 5 = p + q 5 \dfrac{7 + 3\sqrt{5}}{3 + \sqrt{5}} - \dfrac{7 - 3\sqrt{5}}{3 - \sqrt{5}} = p + q\sqrt{5} 3 + 5 7 + 3 5 − 3 − 5 7 − 3 5 = p + q 5
Answer
Since, it is given that
( 7 + 3 5 ) 3 + 5 \dfrac{(7 + 3\sqrt{5})}{3 + \sqrt{5}} 3 + 5 ( 7 + 3 5 ) ( 7 − 3 5 ) ( 3 − 5 ) \dfrac{(7 - 3\sqrt{5})} {(3 - \sqrt{5})} ( 3 − 5 ) ( 7 − 3 5 ) 5 \sqrt{5} 5
On solving,
( 7 + 3 5 ) ( 3 − 5 ) − ( 7 − 3 5 ) ( 3 + 5 ) ( 3 + 5 ) ( 3 − 5 ) ⇒ ( 21 − 7 5 + 9 5 − 15 ) − ( 21 + 7 5 − 9 5 − 15 ) 3 2 − ( 5 ) 2 ⇒ ( 21 − 7 5 + 9 5 − 15 ) − ( 21 + 7 5 − 9 5 − 15 ) 9 − 5 ⇒ 21 − 7 5 + 9 5 − 15 − 21 − 7 5 + 9 5 + 15 9 − 5 ⇒ 18 5 − 14 5 4 ⇒ 4 × 5 4 = 5 ∴ 5 = p + q 5 ⇒ 0 + 1 × 5 = p + q 5 \dfrac{(7 + 3\sqrt{5})(3 - \sqrt{5}) - (7 - 3\sqrt{5})(3 + \sqrt{5})}{(3 + \sqrt{5})(3 - \sqrt{5})} \\[1.5em] \Rightarrow \dfrac{(21 - 7\sqrt{5} + 9\sqrt{5} - 15) - (21 + 7\sqrt{5} - 9\sqrt{5} - 15)}{3^2 - {(\sqrt{5})}^2} \\[1.5em] \Rightarrow \dfrac{(21 - 7\sqrt{5} + 9\sqrt{5} - 15) - (21 + 7\sqrt{5} - 9\sqrt{5} - 15)}{9-5} \\[1.5em] \Rightarrow \dfrac{21 - 7\sqrt{5} + 9\sqrt{5} - 15 - 21 - 7\sqrt{5} + 9\sqrt{5} + 15}{9-5} \\[1.5em] \Rightarrow \dfrac{18{\sqrt{5}} - 14{\sqrt{5}}}{4} \\[1.5em] \Rightarrow \dfrac{4 × {\sqrt{5}}}{4} = \sqrt{5} \\[1.5em] \therefore\sqrt{5} = p + q{\sqrt{5}} \\[1.5em] \Rightarrow 0 + 1 × \sqrt{5} = p + q{\sqrt{5}} ( 3 + 5 ) ( 3 − 5 ) ( 7 + 3 5 ) ( 3 − 5 ) − ( 7 − 3 5 ) ( 3 + 5 ) ⇒ 3 2 − ( 5 ) 2 ( 21 − 7 5 + 9 5 − 15 ) − ( 21 + 7 5 − 9 5 − 15 ) ⇒ 9 − 5 ( 21 − 7 5 + 9 5 − 15 ) − ( 21 + 7 5 − 9 5 − 15 ) ⇒ 9 − 5 21 − 7 5 + 9 5 − 15 − 21 − 7 5 + 9 5 + 15 ⇒ 4 18 5 − 14 5 ⇒ 4 4 × 5 = 5 ∴ 5 = p + q 5 ⇒ 0 + 1 × 5 = p + q 5
Hence, value of p = 0 and q = 1.
Rationalise the denominator of the following and hence evaluate by taking 2 \sqrt{2} 2 3 \sqrt{3} 3
(i) 2 2 + 2 (ii) 1 3 + 2 \begin{matrix} \text{(i)} & \dfrac{\sqrt2}{2+\sqrt2} \\[1.5em] \text{(ii)} & \dfrac{1}{\sqrt3+\sqrt2} \\[1.5em] \end{matrix} (i) (ii) 2 + 2 2 3 + 2 1
Answer
(i) Rationalise the denominator ,
2 2 + 2 = 2 2 + 2 × 2 − 2 2 − 2 ⇒ 2 × ( 2 − 2 ) 2 2 − ( 2 ) 2 ⇒ 2 2 − 2 4 − 2 ⇒ ( 2 − 1 ) \dfrac{\sqrt{2}}{2 + \sqrt{2}} = \dfrac{\sqrt{2}}{2+ \sqrt{2}} × \dfrac{2 - \sqrt{2}}{2 - \sqrt{2}} \\[1.5em] \Rightarrow \sqrt{2} × \dfrac{(2 - \sqrt{2})}{2^2 - (\sqrt{2})^2} \\[1.5em] \Rightarrow \dfrac{2{\sqrt{2}} - 2}{4 - 2} \\[1.5em] \Rightarrow (\sqrt{2} - 1) \\[1.5em] 2 + 2 2 = 2 + 2 2 × 2 − 2 2 − 2 ⇒ 2 × 2 2 − ( 2 ) 2 ( 2 − 2 ) ⇒ 4 − 2 2 2 − 2 ⇒ ( 2 − 1 )
Since, 2 \sqrt{2} 2
⇒ 1.414 − 1 ⇒ 0.414 \Rightarrow {1.414 - 1} \\[1.5em] \Rightarrow\bold{ 0.414} \\[1.5em] ⇒ 1.414 − 1 ⇒ 0.414
( ii ) (\text{ii}) ( ii )
1 3 + 2 = 1 3 + 2 × 3 − 2 3 − 2 ⇒ ( 3 − 2 ) ( 3 ) 2 − ( 2 ) 2 ⇒ 3 − 2 3 − 2 ⇒ ( 3 − 2 ) \dfrac{1}{\sqrt{3} + \sqrt{2}} = \dfrac{1}{\sqrt{3} + \sqrt{2}} × \dfrac{{\sqrt{3} - \sqrt{2}}}{{\sqrt{3} - \sqrt{2}}} \\[1.5em] \Rightarrow \dfrac{(\sqrt{3} - \sqrt{2})}{(\sqrt{3})^2 -(\sqrt{2})^2} \\[1.5em] \Rightarrow \dfrac{\sqrt{3} - \sqrt{2}}{3-2} \\[1.5em] \Rightarrow (\sqrt{3} - \sqrt{2}) \\[1.5em] 3 + 2 1 = 3 + 2 1 × 3 − 2 3 − 2 ⇒ ( 3 ) 2 − ( 2 ) 2 ( 3 − 2 ) ⇒ 3 − 2 3 − 2 ⇒ ( 3 − 2 )
Since, 3 \sqrt{3} 3
⇒ 1.732 − 1.414 ⇒ 0.318 \Rightarrow {1.732 - 1.414} \\[1.5em] \Rightarrow\bold{0.318} \\[1.5em] ⇒ 1.732 − 1.414 ⇒ 0.318
If a = 2 + 3 \sqrt{3} 3 a − 1 a a - \dfrac{1}{a} a − a 1
Answer
Given,
a = 2 + 3 ∴ 1 a = 1 2 + 3 × 2 − 3 2 − 3 ⇒ 2 − 3 2 2 − ( 3 ) 2 ⇒ 2 − 3 4 − 3 ⇒ ( 2 − 3 ) ∴ a − 1 a = 2 + 3 − 2 + 3 ⇒ a − 1 a = 2 3 a = 2 + \sqrt{3} \\[1.5em] \therefore \dfrac{1}{a} = \dfrac{1}{2 + \sqrt{3}} ×\dfrac{2 - \sqrt{3}}{2 - \sqrt{3}} \\[1.5em] \Rightarrow \dfrac{2 - \sqrt{3}}{2^2 - (\sqrt{3})^2} \\[1.5em] \Rightarrow \dfrac{2 - \sqrt{3}}{4 - 3} \\[1.5em] \Rightarrow (2 - \sqrt{3}) \\[1.5em] \therefore a - \dfrac{1}{a} = 2 + \sqrt{3} - 2 + \sqrt{3} \\[1.5em] \Rightarrow\bold{ a - \dfrac{1}{a} = 2\sqrt{3}} \\[1.5em] a = 2 + 3 ∴ a 1 = 2 + 3 1 × 2 − 3 2 − 3 ⇒ 2 2 − ( 3 ) 2 2 − 3 ⇒ 4 − 3 2 − 3 ⇒ ( 2 − 3 ) ∴ a − a 1 = 2 + 3 − 2 + 3 ⇒ a − a 1 = 2 3
If x = 1 − 2 x = 1-\sqrt{2} x = 1 − 2 ( x − 1 x ) 4 {\Big(x-\dfrac{1}{x}\Big)}^4 ( x − x 1 ) 4
Answer
Given,
x = 1 − 2 ∴ 1 x = 1 1 − 2 × 1 + 2 1 + 2 ⇒ 1 + 2 1 − ( 2 ) 2 ⇒ 1 + 2 1 − 2 ⇒ − ( 1 + 2 ) ∴ x − 1 x = 1 − 2 + 1 + 2 ⇒ x − 1 x = 2 ∴ ( x − 1 x ) 4 = 2 4 = 16 x = 1 - \sqrt{2} \\[1.5em] \therefore \dfrac{1}{x} = \dfrac{1}{1 - \sqrt{2}} × \dfrac{1 + \sqrt{2}}{1 + \sqrt{2}} \\[1.5em] \Rightarrow \dfrac{1 + \sqrt{2}}{1 - (\sqrt{2})^2} \\[1.5em] \Rightarrow \dfrac{1 + \sqrt{2}}{1 - 2} \\[1.5em] \Rightarrow - (1 +\sqrt{2}) \\[1.5em] \therefore x - \dfrac{1}{x} = 1 - \sqrt{2} + 1 + \sqrt{2} \\[1.5em] \Rightarrow x - \dfrac{1}{x} = 2 \\[1.5em] \therefore \bold{{\Big(x-\dfrac{1}{x}\Big)}^4 = 2 ^4 = 16} x = 1 − 2 ∴ x 1 = 1 − 2 1 × 1 + 2 1 + 2 ⇒ 1 − ( 2 ) 2 1 + 2 ⇒ 1 − 2 1 + 2 ⇒ − ( 1 + 2 ) ∴ x − x 1 = 1 − 2 + 1 + 2 ⇒ x − x 1 = 2 ∴ ( x − x 1 ) 4 = 2 4 = 16
If x = 5 − 2 6 5 - 2\sqrt{6} 5 − 2 6 x 2 + 1 x 2 x^2 + \dfrac{1}{x^2} x 2 + x 2 1
Answer
Given x = 5 − 2 6 5 - 2\sqrt{6} 5 − 2 6
∴ 1 x = 1 5 − 2 6 = 1 5 − 2 6 × 5 + 2 6 5 + 2 6 ⇒ 5 + 2 6 ( 5 ) 2 − ( 2 6 ) 2 ⇒ 5 + 2 6 25 − 24 = 5 + 2 6 1 ⇒ 1 x = 5 + 2 6 ∴ ( x + 1 x ) = ( 5 − 2 6 ) + ( 5 + 2 6 ) = 10 ....(i) \therefore \dfrac{1}{x} = \dfrac{1}{5 - 2\sqrt{6}} = \dfrac{1}{5 - 2\sqrt{6}} × \dfrac{5 + 2\sqrt{6}}{5 + 2\sqrt{6}} \\[1.5em] \Rightarrow \dfrac{5 + 2\sqrt{6}}{(5)^2 - (2\sqrt{6})^2} \\[1.5em] \Rightarrow \dfrac{5 + 2\sqrt{6}}{25 - 24} = \dfrac{5 + 2\sqrt{6}}{1} \\[1.5em] \Rightarrow\dfrac{1}{x} = 5+2\sqrt6 \\[1.5em] \therefore (x + \dfrac{1}{x}) = (5 - 2\sqrt6) + (5 + 2\sqrt6) = 10 \qquad \text{....(i)} ∴ x 1 = 5 − 2 6 1 = 5 − 2 6 1 × 5 + 2 6 5 + 2 6 ⇒ ( 5 ) 2 − ( 2 6 ) 2 5 + 2 6 ⇒ 25 − 24 5 + 2 6 = 1 5 + 2 6 ⇒ x 1 = 5 + 2 6 ∴ ( x + x 1 ) = ( 5 − 2 6 ) + ( 5 + 2 6 ) = 10 ....(i)
We know that ( x + 1 x ) 2 = x 2 + 1 x 2 + 2 {\Big(x + \dfrac{1}{x}\Big)}^2 = x^2 + \dfrac{1}{x^2} + 2 ( x + x 1 ) 2 = x 2 + x 2 1 + 2
⇒ x 2 + 1 x 2 = ( x + 1 x ) 2 − 2 ⇒ x 2 + 1 x 2 = 10 2 − 2..... using(i) ⇒ x 2 + 1 x 2 = 100 − 2 x 2 + 1 x 2 = 98 \Rightarrow x^2 + \dfrac{1}{x^2} = {\Big(x + \dfrac{1}{x}\Big)}^2 -2 \\[1.5em] \Rightarrow x^2 + \dfrac{1}{x^2} = 10^2 - 2 ..... \text{using(i)} \\[1.5em] \Rightarrow x^2 + \dfrac{1}{x^2} = 100 - 2 \\[1.5em] \bold{x^2 + \dfrac{1}{x^2} = 98} ⇒ x 2 + x 2 1 = ( x + x 1 ) 2 − 2 ⇒ x 2 + x 2 1 = 1 0 2 − 2..... using(i) ⇒ x 2 + x 2 1 = 100 − 2 x 2 + x 2 1 = 98
If p = 2 − 5 2 + 5 \dfrac{2-\sqrt{5}}{2+\sqrt{5}} 2 + 5 2 − 5 2 + 5 2 − 5 \dfrac{2+\sqrt{5}}{2-\sqrt{5}} 2 − 5 2 + 5
(i) p + q
(ii) p - q
(iii) p2 + q2
(iv) p2 - q2
Answer
(i)
p + q = 2 − 5 2 + 5 + 2 + 5 2 − 5 ⇒ ( 2 − 5 ) 2 + ( 2 + 5 ) 2 ( 2 − 5 ) ( 2 + 5 ) ⇒ 4 + 5 − 4 5 + 4 + 5 + 4 5 ( 2 ) 2 − ( 5 ) 2 p + q = − 18 ....(i) p+q = \dfrac{2-\sqrt{5}}{2+\sqrt{5}} +\dfrac{2+\sqrt{5}}{2-\sqrt{5}} \\[1.5em] \Rightarrow\dfrac{(2-\sqrt{5})^2 + (2+\sqrt{5})^2}{(2-\sqrt{5})(2+\sqrt{5})} \\[1.5em] \Rightarrow\dfrac{4 + 5 - 4\sqrt{5} + 4 + 5 + 4\sqrt{5}}{(2)^2 -(\sqrt{5})^2} \\[1.5em] \bold{p+q = -18} \qquad \text{....(i)} \\[1.5em] p + q = 2 + 5 2 − 5 + 2 − 5 2 + 5 ⇒ ( 2 − 5 ) ( 2 + 5 ) ( 2 − 5 ) 2 + ( 2 + 5 ) 2 ⇒ ( 2 ) 2 − ( 5 ) 2 4 + 5 − 4 5 + 4 + 5 + 4 5 p + q = − 18 ....(i)
