## anonymous 5 years ago Does this seem legit?

1. anonymous

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. $$\blacksquare$$

2. JamesJ

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.

3. anonymous

once you prove the product of irrational and rational is irrational

4. JamesJ

of course

5. JamesJ

...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?

6. anonymous

Hi pre-algebra :) wanna chat

7. anonymous

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

8. JamesJ

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

9. JamesJ

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.

10. anonymous

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.

11. anonymous

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