Who's good with proofs 2.

[Sam Finch]

I've just noticed an assumption I've made in Who's good with proofs. I have assumed that


if a*b = 0 then either a is 0 or b is 0.

this is true for real and complex numbers, but not for all numbers, (it's not true for 2x2 matricies for example)


(1 0) (0 0) = 0
(0 0) (0 1)



I think it holds true if multiplication is commutitive, but I'm not sure, does anyone know?


Again a proof or counterexample would be a great help (actually a counterexample would be a real arse but still)

Thanks.

[Guv]

Quaternion multiplication is not commutative, but Q*P = 0 implies either q or p or both zero for
Quaternions. At least I think so. I have not really analyzed this.

[Guv]

I do not feel so good about my claim that zero cannot be the product of two non zero quaternions.

I am too lazy to look for an example, but after looking at the problem just a little, it looks like there could be many pairs of non zero quaternions whose product is zero.

[Sam Finch]

Actually guv I think you're right, I've changed tact again on my route through this and I havn't got up to quaternions yet, but I'm pretty sure there aren't any that multiply to make 0.

In any case I've found a counterexample.

the set of matricies in the form

(a b)
(b a)

is comutitive under multiplication (fairly easy to prove)


but to 2 matricies

(1 1)
(1 1)

and

(1 -1)
(-1 1)

multiply to make 0.


but thanks for posting again anyway, after trying to prove the statements in the other proofs post I realized that you couldn't assume that much about what was 0.


Basicly in any set of numbers there is a set of numbers with no inverse.

in the real numbers anything in the form x - x

if this set is closed under addittion then anything in the set can for all intents and purposes be called 0. but if it isn't closed under addition then each element of the set must be treated differently.

What I'm looking at now is introducing a second number j to the field of real numbers, where j^2 = 1


and we get (1 + j) (1 - j) = 0

this means that the set of points with no inverse is not closed under addition, and can't all be called 0.


I'm trying to decide whether it's still valid to say (x - x) = 0. (I havn't really started yet, but that's what I'm going to do after I've got bored with infinite cyclic groups) Actually that matricies example has just decided it for me, I just have to prove it.

[Guv]

Sam: after playing with quaternion multiplication for about 10-15 minutes, I am on the fence about the conjecture relating to the product of two quaternions.

The notation I am using has two sets of unknowns: (A, B, C, D) and (a, b, c, d).

There are 4 equations in the 8 unknowns. Aa - Bb - Cc - Dd = 0 Ab + Ba + Cd - Dc = 0 Ac + Ca + Db - Bd = 0 Ad + Da + Bc - cB = 0Find a solution to the above equations such that neither (A, B, C, D) nor (a, b, c, d) is (0, 0, 0, 0) and you have two non zero quaternions whose product is zero.

4 equations in 8 unknowns would seem to have many solutions, even with the restrictions given. There seems to be a lot of degrees of freedom here to play with. This makes me think there are solutions meeting the conditions.

After working for about 10-15 minutes, it seems as though there are symmetries to the equations which start forcing unknowns to be zero. This makes me think there are no solutions meeting the conditions.

Perhaps you or somebody else would like to spend some time on this. My lazy tendencies have won out over my obsessive-compulsive personality.

[Guv]

Sam: I woke up in the middle of the night and realized that the product of two quaternions cannot be zero unless one of quaternions used as a multiplier is zero. I feel stupid for not realizing it earlier.

In my previous post, I showed 4 equations in 8 unknowns which must be satisfied for a quaternion product to be zero, with the condition that neither of the following sets be equal to {0, 0, 0, 0}: {A, B, C, D}, {a, b, c, d}. The equations are as follows.

Aa - Bb - Cc - Dd = 0
Ab + Ba + Cd - Dc = 0
Ac + Ca + Db - Bd = 0
Ad + Da + Bc - cB = 0

If you assign arbitrary nonzero values for one of the sets, you get 4 equations in 4 unknowns. You can view the equations as a matrix times the vector: [0, 0, 0, 0]. The solution vector is the inverse of the matrix times [0, 0, 0, 0]. Thus the solution vector must be zero for any arbitrary choice of values for the first set of values. Thus the conditions cannot be met.