# Quaternion ring

by Gtay
Tags: quaternion, ring
 P: 14 1. The problem statement, all variables and given/known data Show that the quaternion division ring H has infinitely many u satisfying u$$^{2}$$=-1 2. Relevant equations Elements of H is of the form a.1 +bi+cj+dk where a, b, c, d in $$\textsl{R}$$ ( reals) and i$$^{2}$$= j$$^{2}$$= k$$^{2}$$=ijk = -1. 3. The attempt at a solution Let u = a.1 +bi+cj+dk then u$$^{2}$$=a$$^{2}$$-b$$^{2}$$-c$$^{2}$$-d$$^{2}$$+2a(bi+cj+dk)+2bicj+2bidk+2cjdk and this = -1 provided a=0 and -b$$^{2}$$-c$$^{2}$$-d$$^{2}$$=1 and 2bicj+2bidk+2cjdk= 0 but I do not see how 2bicj+2bidk+2cjdk= 0.
 Emeritus Sci Advisor PF Gold P: 16,091 Two comments: (1) You do know what the product $i \cdot j$ is, right? (2) -bē-cē-dē=1 can never be satisfied.
 P: 14 Thanks for the comments (1) The product of i.j =k but this ring is not a commutative ring and so I do not have 2bicj = 2bcij=2bck. ij=k, jk=i, ki=j, ji=-k, kj=-i, ik=-j (2) My bad. I meant -bē-cē-dē=-1 But still, I do not see why 2bicj+2bidk+2cjdk= 0. Is there anything that I am missing here?
 P: 641 Quaternion ring if its not a commutative ring then how do you have 2bicj? Shouldnt that term then be : i j (b*c) + j i (c*b) If ij=-ji, and c*b=b*c then these cancel, right?
 P: 641 I guess the question is for your group what is [bi,cj] = ? or even {bi,cj} = ?
 P: 14 Yes!!! Thank you very much. I see it now. Poor me.
P: 14
 Quote by K.J.Healey I guess the question is for your group what is [bi,cj] = ? or even {bi,cj} = ?
I do not know what [bi, cj] or {bi, cj} stand for.
Emeritus
PF Gold
P: 16,091
 Quote by Gtay (1) The product of i.j =k but this ring is not a commutative ring and so I do not have 2bicj = 2bcij=2bck.
But fortunately, the real numbers are in the center of the ring -- a fact that is usually explicitly given by describing the quaternions as an algebra over the reals. However (unless I made an error), you can actually derive this fact from the identities given.

(I note that you already assumed this fact when you simplified bibi to -bē)
P: 641
 Quote by Gtay I do not know what [bi, cj] or {bi, cj} stand for.
commutator and anticommutator:
[A,B] = AB-BA
{A,B} = AB+BA

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