### Chapter 1: Real Numbers

Q
##### Real Numbers Solutions

Question:

Prove that the following are irrationals : (i) 1/√2 (ii) 7√5 (iii) 6 + √2

1.     $\frac{1}{\sqrt{2}}$

To prove that  $\frac{1}{\sqrt{2}}$  is an irrational number.

Let us assume that  $\frac{1}{\sqrt{2}}$   is rational. Then  $\frac{1}{\sqrt{2}}$   =   $\frac{a}{b}$

Since a and b are co primes with only 1 common factor and b≠0.

⇒ $\frac{1}{\sqrt{2}}$   =   $\frac{a}{b}$

$\frac{a}{b}$  is a rational number, so  $\sqrt{2}$ ​ should be a rational number.

But   $\sqrt{2}$​ is an irrational number, so it is contradicted.

Therefore,  $\frac{1}{\sqrt{2}}$  is an irrational number.

(ii) 7$\sqrt{5}$

To prove that 7$\sqrt{5}$ is an irrational number

Let us assume that  7$\sqrt{5}$  is rational. Then 7$\sqrt{5}$  =  $\frac{a}{b}$

Since a and b are co primes with only 1 common factor and b≠0.

⇒ 7$\sqrt{5}$  =  $\frac{a}{b}$

Now divide by 7 into both sides,  we get

$\sqrt{5}$  =  $\frac{a}{7b}$

Here a and b are integer so $\frac{a}{b}$  is a rational number, so $\sqrt{5}$​ should be a rational number.

But  $\sqrt{5}$ ​ is an irrational number, so it is contradicted.
Therefore, 7$\sqrt{5}$ is an irrational number.

(iii) 6 + $\sqrt{2}$

To prove that 6 + $\sqrt{2}$ is an irrational number

Let us assume that 6 + $\sqrt{2}$  is rational. Then 6 + $\sqrt{2}$  = $\frac{a}{b}$

Since a and b are co primes with only 1 common factor and b≠0.

⇒ 6 + $\sqrt{2}$  =    $\frac{a}{b}$

$\sqrt{2}$ = $\frac{a}{b}$   – 6

$\sqrt{2}$ =    $\frac{a-6b}{b}$

Here a and b are integer so $\frac{a-6b}{b}$   is a rational number, so  $\sqrt{2}$ ​ should be a rational number.

But  $\sqrt{2}$ ​ is an irrational number, so it is contradicted.

Therefore, 6 +
$\sqrt{2}$  is an irrational number.

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