# Proof of Addition Reversibility

• I
How can you prove that

##f(x)=g(x) \Leftrightarrow f(x)+C=g(x)+C##

## Answers and Replies

fresh_42
Mentor
I can't as long as I don't know where your elements are taken from.

• FAS1998
I can't as long as I don't know where your elements are taken from.
What do you mean by "elements"?

fresh_42
Mentor
What do you mean by "elements"?
##f(x),g(x),C##

If they were, as usual, from ##\mathbb{R}##, then the answer would be: because ##(\mathbb{R},+)## is a group. But if you had defined addition differently on some set, then there is not enough information about it.

E.g. ##1+1=2## and ##1+1=0## are both true, just not in the same set.

• FAS1998
##f(x),g(x),C##

If they were, as usual, from ##\mathbb{R}##, then the answer would be: because ##(\mathbb{R},+)## is a group. But if you had defined addition differently on some set, then there is not enough information about it.
This is what I meant.

I just looked over the wikipedia page for groups and now understand why ##(\mathbb{R},+)## is a group, but why does belonging to a group imply reversibility?

jbriggs444
Science Advisor
Homework Helper
I just looked over the wikipedia page for groups and now understand why ##(\mathbb{R},+)## is a group, but why does belonging to a group imply reversibility?
Being in a group means existence of an additive inverse, -C.
So given that ##f(x)+C=g(x)+C## you can write down ##f(x)+C+ -C=g(x)+C+ -C##.

Given associativity, the definition of an additive inverse and the definition of zero, it is all downhill from there.

• FAS1998
How would we prove the reversibility of other operations such as exponentiation (for values >= 0), that don't belong to groups?

fresh_42
Mentor
How would we prove the reversibility of other operations such as exponentiation (for values >= 0), that don't belong to groups?
##x \longmapsto e^x## or ##x \longmapsto x^n## e.g. by the theorem of invertible functions or in general step by step. They aren't operations anymore, just other functions.

• FAS1998