( ii ) (\text{ii}) ( ii )
p − q = 2 − 5 2 + 5 − 2 + 5 2 − 5 ⇒ ( 2 − 5 ) 2 − ( 2 + 5 ) 2 ( 2 − 5 ) ( 2 + 5 ) ⇒ 4 + 5 − 4 5 − 4 − 5 − 4 5 ( 2 ) 2 − ( 5 ) 2 ⇒ p − q = 8 5 ....(ii) p-q = \dfrac{2-\sqrt{5}}{2+\sqrt{5}} -\dfrac{2+\sqrt{5}}{2-\sqrt{5}} \\[1.5em] \Rightarrow\dfrac{(2-\sqrt{5})^2 - (2+\sqrt{5})^2}{(2-\sqrt{5})(2+\sqrt{5})} \\[1.5em] \Rightarrow\dfrac{4 + 5 - 4\sqrt{5} - 4 - 5 - 4\sqrt{5}}{(2)^2 - (\sqrt{5})^2} \\[1.5em] \Rightarrow\bold {p - q = 8\sqrt{5}} \qquad \text{....(ii)} p − q = 2 + 5 2 − 5 − 2 − 5 2 + 5 ⇒ ( 2 − 5 ) ( 2 + 5 ) ( 2 − 5 ) 2 − ( 2 + 5 ) 2 ⇒ ( 2 ) 2 − ( 5 ) 2 4 + 5 − 4 5 − 4 − 5 − 4 5 ⇒ p − q = 8 5 ....(ii)
( iii ) (\text{iii}) ( iii )
( p + q ) 2 = ( p ) 2 + ( q ) 2 + 2 p q ⇒ ( p ) 2 + ( q ) 2 = ( p + q ) 2 − 2 p q ....(iii) ⇒ p q = 2 − 5 2 + 5 × 2 + 5 2 − 5 = 1 ....(iv) (p+q)^2 = (p)^2 + (q)^2 +2pq \\[1.5em] \Rightarrow(p)^2+(q)^2 =(p+q)^2 -2pq \qquad \text{....(iii)} \\[1.5em] \Rightarrow pq = \dfrac{2-\sqrt{5}}{2+\sqrt{5}} ×\dfrac{2+\sqrt{5}} {2-\sqrt{5}} = 1 \qquad \text{....(iv)} \\[1.5em] ( p + q ) 2 = ( p ) 2 + ( q ) 2 + 2 pq ⇒ ( p ) 2 + ( q ) 2 = ( p + q ) 2 − 2 pq ....(iii) ⇒ pq = 2 + 5 2 − 5 × 2 − 5 2 + 5 = 1 ....(iv)
substituting value of (i) and (iv) in (iii) :
p 2 + q 2 = ( − 18 ) 2 − 2 = 324 − 2 = 322 \bold{p^2+q^2 = (-18)^2 - 2 = 324-2 = 322} p 2 + q 2 = ( − 18 ) 2 − 2 = 324 − 2 = 322
( iv ) (\text{iv}) ( iv )
( p ) 2 − ( q ) 2 = ( p + q ) ( p − q ) ....(v) (p)^2 - (q)^2 = (p+q)(p-q) \qquad \text{....(v)} ( p ) 2 − ( q ) 2 = ( p + q ) ( p − q ) ....(v)
Using (i) and (ii) in (v) we get,
p 2 − q 2 = − 18 × 8 5 = − 144 5 \bold{p^2 - q^2 = -18 × 8\sqrt{5} = -144\sqrt{5}} p 2 − q 2 = − 18 × 8 5 = − 144 5
Multiple Choice Questions
Choose the correct statement :
Reciprocal of every rational number is a rational number. The square roots of all positive integers are irrational numbers. The product of a rational and an irrational number is an irrational number. The difference of a rational number and an irrational number is an irrational number. Answer
The difference of a rational number and an irrational number is an irrational number.
For example, 2 is rational number, 3 \sqrt{3} 3 3 \sqrt{3} 3
∴ Option 4, is the correct option.
Every rational number is
a natural number an integer a real number a whole number Answer
Every rational number is a real number.
∴ Option 3, is the correct option.
Between two rational numbers
there is no rational number there is exactly one rational number there are infinitely many rational numbers there are only rational numbers and no irrational numbers. Answer
Between two rational numbers there are infinitely many rational numbers.
∴ Option 3, is the correct option.
Decimal representation of a rational number cannot be
terminating non-terminating non-terminating repeating non-terminating non-repeating Answer
Decimal representation of a rational number is terminating or non-terminating repeating but not non-terminating non-repeating.
∴ Option 4, is the correct option.
The product of any two irrational numbers is
always an irrational number always a rational number always an integer sometimes rational, sometimes irrational Answer
The product of any two irrational numbers is sometimes rational, sometimes irrational
For example, 2 3 2\sqrt{3} 2 3 3 3 3\sqrt{3} 3 3
Their product 2 3 2\sqrt{3} 2 3 3 3 3\sqrt{3} 3 3 ( 3 ) 2 (\sqrt{3})^2 ( 3 ) 2
Again, let 2 2 2\sqrt{2} 2 2 3 3 3\sqrt{3} 3 3
Their product 2 2 2\sqrt{2} 2 2 3 3 3\sqrt{3} 3 3 2 \sqrt{2} 2 3 \sqrt{3} 3 6 6 6\sqrt{6} 6 6
∴ Option 4, is the correct option.
The division of two irrational numbers is
a rational number an irrational number either a rational number or an irrational number neither rational number nor irrational number Answer
The division of two irrational numbers is either a rational number or an irrational number.
For example, let 2 3 2\sqrt{3} 2 3 3 3 3\sqrt{3} 3 3
2 3 3 3 \dfrac{2\sqrt{3}}{3\sqrt{3}} 3 3 2 3 2 3 \dfrac{2}{3} 3 2
Let 2 3 2\sqrt{3} 2 3 3 5 3\sqrt{5} 3 5
2 3 3 5 \dfrac{2\sqrt{3}}{3\sqrt{5}} 3 5 2 3
∴ Option 3, is the correct option.
Which of the following is an irrational number
4 9 \sqrt{\dfrac{4}{9}} 9 4
12 3 \dfrac{\sqrt{12}}{\sqrt{3}} 3 12
7 \sqrt{7} 7
81 \sqrt{81} 81
Answer
4 9 \sqrt{\dfrac{4}{9}} 9 4 4 9 \dfrac{\sqrt{4}}{\sqrt{9}} 9 4 ( 2 ) 2 ( 3 ) 2 \dfrac{(\sqrt{2})^2}{(\sqrt{3})^2} ( 3 ) 2 ( 2 ) 2 2 3 \dfrac{2}{3} 3 2
12 3 \dfrac{\sqrt{12}}{\sqrt{3}} 3 12 2 × 2 × 3 3 \dfrac{\sqrt{2 × 2 × 3}}{\sqrt{3}} 3 2 × 2 × 3 2 3 3 \dfrac{2\sqrt{3}}{\sqrt{3}} 3 2 3
7 \sqrt{7} 7
81 \sqrt{81} 81 9 × 9 \sqrt{9 × 9} 9 × 9
∴ Option 3, is the correct option.
Which of the following numbers has terminating decimal representation ?
3 7 \dfrac{3}{7} 7 3
3 5 \dfrac{3}{5} 5 3
1 3 \dfrac{1}{3} 3 1
3 11 \dfrac{3}{11} 11 3
Answer
3 7 \dfrac{3}{7} 7 3
3 5 \dfrac{3}{5} 5 3
1 3 \dfrac{1}{3} 3 1
3 11 \dfrac{3}{11} 11 3
∴ Option 2, is the correct option.
Which of the following is an irrational number ?
0.14 0.14 16 ‾ 0.14\overline{16} 0.14 16 0. 1416 ‾ 0.\overline{1416} 0. 1416 0.4014001400014... Answer
0.14 is terminating decimal number hence, rational number.
0.14 16 ‾ 0.14\overline{16} 0.14 16
0. 1416 ‾ 0.\overline{1416} 0. 1416
0.4014001400014... is non-terminating, non-repeating decimal number hence, it is an irrational number.
∴ Option 4, is the correct option.
Which of the following numbers has non-terminating repeating decimal expansion ?
11 80 \dfrac{11}{80} 80 11
17 160 \dfrac{17}{160} 160 17
63 240 \dfrac{63}{240} 240 63
93 420 \dfrac{93}{420} 420 93
Answer
11 80 \dfrac{11}{80} 80 11
17 160 \dfrac{17}{160} 160 17
63 240 \dfrac{63}{240} 240 63
93 420 \dfrac{93}{420} 420 93
∴ Option 4, is the correct option.
A rational number between 2 \sqrt{2} 2 3 \sqrt{3} 3
2 + 3 2 \dfrac{\sqrt{2} + \sqrt{3}}{2} 2 2 + 3
2 × 3 2 \dfrac{\sqrt{2} × \sqrt{3}}{2} 2 2 × 3
1.5
1.8
Answer
Consider the squares of 2 \sqrt{2} 2 3 \sqrt{3} 3
( 2 ) 2 {(\sqrt2)^2} ( 2 ) 2 ( 3 ) 2 {(\sqrt3)^2} ( 3 ) 2
Take any rational number between 2 and 3 which is a perfect square of a rational number,
One such number is 2.25 and
2.25 = ( 1.5 ) 2 (1.5)^2 ( 1.5 ) 2
2.25 \sqrt{2.25} 2.25
As, 2 < 2.25 < 3 2 \lt 2.25 \lt 3 2 < 2.25 < 3
2 < 2.25 < 3 \sqrt{2} \lt \sqrt{2.25} \lt \sqrt{3} 2 < 2.25 < 3
2 < 1.5 < 3 \sqrt{2} \lt 1.5 \lt \sqrt{3} 2 < 1.5 < 3
Hence, one rational number between 2 \sqrt{2} 2 3 \sqrt{3} 3 .
∴ Option 3, is the correct option.
The decimal expansion of 2 - 3 \sqrt{3} 3
terminating and non-repeating terminating and repeating non-terminating and non-repeating non-terminating and repeating Answer
The decimal expansion of 2 - 3 \sqrt{3} 3 3 \sqrt{3} 3 3 \sqrt{3} 3
∴ Option 3, is the correct option.
