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Let $5=\frac ab$ $\forall\ a,b\ \epsilon\ N$. And $(a,b)=1$
Squaring both sides,
$25b^2=a^2$
Thus, $25|a^2$; $25|a$
So $a=25m$
Substituting, $25b^2=25^2m^2$
So $b^2=25m^2$
So $25|b$ (By the same logic used before).
But are assumption is proved to be wrong, because $25$ comes to be the common factor. So contradiction, proving that $5$ is not rational. So how is it possible?

Aditya Agarwal
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    Generally with fake proofs like this you should take a known counterexample and see where your proof fails. Clearly $a=5,b=1$ is a solution, so work from there and see where you come to a false conclusion. In this case, you see that the implication that if $25\mid a^2$, then $25\mid a$ is false. – msinghal Jul 28 '15 at 11:07

2 Answers2

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If 25 divides $a^2$ all you can conclude is that $5$ divides $a.$

kodlu
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But you can't infer from 25 divides $a^2$ that 25 divides $a$.

Countereaxample: $a = 5$.

Peter Smith
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