Prove that S has no right inverse, but that

[Jamin2112]

Homework Statement

PLEASE HELP. HOMEWORK DUE TOMORROW.

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.

Homework 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.

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?

 

[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.

 

[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?

 

[Jamin2112]

Hey, I'm working on the same problem

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

 

[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 textbook and I'm in the same chapter.

 

[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. =]

 

[HallsofIvy]

Homework Statement

PLEASE HELP. HOMEWORK DUE TOMORROW.

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.

Homework 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.

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.

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?

 

[Jamin2112]

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. =]

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

 

[Jamin2112]

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.

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.