Proving the Associative PropertyDate: 02/24/2001 at 15:15:07 From: Cesar Valverde Subject: Associative properties In a binary operation, for example: # | a b c ---+--------- a | b c a b | c b b c | a b a I understand that I can prove it is commutative (i.e. a#b = b#a) because there's symmetry along the diagonal (the one that goes "\"). But how can I prove that it is associative, i.e. a#(b#c) = (a#b)#c? Date: 02/26/2001 at 17:24:54 From: Doctor Achilles Subject: Re: Associative properties Hi Cesar, Thanks for writing to Dr. Math. You're right that you can prove the commutative property holds because of the diagonal. However, you actually are very close to a proof that the associative property DOES NOT hold. The associative property of an operation says that for all values of x, y and z, (x*(y*z)) = ((x*y)*z). To prove it, you need an argument that shows that it always holds. To disprove it, you need one counterexample, i.e. one time when (x*(y*z)) = ((x*y)*z) is false. (a*(b*c)) in your operation can be evaluated like this. Write the expression to evaluate: (a*(b*c)) Just like anything else, we have to do what's inside the parentheses first: (a*b) Now we can evaluate the answer: c Try the same thing for ((a*b)*c). Remember to evaluate what's inside the parentheses first. See what you come out with. If it's not c, then you have a counterexample. You can then disprove the associative property by simply saying that if it did hold, then the two expressions would have the same value, since they have different values, it doesn't hold. Hope this helps. If you have any other questions about this or any other math topics, please write back. - Doctor Achilles, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/