Refute the assertion that $a^b$ is irrational for all irrational real numbers $a$,$b$.

Question

I found this exercise in the first tome of Ken Binmore's Foundations of Analysis.

I know that similar questions exist in MSE so I will try to avoid the duplicate. I know also Eugenia Cheng (2004) found this proof unsatisfying.

Nevertheless I have two questions: Why the autor chose to work with $\sqrt{2}$ in the first place in his search of counterexample? I have never seen similar non-constructive reasoning. What other proofs or refutations follow such an approach?

Answer 1

The point of the proof is not in taking $\sqrt 2$ exactly. The point is in taking two irrational numbers that multiply into a rational number, and that there exist irrational numbers that, taken to a rational power, are rational.

In particular, take any irrational $\alpha_1, \beta_1, \beta_2$ with the property that $\alpha_1^{\beta_1\cdot \beta_2} \in\mathbb Q$.

Then the proof can be done in the same way, by separating two cases:

1. If $\alpha_1^{\beta_1}$ is rational, then $\alpha_1, \beta_1$ provide a counterexample.
2. If $\alpha_1^{\beta_1}$ is irrational, then take $\alpha_2 = \alpha_1^{\beta_1}$, and $\alpha_2, \beta_2$ provide a counterexample, since$$\alpha_2^{\beta_2} = \alpha_1^{\beta_1\cdot \beta_2}\in\mathbb Q$$

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Note, however, that in many ways, $\alpha_1=\beta_1=\beta_2\sqrt 2$ is the simplest possible version of the counterexample. But you could also take:

• $\alpha_1=\sqrt[23976643]{290734601983798456321}$
• $\beta_1=e^{187638932679} - 23438972\pi + 39232186\cdot \sqrt{89327467}$
• $\beta_2=\frac{23976643}{e^{187638932679} - 23438972\pi + 39232186\cdot \sqrt{89327467}}$

and the proof would work just as well.

Answer 2

Well, there are other counterexamples, but they might be harder for one reason or another. Take $\sqrt[3]{3}$. Then, $$\left(\left(\left(\sqrt[3]{3}\right)^{\sqrt[3]{3}}\right)^{\sqrt[3]{3}}\right)^{\sqrt[3]{3}}=\left(\left(\sqrt[3]{3}\right)^{\sqrt[3]{3}}\right)^{\sqrt[3]{3^2}}=\left(\sqrt[3]{3}\right)^{3}=3.$$ However, here we have an extra exponent, and there are more numbers to consider as candidates for rationality and irrationality. Also, it's slightly harder to prove that $\sqrt[3]{3}$ is irrational. For this style of proof, $\sqrt{2}$ seems to be the simplest one because it leaves us only with the two possibilites presented.

For an easier-looking example, we could take the numbers $e$ and $\ln(5)$. It is know that these are both irrational numbers, but $e^{\ln(5)}=5$ is rational. Why not choose this example? In my opinion, it is also because showing that $e$ is irrational takes a lot more effort than showing the same for $\sqrt{2}$ does.

As for proofs similar to this one, I don't remember to have seen any. However, there are many non-constructive elementary proofs and theorems. An example that comes to mind is the Intermediate Value Theorem, which states that if a function is continuous on the interval $[a,b]$, and $y$ is a value between $f(a)$ and $f(b)$, then there exists $c\in(a,b)$ such that $y=f(c)$. What is $c$? Who knows, but one can prove it exists without constructing it.