22
$\begingroup$

I wanted to factorize $a^2+b^2+c^2$ into two factors in a similar way to $$a^2+b^2 = (a+ib)(a-ib)$$ This doesn't seem to be possible using real or complex numbers. However I came up with the following idea $$ (a + ib + jc) (a -ib -jc) = a^2+b^2+c^2$$ if we define $$i^2=j^2=-1$$ and $$ ij = -ji.$$ So instead of using complex numbers I had to define a kind of double complex number obeying a non-commutative multiplication rule to solve my factorization problem. Does my definition of this numbers make sense or is it somehow inconsistent? Do these numbers already have a name in mathematics and are they used? Is there any literature about them?

$\endgroup$
5
  • 5
    $\begingroup$ Yes this works. but you are really working in quaternions, which has $3$ 'imaginary numbers,' $i,j,k$. en.wikipedia.org/wiki/Quaternion. Basically, $ij=k$. $\endgroup$ May 19, 2012 at 21:20
  • 13
    $\begingroup$ Congratulations on rediscovering the quaternions! $\endgroup$ May 20, 2012 at 1:20
  • 3
    $\begingroup$ I agree, you have had a very nice insight there. Hamilton searched for a long time -- it was a running joke among his kids every morning -- for a way to generalize complex numbers into 3D space. He "got it" while walking with his wife over a bridge, realizing that he needed three $i$-like quantities, not just two, to make everything work. Despite quaternions falling out of favor later on, they were instrumental both the the development of vectors and concepts such as dot and cross products, all of which are quaternions subsets. Maxwell's original equations use quaternions, not vectors. $\endgroup$ May 20, 2012 at 4:01
  • $\begingroup$ @QiaochuYuan Thank you. "One cannot invent useful things, one can only rediscover them." ;-) $\endgroup$
    – asmaier
    May 20, 2012 at 13:00
  • $\begingroup$ I should add on @Terry's comment that the bridge mentioned has an inscription about this realization. One can still see this in Dublin (if one is not too hazy from drinking). $\endgroup$
    – Asaf Karagila
    May 22, 2012 at 22:15

3 Answers 3

18
$\begingroup$

You have come across the quaternions. They are numbers of the form $$a+bi+cj+dk$$ where $a,b,c,d\in\mathbb{R}$ and $i$, $j$, and $k$ are symbols satisfying $$i^2=j^2=k^2=ijk=-1$$ $$ij=k,\quad jk=i,\quad ki=j$$ $$ji=-k,\quad kj=-i, \quad ik=-j$$ Multiplication of quaternions is non-commutative in general, but it is still associative.

The conjugate of a quaternion $q=a+bi+cj+dk$ is defined to be $$q^*=a-bi-cj-dk,$$ and the norm of a quaternion is defined to be. $$\|q\|=\sqrt{qq^*}.$$ After expanding everything out, one can see that $$\|a+bi+cj+dk\|=\sqrt{a^2+b^2+c^2+d^2}.$$ So, to factor the expression $a^2+b^2+c^2$, you were just using quaternions for which the coefficient of $k$ is equal to $0$: $$(a+bi+cj+0k)(a-bi-cj-0k)=a^2+b^2+c^2+0^2=a^2+b^2+c^2.$$ There are other extensions of the complex numbers that are possible: the other one usually mentioned are the octonions, which are $8$-dimensional over the real numbers.

$\endgroup$
3
  • $\begingroup$ Is there a special name for quaternions which have the coefficient of $k$ set to zero? And can't there also be quintions which could then factorize $a^2+b^2+c^2+d^2+e^2$ into two factors ? $\endgroup$
    – asmaier
    May 20, 2012 at 12:55
  • $\begingroup$ No, there are no quintions. The next number system is octonions, which are not associative. $\endgroup$ May 20, 2012 at 14:37
  • $\begingroup$ @asmaier Quaternions with their $k$ component zero have no special name because they aren't closed under multiplication; I've added an answer that explains the details on this. $\endgroup$ May 22, 2012 at 22:32
3
$\begingroup$

It's worth pointing out that you need to go up to the quaternions - a four-dimensional space - because a factorization of the sort you're describing doesn't 'work' in just three dimensions. Suppose you had $z=a+bi+cj$ (and so $\bar{z}=a-bi-cj$) and $w=d+ei+fj$ (with $\bar{w}=d-ei-fj$). Then $x=zw$ would be of the form $r+si+tj$, with $r$, $s$ and $t$ each expressions in $a\ldots f$, and $\bar{x}$ would be $r-si-tj$; taking the norms (or in other words looking at $|x|^2 = x\bar{x} = |z|^2|w|^2$) would give you an identity of the form $(a^2+b^2+c^2)\cdot(d^2+e^2+f^2) = r^2+s^2+t^2$. But consider $(1^2+1^2+1^2)\cdot(4^2+2^2+1^2) = 3\cdot 21 = 63$; this number is of the form $8n+7$ and so by the Sum Of Three Squares theorem it can't be expressed as $r^2+s^2+t^2$ for any values of $r,s,t$. This means the three-squares identity (or in other words, the three-dimensional product you're looking for) can't exist.

$\endgroup$
1
$\begingroup$

Going from complex numbers to the quaternions you lose commutativity. Going from quaternions to octonions you lose associativity. What’s next? There is very nice and elegant mathematics around these hypercomplex numbers. Adolf Hurwitz proved it in 1898 that every normed division algebra with an identity is isomorphic to one of the following four algebras: $\mathbb{R}$, $\mathbb{C}$, $\mathbb{H}$ and $\mathbb{O}$, that is the real numbers, the complex numbers, the quaternions and the octonions. And that’s it. No more, no less.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .