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How can you prove that

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

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
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