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Chapter 1

Rational and Irrational Numbers

Class - 9 RS Aggarwal Mathematics Solutions

Exercise 1(A)

Question 1

Without actual division, find which of the following fractions are terminating decimals :

(i) 925\dfrac{9}{25}

(ii) 712\dfrac{7}{12}

(iii) 1316\dfrac{13}{16}

(iv) 25128\dfrac{25}{128}

(v) 950\dfrac{9}{50}

(vi) 121125\dfrac{121}{125}

(vii) 1955\dfrac{19}{55}

(viii) 3778\dfrac{37}{78}

(ix) 2380\dfrac{23}{80}

(x) 1930\dfrac{19}{30}

Answer

In rational numbers, if the denominator of the fraction can be expressed in the form of 2m. × 5n., then it is a terminating decimal.

(i) So, 25 can be expressed as 20. × 52., which is in the form of 2m. × 5n..

Hence, 925\dfrac{9}{25} is a terminating decimal number.

(ii) So, 12 can be expressed as 3 × 22. × 50., which is not in the form of 2m. × 5n..

Hence, 712\dfrac{7}{12} is not a terminating decimal number.

(iii) So, 16 can be expressed as 24. × 50., which is in the form of 2m. × 5n..

Hence, 1316\dfrac{13}{16} is a terminating decimal number.

(iv) So, 128 can be expressed as 27. × 50., which is in the form of 2m. × 5n..

Hence, 25128\dfrac{25}{128} is a terminating decimal number.

(v) So, 50 can be expressed as 21. × 52., which is in the form of 2m. × 5n..

Hence, 950\dfrac{9}{50} is a terminating decimal number.

(vi) So, 125 can be expressed as 20. × 53., which is in the form of 2m. × 5n..

Hence, 121125\dfrac{121}{125} is a terminating decimal number.

(vii) So, 55 can be expressed as 11 × 20. × 51., which is not in the form of 2m. × 5n..

Hence, 1955\dfrac{19}{55} is not a terminating decimal number.

(viii) So, 78 can be expressed as 39 × 21. × 50., which is not in the form of 2m. × 5n..

Hence, it is not a terminating decimal number.

(ix) So, 80 can be expressed as 24. × 51., which is in the form of 2m. × 5n..

Hence, 2380\dfrac{23}{80} is a terminating decimal number.

(x) So, 30 can be expressed as 3 × 21. × 51., which is not in the form of 2m. × 5n..

Hence, 1930\dfrac{19}{30} is not a terminating decimal number.

Question 2

Convert each of the following decimals into vulgar fraction in its lowest terms :

(i) 0.65

(ii) 1.08

(iii) 0.075

(iv) 2.016

(v) 1.732

Answer

(i) Given,

⇒0.65=65100\Rightarrow 0.65 = \dfrac{65}{100}

The H.CF. of 65 and 100 is 5, dividing the numerator and denominator by 5,

⇒65÷5100÷5⇒1320.\Rightarrow \dfrac{65 ÷ 5}{100 ÷ 5} \\[1em] \Rightarrow \dfrac{13}{20}.

Hence, 0.65 = 1320\dfrac{13}{20}.

(ii) Given,

⇒1.08=108100\Rightarrow 1.08 = \dfrac{108}{100}

The H.CF. of 108 and 100 is 4, dividing the numerator and denominator by 4,

⇒108÷4100÷4⇒2725\Rightarrow \dfrac{108 ÷ 4}{100 ÷ 4} \\[1em] \Rightarrow \dfrac{27}{25}

Hence, 1.08 = 2725\dfrac{27}{25}.

(iii) Given,

⇒0.075=751000\Rightarrow 0.075 = \dfrac{75}{1000}

The H.CF. of 75 and 1000 is 25, dividing the numerator and denominator by 25,

⇒75÷251000÷25⇒340\Rightarrow \dfrac{75 ÷ 25}{1000 ÷ 25} \\[1em] \Rightarrow \dfrac{3}{40}

Hence, 0.075 = 340\dfrac{3}{40}.

(iv) Given,

⇒2.016=20161000\Rightarrow 2.016 = \dfrac{2016}{1000}

The H.CF. of 2016 and 1000 is 8, dividing the numerator and denominator by 8,

⇒2016÷81000÷8⇒252125\Rightarrow \dfrac{2016 ÷ 8}{1000 ÷ 8} \\[1em] \Rightarrow \dfrac{252}{125}

Hence, 2.016 = 252125\dfrac{252}{125}.

(v) Given,

⇒1.732=17321000\Rightarrow 1.732 = \dfrac{1732}{1000}

The H.CF. of 1732 and 1000 is 4, dividing the numerator and denominator by 4

⇒1732÷41000÷4⇒433250\Rightarrow \dfrac{1732 ÷ 4}{1000 ÷ 4} \\[1em] \Rightarrow \dfrac{433}{250}

Hence, 1.732 = 433250\dfrac{433}{250}.

Question 3

Convert each of the following fractions into a decimal :

(i) 18\dfrac{1}{8}

(ii) 332\dfrac{3}{32}

(iii) 449\dfrac{44}{9}

(iv) 1124\dfrac{11}{24}

(v) 1213\dfrac{12}{13}

(vi) 2744\dfrac{27}{44}

(vii) 25122\dfrac{5}{12}

(viii) 131551\dfrac{31}{55}

Answer

(i) 18\dfrac{1}{8}

On dividing,

18=0.125\dfrac{1}{8} = 0.125

Hence, 18=0.125\dfrac{1}{8} = 0.125.

(ii) 332\dfrac{3}{32}

On dividing,

332=0.09375\dfrac{3}{32} = 0.09375

Hence, 332=0.09375\dfrac{3}{32} = 0.09375.

(iii) 449\dfrac{44}{9}

On dividing,

449=4.8‾\dfrac{44}{9} = 4.\overline{8}

Hence, 449=4.8‾\dfrac{44}{9} = 4.\overline{8}.

(iv) 1124\dfrac{11}{24}

On dividing,

1124=0.4583‾\dfrac{11}{24} = 0.458\overline{3}

Hence, 1124=0.4583‾\dfrac{11}{24} = 0.458\overline{3}.

(v) 1213\dfrac{12}{13}

On dividing,

1213=0.923076‾\dfrac{12}{13} = 0.\overline{923076}

Hence, 1213=0.923076‾\dfrac{12}{13} = 0.\overline{923076}.

(vi) 2744\dfrac{27}{44}

On dividing,

2744=0.6136‾\dfrac{27}{44} = 0.61\overline{36}

Hence, 2744=0.6136‾\dfrac{27}{44} = 0.61\overline{36}.

(vii) 25122\dfrac{5}{12}

On dividing,

2512=2912=2.416‾2\dfrac{5}{12} = \dfrac{29}{12} = 2.41\overline{6}

Hence, 2912=2.416‾\dfrac{29}{12} = 2.41\overline{6}.

(viii) 131551\dfrac{31}{55}

On dividing,

13155=8655=1.563‾1\dfrac{31}{55} = \dfrac{86}{55} = 1.5\overline{63}

Hence, 13155=1.563‾.1\dfrac{31}{55} = 1.5\overline{63}..

Question 4

Express 1556\dfrac{15}{56} as a decimal, correct to four decimal places.

Answer

On dividing,

1556=0.26785\dfrac{15}{56} = 0.26785

Hence, 1556=0.2679\dfrac{15}{56} = 0.2679.

Question 5

Express 1334\dfrac{13}{34} as a decimal, correct to three decimal places.

Answer

On dividing,

1334=0.3823\dfrac{13}{34} = 0.3823

Hence, 1334=0.382\dfrac{13}{34} = 0.382.

Question 6

By actual division show that :

(i) 119=1.2‾\dfrac{11}{9} = 1.\overline{2}

(ii) 4311=3.90‾\dfrac{43}{11} = 3.\overline{90}

(iii) 10745=2.37‾\dfrac{107}{45} = 2.3\overline{7}

(iv) 2155=0.381‾\dfrac{21}{55} = 0.3\overline{81}

Answer

(i) On dividing,

119=1.222......\dfrac{11}{9} = 1.222......

Hence, proved that 119=1.2‾\dfrac{11}{9} = 1.\overline{2}.

(ii) On dividing,

4311=3.9090...\dfrac{43}{11} = 3.9090...

Hence, proved that 4311=3.90‾\dfrac{43}{11} = 3.\overline{90}.

(iii) On dividing,

10745=2.3777...\dfrac{107}{45} = 2.3777...

Hence, proved that 10745=2.37‾\dfrac{107}{45} = 2.3\overline{7}.

(iv) On dividing,

2155=0.38181...\dfrac{21}{55} = 0.38181...

Hence, proved that 2155=0.381‾\dfrac{21}{55} = 0.3\overline{81}.

Question 7

Express each of following as a vulgar fraction in simplest form :

(i) 0.5‾0.\overline{5}

(ii) 0.43‾0.\overline{43}

(iii) 0.158‾0.\overline{158}

(iv) 1.3‾1.\overline{3}

(v) 4.17‾4.\overline{17}

(vi) 0.12‾0.\overline{12}

(vii) 0.136‾0.1\overline{36}

(viii) 1.57‾1.5\overline{7}

Answer

(i) 0.5‾0.\overline{5}

Let x = 0.5‾0.\overline{5}.

⇒ x = 0.555..     ...........(1)

⇒ 10x = 5.555..     ..........(2)

On subtracting (1) from (2), we get :

⇒ 10x - x = 5.555... - 0.555...

⇒ 9x = 5

⇒ x = 59\dfrac{5}{9}

Hence, 0.5‾=590.\overline{5} = \dfrac{5}{9}.

(ii) 0.43‾0.\overline{43}

Let x = 0.43‾0.\overline{43}.

⇒ x = 0.434343..     .........(1)

⇒ 100x = 43.4343..     ........(2)

On subtracting (1) from (2), we get :

⇒ 100x - x = 43.4343.... - 0.4343......

⇒ 99x = 43

⇒ x = 4399\dfrac{43}{99}

Hence, 0.43‾=43990.\overline{43} = \dfrac{43}{99}.

(iii) 0.158‾0.\overline{158}

Let x = 0.158‾0.\overline{158}

⇒ x = 0.158158..     ......(1)

⇒ 1000x = 158.158158..     ......(2)

On subtracting (1) from (2), we get :

⇒ 1000x - x = 158.158158.... - 0.158158.....

⇒ 999x = 158

⇒ x = 158999\dfrac{158}{999}

Hence, 0.158‾=1589990.\overline{158} = \dfrac{158}{999}.

(iv) 1.3‾1.\overline{3}

Let x = 1.3‾1.\overline{3}

⇒ x = 1.3333..     ......(1)

⇒ 10x = 13.333..     ......(2)

On subtracting (1) from (2), we get :

⇒ 10x - x = 13.333.. - 1.333..

⇒ 9x = 12

⇒ x = 129\dfrac{12}{9}

⇒ x = 43\dfrac{4}{3}

Hence, 1.3‾=431.\overline{3} = \dfrac{4}{3}.

(v) 4.17‾4.\overline{17}

Let x = 4.17‾4.\overline{17}

⇒ x = 4.1717..     ........(1)

⇒ 100x = 417.1717..     .......(2)

On subtracting (1) from (2), we get :

⇒ 100x - x = 417.1717..... - 4.1717.....

⇒ 99x = 413

⇒ x = 41399\dfrac{413}{99}

Hence, 4.17‾=413994.\overline{17} = \dfrac{413}{99}.

(vi) 0.12‾0.\overline{12}

Let x = 0.12‾0.\overline{12}

⇒ x = 0.1222..     ...........(1)

⇒ 100x = 12.222...(ii)

On subtracting (i) from (ii), we get

⇒ 99x = 12

⇒ x = 1299\dfrac{12}{99}

⇒ x = 12÷399÷3\dfrac{12÷3}{99÷3}

⇒ x = 433\dfrac{4}{33}

Hence, 0.12... = 433\dfrac{4}{33}.

(vii) 0.136‾0.1\overline{36}

Let x = 0.136‾0.1\overline{36}

⇒ x = 0.13636..     ...........(1)

⇒ 10x = 1.3636..     ...........(2)

⇒ 1000x = 136.3636..     ...........(3)

On subtracting (2) from (3), we get

⇒ 1000x - 10x = 136.3636.. - 1.3636..

⇒ 990x = 135

⇒ x = 135990\dfrac{135}{990}

⇒ x = 135÷45990÷45\dfrac{135 ÷ 45}{990 ÷ 45}

⇒ x = 322\dfrac{3}{22}

Hence, 0.136‾3220.1\overline{36}\dfrac{3}{22}.

(viii) 1.57‾1.5\overline{7}

Let x = 1.57‾1.5\overline{7}.

⇒ x = 1.5777..     ...........(1)

⇒ 10x = 15.777..     ...........(2)

⇒ 100x = 157.7777..     ...........(3)

On subtracting (i) from (iii), we get

⇒ 100x - 10x = 157.7777.. - 15.777......

⇒ 90x = 142

⇒ x = 14290\dfrac{142}{90}

⇒ x = 142÷290÷2\dfrac{142 ÷ 2}{90 ÷ 2}

⇒ x = 7145\dfrac{71}{45}

Hence, 1.57‾=71451.5\overline{7} = \dfrac{71}{45}.

Exercise 1(B)

Question 1

Write the additive inverse of :

(i) 5

(ii) -7

(iii) 59\dfrac{5}{9}

(iv) −317\dfrac{-3}{17}

(v) 0

(vi) 1151711\dfrac{5}{17}

(vii) −538-5\dfrac{3}{8}

(viii) -37

(ix) 1

Answer

The additive inverse of a number is the number which, when added to the original number, results in zero.

(i) Let x be the additive inverse of 5, then :

⇒ 5 + x = 0

⇒ x = -5.

Hence, additive inverse of 5 = -5.

(ii) Let x be the additive inverse of -7, then :

⇒ -7 + x = 0

⇒ x = 7.

Hence, additive inverse of -7 = 7.

(iii) Let x be the additive inverse of 59\dfrac{5}{9}, then :

⇒ 59\dfrac{5}{9} + x = 0

⇒ x = −59-\dfrac{5}{9}.

Hence,additive inverse of 59=−59\dfrac{5}{9} = -\dfrac{5}{9}.

(iv) Let x be the additive inverse of −317-\dfrac{3}{17}, then :

⇒ −317-\dfrac{3}{17} + x = 0

⇒ x = 317\dfrac{3}{17}.

Hence,additive inverse of −317=317-\dfrac{3}{17} = \dfrac{3}{17}.

(v) Let x be the additive inverse of 0, then :

⇒ 0 + x = 0

⇒ x = 0.

Hence, additive inverse of 0 is 0.

(vi) Let x be the additive inverse of 1151711\dfrac{5}{17}, then :

⇒ 1151711\dfrac{5}{17} + x = 0

⇒ 19217\dfrac{192}{17} + x = 0

⇒ x = −19217-\dfrac{192}{17}.

Hence, additive inverse of 11517=−1921711\dfrac{5}{17} = -\dfrac{192}{17}.

(vii) Let x be the additive inverse of −538-5\dfrac{3}{8}, then :

⇒ −538-5\dfrac{3}{8} + x = 0

⇒ −438-\dfrac{43}{8} + x = 0

⇒ x = 438\dfrac{43}{8}.

Hence, additive inverse of −538=438-5\dfrac{3}{8} = \dfrac{43}{8}.

(viii) Let x be the additive inverse of -37, then :

⇒ -37 + x = 0

⇒ x = 37.

Hence, additive inverse of -37 = 37.

