Question

Prove that the following are irrationals :

1. $\dfrac{1}{\sqrt 2}$
2. $7\sqrt 5$
3. $6 + \sqrt 2$

Answer 1

(i) $\dfrac{1}{\sqrt2}$

Let $\dfrac{1}{\sqrt2}$ is rational.
Therefore, we can find two co-prime integers $a, b \ (b ≠ 0)$ such that
\dfrac{1}{\sqrt2}$$=\dfrac{𝑎}{b}
Or $\sqrt 2=\dfrac{b}{a}$
$\frac ba$ is rational as a and b are integers.
Therefore, $\sqrt 2$ is rational which contradicts the fact that $\sqrt 2$ is irrational.

Hence, our assumption is false and $\dfrac{1}{\sqrt 2}$ is irrational.

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(ii) $7\sqrt 5$

Let $7\sqrt 5$ is rational.
Therefore, we can find two co-prime integers a, b (b ≠ 0) such that
7\sqrt 5$$=\dfrac{𝑎}{𝑏}
$⇒\sqrt 5=\dfrac{𝑎}{7𝑏}$
$\dfrac{𝑎}{7𝑏}$ is rational as a and b are integers.
Therefore, $\sqrt 5$ should be rational.

This contradicts the fact that $\sqrt 5$ is irrational. Therefore, our assumption is that $7\sqrt 5$ is rational is false. Hence, $7\sqrt 5$ is irrational.

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(iii)** ** $6+\sqrt 2$

Let $6+\sqrt 2$ be rational.
Therefore, we can find two co-prime integers $a, b\ (b ≠ 0)$ such that

$6+\sqrt 2=\dfrac{𝑎}{𝑏}$

$⇒\sqrt 2=\dfrac{𝑎}{𝑏} -6$
Since a and b are integers,$(\dfrac{𝑎}{𝑏} -6)$ is also rational, and hence, $\sqrt 2$ should be rational.
This contradicts the fact that $\sqrt 2$ is irrational.
Therefore, our assumption is false, and hence, $6+\sqrt 2$ is irrational.