# Does commutativity imply associativity?

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!

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".

lurflurf
Homework Helper
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

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!!!)

lurflurf
Homework Helper
Never played rock paper scissors?
R=rock
P=paper
S=scissors

RR=R
RP=P
RS=R
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
[corected I had RS=S before clearly false]

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.