Rational and Irrational Numbers

Let zero = $\dfrac{p}{q}$, where p and q are integers. What additional condition will make 0 = $\dfrac{p}{q}$ a rational number :

1. q = 0

2. p ≠ 0

3. q ≠ 0

4. p ≠ 0 and q ≠ 0.

Answer

$\dfrac{p}{q} = 0$ is a rational number. If,

p and q are integers and q ≠ 0.

Hence, Option 3 is the correct option.

Every non-terminating decimal number is a :

1. recurring decimal

2. real number

3. non-recurring decimal

4. circulating decimal

Answer

Every non-terminating decimal number is a real number.

Hence, Option 2 is the correct option.

7.478478.... is a :

1. terminating rational

2. recurring

3. neither rational non-terminating

4. not real

Answer

7.478478 is a recurring decimal.

Hence, Option 2 is the correct option.

$\dfrac{71}{75}$ is :

1. terminating

2. non-terminating

3. periodic decimal

4. not a rational number

Answer

On solving,

$\dfrac{71}{75}$ = 0.94666.......

∴ $\dfrac{71}{75}$ is a recurring or periodic decimal.

Hence, Option 3 is the correct option.

Which of the following is terminating:

$\dfrac{13}{85}, \dfrac{51}{405} \text{ and } \dfrac{9}{524}$

1. $\dfrac{13}{85}$

2. $\dfrac{9}{524}$

3. $\dfrac{51}{405}$

4. None

Answer

On solving,

$\dfrac{13}{85} = 0.152941$........

$\dfrac{51}{405} = 0.1259259$......

$\dfrac{9}{524}$ = 0.0171755......

None of the above fraction is terminating.

Hence, Option 4 is the correct option.

Are the following statements true or false ? Give reasons for your answers ?

(i) Every whole number is a natural number.

(ii) Every whole number is a rational number.

(iii) Every integer is a rational number.

(iv) Every rational number is a whole number.

Answer

(i) False, as zero is a whole number but not a natural number.

(ii) True

(iii) True

(iv) False, as $\dfrac{2}{5}$ is a rational number but not a whole number.

Arrange $-\dfrac{5}{9}, \dfrac{7}{12}, -\dfrac{2}{3} \text{ and } \dfrac{11}{18}$ in the ascending order of their magnitudes.

Also, find the difference between the largest and the smallest of these rational numbers. Express this difference as a decimal fraction correct to one decimal place.

Answer

L.C.M. of 9, 12, 3 and 18 is 36.

So, converting denominator of each fraction $-\dfrac{5}{9}, \dfrac{7}{12}, -\dfrac{2}{3} \text{ and } \dfrac{11}{18}$ into 36.

$\Rightarrow -\dfrac{5}{9} \times \dfrac{4}{4} = -\dfrac{20}{36} \\[1em] \Rightarrow \dfrac{7}{12} \times \dfrac{3}{3} = \dfrac{21}{36} \\[1em] \Rightarrow -\dfrac{2}{3} \times \dfrac{12}{12} = -\dfrac{24}{36} \\[1em] \Rightarrow \dfrac{11}{18} \times \dfrac{2}{2} = \dfrac{22}{36}.$

Since, -24 < -20 < 21 < 22

∴ -$\dfrac{24}{36} \lt -\dfrac{20}{36} \lt \dfrac{21}{36} \lt \dfrac{22}{36}$

$\Rightarrow -\dfrac{2}{3} \lt -\dfrac{5}{9} \lt \dfrac{7}{12} \lt \dfrac{11}{18}$

Difference between largest and smallest fraction :

$\Rightarrow \dfrac{11}{18} - \Big(-\dfrac{2}{3}\Big) \\[1em] \Rightarrow \dfrac{11}{18} + \dfrac{2}{3} \\[1em] \Rightarrow \dfrac{11}{18} + \dfrac{12}{18} \\[1em] \Rightarrow \dfrac{23}{18} \\[1em] \Rightarrow 1.3$

Hence, fractions in ascending order are $-\dfrac{2}{3} \lt -\dfrac{5}{9} \lt \dfrac{7}{12} \lt \dfrac{11}{18}$ and required difference = 1.3

Arrange $\dfrac{5}{8}, -\dfrac{3}{16}, -\dfrac{1}{4} \text{ and } \dfrac{17}{32}$ in the descending order of their magnitudes.

Also, find the sum of the lowest and the largest of these rational numbers. Express the result obtained as a decimal fraction correct to two decimal places.

Answer

L.C.M. of 4, 8, 16 and 32 is 32.

So, converting denominator of each fraction $\dfrac{5}{8}, -\dfrac{3}{16}, -\dfrac{1}{4} \text{ and } \dfrac{17}{32}$ into 32.

$\Rightarrow \dfrac{5}{8} \times \dfrac{4}{4} = \dfrac{20}{32} \\[1em] \Rightarrow -\dfrac{3}{16} \times \dfrac{2}{2} = -\dfrac{6}{32} \\[1em] \Rightarrow -\dfrac{1}{4} \times \dfrac{8}{8} = -\dfrac{8}{32} \\[1em] \Rightarrow \dfrac{17}{32} \times \dfrac{1}{1} = \dfrac{17}{32}.$

Since, -8 < -6 < 17 < 20.

$\therefore -\dfrac{8}{32} \lt -\dfrac{6}{32} \lt \dfrac{17}{32} \lt \dfrac{20}{32}$

$\therefore -\dfrac{1}{4} \lt -\dfrac{3}{16} \lt \dfrac{17}{32} \lt \dfrac{5}{8}$

So, in descending order.

