# Puzzle: A little properties challenge

**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.

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)?

Nice question. I feel very uninquisitive for not thinking about this myself.

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?

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 the couterexample.

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).

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

Rock/Paper/Scissors

Who’s next?