Who's good with proofs?

[Sam Finch]

I'm doing some wierd maths, trying to map groups onto sets of hypercomplex numbers, (If you don't know what a hypercomplex number is don't wory you can still help)

I need some help trying to prove this statement

For any number r where (r^n = 1) any number expressible as a polynomial in r (ie - a + br + c(r^2) + ... + d(r^(n-1)) can be expressed in the form (x + yr)


where a,b,c,d and x and y are real numbers.


r is not neccecerally a real or complex number (but you may assume addittion and multiplication are both asossiative and commutitive)

I've attatched a gif with the formal statement in it, and I can explain a bit more If it's unclear, I don't actually know if it's true or not, so a counterexample will do fine.

[Sam Finch]

Forgot to mention.


It can be shown that if r is not equal to 1 then 1 + r + r^2 + ... + r^n = 0


I've proved it for any n with no prime factor higher than 3.

[Guv]

Sam: If I understand what you are saying, a proof hardly seems necessary.

If r^n = 1, then r is a complex number, and the value (say Q) of any polynomial in r is some complex number.

So Q can be represented as A + iB.

Now can X, Y be found such that X + Y*r = A + iB ?

With A, B, & r known, the above is one equation in two unknowns (X & Y). There should be an unbounded number of solutions for X & Y.

[Sam Finch]

Guv,

I think you've got pretty much what I'm saying, it's just that r isn't nesseserally a complex number, (I did mention it somewhere at the bottom, if not very clearly)

r could be a matrix, a quaternion, even a function, in fact anything with a binary operation which I can treat as multiplication (I'm going to try and define the addition operation later, but for the moment I'm assuming there is one)


The theory is that if you have some elements of some sort, and a binary operation, so that you can form a group from them, then we can find a set of Hypercomplex numbers Isomorphic to that group, from this we can work out other operations and generally increase the scope of what we can do with our elements.


That's the theory anyway, whether it will work or not I don't know. At the moment I'm just looking at cyclic groups, and I'm trying to prove that any cyclic group can be represented by points in the complex plane. Obviously maths being picky I have to show it for all elements of any cyclic group, which is why it requires such a formal proof and I can't just assume they are complex.

but If I can proove it then it looks like that's enough, and I can move onto other abelean groups. Then Try to do the non Abelian ones.

I've proved another couple of statements, which are making them look even more like complex numbers, (actually the first one's a conjectire, but I'm sure I can proove it, I'm just looking for a slightly less messy way than the one I have in mind)

[Guv]

Sam: I did some messy algebra with quaternions. Assuming I did not screw up somewhere (always a possibility), some conclusions I got were interesting.

For Q = A + Bi + Cj + Dk, I got the following results.

Q^2 = (A^2 - B^2 - C^2 - D^2) + 2*A*(Bi + Cj + Dk)

Q^3 =  A^3 - 3*A*(B^2 + C^2 + D^2)
             + (3*A^2 - B^2 - C^2 -D^2)*(Bi + Cj + Dk)

If Q^2 = 1, A cannot be zero, else the square is a negative number. If A is not zero, then B, C, and D must be zero to eliminate the Quaternion part of the square. Hence, only the ordinary real values satisfy Q^2 = 1.

Now, think about Q^2 = -1. A must equal zero to eliminate the Quaternion part of the square.
Now, Q^2 = -B^2 - C^2 - D^2. Obviously: Q = i, Q = j, and Q = k (as expected) satisfy the conditions. Note that there is an unbounded number of Quaternions which satisfy Q^2 = -1
The following are a few.

sqr(1/3)i + sqr(1/3)j + sqr(1/3)k
sqr(½)j + sqr(½)k
sqr(½)i + sqr(1/3)j + sqr(1/6)k

Think about Q^3 = 1.

If (3*A^2 - B^2 - C^2 - D^2) = 0, then the Quaternion part vanishes (necessary for Q^3 = 1).

This implies: 3*A^2 = B^2 + C^ 2 + D^2

Substituting in the real part of Q^3 gives the following (after just a little work).

-8*A^3 = 1 results in Q^3 = 1

Hence A = -1/2 is necessary for Q^3 = 1

Now 3A^2 = B^2 + C^2 + D^2 (From previous analysis).

Or B^2 + C^2 + D^ 2 = 3/4 is necessary and sufficient for the Quaternion part to vanish.

Hence Q^3 = 1 is satisfied by an unbounded number of Quaternions. -1/2 is always the real part.

From the above, my intuition tells me that for n > 2, your theorem is not going to be valid for Quaternions. With all the values satisfying Q^3 = 1, it just seems as though it should be possible to construct a counter example. There are so many degrees of freedom here that constructing a counter example should be possible.

Q^3 = 1 seems to be satisfied by an unbounded number of quaternions. There are no restrictions on the polynomial coefficients. It seems as though you could find some quaternion not expressible as X + Y*r for some r. Then choose polynomial coefficients which would produce that value of X + Y*r.

for n > 3, I would expect similar results. Q^n = 1 is satisfied by an unbounded number of values.