$\Rightarrow \dfrac{5}{8} \gt \dfrac{17}{32} \gt -\dfrac{3}{16} \gt -\dfrac{1}{4}$

Sum of largest and lowest :

$= \dfrac{5}{8} + \Big(-\dfrac{1}{4}\Big) \\[1em] = \dfrac{5}{8} - \dfrac{1}{4} \\[1em] = \dfrac{5 \times 4}{32} - \dfrac{1 \times 8}{32} \\[1em] = \dfrac{20 - 8}{32} \\[1em] = \dfrac{12}{32} \\[1em] = \dfrac{3}{8} \\[1em] = 0.38$

Hence, fractions in descending order are : $\dfrac{5}{8} \gt \dfrac{17}{32} \gt -\dfrac{3}{16} \gt -\dfrac{1}{4}$ and required sum = 0.38

Without doing any actual division, find which of the following rational numbers have terminating decimal representation :

(i) $\dfrac{7}{16}$

(ii) $\dfrac{23}{125}$

(iii) $\dfrac{9}{14}$

(iv) $\dfrac{32}{45}$

(v) $\dfrac{43}{50}$

Answer

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

(i) $\dfrac{7}{16} = \dfrac{7}{2^4}$

So, 16 can be expressed as 2⁴ × 5⁰.

Hence, it is a terminating decimal number.

(ii) $\dfrac{23}{125} = \dfrac{23}{5^3}$

So, 125 can be expressed as 2⁰ × 5³.

Hence, it is a terminating decimal number.

(iii) $\dfrac{9}{14} = \dfrac{9}{2 \times 7}$

So, 14 cannot be expressed in form of 2^m × 5ⁿ.

Hence, it is not a terminating decimal number.

(iv) $\dfrac{32}{45} = \dfrac{32}{3^2 \times 5}$

So, 45 cannot be expressed in form of 2^m × 5ⁿ.

Hence, it is not a terminating decimal number.

(v) $\dfrac{43}{50} = \dfrac{43}{2 \times 5^2}$

So, 50 can be expressed in form of 2¹ × 5².

Hence, it is a terminating decimal number.

The negative of an irrational number is :

1. a rational number

2. an irrational number

3. a rational number and an irrational number

4. a whole number

Answer

Negative of an irrational number is also an irrational number.

Hence, Option 2 is the correct option.

$\sqrt{8}(\sqrt{8} - 1)$ is always :

1. rational

2. irrational

3. whole number

4. natural number

Answer

Given,

$\Rightarrow \sqrt{8}(\sqrt{8} - 1) \\[1em] \Rightarrow 8 - \sqrt{8} \\[1em] \Rightarrow 8 - 2.82... \\[1em] \Rightarrow 5.17....$

which is an irrational number.

Hence, Option 2 is the correct option.

For the given figure length of OA is :

1. $\sqrt{5}$

2. $\sqrt{3}$

3. $\sqrt{5} \text{ or } \sqrt{3}$

4. neither $\sqrt{5}$ nor $\sqrt{3}$

Answer

By pythagoras theorem,

Hypotenuse² = Perpendicular² + Base²

⇒ OA² = AB² + OB²

⇒ OA² = 2² + 1²

⇒ OA² = 4 + 1

⇒ OA = 5

⇒ OA = $\sqrt{5}$.

Hence, Option 1 is the correct option.

$2\sqrt{3} \times 3\sqrt{8}$ is :

1. rational

2. irrational

3. neither rational nor irrational

4. 12 $\times \sqrt{5}$

Answer

Given,

$\Rightarrow 2\sqrt{3} \times 3\sqrt{8} \\[1em] \Rightarrow 6\sqrt{24} \\[1em] \Rightarrow 6 \times 2\sqrt{6} \\[1em] \Rightarrow 12 \times \sqrt{6} \\[1em] \Rightarrow 12 \times 2.449..... \\[1em] \Rightarrow 29.393.......$

which is an irrational number.

Hence, Option 2 is the correct option.

Two irrational numbers between 8 and 11 are :

1. $\sqrt{65} \text{ and } \sqrt{120}$

2. $\sqrt{69}$ and 10.5

3. 8.2 and $\sqrt{125}$

4. 3 and $\sqrt{110}$

Answer

$\sqrt{65}$ = 8.0622.... and $\sqrt{120}$ = 10.954......

Since, the above two nos. are non-terminating as well as non-recurring.

∴ They are irrational and in between 8 and 11.

Hence, Option 1 is the correct option.

State whether the following numbers are rational or not :

(i) $(2 + \sqrt{2})^2$

(ii) $(3 - \sqrt{3})^2$

(iii) $(5 + \sqrt{5})(5 - \sqrt{5})$

(iv) $(\sqrt{3} - \sqrt{2})^2$

Answer

(i) Given,

$\Rightarrow (2 + \sqrt{2})^2 \\[1em] \Rightarrow 4 + 2 + 4\sqrt{2} \\[1em] \Rightarrow 6 + 4\sqrt{2} \\[1em] \Rightarrow 6 + 5.658.... \\[1em] \Rightarrow 11.658....$

which is irrational.

Hence, $(2 + \sqrt{2})^2$ is not a rational number.

(ii) Given,

$\Rightarrow (3 - \sqrt{3})^2 \\[1em] \Rightarrow 9 + 3 - 6\sqrt{3} \\[1em] \Rightarrow 12 - 6\sqrt{3} \\[1em] \Rightarrow 12 - 10.392..... \\[1em] \Rightarrow 1.607......$

which is irrational.

Hence, $(3 - \sqrt{3})^2$ is not a rational number.

(iii) Given,

$\Rightarrow (5 + \sqrt{5})(5 -\sqrt{5}) \\[1em] \Rightarrow 25 - 5\sqrt{5} + 5\sqrt{5} - 5 \\[1em] \Rightarrow 20$

which is rational.

Hence, $(5 + \sqrt{5})(5 -\sqrt{5})$ is a rational number.

(iv) Given,

$\Rightarrow (\sqrt{3} - \sqrt{2})^2 \\[1em] \Rightarrow 3 + 2 - 2 \times \sqrt{3} \times \sqrt{2} \\[1em] \Rightarrow 5 - 2\sqrt{6} \\[1em] \Rightarrow 5 - 2 \times 2.449... \\[1em] \Rightarrow 5 - 4.898..... \\[1em] \Rightarrow 0.101.....$

which is irrational.

Hence, $(\sqrt{3} - \sqrt{2})^2$ is not a rational number.

