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LostInParadise's avatar

Can you solve this math problem, which is much simpler than it may at first appear?

Asked by LostInParadise (17690 points ) 2 months ago

This is a classic recreational math problem. If you have high school students at home, you may want to share this with them.

We want to show the existence of two irrational numbers a and b such that a^b is rational. To show that two such numbers exist, consider (sqrt(2)^sqrt(2))^sqrt(2)

All you need to know is:
1. (a^b)^c = a^(bc)
2. sqrt(2) is irrational
3. Whole numbers and fractions are rational

Please wait until tomorrow before answering to give others a chance. If nobody answers I will give the solution tomorrow evening.

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7 Answers

Mariah's avatar

I’m a little confused? The point of the problem is to show existence (>= 1 example that works) but you gave an example already in the problem statement? Doesn’t that answer it already?

bolwerk's avatar

ROT13 encoded answer: gur nafjre vf gjb

Here is the chicken scratch, but deliberately left incomplete.

a = √(2)
b = √(2)
c = √(2)

aᵇᶜ = aˢʳᵗ⁽²⁾ ˣ ˢʳᵗ⁽²⁾ = a² = √(2) × √(2)

srt = superscript squareroot

LostInParadise's avatar

@Mariah , You have to be able to demonstrate that a and b are irrational and that a^b is rational. The only assumption allowed that does not follow from the definition of the terms is that √(2) is irrational..

Mariah's avatar

Ahh, gotcha.

LostInParadise's avatar

As @bolwerk indicated, we get (√(2)^√(2))^√(2) = 2. It is interesting how all these square roots of 2 come together to equal 2. It is tempting to say that a=(√(2)^√(2)) and b =√(2). That would provide a solution to the puzzle if (√(2)^√(2)) is irrational. I know it certainly seems to be, and I read that there is long proof somewhere that it is true, but for the purposes of this problem it does not matter whether it is, because the only other possibility is that (√(2)^√(2)) is rational. In that case a=√(2) and b = √(2) and we have found a case where a^b is rational for irrational a and b.

Summarizing, (√(2)^√(2)) is either irrational or rational and in either case we get a solution to the problem.

Mariah's avatar

Ah, cool trick!

gasman's avatar

By coincidence I just came across something about this problem in this book from 1966 that I found recently at an estate sale. On page 63 they say:

Does 2^sqrt(2) have any meaning?...One is not able to tell by looking at them which irrational exponents produce rational quantities…sqrt(2)^sqrt(2) is either rational or irrational. If it is rational we need go no farther. If it is irrational, then…

After presenting same argument given by @LostInParadise they continue:

It turns out that the second alternative is the correct one. After long efforts of many mathematicians it was shown that 2^sqrt(2) is transcendental. Therefore [a little more calculation] sqrt(2)^sqrt(2) is transcendental…since x=2 we have an integer expressed as an irrational power of a transcendental number.

Apologies to Dover Books.
Thanks for a great math topic!

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