Question

prove that √5 irrational number​

Answer 1

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Answer 2

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$\sqrt{5} \: is \: a \: irrational \: number$

$prove↓↓↓$

Let √5 be a rational number.

$then \: it \: must \: be \: in \: form \: of \: \frac{p}{q} \: where \: , q ≠ 0 \: ( p \: and \: q \: are \: co-prime)$

$\sqrt{5} = \frac{p}{q}$

$\implies{\sqrt{5} \times q =p}$

Squaring on both sides,

$5q² = p² \: –––(1)$

p² is divisible by 5.

So, p is divisible by 5.

$p=5c$

Squaring on both sides,

$p²=25c² \: –––(2)$

Put p² in equation(1)

$5q² = 25(c)²$

$\implies{q² = 5c²}$

So, q is divisible by 5.

Thus p and q have a common factor of 5.

So, there is a contradiction as per our assumption.

We have assumed p and q are co-prime but here they a common factor of 5.

The above statement contradicts our assumption.

Therefore, √5 is an irrational number.

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