Prove or disprove that there is a rational number x and an i

[r0bHadz]

Homework Statement

Prove or disprove that there is a rational number x and an
irrational number y such that x^y is irrational

Homework Equations

The Attempt at a Solution

Please guys do not give me an answer. My only question is: what type of proof would you use?

It seems like with irrational numbers, proof by contradiction seems to be the best option in every proof I have encountered so far. The thing is, this course has a requirement of just calculus 1 and 2, and using Stewarts book I do not think he taught us anything about series expansions of irrational numbers unless I'm mistaken. I don't see how you can possibly answer this without that knowledge.

Like I said guys I really want to answer this my question myself. I do not want an answer, I found one on google already but I didnt read it. just simply: what proof would you use? My bet is proof by contradiction.

I'm going to sleep now, and I'm just interested in everyones responses, I will be back on this question in the morning.

As for the question itself, I don't see why it shouldn't be irrational. Sure, any irrational number is between two rational numbers, but I was taught that a irrational number never terminates. How could it possibly be rational if it's being raised to a number that never ends?

Anyways thanks guys, peace.

 

[LCKurtz]

Homework Statement

Prove or disprove that there is a rational number x and an
irrational number y such that x^y is irrational

Homework Equations

The Attempt at a Solution

How could it possibly be rational if it's being raised to a number that never ends?

I'm just replying to the last line. $\frac 1 3 = .333\dots$ which "never ends". What about $8^{\frac 1 3}$?

 

[YoungPhysicist]

How could it possibly be rational if it's being raised to a number that never ends?

Irrational numbers are numbers that can’t be a fraction, not never ends.

 

[RPinPA]

As for the question itself, I don't see why it shouldn't be irrational. Sure, any irrational number is between two rational numbers, but I was taught that a irrational number never terminates. How could it possibly be rational if it's being raised to a number that never ends?

Sure, your intuition leads you to expect the result is always irrational. But that's not a proof.

Proving the statement "there exists an $x$ and a $y$" means finding one example such that $x^y$ is irrational. So you'd make a choice of $x$ and $y$ and then go to a proof of irrationality, which might be amenable to proof by contradiction.

Disproving it means establishing that you can NOT make $x^y$ irrational. Thus you'd have to show that given any rational $x$ and irrational $y$, the result is rational.

My intuition, like yours, leads me to suspect that in most cases the result is irrational, so the statement is true, so the job is to find an example which is provable.

 

[haruspex]

Prove or disprove

proof by contradiction seems to be the best option

It rather depends which you are trying to do - prove or disprove.
If you want to prove the existence of something, the easiest, in general, is by construction. There are cases where non-constructive proofs are easier, but they are rare.
If you want to disprove an existence (prove a nonexistence) then, yes, contradiction is generally a good choice; assume such a thing does exist and arrive at a contradiction.

Though not for the reason you give, I also suspect such a pair exists, so I would start by trying to construct an example. If that suspicion is wrong, you will fail, but by the attempts may gain insight into why you fail.

 

[Dick]

The general case follows from the Gelfond-Schneider theorem. But you don't need that to prove an example exists if you know there are transcendental numbers. E.g. you know there is a number $x$ such that $2^x=e$. Use that $e$ is transcendental.