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

1. ## 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!

2. 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. $a^{-1}\implies -a$)

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

Does this make sense?

3. 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. $a^{-1}\implies -a$)

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

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

4. Hmm so this state is proven to be true right?

5. 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

you are reading something wrong. please review your text

Originally Posted by Kitizhi
Hmm so this state is proven to be true right?
what Chris did is correct. review it carefully

6. oops for some reason i kept thinking multiplication.
sorry my mistake

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