Are there an endless number of imaginary square roots for negative one?

Question

When i learned about imaginary numbers, I learned that i represents the square root of negative one, as does i's counterpart below zero, "-i."

But then I learned the same is true of the quaternion imginary units j and k.

And then I leaned the same is true of the octonion imaginary units e1, e2, e3, e4, e5, e6 and e7.

I couold not find any information on the specifics of the sedonion imaginary units but I assume the same is true of all of them.

And then I learned about the Cayley Dickson Construction which extends these patterns of numbers based on imaginary units endlessly.

Am I wrong in assuming there are therefore an endless number of imaginary square roots to negative one and that this is proven by the Cayley Dickson Construction, but that there are just two real square roots (+1, -1) to positive one?

I'm just an interested math novice so apologies if I've made any errors and please do your best wherever possible to answer as if speaking to a ten year old. Thank you. :-)

Answer 1

It only makes sense to ask how many roots a given number has within a specific number system. In particular, here is the situation with respect to $1$ and $-1$:

• In $\mathbb{R}$, $1$ has two square roots and $-1$ has no square roots.

• In $\mathbb{C}$, both $1$ and $-1$ have two square roots.

• In $\mathbb{H}$ (= the quaternions), every negative real number has infinitely many square roots! For example, not only are $i,j,k$ each square roots of $-1$ in $\mathbb{H}$, but so is ${1\over\sqrt{2}}(i+j)$ since $$(i+j)^2=i^2+ij+ji+j^2=-2.$$ (So there's no need to go any further with the Cayley-Dickson process.)

The term "imaginary" usually refers specifically to $\mathbb{C}$, so I would not say your claim is justified as phrased. Rather, I would say that what the Cayley-Dickson process demonstrates is that once we loosen our sense of what "number" is to allow general "reasonably nice rings," there is nothing preventing everything from having lots of square roots.

Answer 2

It may help to re-imagine what "$i$" is: it is a number - in whatever context you are working in - that satisfies $$i^2 = -1.$$ Or equivalently, it is a root of the equation $$x^2 +1 = 0.$$ Note that this definition doesn't imply existence or uniqueness of $i$! As Noah mentions, over the $\mathbb{R}$, no such solution exists (the polynomial $x^2 + 1$ is in fact irreducible). Over $\mathbb{C}$ there are two solutions; by convention we call one of them $i$ and the other $-i$. In other number systems, such as quaternions, there are more solutions.

However if we are working over a field (an especially nice ring, where multiplication is commutative and nonzero elements have multiplicative inverses), we can in fact say there are at most 2 solutions to the polynomial $x^2 + 1$. So to answer your question: if we are interested in "imaginary square roots" contained in a field, then there can be at most two. $\mathbb{C}$ and $\mathbb{R}$ are fields, while $\mathbb{H}$ is not.

This is useful because it extends to things like finite fields. In the finite field of 5 elements, $\mathbb{Z}/5\mathbb{Z}$, we have $2^2 = 4 \equiv -1 \pmod{5}$, so perhaps 2 deserves to be called $i$ in this setting? This is not standard notation (we would instead call -1 a quadratic residue modulo 5), but illustrates that in other number systems we can find things that satisfy similar properties. On the other hand, the finite field with 3 elements, $\mathbb{Z}/3\mathbb{Z}$, -1 is not a quadratic residue, i.e. there is no element deserving of the name $i$.

One final point: in some sense, this approach doesn't allow us to distinguish between $+i$ and $-i$. After all, both are solutions to $x^2 + 1$. In fact, swapping these labels is precisely complex conjugation, which extends to an automorphism of $\mathbb{C}$ sending $$a + bi \mapsto a - bi.$$ This isn't usually an issue, but to avoid confusion sometimes you'll see people say "fix an algebraic closure of the rational numbers $\mathbb{Q}$." This makes sure we have all agreed on which is $+i$ and which is $-i$ up front (and all the other roots of all the other polynomials!).

Answer 3

Besides Cayley-Dickson Construction, you can have tessarines of $2^n$ dimension. In each such system there is one real unit, $1$, $2^n/2-1$ hypernbolic units ($2^n-2$ non-real roots of 1), and $2^n/2$ imaginary units ($2^n$ non-real roots of $1$). Each such system, regardless of dimension is associative and commutative.

Alternatively, you can consider $\mathbb{C}^n$ with Hadamard (element-wise) product. It is $2n$-dimensional and has $2^n$ roots of $-1$. Each such system is also commutative and associative.

For instance, in $9$-dimensional case, one root of $-1$ is $(i,-i,-i,i,i,i-i,-i,i)$:

$(i,-i,-i,i,i,i-i,-i,i)^2=(-1,-1,-1,-1,-1,-1,-1,-1,-1)=-1$