The decimal expansion of the rational number 33 2 2 × 5 \dfrac{33}{2^2 × 5} 2 2 × 5 33
one decimal place two decimal places three decimal places four decimal places Answer
33 2 2 × 5 \dfrac{33}{2^2 × 5} 2 2 × 5 33 33 4 × 5 \dfrac{33}{4 × 5} 4 × 5 33 33 20 \dfrac{33}{20} 20 33
Hence, decimal expansion of the rational number 33 2 2 × 5 \dfrac{33}{2^2 × 5} 2 2 × 5 33
∴ Option 2, is the correct option.
10 × 15 \sqrt{10} × \sqrt{15} 10 × 15
6 5 6\sqrt{5} 6 5 5 6 5\sqrt{6} 5 6 25 \sqrt{25} 25 10 5 10\sqrt{5} 10 5 Answer
10 × 15 \sqrt{10} × \sqrt{15} 10 × 15 2 × 5 × 3 × 5 \sqrt{2 × 5} × \sqrt{3 × 5} 2 × 5 × 3 × 5 2 × 5 × 3 × 5 \sqrt{2 × 5 × 3 × 5} 2 × 5 × 3 × 5 5 6 5\sqrt{6} 5 6
∴ Option 2, is the correct option.
2 3 + 3 2\sqrt{3} + \sqrt{3} 2 3 + 3
2 6 2\sqrt{6} 2 6 6 3 3 3\sqrt{3} 3 3 4 6 4\sqrt{6} 4 6 Answer
2 3 + 3 2\sqrt{3} + \sqrt{3} 2 3 + 3 3 ( 2 + 1 ) \sqrt{3}(2 + 1) 3 ( 2 + 1 ) 3 3 3\sqrt{3} 3 3
∴ Option 3, is the correct option.
The value of 8 + 18 \sqrt{8} + \sqrt{18} 8 + 18
26 \sqrt{26} 26 2 ( 2 + 3 ) 2(\sqrt{2} + \sqrt{3}) 2 ( 2 + 3 ) 5 2 5\sqrt{2} 5 2 6 2 6\sqrt{2} 6 2 Answer
8 + 18 \sqrt{8} + \sqrt{18} 8 + 18 2 × 2 × 2 + 2 × 3 × 3 \sqrt{2 × 2 × 2} + \sqrt{2 × 3 × 3} 2 × 2 × 2 + 2 × 3 × 3 2 2 + 3 2 2\sqrt{2} + 3\sqrt{2} 2 2 + 3 2 ( 2 + 3 ) 2 (2 + 3)\sqrt{2} ( 2 + 3 ) 2 5 2 5\sqrt{2} 5 2
∴ Option 3, is the correct option.
The number ( 2 − 3 ) 2 (2 - \sqrt{3})^2 ( 2 − 3 ) 2
a natural number an integer a rational number an irrational number Answer
( 2 − 3 ) 2 (2 - \sqrt{3})^2 ( 2 − 3 ) 2 ( 2 ) 2 + ( 3 ) 2 − 2 × 2 × 3 (2)^2 + (\sqrt{3})^2 - 2 × 2 × \sqrt{3} ( 2 ) 2 + ( 3 ) 2 − 2 × 2 × 3 4 3 4\sqrt{3} 4 3 4 3 4\sqrt{3} 4 3
Since, 7 - 4 3 4\sqrt{3} 4 3 ( 2 − 3 ) 2 (2 - \sqrt{3})^2 ( 2 − 3 ) 2
∴ Option 4, is the correct option.
If x is a positive rational number which is not a perfect square, then − 5 x -5\sqrt{x} − 5 x
a negative integer an integer a rational number an irrational number Answer
x is a positive rational number and x is not a perfect square.
Then, x \sqrt{x} x
Therefore , − 5 x -5\sqrt{x} − 5 x
∴ Option 4, is the correct option.
If x, y are both positive rational numbers, then ( x + y ) ( x − y ) (\sqrt{x} + \sqrt{y})(\sqrt{x} - \sqrt{y}) ( x + y ) ( x − y )
a rational number an irrational number neither rational nor irrational number both rational as well as irrational number Answer
( x + y ) ( x − y ) = x × x − x × y − x × y − y × y ⇒ ( x ) 2 − x y + x y − ( y ) 2 = x − y (\sqrt{x} + \sqrt{y})(\sqrt{x} - \sqrt{y}) = \sqrt{x} × \sqrt{x} - \sqrt{x} × \sqrt{y} - \sqrt{x} × \sqrt{y} - \sqrt{y} × \sqrt{y} \\[1.5em] \Rightarrow (\sqrt{x})^2 - \sqrt{xy} + \sqrt{xy} -(\sqrt{y})^2 = x - y \\[1.5em] ( x + y ) ( x − y ) = x × x − x × y − x × y − y × y ⇒ ( x ) 2 − x y + x y − ( y ) 2 = x − y
Since, x, y are both positive rational numbers, so the difference of two positive rational numbers is also a rational number .
Therefore, x − y x - y x − y ( x + y ) ( x − y ) (\sqrt{x} + \sqrt{y})(\sqrt{x} - \sqrt{y}) ( x + y ) ( x − y )
∴ Option 1, is the correct option.
After rationalising the denominator of 7 3 3 − 2 2 \dfrac{7}{3\sqrt3 - 2\sqrt{2}} 3 3 − 2 2 7
13 19 5 35 Answer
7 3 3 − 2 2 \dfrac{7}{3\sqrt3 - 2\sqrt{2}} 3 3 − 2 2 7
Let us rationalise the denominator,
Then,
7 3 3 − 2 2 = 7 3 3 − 2 2 × 3 3 + 2 2 3 3 + 2 2 ⇒ 7 ( 3 3 + 2 2 ) ( 3 3 ) 2 − ( 2 2 ) 2 ⇒ 7 × 3 3 + 14 × 2 27 − 8 ⇒ 21 3 + 14 2 19 \dfrac{7}{3\sqrt3 - 2\sqrt{2}} = \dfrac{7}{3\sqrt{3} - 2\sqrt{2}} × \dfrac{3\sqrt{3} + 2\sqrt{2}}{3\sqrt{3} + 2\sqrt{2}} \\[1.5em] \Rightarrow \dfrac{7({3\sqrt{3} + 2\sqrt{2}})}{(3\sqrt{3})^2 - (2\sqrt{2})^2} \\[1.5em] \Rightarrow \dfrac{7 × 3\sqrt{3} + 14 × \sqrt{2}}{27 - 8} \\[1.5em] \Rightarrow \dfrac{21\sqrt{3}+ 14\sqrt{2}}{19} \\[1.5em] 3 3 − 2 2 7 = 3 3 − 2 2 7 × 3 3 + 2 2 3 3 + 2 2 ⇒ ( 3 3 ) 2 − ( 2 2 ) 2 7 ( 3 3 + 2 2 ) ⇒ 27 − 8 7 × 3 3 + 14 × 2 ⇒ 19 21 3 + 14 2
∴ Option 2, is the correct option.
The number obtained on rationalising the denominator of 1 7 − 2 {\dfrac{1}{\sqrt{7} - 2}} 7 − 2 1
7 + 2 3 {\dfrac{\sqrt{7} + 2}{3}} 3 7 + 2
7 − 2 3 {\dfrac{\sqrt{7} - 2}{3}} 3 7 − 2
7 + 2 5 {\dfrac{\sqrt{7} + 2}{5}} 5 7 + 2
7 + 2 45 {\dfrac{\sqrt{7} + 2}{45}} 45 7 + 2
Answer
Given,
1 7 − 2 {\dfrac{1}{\sqrt{7} - 2}} 7 − 2 1
Let us rationalise the denominator,
Then,
1 7 − 2 = 1 7 − 2 × 7 + 2 7 + 2 ⇒ 7 + 2 ( 7 ) 2 − 2 2 ⇒ 7 + 2 7 − 4 ⇒ 7 + 2 3 {\dfrac{1}{\sqrt{7} - 2}} = {\dfrac{1}{\sqrt{7} - 2}} × \dfrac{\sqrt{7} + 2}{\sqrt{7} + 2} \\[1.5em] \Rightarrow \dfrac{\sqrt{7} + 2}{(\sqrt{7})^2 - {2}^2} \\[1.5em] \Rightarrow \dfrac{\sqrt{7} + 2}{7 - 4} \\[1.5em] \Rightarrow \dfrac{\sqrt{7} + 2}{3} \\[1.5em] 7 − 2 1 = 7 − 2 1 × 7 + 2 7 + 2 ⇒ ( 7 ) 2 − 2 2 7 + 2 ⇒ 7 − 4 7 + 2 ⇒ 3 7 + 2
∴ Option 1, is the correct option.
The number 0. 25 ‾ 0.\overline{25} 0. 25
65 99 \dfrac{65}{99} 99 65
37 99 \dfrac{37}{99} 99 37
5 9 \dfrac{5}{9} 9 5
25 99 \dfrac{25}{99} 99 25
Answer
Let x = 0. 25 ‾ 0.\overline{25} 0. 25
x = 0.252525..... .........(1)
Multiplying both side by 100, we get
100x = 25.252525... .........(2)
Subtracting equation (2) from equation (1), we get :
⇒ 100x - x = 25.252525..... - 0.252525......
⇒ 99x = 25
⇒ x = 25 99 \dfrac{25}{99} 99 25
Hence, option 4 is the correct option.
The value of 1.99 9 ‾ 1.99\overline{9} 1.99 9 p q \dfrac{p}{q} q p
19 20 \dfrac{19}{20} 20 19
1999 1000 \dfrac{1999}{1000} 1000 1999
2
1 9 \dfrac{1}{9} 9 1
Answer
Let x = 1.99 9 ‾ 1.99\overline{9} 1.99 9
x = 1.999999..... .........(1)
Multiplying both side with 10, we get
10x = 19.99999... .........(2)
Subtracting equation (2) from equation (1), we get
⇒ 10x - x = 19.99999..... - 1.99999......
⇒ 9x = 18
⇒ x = 18 9 \dfrac{18}{9} 9 18
Hence, option 3 is the correct option.
Consider the following two statements:
Statement 1: 2m x 3n = (2 + 3)m + n , where m, n are positive integers.
Statement 2: If a is a rational number, and m, n are integers, then am .an = am + n
Which of the following is valid?
Both the Statements are true.
Both the Statements are false.
Statement 1 is true, and Statement 2 is false.
Statement 1 is false, and Statement 2 is true.
Answer
According to statement 1 :
2m x 3n = (2 + 3)m + n , where m, n are positive integers.
This statement does not reflect any general law of exponents.
∴ Statement 1 is false.
According to statement 2 :
If a is a rational number, and m, n are integers, then am .an = am + n
This is one of the fundamental laws of exponents. It holds for any rational base, and all integer exponents.
∴ Statement 2 is true.
Hence, option 4 is the correct option.
Assertion Reason Type Questions
Assertion (A): − 2 7 -\dfrac{2}{7} − 7 2
Reason (R): Any number that can be expressed in the form p q \dfrac{p}{q} q p
Assertion (A) is true, Reason (R) is false.
Assertion (A) is false, Reason (R) is true.
Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).
Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason (or explanation) for Assertion (A).
Answer
Any number that can be expressed in the form p q \dfrac{p}{q} q p
The conditions are that p and q must be integers, and q must not be equal to zero.
∴ Reason (R) is false.
In the number − 2 7 -\dfrac{2}{7} − 7 2
Therefore, − 2 7 -\dfrac{2}{7} − 7 2
∴ Assertion (A) is true.
Hence, option 1 is the correct option.
Assertion (A): -10 + π is an irrational number.
Reason (R): Sum of a non-zero rational number and an irrational number is an irrational number.
Assertion (A) is true, Reason (R) is false.
Assertion (A) is false, Reason (R) is true.
Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).
Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason (or explanation) for Assertion (A).
Answer
Sum of a non-zero rational number and an irrational number is always an irrational number.
This is a fundamental property of irrational and rational numbers.
∴ Reason (R) is true.
π is an irrational number, -10 is a rational number.
So, their sum -10 + π, will be an irrational number.
∴ Assertion (A) is true.
∴ Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).
Hence, option 3 is the correct option.
Assertion (A): 0. 36 ‾ 0.\overline{36} 0. 36
Reason (R): Any real number that can be expressed in the form of p q \dfrac{p}{q} q p
Assertion (A) is true, Reason (R) is false.
Assertion (A) is false, Reason (R) is true.
Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).
Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason (or explanation) for Assertion (A).
Answer
Let x = 0. 36 ‾ 0.\overline{36} 0. 36
x = 0.363636..... .........(1)
Multiplying both side by 100, we get
100x = 36.363636... .........(2)
Subtracting equation (1) from equation (2), we get
⇒ 100x - x = 36.363636..... - 0.363636......
⇒ 99x = 36
⇒ x = 36 99 = 4 11 \dfrac{36}{99} = \dfrac{4}{11} 99 36 = 11 4
Thus, 0. 36 ‾ 0.\overline{36} 0. 36
∴ Assertion (A) is false.