Find the square of :

(i) $\dfrac{3\sqrt{5}}{5}$

(ii) $\sqrt{3} + \sqrt{2}$

(iii) $\sqrt{5} - 2$

(iv) $3 + 2\sqrt{5}$

Answer

(i) Squaring,

$\Rightarrow \Big(\dfrac{3\sqrt{5}}{5}\Big)^2 \\[1em] \Rightarrow \dfrac{3\sqrt{5} \times 3\sqrt{5}}{5 \times 5} \\[1em] \Rightarrow \dfrac{45}{25} \\[1em] \Rightarrow \dfrac{9}{5} \\[1em] \Rightarrow 1\dfrac{4}{5}.$

Hence, square of $\dfrac{3\sqrt{5}}{5} = 1\dfrac{4}{5}$.

(ii) Squaring,

$\Rightarrow (\sqrt{3} + \sqrt{2})^2 \\[1em] \Rightarrow (\sqrt{3})^2 + (\sqrt{2})^2 + 2 \times \sqrt{3} \times \sqrt{2} \\[1em] \Rightarrow 3 + 2 + 2\sqrt{6} \\[1em] \Rightarrow 5 + 2\sqrt{6}.$

Hence, square of $\sqrt{3} + \sqrt{2} = 5 + 2\sqrt{6}$.

(iii) Squaring,

$\Rightarrow (\sqrt{5} - 2)^2 \\[1em] \Rightarrow (\sqrt{5})^2 + (2)^2 - 2 \times \sqrt{5} \times 2 \\[1em] \Rightarrow 5 + 4 - 4\sqrt{5} \\[1em] \Rightarrow 9 - 4\sqrt{5}.$

Hence, square of $\sqrt{5} - 2 = 9 - 4\sqrt{5}$.

(iv) Squaring,

$\Rightarrow (3 + 2\sqrt{5})^2 \\[1em] \Rightarrow (3)^2 + (2\sqrt{5})^2 + 2 \times 3 \times 2\sqrt{5} \\[1em] \Rightarrow 9 + 20 + 12\sqrt{5} \\[1em] \Rightarrow 29 + 12\sqrt{5}.$

Hence, square of $3 + 2\sqrt{5} = 29 + 12\sqrt{5}$.

State in each case, whether true or false :

(i) $\sqrt{2} + \sqrt{3} = \sqrt{5}$

(ii) $2\sqrt{4} + 2 = 6$

(iii) $3\sqrt{7} - 2\sqrt{7} = \sqrt{7}$

(iv) $\dfrac{2}{7}$ is an irrational number.

(v) $\dfrac{5}{11}$ is a rational number.

(vi) All rational numbers are real numbers.

(vii) All real numbers are rational numbers.

(viii) Some real numbers are rational numbers.

Answer

(i) $\sqrt{2}$ = 1.41, $\sqrt{3}$ = 1.732 and $\sqrt{5}$ = 2.24

$\sqrt{2} + \sqrt{3}$ = 3.14, which is not equal to 2.24.

$\therefore \sqrt{2} + \sqrt{3} \ne \sqrt{5}$

Hence, above statement is false.

(ii) Given,

$2\sqrt{4} + 2 = 6$

Solving, L.H.S. :

$\Rightarrow 2\sqrt{4} + 2 \\[1em] \Rightarrow 2 \times 2 + 2 \\[1em] \Rightarrow 6.$

Since, L.H.S. = R.H.S.

Hence, above statement is true.

(iii) Given,

$3\sqrt{7} - 2\sqrt{7} = \sqrt{7}$

Solving L.H.S. :

$\Rightarrow 3\sqrt{7} - 2\sqrt{7} \\[1em] \Rightarrow \sqrt{7}(3 - 2) \\[1em] \Rightarrow \sqrt{7} \times 1 \\[1em] \Rightarrow \sqrt{7}.$

Since, L.H.S. = R.H.S.

Hence, above statement is true.

(iv) Since, in $\dfrac{2}{7}$ denominator is not equal to zero.

2 and 7 have no common factor.

∴ $\dfrac{2}{7}$ is rational number.

Hence, above statement is false.

(v) Since, in $\dfrac{5}{11}$ denominator is not equal to zero.

5 and 11 have no common factor.

∴ $\dfrac{5}{11}$ is rational number.

Hence, above statement is true.

(vi) Both, rational and irrational numbers are real numbers.

Hence, above statement is true.

(vii) Real numbers are both rational as well as irrational number.

Hence, above statement is false.

(viii) Some rational numbers are also real numbers.

Hence, above statement is true.

Given universal set

= ${-6, -5\dfrac{3}{4}, -\sqrt{4}, -\dfrac{3}{5}, -\dfrac{3}{8}, 0, \dfrac{4}{5}, 1, 1\dfrac{2}{3}, \sqrt{8}, 3.01, π, 8.47}$

From the given set, find :

(i) set of rational numbers

(ii) set of irrational numbers

(iii) set of integers

(iv) set of non-negative integers

Answer

(i) We need to find the set of rational numbers.

Rational numbers are :

1. Of form $\dfrac{p}{q}$, where q ≠ 0.

2. Integers as well as terminating and recurring decimals are rational numbers.

From the universal set

Set of rational numbers

= ${-6, -5\dfrac{3}{4}, -\sqrt{4}, -\dfrac{3}{5}, -\dfrac{3}{8}, 0, \dfrac{4}{5}, 1, 1\dfrac{2}{3}, 3.01, 8.47}$

(ii) Since,

$\sqrt{8} = 2.82.... \text{ and π} = 3.142....$

Since, the above numbers are neither terminating nor recurring, hence they are irrational.

From the universal set

Set of irrational numbers = ${\sqrt{8}, π}$

(iii) From the universal set

Set of integers = ${-6, -\sqrt{4}, 0, 1}$

(iv) From the universal set

Set of non-negative integers = {0, 1}.

Prove that each of the following numbers is irrational:

(i) $\sqrt{3} + \sqrt{2}$

(ii) 3 - $\sqrt{2}$

Answer

(i) Let us assume $\sqrt{3} + \sqrt{2}$ is a rational number.

