Rational & Irrational Numbers — Test Yourself

Since, 90 = 2 x 3 x 3 x 5, $\dfrac{23}{90}$ is not a terminating decimal;

1. True

2. False

3. none of these

Answer

A fraction has a terminating decimal if, when simplified, its denominator consists solely of the prime factors 2 and/or 5.

90 = 2 x 3 x 3 x 5

Since, 90 consists 2, 3 and 5.

So, $\dfrac{23}{90}$ is not a terminating decimal.

Hence, option 1 is the correct option.

$\sqrt{27}$ is irrational and $\sqrt{3}$ is also irrational, then which of the following is rational:

1. $\sqrt{27} - \sqrt{3}$

2. $\sqrt{27} + \sqrt{3}$

3. $\sqrt{27} \times \sqrt{3}$

4. none of these

Answer

As we know that

$\Rightarrow \sqrt{27} \times \sqrt{3}\\[1em] = \sqrt{81}\\[1em] = 9.$

9 is rational number. So, $\sqrt{27} \times \sqrt{3}$ is also a rational number.

Hence, option 3 is the correct option.

If x = $\sqrt{6} - \sqrt{5}$, then $x - \dfrac{1}{x}$ is equal to:

1. 1

2. 11

3. $2\sqrt{6}$

4. -2 $\sqrt{5}$

Answer

Given, x = $\sqrt{6} - \sqrt{5}$

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

Now,

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

Hence, option 4 is correct option.

$\dfrac{2 + \sqrt{3}}{2 - \sqrt{3}} - \dfrac{2 - \sqrt{3}}{2 + \sqrt{3}}$ is equal to:

1. 4

2. $2\sqrt{3}$

3. 1

4. $8\sqrt{3}$

Answer

Given, $\dfrac{2 + \sqrt{3}}{2 - \sqrt{3}} - \dfrac{2 - \sqrt{3}}{2 + \sqrt{3}}$

Solving,

$\Rightarrow \dfrac{(2 + \sqrt{3}) \times (2 + \sqrt{3})}{(2 - \sqrt{3}) \times (2 + \sqrt{3})} - \dfrac{(2 - \sqrt{3})\times (2 - \sqrt{3})}{(2 + \sqrt{3}) \times (2 - \sqrt{3})}\\[1em] \Rightarrow \dfrac{(2 + \sqrt{3})^2}{(2)^2 - (\sqrt{3})^2} - \dfrac{(2 - \sqrt{3})^2}{(2)^2 - (\sqrt{3})^2}\\[1em] \Rightarrow \dfrac{(2)^2 + (\sqrt{3})^2 + 2 \times 2 \times \sqrt{3}}{4 - 3} - \dfrac{(2)^2 + (\sqrt{3})^2 - 2 \times 2 \times \sqrt{3}}{4 - 3}\\[1em] \Rightarrow \dfrac{4 + 3 + 4\sqrt{3}}{1} - \dfrac{4 + 3 - 4\sqrt{3}}{1}\\[1em] \Rightarrow 7 + 4\sqrt{3} - (7 - 4\sqrt{3})\\[1em] \Rightarrow 7 + 4\sqrt{3} - 7 + 4\sqrt{3}\\[1em] \Rightarrow 8\sqrt{3}.$

Hence, option 4 is correct option.

Statement 1: If a = $3\sqrt{3}$ and b = $\dfrac{5}{\sqrt{12}}$, then a x b is irrational.

Statement 2: a x b = $3 \sqrt{3} \times \dfrac{5}{\sqrt{12}} = \dfrac{15 \times \sqrt{3}}{2\sqrt{3}} = \dfrac{15}{2}$

1. Both the statements are true.

2. Both the statements are false.

3. Statement 1 is true, and statement 2 is false.

4. Statement 1 is false, and statement 2 is true.

Answer

Given, a = $3\sqrt{3}$ and b = $\dfrac{5}{\sqrt{12}}$

$\Rightarrow a \times b = 3 \sqrt{3} \times \dfrac{5}{\sqrt{12}}\\[1em] = 3 \sqrt{3} \times \dfrac{5}{\sqrt{4 \times 3}}\\[1em] = 3 \sqrt{3} \times \dfrac{5}{\sqrt{4} \times \sqrt{3}}\\[1em] = 3 \sqrt{3} \times \dfrac{5}{2\sqrt{3}}\\[1em] = \dfrac{5 \times 3 \sqrt{3}}{2\sqrt{3}}\\[1em] = \dfrac{15 \sqrt{3}}{2\sqrt{3}}\\[1em] = \dfrac{15}{2}\\[1em] = 7.5$

7.5 is rational number. Thus, a x b is a rational number.

