**Q.1: Without actual division, show that each of the following rational numbers is a terminating decimal. Express each in decimal form.**

**(i) \(\frac{ 23 }{ (2 ^{ 2 } \times 5 ^{ 2 } ) }\) **

**(ii) \(\frac{ 24 }{ 125 }\)**

**(iii) \(\frac{ 171 }{ 800 }\) **

**(iv) \(\frac{ 15 }{ 1600 }\)**

**(v) \(\frac{ 17 }{ 320 }\) **

**(vi) \(\frac{ 19 }{ 3125 }\)**

**Sol:**

**i)** \(\frac{ 23 }{ 2 ^{ 3 } \times 5 ^{ 2 } } = \frac{ 23 \times 5 }{ 2 ^{ 3 } \times 5 ^{ 3 } }\)

We know either 2 or 5 is not a factor of 23, so it is in its simplest form.

Moreover, it is in the form of (\(2^{m} X 5^{n}\)

Hence, the given rational is terminating.

**ii)** \(\frac{24}{125}= \frac{24}{5^{3}}=\frac{24 X 2^{3}}{5^{3} X 2^{3}}=\frac{192}{1000}\)

We know 5 is not a factor of 23, so it is in its simplest form.

Moreover, it is in the form of (\(2^{m} X 5^{n}\)

Hence, the given rational is terminating.

**(iii)** \(\frac{171}{800}= \frac{171}{5^{2} X 2^{5}}=\frac{171 X 5^{3}}{2^{5} X 2^{5}}=\frac{21375}{100000}\)

We know either 2 or 5 is not a factor of 171, so it is in its simplest form.

Moreover, it is in the form of (\(2^{m} X 5^{n}\)

Hence, the given rational is terminating.

**iv)** \(\frac{15}{1600}= \frac{15}{5^{2} X 2^{6}}=\frac{15 X 5^{4}}{2^{6} X 5^{6}}=\frac{9375}{1000000}\)

We know either 2 or 5 is not a factor of 15, so it is in its simplest form.

Moreover, it is in the form of (\(2^{m} X 5^{n}\)

Hence, the given rational is terminating.

**v)** \(\frac{17}{320}= \frac{17}{5 X 2^{6}}=\frac{17 X 5^{5}}{2^{6} X 5^{6}}=\frac{53125}{1000000}\)

We know either 2 or 5 is not a factor of 17, so it is in its simplest form.

Moreover, it is in the form of (\(2^{m} X 5^{n}\)

Hence, the given rational is terminating.

vi) \(\frac{19}{3125}=\frac{19}{5^{5}} = \frac{19 X 2^{5}}{5^{5}X 2^{5}}=\frac{608}{100000}\)

We know either 2 or 5 is not a factor of 19, so it is in its simplest form.

Moreover, it is in the form of (\(2^{m} X 5^{n}\)

Hence, the given rational is terminating.

**Q.2: Without actual division, show that each of the following rational numbers is a non-terminating repeating decimal.**

**(i) \(\frac{ 11 }{ ( 2 ^{ 3 } \times 3) }\) **

**(ii)** **\(\frac{ 73 }{ ( 2 ^{ 2 } \times 3 ^{ 3 } \times 5) }\)**

**(iii) \(\frac{ 129 }{ ( 2 ^{ 2 } \times 5 ^{ 3 } \times 7 ^{ 2 } ) }\)**

**(iv) \(\frac{ 9 }{ 35 }\) **

**(v) \(\frac{ 77 }{ 210 }\) **

**(vi) \(\frac{ 32 }{ 147 }\) **

**(vii) \(\frac{ 29 }{ 343 }\) **

**(viii) \(\frac{ 64 }{ 455 }\)**

**Sol:**

**(i)** \(\frac{11}{2^{3}X3}\)

We know either 2 or 3 is not a factor of 11, so it is in its simplest form.

Moreover, (2^{3} x 3) \(\neq\)^{m} x 5^{n})

Hence, the given rational is non-terminating repeating decimal.

**(ii)** We know 2, 3 or 5 is not a factor of 73, so it is in its simplest form.

Moreover, (2^{2} x 3^{3} x 5) \(\neq\)^{m} x 5^{n})

Hence, the given rational is non-terminating repeating decimal.

**(iii)** We know 2, 5 or 7 is not a factor of 129, so it is in its simplest form.

Moreover, (2^{2} x 5^{7} x 7^{5}) \(\neq\)^{m} x 5^{n})

Hence, the given rational is non-terminating repeating decimal.

**(iv)** \(\frac{9}{35}\)

We know either 5 or 7 is not a factor of 9, so it is in its simplest form. Moreover, (5 x 7) \(\neq\)^{m} x 5^{n})

Hence, the given rational is non-terminating repeating decimal.

**(v)** \(\frac{77}{210}\)

\(\frac{11}{2X3X5}\)

We know 2, 3 or 5 is not a factor of 11, so \(\frac{11}{30}\)

Moreover, (2 x 3 x 5) \(\neq\)^{m} x 5^{n})

Hence, the given rational is non-terminating repeating decimal.

