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Isn't 2^(0.5) rational?

Method for proving: Contradiction. So show not P:2^(0.5) is rational.

Yasmina Anand
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  • $\sqrt7$ and $\sqrt2$ are irrational. And not every pair of irrational numbers has irrational difference (counterexample: $(\sqrt2,\sqrt2+1)$) – user_194421 Feb 08 '18 at 04:36
  • https://math.stackexchange.com/questions/457382/can-sqrtn-sqrtm-be-rational-if-neither-n-m-are-perfect-squares/457384#457384 – lab bhattacharjee Feb 08 '18 at 04:37

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Let $7^{\frac 12} = a$ and $2^{\frac 12} = b$.

If you're allowed to assume (or you can show, and it's quite easy) that both $a$ and $b$ are irrational, then the problem becomes very easy.

Assume to the contrary that $a-b = X$ is rational.

Then $a+b = Y$ has to be irrational because $X+Y = 2a$ which is irrational (as assumed or previously shown).

Now consider $XY = (a+b)(a-b) = a^2 - b^2 = 7-2 = 5$.

This is rational. But the product of a rational ($X$, by assumption) and an irrational ($Y$, as deduced) number cannot be rational. We have arrived at a contradiction. Hence $X$ is irrational.

Deepak
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