# Irrational number to an irrational power may be rational

This is an interesting problem. To solve it I should find two irrational numbers r and s such that r^{s} is rational.

I am not sure I am able to do that. However, I am confident that the following argument does solve the problem.

As we know, √2 is irrational. In particular, √2 is real and also positive. Then √2^{√2} is also real. Which means that it is either rational or irrational.

If it's rational, the problem is solved with r = √2 and s = √2.

Assume √2^{√2} is irrational. Let r = √2^{√2} and s = √2. Then r^{s} = (√2^{√2})^{√2} = √2^{√22} = √2^{2} = 2. Which is clearly rational.

Either way, we have a pair of irrational numbers r and s such that r^{s} is rational. Or do we? If we do, which is that?

(There's an interesting related problem.)

Chris Reineke came up with an additional example. This one is more direct and constructive. Both log(4) and √10 are irrational. However

√10^{ log(4)} = 10^{log(2)} = 2.

Stan Dolan showed that every rational number greater than 1/e^{1/e} is in the form a^{a} where a is either a positive integer or an irrational number.

### References

- T. Gowers (ed.),
*The Princeton Companion to Mathematics*, Princeton University Press, 2008, p. 157

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Copyright © 1996-2018 Alexander BogomolnySubject: | Irrational number to an irrational power may be rational |
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Date: | Fri, 27 Jun 2003 18:18:19 -0400 |

From: | Mario Bourgoin |

Dear Professor Bogomolny,

In your web page:

https://www.cut-the-knot.org/do_you_know/irrat.shtmltitled "Irrational number to an irrational power may be rational" you seek an irrational number which when raised to an irrational power is rational. In your proof of existence, you raise the square root of 2 to the square root of 2, and ask whether it is rational or irrational. This is answered on page 216 of Herstein's "Topics in Algebra", 2nd edition where we learn that in 1934, Gelfond and Schneider both proved (independently) that if numbers a and b are both algebraic and b is irrational, then a to the b is transcendental. Since you no doubt knew this, there remains to ask why you didn't mention this in your statement.

Sincerely,

--Mario Bourgoin

Well, this is a hard question. I am not sure if that's all that remains to ask. Certainly, with some effort, more questions could be raised. The site is not a text book, but a miscellany. I do not work on it on a regular basis. In all likelihood, an average page appearance and contents are due as much to my mood at the time of writing as to my assumed knowledge of the subject. Thank you for bringing that up. As you see, pages have both a spacial and a temporal life of their own.

To make it clear, what Mario says implies that √2^{√2} is transcendental because √2 is algebraic: it's one of the roots of the polynomial equation

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