Let $\sqrt{3} + \sqrt{2}$ = x

Squaring both sides, we get;

$\Rightarrow (\sqrt{3} + \sqrt{2})^2 = x^2\\[1em] \Rightarrow (\sqrt{3})^2 + (\sqrt{2})^2 + 2 \times \sqrt{3} \times \sqrt{2} = x^2\\[1em] \Rightarrow 3 + 2 + 2\sqrt{6} = x^2\\[1em] \Rightarrow 5 + 2\sqrt{6} = x^2\\[1em] \Rightarrow 2\sqrt{6} = x^2 - 5 \\[1em] \Rightarrow \sqrt{6} = \dfrac{x^2 - 5}{2}$

Here, x is rational,

∴ x² is rational ..................(1)

⇒ x² - 5 is rational

So, $\dfrac{x^2 - 5}{2}$ is rational.

But $\sqrt{6}$ is irrational, as it is square root of non-perfect square.

⇒ $\dfrac{x^2 - 5}{2}$ is irrational i.e. x² - 5 is irrational and so x² is irrational ....................(2)

From (1), x² is rational, and

From (2), x² is irrational

∴ We arrive at a contradiction.

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

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

(ii) Let us assume 3 - $\sqrt{2}$ is a rational number.

Let, 3 - $\sqrt{2}$ = x

Squaring both sides, we get;

$\Rightarrow (3 - \sqrt{2})^2 = x^2 \\[1em] \Rightarrow (3)^2 + (\sqrt{2})^2 - 2 \times 3 \times \sqrt{2} = x^2 \\[1em] \Rightarrow 9 + 2 - 6\sqrt{2} = x^2 \\[1em] \Rightarrow 11 - 6\sqrt{2} = x^2 \\[1em] \Rightarrow 6\sqrt{2} = 11 - x^2 \\[1em] \Rightarrow \sqrt{2} = \dfrac{11 - x^2}{6}$

Here, x is rational,

∴ x² is rational ..................(1)

⇒ 11 - x² is rational

So, $\dfrac{11 - x^2}{6}$ is rational.

But $\sqrt{2}$ is irrational, as it is a square root of non-perfect square.

⇒ $\dfrac{11 - x^2}{6}$ is irrational i.e. 11 - x² is irrational and so x² is irrational ....................(2)

From (1), x² is rational, and

From (2), x² is irrational

∴ We arrive at a contradiction.

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

Hence, 3 - $\sqrt{2}$ is an irrational number.

Write a pair of irrational numbers whose sum is irrational.

Answer

Let $\sqrt{3} + 2 \text{ and } \sqrt{2} - 3$ be two irrational numbers.

Sum of numbers

$= \sqrt{3} + 2 + \sqrt{2} - 3 \\[1em] = \sqrt{3} + \sqrt{2} - 1.$

$\sqrt{3} + \sqrt{2} - 1$ is an irrational number.

Hence, required pair = $\sqrt{3} + 2 \text{ and } \sqrt{2} - 3$.

Write a pair of irrational numbers whose sum is rational.

Answer

Let $\sqrt{3} + 2 \text{ and } 5 - \sqrt{3}$ be two irrational numbers.

Sum of numbers

$= \sqrt{3} + 2 + 5 - \sqrt{3} \\[1em] = 7.$

7 is a rational number.

Hence, required pair = $\sqrt{3} + 2 \text{ and } 5 - \sqrt{3}$.

Write a pair of irrational numbers whose difference is irrational.

Answer

Let $\sqrt{5} + 5 \text{ and } \sqrt{2} + 5$ be two irrational numbers.

Difference of numbers

$= \sqrt{5} + 5 - (\sqrt{2} + 5) \\[1em] = \sqrt{5} - \sqrt{2} + 5 - 5 \\[1em] = \sqrt{5} - \sqrt{2}.$

$\sqrt{5} - \sqrt{2}$ is an irrational number.

Hence, required pair = $\sqrt{5} + 5 \text{ and } \sqrt{2} + 5$.

Write a pair of irrational numbers whose difference is rational.

Answer

Let $\sqrt{3} + 5 \text{ and } \sqrt{3} + 2$ be two irrational numbers.

Difference of numbers

$= \sqrt{3} + 5 - (\sqrt{3} + 2) \\[1em] = \sqrt{3} - \sqrt{3} + 5 - 2 \\[1em] = 3.$

3 is a rational number.

Hence, required pair = $\sqrt{3} + 5 \text{ and } \sqrt{3} + 2$.

Write a pair of irrational numbers whose product is irrational.

Answer

Let $\sqrt{2} \text{ and } \sqrt{3}$ be two irrational numbers.

Product of numbers

$= \sqrt{2} \times \sqrt{3} \\[1em] = \sqrt{6}.$

$\sqrt{6}$ is an irrational number.

Hence, required pair = $\sqrt{2} \text{ and } \sqrt{3}$.

Write a pair of irrational numbers whose product is rational.

Answer

Let $5 + \sqrt{2} \text{ and } 5 - \sqrt{2}$ be two irrational numbers.

Product of numbers

$= (5 + \sqrt{2})(5 - \sqrt{2}) \\[1em] = 25 - 5\sqrt{2} + 5\sqrt{2} - 2 \\[1em] = 23.$

23 is a rational number.

Hence, required pair = $5 + \sqrt{2} \text{ and } 5 - \sqrt{2}$.

If x = $\sqrt{5} - 2, x + \dfrac{1}{x}$ is equal to :

1. $2\sqrt{5}$

2. 4

3. $4\sqrt{5}$

4. $\sqrt{-4}$

Answer

Given,

x = $\sqrt{5} - 2$

$\dfrac{1}{x} = \dfrac{1}{\sqrt{5} - 2}$

Rationalizing,

$\Rightarrow \dfrac{1}{x} = \dfrac{1}{\sqrt{5} - 2} \times \dfrac{\sqrt{5} + 2}{\sqrt{5} + 2} \\[1em] = \dfrac{\sqrt{5} + 2}{(\sqrt{5})^2 - (2)^2} \\[1em] = \dfrac{\sqrt{5} + 2}{5 - 4} \\[1em] = \dfrac{\sqrt{5} + 2}{1} \\[1em] = \sqrt{5} + 2. \\[1em] \Rightarrow x + \dfrac{1}{x} = \sqrt{5} - 2 + \sqrt{5} + 2 \\[1em] = 2\sqrt{5}.$

Hence, Option 1 is the correct option.

