Directly proving $(-a)^2 = a^2$

Question

I know one can easily prove that $(-a)^2 = a^2$ from $(-a)*0 = (-a)(a+(-a)) = (-a)^2+(-a)a$ and one can also show that $(-a)a = -a^2$ (or even $(-a) = (-1)*a$), however, is there a way of starting with $(-a)^2$ and end with $a^2$ directly, simply in terms of the axioms?

Answer 1

Aren't your arguments only using axioms?

Assuming field axioms:

$a*0 + a*0 = a(0+0)$ by distribution

$a*0 + a*0 = a(0)$ by definition of 0

$a*0 + a*0 + (-a*0) = a(0) + (-a*0)$ by axiom of that all elements have additive inverses.

$a*0 + 0 = 0$

so

$a*0 = 0$ for all $a$.

So $0 = a*0 = a(1 + (-1)) = a*1 + a*(-1) = a + a*(-1)$ by above and distribution.

By existance of additive inverse axiom.

$-a + 0 = -a + a + a*(-1) = 0 + a*(-1) = a*(-1)$

so $a*(-1) = -a$ for all $a$.

Further more $-(-a) + (-a) = 0$ so $-(-a) + (-a) + a = 0 + a$ so $-(-a) + 0 = -(-a) = a$ for all $a$. by axiom of additive inverses and binary nature of addition.

So $(-a)^2 = (a*-1)*(a*-1) = a^2*(-1)^2$ by associativity and commutativity.

So $(-a)^2 = a^2*(-1)^2 = a^2*[(-1)*(-1)] = a^2*[-(-1)] = a^2 *1 = a^2$ by results proven above.

That's assuming your axioms were the field axioms.

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Postscript:

It occurs to me maybe the OP wants to prove the result directly from axioms without using any "inbetween" propositions such as $0x = 0$ and $-(-x) = x$.

Well, seems silly but here goes...

$(-a)^2 = (-a)^2$

$(-a)^2 + (-a)a = (-a)^2 + (-a)a$

$-a(-a + a) =(-a)^2 + (-a)a$

$-a*0 = (-a)^2 + (-a)a$

$-a*0 + (-a)*0 = (-a)^2 + (-a)a + (-a)*0$

$-a*0 + (-a)*0 = (-a)^2 + (-a)a + (-a)*0$

$-a(0 + 0) = (-a)^2 + (-a)a + (-a)*0$

$-a*0 = (-a)^2 + (-a)a + (-a)*0$

$-a*0 + [-(-a*0)] = (-a)^2 +(-a)a + (-a)*0 + [-(-a*0)]$

$0 = (-a)^2 + (-a)a$

$a^2 = (-a)^2 + (-a)a + a^2$

$a^2 = (-a)^2 + (-a + a)a$

$a^2 = (-a)^2 + 0*a$

$a^2 + 0*a = (-a)^2 + 0*a + 0*a$

$a^2 + 0*a = (-a)^2 + (0+0)a$

$a^2 + 0*a = (-a)^2 + 0*a$

$a^2 + 0*a + (-0*a) = (-a)^2 + 0*a + (-0*a)$

$a^2 + 0 = (-a)^2 + 0$

$a^2 = (-a)^2$

So $(-a)^2 = a^2$.

ONLY axioms and definitions were used!

Silly, but I guess it was good for me...