Question

proof that route 2 is irrational

Answer 1

substituting for a we get to b² is equals to 4c² that is b² is equals to 2 c² this means that 2 divides b square so to divides b ,a and b have at least 2 to as a common factor but this contradicts the fact that a and b have no common factor other than 1 as they are co-prime this contradiction has a rise in because our in correct assumption that that is root 2 is irrational so we conclude that root 2 is irrational

Answer 2

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PROOF:

. let us assume that$\sqrt{2}$ is not an irrational.

$\sqrt{2}$ is rational.

$\sqrt{2}$ = $\frac{p}{q}$ where p,q= 0 q is $\cancel{=}$ 0. p,q are co - primes.

squaring on both sides.

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

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

${p}^2$= ${2×{q}^2}$

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

${p}$ is also divisible by 2.

$\boxed{let p= 2k.....1}$

from equation 1

$({2k}^2)$ = ${2×{q}^2}$

$\cancel{4}$${k}^2$= $\cancel{2}$${q}^2$

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

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

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

${q}$ is also divisible by 2.

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

It is contradiction to our assumption

oure assumption is wrong.

$\sqrt{2}$ is not a rational

$\sqrt{2}$ is an irrational.

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