• Jul 1st 2009, 07:53 AM
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
• Jul 1st 2009, 08:03 AM
malaygoel
Quote:

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 \$\displaystyle 0^,\$ be a number other than zero such that for all a,

\$\displaystyle a+0^,=a\$

then, since zero is number

\$\displaystyle 0+0^,=0\$

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

which implies \$\displaystyle 0^,\$ is same as \$\displaystyle 0\$....hence, zero is unique.
• Jul 3rd 2009, 05:17 PM
Referos
2) The inverse of \$\displaystyle a\$ is \$\displaystyle -a\$. Assume that there exists another inverse element denoted as \$\displaystyle a^{-1}\$ such that \$\displaystyle a^{-1}\neq-a\$

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

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

But from the third axiom, we have:

\$\displaystyle a+(-a)=0\$

Ergo:

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

\$\displaystyle a^{-1}=-a\$

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