By definition,
Any real number that can be expressed in the form of p q \dfrac{p}{q} q p
∴ Reason (R) is true.
∴ Assertion (A) is false, Reason (R) is true.
Hence, option 2 is the correct option.
Assertion (A): All surds are irrational numbers.
Reason (R): All irrational numbers are surds.
Assertion (A) is true, Reason (R) is false.
Assertion (A) is false, Reason (R) is true.
Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).
Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason (or explanation) for Assertion (A).
Answer
A surd is a number that cannot be expressed as a simple fraction and is the root of a rational number (like 2 , 5 3 \sqrt{2}, \sqrt[3]{5} 2 , 3 5
∴ Assertion (A) is true.
While all surds are irrational, not all irrational numbers are surds.
For example, pi (π) is irrational number but are not roots of rational numbers and therefore are not surds.
∴ Reason (R) is false.
∴ Assertion (A) is true, Reason (R) is false.
Hence, option 1 is the correct option.
Assertion (A): The rationalising factor of 2 + 3 \sqrt{3} 3 3 \sqrt{3} 3
Reason (R): Both 2 + 3 \sqrt{3} 3 3 \sqrt{3} 3
Assertion (A) is true, Reason (R) is false.
Assertion (A) is false, Reason (R) is true.
Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).
Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason (or explanation) for Assertion (A).
Answer
Multiplying 2 + 3 \sqrt{3} 3 3 \sqrt{3} 3
⇒ ( 2 + 3 ) ( 2 − 3 ) ⇒ 2 ( 2 − 3 ) + 3 ( 2 − 3 ) ⇒ 4 − 2 3 + 2 3 − ( 3 ) 2 ⇒ 4 − 3 ⇒ 1. \Rightarrow (2 + \sqrt{3})(2 - \sqrt{3}) \\[1em] \Rightarrow 2(2 - \sqrt{3}) + \sqrt{3}(2 - \sqrt{3}) \\[1em] \Rightarrow 4 - 2\sqrt{3} + 2\sqrt{3} - (\sqrt{3})^2 \\[1em] \Rightarrow 4 - 3 \\[1em] \Rightarrow 1. ⇒ ( 2 + 3 ) ( 2 − 3 ) ⇒ 2 ( 2 − 3 ) + 3 ( 2 − 3 ) ⇒ 4 − 2 3 + 2 3 − ( 3 ) 2 ⇒ 4 − 3 ⇒ 1.
Since, 1 is a rational number.
Thus, we can say that the rationalising factor of 2 + 3 \sqrt{3} 3 3 \sqrt{3} 3
∴ Assertion (A) is true.
A surd is an irrational root of a rational number.
Thus, 2 + 3 \sqrt{3} 3 3 \sqrt{3} 3 ( 3 ) (\sqrt{3}) ( 3 )
∴ Reason (R) is false.
Hence, option 1 is the correct option.
Without actual division, find whether the following rational numbers are terminating decimals or recurring decimals :
(i) 13 45 (ii) − 5 56 (iii) 7 125 (iv) − 23 80 (v) − 15 66 \begin{matrix} \text{(i)} & \dfrac{13}{45} \\[1.5em] \text{(ii)} & -\dfrac{5}{56} \\[1.5em] \text{(iii)} & \dfrac{7}{125} \\[1.5em] \text{(iv)} & -\dfrac{23}{80} \\[1.5em] \text{(v)} & -\dfrac{15}{66} \\[1.5em] \end{matrix} (i) (ii) (iii) (iv) (v) 45 13 − 56 5 125 7 − 80 23 − 66 15
In case of terminating decimals, write their decimal expansions.
Answer
(i) 13 45 \text{(i) } \dfrac{13}{45} (i) 45 13
The given number 13 45 \dfrac{13}{45} 45 13
Prime factorization of denominator 45:
3 45 3 15 5 5 1 \begin{array}{l|l} 3 & 45 \\ \hline 3 & 15 \\ \hline 5 & 5 \\ \hline & 1 \end{array} 3 3 5 45 15 5 1
45 = 3 x 3 x 5 x 12 x 5 x 1
Denominator is not of the form 2m x 5n , where m, n are non-negative integers.
∴ The given number 13 45 \dfrac{13}{45} 45 13
(ii) − 5 56 \text{(ii) } -\dfrac{5}{56} (ii) − 56 5
The given number − 5 56 -\dfrac{5}{56} − 56 5
Prime factorization of denominator 56:
2 56 2 28 2 14 7 7 1 \begin{array}{l|l} 2 & 56 \\ \hline 2 & 28 \\ \hline 2 & 14 \\ \hline 7 & 7 \\ \hline & 1 \end{array} 2 2 2 7 56 28 14 7 1
56 = 2 x 2 x 2 x 7 x 13 x 7 x 1
Denominator is not of the form 2m x 5n , where m, n are non-negative integers.
∴ The given number − 5 56 -\dfrac{5}{56} − 56 5
(iii) 7 125 \text{(iii) } \dfrac{7}{125} (iii) 125 7
The given number 7 125 \dfrac{7}{125} 125 7
Prime factorization of denominator 125:
5 125 5 25 5 5 1 \begin{array}{l|l} 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array} 5 5 5 125 25 5 1
125= 5 x 5 x 5 x 13 x 13 x 20
Denominator is of the form 2m x 5n , where m, n are non-negative integers.
7 125 = 7 2 0 × 5 3 = 7 × 2 3 2 3 × 5 3 = 56 ( 2 × 5 ) 3 = 56 ( 10 ) 3 = 56 1000 = 0.056 \dfrac{7}{125} = \dfrac{7}{2^0 × 5^3} \\[1.5em] = \dfrac{7 × 2^3}{2^3 × 5^3} \\[1.5em] = \dfrac{56}{(2 × 5)^3} \\[1.5em] = \dfrac{56}{(10)^3} \\[1.5em] = \dfrac{56}{1000} \\[1.5em] = 0.056 125 7 = 2 0 × 5 3 7 = 2 3 × 5 3 7 × 2 3 = ( 2 × 5 ) 3 56 = ( 10 ) 3 56 = 1000 56 = 0.056
∴ The given number 7 125 \dfrac{7}{125} 125 7
(iv) − 23 80 \text{(iv) } \dfrac{-23}{80} (iv) 80 − 23
The given number − 23 80 -\dfrac{23}{80} − 80 23
Prime factorization of denominator 80:
2 80 2 40 2 20 2 10 5 5 1 \begin{array}{l|l} 2 & 80 \\ \hline 2 & 40 \\ \hline 2 & 20 \\ \hline 2 & 10 \\ \hline 5 & 5 \\ \hline & 1 \end{array} 2 2 2 2 5 80 40 20 10 5 1
80 = 2 x 2 x 2 x 2 x 5 x 14 x 51
Denominator is of the form 2m x 5n , where m, n are non-negative integers.
− 23 80 = − 23 2 4 × 5 1 = − 23 × 5 3 2 4 × 5 4 = − 23 × 125 ( 2 × 5 ) 4 = − 2875 ( 10 ) 4 = − 2875 10000 = − 0.2875 -\dfrac{23}{80} = -\dfrac{23}{2^4 × 5^1} \\[1.5em] = -\dfrac{23 × 5^3} {2^4 × 5^4} \\[1.5em] = -\dfrac{23 × 125}{(2 × 5)^4} \\[1.5em] = -\dfrac{2875}{(10)^4} \\[1.5em] = -\dfrac{2875}{10000} = -0.2875 − 80 23 = − 2 4 × 5 1 23 = − 2 4 × 5 4 23 × 5 3 = − ( 2 × 5 ) 4 23 × 125 = − ( 10 ) 4 2875 = − 10000 2875 = − 0.2875
∴ The given number − 23 80 -\dfrac{23}{80} − 80 23
(v) − 15 66 \text{(v) } -\dfrac{15}{66} (v) − 66 15
The given number − 15 66 -\dfrac{15}{66} − 66 15
Prime factorization of denominator 66:
2 66 3 33 11 11 1 \begin{array}{l|l} 2 & 66 \\ \hline 3 & 33 \\ \hline 11 & 11 \\ \hline & 1 \end{array} 2 3 11 66 33 11 1
66 = 2 x 3 x 11 x 1
Denominator is not of the form 2m x 5n , where m, n are non-negative integers.
∴ The given number − 15 66 -\dfrac{15}{66} − 66 15
Express the following recurring decimals as vulgar fractions :
(i) 1.3 45 ‾ 1.3\overline{45} 1.3 45
(ii) 2. 357 ‾ 2.\overline{357} 2. 357
Answer
(i) Let x = 1.3 45 ‾ 1.3\overline{45} 1.3 45 ....(i) \qquad \text{....(i)} ....(i)
So multiplying both sides of (i) by 10
we get,
10x = 13.4545.......(ii) \qquad \text{....(ii)} ....(ii)
Again multiply by 100 on both sides ,
1000x =1345.4545.........(iii) \qquad \text{....(iii)} ....(iii)
Subtracting (ii) from (iii), we get
1000x - 10x = 1345.4545... - 13.4545...
990x = 1332
x = 1332 990 \dfrac{1332}{990} 990 1332 74 55 \bold{\dfrac{74}{55}} 55 74
which is in the form of p q \dfrac{p}{q} q p
(ii) Let x = 2. 357 ‾ 2.\overline{357} 2. 357 ....(i) \qquad \text{....(i)} ....(i)
So multiplying both sides of (i) by 1000,
we get,
1000x = 2357.357357.......(ii) \qquad \text{....(ii)} ....(ii)
Subtracting (i) from (ii), we get
1000x - x = 2357.357357... - 2.357357...
999x = 2355
x = 2355 999 \bold{\dfrac{2355}{999}} 999 2355
which is in the form of p q \dfrac{p}{q} q p
Insert a rational number between 5 9 \dfrac{5}{9} 9 5 7 13 \dfrac{7}{13} 13 7
Answer
The L.C.M. of 9 and 13 is 117.
5 9 = 5 × 13 9 × 13 = 65 117 7 13 = 7 × 9 13 × 9 = 63 117 Since 9 < 13 , 1 13 < 1 9 \dfrac{5}{9} = \dfrac{5 \times 13}{9 \times 13} = \dfrac{65}{117} \\[1.5em] \dfrac{7}{13} = \dfrac{7 \times 9}{13 \times 9} = \dfrac{63}{117} \\[1.5em] \text{Since } 9 \lt 13, \dfrac{1}{13} \lt \dfrac{1}{9} 9 5 = 9 × 13 5 × 13 = 117 65 13 7 = 13 × 9 7 × 9 = 117 63 Since 9 < 13 , 13 1 < 9 1
A rational number between 5 9 \dfrac{5}{9} 9 5 7 13 \dfrac{7}{13} 13 7
= 5 9 + 7 13 2 = 65 + 63 117 2 = 128 117 × 2 = 64 117 = \dfrac{\dfrac{5}{9} + \dfrac{7}{13}}{2} \\[1.5em] = \dfrac{\dfrac{65 + 63}{117}}{2} \\[1.5em] = \dfrac{128}{117 × 2} \\[1.5em] = \bold{\dfrac{64}{117}} \\[1.5em] = 2 9 5 + 13 7 = 2 117 65 + 63 = 117 × 2 128 = 117 64
∴ 7 13 < 64 117 < 5 9 \therefore\dfrac{7}{13} \lt \dfrac{64}{117} \lt \dfrac{5}{9} ∴ 13 7 < 117 64 < 9 5
Hence, numbers in ascending order are:
7 13 , 64 117 , 5 9 \bold{\dfrac{7}{13}}, \bold{\dfrac{64}{117}}, \bold{\dfrac{5}{9}} 13 7 , 117 64 , 9 5
Insert four rational numbers between 4 5 \dfrac{4}{5} 5 4 5 6 \dfrac{5}{6} 6 5
Answer
The L.C.M of 5 and 6 is 30.
