Irrational Powers

Number Theory Level 3

True or False? If $a$ and $b$ are positive irrational numbers, then $a^b$ must be irrational as well.

True if we accept the Axiom of Choice False Cannot be decided True

1 solution

Brian Charlesworth
May 9, 2015

The answer is "False".

Proof by counterexample: Both $e$ and $\ln(2)$ , are positive and irrational, (transcendental, in fact), but $\large e^{\ln(2)} = 2$ is rational.

Existence proof: $\sqrt{2}$ is irrational. Now $\large \sqrt{2}^{\sqrt{2}}$ can be either rational or irrational. If it is rational then we are done. If it is irrational, we then have that

$\large (\sqrt{2}^{\sqrt{2}})^{\sqrt{2}} = \sqrt{2}^{2} = 2$ ,

i.e., an irrational raised to an irrational yielding a rational number.

Thank you for indulging me in this silly little problem!

I like your second proof better, since we can all prove with ease (as Plato could) that $\sqrt{2}$ is irrational. I must confess that I would be hard pressed to prove that $\ln{2}$ is irrational without knowing that $e$ is transcendental (which I would also be hard pressed to prove without refreshing my memory).

I'm surprised that you call the second proof "constructive"... I would have termed it an existence proof.

Otto Bretscher - 6 years, 1 month ago

Although the proofs of the irrationality of $e$ and $\ln(2)$ are indeed less "familiar" than those for $\sqrt{2},$ my counterexample was too compact not to mention. :)

And yes, you're right, my second proof is more an "existence" than "constructive" proof, so I'll make that edit. The proof doesn't actually tell us whether or not $\sqrt{2}^{\sqrt{2}}$ is rational, just that one way or the other there exists a counterexample to the given statement. I suppose that, given the definition , the second proof is not of the pure existence variety since I have constructed a "method" to provide an example, although not in the truest sense.

Brian Charlesworth - 6 years, 1 month ago

Another counterexample: $a = \sqrt 2, b = \log_2 9$ .

Pi Han Goh - 6 years, 1 month ago

Very nice! Here it's easy to see that both a and b are irrational! Thanks!

Otto Bretscher - 6 years, 1 month ago

CHALLENGE MASTER NOTE: For algebraic numbers $a,b$ with $a\ne 0,1$ and $b$ irrational, find the value(s) of $a$ and $b$ such that $a^b$ is rational.

Pi Han Goh - 6 years ago

@Pi Han Goh – That is not going to happen, by Gelfond-Schneider

Otto Bretscher - 6 years ago

@Otto Bretscher – Oh, I thought I can trick you =C

Pi Han Goh - 6 years ago

@Pi Han Goh – You can, often, but not this time... ;)

Otto Bretscher - 6 years ago

@Otto Bretscher – Cool. That's definitely my new theorem for the day. :)

Brian Charlesworth - 6 years ago

If $a=\sqrt { 7 } ,b=\sqrt { 2 }$ , then how could we prove that ${ a }^{ b }$ is rational

Vighnesh Raut - 6 years, 1 month ago

What makes you think that it's rational?

Pi Han Goh - 6 years, 1 month ago

This shows that the statement in the question is sometimes true as well.

Vighnesh Raut - 6 years, 1 month ago

@Vighnesh Raut – You misinterpreted the question. It didn't say that all irrational $a,b$ produces irrational $a^b$ . It's like asking: True/False: "If, $c,d$ are prime numbers, then $c+d$ MUST be an even number".

Pi Han Goh - 6 years, 1 month ago

@Pi Han Goh – What if $c=2$ and $d$ is some other odd prime number, then $c+d$ is odd as well :D It isn't necessary that ${ a }^{ b }$ is always rational for irrational $a$ and $b$ . So, according to me , answer should be "CANNOT BE DECIDED" because it depends on what $a$ and $b$ we choose. Please correct me if I am wrong.

Vighnesh Raut - 6 years, 1 month ago

@Vighnesh Raut – You still misinterpreted it. The question asked can be rephrased as such: "Is it always true that no matter what irrational numbers I choose, say $a$ and $b$ , the resultant number $a^b$ will ALWAYS be an irrational number?"

Same goes with my other question. "Is it always true that no matter what two prime numbers I choose, their sum will ALWAYS be an even number?"

Pi Han Goh - 6 years, 1 month ago

@Pi Han Goh – Oh...thanks.... I interpreted must be as may be which was wrong. Thanks for helping me out.

Vighnesh Raut - 6 years, 1 month ago

@Vighnesh Raut – With a statement that says something is $always$ the case, it is sufficient to provide a counterexample, (either direct or existential), to prove that the statement is false.

Brian Charlesworth - 6 years, 1 month ago

Irrational Powers

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