If x = 1 + $\sqrt{2}, \text{ then } \Big(x + \dfrac{1}{x}\Big)^2$ is :

1. $2\sqrt{2}$

2. 8

3. 4

4. $4\sqrt{2}$

Answer

Given,

x = $1 + \sqrt{2}$

$\dfrac{1}{x} = \dfrac{1}{1 + \sqrt{2}}$

Rationalizing,

$\Rightarrow \dfrac{1}{1 + \sqrt{2}} \times \dfrac{1 - \sqrt{2}}{1 - \sqrt{2}} \\[1em] = \dfrac{1 - \sqrt{2}}{(1)^2 - (\sqrt{2})^2} \\[1em] = \dfrac{1 - \sqrt{2}}{1 - 2} \\[1em] = \dfrac{1 - \sqrt{2}}{-1} \\[1em] = -1 + \sqrt{2}. \\[1em] \Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 = [1 + \sqrt{2} + (-1 + \sqrt{2})]^2 \\[1em] = [1 - 1 + \sqrt{2} + \sqrt{2}]^2 \\[1em] = [2\sqrt{2}]^2 \\[1em] = 8.$

Hence, Option 2 is the correct option.

$\dfrac{2\sqrt{27} + 3\sqrt{12}}{4\sqrt{3}}$ is equal to :

1. $2\sqrt{3}$

2. $3\sqrt{2}$

3. 3

4. $\sqrt{3} + \sqrt{2}$

Answer

Solving,

$\Rightarrow \dfrac{2\sqrt{27} + 3\sqrt{12}}{4\sqrt{3}} \\[1em] \Rightarrow \dfrac{2 \times 3\sqrt{3} + 3 \times 2\sqrt{3}}{4\sqrt{3}} \\[1em] \Rightarrow \dfrac{6\sqrt{3} + 6\sqrt{3}}{4\sqrt{3}} \\[1em] \Rightarrow \dfrac{12\sqrt{3}}{4\sqrt{3}} \\[1em] \Rightarrow 3.$

Hence, Option 3 is the correct option.

$(\sqrt{5} - \sqrt{3})^2$ is :

1. $8 + 2\sqrt{15}$

2. $8 + \sqrt{15}$

3. $8 - \sqrt{15}$

4. $8 - 2\sqrt{15}$

Answer

Solving,

$\Rightarrow (\sqrt{5} - \sqrt{3})^2 \\[1em] \Rightarrow (\sqrt{5})^2 + (\sqrt{3})^2 - 2 \times \sqrt{5} \times \sqrt{3} \\[1em] \Rightarrow 5 + 3 - 2\sqrt{15} \\[1em] \Rightarrow 8 - 2\sqrt{15}.$

Hence, Option 4 is the correct option.

$\dfrac{3}{4 + \sqrt{7}}$ is equal to :

1. $\dfrac{1}{3}(4 - \sqrt{7})$

2. $3(4 - \sqrt{7})$

3. $\dfrac{1}{3}(4 + \sqrt{7})$

4. $3(4 + \sqrt{7})$

Answer

Rationalizing,

$\Rightarrow \dfrac{3}{4 + \sqrt{7}} \times \dfrac{4 - \sqrt{7}}{4 - \sqrt{7}} \\[1em] \Rightarrow \dfrac{3(4 - \sqrt{7})}{(4)^2 - (\sqrt{7})^2} \\[1em] \Rightarrow \dfrac{3(4 - \sqrt{7})}{16 - 7} \\[1em] \Rightarrow \dfrac{3(4 - \sqrt{7})}{9} \\[1em] \Rightarrow \dfrac{1}{3}(4 - \sqrt{7}).$

Hence, Option 1 is the correct option.

$\dfrac{1}{7 - \sqrt{5}}$ is equal to :

1. $4(7 + \sqrt{5})$

2. $\dfrac{1}{44}(7 + \sqrt{5})$

3. $\dfrac{1}{44}(7 - \sqrt{5})$

4. $4(7 - \sqrt{5})$

Answer

Rationalizing,

$\Rightarrow \dfrac{1}{7 - \sqrt{5}} \times \dfrac{7 + \sqrt{5}}{7 + \sqrt{5}} \\[1em] \Rightarrow \dfrac{7 + \sqrt{5}}{(7)^2 - (\sqrt{5})^2} \\[1em] \Rightarrow \dfrac{7 + \sqrt{5}}{49 - 5} \\[1em] \Rightarrow \dfrac{1}{44}(7 + \sqrt{5}).$

Hence, Option 2 is the correct option.

If x = $\sqrt{2} - 1, \text{ then } \Big(x - \dfrac{1}{x}\Big)^2$ is :

1. $2\sqrt{2}$

2. 2

3. 4

4. $2 - \sqrt{2}$

Answer

Given,

x = $\sqrt{2} - 1$

$\dfrac{1}{x} = \dfrac{1}{\sqrt{2} - 1}$

Rationalizing,

$\Rightarrow \dfrac{1}{x} = \dfrac{1}{\sqrt{2} - 1} \times \dfrac{\sqrt{2} + 1}{\sqrt{2} + 1} \\[1em] = \dfrac{\sqrt{2} + 1}{(\sqrt{2})^2 - (1)^2} \\[1em] = \dfrac{\sqrt{2} + 1}{2 - 1} \\[1em] = \dfrac{\sqrt{2} + 1}{1} \\[1em] = \sqrt{2} + 1. \\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = [\sqrt{2} - 1 - (\sqrt{2} + 1)]^2 \\[1em] = [\sqrt{2} - \sqrt{2} - 1 - 1]^2 \\[1em] = [-2]^2 \\[1em] = 4.$

Hence, Option 3 is the correct option.

