Another proof I have a question about

[Ed Quanta]

Where x is an element of integers, show that for x=x^2 that 0 or 1 are the only solutions to this equation. I have shown that 0 and 1 are solutions to this, but I am trying to show that no other solutions are possible. My plan was to show that no negative number could be a solution to this problem because two negatives would make a positive and this could never equal a negative number. And I also showed that for any number greater than 1 that this equation could never be true because the right hand side would always be greater. I am having trouble doing this expressing the first five axioms, or if it is not possible to prove this with the first five, then maybe the first 6. I am not sure how to account for greater than or less than relations yet. Thanks for reading this whole thing.

[Hurkyl]

What axioms are you allowed to use?

[Ed Quanta]

Associative Property, Commutative property, Distributive Property, If m doesn't equal 0 then mn=np implies n=p, identity element for addition is 0, identity element for multiplication is 1, each integer has an additive inverse.

[Hurkyl]

Well, let's think how we would "normally" attack this problem:


\begin{equation*}\begin{split}
x^2 = x \\
x^2 - x = 0 \\
x(x-1) = 0 \\
x = 0 \vee x = 1
\end{split}\end{equation*}


So our first thought should be if we can write this derivation in terms of the axioms. The only difficulty here is going from the third to the fourth step, and I'll give you a hint for it:

Can you prove that:


a b = 0 \Leftrightarrow a = 0 \vee b = 0


?



P.S. you have a typo in your cancellation law

P.P.S. this thing I asked you to prove, the "no zero divisors law" (I don't think that's a standard name), is an important one to remember; whether or not it is true has a big impact on the algebra of a system. For example, when you're working over the integers mod 6, this law fails (2 * 3 = 0), and your equation has four solutions (0, 1, 3, and 4).

[Ed Quanta]

Thanks man, I'm pretty sure I got it now. I wasn't sure that the axioms allowed for the use of the distributive property backwards. I knew you could go from a(b+c) to ab +ac, but I didnt know you could go from ab +ac to a(b+c) as stupid as that may sound. So I relied on just showing that 1 and 0 satisfied the solution and that any other number could not work. But thanks Hurkyl, you the man.