(ix) Let x be the additive inverse of 1, then :

⇒ 1 + x = 0

⇒ x = -1.

Hence, additive inverse of 1 = -1.

Question 2

Write the multiplicative inverse of :

(i) 9

(ii) -1

(iii) 1116\dfrac{11}{16}

(iv) 5145\dfrac{1}{4}

(v) −23\dfrac{-2}{3}

(vi) 1732017\dfrac{3}{20}

(vii) –1812–18\dfrac{1}{2}

(viii) –5

(ix) −2041\dfrac{-20}{41}

Answer

The multiplicative inverse of a number is defined as a number that when multiplied by the original number gives the product as 1.

(i) Let the multiplicative inverse of 9, be x.

⇒ 9 × x = 1

⇒ x = 19\dfrac{1}{9}.

Hence, multiplicative inverse of 9 = 19\dfrac{1}{9}.

(ii) Let the multiplicative inverse of -1, be x.

⇒ -1 × x = 1

⇒ x = −11-\dfrac{1}{1} = -1.

Hence, multiplicative inverse of -1 = -1.

(iii) Let the multiplicative inverse of 1116\dfrac{11}{16}, be x.

⇒ 1116\dfrac{11}{16} × x = 1

⇒ x = 1611\dfrac{16}{11}.

Hence, multiplicative inverse of 1116=1611\dfrac{11}{16} = \dfrac{16}{11}.

(iv) Let the multiplicative inverse of 5145\dfrac{1}{4}, be x.

⇒ 5145\dfrac{1}{4} × x = 1

⇒ 214\dfrac{21}{4} × x = 1

⇒ x = 421\dfrac{4}{21}.

Hence, multiplicative inverse of 514=4215\dfrac{1}{4} = \dfrac{4}{21}.

(v) Let the multiplicative inverse of −23-\dfrac{2}{3}, be x.

⇒ −23-\dfrac{2}{3} × x = 1

⇒ x = −32-\dfrac{3}{2}.

Hence, multiplicative inverse of −23=−32-\dfrac{2}{3} = -\dfrac{3}{2}.

(vi) Let the multiplicative inverse of 1732017\dfrac{3}{20}, be x.

⇒ 1732017\dfrac{3}{20} × x = 1

⇒ 34320\dfrac{343}{20} × x = 1

⇒ x = 20343\dfrac{20}{343}.

Hence, multiplicative inverse of 17320=2034317\dfrac{3}{20} = \dfrac{20}{343}.

(vii) Let the multiplicative inverse of –1812–18\dfrac{1}{2}, be x.

⇒ –1812–18\dfrac{1}{2} × x = 1

⇒ −372-\dfrac{37}{2} × x = 1

⇒ x = –237–\dfrac{2}{37}.

Hence, multiplicative inverse of −1812=−237-18\dfrac{1}{2} = -\dfrac{2}{37}.

(viii) Let the multiplicative inverse of –5, be x.

⇒ –5 × x = 1

⇒ x = −15-\dfrac{1}{5}.

Hence, multiplicative inverse of −5=−15-5 = -\dfrac{1}{5}.

(ix) Let the multiplicative inverse of −2041\dfrac{-20}{41}, be x.

⇒ −2041\dfrac{-20}{41} × x = 1

⇒ x = −4120-\dfrac{41}{20}.

Hence, multiplicative inverse of −2041=−4120-\dfrac{20}{41} = -\dfrac{41}{20}.

Question 3

Represent each of the following on the number line :

(i) 37\dfrac{3}{7}

(ii) 165\dfrac{16}{5}

(iii) −49-\dfrac{4}{9}

(iv) −1811-\dfrac{18}{11}

(v) −316-3\dfrac{1}{6}

Answer

(i) On dividing,

37\dfrac{3}{7} = 0.428

(ii) On dividing,

165\dfrac{16}{5} = 3.2

(iii) On dividing,

−49-\dfrac{4}{9} = -0.4444..

(iv) On dividing,

−1811-\dfrac{18}{11} = -1.6363..

(v) On dividing,

−316=−196-3\dfrac{1}{6} = -\dfrac{19}{6} = -3.166..

Represent each of the following on the number line: R.S. Aggarwal Mathematics Solutions ICSE Class 9.

Question 4

Find a rational number between 35\dfrac{3}{5} and 79\dfrac{7}{9}.

Answer

Let x be a rational number between 35\dfrac{3}{5} and 79\dfrac{7}{9}.

⇒x=12(35+79)⇒x=12(27+3545)⇒x=12(6245)⇒x=3145\Rightarrow x = \dfrac{1}{2}\Big(\dfrac{3}{5} + \dfrac{7}{9}\Big) \\[1em] \Rightarrow x = \dfrac{1}{2}\Big(\dfrac{27 + 35}{45} \Big) \\[1em] \Rightarrow x = \dfrac{1}{2}\Big(\dfrac{62}{45} \Big) \\[1em] \Rightarrow x = \dfrac{31}{45} \\[1em]

Hence, a rational number between 35 and 79 is 3145\dfrac{3}{5} \text{ and } \dfrac{7}{9} \text{ is } \dfrac{31}{45}.

Question 5

Find two rational numbers between :

(i) 2 and 3

(ii) 13 and 25\dfrac{1}{3} \text{ and } \dfrac{2}{5}

(iii) 34 and 115\dfrac{3}{4} \text{ and } 1\dfrac{1}{5}

(iv) –2 and 1

Answer

(i) Let the first rational number between 2 and 3 be x.

⇒x=12(2+3)⇒x=12×5⇒x=52\Rightarrow x = \dfrac{1}{2}\Big(2 + 3\Big) \\[1em] \Rightarrow x = \dfrac{1}{2} \times 5 \\[1em] \Rightarrow x = \dfrac{5}{2} \\[1em]

Let the second rational number be y.

⇒y=12(52+3)⇒y=12(5+62)⇒y=12(112)⇒y=114\Rightarrow y = \dfrac{1}{2} \Big(\dfrac{5}{2} + 3\Big) \\[1em] \Rightarrow y = \dfrac{1}{2} \Big(\dfrac{5+6}{2}\Big) \\[1em] \Rightarrow y = \dfrac{1}{2} \Big(\dfrac{11}{2}\Big) \\[1em] \Rightarrow y = \dfrac{11}{4} \\[1em]

Hence, two rational numbers between 2 and 3 are 52 and 114\dfrac{5}{2} \text{ and } \dfrac{11}{4}.

(ii) Let the first rational number between 13\dfrac{1}{3} and 25\dfrac{2}{5} be x.

⇒x=12(13+25)⇒x=12(5+615)⇒x=12(1115)⇒x=1130\Rightarrow x = \dfrac{1}{2} \Big(\dfrac{1}{3} + \dfrac{2}{5}\Big) \\[1em] \Rightarrow x = \dfrac{1}{2} \Big(\dfrac{5 + 6}{15}\Big) \\[1em] \Rightarrow x = \dfrac{1}{2} \Big(\dfrac{11}{15} \Big) \\[1em] \Rightarrow x = \dfrac{11}{30}

Let the second rational number be y.

⇒y=12(1130+25)⇒y=12(11+1230)⇒y=12(2330)⇒y=2360\Rightarrow y = \dfrac{1}{2} \Big(\dfrac{11}{30} + \dfrac{2}{5}\Big) \\[1em] \Rightarrow y = \dfrac{1}{2} \Big(\dfrac{11 + 12}{30}\Big) \\[1em] \Rightarrow y = \dfrac{1}{2} \Big(\dfrac{23}{30}\Big) \\[1em] \Rightarrow y = \dfrac{23}{60} \\[1em]

Hence, two rational numbers between 13\dfrac{1}{3} and 25\dfrac{2}{5} are 1130 and 2360\dfrac{11}{30} \text{ and } \dfrac{23}{60} .

(iii) Let the first rational number between 34\dfrac{3}{4} and 1151\dfrac{1}{5} be x.

⇒x=12(34+115)⇒x=12(34+65)⇒x=12(15+2420)⇒x=12(3920)⇒x=3940\Rightarrow x = \dfrac{1}{2}\Big(\dfrac{3}{4} + 1\dfrac{1}{5}\Big) \\[1em] \Rightarrow x = \dfrac{1}{2}\Big(\dfrac{3}{4} + \dfrac{6}{5}\Big) \\[1em] \Rightarrow x = \dfrac{1}{2}\Big(\dfrac{15 + 24}{20}\Big) \\[1em] \Rightarrow x = \dfrac{1}{2}\Big(\dfrac{39}{20} \Big) \\[1em] \Rightarrow x = \dfrac{39}{40}

Let the second rational number be y.

⇒y=12(3940+65)⇒y=12(39+4840)⇒y=12(8740)⇒y=8780\Rightarrow y = \dfrac{1}{2}\Big(\dfrac{39}{40} + \dfrac{6}{5}\Big) \\[1em] \Rightarrow y = \dfrac{1}{2}\Big(\dfrac{39 + 48}{40}\Big) \\[1em] \Rightarrow y = \dfrac{1}{2}\Big(\dfrac{87}{40}\Big) \\[1em] \Rightarrow y = \dfrac{87}{80} \\[1em]

Hence, two rational numbers between 34\dfrac{3}{4} and 1151\dfrac{1}{5} are 3940 and 8780\dfrac{39}{40} \text{ and } \dfrac{87}{80} .

(iv) Let the first rational number between -2 and 1 be x.

⇒x=12(−2+1)⇒x=12×−1⇒x=−12\Rightarrow x = \dfrac{1}{2}(-2 + 1) \\[1em] \Rightarrow x = \dfrac{1}{2} \times -1 \\[1em] \Rightarrow x = -\dfrac{1}{2} \\[1em]

Let the second rational number be y.

⇒y=12[−12+(−2)]⇒y=12(−1−42)⇒y=12×−52⇒y=−54.\Rightarrow y = \dfrac{1}{2}\Big[-\dfrac{1}{2} + (-2)\Big] \\[1em] \Rightarrow y = \dfrac{1}{2}\Big(\dfrac{-1-4}{2}\Big) \\[1em] \Rightarrow y = \dfrac{1}{2} \times -\dfrac{5}{2} \\[1em] \Rightarrow y = -\dfrac{5}{4}.

Hence, two rational numbers between -2 and 1 are −12 and −54-\dfrac{1}{2} \text{ and } -\dfrac{5}{4}.

Question 6

Find three rational numbers between :

(i) 4 and 5

(ii) 12 and 35\dfrac{1}{2} \text{ and }\dfrac{3}{5}

(iii) –1 and 1

(iv) 213 and 3232\dfrac{1}{3} \text{ and }3\dfrac{2}{3}

(v) −12 and 13-\dfrac{1}{2} \text{ and }\dfrac{1}{3}

(vi) −13 and 14-\dfrac{1}{3} \text{ and }\dfrac{1}{4}

Answer

(i) Let the first rational number between 4 and 5 be x.

⇒x=12(4+5)⇒x=12×9⇒x=92.\Rightarrow x = \dfrac{1}{2}(4 + 5) \\[1em] \Rightarrow x = \dfrac{1}{2} \times 9 \\[1em] \Rightarrow x = \dfrac{9}{2}.

Let the second rational number be y.

⇒y=12(92+5)⇒y=12(9+102)⇒y=12(192)⇒y=194\Rightarrow y = \dfrac{1}{2}\Big(\dfrac{9}{2} + 5\Big) \\[1em] \Rightarrow y = \dfrac{1}{2}\Big(\dfrac{9 + 10}{2}\Big) \\[1em] \Rightarrow y = \dfrac{1}{2}\Big(\dfrac{19}{2}\Big) \\[1em] \Rightarrow y = \dfrac{19}{4}

Let the third rational number be z.

⇒z=12(92+4)⇒z=12(9+82)⇒z=12(172)⇒z=174\Rightarrow z = \dfrac{1}{2}\Big(\dfrac{9}{2} + 4\Big) \\[1em] \Rightarrow z = \dfrac{1}{2}\Big(\dfrac{9 + 8}{2}\Big) \\[1em] \Rightarrow z = \dfrac{1}{2}\Big(\dfrac{17}{2}\Big) \\[1em] \Rightarrow z = \dfrac{17}{4}

Hence, three rational numbers between 4 and 5 are 174,92 and 194\dfrac{17}{4}, \dfrac{9}{2} \text{ and }\dfrac{19}{4}.

(ii) Let the first rational number between 12\dfrac{1}{2} and 35\dfrac{3}{5} be x.

⇒x=12(12+35)⇒x=12(5+610)⇒x=12×1110⇒x=1120.\Rightarrow x = \dfrac{1}{2}\Big(\dfrac{1}{2} + \dfrac{3}{5}\Big)\\[1em] \Rightarrow x = \dfrac{1}{2}\Big(\dfrac{5 + 6}{10}\Big)\\[1em] \Rightarrow x = \dfrac{1}{2} \times \dfrac{11}{10} \\[1em] \Rightarrow x = \dfrac{11}{20}.

Let the second rational number be y.

⇒y=12(1120+35)⇒y=12(11+1220)⇒y=12(2320)⇒y=2340.\Rightarrow y = \dfrac{1}{2}\Big(\dfrac{11}{20} + \dfrac{3}{5}\Big) \\[1em] \Rightarrow y = \dfrac{1}{2}\Big(\dfrac{11 + 12}{20}\Big) \\[1em] \Rightarrow y = \dfrac{1}{2} \Big(\dfrac{23}{20}\Big) \\[1em] \Rightarrow y = \dfrac{23}{40}.

Let the third rational number be z.

⇒z=12(1120+12)⇒z=12(11+1020)⇒z=12×2120⇒z=2140\Rightarrow z = \dfrac{1}{2}\Big(\dfrac{11}{20} + \dfrac{1}{2}\Big) \\[1em] \Rightarrow z = \dfrac{1}{2}\Big(\dfrac{11 + 10}{20}\Big) \\[1em] \Rightarrow z = \dfrac{1}{2} \times \dfrac{21}{20} \\[1em] \Rightarrow z = \dfrac{21}{40} \\[1em]

Hence, three rational numbers between 12\dfrac{1}{2} and 35\dfrac{3}{5} are 2140,1120 and 2340\dfrac{21}{40}, \dfrac{11}{20} \text{ and } \dfrac{23}{40}.

(iii) Let the first rational number between -1 and 1 be x.

⇒x=12(−1+1)⇒x=12×0⇒x=0.\Rightarrow x = \dfrac{1}{2}(-1 + 1) \\[1em] \Rightarrow x = \dfrac{1}{2} \times 0 \\[1em] \Rightarrow x = 0.

Let the second rational number be y.

⇒y=12(0+1)⇒y=12×1⇒y=12.\Rightarrow y = \dfrac{1}{2}(0 + 1) \\[1em] \Rightarrow y = \dfrac{1}{2} \times 1 \\[1em] \Rightarrow y = \dfrac{1}{2}.

Let the third rational number be z.

⇒z=12[0+(−1)]⇒z=12×−1⇒z=−12\Rightarrow z = \dfrac{1}{2}[0 + (-1)] \\[1em] \Rightarrow z = \dfrac{1}{2} \times -1 \\[1em] \Rightarrow z = -\dfrac{1}{2}

Hence, three rational numbers between -1 and 1 are −12,0 and 12-\dfrac{1}{2}, 0 \text{ and } \dfrac{1}{2}.

(iv) Let the first rational number between 2132\dfrac{1}{3} and 3233\dfrac{2}{3} be x.

