anonymous
  • anonymous
Need help with a discrete math problem.
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
@mathstudent55
DebbieG
  • DebbieG
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?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

DebbieG
  • DebbieG
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.
anonymous
  • anonymous
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
anonymous
  • anonymous
To answer the question though, would it show that the function is unique?
DebbieG
  • DebbieG
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. :)
anonymous
  • anonymous
Hmmm, I think that does make sense. :)
anonymous
  • anonymous
I will read over it some more to make sure. Thank you so much for the help!
DebbieG
  • DebbieG
You're welcome. :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.