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Prove that there exist irrational numbers $a$ and $b$ such that $a^{b}$ is rational.

What i tried

Prove by contradiction

I assume the statement

For all rational numbers $a$ and $b$ such that $a^{b}$ is rational

Since $a$ and $b$ are assumed to be rational, they can be expressed in the form

$$a=\frac{c}{d}$$

$$b=\frac{x}{y}$$

then $$a^{b}=(\frac{c}{d})^\frac{x}{y}$$

From here we know that $$(\frac{c}{d})^\frac{x}{y}$$

doesent necessarily have to be rational as can be seen from the example if $x=1$ and $y=2$, hence $a^{b}$ dosent necessarily have to be rational. But from our assumption, we assumed that $a^{b}$ have to be rational. Hence a contradiction and thus proving the original statement. Is my proof correct. Could anyone explain. Thanks

ys wong
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  • See http://math.stackexchange.com/questions/104119/can-an-irrational-number-raised-to-an-irrational-power-be-rational and http://www.cut-the-knot.org/do_you_know/irrat.shtml – lab bhattacharjee Aug 28 '15 at 16:58
  • If you "know that $(c/d)^{x/y}$ doesn't have to be rational" and you also give an example, why don't you give just an counterexample as proof? – Antoine Aug 28 '15 at 17:00
  • I hate to tell this, but your argument is almost totally unrelated to the claim, as it begins with an erroneous contraposition – Hagen von Eitzen Aug 28 '15 at 17:07

2 Answers2

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This is a classic problem:

Note that $\sqrt{2}$ is irrational.

If $\sqrt{2}^\sqrt{2}$ is rational then you are done.

If $\sqrt{2}^\sqrt{2}$ is irrational, then $(\sqrt{2}^\sqrt{2})^\sqrt{2} = \sqrt{2}^2 = 2$ is rational, and satisfies the desired constraint of "irrational to the power of an irrational... is rational."

Benjamin Dickman
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This is a very popular problem. Here is the solution.

If $\sqrt{2}^{\sqrt{2}}$ is rational, you are done. Otherwise, it is irrational.

Then take $$\left( \sqrt{2}^{\sqrt{2}} \right)^{\sqrt{2}} = \sqrt{2}^{\sqrt{2}\sqrt{2}}= \sqrt{2}^2=2$$ which is rational.

In any case, you have that there exist some two irrational numbers $a,b$ such that $a^b$ is rational.

Crostul
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  • Oh Thanks but is my way of doing also considered correct? – ys wong Aug 28 '15 at 17:03
  • No, your argument is totally incorrect. – Crostul Aug 28 '15 at 17:16
  • Oh but why is my argument incorect? – ys wong Aug 28 '15 at 17:18
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    As it is stated in the comments above, in order to prove your statement by contradiction, you should begin assuming that "for all $a,b>0$ irrational, $a^b$ is irrational" and get some contradiction. In other words, you made a mistake in taking the negation of your statement. – Crostul Aug 28 '15 at 17:22