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## Main Question or Discussion Point

**proving that -(-a)=a???**

Well, i was trying to prove from the axioms of the field of reals that -(-a)=a, but i am not sure that what i did is valid and follows from the axioms, since i have not seen the actual proof before.

From the axiom of the existence of the inverse we know that

a+(-a)=0 , now adding on both sides -(-a) from the left we get

-(-a)+(a+(-a))=-(-a)+0 now on the left hand side from the commutative axiom of addition we get

-(-a)+((-a)+a)=-(-a)+0 now on the right hand side, from the axiom of existence of the neutral element we get:

-(-a)+((-a)+a)=-(-a) now from the associative axiom we get on the left hand side:

(-(-a)+(-a))+a=-(-a) we know from the axom of the existence of the inverse of addition that the inverse of (-a) is -(-a) so we get:

0+a=-(-a) now from the axiom of the existence of the neutral element we get

a=-(-a) what we wanted to prove.

Is this kind of proof correct, or is there any more rigorous proof to show this. I mean a proof that follows from the axioms of the field of the real numbers, because i think there is another kind of proof usind classes of equivalence or sth.

Thnx in advance.