**(vi)** We know either 3 or 7 is not a factor of 32, so it is in its simplest form.

Moreover, (3 x 7^{2}) \(\neq\)^{m} x 5″)

Hence, the given rational is non-terminating repeating decimal.

**(vii)** We know 7 is not a factor of 29, so it is in its simplest form.

Moreover, 7^{3}\(\neq\)^{m} x 5″)

Hence, the given rational is non-terminating repeating decimal.

**(viii)** \(\frac{64}{455}\)

We know 5, 7 or 13 is not a factor of 64, so it is in its simplest form.

Moreover, (5 x 7 x 13) \(\neq\)^{m} x 5″)

Hence, the given rational is non-terminating repeating decimal.

**Q.3: Express each of the following as a fraction in simplest form.**

**(i) \(0.\overline{ 8 }\) **

**(ii) \(2.\overline{ 4 }\)**

**(iii) \(0.\overline{ 24 }\) **

**(iv) \(0.1 \overline{ 2 }\)**

**(v) \(2.2\overline{ 4 }\) **

**(vi) \(0.\overline{ 365 }\)**

**Sol:**

**(i)** Let x = \(\bar{0.8}\)

Therefore, x = 0.888 . . . . . . . . (1)

10x = 8.888 . . . . . . (2)

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

9x = 8\(\Rightarrow\)

**(ii)** Let x = \(\bar{2.4}\)

Therefore, x =2.444 … (1)

10x = 24.444 . . . . . . . . (2)

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

9x = 22 \(\Rightarrow\)

Therefore, 2.4 = \(\bar{\frac{22}{9}}\)

**(iii)** Let x = \(\bar{0.24}\)

Therefore, x = 0.2424 …(1)

100x = 24.2424 …(2)

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

99x = 24 \(\Rightarrow\)

Therefore, 0.24 = \(\bar{\frac{8}{33}}\)

**(iv)** Let x = \(\bar{0.12}\)

Therefore, x = 0.1212 …….(1)

100x = 12.1212 …….(2)

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

99x=12 \(\Rightarrow\)

Therefore, 0.12 = \(\bar{\frac{4}{33}}\)

**(v)** Let x = \(\bar{2.24}\)

Therefore, x = 2.2444 –(1)

10x = 22.444 …(2)

100x = 224.444 …(3)

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

90x = 202 \(\Rightarrow\)

Hence, \(\bar{2.24}\)

**(vi)** Let x = \(\bar{0.365}\)

Therefore, x = 0.3656565 …(1)

10x = 3.656565 …(2)

1000x = 365.656565 ….(3)

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

990x = 362 = x = \(\frac{362}{990}\)

Hence, \(\bar{0.365}\)

**Q.5: Define (i) rational numbers (ii) irrational numbers (iii) real numbers.**

**Sol:**

Rational numbers: The numbers of the form \(\frac{p}{q}\)

Example: \(\frac{2}{3}\)

Irrational numbers: The numbers which when expressed in decimal form are expressible as non-terminating and non-repeating decimals are called irrational numbers.

Example: \(\sqrt{2}\)

Real numbers: The numbers which are positive or negative, whole numbers or decimal numbers and rational number or irrational number are called real numbers.

Example: 2, \(\frac{1}{3}\)

**Q.6: Classify the following numbers as rational and irrational:**

**(i) \(\frac{ 22 }{ 7 }\) **

**(ii) 3.141**

**iii) \(\pi\) **

**(iv) \(3.\overline{ 142857 }\) **

**(v) 5.636363 … **

**(vi) 2.040040004 …**

**(vii) 1.535335333 … **

**(viii) 3.121221222 … **

**(ix) \(\sqrt{ 21 }\)**

**(x) \(\sqrt[ 3 ]{ 3 }\)**

**Sol:**

**(i)** \(\frac{ 22 }{ 7 }\)

**(ii)** 3.1416 is a rational number because it is a terminating decimal.

**(iii)** \(\pi\)

**(iv)** \(3.\bar{1}\bar{4}\bar{2}\bar{8}\bar{5}\bar{7}\)

**(v)** 5.636363… is a rational number because it is a non-terminating, repeating decimal.

**(vi)** 2.040040004… is an irrational number because it is a non-terminating and non-repeating decimal.

**(vii)** 1.535335333… is an irrational number because it is a non-terminating and non-repeating decimal.

**(viii)** 3.121221222… is an irrational number because it is a non-terminating and non-repeating decimal.

**(ix)** \(\sqrt{21}= \sqrt{3} X \sqrt{7}\)

**(x)** \(\sqrt[3]{3}\)

**Q.7: Prove that each of the following numbers is irrational.**

**(i) \(\sqrt{ 6 }\) **

**(ii) \(\left (2 – \sqrt{ 3 } \right )\)**

**(iii) \(\left (3 + \sqrt{ 2 } \right )\) **

**(iv)** **\(\left (2 + \sqrt{ 5 } \right )\)**

**(v) \(\left (5 + 3 \sqrt{ 2 } \right )\)**

**(vi) \(3 \sqrt{ 7 }\)**

**(vii) \(\frac{ 3 }{ \sqrt{ 5 }}\)**

**(viii) \(\left (2 – 3 \sqrt{ 5 } \right )\)**

**(ix) \(\left ( \sqrt{ 3 } + \sqrt{ 5 } \right )\)**

**Sol:**

**(i)** Let \(\sqrt{6}\)

Hence, \(\sqrt{2}\)

