Question

Proof that √5 is a irrational numbet

Answer 1

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PROOF:

. let us assume that$\sqrt{5}$ is not an irrational.

$\sqrt{5}$ is rational.

$\sqrt{5}$ = $\frac{p}{q}$ where p,q= 0 q is $\cancel{=}$ 0. p,q are co - primes.

squaring on both sides.

$(\sqrt{5})^2$ = ($\frac{p}{q})$${^2}$

5= $\frac{p{^2}}{q{^2}}$

${p}^2$= ${5×{q}^2}$

${p}^2$ is divisible by 5.

${p}$ is also divisible by 5.

$\boxed{let p= 5k.....1}$

from equation 1

$({5k}^2)$ = ${5×{q}^2}$

$\cancel{25}$${k}^2$= $\cancel{5}$${q}^2$

${5{k}^2}$ = ${q}^2}$

= ${q}^2}$ =${5{k}^2}$

${q}^2$ is divisible by 5 .

${q}$ is also divisible by 5

5 is the common factor for both p,q but p,q are co - primes

It is contradiction to our assumption

oure assumption is wrong.

$\sqrt{5}$ is not a rational

$\sqrt{5}$ is an irrational.

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