# For all a,b,c ϵ B | a+b = a+c -> b = c

• Feb 26th 2009, 06:53 PM
Kitizhi
For all a,b,c ϵ B | a+b = a+c -> b = c
Can some tell me if this proof works and if not correct me please.

a+b = a+c -> b=c
~a·a+b = ~a·a+c -> b=c
b=c -> b=c

Proved!
• Feb 26th 2009, 07:02 PM
Chris L T521
Quote:

Originally Posted by Kitizhi
Can some tell me if this proof works and if not correct me please.

a+b = a+c -> b=c
~a·a+b = ~a·a+c -> b=c
b=c -> b=c

Proved!

Your proof is not correct. You need to keep in mind that we are dealing in additive notation, so the inverse must be expressed in additive notation (i.e. \$\displaystyle a^{-1}\implies -a\$)

It should be \$\displaystyle a+b=a+c\implies (-a)+a+b=(-a)+a+c\implies(-a+a)+b=(-a+a)+c\implies b=c\$

Does this make sense?
• Feb 26th 2009, 07:06 PM
Kitizhi
Quote:

Originally Posted by Chris L T521
Your proof is not correct. You need to keep in mind that we are dealing in additive notation, so the inverse must be expressed in additive notation (i.e. \$\displaystyle a^{-1}\implies -a\$)

It should be \$\displaystyle a+b=a+c\implies (-a)+a+b=(-a)+a+c\implies(-a+a)+b=(-a+a)+c\implies b=c\$

Does this make sense?

Hmm kinda but i dont know of additive notation.
The identity law states -a + a = 1
so shouldnt it be 1+b = 1+c
• Feb 26th 2009, 09:35 PM
Kitizhi
Hmm so this state is proven to be true right?
• Feb 26th 2009, 09:47 PM
Jhevon
Quote:

Originally Posted by Kitizhi
Hmm kinda but i dont know of additive notation.
The identity law states -a + a = 1
so shouldnt it be 1+b = 1+c

what??

no, -a + a = 0