# Between complex numbers and quaternions?

Complex numbers are $a+ b i$; Quaternions are $a + b i + c j + d k$. So, do there exist numbers like $a + b i + c j$? Here $a$, $b$, $c$, $d$ are all real.

Did Hamilton consider such a case?

• No, there is no real division algebra of dimension $3$, see here. But Hamilton tried for a long time to find such numbers. Sep 1, 2015 at 14:42
• So Hamilton knew this fact? Sep 1, 2015 at 14:44
• Yes, Hamilton sort of knew it, i.e., realized it the hard way. Sep 1, 2015 at 14:44

In fact Hamilton originally tried to construct precisely such a theory -- a three-dimensional analogue for the two-dimensional representation of the complex numbers. In order to define multiplication in such a system, you have to specify what the product of $i$ and $j$ should be. One obvious choice is to say that $ij=0$, but then you have a problem: the system you are working with is not a field (nor even an integral domain), and Hamilton wanted to construct a three-dimensional field. Hamilton tried various combinations, but in none of them did the geometric interpretation of the 3-vectors match up properly with the algebraic computations. He finally concluded that the only way to make everything work out was to introduce a fourth "direction", $k$, and to make multiplication non-commutative.
But of course you don't need to feel tied down to Hamilton's criteria; in particular, if you don't care about whether your algebraic structure is an integral domain then you can define lots of different multiplicative structures on the set of objects of the form $a + bi + cj$, corresponding to different choices of the products $ij$ and $ji$.
Compare $b$i+$c$j+$d$k with the vector $(b,c,d)$ and for multiplication of such entities compare quaternion multiplication with the vector product. You will see that there is a clear relationship between the two.