
#1
Mar1108, 10:45 PM

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[tex]^{2}[/tex]=1 2. Relevant equations Elements of H is of the form a.1 +bi+cj+dk where a, b, c, d in [tex]\textsl{R}[/tex] ( reals) and i[tex]^{2}[/tex]= j[tex]^{2}[/tex]= k[tex]^{2}[/tex]=ijk = 1. 3. The attempt at a solution Let u = a.1 +bi+cj+dk then u[tex]^{2}[/tex]=a[tex]^{2}[/tex]b[tex]^{2}[/tex]c[tex]^{2}[/tex]d[tex]^{2}[/tex]+2a(bi+cj+dk)+2bicj+2bidk+2cjdk and this = 1 provided a=0 and b[tex]^{2}[/tex]c[tex]^{2}[/tex]d[tex]^{2}[/tex]=1 and 2bicj+2bidk+2cjdk= 0 but I do not see how 2bicj+2bidk+2cjdk= 0. 



#2
Mar1108, 10:50 PM

Emeritus
Sci Advisor
PF Gold
P: 16,101

Two comments:
(1) You do know what the product [itex]i \cdot j[/itex] is, right? (2) bēcēdē=1 can never be satisfied. 



#3
Mar1208, 12:31 AM

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 bēcēdē=1 But still, I do not see why 2bicj+2bidk+2cjdk= 0. Is there anything that I am missing here? 



#4
Mar1208, 01:02 AM

P: 640

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? 



#5
Mar1208, 01:03 AM

P: 640

I guess the question is for your group what is [bi,cj] = ? or even {bi,cj} = ?




#6
Mar1208, 01:08 AM

P: 14

Yes!!! Thank you very much. I see it now. Poor me.




#7
Mar1208, 01:09 AM

P: 14





#8
Mar1208, 09:29 AM

Emeritus
Sci Advisor
PF Gold
P: 16,101

(I note that you already assumed this fact when you simplified bibi to bē) 



#9
Mar1208, 12:22 PM

P: 640

[A,B] = ABBA {A,B} = AB+BA 


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