Question

Prove that the following are irrational:
(i) $\frac{1}{\sqrt{2}}$ (ii) $7\sqrt{5}$ (iii) $6+\sqrt{2}$

Answer 1

(i) $\frac{1}{\sqrt{2}}$

Let us assume $\frac{1}{\sqrt{2}}$ is rational.

So we can write this number as

$\frac{1}{\sqrt{2}}=\frac{a}{b}$ ---- (1)

Here, $a$ and $b$ are two co-prime numbers and $b$ is not equal to zero.

Simplify the equation (1) multiply by $\sqrt{2}$ both sides, we get

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

Now, divide by $b$, we get

$b=a\sqrt{2}$ or $\frac{b}{a}=\sqrt{2}$

Here, $a$ and $b$ are integers so, $\frac{b}{a}$ is a rational number,

so $\sqrt{2}$ should be a rational number.

But $\sqrt{2}$ is a irrational number, so it is contradictory.

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

(ii) $7\sqrt{5}$

Let us assume $7\sqrt{5}$ is rational.

So, we can write this number as

$7\sqrt{5}=\frac{a}{b}$ ---- (1)

Here, $a$ and $b$ are two co-prime numbers and $b$ is not equal to zero.

Simplify the equation (1) divide by 7 both sides, we get

$\sqrt{5}=\frac{a}{7b}$

Here, $a$ and $b$ are integers, so $\frac{a}{7b}$ is a rational

number, so $\sqrt{5}$ should be a rational number.

But $\sqrt{5}$ is a irrational number, so it is contradictory.

Therefore, $7\sqrt{5}$ is irrational number.

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

Let us assume $6+\sqrt{2}$ is rational.

So we can write this number as

$6+\sqrt{2}=\frac{a}{b}$ ---- (1)

Here, $a$ and $b$ are two co-prime number and $b$ is not equal to zero.

Simplify the equation (1) subtract 6 on both sides, we get

$\sqrt{2}=\frac{a}{b}-6$

$\sqrt{2}=\frac{a-6b}{b}$

Here, $a$ and $b$ are integers so, $\frac{a-6b}{b}$ is a rational

number, so $\sqrt{2}$ should be a rational number.

But $\sqrt{2}$ is a irrational number, so it is contradictory.

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