This contradicts the fact that \(\sqrt{2}\)

The contradiction arises by assuming \(\sqrt{6}\)

Hence, \(\sqrt{6}\)

**(ii)** Let \(2-\sqrt{3}\)

Hence, 2 and \(2-\sqrt{3}\)

(2 – \(2+\sqrt{3}\)

This contradicts the fact that \( \sqrt{3}\)

The contradiction arises by assuming \( 2- \sqrt{3}\)

Hence, \( 2- \sqrt{3}\)

**(iii)** Let, \( 3+ \sqrt{2}\)

Hence, 3 and \( 3+ \sqrt{2}\)

\( 3+ \sqrt{2}\)

This contradicts the fact that \(\sqrt{2}\)

The contradiction arises by assuming 3 + \(\sqrt{2}\)

Hence, 3 + \(\sqrt{2}\)

**(iv)** Let 2 + .VS be rational. Hence, 2 and .5 are rational. (2 + – 2 = 2 + f – 2 = f = rational [v Difference of two rational is rational] this contradicts the fact that VE is irrational. The contradiction arises by assuming 2 – is rational. Hence, 2 – is irrational.

**(v)** Let, 5 + 3f2 be rational. Hence, 5 and 5 + 3.V2 are rational. (5 + 3f – 5) = 3/2. = rational [v Difference of two rational is rational] x 30 = 1,/2 = rational Product of two rational is rational] This contradicts the fact that V2 is irrational. The contradiction arises by assuming 5 + 3f is rational. Hence, 5 3f is irrational.

**(vi)** Let 30 be rational. x 3f = 1/77 = rational [v Product of two rational is rational] This contradicts the fact that 1/7 is irrational. The contradiction arises by assuming 30 is rational. Hence, 30 are irrational.

**(vii)** Let 3 — be rational. 3 1 VS : – ,/5 3 x – = – = rational This contradicts the fact that 1 is irrational.

**(viii)** [Product of two rational is rational]

So, if is rational, then Nig is rational. 5 x 115 = f= rational Hence, 71/- is irrational.

[Product of two rational is rational]

3 The contradiction arises by assuming VS — is rational.

3 Hence, VS — is irrational

**(viii)** Let 2 — 3N5 be rational. Hence 2 and 2 — 3 f are rational. :. 2 — (2 — 3A/g = 2 — 2 + 3.‘,/g = 3.5 = rational [v Difference of two rational is rational] x 3 Vg = f = rational Product of two rational is rational] This contradicts the fact that -N5 is irrational. The contradiction arises by assuming 2 — 3, 5 is rational. Hence, 2 — 3, 73 is irrational.

**(ix)** Let \(\sqrt{3} + \sqrt{5}\)

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

Therefore, \(\sqrt{3} \)

On squaring both sides of equation (1), we get

3 = \((a-\sqrt{5})^{2}\)

This is impossible because right-hand side is rational, whereas the left-hand side is irrational.

This is a contradiction.

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

**Q.8: Prove that \(\frac{ 1 }{ \sqrt{ 3 } }\) is irrational.**

**Sol:**

Let \(\frac{ 1 }{ \sqrt{ 3 } }\)

Therefore, \(\frac{ 1 }{\sqrt{ 3 } }\)

Therefore, \(\sqrt{ 3 }\)

Since a, b are non-zero integers, = \(\frac{ b }{ a }\)

Thus, equation (1) shows that \(\sqrt{ 3 }\)

This contradicts the fact that \(\sqrt{ 3 } \)

The contradiction arises by assuming \(\sqrt{ 3 }\)

Hence, \(\frac{ 1 }{ \sqrt{ 3 }}\)

**Q.9: (i) Given an example of two irrationals whose sum is rational.**

**(ii) Give an example of two irrationals whose product is rational.**

**Hint**

**(i) Take ( 2 + \(\sqrt{ 3 }\) ) and 2 – \(\sqrt{ 3 }\) )**

**(ii) Take ( 2 + \(\sqrt{ 2 }\) and ( 3 – \(\sqrt{ 3 }\))**

**Sol:**

**(i)** Let (2 +\(\sqrt{ 3 }\)

Therefore, (2 + \(\sqrt{3}\)

**(ii)** Let 2\(\sqrt{3}\)

Therefore, 2\(\sqrt{3}\)

**Q.10: State whether the given statement is true or false.**

**(i) The sum of two rationales’ is always rational.**

**(ii) The product of two rationales’ is always rational. **

**(iii) The sum of two irrationals is always irrational.**

**(iv) The product of two irrationals is always irrational.**

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

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

**Sol:**

**(i)** True

**(ii)** True

**(iii)** False

Counter example: 2 +\(\sqrt{3}\)

**(iv)** False

Counter example: 2\(\sqrt{3}\)

**(v)** True

**(vi)** True