Question

prove that √5 is irrational​

Answer 1

Answer:

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Answer 2

$\large\bf\underline\red{Answer \: :-}$

Let √5 be a rational number.

then it must be form p/q

Where,

$\sf{p \neq q \: \: \: ( p \: and \: q \: are \: co \: prime) }$

$\longmapsto\sf{ \sqrt{5} = \frac{p}{q} } \: \: \: \: \: \: \: \: \\ \\ \longmapsto\sf{ \sqrt{5} \times q = p }$

Squaring of the both sides,

$\bigstar \: \: \bf\purple{5q {}^{2} = p {}^{2} \: \: \: \: \: \: \: \: \: \: ...(1) }$

p² is divisible by 5

So,

p = 5c

Squaring of the both sides,

$\bigstar \: \: \bf\purple{p {}^{2} = 25c {}^{2} \: \: \: \: \: \: \: \: \: \: ...(2) }$

Put p² in equation (1)

$\longmapsto\sf{5q {}^{2} = 25(c) {}^{2} } \\ \longmapsto\sf{q {}^{2} = 5c {}^{2} } \: \: \: \: \: \: \: \:$

So, q is divisible by 5

Thus p and q have 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.

$\therefore\bf{\sqrt{5} \: is \: an \: irrational \: number.}$

Hence Proved

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