⇒x=12(213+323)⇒x=12(73+113)⇒x=12×183⇒x=12×6⇒x=3\Rightarrow x = \dfrac{1}{2}\Big(2\dfrac{1}{3} + 3\dfrac{2}{3}\Big) \\[1em] \Rightarrow x = \dfrac{1}{2}\Big(\dfrac{7}{3} + \dfrac{11}{3}\Big) \\[1em] \Rightarrow x = \dfrac{1}{2} \times \dfrac{18}{3} \\[1em] \Rightarrow x = \dfrac{1}{2} \times 6 \\[1em] \Rightarrow x = 3

Let the second rational number be y.

⇒y=12(3+113)⇒y=12(9+113)⇒y=12×203⇒y=103.\Rightarrow y = \dfrac{1}{2}\Big(3 + \dfrac{11}{3}\Big) \\[1em] \Rightarrow y = \dfrac{1}{2}\Big(\dfrac{9 + 11}{3}\Big) \\[1em] \Rightarrow y = \dfrac{1}{2} \times \dfrac{20}{3} \\[1em] \Rightarrow y = \dfrac{10}{3}.

Let the third rational number be z.

⇒z=12(3+73)⇒z=12(9+73)⇒z=12×163⇒z=83.\Rightarrow z = \dfrac{1}{2}\Big(3 + \dfrac{7}{3}\Big) \\[1em] \Rightarrow z = \dfrac{1}{2}\Big(\dfrac{9 + 7}{3}\Big) \\[1em] \Rightarrow z = \dfrac{1}{2} \times \dfrac{16}{3} \\[1em] \Rightarrow z = \dfrac{8}{3}.

Hence, three rational numbers between 73\dfrac{7}{3} and 113\dfrac{11}{3} are 83,3 and 103\dfrac{8}{3}, 3 \text{ and } \dfrac{10}{3}.

(v) Let the first rational number between −12-\dfrac{1}{2} and 13\dfrac{1}{3} be x.

⇒x=12(−12+13)⇒x=12(−3+26)⇒x=12(−16)⇒x=−112\Rightarrow x = \dfrac{1}{2}\Big(-\dfrac{1}{2} + \dfrac{1}{3}\Big) \\[1em] \Rightarrow x = \dfrac{1}{2}\Big(\dfrac{-3 + 2}{6}\Big) \\[1em] \Rightarrow x = \dfrac{1}{2}\Big(\dfrac{-1}{6} \Big) \\[1em] \Rightarrow x = \dfrac{-1}{12}

Let the second rational number be y.

⇒y=12(−112+−12)⇒y=12(−1−612)⇒y=12(−712)⇒y=−724\Rightarrow y = \dfrac{1}{2}\Big(\dfrac{-1}{12} + \dfrac{-1}{2}\Big) \\[1em] \Rightarrow y = \dfrac{1}{2}\Big(\dfrac{-1 - 6}{12}\Big) \\[1em] \Rightarrow y = \dfrac{1}{2}\Big(\dfrac{-7}{12}\Big) \\[1em] \Rightarrow y = -\dfrac{7}{24} \\[1em]

Let the third rational number be z.

⇒z=12(−112+13)⇒z=12(−1+412)⇒z=12(312)⇒z=18\Rightarrow z = \dfrac{1}{2}\Big(\dfrac{-1}{12} + \dfrac{1}{3}\Big) \\[1em] \Rightarrow z = \dfrac{1}{2}\Big(\dfrac{-1 + 4}{12}\Big) \\[1em] \Rightarrow z = \dfrac{1}{2}\Big(\dfrac{3}{12}\Big) \\[1em] \Rightarrow z = \dfrac{1}{8} \\[1em]

Hence, three rational numbers between −12\dfrac{-1}{2} and 13\dfrac{1}{3} are −724,−112 and 18\dfrac{-7}{24}, \dfrac{-1}{12} \text{ and } \dfrac{1}{8}.

(vi) Let the first rational number between −13-\dfrac{1}{3} and 14\dfrac{1}{4} be x.

⇒x=12(−13+14)⇒x=12(−4+312)⇒x=12(−112)⇒x=−124.\Rightarrow x = \dfrac{1}{2}\Big(-\dfrac{1}{3} + \dfrac{1}{4}\Big) \\[1em] \Rightarrow x = \dfrac{1}{2}\Big(\dfrac{-4 + 3}{12}\Big) \\[1em] \Rightarrow x = \dfrac{1}{2}\Big(\dfrac{-1}{12} \Big) \\[1em] \Rightarrow x = -\dfrac{1}{24}.

Let the second rational number be y.

⇒y=12(−124+−13)⇒y=12(−1−824)⇒y=12×−924⇒y=−948=−316.\Rightarrow y = \dfrac{1}{2} \Big(\dfrac{-1}{24} + \dfrac{-1}{3}\Big) \\[1em] \Rightarrow y = \dfrac{1}{2}\Big(\dfrac{-1 - 8}{24}\Big) \\[1em] \Rightarrow y = \dfrac{1}{2} \times \dfrac{-9}{24} \\[1em] \Rightarrow y = -\dfrac{9}{48} = -\dfrac{3}{16}.

Let the third rational number be z.

⇒z=12(−124+14)⇒z=12(−1+624)⇒z=12×524⇒z=548\Rightarrow z = \dfrac{1}{2}\Big(\dfrac{-1}{24} + \dfrac{1}{4}\Big) \\[1em] \Rightarrow z = \dfrac{1}{2}\Big(\dfrac{-1+6}{24} \Big) \\[1em] \Rightarrow z = \dfrac{1}{2} \times \dfrac{5}{24} \\[1em] \Rightarrow z = \dfrac{5}{48}

Hence, three rational numbers between −13-\dfrac{1}{3} and 14\dfrac{1}{4} are −316,−124 and 548-\dfrac{3}{16}, -\dfrac{1}{24} \text{ and } \dfrac{5}{48}.

Question 7

Find four rational numbers between 4 and 4.5.

Answer

Let a = 4, b = 4.5 and n = 4

Difference between consecutive rational numbers =

b−an+1=4.5−44+1=0.55=0.1\dfrac{b - a}{n + 1} = \dfrac{4.5 - 4}{4 + 1} = \dfrac{0.5}{5} = 0.1

Rational numbers between 4 and 4.5 are :

⇒ a + d, a + 2d, a + 3d, a + 4d

⇒ 4 + 0.1, 4 + 0.2, 4 + 0.3, 4 + 0.4

⇒ 4.1, 4.2, 4.3, 4.4

Hence, four rational numbers between 4 and 4.5 are 4.1, 4.2, 4.3 and 4.4.

Question 8

Find six rational numbers between 3 and 4.

Answer

Let a = 3, b = 4 and n = 9

Difference between consecutive rational numbers =

b−an+1=4−39+1=110=0.1\dfrac{b - a}{n + 1} = \dfrac{4 - 3}{9 + 1} = \dfrac{1}{10} = 0.1

Rational numbers between 3 and 4 are :

⇒ a + d, a + 2d, a + 3d, a + 4d, a + 5d, a + 6d

⇒ 3 + 0.1. 3 + 0.2, 3 + 0.3, 3 + 0.4, 3 + 0.5, 3 + 0.6

⇒ 3.1, 3.2, 3.3, 3.4, 3.5, 3.6

Hence, six rational numbers between 3 and 4 are 3.1, 3.2, 3.3, 3.4, 3.5, 3.6.

Exercise 1(C)

Question 1

Classify the rational and irrational numbers from the following :

(i) 5

(ii) 914\dfrac{9}{14}

(iii) 3\sqrt{3}

(iv) π

(v) 3.1416

(vi) 4\sqrt{4}

(vii) −5-\sqrt{5}

(viii) 83\sqrt[3]{8}

(ix) 33\sqrt[3]{3}

(x) 262\sqrt{6}

(xi) 0.36‾0.\overline{36}

(xii) 0.202202220...

(xiii) 23\dfrac{2}{\sqrt{3}}

(xiv) 227\dfrac{22}{7}

Answer

(i) 5 can be expressed in the form pq\dfrac{p}{q}, where p and q are integers and q ≠ 0.

Hence, 5 is a rational number.

(ii) 914\dfrac{9}{14} can be expressed in the form pq\dfrac{p}{q}, where p and q are integers and q ≠ 0.

Hence, 914\dfrac{9}{14} is a rational number.

(iii) 3\sqrt{3} is square root of non-perfect square i.e. 3.

Hence, 3\sqrt{3} is an irrational number.

(iv) π is a non-terminating and non-repeating decimal.

Hence, π is a irrational number.

(v) 3.1416 is a terminating decimal, so it can be expressed in the form of pq\dfrac{p}{q}, where p and q are integers and q ≠ 0.

Hence, 3.1416 is a rational number.

(vi) 4\sqrt{4} is square root of perfect square i.e. 4.

4=2=21\sqrt{4} = 2 = \dfrac{2}{1}.

Hence, 4\sqrt{4} is a rational number.

(vii) −5-\sqrt{5} is square root of non-perfect square.

Hence, −5-\sqrt{5} is an irrational number.

(viii) Given,

83=2=21\sqrt[3]{8} = 2 = \dfrac{2}{1}.

∴ 83\sqrt[3]{8} can be expressed in the form of pq\dfrac{p}{q}, where p and q are integers and q ≠ 0.

Hence, 83\sqrt[3]{8} is a rational number.

(ix) 33\sqrt[3]{3} is cube root of non-perfect cube.

Hence, 33\sqrt[3]{3} is an irrational number.

(x) 262\sqrt{6}

Here, 6\sqrt{6} is square root of a non-perfect square i.e. 6, thus it is an irrational number.

The product of a non-zero rational number and an irrational number is always an irrational number.

Hence, 262\sqrt{6} is an irrational number.

(xi) 0.36‾0.\overline{36} is a repeating decimal.

Thus, 0.36‾0.\overline{36} can be expressed as a fraction with an integer numerator and a non-zero integer denominator.

Hence, 0.36‾0.\overline{36} is a rational number.

(xii) 0.2022022220... is a non-terminating and non-repeating decimal.

Hence, 0.2022022220... is an irrational number.

(xiii) 23\dfrac{2}{\sqrt{3}}.

2 is rational number and 3\sqrt{3} is an irrational number.

Since, on dividing a rational number by irrational number the solution is always an irrational number.

Hence, 23\dfrac{2}{\sqrt{3}} is an irrational number.

(xiv) 227\dfrac{22}{7} can be expressed in the form of pq\dfrac{p}{q}, where p and q are integers and q ≠ 0.

Hence, 227\dfrac{22}{7} is a rational number.

Question 2

Separate the rationals and irrationals from among the following numbers :

(i) -8

(ii) 25\sqrt{25}

(iii) −35\dfrac{-3}{5}

(iv) 8\sqrt{8}

(v) 0

(vi) π

(vii) 53\sqrt[3]{5}

(viii) 2.4‾2.\overline{4}

(ix) −3-\sqrt{3}

Answer

(i) -8 can be expressed in the form of pq\dfrac{p}{q}, where p and q are integers and q ≠ 0.

Hence, -8 is a rational number.

(ii) 25=5=51\sqrt{25} = 5 = \dfrac{5}{1}

Thus, 25\sqrt{25} can be expressed in the form of pq\dfrac{p}{q}, where p and q are integers and q ≠ 0.

Hence, 25\sqrt{25} is a rational number.

(iii) −35\dfrac{-3}{5} can be expressed in the form of pq\dfrac{p}{q}, where p and q are integers and q ≠ 0.

Hence, −35\dfrac{-3}{5} is a rational number.

(iv) 8\sqrt{8} is square root of non-perfect square i.e. 8.

Hence, 8\sqrt{8} is an irrational number.

(v) 0 can be expressed in the form of pq\dfrac{p}{q}, where p and q are integers and q ≠ 0.

Hence, 0 is a rational number.

(vi) π is a non-terminating and non-repeating decimal.

Hence, π is an irrational number.

(vii) 53\sqrt[3]{5} is cube root of non-perfect cube i.e. 5.

Hence, 53\sqrt[3]{5} is an irrational number.

(viii) 2.4‾2.\overline{4} is a repeating decimal.

Hence, 2.4‾2.\overline{4} is a rational number.

(ix) −3-\sqrt{3} is square root of non-perfect square i.e. 3.

Hence, −3-\sqrt{3} is an irrational number.

Question 3

Represent each of the following on the real number line.

(i) 3\sqrt{3}

(ii) 5\sqrt{5}

(iii) 6\sqrt{6}

(iv) 10\sqrt{10}

Answer

(i) 3\sqrt{3} = 1.732..

(ii) 5\sqrt{5} = 2.236..

(iii) 6\sqrt{6} = 2.449..

(iv) 10\sqrt{10} = 3.162..

Represent each of the following on the real number line: Rational and Irrational Numbers, R.S. Aggarwal Mathematics Solutions ICSE Class 9.

Question 4

Write down the values of :

(i) (23)2(2\sqrt{3})^2

(ii) (322)2\Big(\dfrac{3}{2}\sqrt{2}\Big)^2

(iii) (5+3)2(5 + \sqrt{3})^2

(iv) (6−3)2(\sqrt{6} - 3)^2

(v) (3+25)2(3 + 2\sqrt{5})^2

(vi) (5+6)2(\sqrt{5} + \sqrt{6})^2

(vii) (322)2\Big(\dfrac{3}{2\sqrt{2}}\Big)^2

(viii) (5−63)2(5 - 6\sqrt{3})^2

Answer

(i) Solving,

⇒(23)2\Rightarrow (2\sqrt{3})^2

⇒23×23\Rightarrow 2\sqrt{3} \times 2\sqrt{3}

⇒ 4 × 3

⇒ 12.

Hence, (23)2(2\sqrt{3})^2 = 12.

(ii) Solving,

⇒(322)2⇒322×322⇒94×2⇒92.\Rightarrow \Big(\dfrac{3}{2}\sqrt{2}\Big)^2 \\[1em] \Rightarrow \dfrac{3}{2}\sqrt{2} \times \dfrac{3}{2}\sqrt{2} \\[1em] \Rightarrow \dfrac{9}{4} \times 2 \\[1em] \Rightarrow \dfrac{9}{2}.

Hence, (322)2=92\Big(\dfrac{3}{2}\sqrt{2}\Big)^2 = \dfrac{9}{2}.

(iii) Solving,

⇒(5+3)2⇒(5)2+(3)2+2×5×3⇒25+3+103⇒28+103\Rightarrow (5 + \sqrt{3})^2 \\[1em] \Rightarrow (5)^2 + (\sqrt{3})^2 + 2 \times 5 \times \sqrt{3} \\[1em] \Rightarrow 25 + 3 + 10\sqrt{3} \\[1em] \Rightarrow 28 + 10\sqrt{3}

Hence, (5+3)2=28+103(5 + \sqrt{3})^2 = 28 + 10\sqrt{3}.

(iv) Solving,

⇒(6−3)2⇒(6)2+(3)2−2×3×6⇒6+9−66⇒15−66\Rightarrow (\sqrt{6} - 3)^2 \\[1em] \Rightarrow (\sqrt{6})^2 + (3)^2 - 2 \times 3 \times \sqrt{6} \\[1em] \Rightarrow 6 + 9 - 6\sqrt{6} \\[1em] \Rightarrow 15 - 6\sqrt{6}

Hence, (6−3)2=15−66(\sqrt{6} - 3)^2 = 15 - 6\sqrt{6}.

(v) Solving,

⇒(3+25)2⇒(3)2+(25)2+2×3×25⇒9+4×5+125⇒9+20+125⇒29+125\Rightarrow (3 + 2\sqrt{5})^2 \\[1em] \Rightarrow (3)^2 + (2\sqrt{5})^2 + 2 \times 3 \times 2\sqrt{5} \\[1em] \Rightarrow 9 + 4 \times 5 + 12\sqrt{5} \\[1em] \Rightarrow 9 + 20 + 12\sqrt{5} \\[1em] \Rightarrow 29 + 12\sqrt{5}

Hence, (3+25)2=29+125(3 + 2\sqrt{5})^2 = 29 + 12\sqrt{5}.

