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

Staff Emeritus
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

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