Proof of Only One Identity Properity for Binary OperationsDate: 10/31/2001 at 12:15:42 From: Tara Saddig Subject: Proof of only one identity property for binary operations Dear Dr. Math, I am trying to prove for a college geometry class that there is one and only one identity property for every operation. This is what I have so far: (* is star, an unknown binary operation, not times) Suppose 1 and 1' are both identities in a group a*1 = 1*a = a a*1' = 1'*a = a a*1 = a*1' = a (I want to just divide by a and get that 1 = 1', but since we're working with * not times, I can't do this, and therein lies my problem.) I also used the inverse properity to get some more stuff, but no proof: a*b = 1 a'*b' = 1' a*1 = a*1' a*(a*b) = a*(a'*b') (a*1)*(a*b) = (a*1')*(a'*b') I don't know where to go from here. I feel as if I have a lot of information, but don't know how to prove that there's only one identity for operation * without simply dividing both sides by a. Do you have any suggestions for finishing this proof? Thanks so much! Date: 10/31/2001 at 13:49:36 From: Doctor Paul Subject: Re: Proof of only one identity property for binary operations You've done a lot of nice work, but it's not really going to lead you where you need to go. The first portion of what you have above is good, but we need to take a different approach: If 1 and 1' are both identity elements in a group, then the following hold: 1 * 1' = 1' * 1 = 1 (since 1' is an identity element) But notice also that: 1' * 1 = 1 * 1' = 1' (since 1 is an identity element) Then 1 * 1' = 1 and 1 * 1' = 1' so it follows that 1 = 1' - Doctor Paul, The Math Forum http://mathforum.org/dr.math/ |
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