Irrational number to an irrational power

[Mabs]

Homework Statement

if a and b are irrational numbers, is a^b necessarily an irrational number ? prove it.

The Attempt at a Solution

this is an question i got from my first maths(real analysis) class (college) , and have to say, i have only little knowledge about rational number, i would like to give it a try to find a solution but need some help, i know that a^b is not necessarily an irrational number cause if a = √2^√2 and b = √2 it becomes rational.

but i don't think its the proof that the question is expecting as its on real analysis. any help would be greatly appreciated and sorry for my English too.

 

[CAF123]

You cannot prove that statement since, as you have shown via a counterexample, it is false. The question might have said:

'If a and b are irrational numbers, is a^b necessarily irrational? Find a counterexample or prove it.​

 

[Mabs]

sorry for that mate it actually says to prove my answer , yes or no. i know its no but i have no idea how to prove it.

 

[Borek]

Showing the counterexample you have proven the statement to be false. So the answer is "no" and the counterexample is the proof.

 

[collinsmark]

Showing the counterexample you have proven the statement to be false. So the answer is "no" and the counterexample is the proof.

Assuming that one also starts with the proof that (√2)^√2 is irrational.

 

[Borek]

Good point

 

[epenguin]

Assuming that one also starts with the proof that (√2)^√2 is irrational.

Don't need to prove that, since if it is rational we have found our counterexample. However if it is irrational, I don't see the proof that (√2)^(√2)(√2) is rational.

 

[dextercioby]

It's not the tower power of sqrt(2), rather

$$\left(\sqrt{2}^{\sqrt{2}}\right) ~ ^{\sqrt{2}} $$

which gives you the desired counterexample to the question in the problem, iff you manage to prove that the nr. inside the bracket is irrational.

 

[collinsmark]

Don't need to prove that, since if it is rational we have found our counterexample.

Also a good point.

 

[epenguin]

Also a good point.

Independently of whether or not (√2)^(√2)(√2) and ((√2)^(√2))^(√2) are the same thing.

It's not the tower power of sqrt(2), rather

$$\left(\sqrt{2}^{\sqrt{2}}\right) ~ ^{\sqrt{2}} $$

which gives you the desired counterexample to the question in the problem, iff you manage to prove that the nr. inside the bracket is irrational.

If you did, how would it follow that your expression here is rational?

 

[Mabs]

thanks guys , proving sqrt2 ^sqrt2 is irrational , need to try on that one . thanks for the help.

 

[Mentallic]

If you did, how would it follow that your expression here is rational?

$$(\sqrt{2}^{\sqrt{2}})^{\sqrt{2}}=(\sqrt{2})^{\sqrt{2}\cdot \sqrt{2}}=(\sqrt{2})^2=2$$

 

[Curious3141]

thanks guys , proving sqrt2 ^sqrt2 is irrational , need to try on that one . thanks for the help.

That's a toughie. Look up the Gelfond-Schneider theorem. With the theorem, it's trivial, but proving that theorem is very hard.

But as for the original question, this is a well-known proof that's often cited as a classic non-constructive proof. The proposition is that an irrational raised to an irrational power can be rational.

So we consider $x = \sqrt{2}^\sqrt{2}$. Either $x$ is rational or irrational. If it's the former, our work is done. If the latter, then let $y = x^\sqrt{2} = \sqrt{2}^2 = 2$. In this case, we've taken an irrational ($x$), raised it to an irrational power ($\sqrt{2}$) and the result is rational. QED.

It's a non-constructive proof because we didn't actually commit ourselves as to whether $x$ was rational or irrational. But we "covered our bases" either way.

 

[Mabs]

thanks mate, can anyone recommend me some books for additional reading on rational numbers ( from the basics ) , it would be great .