∴ Statement 1 is false, and statement 2 is true.

Hence, option 4 is correct option.

Statement 1: If x = $\sqrt{5} + 2$, then $x - \dfrac{1}{x} = 4.$

Statement 2: $\dfrac{1}{x} = \dfrac{1}{\sqrt{5} + 2} \times \dfrac{\sqrt{5} - 2}{\sqrt{5} - 2} = \sqrt{5} - 2$

1. Both the statements are true.

2. Both the statements are false.

3. Statement 1 is true, and statement 2 is false.

4. Statement 1 is false, and statement 2 is true.

Answer

Given, x = $\sqrt{5}$ + 2

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

So, statement 2 is true.

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

So, statement 1 is true.

∴ Both statements are true.

Hence, option 1 is correct option.

Assertion (A): $x + \dfrac{1}{x} = 4 \text{ and } \dfrac{1}{x} = 2 + \sqrt{3}, \text{ then x } = 2 - \sqrt{3}$

Reason (R):
$x + \dfrac{1}{x} = 4 \\[1em] \Rightarrow x + 2 + \sqrt{3} = 4 \\[1em] \Rightarrow x = 2 - \sqrt{3}$

1. A is true, but R is false.

2. A is false, but R is true.

3. Both A and R are true, and R is the correct reason for A.

4. Both A and R are true, and R is the incorrect reason for A.

Answer

Given, $\dfrac{1}{x} = 2 + \sqrt{3}\text{ and } x + \dfrac{1}{x} = 4$

And,

$\Rightarrow x + \dfrac{1}{x} = 4\\[1em] = x + 2 + \sqrt{3} = 4\\[1em] = x = 4 - (2 + \sqrt{3})\\[1em] = x = 4 - 2 - \sqrt{3}\\[1em] = x = 2 - \sqrt{3}.$

So, assertion (A) is true.

If, $\dfrac{1}{x} = 2 + \sqrt{3}$

$\Rightarrow x = \dfrac{1}{2 + \sqrt{3}}\\[1em] = \dfrac{1}{2 + \sqrt{3}} \times \dfrac{2 - \sqrt{3}}{2 - \sqrt{3}} \text{ (On rationalizing)} \\[1em] = \dfrac{2 - \sqrt{3}}{(2)^2 - (\sqrt{3})^2}\\[1em] = \dfrac{2 - \sqrt{3}}{4 - 3}\\[1em] = \dfrac{2 - \sqrt{3}}{1}\\[1em] = 2 - \sqrt{3}.$

So, reason (R) is true.

∴ Both A and R are true, and R is the correct reason for A.

Hence, option 3 is correct option.

Assertion (A): $\sqrt{22}, \sqrt{23}, \sqrt{24}, \sqrt{25}, \sqrt{26}$ and $\sqrt{27}$ are irrational numbers between $\sqrt{21}$ and $\sqrt{28}$.

Reason (R): $\sqrt{25}$ is not an irrational number as $\sqrt{25}$ = 5; which is rational number.

1. A is true, but R is false.

2. A is false, but R is true.

3. Both A and R are true, and R is the correct reason for A.

4. Both A and R are true, and R is the incorrect reason for A.

Answer

Though all the given values $\sqrt{22}, \sqrt{23}, \sqrt{24}, \sqrt{25}, \sqrt{26}, \sqrt{27}$ lie between $\sqrt{21}$ and $\sqrt{28}$

But $\sqrt{25}$ is not an irrational number, as $\sqrt{25}$ = 5, which is a rational number.

∴ A is false, but R is true.

Hence, option 2 is the correct option.

Simplify :

$\dfrac{\sqrt{x^2 + y^2} - y}{x - \sqrt{x^2 - y^2}} ÷ \dfrac{\sqrt{x^2 - y^2} + x}{\sqrt{x^2 + y^2} + y}$

Answer

Solving,

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

Hence, $\dfrac{\sqrt{x^2 + y^2} - y}{x - \sqrt{x^2 - y^2}} ÷ \dfrac{\sqrt{x^2 - y^2} + x}{\sqrt{x^2 + y^2} + y} = \dfrac{x^2}{y^2}$.