4 5 = 4 × 6 5 × 6 = 24 30 5 6 = 5 × 5 6 × 5 = 25 30 Since 24 < 25 , ∴ 4 5 < 5 6 \dfrac{4}{5} = \dfrac{4 \times 6}{5 \times 6} = \dfrac{24}{30} \\[1.5em] \dfrac{5}{6} = \dfrac{5 \times 5}{6 \times 5} = \dfrac{25}{30} \\[1.5em] \text{Since } 24 \lt 25, \\[1.5em] \therefore \dfrac{4}{5} \lt \dfrac{5}{6} 5 4 = 5 × 6 4 × 6 = 30 24 6 5 = 6 × 5 5 × 5 = 30 25 Since 24 < 25 , ∴ 5 4 < 6 5
To find four rational number between 4 5 \dfrac{4}{5} 5 4 5 6 \dfrac{5}{6} 6 5 24 30 \dfrac{24}{30} 30 24 25 30 \dfrac{25}{30} 30 25 120 150 \dfrac{120}{150} 150 120 125 150 \dfrac{125}{150} 150 125
Since , 120 < 121 < 122 < 123 < 124 < 125 ⇒ 120 150 < 121 150 < 122 150 < 123 150 < 124 150 < 125 150 ⇒ 4 5 < 121 150 < 61 75 < 123 150 < 62 75 < 5 6 \text{Since}, 120 \lt 121 \lt 122 \lt 123 \lt 124 \lt 125 \\[1.5em] \Rightarrow\dfrac{120}{150} \lt \dfrac{121}{150} \lt \dfrac{122}{150} \lt \dfrac{123}{150} \lt \dfrac{124}{150} \lt \dfrac{125}{150} \\[1.5em] \Rightarrow\dfrac{4}{5} \lt \dfrac{121}{150} \lt \dfrac{61}{75} \lt \dfrac{123}{150} \lt \dfrac{62}{75} \lt \dfrac{5}{6} \\[1.5em] Since , 120 < 121 < 122 < 123 < 124 < 125 ⇒ 150 120 < 150 121 < 150 122 < 150 123 < 150 124 < 150 125 ⇒ 5 4 < 150 121 < 75 61 < 150 123 < 75 62 < 6 5
Four rational number between 4 5 \dfrac{4}{5} 5 4 5 6 \dfrac{5}{6} 6 5
121 150 , 61 75 , 123 150 , 62 75 \bold{\dfrac{121}{150}}, \bold{\dfrac{61}{75}}, \bold{\dfrac{123}{150}}, \bold{\dfrac{62}{75}} 150 121 , 75 61 , 150 123 , 75 62
Prove that the reciprocal of an irrational number is irrational.
Answer
Let us consider, x as an irrational number.
Reciprocal of x is 1 x \dfrac{1}{x} x 1
Let us consider 1 x \dfrac{1}{x} x 1
Then, x × 1 x x × \dfrac{1}{x} x × x 1
But x × 1 x x × \dfrac{1}{x} x × x 1
Hence, our supposition is wrong. So, 1 x \dfrac{1}{x} x 1
Prove that the following numbers are irrational:
(i) 8 \sqrt{8} 8
(ii) 14 \sqrt{14} 14
(iii) 2 3 \sqrt[3]{2} 3 2
Answer
(i) 8 \sqrt{8} 8 2 2 2\sqrt{2} 2 2 2 \sqrt{2} 2
Let 2 \sqrt{2} 2
2 = p q , \sqrt{2} = \dfrac{p}{q}, 2 = q p ,
where p, q are integers, q ≠ 0 and p, q have no common factors (except 1)
⇒ 2 = p 2 q 2 ⇒ p 2 = 2 q 2 ....(i) \Rightarrow 2 = \dfrac{p^2}{q^2} \\[1.5em] \Rightarrow p^2 = 2q^2 \qquad \text{....(i)} ⇒ 2 = q 2 p 2 ⇒ p 2 = 2 q 2 ....(i)
As 2 divides 2q2 , so 2 divides p2 but 2 is prime
⇒ 2 divides p (Theorem 1) \Rightarrow 2 \text{ divides } p \qquad \text{(Theorem 1)} ⇒ 2 divides p (Theorem 1)
Let p = 2m, where m is an integer.
Substituting this value of p in (i), we get
( 2 m ) 2 = 2 q 2 ⇒ 4 m 2 = 2 q 2 ⇒ 2 m 2 = q 2 (2m)^2 = 2q^2 \\[1.5em] \Rightarrow 4m^2 = 2q^2 \\[1.5em] \Rightarrow 2m^2 = q^2 \\[1.5em] ( 2 m ) 2 = 2 q 2 ⇒ 4 m 2 = 2 q 2 ⇒ 2 m 2 = q 2
As 2 divides 2m2 , so 2 divides q2 but 2 is prime
⇒ 2 divides q (Theorem 1) \Rightarrow 2 \text{ divides } q \qquad \text{(Theorem 1)} ⇒ 2 divides q (Theorem 1)
Thus, p and q have a common factor 2. This contradicts that p and q have no common factors (except 1).
Hence, 2 \sqrt{2} 2 2 \sqrt{2} 2
Since, product of non-zero rational number and an irrational number is an irrational number.
And 2 \sqrt{2} 2 2 2 2\sqrt{2} 2 2 8 \bold{\sqrt{8}} 8 is an irrational number .
(ii) Suppose that 14 \sqrt{14} 14
14 = p q , \sqrt{14} = \dfrac{p}{q}, 14 = q p ,
where p, q are integers, q ≠ 0 and p, q have no common factors (except 1)
⇒ 14 = p 2 q 2 ⇒ p 2 = 14 q 2 ....(i) \Rightarrow 14 = \dfrac{p^2}{q^2} \\[1.5em] \Rightarrow p^2 = 14q^2 \qquad \text{....(i)} ⇒ 14 = q 2 p 2 ⇒ p 2 = 14 q 2 ....(i)
As 2 divides 14q2 , so 2 divides p2 but 2 is prime
⇒ 2 divides p (Theorem 1) \Rightarrow 2 \text{ divides } p \qquad \text{(Theorem 1)} ⇒ 2 divides p (Theorem 1)
Let p = 2k, where k is some integer.
Substituting this value of p in (i), we get
( 2 k ) 2 = 14 q 2 ⇒ 4 k 2 = 14 q 2 ⇒ 2 k 2 = 7 q 2 (2k)^2 = 14q^2 \\[1.5em] \Rightarrow 4k^2 = 14q^2 \\[1.5em] \Rightarrow 2k^2 = 7q^2 \\[1.5em] ( 2 k ) 2 = 14 q 2 ⇒ 4 k 2 = 14 q 2 ⇒ 2 k 2 = 7 q 2
As 2 divides 2k2 , so 2 divides 7q2
⇒ \Rightarrow ⇒ 2
But 2 does not divide 7, therefore, 2 divides q2
⇒ \Rightarrow ⇒
Thus, p and q have a common factor 2. This contradicts that p and q have no common factors (except 1).
Hence, our supposition is wrong. Therefore, 14 \sqrt{14} 14 14 \bold{\sqrt{14}} 14 .
(iii) Suppose that 2 3 \sqrt[3]{2} 3 2 p q \dfrac{p}{q} q p
⇒ 2 = ( p q ) 3 ⇒ p 3 = 2 q 3 ....(i) \Rightarrow 2 = \Big(\dfrac{p}{q}\Big)^3 \\[1.5em] \Rightarrow p^3 = 2q^3 \qquad \text{....(i)} ⇒ 2 = ( q p ) 3 ⇒ p 3 = 2 q 3 ....(i)
As 2 divides 2q3 ⇒ \Rightarrow ⇒ 3
⇒ \Rightarrow ⇒
Let p = 2k , where k is an integer.
Substituting this value of p in (i), we get
⇒ \phantom{\Rightarrow} ⇒ 3 = 2q3 ⇒ \Rightarrow ⇒ 3 = 2q3 ⇒ \Rightarrow ⇒ 3 = q3
As 2 divides 4k3 ⇒ \Rightarrow ⇒ 3
⇒ \Rightarrow ⇒
Thus, p and q have a common factor 2. This contradicts that p and q have no common factors (except 1).
Hence, our supposition is wrong. It follows that 2 3 \sqrt[3]{2} 3 2 p q \dfrac{p}{q} q p
∴ 2 3 \bold{\sqrt[3]{2}} 3 2
Prove that 3 \sqrt{3} 3 3 \sqrt{3} 3
Answer
Let 3 \sqrt{3} 3
3 = p q , \sqrt{3} = \dfrac{p}{q}, 3 = q p ,
where p, q are integers, q ≠ 0 and p, q have no common factors (except 1)
⇒ 3 = p 2 q 2 ⇒ p 2 = 3 q 2 ....(i) \Rightarrow 3 = \dfrac{p^2}{q^2} \\[1.5em] \Rightarrow p^2 = 3q^2 \qquad \text{....(i)} ⇒ 3 = q 2 p 2 ⇒ p 2 = 3 q 2 ....(i)
As 3 divides 3q2 , so 3 divides p2 but 3 is prime
⇒ 3 divides p (Theorem 1) \Rightarrow 3 \text{ divides } p \qquad \text{(Theorem 1)} ⇒ 3 divides p (Theorem 1)
Let p = 3m, where m is an integer.
Substituting this value of p in (i), we get
( 3 m ) 2 = 3 q 2 ⇒ 9 m 2 = 3 q 2 ⇒ 3 m 2 = q 2 (3m)^2 = 3q^2 \\[1.5em] \Rightarrow 9m^2 = 3q^2 \\[1.5em] \Rightarrow 3m^2 = q^2 \\[1.5em] ( 3 m ) 2 = 3 q 2 ⇒ 9 m 2 = 3 q 2 ⇒ 3 m 2 = q 2
As 3 divides 3m2 , so 3 divides q2 but 3 is prime
⇒ 3 divides q (Theorem 1) \Rightarrow 3 \text{ divides } q \qquad \text{(Theorem 1)} ⇒ 3 divides q (Theorem 1)
Thus, p and q have a common factor 3. This contradicts that p and q have no common factors (except 1).
Hence, 3 \sqrt{3} 3 3 \sqrt{3} 3
Suppose that 5 − 3 5 - \sqrt{3} 5 − 3
Then, 5 − 3 5 - \sqrt{3} 5 − 3
⇒ − 3 = r − 5 ⇒ 3 = 5 − r \Rightarrow - \sqrt{3} = r - 5 \\[1.5em] \Rightarrow \sqrt{3} = 5 - r \\[1.5em] ⇒ − 3 = r − 5 ⇒ 3 = 5 − r
As r is rational and r ≠ 0, so 5 - r is rational
⇒ 3 \Rightarrow \sqrt{3} ⇒ 3
But this contradicts that 3 \sqrt{3} 3
∴ 5 − 3 \bold{5 - \sqrt{3}} 5 − 3 .
Prove that the following numbers are irrational:
(i) 3 + 5 3 + \sqrt{5} 3 + 5
(ii) 15 − 2 7 15 - 2\sqrt{7} 15 − 2 7
(iii) 1 3 − 5 \dfrac{1}{3 - \sqrt5} 3 − 5 1
Answer
(i) 3 + 5 \text{(i) } 3 + \sqrt{5} (i) 3 + 5
Let us assume that 3 + 5 3 + \sqrt{5} 3 + 5
Then,
3 + 5 = r ⇒ 5 = r − 3 3 + \sqrt{5} = r \\[1.5em] \Rightarrow \sqrt{5} = r - 3 3 + 5 = r ⇒ 5 = r − 3
As r is rational, r - 3 is rational
⇒ 5 \Rightarrow \sqrt{5} ⇒ 5
But this contradicts the fact that 5 \sqrt{5} 5
Hence, our assumption is wrong.
∴ 3 + 5 \bold{3 + \sqrt{5}} 3 + 5
(ii) 15 − 2 7 \text{(ii) } 15 - 2\sqrt{7} (ii) 15 − 2 7
Let us assume that 15 − 2 7 15 - 2\sqrt{7} 15 − 2 7
Then,
15 − 2 7 = r ⇒ 2 7 = 15 − r ⇒ 7 = 15 − r 2 15 - 2\sqrt{7} = r \\[1.5em] \Rightarrow 2\sqrt{7} = 15 - r \\[1.5em] \Rightarrow \sqrt{7} = \dfrac{15 - r}{2} \\[1.5em] 15 − 2 7 = r ⇒ 2 7 = 15 − r ⇒ 7 = 2 15 − r
As r is rational, 15 - r is rational
⇒ 15 − r 2 \Rightarrow \dfrac{15 - r}{2} ⇒ 2 15 − r
⇒ 7 \Rightarrow \sqrt{7} ⇒ 7
But this contradicts the fact that 7 \sqrt{7} 7
Hence, our assumption is wrong.
∴ 15 − 2 7 \bold{15 - 2\sqrt{7}} 15 − 2 7
(iii) 1 3 − 5 \text{(iii) }\dfrac{1}{3 - \sqrt5} (iii) 3 − 5 1
Let us rationalise the denominator
1 3 − 5 = 1 3 − 5 × 3 + 5 3 + 5 = 3 + 5 ( 3 ) 2 − ( 5 ) 2 = 3 + 5 9 − 5 = 3 + 5 4 \dfrac{1}{3 - \sqrt5} = \dfrac{1}{3 - \sqrt5} × \dfrac{3 + \sqrt5}{3 + \sqrt5} \\[1.5em] = \dfrac{3 + \sqrt5}{(3)^2 - (\sqrt5)^2} \\[1.5em] = \dfrac{3 + \sqrt5}{9 - 5} \\[1.5em] = \dfrac{3 + \sqrt5}{4} \\[1.5em] 3 − 5 1 = 3 − 5 1 × 3 + 5 3 + 5 = ( 3 ) 2 − ( 5 ) 2 3 + 5 = 9 − 5 3 + 5 = 4 3 + 5
Let us assume that 3 + 5 4 \dfrac{3 + \sqrt5}{4} 4 3 + 5
Then,
3 + 5 4 = r ⇒ 3 + 5 = 4 r ⇒ 5 = 4 r − 3 \dfrac{3 + \sqrt5}{4} = r \\[1.5em] \Rightarrow 3 +\sqrt{5} = 4r \\[1.5em] \Rightarrow \sqrt{5} = 4r - 3 \\[1.5em] 4 3 + 5 = r ⇒ 3 + 5 = 4 r ⇒ 5 = 4 r − 3
As r is rational, 4r is rational
⇒ \Rightarrow ⇒
⇒ 5 \Rightarrow \sqrt{5} ⇒ 5
But this contradicts the fact that 5 \sqrt{5} 5
Hence, our assumption is wrong.