(vi) Solving,

⇒(5+6)2⇒(5)2+(6)2+2×5×6⇒5+6+230⇒11+230\Rightarrow (\sqrt{5} + \sqrt{6})^2 \\[1em] \Rightarrow (\sqrt{5})^2 + (\sqrt{6})^2 + 2 \times \sqrt{5} \times \sqrt{6} \\[1em] \Rightarrow 5 + 6 + 2\sqrt{30} \\[1em] \Rightarrow 11 + 2\sqrt{30}

Hence, (5+6)2=11+230(\sqrt{5} + \sqrt{6})^2 = 11 + 2\sqrt{30}.

(vii) Solving,

⇒(322)2⇒322×322⇒94×2⇒98\Rightarrow \Big(\dfrac{3}{2\sqrt{2}}\Big)^2 \\[1em] \Rightarrow \dfrac{3}{2\sqrt{2}} \times \dfrac{3}{2\sqrt{2}} \\[1em] \Rightarrow \dfrac{9}{4 \times 2} \\[1em] \Rightarrow \dfrac{9}{8}

Hence, (322)2=98\Big(\dfrac{3}{2\sqrt{2}}\Big)^2 = \dfrac{9}{8}.

(viii) Solving,

⇒(5−63)2⇒(5)2+(63)2−2×5×63⇒25+108−603⇒133−603\Rightarrow (5 - 6\sqrt{3})^2 \\[1em] \Rightarrow (5)^2 + (6\sqrt{3})^2 - 2 \times 5 \times 6\sqrt{3} \\[1em] \Rightarrow 25 + 108 - 60\sqrt{3} \\[1em] \Rightarrow 133 - 60\sqrt{3}

Hence, (5−63)2=133+603(5 - 6\sqrt{3})^2 = 133 + 60\sqrt{3}.

Question 5

State, giving reason, wether the given number is rational or irrational:

(i) (3+5)(3 + \sqrt{5})

(ii) (−1+3)(-1 + \sqrt{3})

(iii) 565\sqrt{6}

(iv) −7-\sqrt{7}

(v) 64\dfrac{\sqrt{6}}{4}

(vi) 32\dfrac{3}{\sqrt{2}}

(vii) (3+3)(3−3)(3 + \sqrt{3}) (3 - \sqrt{3})

Answer

(i) Given,

(3+5)(3 + \sqrt{5})

3 is a rational number as it can be expressed in the form of pq\dfrac{p}{q}, where p and q are integers and q ≠ 0.

5\sqrt{5} is an irrational number as it is a square root of a non-perfect square i.e. 5.

The sum of a rational number and an irrational number is always irrational.

Hence, (3+5)(3 + \sqrt{5}) is a irrational number.

(ii) Given,

(−1+3)(-1 + \sqrt{3})

-1 is a rational number as it can be expressed in the form of pq\dfrac{p}{q}, where p and q are integers and q ≠ 0.

3\sqrt{3} is an irrational number as it is a square root of a non-perfect square i.e. 3.

The sum of a rational number and an irrational number is always irrational.

Hence, (−1+3)(-1 + \sqrt{3}) is a irrational number.

(iii) Given,

565\sqrt{6}

5 is a rational number as it can be expressed in the form of pq\dfrac{p}{q}, where p and q are integers and q ≠ 0.

6\sqrt{6} is an irrational number as it is a square root of a non-perfect square i.e. 6.

The product of a rational number and an irrational number is always irrational.

Hence, 565\sqrt{6} is an irrational number.

(iv) Given,

7\sqrt{7}, is an irrational number as it is a square root of a non-perfect square i.e. 7.

∴ −7-\sqrt{7} is an irrational number.

Hence, −7-\sqrt{7} is an irrational number.

(v) Given,

64\dfrac{\sqrt{6}}{4}

4 is a rational number as it can be expressed in the form of pq\dfrac{p}{q}, where p and q are integers and q ≠ 0.

6\sqrt{6} is an irrational number as it is a square root of a non-perfect square i.e. 6.

The division of a rational number and an irrational number is always irrational.

Hence, 64\dfrac{\sqrt{6}}{4} is an irrational number.

(vi) Given,

32\dfrac{3}{\sqrt{2}}

3 is a rational number as it can be expressed in the form of pq\dfrac{p}{q}, where p and q are integers and q ≠ 0.

2\sqrt{2} is an irrational number as it is a square root of a non-perfect square i.e. 2.

The division of a rational number and an irrational number is always irrational.

Hence, 32\dfrac{3}{\sqrt{2}} is an irrational number.

(vii) Given,

⇒(3+3)(3−3)⇒(3)2−(3)2⇒9−3⇒3\Rightarrow (3 + \sqrt{3}) (3 - \sqrt{3}) \\[1em] \Rightarrow (3)^2- (\sqrt{3})^2 \\[1em] \Rightarrow 9 - 3 \\[1em] \Rightarrow 3

3 is a rational number as it can be expressed in the form of pq\dfrac{p}{q}, where p and q are integers and q ≠ 0.

Hence, (3+3)(3−3)(3 + \sqrt{3})(3 - \sqrt{3}) is a rational number.

Question 6

Show that each of the following is irrational :

(i) (2+5)2\Big(2 + \sqrt{5}\Big)^2

(ii) (3−3)2\Big(3 - \sqrt{3}\Big)^2

(iii) (5+3)2\Big(\sqrt{5} + \sqrt{3}\Big)^2

(iv) 63\dfrac{6}{\sqrt{3}}

Answer

(i) Given,

⇒(2+5)2⇒(2)2+(5)2+2×2×5⇒4+5+45⇒9+45.\Rightarrow \Big(2 + \sqrt{5}\Big)^2 \\[1em] \Rightarrow (2)^2 + (\sqrt{5})^2 + 2 \times 2 \times \sqrt{5} \\[1em] \Rightarrow 4 + 5 + 4\sqrt{5} \\[1em] \Rightarrow 9 + 4\sqrt{5}.

9 is a rational number as it can be expressed in the form of pq\dfrac{p}{q}, where p and q are integers and q ≠ 0.

4 is a rational number as it can be expressed in the form of pq\dfrac{p}{q}, where p and q are integers and q ≠ 0.

5\sqrt{5} is an irrational number as it is a square root of a non-perfect square i.e. 5.

The product of a rational number and an irrational number is always irrational. i.e. 454\sqrt{5}

The sum of a rational number and an irrational number is always irrational. i.e. 9+459 + 4\sqrt{5}

Hence, (2+5)2\Big(2 + \sqrt{5}\Big)^2 is an irrational number.

(ii) Given,

⇒(3−3)2⇒(3)2+(3)2−2×3×3⇒9+3−63⇒12−63\Rightarrow \Big(3 - \sqrt{3}\Big)^2 \\[1em] \Rightarrow (3)^2 + (\sqrt{3})^2 - 2 \times 3 \times \sqrt{3} \\[1em] \Rightarrow 9 + 3 - 6 \sqrt{3} \\[1em] \Rightarrow 12 - 6\sqrt{3} \\[1em]

12 is a rational number as it can be expressed in the form of pq\dfrac{p}{q}, where p and q are integers and q ≠ 0.

6 is a rational number as it can be expressed in the form of pq\dfrac{p}{q}, where p and q are integers and q ≠ 0.

3\sqrt{3} is an irrational number as it is a square root of a non-perfect square i.e. 3.

The product of a rational number and an irrational number is always irrational. i.e. 636\sqrt{3}

The difference between a rational number and an irrational number is always irrational. i.e. 12−6312 - 6\sqrt{3}.

Hence, (3−3)2\Big(3 - \sqrt{3}\Big)^2 is an irrational number.

(iii) Given,

⇒(5+3)2⇒(5)2+(3)2+2×5×3⇒5+3+215⇒8+215\Rightarrow \Big(\sqrt{5} + \sqrt{3}\Big)^2 \\[1em] \Rightarrow (\sqrt{5})^2 + (\sqrt{3})^2 + 2 \times \sqrt{5} \times \sqrt{3} \\[1em] \Rightarrow 5 + 3 + 2\sqrt{15} \\[1em] \Rightarrow 8 + 2\sqrt{15} \\[1em]

8 is a rational number as it can be expressed in the form of pq\dfrac{p}{q}, where p and q are integers and q ≠ 0.

2 is a rational number as it can be expressed in the form of pq\dfrac{p}{q}, where p and q are integers and q ≠ 0.

15\sqrt{15} is an irrational number as it is a square root of a non-perfect square i.e. 15.

The product of a rational number and an irrational number is always irrational. i.e. 2152\sqrt{15}

The sum of a rational number and an irrational number is always irrational. i.e. 8+2158 + 2\sqrt{15}

Hence, (5+3)2\Big(\sqrt{5} + \sqrt{3}\Big)^2 is an irrational number.

(iv) Given,

63\dfrac{6}{\sqrt{3}}

6 is a rational number as it can be expressed in the form of pq\dfrac{p}{q}, where p and q are integers and q ≠ 0.

3\sqrt{3} is an irrational number as it is a square root of a non-perfect square i.e. 3.

The division of a rational number and an irrational number is always irrational. i.e. 63\dfrac{6}{\sqrt{3}}

Hence, 63\dfrac{6}{\sqrt{3}} is an irrational number.

Question 7

Prove that 5\sqrt{5} is irrational number.

Answer

Let 5\sqrt{5} be rational.

Thus, 5\sqrt{5} can be expressed in the form of pq\dfrac{p}{q}.

⇒5=pq⇒5q=pSquaring both sides, we get : ⇒(5q)2=p2⇒5q2=p2 .......(1)\Rightarrow \sqrt{5} = \dfrac{p}{q} \\[1em] \Rightarrow \sqrt{5}q = p \\[1em] \text{Squaring both sides, we get : } \\[1em] \Rightarrow (\sqrt{5}q)^2 = p^2 \\[1em] \Rightarrow 5q^2 = p^2 \text{ .......(1)}

As 5 divides 5q2, so 5 divides p2 but 5 is prime,

Thus, 5 divides p.

Let p = 5m for some positive integer m.

Then, p = 5m

Substituting this value of p in (1), we get :

⇒5q2=(5m)2⇒5q2=25m2⇒q2=5m2\Rightarrow 5q^2 = (5m)^2 \\[1em] \Rightarrow 5q^2 = 25m^2 \\[1em] \Rightarrow q^2 = 5m^2

As 5 divides 5m2, so 5 divides q2 but 5 is prime.

Thus, 5 divides q.

This shows that 5 is a common factor of p and q. This contradicts the hypothesis that p and q have no common factor, other than 1.

∴5\therefore \sqrt{5} is not a rational number.

Hence, proved that 5\sqrt{5} is a irrational number.

Question 8

Write down the examples of 4 distinct irrational numbers.

Answer

We know that,

Square root of non-perfect squares are irrational numbers.

Since, 2, 3, 5 and 6 are non-perfect squares.

∴ 2,3,5,6\sqrt{2}, \sqrt{3}, \sqrt{5}, \sqrt{6} are irrational numbers.

Hence, 2,3,5,6\sqrt{2}, \sqrt{3}, \sqrt{5}, \sqrt{6} are 4 distinct irrational numbers.

Question 9

Prove that (3+7)\Big(\sqrt{3} + \sqrt{7}\Big) is irrational.

Answer

Let us assume 3+7\sqrt{3} + \sqrt{7} is a rational number.

Let, 3+7=x\sqrt{3} + \sqrt{7} = x

Squaring both sides, we get :

⇒(3+7)2=x2⇒(3)2+(7)2+2×3×7=x2⇒3+7+221=x2⇒10+221=x2⇒21=x2−102.\Rightarrow (\sqrt{3} + \sqrt{7})^2 = x^2 \\[1em] \Rightarrow (\sqrt{3})^2 + (\sqrt{7})^2 + 2 \times \sqrt{3} \times \sqrt{7} = x^2 \\[1em] \Rightarrow 3 + 7 + 2\sqrt{21} = x^2 \\[1em] \Rightarrow 10 + 2\sqrt{21} = x^2 \\[1em] \Rightarrow \sqrt{21} = \dfrac{x^2 -10}{2}.

Here, x is rational,

∴ x2 is rational .........(1)

⇒ x2 - 10 is rational (As difference between two rational numbers is always rational)

⇒ x2−102\dfrac{x^2 - 10}{2} is rational (Dividing two rational numbers results in a rational number)

But, 21\sqrt{21} is irrational.

∴x2−102\therefore \dfrac{x^2 - 10}{2} is irrational.

Thus, x2 - 10 is irrational and so x2 is irrational ........(2)

(1) and (2) do not match with each other.

∴ We arrive at a contradiction.

So, our assumption that 3+7\sqrt{3} + \sqrt{7} is a rational number is wrong.

∴ 3+7\sqrt{3} + \sqrt{7} is irrational.

Hence, proved that 3+7\sqrt{3} + \sqrt{7} is an irrational number.

Question 10

Prove that (2+3)(\sqrt{2} + \sqrt{3}) is irrational.

Answer

Let us assume 2+3\sqrt{2} + \sqrt{3} is a rational number.

Let, (2+3)=x(\sqrt{2} + \sqrt{3}) = x

Squaring on both sides, we get :

⇒(2+3)2=x2⇒(2)2+(3)2+2×2×3=x2⇒2+3+26=x2⇒5+26=x2⇒26=x2−5⇒6=x2−52.\Rightarrow (\sqrt{2}+\sqrt{3})^2 = x^2 \\[1em] \Rightarrow (\sqrt{2})^2 + (\sqrt{3})^2 + 2 \times \sqrt{2} \times \sqrt{3} = x^2 \\[1em] \Rightarrow 2 + 3 + 2\sqrt{6} = x^2 \\[1em] \Rightarrow 5 + 2\sqrt{6} = x^2 \\[1em] \Rightarrow 2\sqrt{6} = x^2 - 5 \\[1em] \Rightarrow \sqrt{6} = \dfrac{x^2 - 5}{2}.

Here, x is rational,

∴ x2 is rational .........(1)

⇒ x2 - 5 is rational (Difference between two rational numbers is always rational)

So, x2−52\dfrac{x^2 - 5}{2} is rational (Dividing two rational numbers results in a rational number)

But 6\sqrt{6} is irrational,

∴x2−52\therefore \dfrac{x^2 - 5}{2} is irrational

Thus, x2 - 5 is irrational and so x2 is irrational ........(2)

(1) and (2) do not match with each other.

∴ We arrive at a contradiction.

So, our assumption that 2+3\sqrt{2} + \sqrt{3} is a rational number is wrong.

∴ 2+3\sqrt{2} + \sqrt{3} is irrational.

Hence, proved that 2+3\sqrt{2} + \sqrt{3} is an irrational number.

Question 11

Write two irrational numbers between 14 and 19\sqrt{14} \text{ and } \sqrt{19}

Answer

We want two irrational numbers between 14 and 19\sqrt{14} \text{ and } \sqrt{19}.

Consider any two numbers between 14 and 19 such that they are not perfect squares.

Let us take 15 and 17 as they are not perfect squares.

We know that square root of a non-perfect square is an irrational number.

∴ 15 and 17\sqrt{15} \text{ and } \sqrt{17} are irrational numbers.

