# Prove that S has no right inverse, but that

1. Sep 29, 2011

### Jamin2112

1. The problem statement, all variables and given/known data

Let N denote the set {1, 2, 3, ...} of natural numbers, and let S:N-->N be the shift map, defined by S(n) = n + 1. Prove that S be no right inverse, but it has infinitely many left inverses.

2. Relevant equations

Some definitions.

If an element a has both a left inverse L and a right inverse R, i.e., La = 1 and aR = 1, then L = R, a is invertible, R is its inverse.

3. The attempt at a solution

My first time doing senior-level algebra.

So, supposedly there can not be a number R such that (n + 1) * R = 1, and I'm supposed to prove that. Am I supposed to think of this as a law of composition on the natural numbers, N x N ---> N where we're combining n and 1 to get this other natural number n + 1? Or what should I be doing? Can I get a hint?

2. Sep 29, 2011

### FunkReverend

Hey, I'm working on the same problem, and equally stuck. Here's a line of thought I think might be the key.

We aren't looking for an inverse such that s*R = 1, because s isn't a number, it's a map. The definition of s*R = 1 is for an element s. What s is an element of is the maps N -> N, so when we are looking for an inverse of s we are looking for a s(n)*R = n, as the identity map is the map that returns everything to it's original value.

I still am not sure how to solve this problem, but this is as far as I've gotten.

Last edited: Sep 29, 2011
3. Sep 29, 2011

### FunkReverend

Alright, I've just about got a solution now, this should get you started on the right track:

I was right previously, we are not looking for a number r such that s(n)*r = 1, but for a map. The right inverse of s would mean that s(r(n)) = n for all n in the natural numbers. Can you prove that r cannot exist?

4. Sep 30, 2011

### Jamin2112

[PLAIN]http://depts.washington.edu/alumni/blogs/bdtw/files/2011/01/UW-logo-20011.jpg [Broken] ?????????????????????

Last edited by a moderator: May 5, 2017
5. Sep 30, 2011

### FunkReverend

What? I can only assume you're asking if I go to Washington. No, It's just the problem comes straight from a common algebra text book and I'm in the same chapter.

Last edited: Sep 30, 2011
6. Sep 30, 2011

### BananaNeil

haha, i'm stuck on the same question, and yesh i do go to UW.

I can only assume that we are in the same class. So i'll see you there. =]

7. Sep 30, 2011

### HallsofIvy

Staff Emeritus
No. You are completely missing the point. We are not talking about multiplication nor about an operation on N, we are talking about a mapping and inverse mapping. If S is a mapping from N to itself, then its "right inverse" would have to be a mapping,
R, again from N to itself, such that SR(x)= x. Here, S(x)= x+ 1 so R(x) would have to map n into x, such that SR(n)= S(x)= x+ 1= n. What is x? Why is that impossible (why is it NOT a mapping of N to itself)?

A left inverse, L, is a mapping from N to itself, such that LR(x)= x. Now, R(x)= x+ 1 so that says L(x+1)= x. What function does that? Why are there an infinite number of such functions? Be careful about the domain of L.

Last edited: Sep 30, 2011
8. Sep 30, 2011

### Jamin2112

Yeah ...... we're gonna need to start a study group. gorudy@uw.edu: Email me before the next assignment is due

9. Sep 30, 2011

### Jamin2112

It seems like you're drawing on some definitions that state (but maybe implied?) in my book.

If S has a right inverse, that means for all elements in the codomain ---- x + 1, where x is in N ---- there exists an element in the domain ---- n, where n is in N ---- such that n = x + 1. Is that right? And does the domain and codomain of S have to contain all the natural numbers? I've inferred no such restriction (and again, this isn't in my book). Of course, x = n - 1 would mean that x would be 0 if n = 1, meaning we don't have a right inverse.