function; strange notation i've never seen

[mikeczyz]

the problem is written as such:

a # b =a^2 +2ab +b^2; find x#2

what does x have to do with any of this?

edit: i think i just got it...substitute "x" for "a" and "2" for "b"
is that right?

[flipdeadshot22]

correct

[PairTheBoard]

They are defining # to be a binary operator like + or * .

Looks like # is commutative. Is is associative?

PairTheBoard

[mikeczyz]

i don't really understand what your question is. however, i typed up the question as I found it in the workbook...nothing was omitted from my original post.

[blah_blah]

[ QUOTE ]
Is is associative?

[/ QUOTE ]

no.

[PairTheBoard]

[ QUOTE ]
i don't really understand what your question is. however, i typed up the question as I found it in the workbook...nothing was omitted from my original post.

[/ QUOTE ]

I observed that the binary operation a#b is commutative, ie.

a#b = b#a

I wondered if it was associative, ie

Is a#(b#c) = (a#b)#c ?

blah blah says he's worked it out and the answer is No.

Notice + and * are commutative and associative.

PairTheBoard

[bigpooch]

If in the original post, + and . are taken in the usual
sense on the set of complex numbers, then clearly, # is
commutative and won't be associative ((a#b)#c will have a
term a^4, but a#(b#c) will have an a^2 term).

Now, if the "+" operation is commutative (as is usually the
case; it's usually the "addition" on some ring) and the "."
operation is ANY binary operation, then the operation "#" is
not commutative: (a#b)-(b#a) = 2[a.b-b.a]. There IS a
difference between the operation "#" and the operation "~",
if by definition, a~b = (a+b)^2 = (a+b).(a+b). Then, ~ is
commutative regardless of any properties of "." or not as
"pathological" as "#".

This leads me to wonder why "#" was explicitly defined as it
was, and not in the same way as "~" is defined above.