Thus, we have :

⇒14<15<17<19\Rightarrow \sqrt{14} \lt \sqrt{15} \lt \sqrt{17} \lt \sqrt{19}

Hence, two irrational numbers between

14 and 19 are 15 and 17\sqrt{14} \text{ and } \sqrt{19} \text{ are } \sqrt{15} \text{ and } \sqrt{17}.

Question 12

Write three irrational numbers between 2 and 7\sqrt{2} \text{ and } \sqrt{7}

Answer

We want three irrational numbers between 2 and 7\sqrt{2} \text{ and } \sqrt{7}.

Consider any three numbers between 2 and 7 such that they are not perfect squares.

Let us take 3, 5 and 6 as they are not perfect squares.

We know that square root of a non-perfect square is an irrational number.

∴ 3,5 and 6\sqrt{3}, \sqrt{5} \text{ and } \sqrt{6} are irrational numbers.

Thus, we have :

⇒2<3<5<6<7\Rightarrow \sqrt{2} \lt \sqrt{3} \lt \sqrt{5} \lt \sqrt{6} \lt\sqrt{7}

Hence, three irrational numbers between

2 and 7 are 3,5 and 6\sqrt{2} \text{ and } \sqrt{7} \text{ are } \sqrt{3}, \sqrt{5} \text{ and } \sqrt{6}.

Question 13

State in each case, whether true or false :

(i) The sum of two rationals is a rational.

(ii) The sum of two irrationals is an irrational.

(iii) The product of two rationals is a rational.

(iv) The product of two irrationals is an irrational.

(v) The sum of a rational and an irrational is an irrational.

(vi) The product of a rational and an irrational is a rational.

Answer

(i) Adding two rational numbers will always result in rational number.

Hence, above statement is true.

(ii) Adding two irrational numbers can be rational as well as irrational.

Hence, above statement is false.

(iii) Multiplying two rational numbers will always result in rational number.

Hence, above statement is true.

(iv) Multiplying two irrational numbers can be rational as well as irrational.

Hence, above statement is false.

(v) Adding a rational and an irrational will always result irrational number.

Hence, above statement is true.

(iv) Multiplying a rational and an irrational can be rational as well as irrational.

Hence, above statement is false.

Question 14

What are rational numbers? Give ten examples.

Answer

The numbers of the form pq\dfrac{p}{q}, where p and q are integers and q is not equal to zero, are called rational numbers.

Examples:

(i) 23\dfrac{2}{3}

(ii) −58-\dfrac{5}{8}

(iii) 7

(iv) 0

(v) -9

(vi) 112\dfrac{11}{2}

(vii) -1.25

(viii) 3.75

(ix) 10000

(x) −167\dfrac{-16}{7}

Question 15

What are irrational numbers? Give ten examples.

Answer

A number which when expressed in decimal form is expressible as a non-terminating and non-repeating decimal, is called an irrational number.

Examples:

  1. π

  2. 2\sqrt{2}

  3. 3\sqrt{3}

  4. 5\sqrt{5}

  5. 6\sqrt{6}

  6. 7\sqrt{7}

  7. 8\sqrt{8}

  8. 10\sqrt{10}

  9. 11\sqrt{11}

  10. 12\sqrt{12}

Exercise 1(D)

Question 1

Rationalize the denominator:

26\dfrac{2}{\sqrt{6}}

Answer

(i) Rationalizing the denominator,

⇒26×66⇒2(6)(6)2⇒266⇒63\Rightarrow \dfrac{2}{\sqrt{6}} \times \dfrac{ \sqrt{6}}{\sqrt{6}} \\[1em] \Rightarrow \dfrac{2(\sqrt{6})}{(\sqrt{6})^2} \\[1em] \Rightarrow \dfrac{2\sqrt{6}}{6} \\[1em] \Rightarrow \dfrac{\sqrt{6}}{3}

Hence, on rationalizing 26=63\dfrac{2}{\sqrt{6}} = \dfrac{\sqrt{6}}{3}.

Question 2

Rationalize the denominator:

223\dfrac{\sqrt{2}}{2\sqrt{3}}

Answer

Rationalizing the denominator, ⇒223×2323⇒2(6)(23)2⇒26(4×3)⇒2612⇒66.\Rightarrow \dfrac{\sqrt{2}}{2\sqrt{3}} \times \dfrac{2\sqrt{3}}{2\sqrt{3}} \\[1em] \Rightarrow \dfrac{2(\sqrt{6})}{(2\sqrt{3})^2} \\[1em] \Rightarrow \dfrac{2\sqrt{6}}{(4 \times 3)} \\[1em] \Rightarrow \dfrac{2\sqrt{6}}{12} \\[1em] \Rightarrow \dfrac{\sqrt{6}}{6}.

Hence, on rationalizing = 223=66\dfrac{\sqrt{2}}{2\sqrt{3}} = \dfrac{\sqrt{6}}{6}.

Question 3

Rationalize the denominator:

1(3+5)\dfrac{1}{(3 + \sqrt{5})}.

Answer

Rationalizing the denominator,

⇒1(3+5)×(3−5)(3−5)⇒(3−5)(3)2−(5)2⇒(3−5)9−5⇒(3−5)4\Rightarrow \dfrac{1}{(3 + \sqrt{5})} \times \dfrac{(3 - \sqrt{5})} {(3 - \sqrt{5})} \\[1em] \Rightarrow \dfrac{(3 - \sqrt{5})} {(3)^2 - (\sqrt{5})^2} \\[1em] \Rightarrow \dfrac{(3 - \sqrt{5})} {9 - 5} \\[1em] \Rightarrow \dfrac{(3 - \sqrt{5})} {4} \\[1em]

Hence, on rationalizing 1(3+5)=(3−5)4\dfrac{1}{(3 + \sqrt{5})} = \dfrac{(3 - \sqrt{5})} {4}.

Question 4

Rationalize the denominator:

1(3−1)\dfrac{1}{(\sqrt{3} - 1)}

Answer

Rationalizing on denominator,

⇒1(3−1)×(3+1)(3+1)⇒(3+1)(3)2−(1)2⇒(3+1)3−1⇒(3+1)2\Rightarrow \dfrac{1}{(\sqrt{3} - 1)} \times \dfrac{(\sqrt{3} + 1)}{(\sqrt{3} + 1)} \\[1em] \Rightarrow \dfrac{(\sqrt{3} + 1)}{(\sqrt{3})^2 - (1)^2} \\[1em] \Rightarrow \dfrac{(\sqrt{3} + 1)}{3 - 1} \\[1em] \Rightarrow \dfrac{(\sqrt{3} + 1)}{2} \\[1em]

Hence, on rationalizing 1(3−1)=(3+1)2\dfrac{1}{(\sqrt{3} - 1)} = \dfrac{(\sqrt{3} + 1)}{2}.

Question 5

Rationalize the denominator:

1(4+23)\dfrac{1}{(4 + 2\sqrt{3})}

Answer

Rationalizing the denominator,

⇒1(4+23)×(4−23)(4−23)⇒(4−23)(4)2−(23)2⇒(4−23)16−12⇒2(2−3)4⇒(2−3)2\Rightarrow \dfrac{1}{(4 + 2\sqrt{3})} \times \dfrac{(4 - 2\sqrt{3})} {(4 - 2\sqrt{3})} \\[1em] \Rightarrow \dfrac{(4 - 2\sqrt{3})} {(4)^2 - (2\sqrt{3})^2} \\[1em] \Rightarrow \dfrac{(4 - 2\sqrt{3})} {16 - 12} \\[1em] \Rightarrow \dfrac{2(2 - \sqrt{3})} {4} \\[1em] \Rightarrow \dfrac{(2 - \sqrt{3})} {2} \\[1em]

Hence, on rationalizing 1(4+23)=(2−3)2\dfrac{1}{(4 + 2\sqrt{3})} = \dfrac{(2 - \sqrt{3})} {2}.

Question 6

Rationalize the denominator:

1(6−3)\dfrac{1}{(\sqrt{6} - \sqrt{3})}

Answer

Rationalizing the denominator,

⇒1(6−3)×(6+3)(6+3)⇒(6+3)(6)2−(3)2⇒(6+3)6−3⇒(6+3)3\Rightarrow \dfrac{1}{(\sqrt{6} - \sqrt{3})} \times \dfrac{(\sqrt{6} + \sqrt{3})}{(\sqrt{6} + \sqrt{3})} \\[1em] \Rightarrow \dfrac{(\sqrt{6} + \sqrt{3})}{(\sqrt{6})^2 - (\sqrt{3})^2} \\[1em] \Rightarrow \dfrac{(\sqrt{6} + \sqrt{3})}{6 - 3} \\[1em] \Rightarrow \dfrac{(\sqrt{6} + \sqrt{3})}{3} \\[1em]

Hence, on rationalizing 1(6−3)=(6+3)3\dfrac{1}{(\sqrt{6} - \sqrt{3})} = \dfrac{(\sqrt{6} + \sqrt{3})}{3}.

Question 7

Rationalize the denominator:

3−13+1\dfrac{\sqrt{3} - 1}{\sqrt{3} + 1}.

Answer

Rationalizing the denominator,

⇒3−13+1×3−13−1⇒(3−1)2(3)2−(1)2⇒(3)2+(1)2−2×3×13−1⇒3+1−232⇒(4−23)2⇒2(2−3)2⇒(2−3)\Rightarrow \dfrac{\sqrt{3} - 1}{\sqrt{3} + 1} \times \dfrac{\sqrt{3} - 1}{\sqrt{3} - 1} \\[1em] \Rightarrow \dfrac{(\sqrt{3} - 1)^2}{(\sqrt{3})^2 - (1)^2} \\[1em] \Rightarrow \dfrac{(\sqrt{3})^2 + (1)^2 - 2 \times \sqrt{3} \times 1 }{3 - 1} \\[1em] \Rightarrow \dfrac{3 + 1 - 2\sqrt{3}}{2} \\[1em] \Rightarrow \dfrac{(4 - 2\sqrt{3})}{2} \\[1em] \Rightarrow \dfrac{2(2 - \sqrt{3})}{2} \\[1em] \Rightarrow (2 - \sqrt{3})

Hence, on rationalizing = 3−13+1=(2−3)\dfrac{\sqrt{3} - 1}{\sqrt{3} + 1} = (2 - \sqrt{3}).

Question 8

Rationalize the denominator:

3−223+22\dfrac{3 - 2\sqrt{2}}{3 + 2\sqrt{2}}

Answer

Rationalizing the denominator,

⇒3−223+22×3−223−22⇒(3−22)2(3)2−(22)2⇒(3)2+(22)2−2×3×229−8⇒(9+8−122)1⇒(17−122)\Rightarrow \dfrac{3 - 2\sqrt{2}}{3 + 2\sqrt{2}} \times \dfrac{3 - 2\sqrt{2}}{3 - 2\sqrt{2}} \\[1em] \Rightarrow \dfrac{(3 - 2\sqrt{2})^2} {(3)^2 - ( 2\sqrt{2})^2} \\[1em] \Rightarrow \dfrac{(3)^2 + ( 2\sqrt{2})^2 -2 \times 3 \times 2\sqrt{2} } {9-8} \\[1em] \Rightarrow \dfrac{(9 + 8 - 12\sqrt{2})} {1} \\[1em] \Rightarrow (17 - 12\sqrt{2}) \\[1em]

Hence, on rationalizing 3−223+22=(17−122)\dfrac{3 - 2\sqrt{2}}{3 + 2\sqrt{2}} = (17 - 12\sqrt{2}).

Question 9

Rationalize the denominator:

1(25−3)\dfrac{1}{(2\sqrt{5} - \sqrt{3})}.

Answer

Rationalizing the denominator,

⇒1(25−3)×(25+3)(25+3)⇒(25+3)(25)2−(3)2⇒(25+3)20−3⇒(25+3)17\Rightarrow \dfrac{1}{(2\sqrt{5} - \sqrt{3})} \times \dfrac{(2\sqrt{5} + \sqrt{3})} {(2\sqrt{5} + \sqrt{3})} \\[1em] \Rightarrow \dfrac{(2\sqrt{5} + \sqrt{3})} {(2\sqrt{5})^2 - (\sqrt{3})^2} \\[1em] \Rightarrow \dfrac{(2\sqrt{5} + \sqrt{3})} {20 - 3} \\[1em] \Rightarrow \dfrac{(2\sqrt{5} + \sqrt{3})} {17} \\[1em]

Hence, on rationalizing 1(25−3)=(25+3)17\dfrac{1}{(2\sqrt{5} - \sqrt{3})} = \dfrac{(2\sqrt{5} + \sqrt{3})}{17}.

Question 10

Rationalize the denominator:

1(1+5+3)\dfrac{1}{(1 + \sqrt{5} + \sqrt{3})}.

Answer

Rationalizing the denominator,

⇒1(1+5+3)×(1+5−3)(1+5−3)⇒(1+5−3)(1+5+3)×(1+5−3)⇒(1+5−3)(1+5)2−(3)2⇒(1+5−3)1+5+25−3⇒(1+5−3)(3+25)\Rightarrow \dfrac{1}{(1 + \sqrt{5} + \sqrt{3})} \times \dfrac{(1 + \sqrt{5} - \sqrt{3})} {(1 + \sqrt{5} - \sqrt{3})} \\[1em] \Rightarrow \dfrac{(1 + \sqrt{5} - \sqrt{3})}{(1 + \sqrt{5} + \sqrt{3}) \times (1 + \sqrt{5} - \sqrt{3})} \\[1em] \Rightarrow \dfrac{(1 + \sqrt{5} - \sqrt{3})}{(1 + \sqrt{5})^2 - (\sqrt{3})^2} \\[1em] \Rightarrow \dfrac{(1 + \sqrt{5} - \sqrt{3})}{1 + 5 + 2\sqrt{5} - 3} \\[1em] \Rightarrow \dfrac{(1 + \sqrt{5} - \sqrt{3})}{(3 + 2\sqrt{5})}

Rationalizing the denominator again,

⇒(1+5−3)(3+25)×(3−25)(3−25)⇒3−25+35−10−33+215(3)2−(25)2⇒−7+5−33+2159−20⇒−(7−5+33−215)−11⇒(7−5+33−215)11\Rightarrow \dfrac{(1 + \sqrt{5} - \sqrt{3})}{(3 + 2\sqrt{5})} \times \dfrac {(3 - 2\sqrt{5})}{(3 - 2\sqrt{5})} \\[1em] \Rightarrow \dfrac{3 - 2\sqrt{5} + 3\sqrt{5} - 10 - 3\sqrt{3} + 2\sqrt{15}}{(3)^2 - (2\sqrt{5})^2} \\[1em] \Rightarrow \dfrac{-7 + \sqrt{5} - 3\sqrt{3} + 2\sqrt{15}}{9 - 20} \\[1em] \Rightarrow \dfrac{-(7 - \sqrt{5} + 3\sqrt{3} - 2\sqrt{15})}{-11} \\[1em] \Rightarrow \dfrac{(7 - \sqrt{5} + 3\sqrt{3} - 2\sqrt{15})}{11}

Hence, on rationalizing 1(1+5+3)=(7−5+33−215)11\dfrac{1}{(1 + \sqrt{5} + \sqrt{3})} = \dfrac{(7 - \sqrt{5} + 3\sqrt{3} - 2\sqrt{15})} {11}.

Question 11

Rationalize the denominator:

2(2+3−5)\dfrac{\sqrt{2}}{(\sqrt{2} + \sqrt{3} - \sqrt{5})}.

