# I Proof of Addition Reversibility

#### FAS1998

How can you prove that

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

#### fresh_42

Mentor
2018 Award
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
2018 Award
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
2018 Award
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.

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