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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?

Frank Vel
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    The proof depends of the book you are reading and what is proved about multiplication by now. – Masacroso Aug 02 '16 at 15:41
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    Which axioms? Axioms of a ring? a field? Peano axioms? – Plutoro Aug 02 '16 at 15:45
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    Your arguments in your post are basic and use results provable by axioms. So these are arguments from axioms. Do you mean how to prove directly from axioms without any "in between" propositions such as Prop: ax0 = 0. -(-a) = a. And (-1)a = -a? Um... not really... as you have no axiom that says (-a)^2 = a^2 you will need to prove inbetween statements. I suppose you don't need to make general propositions but ... seems tedious and pointless. ... But maybe I'll try.... – fleablood Aug 02 '16 at 16:33

1 Answers1

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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...

fleablood
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