Prove root 2 i s irrational
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Sia ? 6 years, 4 months ago
Suppose {tex}\sqrt2{/tex} is a rational number. That is , {tex}\sqrt2{/tex} = {tex}\frac{p}{q}{/tex} for some p{tex}\in{/tex}Z and q {tex}\in{/tex}Z. We can assume the fraction is in lowest fraction, That is p and q shares no common factors.
Then {tex}\sqrt2q=p{/tex}
Squaring both side we get,
{tex}2q^2=p^2{/tex}
So {tex}p^2{/tex} is a multiple of 2,
let's assume {tex}p=2m{/tex}
Then, {tex}2q^2=\left(2m\right)^2{/tex}
{tex}2q^2=4m^2{/tex}
Or {tex}q^2=2m^2{/tex}
So {tex}q^2{/tex} is a multiple of 2,
{tex}\therefore{/tex} q is multiple of 2
Thus p and q shares a common factor.This is contradiction.
{tex}\Rightarrow {/tex}{tex}\sqrt { 2 }{/tex} is an irrational number.
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