Evaluate, correct to one place of decimal, the expression $\dfrac{5}{\sqrt{20} - \sqrt{10}}, \text{ if } \sqrt{5} = 2.2 \text{ and } \sqrt{10} = 3.2$

Answer

Solving,

$\Rightarrow \dfrac{5}{\sqrt{20} - \sqrt{10}} \\[1em] \Rightarrow \dfrac{5}{2\sqrt{5} - \sqrt{10}} \\[1em] \Rightarrow \dfrac{5}{2 \times 2.2 - 3.2} \\[1em] \Rightarrow \dfrac{5}{4.4 - 3.2} \\[1em] \Rightarrow \dfrac{5}{1.2} \\[1em] \Rightarrow \dfrac{50}{12} \\[1em] \Rightarrow \dfrac{25}{6} = 4.2$

Another method of solving is by rationalizing,

$\Rightarrow \dfrac{5}{\sqrt{20} - \sqrt{10}} \times \dfrac{\sqrt{20} + \sqrt{10}}{\sqrt{20} + \sqrt{10}} \\[1em] \Rightarrow \dfrac{5(\sqrt{20} + \sqrt{10})}{(\sqrt{20})^2 - (\sqrt{10})^2} \\[1em] \Rightarrow \dfrac{5(\sqrt{20} + \sqrt{10})}{20 - 10} \\[1em] \Rightarrow \dfrac{5(\sqrt{20} + \sqrt{10})}{10} \\[1em] \Rightarrow \dfrac{(2\sqrt{5} + \sqrt{10})}{2} \\[1em] \Rightarrow \dfrac{2 \times 2.2 + 3.2}{2} \\[1em] \Rightarrow \dfrac{4.4 + 3.2}{2} \\[1em] \Rightarrow \dfrac{7.6}{2} \\[1em] \Rightarrow 3.8$

Hence, $\dfrac{\sqrt{x^2 + y^2} - y}{x - \sqrt{x^2 - y^2}} ÷ \dfrac{\sqrt{x^2 - y^2} + x}{\sqrt{x^2 + y^2} + y}$ = 3.8 or 4.2

If x = $\sqrt{3} - \sqrt{2}$, find the value of :

(i) $x + \dfrac{1}{x}$

(ii) $x^2 + \dfrac{1}{x^2}$

(iii) $x^3 + \dfrac{1}{x^3}$

(iv) $x^3 + \dfrac{1}{x^3} - 3\Big(x^2 + \dfrac{1}{x^2}\Big) + x + \dfrac{1}{x}$

Answer

(i) Given,

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

$\dfrac{1}{x} = \dfrac{1}{\sqrt{3} - \sqrt{2}}$

Rationalizing,

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

Hence, $x + \dfrac{1}{x} = 2\sqrt{3}$.

(ii) Solving,

$\Rightarrow x^2 + \dfrac{1}{x^2} = (\sqrt{3} - \sqrt{2})^2 + (\sqrt{3} + \sqrt{2})^2 \\[1em] = (\sqrt{3})^2 + (\sqrt{2})^2 - 2 \times \sqrt{3} \times \sqrt{2} + (\sqrt{3})^2 + (\sqrt{2})^2 + 2 \times \sqrt{3} \times \sqrt{2} \\[1em] \Rightarrow 3 + 2 - 2\sqrt{6} + 2 + 3 + 2\sqrt{6} \\[1em] \Rightarrow 10.$

Hence, $x^2 + \dfrac{1}{x^2} = 10$.

(iii) Solving,

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

Hence, $x^3 + \dfrac{1}{x^3} = 18\sqrt{3}$.

(iv) Substituting values from part (i), (ii) and (iii), we get :

$\Rightarrow x^3 + \dfrac{1}{x^3} - 3\Big(x^2 + \dfrac{1}{x^2}\Big) + x + \dfrac{1}{x} = 18\sqrt{3} - 3 \times 10 + 2\sqrt{3} \\[1em] = 20\sqrt{3} - 30 \\[1em] = 10(2\sqrt{3} - 3).$

Hence, $x^3 + \dfrac{1}{x^3} - 3\Big(x^2 + \dfrac{1}{x^2}\Big) + x + \dfrac{1}{x} = 10(2\sqrt{3} - 3)$.

State true or false :

(i) Negative of an irrational number is irrational.

(ii) The product of a non-zero rational number and an irrational number is a rational number.

Answer

(i) True

For example, $\sqrt{3}$ is irrational and $-\sqrt{3}$ is also irrational.

