Confirmation of irrational proof

[Jbreezy]

Homework Statement

Prove $\sqrt n$ is irrational

Homework Equations

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.

 

[micromass]

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

 

[HallsofIvy]

Homework Statement

Prove $\sqrt n$ is irrational

Homework Equations

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.

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?

 

[Jbreezy]

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

Would if fail with this assumption right off the bat?

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

 

[micromass]

Would if fail with this assumption right off the bat?

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

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

 

[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

 

[Ray Vickson]

Homework Statement

Prove $\sqrt n$ is irrational

Homework Equations

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.

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.

 

[Jbreezy]

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.

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?

 

[Jbreezy]

This is crap. I need to think