Answer

Rationalizing the denomaintor,

⇒2(2+3−5)×(2+3+5)(2+3+5)⇒2×(2+3+5)(2+3−5)×(2+3+5)⇒(2+6+10)(2+6+10+6+3+15−10−15−5)⇒(2+6+10)(26)\Rightarrow \dfrac{\sqrt{2}}{(\sqrt{2} + \sqrt{3} - \sqrt{5})} \times \dfrac{(\sqrt{2} + \sqrt{3} + \sqrt{5})}{(\sqrt{2} + \sqrt{3} + \sqrt{5})} \\[1em] \Rightarrow \dfrac{ \sqrt{2} \times (\sqrt{2}+ \sqrt{3} + \sqrt{5})} {(\sqrt{2}+ \sqrt{3} - \sqrt{5}) \times (\sqrt{2} + \sqrt{3} + \sqrt{5})} \\[1em] \Rightarrow \dfrac{(2 + \sqrt{6} + \sqrt{10})} {(2 + \sqrt{6} + \sqrt{10} + \sqrt{6} + 3 + \sqrt{15} - \sqrt{10} - \sqrt{15} - 5)} \\[1em] \Rightarrow \dfrac{(2 + \sqrt{6} + \sqrt{10})} {(2\sqrt{6})}

Rationalizing again,

⇒(2+6+10)(26)×66⇒(2+6+10)×6(26)×6⇒(26+6+60)12⇒(26+6+15×4)12⇒26+6+21512⇒2(6+3+15)12⇒(6+3+15)6.\Rightarrow \dfrac{(2 + \sqrt{6} + \sqrt{10})} {(2\sqrt{6})} \times \dfrac{\sqrt{6}}{\sqrt{6}} \\[1em] \Rightarrow \dfrac{(2 + \sqrt{6} + \sqrt{10}) \times \sqrt{6}} {(2\sqrt{6})\times \sqrt{6}} \\[1em] \Rightarrow \dfrac{(2\sqrt{6} + 6 + \sqrt{60})} {12} \\[1em] \Rightarrow \dfrac{(2\sqrt{6} + 6 + \sqrt{15 \times 4})}{12} \\[1em] \Rightarrow \dfrac{2\sqrt{6} + 6 + 2\sqrt{15}}{12} \\[1em] \Rightarrow \dfrac{2(\sqrt{6} + 3 + \sqrt{15})}{12} \\[1em] \Rightarrow \dfrac{(\sqrt{6} + 3 + \sqrt{15})}{6}.

Hence, on rationalizing (6+3+15)6=(6+3+15)6\dfrac{(\sqrt{6} + 3 + \sqrt{15})}{6} = \dfrac{(\sqrt{6} + 3 + \sqrt{15})}{6}.

Question 12

If 3+13−1=a+b3\dfrac{\sqrt{3} + 1}{\sqrt{3} - 1} = a + b\sqrt{3} ,find the values of 'a' and 'b'.

Answer

Given,

Equation : 3+13−1=a+b3\dfrac{\sqrt{3} + 1}{\sqrt{3} - 1} = a + b\sqrt{3}

Rationalizing the denominator of L.H.S. of the above equation :

⇒3+13−1×3+13+1⇒(3+1)2(3)2−12⇒(3)2+12+2×3×13−1⇒3+1+232⇒4+232⇒2(2+3)2⇒2+3.\Rightarrow \dfrac{\sqrt{3} + 1}{\sqrt{3} - 1} \times \dfrac{\sqrt{3} + 1}{\sqrt{3} + 1} \\[1em] \Rightarrow \dfrac{(\sqrt{3} + 1)^2}{(\sqrt{3})^2 - 1^2} \\[1em] \Rightarrow \dfrac{(\sqrt{3})^2 + 1^2 + 2 \times \sqrt{3} \times 1}{3 - 1} \\[1em] \Rightarrow \dfrac{3 + 1 + 2\sqrt{3}}{2} \\[1em] \Rightarrow \dfrac{4 + 2\sqrt{3}}{2} \\[1em] \Rightarrow \dfrac{2(2 + \sqrt{3})}{2} \\[1em] \Rightarrow 2 + \sqrt{3}.

Comparing 2+3 with a+b32 + \sqrt{3} \text{ with } a + b\sqrt{3}, we get :

a = 2 and b = 1.

Hence, a = 2 and b = 1.

Question 13

If 3+23−2=a+b2\dfrac{3 + \sqrt{2}}{3 - \sqrt{2}} = a + b\sqrt{2}, find the values of 'a' and 'b'.

Answer

Given,

Equation : 3+23−2=a+b2\dfrac{3 + \sqrt{2}}{3 - \sqrt{2}} = a + b\sqrt{2}

Rationalizing L.H.S. of the above equation :

⇒3+23−2×3+23+2⇒(3+2)2(3)2−(2)2⇒(3)2+(2)2+2×3×29−2⇒9+2+627⇒11+627⇒117+627\Rightarrow \dfrac{3 + \sqrt{2}}{3 - \sqrt{2}} \times \dfrac{3 + \sqrt{2}}{3 + \sqrt{2}} \\[1em] \Rightarrow \dfrac{(3 + \sqrt{2})^2}{(3)^2 - (\sqrt{2})^2} \\[1em] \Rightarrow \dfrac{(3)^2 + (\sqrt{2})^2 + 2 \times 3 \times \sqrt{2}}{9 - 2} \\[1em] \Rightarrow \dfrac{9 + 2 + 6\sqrt{2}}{7} \\[1em] \Rightarrow \dfrac{11 + 6\sqrt{2}}{7} \\[1em] \Rightarrow \dfrac{11}{7} + \dfrac{6\sqrt{2}}{7}

Comparing 117+672 with a+b2\dfrac{11}{7} + \dfrac{6}{7}\sqrt{2} \text{ with } a + b\sqrt{2}, we get :

a=117 and b=67.a = \dfrac{11}{7} \text{ and } b = \dfrac{6}{7}.

Hence, a=117 and b=67a = \dfrac{11}{7} \text{ and } b = \dfrac{6}{7}.

Question 14

If 5−65+6=a−b6\dfrac{5 - \sqrt{6}}{5 + \sqrt{6}} = a - b\sqrt{6}, find the values of 'a' and 'b'.

Answer

Given,

Equation : 5−65+6=a−b6\dfrac{5 - \sqrt{6}}{5 + \sqrt{6}} = a - b\sqrt{6}

Rationalizing the denomiantor of L.H.S. of the above equation :

⇒5−65+6×5−65−6⇒(5−6)2(5)2−(6)2⇒(5)2+(6)2−2×5×625−6⇒25+6−10619⇒31−10619⇒3119−10619\Rightarrow \dfrac{5 - \sqrt{6}}{5 + \sqrt{6}} \times \dfrac{5 - \sqrt{6}}{5 - \sqrt{6}} \\[1em] \Rightarrow \dfrac{(5 - \sqrt{6})^2}{(5)^2 - (\sqrt{6})^2} \\[1em] \Rightarrow \dfrac{(5)^2 + (\sqrt{6})^2 - 2 \times 5 \times \sqrt{6}}{25 - 6} \\[1em] \Rightarrow \dfrac{25 + 6 - 10\sqrt{6}}{19} \\[1em] \Rightarrow \dfrac{31 - 10\sqrt{6}}{19} \\[1em] \Rightarrow \dfrac{31}{19} - \dfrac{10\sqrt{6}}{19}

Comparing 3119−10196 with a−b6\dfrac{31}{19} - \dfrac{10}{19}\sqrt{6} \text{ with } a - b\sqrt{6}, we get :

a=3119 and b=1019.a = \dfrac{31}{19} \text{ and } b = \dfrac{10}{19}.

Hence, a=3119 and b=1019a = \dfrac{31}{19} \text{ and } b = \dfrac{10}{19}.

Question 15

If 5+237+43=a−b3\dfrac{5 + 2\sqrt{3}}{7 + 4\sqrt{3}} = a - b\sqrt{3}, find the values of 'a' and 'b'.

Answer

Given,

Equation : 5+237+43=a−b3\dfrac{5 + 2\sqrt{3}}{7 + 4\sqrt{3}} = a - b\sqrt{3}

Rationalizing the denominator of L.H.S. of the above equation :

⇒5+237+43×7−437−43⇒(5+23)×(7−43)(7)2−(43)2⇒35−203+143−8×349−48⇒35−63−241⇒11−63\Rightarrow \dfrac{5 + 2\sqrt{3}}{7 + 4\sqrt{3}} \times \dfrac{7 - 4\sqrt{3}}{7 - 4\sqrt{3}} \\[1em] \Rightarrow \dfrac{(5 + 2\sqrt{3}) \times (7 - 4\sqrt{3})}{(7)^2 - (4\sqrt{3})^2} \\[1em] \Rightarrow \dfrac{35 - 20\sqrt{3} + 14\sqrt{3} - 8 \times 3}{49 - 48} \\[1em] \Rightarrow \dfrac{35 - 6\sqrt{3} - 24}{1} \\[1em] \Rightarrow 11 - 6\sqrt{3} \\[1em]

Comparing, 11−63 with a−b311 - 6\sqrt{3} \text{ with } a - b\sqrt{3}, we get :

a = 11 and b = 6.

Hence, a = 11 and b = 6.

Question 16

Simplify : 5+35−3+5−35+3\dfrac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}} + \dfrac{\sqrt{5} - \sqrt{3}}{\sqrt{5} + \sqrt{3}}

Answer

Given,

Equation : 5+35−3+5−35+3\dfrac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}} + \dfrac{\sqrt{5} - \sqrt{3}}{\sqrt{5} + \sqrt{3}}

Simplifying the above equation :

⇒(5+3)2+(5−3)2(5−3)(5+3)⇒(5)2+(3)2+2×5×3+(5)2+(3)2−2×5×3(5)2−(3)2⇒5+3+215+5+3−2155−3⇒162⇒8\Rightarrow \dfrac{(\sqrt{5} + \sqrt{3})^2 + (\sqrt{5} - \sqrt{3})^2}{(\sqrt{5} - \sqrt{3})(\sqrt{5} + \sqrt{3})} \\[1em] \Rightarrow \dfrac{(\sqrt{5})^2 + (\sqrt{3})^2 + 2 \times \sqrt{5} \times \sqrt{3} + (\sqrt{5})^2 + (\sqrt{3})^2 - 2 \times \sqrt{5} \times \sqrt{3}}{(\sqrt{5})^2 - (\sqrt{3})^2} \\[1em] \Rightarrow \dfrac{5 + 3 + 2\sqrt{15} + 5 + 3 - 2\sqrt{15}}{5 - 3} \\[1em] \Rightarrow \dfrac{16}{2} \\[1em] \Rightarrow 8

Hence, 5+35−3+5−35+3\dfrac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}}+\dfrac{\sqrt{5} - \sqrt{3}}{\sqrt{5} + \sqrt{3}} = 8.

Question 17

Simplify : 7+353+5−7−353−5\dfrac{7 + 3\sqrt{5}}{3 + \sqrt{5}} - \dfrac{7 - 3\sqrt{5}}{3 - \sqrt{5}}

Answer

Given,

Equation : 7+353+5−7−353−5\dfrac{7 + 3\sqrt{5}}{3 + \sqrt{5}} - \dfrac{7 - 3\sqrt{5}}{3 - \sqrt{5}}

Simplifying the above equation :

⇒(7+35)×(3−5)−(7−35)×(3+5)(3+5)×(3−5)⇒(21−75+95−15)−(21+75−95−15)32−(5)2⇒21−75+95−15−21−75+95+159−5⇒454⇒5\Rightarrow \dfrac{(7 + 3\sqrt{5}) \times (3 - \sqrt{5}) - (7 - 3\sqrt{5}) \times (3 + \sqrt{5})}{(3 + \sqrt{5}) \times (3 - \sqrt{5})} \\[1em] \Rightarrow \dfrac{(21 - 7\sqrt{5} + 9\sqrt{5} - 15)-(21 + 7\sqrt{5} - 9\sqrt{5} - 15)}{3^2 - (\sqrt{5})^2} \\[1em] \Rightarrow \dfrac{21 - 7\sqrt{5} + 9\sqrt{5} - 15 - 21 - 7\sqrt{5} + 9\sqrt{5} + 15}{9 - 5} \\[1em] \Rightarrow \dfrac{4\sqrt{5}}{4} \\[1em] \Rightarrow \sqrt{5}

Hence, 7+353+5−7−353−5=5\dfrac{7 + 3\sqrt{5}}{3 + \sqrt{5}}- \dfrac{7 - 3\sqrt{5}}{3 - \sqrt{5}} = \sqrt{5}.

Question 18

Show that : 1(3−8)+1(7−6)+1(5−2)−1(8−7)−1(6−5)=5\dfrac{1}{(3 - \sqrt{8})} + \dfrac{1}{(\sqrt{7} - \sqrt{6})}+ \dfrac{1}{(\sqrt{5} - 2)} - \dfrac{1}{(\sqrt{8} - \sqrt{7})} - \dfrac{1}{(\sqrt{6} - \sqrt{5})} = 5

Answer

Given,

Equation : 1(3−8)+1(7−6)+1(5−2)−1(8−7)−1(6−5)\dfrac{1}{(3 - \sqrt{8})} + \dfrac{1}{(\sqrt{7} - \sqrt{6})}+ \dfrac{1}{(\sqrt{5} - 2)} - \dfrac{1}{(\sqrt{8} - \sqrt{7})} - \dfrac{1}{(\sqrt{6} - \sqrt{5})}

Simplifying L.H.S. of the above equation :

⇒1(3−8)×(3+8)(3+8)+1(7−6)×(7+6)(7+6)+1(5−2)×(5+2)(5+2)−1(8−7)×(8+7)(8+7)−1(6−5)×(6+5)(6+5)⇒3+832−(8)2+7+6(7)2−(6)2+5+2(5)2−(2)2−8+7(8)2−(7)2−6+5(6)2−(5)2⇒3+89−8+7+67−6+5+25−4−8+78−7−6+56−5⇒3+8+7+6+5+2−(8+7)−(6+5)⇒3+2+8−8+7−7+6−6+5−5⇒5\Rightarrow \dfrac{1}{(3 - \sqrt{8})} \times \dfrac{(3 + \sqrt{8})}{(3 + \sqrt{8})} + \dfrac{1}{(\sqrt{7} - \sqrt{6})} \times \dfrac{(\sqrt{7} + \sqrt{6})}{(\sqrt{7} + \sqrt{6})}+ \dfrac{1}{(\sqrt{5} - 2)} \times \dfrac{(\sqrt{5} + 2)}{(\sqrt{5} + 2)} - \dfrac{1}{(\sqrt{8} - \sqrt{7})} \times \dfrac{(\sqrt{8} + \sqrt{7})}{(\sqrt{8} + \sqrt{7})} - \dfrac{1}{(\sqrt{6} - \sqrt{5})} \times \dfrac{(\sqrt{6} + \sqrt{5})}{(\sqrt{6} + \sqrt{5})} \\[1em] \Rightarrow \dfrac{3 + \sqrt{8}}{3^2 - (\sqrt{8})^2} + \dfrac{\sqrt{7} + \sqrt{6}}{(\sqrt{7})^2 - (\sqrt{6})^2} + \dfrac{\sqrt{5} + 2}{(\sqrt{5})^2 - (2)^2} - \dfrac{\sqrt{8} + \sqrt{7}}{(\sqrt{8})^2 - (\sqrt{7})^2} - \dfrac{\sqrt{6} + \sqrt{5}}{(\sqrt{6})^2 - (\sqrt{5})^2} \\[1em] \Rightarrow \dfrac{3 + \sqrt{8}}{9 - 8} + \dfrac{\sqrt{7} + \sqrt{6}}{7 - 6} + \dfrac{\sqrt{5} + 2}{5 - 4} - \dfrac{\sqrt{8} + \sqrt{7}}{8 - 7} - \dfrac{\sqrt{6} + \sqrt{5}}{6 - 5} \\[1em] \Rightarrow 3 + \sqrt{8} + \sqrt{7} + \sqrt{6} + \sqrt{5} + 2 - (\sqrt{8} + \sqrt{7}) - (\sqrt{6} + \sqrt{5}) \\[1em] \Rightarrow 3 + 2 + \sqrt{8} - \sqrt{8} + \sqrt{7} - \sqrt{7} + \sqrt{6} - \sqrt{6} + \sqrt{5} - \sqrt{5} \\[1em] \Rightarrow 5

Hence, proved that

1(3−8)+1(7−6)+1(5−2)−1(8−7)−1(6−5)=5\dfrac{1}{(3 - \sqrt{8})} + \dfrac{1}{(\sqrt{7} - \sqrt{6})}+ \dfrac{1}{(\sqrt{5} - 2)} - \dfrac{1}{(\sqrt{8} - \sqrt{7})} - \dfrac{1}{(\sqrt{6} - \sqrt{5})} = 5.

