How do you prove the commutative property of multiplication for 4+ factors?


by hamsa0
Tags: commutative, factors, multiplication, property, prove
hamsa0
hamsa0 is offline
#1
May23-11, 09:22 PM
P: 11
I don't know how to construct formal proofs but there is the obvious geometric approach for 2 and 3 factors. However, how do you prove the commutative property holds up for 4+ factors? You end up with a lot of different orders in which you can multiply the factors and you can't just construct a geometric object from them.
Phys.Org News Partner Mathematics news on Phys.org
Researchers help Boston Marathon organizers plan for 2014 race
'Math detective' analyzes odds for suspicious lottery wins
Pseudo-mathematics and financial charlatanism
Mark44
Mark44 is online now
#2
May23-11, 09:52 PM
Mentor
P: 21,019
I don't see where geometry comes into it.
The commutative property of multiplication says that ab = ba. For 3 factors it would be abc = cba. For four factors, I guess you're trying to prove that abcd = dcba.
abcd = (ba)(dc) = (dc)(ba) = dcba
hamsa0
hamsa0 is offline
#3
May23-11, 10:27 PM
P: 11
Ya after I posted this and hopped on the bus I realized it was a retarded question lol. Thanks for the response though man.


Register to reply

Related Discussions
commutative property of numbers and uncertainity General Physics 1
When is matrix multiplication commutative Calculus & Beyond Homework 1
commutative multiplication Calculus & Beyond Homework 2
Why is scalar multiplication on vector spaces not commutative? Linear & Abstract Algebra 3
Show the commutative property with dot product Introductory Physics Homework 5