Proving X^-1 is a Function: Demonstration

[FrostScYthe]

How do you prove this is a function?

g: P(A) > P(A)
g(X) = X^-1


X^-1 means inverse of X

anyway, does anyone know how to do a formal demonstration for this?

 

[HallsofIvy]

You need to be a bit clearer. If this were NOT

 

[HallsofIvy]

You need to be a bit clearer. If this were NOT a "logic" question, I would assume you meant that g is a relation on the Power set of A. However, that would still leave open the question of what is meant by "inverse of X"- a set does not have an inverse.

 

[FrostScYthe]

Alright..

No, I do MEAN PROVE THAT IT IS A FUNCTION. somehow you first have to prove that for the domain there is a solution for every element.. and that there is Unity in the solutions for every element that x contains ;/

Now what I mean that X^c... it´s hard to put the friggin´ notation on that thing but it´s defined something like this

if xRy

then xR^cy means that yRx
that´s all

(x,y) belongs R
then (x,y) belongs R^-1 means (y,x) belongs R

 

[HallsofIvy]

You still haven't told us:

What A is.

What P(A) means.

What X is!

 

[FrostScYthe]

I though you were to know that ;\

but here we go anyway, I think I'm getting it solved from another source, so when I'll get it, I'll post it =d

P(A) - means Parts of set A

so when I say P(A) -> P(A) I simply mean that the function goes from a domain of P(A) to a range of P(A)

g(x) = X^c

That's the function that we're talking about. I believe I explained that one... X is a set by the way

What you have to prove?

1. That under the function X^c any entry and exit belongs to P(A)

2. That there's Unicity that is ...

for all x1 x2(g(x1) != g(x2) -> x1 != x2)