Question 19

If x = (3+8)(3 + \sqrt{8}), find the values of (x2+1x2)\Big(x^2 + \dfrac{1}{x^2}\Big).

Answer

Given,

x = (3+8)(3 + \sqrt{8})

∴1x=1(3+8)\therefore \dfrac{1}{x} = \dfrac{1}{(3 + \sqrt{8})}

Rationalizing,

⇒1x=1(3+8)×(3−8)(3−8)=3−832−(8)2=3−89−8=3−81=3−8.\Rightarrow \dfrac{1}{x} = \dfrac{1}{(3 + \sqrt{8})} \times \dfrac{(3 - \sqrt{8})}{(3 - \sqrt{8})} \\[1em] = \dfrac{3 - \sqrt{8}}{3^2 - (\sqrt{8})^2} \\[1em] = \dfrac{3 - \sqrt{8}}{9 - 8} \\[1em] = \dfrac{3 - \sqrt{8}}{1} \\[1em] = 3 - \sqrt{8}.

By formula,

x2+1x2=(x+1x)2−2x^2 + \dfrac{1}{x^2} = \Big(x + \dfrac{1}{x}\Big)^2 - 2

Substituting values we get :

⇒x2+1x2=(3+8+3−8)2−2=62−2=36−2=34.\Rightarrow x^2 + \dfrac{1}{x^2} = (3 + \sqrt{8} + 3 - \sqrt{8})^2 - 2 \\[1em] = 6^2 - 2 \\[1em] = 36 - 2 \\[1em] = 34.

Hence, x2+1x2x^2 + \dfrac{1}{x^2} = 34.

Question 20

If x = (4−15)(4 - \sqrt{15}), find the values of (x2+1x2)\Big(x^2 + \dfrac{1}{x^2}\Big).

Answer

Given,

x = (4−15)(4 - \sqrt{15})

∴1x=1(4−15)\therefore \dfrac{1}{x} = \dfrac{1}{(4 - \sqrt{15})}

Rationalizing,

⇒1x=1(4−15)×(4+15)(4+15)=4+1542−(15)2=4+1516−15=4+151=4+15.\Rightarrow \dfrac{1}{x} = \dfrac{1}{(4 - \sqrt{15})} \times \dfrac{(4 + \sqrt{15})}{(4 + \sqrt{15})} \\[1em] = \dfrac{4 + \sqrt{15}}{4^2 - (\sqrt{15})^2} \\[1em] = \dfrac{4 + \sqrt{15}}{16 - 15} \\[1em] = \dfrac{4 + \sqrt{15}}{1} \\[1em] = 4 + \sqrt{15}.

By formula,

x2+1x2=(x+1x)2−2x^2 + \dfrac{1}{x^2} = \Big(x + \dfrac{1}{x}\Big)^2 - 2

Substituting values we get :

⇒x2+1x2=(4−15+4+15)2−2=82−2=64−2=62.\Rightarrow x^2 + \dfrac{1}{x^2} = (4 - \sqrt{15} + 4 + \sqrt{15})^2 - 2 \\[1em] = 8^2 - 2 \\[1em] = 64 - 2 \\[1em] = 62.

Hence, x2+1x2x^2 + \dfrac{1}{x^2} = 62.

Multiple Choice Questions

Question 1

Which of the following is a rational number?

  1. π

  2. 2\sqrt{2}

  3. 3.4

  4. 1.010010001..

Answer

3.4 is a terminating decimal, thus it is a rational number.

Hence, Option 3 is the correct option.

Question 2

Which of the following is an irrational number?

  1. 2.7

  2. 2.72‾2.7\overline{2}

  3. 11\sqrt{11}

  4. 27\dfrac{2}{7}

Answer

11\sqrt{11} is square root of non-perfect square i.e 11, thus it is an irrational number.

Hence, Option 3 is the correct option.

Question 3

Which of the following is a prime number?

  1. 51

  2. 57

  3. 71

  4. 81

Answer

71 has only two factors i.e is 1 and 71 itself, thus it is a prime number.

Hence, Option 3 is the correct option.

Question 4

When 8.32‾8.\overline{32} is expressed as a vulgar fraction, then it becomes :

  1. 208824\dfrac{208}{824}

  2. 82499\dfrac{824}{99}

  3. 80099\dfrac{800}{99}

  4. 41645\dfrac{416}{45}

Answer

Let x = 8.32‾8.\overline{32}

⇒ x = 8.32323232..     ..........(1)

Multiplying both sides by 100 (since there are two recurring digits)

⇒ 100x = 832.3232..     ..........(2)

Subtracting equation (1) from (2), we get :

⇒ 100x - x = 832.3232.. - 8.3232..

⇒ 99x = 824

⇒ x = 82499\dfrac{824}{99}

Hence, Option 2 is the correct option.

Question 5

0.3 when expressed as a ratio of two integers, becomes :

  1. 103330\dfrac{103}{330}

  2. 52165\dfrac{52}{165}

  3. 103111\dfrac{103}{111}

  4. 104333\dfrac{104}{333}

Answer

103330=0.312‾\dfrac{103}{330} = 0.3\overline{12}

52165=0.315‾\dfrac{52}{165} = 0.3\overline{15}

103111=0.927‾\dfrac{103}{111} = 0.\overline{927}

104333=0.312‾\dfrac{104}{333} = 0.\overline{312}

Since, 0.312‾0.3\overline{12} is nearest to 0.3

Hence, Option 1 is the correct option.

Question 6

Only by inspecting the prime factors of the denominator, state which of the following fractions will be a terminating decimal?

  1. 712\dfrac{7}{12}

  2. 215\dfrac{2}{15}

  3. 316\dfrac{3}{16}

  4. 421\dfrac{4}{21}

Answer

A rational number will have a terminating decimal expansion if and only if the prime factorization of its denominator contains only powers of 2 and/or 5.

The prime factor of 16 is only 2.

∴ 316\dfrac{3}{16} will have a terminating decimal.

Hence, Option 3 is the correct option.

Question 7

Only by inspecting the prime factors of the denominators, state which of the following fractions will be a recurring decimal?

  1. 716\dfrac{7}{16}

  2. 851\dfrac{8}{51}

  3. 325\dfrac{3}{25}

  4. 1120\dfrac{11}{20}

Answer

A rational number will have a recurring decimal expansion if and only if the prime factorization of its denominator contains any prime factor other than 2 or 5.

The prime factors of 51 are 17 and 3.

Hence, Option 2 is the correct option.

Question 8

The number which is to be subtracted from 72\sqrt{72} to get 32\sqrt{32} is:

  1. 2102\sqrt{10}

  2. 424\sqrt{2}

  3. 323\sqrt{2}

  4. 222\sqrt{2}

Answer

Let the number to be subtarcted be x.

⇒72−x=32⇒x=72−32⇒x=36×2−16×2⇒x=62−42⇒x=22.\Rightarrow \sqrt{72} - x = \sqrt{32} \\[1em] \Rightarrow x = \sqrt{72} - \sqrt{32} \\[1em] \Rightarrow x = \sqrt{36 \times 2} - \sqrt{16 \times 2} \\[1em] \Rightarrow x = 6\sqrt{2} - 4\sqrt{2} \\[1em] \Rightarrow x = 2\sqrt{2}.

Hence, Option 4 is the correct option.

Question 9

If x = 3 + 222\sqrt{2}, then x+1xx + \dfrac{1}{x} equals to :

  1. 424\sqrt{2}

  2. 626\sqrt{2}

  3. 6

  4. 4

Answer

Given,

⇒ x = 3 + 222\sqrt{2}

⇒1x=1(3+22)\Rightarrow \dfrac{1}{x} = \dfrac{1}{(3 + 2\sqrt{2})}

Rationalizing,

⇒1(3+22)×(3−22)(3−22)⇒3−2232−(22)2⇒3−229−8⇒3−22\Rightarrow \dfrac{1}{(3 + 2\sqrt{2})} \times \dfrac{(3 - 2\sqrt{2})}{(3 - 2\sqrt{2})} \\[1em] \Rightarrow \dfrac{3 - 2\sqrt{2}}{3^2 - (2\sqrt{2})^2} \\[1em] \Rightarrow \dfrac{3 - 2\sqrt{2}}{9 - 8} \\[1em] \Rightarrow 3 - 2\sqrt{2}

Adding x and 1x\dfrac{1}{x} we get

⇒x+1x=3+22+3−22⇒6\Rightarrow x + \dfrac{1}{x} = 3 + 2\sqrt{2} + 3 - 2\sqrt{2} \\[1em] \Rightarrow 6

Hence, Option 3 is the correct option.

Question 10

If x = 2 - 2\sqrt{2}, then x−1xx - \dfrac{1}{x} =

  1. 4

  2. -4

  3. 2−322\dfrac{2 - 3\sqrt{2}}{2}

  4. 2+322\dfrac{2 + 3\sqrt{2}}{2}

Answer

Given,

⇒ x = 2 - 2\sqrt{2}

⇒1x=1(2−2)\Rightarrow \dfrac{1}{x} = \dfrac{1}{(2 - \sqrt{2})}

Rationalizing the denominator, we get :

⇒1(2−2)×(2+2)(2+2)⇒2+222−(2)2⇒2+24−2⇒2+22\Rightarrow \dfrac{1}{(2 - \sqrt{2})} \times \dfrac{(2 + \sqrt{2})}{(2 + \sqrt{2})} \\[1em] \Rightarrow \dfrac{2 + \sqrt{2}}{2^2 - (\sqrt{2})^2} \\[1em] \Rightarrow \dfrac{2 + \sqrt{2}}{4 - 2} \\[1em] \Rightarrow \dfrac{2 + \sqrt{2}}{2}

Substituting values in x−1xx - \dfrac{1}{x}, we get :

⇒2−2−2+22⇒2(2−2)−(2+2)2⇒4−22−2−22⇒2−322\Rightarrow 2 - \sqrt{2} - \dfrac{2 + \sqrt{2}}{2} \\[1em] \Rightarrow \dfrac{2(2 - \sqrt{2})-(2 + \sqrt{2})}{2} \\[1em] \Rightarrow \dfrac{4 - 2\sqrt{2} - 2 - \sqrt{2}}{2} \\[1em] \Rightarrow \dfrac{2 - 3\sqrt{2}}{2} \\[1em]

x−1x=2−322x - \dfrac{1}{x} = \dfrac{2 - 3\sqrt{2}}{2}

Hence, Option 3 is the correct option.

Question 11

If x = 5 + 262\sqrt{6}, then x2+1x2x^2 + \dfrac{1}{x^2} =

  1. 98

  2. 142

  3. 49

  4. 138

Answer

Given,

x = (5+26)(5 + 2\sqrt{6})

∴1x=1(5+26)\therefore \dfrac{1}{x} = \dfrac{1}{(5 + 2\sqrt{6})}

Rationalizing,

⇒1x=15+26×(5−26)(5−26)=5−2652−(26)2=5−2625−24=5−261=5−26.\Rightarrow \dfrac{1}{x} = \dfrac{1}{5 + 2\sqrt{6}} \times \dfrac{(5 - 2\sqrt{6})}{(5 - 2\sqrt{6})} \\[1em] = \dfrac{5 - 2\sqrt{6}}{5^2 - (2\sqrt{6})^2} \\[1em] = \dfrac{5 - 2\sqrt{6}}{25 - 24} \\[1em] = \dfrac{5 - 2\sqrt{6}}{1} \\[1em] = 5 - 2\sqrt{6}.

By formula,

x2+1x2=(x+1x)2−2x^2 + \dfrac{1}{x^2} = \Big(x + \dfrac{1}{x}\Big)^2 - 2

Substituting values we get :

⇒x2+1x2=(5+26+5−26)2−2=102−2=100−2=98.\Rightarrow x^2 + \dfrac{1}{x^2} = (5 + 2\sqrt{6} + 5 - 2\sqrt{6})^2 - 2 \\[1em] = 10^2 - 2 \\[1em] = 100 - 2 \\[1em] = 98.

Hence, Option 1 is the correct option.

Question 12

Two rational numbers between −37 and −17-\dfrac{3}{7} \text{ and } -\dfrac{1}{7} is :

  1. 414,314\dfrac{4}{14}, \dfrac{3}{14}

  2. −414,314-\dfrac{4}{14}, \dfrac{3}{14}

  3. 414,−314\dfrac{4}{14}, -\dfrac{3}{14}

  4. −414,−314-\dfrac{4}{14}, -\dfrac{3}{14}

Answer

Let the first rational number between −37 and −17-\dfrac{3}{7} \text{ and } -\dfrac{1}{7} be x.

⇒x=12[−37+(−17)]⇒x=12(−47)⇒x=−414\Rightarrow x = \dfrac{1}{2}\Big[-\dfrac{3}{7} + \Big(-\dfrac{1}{7}\Big)\Big] \\[1em] \Rightarrow x = \dfrac{1}{2}\Big(-\dfrac{4}{7}\Big)\\[1em] \Rightarrow x = -\dfrac{4}{14}

Let the second rational number be y.

⇒y=12[−414+(−17)]⇒y=12(−4−214)⇒y=12(−614)⇒y=−628=−314\Rightarrow y = \dfrac{1}{2}\Big[-\dfrac{4}{14} + \Big(-\dfrac{1}{7}\Big)\Big] \\[1em] \Rightarrow y = \dfrac{1}{2}\Big(\dfrac{-4 - 2}{14}\Big) \\[1em] \Rightarrow y = \dfrac{1}{2}\Big(-\dfrac{6}{14}\Big)\\[1em] \Rightarrow y = -\dfrac{6}{28} = -\dfrac{3}{14}

Hence, Option 4 is the correct option.

Question 13

The correct ascending order of 3,63,74\sqrt{3} , \sqrt[3]{6}, \sqrt[4]{7} is :

  1. 63,74,3\sqrt[3]{6}, \sqrt[4]{7}, \sqrt{3}

  2. 74,3,63\sqrt[4]{7}, \sqrt{3}, \sqrt[3]{6}

  3. 3,74,63\sqrt{3}, \sqrt[4]{7}, \sqrt[3]{6}

  4. 63,3,74\sqrt[3]{6}, \sqrt{3}, \sqrt[4]{7}

Answer

On solving,

⇒ 3\sqrt{3} = 1.732

⇒ 63\sqrt[3]{6} = 1.817

⇒ 74\sqrt[4]{7} = 1.626

∴ 74<3<63\sqrt[4]{7} \lt \sqrt{3} \lt \sqrt[3]{6}

Hence, Option 2 is the correct option.

