# Prove that a^(1/n) is an integer or is irrational

1. Jan 30, 2016

### RJLiberator

1. The problem statement, all variables and given/known data
Let a and n be positive integers. Prove that a^(1/n) is either an integer or is irrational.

2. Relevant equations

3. The attempt at a solution
Proof:
If a^(1/n) = x/y where y divides x, then we have an integer.
If a^(1/n) = x/y where y does not divide x, then
a = (a^(1/n))^n = x^n/y^n is NOT an integer since y^n does not divide x^n. However, this is a contradiction as we declared a to be a positive integer.
Thus, a^(1/n) must be an integer.

However, is neglecting the important part of irrationality.
In my proof, I have convinced myself that a^(1/n) is an integer. But this is obviously not true as 4^(1/3) is irrational.
Where did I go wrong?

Perhaps there is another case?

2. Jan 30, 2016

### Samy_A

Looks ok until :"Thus, a^(1/n) must be an integer." Look more carefully at which hypothesis led to a contradiction.

3. Jan 30, 2016

### RJLiberator

So the hypothesis that led to the contradiction is "if a^(1/n) = x/y where y does not divide x"

This seems to imply that y does divide x. But the first if statement shows that If a^(1/n) = x/y where y divides x, then we have an integer.

Perhaps you are telling me that If a^(1/n) = x/y where y does not divide x, then we have an irrational number.
Is that the connection that I was missing?

4. Jan 30, 2016

### Samy_A

Correct, a^(1/n) = x/y where y does not divide x leads to a contradiction.

Which kind of real numbers can not be expressed as x/y where y does not divide x? (Although you didn't state it explicitly, I assume that x and y are integers.)

5. Jan 30, 2016

### RJLiberator

Irrational numbers!

Proof:
If a^(1/n) = x/y where y divides x, then we have an integer.
If a^(1/n) = x/y where y does not divide x, then
a = (a^(1/n))^n = x^n/y^n is NOT an integer since y^n does not divide x^n. However, this is a contradiction as we declared a to be a positive integer.
And so a^(1/n) cannot be expressed as a rational number or an integer, therefore it is irrational.
end.

So we have covered both possibilities and showed the routes to an integer and to an irrational number.

6. Jan 31, 2016

### WWGD

Hi again, RJ:If you want to generalize this, you can use the proof of the irrationality of $\sqrt 2$ to show that the n-th root of $p/q$ is rational only if $p/q =a^{nj}/b^{nk}$ for integers $a,b,n,j,k$. Or it is just a good result to know, to keep handy, even without a proof.

7. Jan 31, 2016

### RJLiberator

Yes, We went over that proof in class prior to this homework question. I also noticed it elsewhere on the web when I checked in with this problem.

Seems like a highly popular proof to know.

8. Jan 31, 2016

### WWGD

Yes, I see it as a fun combo of number theory and Calculus/Analysis, which one may believe at first have little overlap.