Math Help - proof by contradiction

given a set with the symbols :
" - " for inverse
the constants :
0
AND the axioms:
for all a,b,c : a+(b+c) = (a+b) + c
for all a : a+0 = a
for all a : a +(-a) =0
for all a,b : a+b = b+a
1) THE uniqness of zero
2) THE uniqness of the inverse

2. Originally Posted by alexandros
given a set with the symbols :
" - " for inverse
the constants :
0
AND the axioms:
for all a,b,c : a+(b+c) = (a+b) + c
for all a : a+0 = a
for all a : a +(-a) =0
for all a,b : a+b = b+a
1) THE uniqness of zero
2) THE uniqness of the inverse
1) Let $0^,$ be a number other than zero such that for all a,

$a+0^,=a$

then, since zero is number

$0+0^,=0$

also by our fourth axiom, $0+0^,=0^,+0$ which is $0^,$ by our second axiom.

which implies $0^,$ is same as $0$....hence, zero is unique.

3. 2) The inverse of $a$ is $-a$. Assume that there exists another inverse element denoted as $a^{-1}$ such that $a^{-1}\neq-a$

Now, by the definition of inverse element, we have:

$a+a^{-1}=0$

But from the third axiom, we have:

$a+(-a)=0$

Ergo:

$a+a^{-1}=a+(-a)$

$a^{-1}=-a$

Which contradicts our original assumption $a^{-1}\neq-a$, so the inverse element must be unique.