(ii) False

For example, 2 is rational number and $\sqrt{2}$ is irrational, product of these numbers i.e. $2\sqrt{2}$ is also irrational.

Draw a line segment of length $\sqrt{3}$ cm.

Answer

Steps of construction :

1. Draw a line segment XY.

2. Draw OB = 1 cm which is perpendicular to the line XY at point O.

3. From B draw an arc of 2 cm cutting XY at A.

4. Join BA and OA.

So, OAB is the right angle triangle.

By pythagoras theorem,

⇒ AB² = OB² + OA²

⇒ 2² = 1² + OA²

⇒ OA² = 4 - 1

⇒ OA² = 3

⇒ OA = $\sqrt{3}$ cm.

Hence, OA is the required line of $\sqrt{3}$ cm.

Draw a line segment of length $\sqrt{8}$ cm.

Answer

Steps of construction :

1. Draw a line segment XY.

2. Draw OB = 1 cm which is perpendicular to the line XY at O.

3. From B draw an arc of 3 cm cutting XY at A.

4. Join BA and OA.

So, OAB is the right angle triangle.

By pythagoras theorem,

⇒ AB² = OB² + OA²

⇒ 3² = 1² + OA²

⇒ OA² = 9 - 1

⇒ OA² = 8

⇒ OA = $\sqrt{8}$ cm.

Hence, OA is the required line of $\sqrt{8}$ cm.

Show that :

$x^3 + \dfrac{1}{x^3} = 52$, if x = 2 + $\sqrt{3}$

Answer

(i) Given,

x = 2 + $\sqrt{3}$

$\therefore \dfrac{1}{x} = \dfrac{1}{2 + \sqrt{3}}$

Rationalizing,

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

Substituting value of x and $\dfrac{1}{x} \text{ in } x^3 + \dfrac{1}{x^3}$, we get :

$\Rightarrow x^3 + \dfrac{1}{x^3} = (2 + \sqrt{3})^3 + (2 -\sqrt{3})^3 \\[1em] = 2^3 + (\sqrt{3})^3 + 3 \times 2 \times \sqrt{3} \times (2 + \sqrt{3}) + 2^3 - (\sqrt{3})^3 - 3 \times 2 \times \sqrt{3} \times (2 - \sqrt{3}) \\[1em] = 8 + 3\sqrt{3} + 6\sqrt{3}(2 + \sqrt{3}) + 8 - 3\sqrt{3} - 6\sqrt{3}(2 - \sqrt{3}) \\[1em] = 8 + 8 + 3\sqrt{3} - 3\sqrt{3} + 12\sqrt{3} + 18 - 12\sqrt{3} + 18 \\[1em] = 8 + 8 + 18 + 18 \\[1em] = 52.$

Hence, proved that $x^3 + \dfrac{1}{x^3} = 52$.

Show that :

$x^2 + \dfrac{1}{x^2} = 34, \text{ if x } = 3 + 2\sqrt{2}$.

Answer

Given,

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

$\therefore \dfrac{1}{x} = \dfrac{1}{3 + 2\sqrt{2}}$

Rationalizing,

$\Rightarrow \dfrac{1}{3 + 2\sqrt{2}} \times \dfrac{3 - 2\sqrt{2}}{3 - 2\sqrt{2}} \\[1em] \Rightarrow \dfrac{3 - 2\sqrt{2}}{3^2 - (2\sqrt{2})^2} \\[1em] \Rightarrow \dfrac{3 - 2\sqrt{2}}{9 - 8} \\[1em] \Rightarrow 3 - 2\sqrt{2} \\[1em] \therefore \dfrac{1}{x} = 3 - 2\sqrt{2}$

Substituting value of x and $\dfrac{1}{x} \text{ in } x^2 + \dfrac{1}{x^2}$, we get :

$\Rightarrow x^2 + \dfrac{1}{x^2} = (3 + 2\sqrt{2})^2 + (3 - 2\sqrt{2})^2 \\[1em] \Rightarrow 3^2 + (2\sqrt{2})^2 + 2 \times 3 \times 2\sqrt{2} + 3^2 + (2\sqrt{2})^2 - 2 \times 3 \times 2\sqrt{2} \\[1em] \Rightarrow 9 + 8 + 12\sqrt{2} + 9 + 8 - 12\sqrt{2} \\[1em] \Rightarrow 34.$

Hence, proved that $x^2 + \dfrac{1}{x^2} = 34.$

Show that :

