## Monday, November 12, 2012

### \$ and # operators

I'm considering an operator that looks like this:

This grid means for instance 9\$3 = 33 and 3\$9 = 33 and 3\$(-9) = -33 and -3\$(-9) = 33 and -3\$(9) = -33 etc. In general:
a\$b = b\$a =(-b\$-a) = (-a\$-b) =  -(-b\$a) = -(b\$-a) = -(-a\$b) = -(a\$-b)

An interesting feature is that c = a\$b for a unique pair of a and b. For instance 30 is 6\$7 or -6\$-7 and there is no other way to produce 30 from a pair of a and b, aka b and a.

The operation is transitive over addition:
a\$(b+c) = a\$b + a\$c

but it is not commutative:
a\$(b\$c) != a\$(b\$c)

a\$b is not  f(a,b,+,-,*,/)
simply because x\$x = x^2 and x\$0 = x
the only polynomial with rational coefficients that equals x^2 is x^2 itself and x*0 is 0 not x.

So in a way f(a,b,+,-,*,/,\$) extends the usual algebra.

Every positive integer can form a binary tree like this:
30 = 6\$7
but 6 is 2\$3 and 7 is 1\$5
2 and 3 are both 1\$2 and 1\$3
((1\$2)\$(1\$3))\$(1\$4)
the 1's can be removed.
(2\$3)\$5

Two obvious extensions of the idea are to define \$ over the rationals which looks to be possible as \$ has nice continuity properties, for instance it is always increasing in the a and b directions. Also to make another operator say, #, that acts as the inverse of \$. for instance (a\$b)#a = b and (a\$b)#b = a.