Does this seem legit?

Question

anonymous · Answer

Prove that $\sqrt{8}$ is irrational.

Proof: Assume $\sqrt{8}$ is rational. Then, for some integers $p$ and $q$ where $q\neq0$ and $p$ and $q$ have no common factors, we have $\sqrt{8}=p/q$, $8=p^2/q^2$, $p^2=8q^2$, $p^2=2(4q^2)$. This implies that $p^2$ is an even number, which in turn implies $p$ is an even number. Therefore, for some integer $r$, we can rewrite $p=2r$. Substituting, we obtain $r^2=2q^2$. Similarly, this implies $r$ is an even number, and, for some integer $s$, we can rewrite $r=2s$. Substituting, we obtain $2s^2=q^2$. This implies that $q$ is an even number. Therefore, we have shown that both $p$ and $q$ are even numbers, and must then have the common factor $2$. This is a contradiction of our hypothesis, hence $\sqrt{8}$ must be an irrational number. $\blacksquare$

jamesj · Answer

Yes, it's a straight imitation of the proof $ \sqrt{2} $ is irrational.

Which suggests another proof.  If you know that $ \sqrt{2} $ is irrational, then
$$ \sqrt{8} = 2\sqrt{2} $$
must be irrational also.

anonymous · Answer

once you prove the product of irrational and rational is irrational

jamesj · Answer

of course

jamesj · Answer

...but I know pre-alg has done that already.  

Now, here's a small challenge: what is it about this proof that wouldn't work for $ \sqrt{4} $.  Suppose, in other words, we didn't know this was equal to 2.  What about imitating to proof above fails?

anonymous · Answer

Hi pre-algebra :) wanna chat

anonymous · Answer

not to be a pain, but i always wonder why this is the preferred proof. if you want to prove that the square root of any natural number that is not a perfect square is irrational, the fundamental theorem of arithmetic gets it quickly. the "even/odd" bit seems unnecessarily forced, just because we have a word "even" for divisible by 2, and no such word for "divisible by 7" for example

jamesj · Answer

**correction: what is it about imitating the proof above that fails?

jamesj · Answer

I think it's preferred because it very elementary and doesn't require any other larger results.  It's something you can explain to a HS school student.

anonymous · Answer

in my experience, the problem with it is that after a minute or two the "even/odd" bit gets all muddled up. i am not exactly sure, but i believe all the proofs use the method of infinite descent somewhere.

anonymous · Answer

Well i am soo not intrested in this conversation anyomrre :p