Proving whether irrational powers of irrationals are rational

[mattmns]

Proof: One more irrationality proof :)

Ok, for this one I cannot even start the proof because I do not even know what I am trying to prove

The question states:

Prove of give a counterexample: there do not exist irrational numbers x and y such that x^y is rational.

Ok, let's knock out the counterexample, because I think that there is definitely not one. And now it is a proof, and the statement is an implication. But which way is the implication is what I cannot figure out.

Is it: If x and y are irrational, then x^y is irrational? Or, If x and y are rational, then x^y is rational?

Danke!

 

[StatusX]

The statement is not true. Think about logs.

 

[mattmns]

Ohh, well that makes it so that I do not even have to prove it. Sweet! I will mess around with that. Thanks.

However, what if it were ture. Then how would I write the implication so that I could prove it?

 

[Hurkyl]

there do not exist irrational numbers x and y such that x^y is rational.

That says:

It is false that there exists irrational numbers x and y such that x^y is rational.

right? How does negation propagate through quantifiers?



P.S. it's a simple matter to have figured out how to find the counterexample yourself. You were probably limiting your options!

Note that the counterexample consists of three parts:

An irrational number x
An irrational nubmer y
A rational number x^y

You probably tricked yourself into thinking you need to guess irrational x and y, hoping to get a rational x^y.

It's far easier to guess a rational x^y and an irrational x, and hope for an irrational y.

 

[mattmns]

Then it would be: It is true that for all irrational numbers x and y that x^y is irrational. (Which is going to be false because there is a counter example.

So then the implication would be: If x and y are irrational, then x^y is irrational. Correct? Thanks.

So when I am looking for a counterexample I should think about finding a counterexample of the contrapositive too, or break it into parts as you have.

So there are a infinite counterexamples. When you say it the other way around it is easy

$$\pi^{log_{\pi}42} = 42$$

Whenever I get a chance to make things whatever I want, I am supposed to make them cool numbers right?

Thanks!

 

[matt grime]

you proved thatlog of pi in base 42 is irrational did you? seems like oyu're assuming the answer.

anyway, here is *the* counter example.

conjsider sqrt(2), raise it to the power sqrt(2) what do you get? rational or irrational, if it's rational you've got a counter example, and even if not what can you do now to get a coutner example?