proof by contradiction

alexandros · #1 $Old$ July 1st, 2009, 09:53 AM

given a set with the symbols :
" + " for addition
" - " 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
PROVE using CONTRADICTION
1) THE uniqness of zero
2) THE uniqness of the inverse

malaygoel · #2 $Old$ July 1st, 2009, 10:03 AM

Quote:

Originally Posted by alexandros $View Post$

given a set with the symbols :
" + " for addition
" - " 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
PROVE using CONTRADICTION
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.

Referos · #3 $Old$ July 3rd, 2009, 07:17 PM

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.