Analysis - irrational number or positive integer

[silimay]

Homework Statement

Let n and k be positive integers. Show that k^{1/n} is either a positive integer or an irrational number.

The Attempt at a Solution

I set q = k^{1/n}. Then I set q = \frac{m}{p}. (Where m and p don't have common factors.) Then m^n = k * p^n. So then k is a factor of m^n.

But here I get stuck. In other proofs they usually show that, like, then k must also be a factor of m, (but I don't know how to do that, if it is true), and then so m^n = an integer * k^2, so then k must also be a factor of p^n, which means that m and p do have a common factor.

But I get stuck in the middle.

 

[EnumaElish]

What are you assuming about m and p, are they positive integers? If so, haven't you assumed q is rational?

 

[silimay]

StatusX - Thanks so much for the help. If the factor divides m^n, how do I show that it also divides m?

EnumaElish - I am assuming m and p are positive integers with no common factors. So yeah, I have assumed q is rational. I am trying to prove that it cannot be rational (unless k is an nth power).

 

[JonF]

I think you are close but going about it the wrong way. The easiest way to prove things like this is to assume the opposite.

Suppose q isn’t irrational or an integer.

This means that q can be written in the form of a/b where a and b are relatively prime integers and b is non zero and non 1.

So q = k^(1/n) = a/b.

But this means that k = a^n/b^n

But what did we assume? That a and b are relatively prime, and clearly if a and b are relatively prime so are a^n and b^n. But this means that a^n/b^n isn’t an integer. But we assumed that k is an integer. This leads us to a contradiction, hence our assumption is false. QED.

Now since I gave you a pretty solid outline of the proof, a good way to make sure you really understand it would be to go back and see if you can spot where I used every assumption. Where did I use the fact that a,b and n are integers, where did i use b isn't 1 or 0? etc.

 

[StatusX]

Sorry, I deleted that post because it had a mistake. Just use the prime itself. That is, assume q is a prime factor of k whose power in k is not a multiple of n. Then q must divide m^n, and so clearly also m.

 

[silimay]

JohnF - Thanks, I think I understand better. That proof makes sense to me.

StatusX - I don't understand how q must divide m if it divides m^n.

 

[StatusX]

Because it's a prime. Just think about it a little.

 

[Gib Z]

There seems to be more conditions required (q=4, m=n=2)