Is there a way to introduce quaternions and octonions in a similar way to how we are typically introduced to complex numbers?

Question

So I've been reading a little bit into ideas around quaternions and octonions. I just read the following explanation that introduces them as what happens when you have complex numbers and you then ask "but what if there was another square root of $-1$?":

http://www.askamathematician.com/2015/02/q-quaternions-and-octonions-what/

Now I've skim-read different things and seen a few different ways of introducing and explaining how complex numbers, quaternions and octonions relate to each other, however this particular explanation made me curious.

See, when we get introduced to complex numbers, it's not as a "oh what if there was an extra square root to $-1$" question, but a "there should be a square root to $-1$ but with only real numbers it's undefined". That is, there's an actual equation ${x}^{2} = -1$ that you're trying to solve and can't with only real numbers.

My main question then - is there a similar equation or problem where, with only real and complex numbers, you cannot solve it without introducing quaternions? And similarly again for octonions?

Finally, if instead the way the website above introduces these concepts is, in some sense, fundamental, then what exactly is special about the number $-1$? For instance, why not introduce new number systems based on defining new square roots of some completely different number(s)?

Answer 1

There's nothing too special about looking at square roots of $-1$. It's clear by the fundamental theorem of algebra if you want more than 2 square roots of a number you have to go beyond $\mathbb C$. If we want to get more square roots of various numbers, we need to consider more general rings, algebras, etc., such as the quaternions or matrix algebras.

But, if you want a (finite-dimensional, associative) division ring over $\mathbb R$, then all you have are $\mathbb C$ and $\mathbb H$, Hamilton's quaternions. You can construct $\mathbb H$ by adjoining to $\mathbb C$ another square root $j$ of any negative real number (what's important here is that $j^2$ is not already a square in $\mathbb R$) such that $ij = -ji$.

That said, while looking for more square roots of numbers is an amusing way to introduce $\mathbb H$, there are more motivated ways to introduce $\mathbb H$ (Hamilton's geometric motivations, factoring numbers of the form $x^2+y^2+z^2+w^2$, looking for (skew)fields beyond $\mathbb C$, ...).

Answer 2

There is a nice trick called "Cayley-Dickson doubling" or something like that, which produces a new normed division algebra $D(A)$ from a normed division algebra $A$ equipped with an $\mathbb R$-linear antiautomorphism $x \mapsto \bar{x}$, where $D(A)$ is then also equipped with an antiautomorphism $h \mapsto \bar{h}$.

As a normed vector space, $D(A)\cong A\oplus A$, and $\overline{(a,b)} = (\bar{a},-b)$, and $$ (a,b).(c,d) = (ac-\bar{d}b,b\bar{c}+da). $$ Note that the map $a \mapsto (a,0)$ which identifies $A$ with the subspace $A\oplus \{0\}\subseteq A \oplus A$ is a map of algebras, and thus $A_k$ is normally viewed as a subalgebra of $A_{k+1}$ via this inclusion.

If we set $A_0=\mathbb R$ and the trivial (anti)automorphism $\bar{x}=x$, then we may inductively set $A_{k+1} =D(A_k)$ to get a sequence of algebras $(A_k)_{k \geq 0}$, which begins $A_0 =\mathbb R,A_1=\mathbb C,A_2=\mathbb H,A_3=\mathbb O,...$ where $\mathbb H$ denotes the quaternions and $\mathbb O$ the octonions.

While the sequence continues for all positive integers $k$, at each stage the algebra $A_{k+1}$ is, informally speaking, more unmanageable that $A_k$. Indeed $A_1=\mathbb C$ has a nontrivial antiautomorphism $z\mapsto \bar{z}$, though it is still a field. $A_2 = D(A_1) = \mathbb H$ is no longer commutative (so it is a division algebra), while $A_3 = D(\mathbb H) = \mathbb O$ is no longer associative (though it is "alternative" meaning it is associative on any two-dimensional subspace).

On the other hand, if the quadratic norm $N\colon A_k\to \mathbb R$ on $A_k$ is given by $N(a) = \bar{a}.a = a.\bar{a}$, then $$ \begin{split} (a,b).\overline{(a,b)} &= (a,b).(\bar{a},-b) = (a\bar{a}+\bar{b}b,ba+(-ba))\\ &= (\|a\|^2+\|b\|^2,0) = N(a)+N(b) \end{split} $$ and similarly $$ \begin{split} \overline{(a,b)}(a,b) &= (\bar{a},-b).(a,b) = (\bar{a}a+(-\bar{b}).(-b),b\bar{a}+(-b).\bar{a}))\\ &= (\|a\|^2+\|b\|^2,0) = N(a)+N(b) \end{split} $$ (using our identification of $A_k$ with the subalgebra $A\oplus \{0\}$ of $D(A)$. Thus for all $k\geq 0$ we have $N(a) = a\bar{a}=\bar{a}a$.