Interpretation of complex number in 1d.

Question

In order to solve the equation $$x^2 + 1 = 0$$ we introduce the imaginary number $i$. But to interpret the complex numbers we think of them as laying in the complex plane, which is a 2d plane.

It turns out that instead of introducing the imaginary unit $i$ we can represent the problem with $2\times 2$ matrices. Since they follow the same algebra, if we define $$R= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$ and $$i= \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}.$$

In both cases it seems that we start with an equation in 1d but end up with having to use 2d solution methods.

To me this solution method seems to solve the problem but supply a wrong interpretation. Are there other representations of the solution which does not have this problem?

I'll rephrase my question: Is there a way to solve $$x^2 + 1 = 0$$ without having to introduce objects which has a natural 2d interpretation?

Answer 1

You need to introduce 2-D objects. Like said in the comments, you cannot see complex numbers on a straight line because this would mean they have an ordering (which is proven they do not have). The point here is that you need a mathematical structure in which an element (called $i$ often) is such that $i^2=-1$. So you introduce the imaginary numbers to fulfill this request. Now if you go and define operations, you see that they are well defined and thus you have come up with an algebraic structure in which you can actually solve your equation. Since the element $i$ is not real, you define the basis $\{1,i\}$ for the new object you are dealing with and it is just natural to represent a vector space of dimension 2 with a plane.

As you have pointed out, complex numbers can also be represented through real matrices. In fact, they are just another equivalent way to write complex numbers, because again what is important is the structure your operations define, not the objects you are dealing with.

So I think the answer would be no, because you need to introduce a new element (call it $i$ or whatever you want) which is not a real number, and thus you will always end with a basis of 2 elements.

Answer 2

Here is a model of the complex numbers that you may prefer, which makes the connection with polynomial equations more explicit.

We would like to find some things that behave like numbers in the following ways:

1. They can be added and multiplied
2. They contain a subset that behaves just like the real numbers
3. But they also contain a solution to the equation $x^2+1=0$, which regular numbers do not.

The matrices and the ordered pairs do this. Here is a different kind of object that also works.

We will take the set of all polynomials with real coefficients, and divide it into classes. Every polynomial $P$ will be in exactly one class, which we will write $[P]$. Note that since two different polynomials might be in the same class, we might have $[P]=[Q]$ even when $P \ne Q$. (Just like the way two different people might live in the same country.)

Each class of polynomials will be a new "number" in our new system. To be reasonable "numbers" we need to explain:

1. How can these classes be added and multiplied?
2. Where are the ordinary real numbers among these classes?
3. Which class solves the equation $x^2+1 = 0$?

The answers are deep but simple! $[P]+[Q] = [ P+Q ]$ takes care of addition. $[P]·[Q] = [ PQ ]$ takes care of multiplication. The ordinary real number $r$ appears as $[ r ]$, the class containing the zeroth-degree polynomial whose constant term is $r$. Note that this preserves the behavior of real numbers with respect to ordinary addition and multiplication; for example $[3]+[4] = [7]$ and $[3]·[4] = [12]$ as we want.

If each class contained only one polynomial, nothing interesting would have happened. We would just dealing with the same polynomials as always, but with brackets written around them. And we want to find something that satisfies the equation $x^2+1=0$, but there is no polynomial $P$ for which $[P]^2 + [1] = [0]$. We will fix this by introducing a new rule:

Important Rule: Whenever $P-Q$ is a multiple of $x^2+1$, we will merge the two classes for $P$ and $Q$ into a single class.

So for example, $1, X^2+2, $ and $X^3+X+1$ are all in the same class, and $[1]= [X^2+2]= [ X^3+X+1]$, because we can pick any two of them, and the difference between them is always a multiple of $X^2+1$. (You check this.)

With this important rule, all the multiples of $X^2+1$ itself are in a single class (why?), and in particular, $X^2 + 1$ is in the same class as the polynomial $0$, so $[X^2+1]$ is the exact same class as $[0]$.

In this new system, the equation $x^2+1=0$ has a solution! It is simply $[X]$. Why? Because:

$$\begin{align} [X]^2 + [1] & = [X][X] + [1] \\ & = [X^2] + [1] \\ & = [X^2 + 1] \\ & = [0]. \end{align} $$

The Important Rule plays the crucial role here. Without it, we wouldn't have $[X^2+1] = [0]$ for the last step.

So these polynomial-class-things do what we wanted: we can add and multiply them, and they have a sub-structure that behaves just like the real numbers and for which the addition and multiplication are just the ordinary addition and multiplication of reals. And there is a polynomial-class-thing that is a solution of $x^2+1=0$. (There are two, in fact: $[-X]$ is also a solution, and it is a different solution: $[X]≠[-X]$ because $X-(-X)$ is not a multiple of $X^2+1$.)

What is this new system like? It is not as complicated as it might seem at first. We can prove that for every polynomial $P$, there are some real numbers $a$ and $b$ for which $[P] = [aX+b]$—every polynomial lives in the same country as some first-degree polynomial. This isn't hard to see. First let's note that $[X^2] = [-1]$ (because $X^2 - (-1)$ is a multiple of $X^2+1$). Then multiplying on both sides by $[X]$ we get $[X^3] = [-X]$ (note that $X^3 - (-X)$ is a multiple of $X^2+1$) and then $[X^4] = [1]$ ($X^4-1$ is $(X^2-1)·(X^2+1)$) and so on. Then for example $$\begin{align} [aX^4 + bX^3 + cX^2 + dX + e] & = [a][X^4] + [b][X^3] + [c][X^2] + [d][X] + [e] \\ &= [a][1] + [b][-X] + [c][-1] + [d][X] + [e] \\ &= [a·1 + b·-X + c·-1 + d·X + e] \\ & = [(-b+d)X + (a-c+e)],\end{align}$$ and every other polynomial similarly lives in the same class as some linear polynomial $[pX+q]$.

But each linear polynomial lives in a different class. We never have $[pX+q] = [rX+s]$ because $(pX+q)-(rX+s)$ is never a multiple of $X^2 + 1$. Except when $pX+q = rX+s$, of course.

Each class contains exactly one element of the form $aX+b$, where $a$ and $b$ can be any real number. So the system of the classes is two dimensional, because every $[P]$ can be reduced to $[aX+b]$, and ultimately this is because $X^2+1=0$, the polynomial in the Important Rule, is a second-degree polynomial.

Usually we write $[aX+b]$ as $b + ai$, but that is just a simple change of notation. But the complex number $b+ai$ is essentially the class $[aX+b]$, and it is intrinsically two-dimensional because it is constructed from the space of first-degree polynomials, which is two-dimensional. You said:

it seems that we start with an equation in 1d…

but the set of such equations is not one-dimensional; it is two dimensional.

(Jack D'Aurizio's comment distills this to one sentence, but it is probably opaque if you have not seen the construction. I left out some important considerations though. From my description it might seem like you could do the same thing, but change the important rule to say that that $[P]=[Q]$ whenever $P-Q$ is a multiple of $X^4+1$, instead of $X^2+1$. Then instead of two-dimensional numbers, do we get four-dimensional numbers? Sadly, no, something goes wrong that I quietly swept under the rug in my description above. )