Question 14

The mixed surd for 4323\sqrt[3]{432} is :

  1. 2632\sqrt[3]{6}

  2. 6236\sqrt[3]{2}

  3. 3633\sqrt[3]{6}

  4. 6336\sqrt[3]{3}

Answer

Given,

⇒4323⇒216×23⇒2163×23⇒623\Rightarrow \sqrt[3]{432} \\[1em] \Rightarrow \sqrt[3]{216 \times 2} \\[1em] \Rightarrow \sqrt[3]{216} \times \sqrt[3]{2} \\[1em] \Rightarrow 6\sqrt[3]{2}

Hence, Option 2 is the correct option.

Question 15

What is the pure surd for 5235\sqrt[3]{2} ?

  1. 1253\sqrt[3]{125}

  2. 1503\sqrt[3]{150}

  3. 2503\sqrt[3]{250}

  4. 10003\sqrt[3]{1000}

Answer

Given,

⇒523⇒53×23⇒125×23⇒2503\Rightarrow 5\sqrt[3]{2} \\[1em] \Rightarrow \sqrt[3]{5^3 \times 2} \\[1em] \Rightarrow \sqrt[3]{125 \times 2} \\[1em] \Rightarrow \sqrt[3]{250}

Hence, Option 3 is the correct option.

Case Study Based Questions

Question 1

Case Study

The union of the set of rational numbers (Q) and irrational numbers (P) form the set of real numbers (R). Rational numbers are the set of numbers which can be written in the form of ab\dfrac{a}{b}, where a and b are integers and b is not equal to zero. The decimal expansion of a rational number is either terminating or non-terminating repeating. The number which cannot be expressed in the form ab\dfrac{a}{b} are called irrational numbers. The decimal expansion of irrational numbers is non-terminating non-repeating.

The union of the set of rational numbers (Q) and irrational numbers (P) form the set of real numbers (R). Rational numbers are the set of numbers which can be written in the form of a/b, where a and b are integers and b is not equal to zero. The decimal expansion of a rational number is either terminating or non-terminating repeating. The number which cannot be expressed in the form a/b are called irrational numbers. The decimal expansion of irrational numbers is non-terminating non-repeating.: Rational and Irrational Numbers, R.S. Aggarwal Mathematics Solutions ICSE Class 9.

Based on this information, answer the following questions:

  1. Every rational number is:
    (a) a natural number
    (b) a whole number
    (c) an integer
    (d) a real number

  2. Every real number is:
    (a) an integer
    (b) a rational number
    (c) an irrational number
    (d) either a rational number or an irrational number.

  3. The sum of two irrationals is:
    (a) irrational
    (b) rational
    (c) either rational or irrational
    (d) neither rational or irrational.

  4. The product of a rational and irrational number is:
    (a) an irrational number
    (b) a rational number
    (c) either a rational number or an irrational number
    (d) neither a rational number nor an irrational number.

  5. The number of irrational number is:
    (a) finite
    (b) infinite
    (c) neither finite or infinite
    (d) none of these.

Answer

1. Real numbers include all rational and irrational numbers.

Rational numbers are a subset of real numbers.

Hence, Option (d) is the correct option.

2. Real numbers include all rational and irrational numbers.

Hence, Option (d) is the correct option.

3. The sum of two irrational numbers can be either rational or irrational.

Hence, Option (c) is the correct option.

4. If the rational number is zero, the product is rational.

If the rational number is non-zero, the product is irrational.

Hence, Option (c) is the correct option.

5. The number of irrational number is infinite as the number of real numbers is also infinite.

Hence, Option (b) is the correct option.

Question 2

Case Study

Ms Mehta teaches maths in a school. One day after teaching the lesson of number system, she wanted to check the understanding of the students of her class. So, she wrote two numbers, 311 and 0.52‾\dfrac{3}{11}\text{ and } 0.\overline{52} on the blackboard and asked few questions based on them. You please try to answer the following questions asked by Ms Mehta.

  1. The decimal expansion of 311\dfrac{3}{11} is:
    (a) terminating
    (b) non-terminating
    (c) non-terminating non-repeating
    (d) non-terminating repeating

  2. 0.52‾0.\overline{52} is:
    (a) non-terminating non-repeating
    (b) non-terminating repeating
    (c) non-terminating
    (d) terminating

  3. The decimal form of 311\dfrac{3}{11}:
    (a) 0.27
    (b) 0.2727
    (c) 0.27‾0.\overline{27}
    (d) 0.3

  4. 0.52‾0.\overline{52} as vulgar fraction becomes:

    (a) 5299\dfrac{52}{99}

    (b) 52100\dfrac{52}{100}

    (c) 2625\dfrac{26}{25}

    (d) 1325\dfrac{13}{25}

  5. The sum of 0.52‾ and 3110.\overline{52} \text{ and } \dfrac{3}{11} is :
    (a) 7999\dfrac{79}{99}

    (b) 7099\dfrac{70}{99}

    (c) 5299\dfrac{52}{99}

    (d) 4099\dfrac{40}{99}

Answer

1. On dividing,

311=0.27‾\dfrac{3}{11} = 0.\overline{27}

Hence, Option (d) is the correct option.

2. 0.52‾0.\overline{52} = 0.525252...

Hence, Option (b) is the correct option.

3. 311\dfrac{3}{11} = 0.272727.... = 0.27‾0.\overline{27}

Hence, Option (c) is the correct option.

4. Given,

Let x = 0.525252..     .........(1)

Multiply both side by 100(as two digits are recurring)

⇒ 100x = 52.5252..     .........(2)

Subtracting eqn (1) from eqn (2), we get :

⇒ 100x - x = 52.5252.. - 0.525252

⇒ 99x = 52

⇒ x = 5299\dfrac{52}{99}.

Hence, Option (a) is the correct option.

5. From part 4,

0.52‾=52990.\overline{52} = \dfrac{52}{99}

Adding 5299\dfrac{52}{99} and 311\dfrac{3}{11}

⇒5299+311⇒52+2799⇒7999.\Rightarrow \dfrac{52}{99} + \dfrac{3}{11} \\[1em] \Rightarrow \dfrac{52 + 27}{99} \\[1em] \Rightarrow \dfrac{79}{99}.

Hence, Option (a) is the correct option.

Assertion Reasoning Questions

Question 1

Assertion (A): The number obtained on rationalizing the denominator of 15−2\dfrac{1}{\sqrt{5} - 2} is 2+52 + \sqrt{5}.

Reason (R): If the product of two irrational numbers is rational, then each one is called the rationalizing factor of the other.

  1. A is true, R is false

  2. A is false, R is true

  3. Both A and R are true

  4. Both A and R are false.

Answer

Given,

15−2\dfrac{1}{\sqrt{5} - 2}

Rationalizing the denominator of 15−2\dfrac{1}{\sqrt{5} - 2},

⇒15−2×5+25+2⇒5+2(5)2−(2)2⇒5+25−4⇒5+21⇒2+5\Rightarrow \dfrac{1}{\sqrt{5} - 2} \times \dfrac{\sqrt{5} + 2}{\sqrt{5} + 2} \\[1em] \Rightarrow \dfrac{\sqrt{5} + 2}{(\sqrt{5})^2 - (2)^2}\\[1em] \Rightarrow \dfrac{\sqrt{5} + 2}{5 - 4} \\[1em] \Rightarrow \dfrac{\sqrt{5} + 2}{1} \\[1em] \Rightarrow 2 + \sqrt{5}

∴ Assertion (A) is true.

We know that,

If the product of two irrational numbers is rational, then each one is called the rationalizing factor of the other.

∴ Reason (R) is true.

Hence, Option 3 is the correct option.

Question 2

Assertion (A): Each of the numbers 23,33,43,53,63,73\sqrt[3]{2}, \sqrt[3]{3}, \sqrt[3]{4}, \sqrt[3]{5}, \sqrt[3]{6}, \sqrt[3]{7} is irrational.

Reason (R): The cube roots of all natural numbers is irrational.

  1. A is true, R is false

  2. Both A and R are true

  3. A is false, R is true

  4. Both A and R are false.

Answer

23,33,43,53,63,73\sqrt[3]{2}, \sqrt[3]{3}, \sqrt[3]{4}, \sqrt[3]{5}, \sqrt[3]{6}, \sqrt[3]{7} are irrational, because these are cube roots of not perfect cubes.

∴ Assertion (A) is true.

The cube roots of all perfect cubes are rational. Thus, we cannot say that the cube roots of all natural numbers is irrational.

∴ Reason (R) is false.

Hence, Option 1 is the correct option.

Competency Focused Questions

Question 1

The sum of all rational numbers between 0 and 0.1 is :

  1. finite

  2. infinite

  3. can't say anything

  4. none of these

Answer

Between any two real numbers like 0 and 0.1, there are infinitely many rational numbers. So the sum is infinite.

Hence, Option 2 is the correct option.

Question 2

Four rational numbers p, q, r and s are such that q is the reciprocal of p and s is the reciprocal of r. The value of the expression ([p+1q]÷[r+1s])÷([s+1r]÷[q+1p])\Big([p + \dfrac{1}{q}] ÷ [r + \dfrac{1}{s}]\Big) ÷ \Big([s + \dfrac{1}{r}] ÷ [q + \dfrac{1}{p}]\Big) is equal to:

  1. 1

  2. 0

  3. pr

  4. sq\dfrac{s}{q}

Answer

Given,

⇒q=1p⇒1q=p⇒s=1r⇒1s=r.\Rightarrow q = \dfrac{1}{p} \\[1em] \Rightarrow \dfrac{1}{q} = p \\[1em] \Rightarrow s = \dfrac{1}{r} \\[1em] \Rightarrow \dfrac{1}{s} = r.

Substituting value from above in given equation,

⇒([p+p]÷[r+r])÷([s+s]÷[q+q])⇒(2p2r)÷(2s2q)⇒pr÷sq⇒(pr)×(qs)⇒pr×q×1s⇒pr×1p×r⇒1.\Rightarrow \Big([p + p] ÷ [r + r]\Big)÷ \Big([s + s] ÷ [q + q]\Big) \\[1em] \Rightarrow \Big(\dfrac{2p}{2r}\Big) ÷ \Big(\dfrac{2s}{2q}\Big) \\[1em] \Rightarrow \dfrac{p}{r} ÷ \dfrac{s}{q} \\[1em] \Rightarrow \Big(\dfrac{p}{r}\Big) \times \Big(\dfrac{q}{s}\Big) \\[1em] \Rightarrow \dfrac{p}{r} \times q \times \dfrac{1}{s} \\[1em] \Rightarrow \dfrac{p}{r} \times \dfrac{1}{p} \times r \\[1em] \Rightarrow 1.

Hence, Option 1 is the correct option.

Question 3

0.6+0.7‾+0.47‾0.6 + 0.\overline{7} + 0.4\overline{7} is equal to s:

  1. 15590\dfrac{155}{90}

  2. 14790\dfrac{147}{90}

  3. 16790\dfrac{167}{90}

  4. none of these

Answer

Convert repeating decimals to fraction,

Let, x = 0.7‾0.\overline{7}

⇒ x = 0.7777....     .......(1)

⇒ 10x = 7.777...     .......(2)

Subtracting equation (1) from (2), we get :

⇒ 10x - x = 7.777..... - 0.777.....

⇒ 9x = 7

⇒ x = 79\dfrac{7}{9}

Let, y = 0.47‾0.4\overline{7}

⇒ 10y = 4.7777....     .......(3)

⇒ 100y = 47.777...     .......(4)

Subtracting equation (3) from (4), we get :

⇒ 100y - 10y = 47.777..... - 4.777.....

⇒ 90y = 43

⇒ y = 4390\dfrac{43}{90}

⇒0.6+0.7‾+0.47‾=0.6+79+4390=610+79+4390=54+70+4390=16790.\Rightarrow 0.6 + 0.\overline{7} + 0.4\overline{7} = 0.6 + \dfrac{7}{9} + \dfrac{43}{90} \\[1em] = \dfrac{6}{10} + \dfrac{7}{9} + \dfrac{43}{90} \\[1em] = \dfrac{54 + 70 + 43}{90} \\[1em] = \dfrac{167}{90}.

Hence, Option 3 is the correct option.

Question 4

3234\sqrt[4]{\sqrt[3]{3^2}} can be expressed as :

  1. 363^6

  2. 61/36^{1/3}

  3. 31/123^{1/12}

  4. 31/63^{1/6}

Answer

We know that,

⇒xn=x1n\Rightarrow \sqrt[n]{x} = x^\dfrac{1}{n}

Solving,

⇒3234⇒3234⇒323×4⇒3212=316\Rightarrow \sqrt[4]{\sqrt[3]{3^2}} \\[1em] \Rightarrow \sqrt[4]{3^\dfrac{2}{3}} \\[1em] \Rightarrow 3^\dfrac{2}{3 \times 4} \\[1em] \Rightarrow 3^\dfrac{2}{12} \\[1em] = {3^\dfrac{1}{6}}

Hence, Option 4 is the correct option.

Question 5

1.9‾−1.91.\overline{9} - 1.9 is equal to :

  1. 0

  2. 1

  3. 0.09

  4. 0.1

Answer

Let x = 1.9‾1.\overline{9}

⇒ x = 1.999...     .......(1)

⇒ 10x = 19.999...     .......(2)

Subtracting equation (1) from (2), we get :

⇒ 10x - x = 19.999..... - 1.999....

⇒ 9x = 18

⇒ x = 189\dfrac{18}{9} = 2.

⇒ 1.9‾−1.91.\overline{9} - 1.9

⇒ 2 - 1.9

⇒ 0.1

Hence, Option 4 is the correct option.

Question 6

When written in decimal form, which of the following will be a non-terminating and non-repeating number?

  1. 11/91^{1/9}

  2. 21/92^{1/9}

  3. 2−92^{-9}

  4. 91/29^{1/2}

Answer

A non-terminating and non-repeating decimal is the definition of an irrational number.

Since 2 is not a perfect 9th power of any rational number, its 9th root will be an irrational number.

Hence, Option 2 is the correct option.

Question 7

Observe the values of a, b, c given in the table. If we choose numbers a, b and c from rows a, b and c respectively, what is the maximum possible value of c−ba\dfrac{c - b}{a}?

a246810
b357911
c510152025

Answer

We get the maximum possible value of c−ba\dfrac{c - b}{a}, if c is the largest and a, b are the smallest values.

⇒c−ba⇒25−32⇒222⇒11.\Rightarrow \dfrac{c - b}{a} \\[1em] \Rightarrow \dfrac{25 - 3}{2} \\[1em] \Rightarrow \dfrac{22}{2} \\[1em] \Rightarrow 11.

Hence, the maximum possible value of c−ba\dfrac{c - b}{a} = 11.

Question 8

The portion of a number line between 0 and 4 has been divided into 16 equal parts. Highlight the portion of the number line in which the reciprocal of any rational number is greater than the number itself.

The portion of a number line between 0 and 4 has been divided into 16 equal parts. Highlight the portion of the number line in which the reciprocal of any rational number is greater than the number itself. Rational and Irrational Numbers, R.S. Aggarwal Mathematics Solutions ICSE Class 9.

Answer

As, the complete line is divided in into 16 equal parts. Thus, A = 1, B = 2, C = 3.

Rational number less than 1 and greater than 0, will be the interval in which reciprocal of any rational number is greater than the number itself.

The portion of a number line between 0 and 4 has been divided into 16 equal parts. Highlight the portion of the number line in which the reciprocal of any rational number is greater than the number itself. Rational and Irrational Numbers, R.S. Aggarwal Mathematics Solutions ICSE Class 9.
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