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

[RJLiberator]

Homework Statement

Let a and n be positive integers. Prove that a^(1/n) is either an integer or is irrational.

Homework Equations

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?

 

[Samy_A]

Homework Statement

Let a and n be positive integers. Prove that a^(1/n) is either an integer or is irrational.

Homework Equations

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?

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

 

[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?

 

[Samy_A]

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?

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.)

 

[RJLiberator]

Which kind of real numbers can not be expressed as x/y where y does not divide x?

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.

 

[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.

 

[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.

 

[WWGD]

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.

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