- #1

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How can you prove that

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

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

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- Thread starter FAS1998
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- #1

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How can you prove that

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

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

- #2

fresh_42

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

- #3

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

- #4

fresh_42

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

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.

- #5

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This is what I meant.##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.

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?

- #6

jbriggs444

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Being in a group means existence of an additive inverse, -C.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?

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.

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- #8

fresh_42

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##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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