1. Feb 12, 2008

### lostinmath08

b1. The problem statement, all variables and given/known data[/b]

On the set of real numbers R, the following is defined *:RxR arrow R
(x,y) arrow x*y=a(x+y)-xy
find all the values of the real parameter a such that the operation is associative

2. Relevant equations

associative law states x*(y*z)=(x*y)*z

3. The attempt at a solution

x*y = a (x+y) - xy = ax + ay - xy
(m*n) * o = m * (n*o)

(am + an - mn) * o = m * (an + ao - no)
a(am + an - mn) + ao - o (am + an - mn) = am + a (an + ao - no) - m (an + ao - no)
aam + aan - amn + ao - amo - ano + mno = am + aan + aao - ano - amn - amo + mno
am + o = m + ao
am - ao = m - o
a (m - o) = m - o

a = 1 if m does not equal o, and a does not equal 0 if m = 0

im unsure of my solution...any help would be awesome!

2. Feb 13, 2008

### HallsofIvy

Staff Emeritus
Why switch to m, n, and o? x, y, and z were working fine!

(Oh, and never use "o" as a symbol for a number- it looks too much like 0 and is too confusing.)

Looks to me like you have it! Remember this a must work for all m! If a must be 1 whenever m does not equal to 0 (and certainly there will are numbers that are not equal to 0!) you had better take a= 1.

As far as "a does not equal 0 if m= 0", I see no problem. 1 is not equal to 0!

3. Feb 13, 2008

### lostinmath08

would it be wrong to use m, n and p?
also the way i have presented the answer is it legitimate?

4. Feb 13, 2008

### HallsofIvy

Staff Emeritus
No, it's just that after you have written it in terms of x, y, z, I see no reason to change to other symbols.

Yes, just note that in order that your equations be true for all x, y, z, a must be equal to 1.

5. Feb 28, 2008

### lostinmath08

Thanks so much for your help!