$\dfrac{5 - \sqrt{7}}{5 + \sqrt{7}} - \dfrac{5 + \sqrt{7}}{5 - \sqrt{7}}$ is equal to :

1. $10\sqrt{7}$

2. $\dfrac{1}{9}\sqrt{7}$

3. $\dfrac{10\sqrt{7}}{9}$

4. $-\dfrac{10\sqrt{7}}{9}$

Answer

Given,

$\Rightarrow \dfrac{5 - \sqrt{7}}{5 + \sqrt{7}} - \dfrac{5 + \sqrt{7}}{5 - \sqrt{7}} \\[1em] \Rightarrow \dfrac{(5 - \sqrt{7})^2 - (5 + \sqrt{7})^2}{(5 + \sqrt{7})(5 - \sqrt{7})} \\[1em] \Rightarrow \dfrac{(5)^2 + (\sqrt{7})^2 - 2 \times 5 \times \sqrt{7} - [(5)^2 + (\sqrt{7})^2 + 2 \times 5 \times \sqrt{7}]}{5^2 - (\sqrt{7})^2} \\[1em] \Rightarrow \dfrac{25 + 7 - 10\sqrt{7} - [25 + 7 + 10\sqrt{7}]}{25 - 7} \\[1em] \Rightarrow \dfrac{25 - 25 + 7 - 7 - 10\sqrt{7} - 10\sqrt{7}}{18} \\[1em] \Rightarrow \dfrac{-20\sqrt{7}}{18} \\[1em] \Rightarrow \dfrac{-10\sqrt{7}}{9}.$

Hence, Option 4 is the correct option.

State, with reason, which of the following are surds and which are not :

(i) $\sqrt{180}$

(ii) $\sqrt[4]{27}$

(iii) $\sqrt[5]{128}$

(iv) $\sqrt[3]{64}$

(v) $\sqrt[3]{25}.\sqrt[3]{40}$

(vi) $\sqrt[3]{-125}$

(vii) $\sqrt{π}$

(viii) $\sqrt{3 + \sqrt{2}}$

Answer

(i) Given,

$\sqrt{180} = \sqrt{2 \times 2 \times 3 \times 3 \times 5} = 2 \times 3 \times \sqrt{5} = 6\sqrt{5}$, which is an irrational number.

Since, 180 is a rational number and $\sqrt{180}$ is an irrational number.

Hence, $\sqrt{180}$ is a surd.

(ii) Given,

$\sqrt[4]{27} = \sqrt[4]{3^3} = (3)^{\dfrac{3}{4}}$, which is irrational.

Since, 27 is a rational number and $\sqrt[4]{27}$ is an irrational number.

Hence, $\sqrt[4]{27}$ is a surd.

(iii) Given,

$\sqrt[5]{128} = \sqrt[5]{2^7} = \sqrt[5]{2^5 \times 2^2} \\[1em] = (2^5)^{\frac{1}{5}} \times (2^2)^{\frac{1}{5}} \\[1em] = 2 \times (2)^{\frac{2}{5}}$

which is irrational.

Since, 128 is a rational number and $\sqrt[5]{128}$ is an irrational number.

Hence, $\sqrt[5]{128}$ is a surd.

(iv) Given,

$\sqrt[3]{64} = \sqrt[3]{4^3} = 4^{3 \times \dfrac{1}{3}}$ = 4, which is rational.

Since, 64 is rational and $\sqrt[3]{64}$ is also rational.

Hence, $\sqrt[3]{64}$ is not a surd.

(v) Given,

$\sqrt[3]{25}.\sqrt[3]{40} = \sqrt[3]{1000} = \sqrt[3]{10^3} = 10^{3 \times \dfrac{1}{3}}$ = 10, which is rational.

Since, 1000 is rational and 10 is also rational.

Hence, $\sqrt[3]{25}.\sqrt[3]{40}$ is not a surd.

(vi) Given,

$\sqrt[3]{-125} = \sqrt[3]{(-5)^3} = (-5)^{3 \times \dfrac{1}{3}} = -5$, which is rational,

Since, -125 is rational and -5 is also rational.

Hence, $\sqrt[3]{-125}$ is not a surd.

(vii) Given,

$\sqrt{π}$

Since, π is irrational and $\sqrt{π}$ is also irrational.

Hence, $\sqrt{π}$ is not a surd.

(viii) Given,

$\sqrt{3 + \sqrt{2}}$

Since, $3 + \sqrt{2}$ is irrational.

Hence, $\sqrt{3 + \sqrt{2}}$ is not a surd.

Write the lowest rationalizing factor of :

(i) $5\sqrt{2}$

(ii) $\sqrt{24}$

(iii) $\sqrt{5} - 3$

(iv) $7 - \sqrt{7}$

(v) $\sqrt{18} - \sqrt{50}$

(vi) $\sqrt{5} - \sqrt{2}$

(vii) $\sqrt{13} + 3$

Answer

(i) Given,

$\Rightarrow 5\sqrt{2}$

Rationalizing,

$\Rightarrow 5\sqrt{2} \times \sqrt{2} \\[1em] \Rightarrow 10.$

Hence, lowest rationalizing factor of $5\sqrt{2} = \sqrt{2}$.

(ii) Given,

$\Rightarrow \sqrt{24} \\[1em] \Rightarrow 2\sqrt{6}.$

Rationalizing,

$\Rightarrow 2\sqrt{6} \times \sqrt{6} \\[1em] \Rightarrow 12.$

Hence, lowest rationalizing factor of $\sqrt{24} = \sqrt{6}$.

(iii) Given,

$\Rightarrow \sqrt{5} - 3$

Rationalizing,

$\Rightarrow (\sqrt{5} - 3) \times (\sqrt{5} + 3) \\[1em] \Rightarrow (\sqrt{5})^2 - 3^2 \\[1em] \Rightarrow 5 - 9 \\[1em] \Rightarrow -4.$

Hence, lowest rationalizing factor of $\sqrt{5} - 3 = \sqrt{5} + 3$.

(iv) Given,

$\Rightarrow 7 - \sqrt{7}$

Rationalizing,

$\Rightarrow (7 - \sqrt{7}) \times (7 + \sqrt{7}) \\[1em] \Rightarrow (7)^2 - (\sqrt{7})^2 \\[1em] \Rightarrow 49 - 7 \\[1em] \Rightarrow 42.$

Hence, lowest rationalizing factor of $7 - \sqrt{7} = 7 + \sqrt{7}$.

(v) Given,

$\Rightarrow \sqrt{18} - \sqrt{50}\\[1em] \Rightarrow 3\sqrt{2} - 5\sqrt{2}$

Rationalizing,

$\Rightarrow (3\sqrt{2} - 5\sqrt{2}) \times (\sqrt{2}) \\[1em] \Rightarrow (6 - 10) \\[1em] \Rightarrow -4$

Hence, lowest rationalizing factor of $\sqrt{18} - \sqrt{50} = \sqrt{2}$.

