Question

prove that 7√2 is an irrational number.

Answer 1

Let as assume the opposite 7√2 is rational number.

i.e.

7√2 is a Rational number.

$\sf \frac{7}{ \sqrt{2} } = \frac{a}{b}$

Where a, b ≠ 0 ,

and a, b are Co-prime.

Here,

$7 \sqrt{2} = \frac{a}{b}$

$\sqrt{2} = \frac{1}{7}$

${\bold{\green{\underline{\underline{We \: know}}}}}$

$\sqrt{2}$ is an irrational number , or $\frac{1}{7}$ is a rational number .

Here,

Here,

$\sf \frac{a}{7b \: } \: is \: a \: rational \: number$

But √2 is irrational number

Since, Rational ≠ Irrational

This Contradiction

∴ Our Assumption is incorrect.

Hence 7√2 is an irrational number.

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@Mahor2111

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