# Thread: Prove elementary rules in a field

1. ## Prove elementary rules in a field

Hey.

Given the field F, Prove:
* For any a, b in F, if a*a=b*b, then a=b or a=-b
* For any a, b in F, if a*a*a = b*b*b, then a=b

Given the field F where 1+1=0, Prove:
* For any a in F, -a=a
* For any a, b in F, a+b=a-b

The proofs have to be supported by the axioms of fields rigorously.

I don't know actually how even start proving it.
Thanks.

2. ## Re: Prove elementary rules in a field

I'll start you off on the first one:

As a preliminary Lemma, I will prove that if xy = 0, then x=0 or y=0.

Indeed, if x is not 0, then y = 1y = [(1/x)*x]y = (1/x)(xy) = (1/x)(0) = 0 (You will need to prove that multiplying by 0 in a field always produces 0).

[Notice how I used the existence of a multiplicative identity (namely 1), the multiplicative inverse property, and associativity of multiplication.]

Now, if a*a=b*b, then a*a - b*b = b*b - b*b = 0.

Next, prove that a*a - b*b = (a - b)(a + b) (to prove this you will need to use several field properties and it will require several steps).

By the preliminary Lemma above, a - b = 0 or a + b = 0.

Now, show that this last statement is equivalent to a = b or a = -b.

That should help to get you started.

3. ## Re: Prove elementary rules in a field

For any a, b in F, if a*a*a = b*b*b, then a=b

counterexample:

in the field of complex numbers, let

$\displaystyle a=-\frac{1}{2}+\frac{i \sqrt{3}}{2}$ and $\displaystyle b=1$

$\displaystyle a^3=b^3$ but $\displaystyle a \neq b$

4. ## Re: Prove elementary rules in a field

Originally Posted by Idea
For any a, b in F, if a*a*a = b*b*b, then a=b

counterexample:

in the field of complex numbers, let

$\displaystyle a=-\frac{1}{2}+\frac{i \sqrt{3}}{2}$ and $\displaystyle b=1$

$\displaystyle a^3=b^3$ but $\displaystyle a \neq b$
Thank you!