(vi) Given,

$\Rightarrow \sqrt{5} - \sqrt{2}$

Rationalizing,

$\Rightarrow (\sqrt{5} - \sqrt{2}) \times (\sqrt{5} + \sqrt{2}) \\[1em] \Rightarrow (\sqrt{5})^2 - (\sqrt{2})^2 \\[1em] \Rightarrow 5 - 2 \\[1em] \Rightarrow 3.$

Hence, lowest rationalizing factor of $\sqrt{5} - \sqrt{2} = \sqrt{5} + \sqrt{2}$.

(vii) Given,

$\Rightarrow \sqrt{13} + 3$

Rationalizing,

$\Rightarrow (\sqrt{13} + 3)(\sqrt{13} - 3) \\[1em] \Rightarrow (\sqrt{13})^2 - 3^2 \\[1em] \Rightarrow 13 - 9 \\[1em] \Rightarrow 4.$

Hence, lowest rationalizing factor of $\sqrt{13} + 3 = \sqrt{13} - 3$.

Rationalize the denominators of:

(i) $\dfrac{2\sqrt{3}}{\sqrt{5}}$

(ii) $\dfrac{\sqrt{6} - \sqrt{5}}{\sqrt{6} + \sqrt{5}}$

Answer

(i) Given,

$\dfrac{2\sqrt{3}}{\sqrt{5}}$

Let us rationalise the denominator,

$\Rightarrow \dfrac{2\sqrt{3} \times \sqrt{5}}{\sqrt{5} \times \sqrt{5}} \\[1em] \Rightarrow \dfrac{2\sqrt{15}}{(\sqrt{5})^2}\\[1em] \Rightarrow \dfrac{2\sqrt{15}}{5}$

Hence,$\dfrac{2\sqrt{3}}{\sqrt{5}} = \dfrac{2\sqrt{15}}{5}$

(ii) Given, $\dfrac{\sqrt{6} - \sqrt{5}}{\sqrt{6} + \sqrt{5}}$

Let us rationalise the denominator,

$\Rightarrow \dfrac{(\sqrt{6} - \sqrt{5}) \times (\sqrt{6} - \sqrt{5})}{(\sqrt{6} + \sqrt{5})\times (\sqrt{6} - \sqrt{5})}\\[1em] = \dfrac{(\sqrt{6} - \sqrt{5})^2}{(\sqrt{6})^2 - (\sqrt{5})^2}\\[1em] = \dfrac{(\sqrt{6})^2 + (\sqrt{5})^2 - 2 \times \sqrt{6} \times \sqrt{5}}{6 - 5}\\[1em] = \dfrac{6 + 5 - 2 \times \sqrt{30}}{1}\\[1em] = 11 - 2\sqrt{30}$

Hence, $\dfrac{\sqrt{6} - \sqrt{5}}{\sqrt{6} + \sqrt{5}} = 11 - 2\sqrt{30}$.

Find the values of 'a' and 'b':

$\dfrac{2 + \sqrt{3}}{2 - \sqrt{3}} = a + b\sqrt{3}$

Answer

Given,

Equation : $\dfrac{2 + \sqrt{3}}{2 - \sqrt{3}} = a + b\sqrt{3}$

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

$\Rightarrow \dfrac{2 + \sqrt{3}}{2 - \sqrt{3}} \times \dfrac{2 + \sqrt{3}}{2 + \sqrt{3}} \\[1em] \Rightarrow \dfrac{(2 + \sqrt{3})^2}{2^2 - (\sqrt{3})^2} \\[1em] \Rightarrow \dfrac{2^2 + (\sqrt{3})^2 + 2 \times 2 \times \sqrt{3}}{2^2 - (\sqrt{3})^2} \\[1em] \Rightarrow \dfrac{4 + 3 + 4\sqrt{3}}{4 - 3} \\[1em] \Rightarrow 7 + 4\sqrt{3}.$

Comparing $7 + 4\sqrt{3} \text{ with } a + b\sqrt{3}$, we get :

a = 7 and b = 4.

Hence, a = 7 and b = 4.

Find the values of 'a' and 'b':

$\dfrac{\sqrt{7} - 2}{\sqrt{7} + 2} = a\sqrt{7} + b$

Answer

Given,

Equation : $\dfrac{\sqrt{7} - 2}{\sqrt{7} + 2} = a\sqrt{7} + b$

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

$\Rightarrow \dfrac{\sqrt{7} - 2}{\sqrt{7} + 2} \times \dfrac{\sqrt{7} - 2}{\sqrt{7} - 2} \\[1em] \Rightarrow \dfrac{(\sqrt{7} - 2)^2}{(\sqrt{7})^2 - 2^2} \\[1em] \Rightarrow \dfrac{(\sqrt{7})^2 + 2^2 - 2 \times \sqrt{7} \times 2}{7 - 4} \\[1em] \Rightarrow \dfrac{7 + 4 - 4\sqrt{7}}{3} \\[1em] \Rightarrow \dfrac{11 - 4\sqrt{7}}{3} \\[1em] \Rightarrow -\dfrac{4\sqrt{7}}{3} + \dfrac{11}{3}.$

Comparing $-\dfrac{4\sqrt{7}}{3} + \dfrac{11}{3} \text{ with } a\sqrt{7} + b$, we get :

a = $-\dfrac{4}{3}\text{ and } b = \dfrac{11}{3}$.

Hence, a = $-\dfrac{4}{3}\text{ and } b = \dfrac{11}{3}$.

Find the values of 'a' and 'b':

$\dfrac{3}{\sqrt{3} - \sqrt{2}} = a\sqrt{3} - b\sqrt{2}$

Answer

Given,

Equation : $\dfrac{3}{\sqrt{3} - \sqrt{2}} = a\sqrt{3} - b\sqrt{2}$

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

$\Rightarrow \dfrac{3}{\sqrt{3} - \sqrt{2}} \times \dfrac{\sqrt{3} + \sqrt{2}}{\sqrt{3} + \sqrt{2}} \\[1em] \Rightarrow \dfrac{3(\sqrt{3} + \sqrt{2})}{(\sqrt{3})^2 - (\sqrt{2})^2} \\[1em] \Rightarrow \dfrac{3\sqrt{3} + 3\sqrt{2}}{3 - 2} \\[1em] \Rightarrow \dfrac{3\sqrt{3} + 3\sqrt{2}}{1} \\[1em] \Rightarrow 3\sqrt{3} + 3\sqrt{2}.$

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

a = 3 and b = -3.

