Question

Need help with a discrete math problem.

Answer 1

@mathstudent55

Answer 2

Wow, it's been a long time since I did this stuff... but I'll try to take a stab at it, at least a bit of direction. :)

For 1: Since f is invertible, it is one to one, so each element of S maps to exactly one element of T, and that element in T is mapped to ONLY by that element that you started with in S. So what does that say about the mapping of the inverse, which takes T --> S?

Answer 3

For 2: since f is invertible with inverse g, and we showed in (1) that g is unique, then g is the unique mapping from the T-->S...? As was discussed above, f invertible means that it there is a one to one correspondence of elements in S to elements in T, so there is a one to one correspondence of elements in T to element in S, given by the inverse function g, and hence, g is invertible.

I'm sure there are some others here who could give a much more eloquent statement about it, but that's my "stream of consciousness" on the subject, lol.

Answer 4

So in the first part, you asked what does that say about the mapping of the inverse which takes T--> S? I don't understand what the arrow means in this case

Answer 5

To answer the question though, would it show that the function is unique?

Answer 6

Since f maps from S to T (S-->T) and is invertible, then its inverse maps from T to S (T-->S).

Since the problem is to show that this inverse is unique, I guess my point was this:

f invertible means f is one to one. So if (s,t) is an ordered pair in the function f, then there is no OTHER ordered pair that has (s,t2), nor is there another ordered pair that has (s2, t). (This could be stated MUCH more formally, like I said... this is stream of consciousness... lol).

So (s,t) is the only ordered pair in f that involves those two specific elements of the domain S and the range T, right?

So the inverse, which swaps all the ordered pairs in f, contains (t,s) if f contains (s,t), right? And the uniqueness comes in the fact that, if (s,t) is in f, then (t,s) is in the inverse, and that holds for every unique point (s,t) in the original function f. So there is no "different" function that can be the inverse, because for EVERY point (s,t) in f the point (t,s) must be in the inverse... ??

Or something like that?? lol

sorry, like I said, it's been a while. I do love this kind of stuff, though. :)

Answer 7

Hmmm, I think that does make sense. :)

Answer 8

I will read over it some more to make sure. Thank you so much for the help!

Answer 9

You're welcome. :)