Alright, very simple proof I need help with. Prove x^2 0 for all nonzero x in the reals, using only the properties of an ordered field. I approach the case of x

Question

Alright, very simple proof I need help with. 

Prove x^2 > 0 for all nonzero x in the reals, using only the properties of an ordered field. 

I approach the case of x < 0 first.

so x^2 = -x * -x   = (-1 * x) * (-1 * x)
 = -1 * x * -1 *x = (-1 * -1) * (x * x) 

This is where I need help. Either I have approached this case the wrong way. Or I am on to something, but I dont know how (or if I need to prove) that -1 * -1 =1.

anonymous · Answer

you always want to be working with positive numbers, since we know that if:$$a>0, b>0\Longrightarrow ab>0$$So if x<0, then -x > 0, and you have that:$$(-x)(-x) >0$$