# Analysis Math Questions

1. Jan 20, 2009

### Zero266

Prove using the multiplication axioms that if x is not zero, then 1 / (1/x) is equal to x.

Prove that there is no rational number, p, such that p^2 = 12

I know for all x there exists x^-1 such that xx^-1 = 1 but i don't know how to use that to prove the first one.

For the second one, I understand the proof for p^2 = 2 by contradiction by showing that it was not reduced to lowest term because both a and b in a/b turned out to be even, but i can't seem to duplicate the process for 12.

2. Jan 20, 2009

### snipez90

For the first one, it's a good idea to show that multiplicative inverses are unique, but I don't think you need to (someone else will correct me if I'm wrong).

Basically, you need to first check that if x =/= 0, then (1/x) =/= 0. You can do this by contradiction. Then you basically apply the multiplicative inverse property to (1/x), and you can do this because you have established (1/x) =/= 0.

Eh, the duplication process should be the exact same thing. There are a lot of ways to do this since 12 has a few divisors. Suppose (a/b)^2 = 12, then a must be divisible by 12. So let a = 12k.

EDIT: Actually, it's somewhat important to show that multiplicative inverse are unique. Perhaps the easiest way to do this is by proving that if S is a field and x,y are elements of S with x =/= 0 and xy = 1 for all x,y, then y = x^-1. This follows almost immediately from the axiom.

Last edited: Jan 20, 2009
3. Jan 20, 2009

### Zero266

Thanks a lot. Maybe i'm just stupid, but why is "a" a multiple of 12? (a/b)^2 = 12, so a^2 is certainly a multiple of 12, but i'm not really convinced a is. are you using an axiom? if so which one or ones ? Thanks! ><

4. Jan 20, 2009

### snipez90

Hmmm, well a = 12k implies a2 = 144k2 = 12(12k2) so a2 is divisible by 12. Now we can let $$k \geq 1$$, so that if a2 is divisible by 12, then a2 = 12(12k2) = (12k)2. Then it's easy to see that the converse holds as well.