Hence, a = 3 and b = -3.

Simplify :

$\dfrac{22}{2\sqrt{3} + 1} + \dfrac{17}{2\sqrt{3} - 1}$

Answer

(i) Solving,

$\Rightarrow \dfrac{22}{2\sqrt{3} + 1} + \dfrac{17}{2\sqrt{3} - 1} \\[1em] \Rightarrow \dfrac{22(2\sqrt{3} - 1) + 17(2\sqrt{3} + 1)}{(2\sqrt{3} + 1)(2\sqrt{3} - 1)} \\[1em] \Rightarrow \dfrac{44\sqrt{3} - 22 + 34\sqrt{3} + 17}{(2\sqrt{3})^2 - 1^2} \\[1em] \Rightarrow \dfrac{78\sqrt{3} - 5}{12 - 1} \\[1em] \Rightarrow \dfrac{78\sqrt{3} - 5}{11}.$

Hence, solution = $\dfrac{78\sqrt{3} - 5}{11}.$

Simplify :

$\dfrac{\sqrt{2}}{\sqrt{6} - \sqrt{2}} - \dfrac{\sqrt{3}}{\sqrt{6} + \sqrt{2}}$

Answer

Solving,

$\Rightarrow \dfrac{\sqrt{2}}{\sqrt{6} - \sqrt{2}} - \dfrac{\sqrt{3}}{\sqrt{6} + \sqrt{2}} \\[1em] \Rightarrow \dfrac{\sqrt{2}(\sqrt{6} + \sqrt{2}) - \sqrt{3}(\sqrt{6} - \sqrt{2})}{(\sqrt{6} - \sqrt{2})(\sqrt{6} + \sqrt{2})} \\[1em] \Rightarrow \dfrac{\sqrt{12} + 2 - \sqrt{18} + \sqrt{6}}{(\sqrt{6})^2 - (\sqrt{2})^2} \\[1em] \Rightarrow \dfrac{2\sqrt{3} + 2 - 3\sqrt{2} + \sqrt{6}}{6 - 2} \\[1em] \Rightarrow \dfrac{2\sqrt{3} + 2 - 3\sqrt{2} + \sqrt{6}}{4}.$

Hence, solution = $\dfrac{2\sqrt{3} + 2 - 3\sqrt{2} + \sqrt{6}}{4}$.

If x = $\dfrac{\sqrt{5} - 2}{\sqrt{5} + 2} \text{ and } y = \dfrac{\sqrt{5} + 2}{\sqrt{5} - 2}$; find :

(i) x²

(ii) y²

(iii) xy

(iv) x² + y² + xy

Answer

(i) Substituting value of x, we get :

$\Rightarrow x^2 = \Big(\dfrac{\sqrt{5} - 2}{\sqrt{5} + 2}\Big)^2 \\[1em] = \Big(\dfrac{\sqrt{5}^2 + 2^2 - 2 \times \sqrt{5} \times 2}{(\sqrt{5})^2 + 2^2 + 2 \times \sqrt{5} \times 2}\Big) \\[1em] = \dfrac{5 + 4 - 4\sqrt{5}}{5 + 4 + 4\sqrt{5}} \\[1em] = \dfrac{9 - 4\sqrt{5}}{9 + 4\sqrt{5}}.$

Rationalizing,

$= \dfrac{9 - 4\sqrt{5}}{9 + 4\sqrt{5}} \times \dfrac{9 - 4\sqrt{5}}{9 - 4\sqrt{5}} \\[1em] = \dfrac{(9 - 4\sqrt{5})^2}{9^2 - (4\sqrt{5})^2} \\[1em] = \dfrac{9^2 + (4\sqrt{5})^2 - 2 \times 9 \times 4\sqrt{5}}{81 - 80} \\[1em] = \dfrac{81 + 80 - 72\sqrt{5}}{1} \\[1em] = 161 - 72\sqrt{5}.$

Hence, x² = $161 - 72\sqrt{5}.$

(ii) Substituting value of y, we get :

$\Rightarrow y^2 = \Big(\dfrac{\sqrt{5} + 2}{\sqrt{5} - 2}\Big)^2 \\[1em] = \dfrac{(\sqrt{5})^2 + 2^2 + 2\times \sqrt{5} \times 2}{(\sqrt{5})^2 + 2^2 - 2\times \sqrt{5} \times 2} \\[1em] = \dfrac{5 + 4 + 4\sqrt{5}}{5 + 4 - 4\sqrt{5}} \\[1em] = \dfrac{9 + 4\sqrt{5}}{9 - 4\sqrt{5}}$

Rationalizing,

$= \dfrac{9 + 4\sqrt{5}}{9 - 4\sqrt{5}} \times \dfrac{9 + 4\sqrt{5}}{9 + 4\sqrt{5}} \\[1em] = \dfrac{(9 + 4\sqrt{5})^2}{9^2 - (4\sqrt{5})^2} \\[1em] = \dfrac{9^2 + (4\sqrt{5})^2 + 2 \times 9 \times 4\sqrt{5}}{81 - 80} \\[1em] = \dfrac{81 + 80 + 72\sqrt{5}}{1} \\[1em] = 161 + 72\sqrt{5}.$

Hence, y² = $161 + 72\sqrt{5}.$

(iii) Substituting values of x and y, we get :

$\Rightarrow xy = \dfrac{\sqrt{5} - 2}{\sqrt{5} + 2} \times \dfrac{\sqrt{5} + 2}{\sqrt{5} - 2} \\[1em] = 1.$

Hence, xy = 1.

(iv) Substituting value of x², y² and xy, we get :

$\Rightarrow x^2 + y^2 + xy = 161 - 72\sqrt{5} + 161 + 72\sqrt{5} + 1 \\[1em] = 323.$