# Counterexample to uniqueness of identity element?

1. Jun 25, 2011

### math771

(Hopefully, this question falls under analysis. I was unable to match it well with any of the forums.)
The proof that the identity element of a binary operation, f: X x X $\rightarrow$ X, is unique is simple and quite convincing: for any e and e' belonging to X, e=f(e,e')=f(e',e)=e'.
However, if we take f(m,n)=max(m,n), it appears that any m will have multiple identity elements--the elements of the set of numbers n less than m.
There must be something that I'm missing here. Any help would be appreciated. Thanks!

2. Jun 26, 2011

### HallsofIvy

You are missing the fact that individual elements of the set do not "have" their own identities. In order that "e" be the identity, we must have f(e, a)= f(a, e) for all a in the set. f(a,b)= max(a, b), over some ordered set, has an identity if and only if the set has an largest member.

3. Jun 26, 2011

### math771

Thank you.

4. Jun 26, 2011

### SteveL27

Wouldn't the "max identity" be the smallest element? For example in the natural numbers 0, 1, 2, ..., we have max(0, n) = n for all n, so 0 is the identity of the max operation.

5. Jun 26, 2011

### math771

I was thinking the same thing. Maybe that's what HallsofIvy meant.

6. Jun 26, 2011

### gb7nash

Correct. Assuming the smallest element is e, then max(e,a) = max(a,e) = a for all a in the set. If you're dealing with min, then you do need a largest element in order to have an identity that works for all numbers.

7. Jun 27, 2011

### HallsofIvy

Yes, of course. I don't know why I said "maximum".