The rules: There are lots of answers. Find one or two interesting ones to share.

Find an operation that is commutative but not associative.

Who gets the credit for this one? I don’t know, I think it’s been around forever. But I use it every year in algebra, cute little challenge.

Before they start, the kiddies have done some work with both properties (and closure. Somehow they get to me with no sense of closure). We look at the four commonest: +,-,*,/, but we’ve also begun to look beyond those when the challenge is posed.

Numbers are commutative and associative under both + and ×, and under neither ÷ nor -. As a result, kids often conflate commutativity and associativity. So we work on making these properties distinct for the kidlets.

The immediate inspiration goes to Rolfe, with finder’s fee to Vlorbik.

1. August 17, 2007 pm31 8:44 pm 8:44 pm

averages. use “infix” notation:
xAy := (x+y)/2.
then obviously xAy = yAx
for all x, y (in a given field
of characterisic \not= 2, say).

but, e.g., (4A8)A12= 9
whereas 4A(8A12) = 7.

i figured this one out a long time ago
when i first saw this problem and
it’s the only one that’s stuck.
so maybe i should try to think of another.
umm … lie bracket (mod 2)?

2. August 17, 2007 pm31 10:47 pm 10:47 pm

Not-AND (NAND) is one of my favorite Boolean operations, so I’m glad to see it do the job too. It’s pretty clear that is is commutative: (A NAND B) = (NOT (A AND B)) = (NOT (B AND A)) = (B NAND A).

But associative doesn’t quite work out.

((False NAND True) NAND True) = (True NAND True) = False

But

(False NAND (True NAND True)) = (False NAND False) = True

Hmmm… Is NAND really a Lie Bracket mod 2?

3. August 19, 2007 pm31 8:13 pm 8:13 pm

For associative but not commutative we show them Left(a,b) = a.
Average (thanks Vlorbik) is the first example I use of commutative but not associative.
Clever, Rolfe. I decided to check “if..then” for associativity. I didn’t think so, but it took a moment to find $F \rightarrow (x \rightarrow F) \not= (F \rightarrow x) \rightarrow F$ the couterexample.

4. August 22, 2007 pm31 6:25 pm 6:25 pm

here’s another: “absolute difference”.
define D by aDb := |a-b|.
obviously D is commutative, but
1D(2D3) = 1D1 = 0
whereas
(1D2)D3 = 1D3 = 2.

i came up with this by taking a well-known operation
(subtraction) that’s *neither* commutative *nor* associative
and modifying it to force commutativity. this seems likely
to yield other examples if pushed at all hard. hmmm …
what about aQb := max {a/b, b/a} ? (…. thinking …)
yep. it works [ 2Q(3Q4) \not= (2Q3)Q4, e.g.].
this is fun and easy.

i worked out an operation on {a,b,c} last night
(inspired by rolfe’s use of booleans — operations
on {T,F} — the “minimal counterexample”)
by fiddling around with tables. the first thing
i tried, worked (and i quit looking right there).

aRa = bRb = cRc = a (the main diagonal )
aRb = bRa = a
aRc = cRa =c
bRc = cRb = b.

then i started trying to find some “real world”
interpretation of such an operation
(this is the opposite of mathematical modeling:
the “to a kid with a hammer, everything looks
like a nail” phenomenon).
i even came up with one but it was pretty bogus
(a is the ruling class, b the bourgois, c the proles …
R is “whose opinion dominates” in an encounter …
aR(bRc) = a “because” the yesmen keep the rabble
away from the boss; (aRb)Rc = c because this function
of middlemanagement *hasn’t* been effectively exploited.
that cRc = a is of course a reflection of the need
for “consciousness raising” … oh, never mind).

5. August 22, 2007 pm31 8:31 pm 8:31 pm

I don’t think I’ve ever laughed out loud about a binary relation before. My consciousness has been raised.

6. August 24, 2007 am31 9:57 am 9:57 am

Rock/Paper/Scissors

Who’s next?