Does commutativity imply associativity?

1. Jan 19, 2009

azure kitsune

Hi everyone,

Lately, I've been wondering whether commutativity implies associativity or not. (It's really hard for me to pay attention in my Calculus BC class.) I've never seen an example of a binary operation that is commutative but not associative. It seems intuitively true to me, but I don't know how to prove it (so maybe it's not true?). I also can't find anything on the web about this.

Can anyone help? Thanks!

2. Jan 19, 2009

mutton

Given ab = bc for all a, b, can you prove (ab)c = a(bc) for all a, b, c? Manipulate these equations for a while.

To think of examples, try defining specific binary operations on small sets like {0, 1}.

For an answer that's sort of intuitive, search Wikipedia for "commutative non-associative magma".

3. Jan 19, 2009

lurflurf

Never played rock paper scissors?
R=rock
P=paper
S=scissors

RR=R
RP=P
RS=S
PP=P
PS=S
PR=P
SS=S
SP=S
SR=R

obviously
XY=YX
(XY)Z!=X(YZ)

ie
(RP)S=PS=S
R(PS)=RS=R

4. Jan 20, 2009

azure kitsune

Thanks for the answers. :) It never came across me to define an operation myself but that rock paper scissors example really helped. (Algebra seems so much more complicated now!!!)

5. Jan 20, 2009

lurflurf

[corected I had RS=S before clearly false]

6. Jan 21, 2009

Edgardo

When I was still in school I wanted to (falsely) calculate my average grade in math as following:

First method
I wrote three exams with marks 1,2 and 4. I would take the average of 1 and 2, so I got a 1.5. Then I would take 1.5 and form the average with 4:

(1+2)/2 = 1.5
(1.5+4)/2 = 2.75

Second method
However, had I taken another order the result was different:
I would take 2 and 4 and form the average, so I got a 3. Then I would take 3 and form the average with 1:

(2+4)/2 = 3
(3+1)/2 = 2

I wondered why I had two different results.

Anyways, take the binary operation a°b = (a+b)/2.