# RD Sharma Solutions Class 10 Mathematics Solutions for Real Numbers Exercise 1.5 in Chapter 1 - Real Numbers

Question 1 Real Numbers Exercise 1.5

Show that the following numbers are irrational

(i) 1/√2

(ii) 7√5

(iii) 6 + √2

(iv) 3 − √5

Let’s assume on the contrary that 1/√2 is a rational number. Then, there exist positive integers a

and b such that

1/√2 = a/b where, a and b, are co-primes

\begin{array}{ll} \Rightarrow & (1 / \sqrt{2})^{2}=(a / b)^{2} \\ \Rightarrow & 1 / 2=a^{2} / b^{2} \\ \Rightarrow & 2 a^{2}=b^{2} \\ \Rightarrow & 2 \mid b^{2} \\ \Rightarrow & 2 \mid b \ldots \ldots \ldots \ldots \ldots \end{array}

\Rightarrow \quad 2 \mid \mathrm{b} \ldots \ldots \ldots \ldots \ldots (ii) \Rightarrow \quad b=2 c for some integer c\Rightarrow \quad b^{2}=4 c^{2} \Rightarrow \quad 2 a^{2}=4 c^{2} \Rightarrow \quad \mathrm{a}^{2}=2 \mathrm{c}^{2} \Rightarrow \quad 2 \mid \mathrm{a}^{2} \Rightarrow \quad 2 \mid \mathrm{a} \ldots \ldots \ldots \ldots(\mathrm{i})

From (i) and (ii), we can infer that 2 is a common factor of a and b. But, this contradicts the fact that a and b are co-primes. So, our assumption is incorrect.

Hence, 1/√2 is an irrational number.

(ii) 7√5

Let’s assume on the contrary that 7√5 is a rational number. Then, there exist positive integers a and b such that

7√5 = a/b where, a and b, are co-primes

⇒ √5 = a/7b

⇒ √5 is rational [∵ 7, a and b are integers ∴ a/7b is a rational number]

This contradicts the fact that √5 is irrational. So, our assumption is incorrect. Hence, 1/√2 is an irrational number.

(iii) 6 + √2

Solution:

Let’s assume on the contrary that 6+√2 is a rational number. Then, there exist coprime positive

integers a and b such that 6 + √2 = a/b

⇒ √2 = a/b – 6

⇒ √2 = (a – 6b)/b

⇒ √2 is rational [∵ a and b are integers ∴ (a-6b)/b is a rational number] This contradicts the fact that √2 is irrational. So, our assumption is incorrect.

Hence, 6 + √2 is an irrational number.

(iv) 3 − √5

Let’s assume on the contrary that 3-√5 is a rational number. Then, there exist co prime positive

integers a and b such that 3-√5 = a/b

⇒ √5 = a/b + 3

⇒ √5 = (a + 3b)/b

⇒ √5 is rational [∵ a and b are integers ∴ (a+3b)/b is a rational number] This contradicts the fact that √2 is irrational. So, our assumption is incorrect.

Hence, 3-√5 is an irrational number.

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