⇒ 3 + 5 4 \Rightarrow \dfrac{3 + \sqrt5}{4} ⇒ 4 3 + 5
∴ 1 3 − 5 \bold{\dfrac{1}{3 - \sqrt5}} 3 − 5 1
Rationalise the denominator of the following :
(i) 10 2 2 + 3 (ii) 7 3 − 5 2 48 + 18 (iii) 1 3 − 2 + 1 \begin{matrix} \text{(i)} & \dfrac{10}{2\sqrt{2} + \sqrt{3}} \\[1.5em] \text{(ii)} & \dfrac{7\sqrt{3} - 5\sqrt{2}}{\sqrt{48} + \sqrt{18}} \\[1.5em] \text{(iii)} & \dfrac{1}{\sqrt{3} - \sqrt{2} + 1} \\[1.5em] \end{matrix} (i) (ii) (iii) 2 2 + 3 10 48 + 18 7 3 − 5 2 3 − 2 + 1 1
Answer
(i) \text{(i)} (i)
10 2 2 + 3 \dfrac{10}{2\sqrt{2} + \sqrt{3}} 2 2 + 3 10
Let us rationalise the denominator,
Then,
10 2 2 + 3 = 10 2 2 + 3 × 2 2 − 3 2 2 − 3 = 10 ( 2 2 − 3 ) ( 2 2 ) 2 − ( 3 ) 2 = 10 × 2 2 − 10 × 3 ( 2 2 ) 2 − ( 3 ) 2 = 10 ( 2 2 − 3 ) 8 − 3 = 10 ( 2 2 − 3 ) 5 = 2 ( 2 2 − 3 ) \dfrac{10}{2\sqrt{2} + \sqrt{3}} = \dfrac{10}{2\sqrt{2} + \sqrt{3}} × \dfrac{2\sqrt{2} - \sqrt{3}}{2\sqrt{2} - \sqrt{3}} \\[1.5em] = \dfrac{10({2\sqrt{2} - \sqrt{3}})}{(2\sqrt{2})^2 - (\sqrt{3})^2} \\[1.5em] = \dfrac{10 × 2\sqrt{2} - 10 × \sqrt{3}}{(2\sqrt{2})^2 - (\sqrt{3})^2} \\[1.5em] = \dfrac{10(2\sqrt{2} - \sqrt{3})}{8 - 3} \\[1.5em] = \dfrac{10(2\sqrt{2} - \sqrt{3})}{5} \\[1.5em] \bold{= 2(2\sqrt{2} - \sqrt{3}) } \\[1.5em] 2 2 + 3 10 = 2 2 + 3 10 × 2 2 − 3 2 2 − 3 = ( 2 2 ) 2 − ( 3 ) 2 10 ( 2 2 − 3 ) = ( 2 2 ) 2 − ( 3 ) 2 10 × 2 2 − 10 × 3 = 8 − 3 10 ( 2 2 − 3 ) = 5 10 ( 2 2 − 3 ) = 2 ( 2 2 − 3 )
(ii) \text{(ii)} (ii)
7 3 − 5 2 48 + 18 \dfrac{7\sqrt{3} - 5\sqrt{2}}{\sqrt{48} + \sqrt{18}} 48 + 18 7 3 − 5 2
Let us rationalise the denominator,
7 3 − 5 2 48 + 18 × 48 − 18 48 − 18 = 7 3 × 48 − 7 3 × 18 − 5 2 × 48 + 5 2 × 18 ( 48 ) 2 − ( 18 ) 2 = 7 144 − 7 54 − 5 96 + 5 36 48 − 18 = 7 × 12 − 7 2 × 3 × 3 × 3 − 5 2 × 2 × 2 × 2 × 2 × 3 + 5 2 × 2 × 3 × 3 30 = 84 − 21 6 − 20 6 + 30 30 = 84 − 21 6 − 20 6 + 30 30 = 114 − 41 6 30 = 114 30 − 41 6 30 = 57 15 − 41 6 30 \dfrac{7\sqrt{3} - 5\sqrt{2}}{\sqrt{48} + \sqrt{18}} × \dfrac{\sqrt{48} - \sqrt{18}}{\sqrt{48} - \sqrt{18}} \\[1.5em] = \dfrac{7\sqrt{3} × \sqrt{48} - 7\sqrt{3} × \sqrt{18} - 5\sqrt{2} × \sqrt{48} + 5\sqrt{2} × \sqrt{18} }{(\sqrt{48})^2 -(\sqrt{18})^2} \\[1.5em] = \dfrac{7\sqrt{144} - 7\sqrt{54} - 5\sqrt{96} +5\sqrt{36}}{48 - 18} \\[1.5em] = \dfrac{7 × 12 - 7\sqrt{2 × 3 × 3 × 3 } - 5\sqrt{2 × 2 × 2 × 2 × 2 × 3} +5\sqrt{2 × 2 × 3 × 3}}{30} \\[1.5em] = \dfrac{84 - 21\sqrt{6} - 20\sqrt{6} + 30}{30} \\[1.5em] = \dfrac{84 - 21\sqrt{6} - 20\sqrt{6} + 30}{30} \\[1.5em] = \dfrac{114 - 41{\sqrt6}}{30} \\[1.5em] = \dfrac{114}{30} - \dfrac{41\sqrt{6}}{30} \\[1.5em] = \bold{\dfrac{57}{15} - \dfrac{41\sqrt{6}}{30}} \\[1.5em] 48 + 18 7 3 − 5 2 × 48 − 18 48 − 18 = ( 48 ) 2 − ( 18 ) 2 7 3 × 48 − 7 3 × 18 − 5 2 × 48 + 5 2 × 18 = 48 − 18 7 144 − 7 54 − 5 96 + 5 36 = 30 7 × 12 − 7 2 × 3 × 3 × 3 − 5 2 × 2 × 2 × 2 × 2 × 3 + 5 2 × 2 × 3 × 3 = 30 84 − 21 6 − 20 6 + 30 = 30 84 − 21 6 − 20 6 + 30 = 30 114 − 41 6 = 30 114 − 30 41 6 = 15 57 − 30 41 6
(iii) \text{(iii)} (iii)
1 3 − 2 + 1 \dfrac{1}{\sqrt{3} - \sqrt{2} + 1} 3 − 2 + 1 1
Let us rationalise the denominator,
Then,
1 3 − 2 + 1 = 1 3 − ( 2 − 1 ) × 3 + ( 2 − 1 ) 3 + ( 2 − 1 ) = 3 + 2 − 1 ( 3 ) 2 − ( 2 − 1 ) 2 = 3 + 2 − 1 ( 3 ) 2 − ( ( 2 ) 2 − 2 × 2 + 1 ) = 3 + 2 − 1 3 − ( 2 − 2 2 + 1 ) = 3 + 2 − 1 3 − 2 + 2 2 − 1 = 3 + 2 − 1 2 2 = 3 + 2 − 1 2 2 × 2 2 = 2 ( 3 + 2 − 1 ) 2 2 × 2 = 2 × 3 + 2 × 2 − 2 2 2 × 2 = 6 + 2 − 2 4 = 2 + 6 − 2 4 \dfrac{1}{\sqrt{3} - \sqrt{2} + 1} = \dfrac{1}{\sqrt{3} - (\sqrt{2} - 1)} × \dfrac{\sqrt{3} + (\sqrt{2} - 1)}{\sqrt{3} + (\sqrt{2} - 1)} \\[1.5em] =\dfrac{{\sqrt{3} + \sqrt{2}} - 1}{{(\sqrt{3})^2 }- (\sqrt{2} - 1)^2} \\[1.5em] =\dfrac{\sqrt{3} + \sqrt{2} - 1}{(\sqrt{3})^2 - ( (\sqrt{2})^2 - 2 × \sqrt{2} + 1)} \\[1.5em] =\dfrac{\sqrt{3} + \sqrt{2} - 1}{3 - (2 - 2\sqrt{2} + 1)} \\[1.5em] =\dfrac{\sqrt{3} + \sqrt{2} - 1}{3 - 2 + 2\sqrt{2} -1} \\[1.5em] =\dfrac{\sqrt{3} + \sqrt{2} - 1}{2\sqrt{2}} \\[1.5em] =\dfrac{\sqrt{3} + \sqrt{2} - 1}{2\sqrt{2}} × \dfrac{\sqrt{2}}{\sqrt{2}} \\[1.5em] =\dfrac{\sqrt{2}(\sqrt{3} + \sqrt{2} - 1)}{2\sqrt{2} × \sqrt{2}} \\[1.5em] =\dfrac{\sqrt{2} × \sqrt{3} + \sqrt{2} × \sqrt{2} - \sqrt{2}}{2\sqrt{2} × \sqrt{2}} \\[1.5em] = \dfrac{\sqrt{6} + 2 -\sqrt{2}}{4} \\[1.5em] \bold{{=} \dfrac{2 + \sqrt{6} -\sqrt{2}}{4}} \\[1.5em] 3 − 2 + 1 1 = 3 − ( 2 − 1 ) 1 × 3 + ( 2 − 1 ) 3 + ( 2 − 1 ) = ( 3 ) 2 − ( 2 − 1 ) 2 3 + 2 − 1 = ( 3 ) 2 − (( 2 ) 2 − 2 × 2 + 1 ) 3 + 2 − 1 = 3 − ( 2 − 2 2 + 1 ) 3 + 2 − 1 = 3 − 2 + 2 2 − 1 3 + 2 − 1 = 2 2 3 + 2 − 1 = 2 2 3 + 2 − 1 × 2 2 = 2 2 × 2 2 ( 3 + 2 − 1 ) = 2 2 × 2 2 × 3 + 2 × 2 − 2 = 4 6 + 2 − 2 = 4 2 + 6 − 2
If p, q are rational numbers and p − 15 q = 2 3 − 5 4 3 − 3 5 p - \sqrt{15}q = \dfrac{2\sqrt{3} - \sqrt{5}}{4\sqrt{3} - 3\sqrt{5}} p − 15 q = 4 3 − 3 5 2 3 − 5
Answer
Since, it is given that
2 3 − 5 4 3 − 3 5 = p − 15 q \dfrac{2\sqrt{3} - \sqrt{5}}{4\sqrt{3} - 3\sqrt{5}} = p - \sqrt{15} q 4 3 − 3 5 2 3 − 5 = p − 15 q
On solving,
2 3 − 5 4 3 − 3 5 × 4 3 + 3 5 4 3 + 3 5 = 2 3 × 4 3 + 2 3 × 3 5 − 5 × 4 3 − 5 × 3 5 ( 4 3 ) 2 − ( 3 5 ) 2 = 24 + 6 15 − 4 15 − 15 ( 4 3 ) 2 − ( 3 5 ) 2 = 24 − 15 + 6 15 − 4 15 48 − 45 = 9 + 2 15 3 = 9 3 + 2 15 3 = 3 + 2 3 15 ∴ 3 − ( − 2 3 ) 15 = p − q 15 \dfrac{2\sqrt{3} - \sqrt{5}}{4\sqrt{3} - 3\sqrt{5}} × \dfrac{4\sqrt{3} + 3\sqrt{5}}{4\sqrt{3} + 3\sqrt{5}} \\[1.5em] = \dfrac{2\sqrt{3} × 4\sqrt{3}+ 2\sqrt{3} × 3\sqrt{5} - \sqrt{5} × 4\sqrt{3} - \sqrt{5} × 3\sqrt{5} }{(4\sqrt{3})^2 -(3\sqrt{5})^2} \\[1.5em] = \dfrac{24 + 6\sqrt{15} - 4\sqrt{15} - 15 }{(4\sqrt{3})^2 -(3\sqrt{5})^2} \\[1.5em] = \dfrac{24 - 15 + 6\sqrt{15} - 4\sqrt{15} }{48 - 45} \\[1.5em] = \dfrac{9 + 2{\sqrt{15}}}{3} \\[1.5em] = \dfrac{9}{3} + \dfrac{2\sqrt{15}}{3} \\[1.5em] = 3 + \dfrac{2}{3}{\sqrt{15}} \\[1.5em] \therefore 3 - \Big(-\dfrac{2}{3}\Big){\sqrt{15}} = p - q\sqrt{15} 4 3 − 3 5 2 3 − 5 × 4 3 + 3 5 4 3 + 3 5 = ( 4 3 ) 2 − ( 3 5 ) 2 2 3 × 4 3 + 2 3 × 3 5 − 5 × 4 3 − 5 × 3 5 = ( 4 3 ) 2 − ( 3 5 ) 2 24 + 6 15 − 4 15 − 15 = 48 − 45 24 − 15 + 6 15 − 4 15 = 3 9 + 2 15 = 3 9 + 3 2 15 = 3 + 3 2 15 ∴ 3 − ( − 3 2 ) 15 = p − q 15
Hence, p = 3 and q = − 2 3 -\dfrac{2}{3} − 3 2
If x = 1 3 + 2 2 \dfrac{1}{3 + 2\sqrt2} 3 + 2 2 1 x − 1 x x - \dfrac{1}{x} x − x 1
Answer
Given,
x = 1 3 + 2 2 ....(i) x =\dfrac{1}{3 + 2\sqrt{2}} \qquad \text{....(i)} \\[1.5em] x = 3 + 2 2 1 ....(i)
Let us rationalise the denominator,
x = 1 3 + 2 2 × 3 − 2 2 3 − 2 2 = 3 − 2 2 3 2 − ( 2 2 ) 2 = 3 − 2 2 9 − 8 = ( 3 − 2 2 ) ∴ x = ( 3 − 2 2 ) x = \dfrac{1}{3 + 2\sqrt{2}} ×\dfrac{3 - 2\sqrt{2}}{3 - 2\sqrt{2}} \\[1.5em] = \dfrac{3 - 2\sqrt{2}}{3^2 - (2\sqrt{2})^2} \\[1.5em] = \dfrac{3 - 2\sqrt{2}}{9 - 8} \\[1.5em] = (3 - 2\sqrt{2}) \\[1.5em] \therefore x = (3 - 2\sqrt{2}) x = 3 + 2 2 1 × 3 − 2 2 3 − 2 2 = 3 2 − ( 2 2 ) 2 3 − 2 2 = 9 − 8 3 − 2 2 = ( 3 − 2 2 ) ∴ x = ( 3 − 2 2 )
From (i) we get,
1 x = 3 + 2 2 ∴ x − 1 x = ( 3 − 2 2 ) − ( 3 + 2 2 ) = 3 − 2 2 − 3 − 2 2 ⇒ x − 1 x = − 4 2 \dfrac{1}{x} = 3 + 2\sqrt{2} \\[1.5em] \therefore x - \dfrac{1}{x} = (3 - 2\sqrt{2}) - (3 + 2\sqrt{2}) = 3 - 2\sqrt{2} - 3 - 2\sqrt{2}\\[1.5em] \Rightarrow\bold{ x - \dfrac{1}{x} = -4\sqrt{2}} \\[1.5em] x 1 = 3 + 2 2 ∴ x − x 1 = ( 3 − 2 2 ) − ( 3 + 2 2 ) = 3 − 2 2 − 3 − 2 2 ⇒ x − x 1 = − 4 2
(i) If x = 7 + 3 5 7 − 3 5 \dfrac{7 + 3\sqrt{5}}{7- 3\sqrt{5}} 7 − 3 5 7 + 3 5 x 2 + 1 x 2 x^2 + \dfrac{1}{x^2} x 2 + x 2 1
(ii) If x = 5 − 2 5 + 2 \dfrac{\sqrt{5} - \sqrt{2}}{\sqrt{5} + \sqrt{2}} 5 + 2 5 − 2 5 + 2 5 − 2 \dfrac{\sqrt{5} + \sqrt{2}}{\sqrt{5} - \sqrt{2}} 5 − 2 5 + 2 2 + xy + y2
(iii) If x = 3 − 2 3 + 2 \dfrac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}} 3 + 2 3 − 2 3 + 2 3 − 2 \dfrac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} 3 − 2 3 + 2 3 + y3 .
