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physicsforum7
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I was thinking about identities, and seem to have arrived at a contradiction. I'm sure I'm missing something.
A(n) (two-sided) identity for a binary operation must be unique.
I will reproduce the familiar proof:
Proof: Suppose a is an arbitrary element of a set S, e and e' are both identities, and * is an arbitrary binary operation. Then a*e=e'*a=a. Now take a=e' in the first equation; so e'*e=e'. Take a=e in the second equation; e'*e=e. Thus e'*e=e=e'.
But what about x*y= [itex]\left|x-y\right|[/itex] defined on the set of nonnegative real numbers?
It seems that both 0 and 2x are identities.
Can anybody find my mistake? Thanks ahead of time.
A(n) (two-sided) identity for a binary operation must be unique.
I will reproduce the familiar proof:
Proof: Suppose a is an arbitrary element of a set S, e and e' are both identities, and * is an arbitrary binary operation. Then a*e=e'*a=a. Now take a=e' in the first equation; so e'*e=e'. Take a=e in the second equation; e'*e=e. Thus e'*e=e=e'.
But what about x*y= [itex]\left|x-y\right|[/itex] defined on the set of nonnegative real numbers?
It seems that both 0 and 2x are identities.
Can anybody find my mistake? Thanks ahead of time.