# What was the Hamilton's initial problem that led to him inventing quaternion?

I read wiki topic about history of Quaternion, and it confuses me on why, according to Hamilton, there's a problem with multiplication of triple (i.e. 1+i+j), and somehow the quadruple (or Quaternion) solves that problem.

What was the trouble he faced when he tried to define multiplication for triple? and why did introduction of the 4th element solve the problem?

• "A History of Algebra: From al-Khwārizmī to Emmy Noether" has a few pages that covers this pretty well. – arctic tern Feb 17 at 23:45

Let's say you've just invented a new square root of $$-1$$, called $$j$$, to augment the complex numbers. That's what Hamilton did to describe rotations in $$3$$ dimensions, the way complex numbers work for rotations in $$2$$. Now, you need your number system to be closed under multiplication, but is $$ij$$ of the form $$a+bi+cj$$ with $$a,\,b,\,c\in\Bbb R$$? Sadly not. Indeed, Hurwitz's theorem shows with hindsight what the problem was: Hamilton was trying to do in dimension $$3$$ what can only happen in dimension $$1$$, $$2$$, $$4$$ or $$8$$.

• I wonder how ij is not form of a + bi + cj, but ijk is? Hamilton defines ijk = -1, but why couldn't he define ij = -1? What if I define ij = -1, then is there any problem with such definition? – pbeta Feb 12 at 4:11
• @pbeta Rotations have a matrix representation, so our new numbers need to associate. I invite you to compute $iij$ to show the conjectured form won't work. Since $k:=ij$, $ijk=ijij=-iijj=-1$ by anticommutativity, which allows conjugation to work the expected way. – J.G. Feb 12 at 6:03
• if $ij = -1$, you may multiply both terms of the equation by $-i$ and obtain $j = i$. – mau Feb 12 at 6:13