Does commutativity imply associativity?

[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!

 

[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".

 

[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

 

[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!)

 

[lurflurf]

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]

 

[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 = 2.75Second 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.