$\dfrac{3\sqrt{2} - 2\sqrt{3}}{3\sqrt{2} + 2\sqrt{3}} + \dfrac{2\sqrt{3}}{\sqrt{3} - \sqrt{2}}$ = 11

Answer

Given,

Equation : $\dfrac{3\sqrt{2} - 2\sqrt{3}}{3\sqrt{2} + 2\sqrt{3}} + \dfrac{2\sqrt{3}}{\sqrt{3} - \sqrt{2}}$ = 11

Solving L.H.S. of the equation :

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

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

Hence, proved that $\dfrac{3\sqrt{2} - 2\sqrt{3}}{3\sqrt{2} + 2\sqrt{3}} + \dfrac{2\sqrt{3}}{\sqrt{3} - \sqrt{2}}$ = 11.

Show that x is irrational, if :

(i) x² = 6

(ii) x² = 0.009

(iii) x² = 27

Answer

(i) Given,

⇒ x² = 6

⇒ x = $\sqrt{6}$

⇒ x = 2.449.....

Since, x is non-terminating as well as non-recurring.

Hence, proved that x is irrational.

(ii) Given,

⇒ x² = 0.009

⇒ x = $\sqrt{0.009}$

⇒ x = 0.094.....

Since, x is non-terminating as well as non-recurring.

Hence, proved that x is irrational.

(iii) Given,

⇒ x² = 27

⇒ x = $\sqrt{27}$

⇒ x = 5.196.....

Since, x is non-terminating as well as non-recurring.

Hence, proved that x is irrational.

Show that x is rational, if :

(i) x² = 16

(ii) x² = 0.0004

(iii) x² = $1\dfrac{7}{9}$

Answer

(i) Given,

⇒ x² = 16

⇒ x = $\sqrt{16}$

⇒ x = $\pm 4$.

Integers are considered as rational numbers.

Hence, proved that x is rational.

(ii) Given,

⇒ x² = 0.0004

⇒ x = $\sqrt{0.0004}$

⇒ x = $0.02$.

Terminating decimals are considered as rational numbers.

Hence, proved that x is rational.

(iii) Given,

⇒ x² = $1\dfrac{7}{9}$

⇒ x² = $\dfrac{16}{9}$

⇒ x = $\sqrt{\dfrac{16}{9}}$

⇒ x = $\dfrac{4}{3}$ = 1.333.....

Recurring decimals are considered as rational number.

Hence, proved that x is rational.

Find the value of :

$\dfrac{1}{1 + \sqrt{2}} + \dfrac{1}{\sqrt{2} + \sqrt{3}} + ...+ \dfrac{1}{\sqrt{2025} + \sqrt{2026}}$

Answer

Rationalizing a general term,

$\frac{1}{\sqrt{n} + \sqrt{n+1}}$

To rationalize, we multiply the numerator and denominator by the conjugate, $(\sqrt{n+1} - \sqrt{n})$:

$\dfrac{1}{\sqrt{n+1} + \sqrt{n}} \times \dfrac{\sqrt{n+1} - \sqrt{n}}{\sqrt{n+1} - \sqrt{n}} \\[1em] = \dfrac{\sqrt{n+1} - \sqrt{n}}{(\sqrt{n+1})^2 - (\sqrt{n})^2} \\[1em] = \dfrac{\sqrt{n+1} - \sqrt{n}}{(n+1) - n} \\[1em] = \sqrt{n+1} - \sqrt{n}$

Expanding the series,

First term: $\dfrac{1}{1 + \sqrt{2}} = \sqrt{2} - 1$

Second term: $\dfrac{1}{\sqrt{2} + \sqrt{3}} = \sqrt{3} - \sqrt{2}$

Third term: $\dfrac{1}{\sqrt{3} + \sqrt{4}} = \sqrt{4} - \sqrt{3}$

...

Last term: $\dfrac{1}{\sqrt{2025} + \sqrt{2026}} = \sqrt{2026} - \sqrt{2025}$

Adding all the series,

$(\sqrt{2} - 1) + (\sqrt{3} - \sqrt{2}) + (\sqrt{4} - \sqrt{3}) + \dots + (\sqrt{2026} - \sqrt{2025})$

$\sqrt{2} - 1 + \sqrt{3} - \sqrt{2} + \sqrt{4} - \sqrt{3} + \dots + \sqrt{2026} - \sqrt{2025}$

Everything cancels except the first and last term.

-1 + $\sqrt{2026}$.

Hence, solution = $\sqrt{2026}$ - 1.