Answer
(i) Given x = 7 + 3 5 7 − 3 5 \dfrac{7 + 3\sqrt{5}}{7- 3\sqrt{5}} 7 − 3 5 7 + 3 5
Rationalising the denominator,
7 + 3 5 7 − 3 5 = 7 + 3 5 7 − 3 5 × 7 + 3 5 7 + 3 5 = ( 7 + 3 5 ) 2 ( 7 ) 2 − ( 3 5 ) 2 = 7 2 + 2 × 7 × 3 5 + ( 3 5 ) 2 49 − 45 = 49 + 42 5 + 45 4 = 94 + 42 5 4 = 47 + 21 5 2 ∴ x = 47 + 21 5 2 ⇒ 1 x = 2 47 + 21 5 \dfrac{7 + 3\sqrt{5}}{7- 3\sqrt{5}} = \dfrac{7 + 3\sqrt{5}}{7- 3\sqrt{5}} × \dfrac{7 + 3\sqrt{5}}{7 + 3\sqrt{5}} \\[1.5em] = \dfrac{({7 + 3\sqrt{5}})^2}{(7)^2 - (3\sqrt{5})^2} \\[1.5em] = \dfrac{7^2 + 2 × 7 × 3\sqrt{5} + (3\sqrt{5})^2}{49 - 45} = \dfrac{49 + 42\sqrt{5} + 45}{4} = \dfrac{94 + 42\sqrt{5}}{4} \\[1.5em] = \dfrac{47 + 21\sqrt{5}}{2} \\[1.5em] \therefore x = \dfrac{47 + 21\sqrt{5}}{2} \\[1.5em] \Rightarrow \dfrac{1}{x} = \dfrac{2}{47 + 21\sqrt{5}} \\[1.5em] 7 − 3 5 7 + 3 5 = 7 − 3 5 7 + 3 5 × 7 + 3 5 7 + 3 5 = ( 7 ) 2 − ( 3 5 ) 2 ( 7 + 3 5 ) 2 = 49 − 45 7 2 + 2 × 7 × 3 5 + ( 3 5 ) 2 = 4 49 + 42 5 + 45 = 4 94 + 42 5 = 2 47 + 21 5 ∴ x = 2 47 + 21 5 ⇒ x 1 = 47 + 21 5 2
Rationalising denominator of 1 x \dfrac{1}{x} x 1
1 x = 2 47 + 21 5 × 47 − 21 5 47 − 21 5 = 2 ( 47 − 21 5 ) ( 47 ) 2 − ( 21 5 ) 2 = 2 ( 47 − 21 5 ) 2209 − 2205 = 2 ( 47 − 21 5 ) 4 = ( 47 − 21 5 ) 2 \dfrac{1}{x} = \dfrac{2}{47 + 21\sqrt{5}} × \dfrac{47 - 21\sqrt{5}}{47 - 21\sqrt{5}} \\[1.5em] = \dfrac{{2}(47 - 21\sqrt{5})} {(47)^2 - (21\sqrt{5})^2} \\[1.5em] = \dfrac{{2}(47 - 21\sqrt{5})} {2209 - 2205} \\[1.5em] = \dfrac{{2}(47 - 21\sqrt{5})} {4} \\[1.5em] = \dfrac{(47 - 21\sqrt{5})} {2} \\[1.5em] x 1 = 47 + 21 5 2 × 47 − 21 5 47 − 21 5 = ( 47 ) 2 − ( 21 5 ) 2 2 ( 47 − 21 5 ) = 2209 − 2205 2 ( 47 − 21 5 ) = 4 2 ( 47 − 21 5 ) = 2 ( 47 − 21 5 )
Now,
( x + 1 x ) = 47 + 21 5 2 + ( 47 − 21 5 ) 2 = 47 + 21 5 + 47 − 21 5 2 = 94 2 = 47 ∴ ( x + 1 x ) = 47 ....(i) {\Big(x + \dfrac{1}{x}\Big)} = \dfrac{47 + 21\sqrt{5}}{2} + \dfrac{(47 - 21\sqrt{5})}{2} \\[1.5em] = \dfrac{47 + 21\sqrt{5} + 47 -21\sqrt{5}}{2} \\[1.5em] = \dfrac{94}{2} = 47 \\[1.5em] \therefore {\Big(x + \dfrac{1}{x}\Big)} = 47 \qquad \text{....(i)} ( x + x 1 ) = 2 47 + 21 5 + 2 ( 47 − 21 5 ) = 2 47 + 21 5 + 47 − 21 5 = 2 94 = 47 ∴ ( x + x 1 ) = 47 ....(i)
We know that ( x + 1 x ) 2 = x 2 + 1 x 2 + 2 {\Big(x + \dfrac{1}{x}\Big)}^2 = x^2 + \dfrac{1}{x^2} + 2 ( x + x 1 ) 2 = x 2 + x 2 1 + 2
⇒ x 2 + 1 x 2 = ( x + 1 x ) 2 − 2 ⇒ x 2 + 1 x 2 = ( 47 ) 2 − 2 . . . . . using(i) ⇒ x 2 + 1 x 2 = 2209 − 2 = 2207 x 2 + 1 x 2 = 2207 \Rightarrow x^2 + \dfrac{1}{x^2} = {\Big(x + \dfrac{1}{x}\Big)}^2 -2 \\[1.5em] \Rightarrow x^2 + \dfrac{1}{x^2} = (47)^2 - 2 \qquad ..... \text{using(i)} \\[1.5em] \Rightarrow x^2 + \dfrac{1}{x^2} = 2209 - 2 = 2207 \\[1.5em] \bold{x^2 + \dfrac{1}{x^2} = 2207} \\[1.5em] ⇒ x 2 + x 2 1 = ( x + x 1 ) 2 − 2 ⇒ x 2 + x 2 1 = ( 47 ) 2 − 2 ..... using(i) ⇒ x 2 + x 2 1 = 2209 − 2 = 2207 x 2 + x 2 1 = 2207
(ii) x = 5 − 2 5 + 2 and y = 5 + 2 5 − 2 ∴ x + y = 5 − 2 5 + 2 + 5 + 2 5 − 2 ⇒ ( 5 − 2 ) 2 + ( 5 + 2 ) 2 ( 5 + 2 ) ( 5 − 2 ) ⇒ ( 5 ) 2 − 2 × 2 × 5 + ( 2 ) 2 + ( 5 ) 2 + 2 × 2 × 5 + ( 2 ) 2 ( 5 ) 2 − ( 2 ) 2 ⇒ 5 − 2 10 + 2 + 5 + 2 10 + 2 5 − 2 = 14 3 ⇒ x + y = 14 3 ....(i) Also x y = 5 − 2 5 + 2 × 5 + 2 5 − 2 = 1 ....(ii) \text{(ii) } x = \dfrac{\sqrt{5} - \sqrt{2} }{\sqrt{5} + \sqrt{2}} \text{ and y} = \dfrac{\sqrt{5} + \sqrt{2}}{\sqrt{5} - \sqrt{2}} \\[1.5em] \therefore x+y = \dfrac{\sqrt{5} - \sqrt{2}}{\sqrt{5} + \sqrt{2}} +\dfrac{\sqrt{5} + \sqrt{2}}{\sqrt{5} - \sqrt{2}} \\[1.5em] \Rightarrow\dfrac{(\sqrt{5} - \sqrt{2})^2 + (\sqrt{5} + \sqrt{2})^2}{(\sqrt{5} + \sqrt{2})(\sqrt{5} - \sqrt{2})} \\[1.5em] \Rightarrow\dfrac{(\sqrt{5})^2 - 2 × \sqrt{2} × \sqrt{5} + (\sqrt{2})^2+ (\sqrt{5})^2 + 2 × \sqrt{2} × \sqrt{5} + (\sqrt{2})^2 }{(\sqrt{5})^2 - (\sqrt{2})^2 } \\[1.5em] \Rightarrow\dfrac{5 - 2\sqrt{10} + 2 + 5 + 2\sqrt{10} + 2}{5 - 2} = \dfrac{14}{3} \\[1.5em] \Rightarrow x + y = \dfrac{14}{3} \qquad \text{....(i)} \\[1.5em] \text{Also } xy = \dfrac{\sqrt{5} - \sqrt{2} }{\sqrt{5} + \sqrt{2}} ×\dfrac{\sqrt{5} + \sqrt{2} }{\sqrt{5} - \sqrt{2}} = 1 \qquad \text{....(ii)} \\[1.5em] (ii) x = 5 + 2 5 − 2 and y = 5 − 2 5 + 2 ∴ x + y = 5 + 2 5 − 2 + 5 − 2 5 + 2 ⇒ ( 5 + 2 ) ( 5 − 2 ) ( 5 − 2 ) 2 + ( 5 + 2 ) 2 ⇒ ( 5 ) 2 − ( 2 ) 2 ( 5 ) 2 − 2 × 2 × 5 + ( 2 ) 2 + ( 5 ) 2 + 2 × 2 × 5 + ( 2 ) 2 ⇒ 5 − 2 5 − 2 10 + 2 + 5 + 2 10 + 2 = 3 14 ⇒ x + y = 3 14 ....(i) Also x y = 5 + 2 5 − 2 × 5 − 2 5 + 2 = 1 ....(ii)
We need to find the value of x 2 + x y + y 2 {x^2 + xy + y^2} x 2 + x y + y 2 x 2 + x y + y 2 = x 2 + y 2 + 2 x y − x y ⇒ x 2 + x y + y 2 = ( x + y ) 2 − x y ....(iii) {x^2 + xy + y^2} = x^2 + y^2 + 2xy - xy \\[1.5em] \Rightarrow {x^2 + xy + y^2} = (x+y)^2 - xy \qquad \text{....(iii)} \\[1.5em] x 2 + x y + y 2 = x 2 + y 2 + 2 x y − x y ⇒ x 2 + x y + y 2 = ( x + y ) 2 − x y ....(iii)
Substituting the values from (i) and (ii) in (iii),
x 2 + x y + y 2 = ( 14 3 ) 2 − 1 = 196 9 − 1 = 196 − 9 9 = 187 9 ∴ x 2 + x y + y 2 = 187 9 {x^2 + xy + y^2} = \Big(\dfrac{14}{3}\Big)^2 - 1 = \dfrac{196}{9} - 1 = \dfrac{196 - 9}{9} = \dfrac{187}{9} \\[1.5em] \therefore \bold{x^2 + xy + y^2} = \bold{\dfrac{187}{9}} \\[1.5em] x 2 + x y + y 2 = ( 3 14 ) 2 − 1 = 9 196 − 1 = 9 196 − 9 = 9 187 ∴ x 2 + xy + y 2 = 9 187
(iii) x = 3 − 2 3 + 2 and y = 3 + 2 3 − 2 ∴ x + y = 3 − 2 3 + 2 + 3 + 2 3 − 2 = ( 3 − 2 ) 2 + ( 3 + 2 ) 2 ( 3 + 2 ) ( 3 − 2 ) = ( 3 ) 2 − 2 × 2 × 3 + ( 2 ) 2 + ( 3 ) 2 + 2 × 2 × 3 + ( 2 ) 2 ( 3 ) 2 − ( 2 ) 2 = 3 − 2 6 + 2 + 3 + 2 6 + 2 3 − 2 = 10 1 = 10 ∴ x + y = 10 ....(i) Also x y = 3 − 2 3 + 2 × 3 + 2 3 − 2 = 1 ....(ii) \text{(iii) } x = \dfrac{\sqrt{3} - \sqrt{2} }{\sqrt{3} + \sqrt{2}} \text{ and y} = \dfrac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} \\[1.5em] \therefore x+y = \dfrac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}} +\dfrac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} \\[1.5em] = \dfrac{(\sqrt{3} - \sqrt{2})^2 + (\sqrt{3} + \sqrt{2})^2}{(\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2})} \\[1.5em] = \dfrac{(\sqrt{3})^2 - 2 × \sqrt{2} × \sqrt{3} + (\sqrt{2})^2+ (\sqrt{3})^2 + 2 × \sqrt{2} × \sqrt{3} + (\sqrt{2})^2 }{(\sqrt{3})^2 - (\sqrt{2})^2 } \\[1.5em] = \dfrac{3 - 2\sqrt{6} + 2 + 3 + 2\sqrt{6} + 2}{3 - 2} = \dfrac{10}{1} = 10 \\[1.5em] \therefore x + y = 10 \qquad \text{....(i)} \\[1.5em] \text{Also } xy = \dfrac{\sqrt{3} - \sqrt{2} }{\sqrt{3} + \sqrt{2}} ×\dfrac{\sqrt{3} + \sqrt{2} }{\sqrt{3} - \sqrt{2}} = 1 \qquad \text{....(ii)} \\[1.5em] (iii) x = 3 + 2 3 − 2 and y = 3 − 2 3 + 2 ∴ x + y = 3 + 2 3 − 2 + 3 − 2 3 + 2 = ( 3 + 2 ) ( 3 − 2 ) ( 3 − 2 ) 2 + ( 3 + 2 ) 2 = ( 3 ) 2 − ( 2 ) 2 ( 3 ) 2 − 2 × 2 × 3 + ( 2 ) 2 + ( 3 ) 2 + 2 × 2 × 3 + ( 2 ) 2 = 3 − 2 3 − 2 6 + 2 + 3 + 2 6 + 2 = 1 10 = 10 ∴ x + y = 10 ....(i) Also x y = 3 + 2 3 − 2 × 3 − 2 3 + 2 = 1 ....(ii)
