# Right inverse, left inverse, binary operations

1. Feb 10, 2014

### Panphobia

1. The problem statement, all variables and given/known data
If * is a binary operation on a set B, and the domain of definition is B^2, if * is associative and the neutral element is p. If r and l are elements of b we can say that r is a left inverse of l under * iff r * l = p, and l is a right inverse of r iff l * r = p. Show that if an element of B has a left and right inverse, then they are equal.

3. The attempt at a solution
Does the neutral element have anything to do with finding the answer, also what does associativity have to do with finding the answer? All I can think of is since there is only one neutral element in *, r = p and l = p, but I don't think that is the answer.

Last edited: Feb 10, 2014
2. Feb 10, 2014

### Dick

Sure, you have to assume there is a neutral element p, so that for any x, x*p=p*x=x. Now suppose r is a right inverse of x (so x*r=p) and l is a left inverse of x (so l*x=p). Associativity tells you (l*x)*r=l*(x*r), yes? So?

3. Feb 10, 2014

### Panphobia

Does it matter than the question say l * r = p and r * l = p, where does the x come from? would this mean (l * r) * l = l * (r * l)?

Last edited: Feb 10, 2014
4. Feb 10, 2014

### Dick

The letters were confusing and I changed them. Your writing (l * r) * l = l * (r * l) shows why. The definitions you've got are using 'l' to be a symbol for both the right and left inverse. They are two different things. Those definitions mean something independent of the particular symbols used. Suppose I tell you that b is a right inverse of c. Let's keep the symbol 'p' to mean the neutral element, though I probably would have used the symbol '1' instead. It's more suggestive. What does that mean?

Last edited: Feb 10, 2014
5. Feb 10, 2014

### Panphobia

Then why in the question is l * r = p, and r * l = p defined? btw r and l do not mean right and left inverse, they are arbitrary variables, the question says, "l is left inverse of r under * iff l * r = p, and l is a right inverse of r under * iff r * l = p". It is kind of confusing.

6. Feb 10, 2014

### D H

Staff Emeritus
First off, look at some numbers you know, such as the rationals. With * being the multiplication operator you learned long ago and p being 1, what is the left inverse of 2 ? The right inverse? The answer is obviously 1/2 in both cases: 1/2*2 = 2*1/2 = 1. The goal of this problem is to show that if an element of B has both left and right inverses and if * is associative then there is only one inverse (i.e., the left and right inverses are one and the same).

Note that this result does not necessarily mean that * is commutative. What it does mean is that an element of B commutes with its multiplicative inverse.

Dick took the correct approach by giving that element of B that has both a left and right inverse a new symbol, x.

7. Feb 10, 2014

### Dick

Yes, they are arbitrary variables, and yes, it's kind of confusing. The reason it's confusing is because they made such a poor choice of arbitrary symbols. Try to unscramble it. You didn't answer my question. I'll repeat it. Suppose I tell you that b is a right inverse of c. What does that mean? Bypass the whole 'l' and 'r' thing.

Last edited: Feb 10, 2014
8. Feb 10, 2014

### Panphobia

c * b = p, where p is the neutral element

9. Feb 10, 2014

### Dick

Bingo! Now I'll tell you that d is a left inverse of c. Same question. And then can you show b=d?

Last edited: Feb 10, 2014
10. Feb 10, 2014

### Panphobia

(d * c) * b = d * (c * b)
d * c = c * b = p
p * b = d * p
so this implies d = b, right?

11. Feb 10, 2014

### Dick

Right. Couldn't be clearer. I'm almost wondering if they didn't choose the symbol choice in original definitions to try to test you and throw you off track.

12. Feb 10, 2014

### Panphobia

Thanks for the help!