Confirmation of irrational proof

1. Aug 10, 2013

Jbreezy

1. The problem statement, all variables and given/known data

Prove $\sqrt n$ is irrational

2. Relevant equations

3. The attempt at a solution

Assume $p^2/q^2 = n$ is an irreducible fraction.
If $p^2 = nq^2$, then q is a multiple of n. Call this $p' = nq$
substituting this for our original equation. We get $p'^2 = nq^2$
Implies $(nq)^2 = nq^2$ thus our original assumption is false.

Is this horse crap? Don't just give it away if it is wrong please just indicate where I went wrong.

2. Aug 10, 2013

micromass

Staff Emeritus
Where does this proof fail if $n=4$?

3. Aug 10, 2013

HallsofIvy

Staff Emeritus
Does it bother you at all that the statement you claim to have proved is NOT true? What if, say, n= 4? What happens to your proof in that case?

It is, of course, true if you add the requirement that n is NOT a "perfect square". How have you used that requirement in your proof?

4. Aug 10, 2013

Jbreezy

Would if fail with this assumption right off the bat?

Assume $p/q = \sqrt4$ is an irreducible fraction.
Because $p/q = 2$

5. Aug 10, 2013

micromass

Staff Emeritus
OK, so your proof does fail for some $n$. Your proof should make this clear.

6. Aug 10, 2013

Jbreezy

Yeah, I should of just included the condition because it was given in the book just excluded it in my post.
Thanks for the help

7. Aug 11, 2013

Ray Vickson

Your argument is incomplete. Essentially, you claim that if p^2 is divisible by n then p itself is also divisible by n. You cannot just say it---you need to prove it.

BTW: it is not always true for all integers n, so something about n needs to be specified.

8. Aug 12, 2013

Jbreezy

Yeah, I fooled myself. Let's see if I did it again. What about:

$p/q = \sqrt n$, Assume $(p,q) = 1$ and $n$ is not a perfect square.

Squaring, we have, $p^2 = nq^2$

If $n=/= 1$ There exists a prime integer $k$ such that $k/n$

Substitution gives:

$p^2 = (n/k)q^2.$ Implies that $kp^2 = nq^2$ and since we know that that k divides n, $p^2$ and $q^2$ share a common factor.Thus, our assumption $(p,q) = 1$ is false so $\sqrt n$ is irrational.

What do you think Ray trick myself again?

9. Aug 12, 2013

Jbreezy

This is crap. I need to think