Prove that quaternions are associative

proton
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Homework Statement


prove that quaternions are associative. ie (qr)s = q(rs), where q,r,s are quaternions

This isn't really a HW problem since I'm just wondering if there's a simpler way to prove associativity than the method I tried below


Homework Equations


i^2 = j^2 = k^2 = -1
ij=k=-ji
jk=i=-kj
ki=j=-ik
q=(a,b,c,d), r=(e,f,g,h), s=(l,m,n,p)



The Attempt at a Solution



Its not so much I don't know how to do the problem as it is that my work is tedious and will take forever, unless there's a simpler method

heres my work:
qr = (a,b,c,d)(e,f,g,h) = ae + af*i^2 + ag *ij + ah*k + be*i + bf*i^2 + bg*ij + bh*ik + ce*j + cf*ji + cg *j^2 + ch*jk + de*k + df*k + dg*jk + dh*k^2 = (ae-bf-cg-dh, af+be+ch-dg, ag-bh+ce+df, ah+bg=cf+de)

(qr)s = (ae-bf-cg-dh, af+be+ch-dg, ag-bh+ce+df, ah+bg=cf+de)(l,m,n,p) =
(lae-lbf-lcg-ldh-maf-mbe-mch+mdg-nag+nbh-nce-ndf-pah-pbg+pcf-pde, ...)
I got too tired to work out the i,j,k components

Theres got to be a simpler way to prove associativity
 
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There may be but there doesn't have to be. "Associativity" is always the most tedious property to prove! You are going to have to learn not to get so tired so quickly.
 
it took me a FULL page to prove (qr)s =q(rs)
 
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If one were to show quaternions could be represented as a matrix and the product defined as standard matrix multiplication, would associativity follow as a consequence of the fact that a matrix represents a linear transformation and the matrix product is functional compostion?
 
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If you start with an associative algebra and then modding out by some relations... then the result is automatically an associative algebra.
 
You can first prove the associativity for any triple of 4 basis quaternions 1,i,j,k, and then check that the same property holds for any real linear combination of them.

Eugene.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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