We need to find the value of x 3 + y 3 {x^3 + y^3} x 3 + y 3 x 3 + y 3 = ( x + y ) 3 − 3 x y ( x + y ) ....(iii) {x^3 + y^3} = (x+y)^3 - 3xy(x + y) \qquad \text{....(iii)} \\[1.5em] x 3 + y 3 = ( x + y ) 3 − 3 x y ( x + y ) ....(iii)
Substituting the values from (i) and (ii) in (iii),
x 3 + y 3 = ( 10 ) 3 − 3 × 1 × 10 = 1000 − 30 = 970 ∴ x 3 + y 3 = 970 {x^3+ y^3} = (10)^3 - 3 × 1 × 10 \\[1.5em] = 1000 - 30 \\[1.5em] = 970 \\[1.5em] \therefore \bold{x^3+ y^3 = 970} x 3 + y 3 = ( 10 ) 3 − 3 × 1 × 10 = 1000 − 30 = 970 ∴ x 3 + y 3 = 970
Write the following real numbers in descending order :
2 , 3.5 , 10 , − 5 2 , 5 2 3 \sqrt{2}, 3.5, \sqrt{10}, -\dfrac{5}{\sqrt{2}}, {\dfrac{5}{2}}{\sqrt{3}} 2 , 3.5 , 10 , − 2 5 , 2 5 3
Answer
Write all the numbers as square root under one radical :
2 = 2 3.5 = 12.25 10 = 10 − 5 2 = − 25 2 = − 12.5 5 2 3 = 25 × 3 4 = 75 4 = 18.75 Since , 18.75 > 12.25 > 10 > 2 > − 12.5 ⇒ 18.75 > 12.25 > 10 > 2 > − 12.5 ⇒ 5 2 3 > 3.5 > 10 > 2 > − 5 2 \sqrt{2} = \sqrt{2} \\[1.5em] 3.5 = \sqrt{12.25} \\[1.5em] \sqrt{10} = \sqrt{10} \\[1.5em] {-\dfrac{5}{\sqrt{2}}} = - \sqrt{\dfrac{25}{2}} = -\sqrt{12.5} \\[1.5em] {\dfrac{5}{2}}{\sqrt{3}} = \sqrt{\dfrac{25 × 3}{4}} = \sqrt{\dfrac{75}{4}} = \sqrt{18.75} \\[1.5em] \text{Since} , 18.75 \gt 12.25 \gt 10 \gt 2 \gt - 12.5 \\[1.5em] \Rightarrow \sqrt{18.75} \gt \sqrt{12.25} \gt \sqrt{10} \gt \sqrt{2} \gt -\sqrt{12.5} \\[1.5em] \Rightarrow {\dfrac{5}{2}}{\sqrt{3}}\gt 3.5 \gt \sqrt{10} \gt \sqrt{2} \gt -\dfrac{5}{\sqrt{2}} 2 = 2 3.5 = 12.25 10 = 10 − 2 5 = − 2 25 = − 12.5 2 5 3 = 4 25 × 3 = 4 75 = 18.75 Since , 18.75 > 12.25 > 10 > 2 > − 12.5 ⇒ 18.75 > 12.25 > 10 > 2 > − 12.5 ⇒ 2 5 3 > 3.5 > 10 > 2 > − 2 5
Hence, the given numbers in descending order are 5 2 3 , 3.5 , 10 , 2 , − 5 2 \bold{\dfrac{5}{2}{\sqrt{3}}} , \bold{3.5} , \bold{\sqrt{10}} ,\bold{\sqrt{2}} , \bold{-\dfrac{5}{\sqrt{2}}} 2 5 3 , 3.5 , 10 , 2 , − 2 5
Find a rational number and an irrational number between 3 \sqrt{3} 3 5 \sqrt{5} 5
Answer
Consider the squares of 3 \sqrt{3} 3 5 \sqrt{5} 5
( 3 ) 2 (\sqrt{3})^2 ( 3 ) 2 ( 5 ) 2 (\sqrt{5})^2 ( 5 ) 2
Here, 4 is rational number between 3 and 5
4 \sqrt{4} 4 3 \sqrt{3} 3 5 \sqrt{5} 5
Irrational number between 3 \sqrt{3} 3 5 \sqrt{5} 5 3 + 5 2 \dfrac{\sqrt{3} + \sqrt{5}}{2} 2 3 + 5
Insert three irrational numbers between 2 3 2\sqrt{3} 2 3 2 5 2\sqrt{5} 2 5
Answer
Consider the squares of 2 3 2\sqrt{3} 2 3 2 5 2\sqrt{5} 2 5
( 2 3 ) 2 (2\sqrt{3})^2 ( 2 3 ) 2 ( 2 5 ) 2 (2\sqrt{5})^2 ( 2 5 ) 2
As , 18 > 17 > 15 18 \gt 17 \gt 15 18 > 17 > 15
18 > 17 > 15 \sqrt{18} \gt \sqrt{17} \gt \sqrt{15} 18 > 17 > 15
18 \sqrt{18} 18 17 \sqrt{17} 17 15 \sqrt{15} 15 12 \sqrt{12} 12 20 \sqrt{20} 20 2 3 2\sqrt{3} 2 3 2 5 2\sqrt{5} 2 5
Hence, three irrational number between 2 3 2\sqrt{3} 2 3 2 5 2\sqrt{5} 2 5 18 \sqrt{18} 18 17 \sqrt{17} 17 15 \sqrt{15} 15
Give an example each of two different irrational numbers , whose
(i) sum is an irrational number.
(ii) product is an irrational number.
Answer
Let a = 2 \sqrt{2} 2 3 \sqrt{3} 3
(i) a + b = 2 \sqrt{2} 2 3 \sqrt{3} 3
(ii) a × b = 2 \sqrt{2} 2 3 \sqrt{3} 3 6 \sqrt{6} 6
Give an example of two different irrational numbers , a and b where a b \dfrac{a}{b} b a
Answer
Let a = 2 3 2\sqrt{3} 2 3 5 3 5\sqrt{3} 5 3
Here, a b \dfrac{a}{b} b a 2 3 5 3 \dfrac{2\sqrt{3}}{5\sqrt{3}} 5 3 2 3 2 5 \dfrac{2}{5} 5 2
∴ 2 5 \therefore \bold{\dfrac{2}{5}} ∴ 5 2
If 34.0356 is expressed in the form p q \dfrac{p}{q} q p
Answer
34.0356 can be expressed in the form p q \dfrac{p}{q} q p
This can be written as 34.0356 = 340356 10000 \dfrac{340356}{10000} 10000 340356 85089 2500 \dfrac{85089}{2500} 2500 85089
Here, 85089 and 2500 are coprime integers .
Since , it is terminating decimal
It is Rational number and the prime factors of its denominator q will be 2 or 5 or both .
In each case, state whether the following numbers are rational or irrational. If they are rational and expressed in the form p q \dfrac{p}{q} q p
(i) 279.034 (ii) 76. 17893 ‾ (iii) 3.010010001... (iv) 39.546782 (v) 2.3476817681... (vi) 59.120120012000... \begin{matrix} \text{(i)} & 279.034 \\[1.5em] \text{(ii)} & 76.\overline{17893} \\[1.5em] \text{(iii)} & 3.010010001... \\[1.5em] \text{(iv)} & 39.546782 \\[1.5em] \text{(v)} & 2.3476817681... \\[1.5em] \text{(vi)} & 59.120120012000... \\[1.5em] \end{matrix} (i) (ii) (iii) (iv) (v) (vi) 279.034 76. 17893 3.010010001... 39.546782 2.3476817681... 59.120120012000...
Answer
(i) 279.034
This can be written as 279.034 = 279034 1000 \dfrac{279034}{1000} 1000 279034
Since, it is terminating decimal
It is Rational number and the prime factors of its denominator q will be 2 or 5 or both .
(ii) 76. 17893 ‾ 76.\overline{17893} 76. 17893
Since it is non-terminating recurring decimal,
76. 17893 ‾ 76.\overline{17893} 76. 17893
It is a rational number which is non-terminating and repeating. Its denominator q will have prime factors other than 2 or 5 .
(iii) 3.010010001...
Since, it is non-terminating non-repeating decimal number
∴ It is an Irrational number .
(iv) 39.546782
This can be written as 39.546782 = 39546782 1000000 \dfrac{39546782}{1000000} 1000000 39546782
Since , it is terminating decimal
It is Rational number and the prime factors of its denominator q will be 2 or 5 or both .
(v) 2.3476817681... = 2.34 7681 ‾ 2.34\overline{7681} 2.34 7681
Since, it is a non-terminating repeating decimal number,
∴ It is a Rational number and its denominator q will have prime factors other than 2 or 5.
(vi) 59.120120012000...
Since, it is non-terminating non-repeating decimal number
∴ It is an Irrational number .
