4
$\begingroup$

I've seen the multitude of questions here about defining complex numbers, but I haven't found the answer I'm looking for there:

In elementary algebra, we typically get to $x^2=-1$, define $i=\sqrt{-1}$, and call it a day. In analysis, we define the complex numbers as ordered pairs, along with addition and (a strange-looking) multiplication, then derive the fact that $i^2=-1$. The only places I've seen attempt to justify this simply state it's "more mathematical" to do the latter.

My question is: why do we need to define the complex number system before defining $i$? Why not state $i \triangleq \sqrt{-1}$ and then construct the complex numbers starting from that definition?

$\endgroup$
6
  • 1
    $\begingroup$ a problem that I see is that after you do that you still would need to set $i^2:=-1$, that is, there is no way you can get the identity $i^2=-1$ from just the definition of $i$, indeed you could see natural that $i^2=\sqrt{-1}\cdot \sqrt{-1}=\sqrt{(-1)(-1)}=\sqrt{1}=1$ $\endgroup$
    – Masacroso
    Jan 12 at 3:07
  • $\begingroup$ @Masacroso I think that the situation is not so clear cut. In $\mathbb{R},$ it is true that $\sqrt{a} \times \sqrt{b} = \sqrt{ab}.$ However, it is also true that for $s \geq 0, ~r = \sqrt{s} \implies r^2 = s.$ Personally, my arbitrary instinct is therefore that $i = \sqrt{-1} \implies~$ (by itself) that $~i^2 = -1.$ $\endgroup$ Jan 12 at 3:42
  • 1
    $\begingroup$ @user2661923 the square root is not defined naturally in $\mathbb{R}$, just in the non-negative reals $\endgroup$
    – Masacroso
    Jan 12 at 3:45
  • $\begingroup$ @Masacroso Yes, but once you define $i = \sqrt{-1}$, that definition, (by itself) is a declaration that the square root function has been extended to negative numbers. So, absent any further specification, (i.e. that $i^2 = -1$), it seems reasonable for me to conclude that $i = \sqrt{-1} \implies i^2 = (-1).$ As I say, such a conclusion (of mine) is certainly arbitrary. However, the conclusion seems very natural, as it keys on the fundamental idea behind the square root function. $\endgroup$ Jan 12 at 3:49
  • $\begingroup$ How do you know the properties of the extended square root function hold for negatives the same way they hold for positives? $\endgroup$
    – SenZen
    Jan 12 at 4:10
7
$\begingroup$

Strictly speaking, you can't just make definitions and hope they work. Practically speaking though, mathematicians have a lot of machinery built precisely to make sense of definitions that basically go, "Hm, I wish I had a element that had this property" - and these sorts of definition are ubiquitous in many branches of math. This extends well beyond defining arithmetic on the complex numbers, but let me briefly explain how one may define $\mathbb C$ using the machinery that exists for algebraic structures.

There's a pretty common way to define the complex numbers that basically just says "$\mathbb C$ can be constructed by adding to $\mathbb R$ an extra element $i$ such that $i^2=-1$." You usually avoid square roots in any definition since they're a bit ambiguous. This definition reads as follows: $$\mathbb C = \mathbb R[i]/(i^2+1)$$ This is a ring theoretic construction, so let me unpack it: we're going to construct $\mathbb C$ from $\mathbb R$ in two steps:

  1. Consider first the polynomial ring $\mathbb R[i]$. This consists of every polynomial in some variable $i$ - so has terms such as $i^3+i^2+17i+289$ or $i^{100}$ or just real constants such as $\pi$ - all with the usual rules for adding and multiplying. Polynomial rings capture the idea of "adding" or "adjoining" a new variable to a given ring without specifying any particular properties thereof. If you wanted to add a new variable, but didn't know anything about it, this is what you'd get.

  2. The $/(i^2+1)$ part of the definition says: Take that polynomial ring, and now consider two polynomials $P(i)$ and $Q(i)$ to be equivalent if their difference $P(i)-Q(i)$ is a multiple of $i^2+1$. This is a bit of a tricky thing to get your head around, but note that $i^2$ is equivalent to $-1$ since their difference is $1\cdot (i^2+1)$. Similarly, $i^4$ is equivalent to $1$ since the difference satisfies the relation: $$i^4-1=(i^2-1)(i^2+1)$$ where these are equations of polynomials (and are true of any variable or value as input for $i$).

The second rule, as it turns out, makes every polynomial equivalent to one of the form $a+bi$ - if you have a quadratic or higher term, you can find some multiple of $i^2+1$ with the same leading term as your quadratic, and subtract this multiple from your original polynomial, leading to an equivalent polynomial of lesser degree. Algorithmically, if you wanted to figure out what $i^3+2i^2$ is, you would first subtract $i\cdot (i^2+1)=i^3+i$ from this to get $2i^2-i$, then subtract $2(i^2+1)$ from that to get that your original polynomial $i^3+2i^2$ is equivalent to $-i-2$. This process is known as polynomial long division.

Note that addition and multiplication rules can then be derived - for instance $$(a+bi)+(c+di)=(a+b)+(c+d)i$$ is just ordinary addition of polynomials. For multiplication you get: $$(a+bi)(c+di)=ac+(ad+bc)i+bdi^2$$ as polynomials and then you can subtract $bd(i^2+1)$ (which is equivalent to $0$) to get: $$(a+bi)(c+di)=(ac-bd)+(ad+bc)i$$ in $\mathbb C$.

It turns out that this definition is really the same (in a sense that can be made precise as a universal property in category theory) as saying (where a "ring" is just a structure in which addition and multiplication are sensibly defined):

$\mathbb C$ is a ring containing $\mathbb R$ and an extra element $i$ where the only equalities between elements are those that can be proven from the assumption that $i^2+1=0$.

And the prior equations can be interpreted in this context too - if we knew that $i^2=-1$, deducing that $(a+bi)(c+di)=(ac-bd)+(ad+bc)i$ would simply be done by distribution followed by a substitution. The explicit ring theoretic construction is mostly needed to ensure that there actually is some sensible $\mathbb C$ that realizes this property - however, these sorts of constructions exist in great generality; if all you want to do is add, subtract, and multiply (but not necessarily divide), you can basically add however many variables you like and demand that they satisfy as many polynomial equations as you want, and you'll get, well, something out - though that thing can be badly behaved in many ways (even in ways as extreme as "oops, everything equals zero now"). The complex numbers, however, turn out to be a very well-behaved example of such a construction, fundamentally because $i^2+1$ cannot be factored as a real polynomial.

It's a bit more tricky when you want to talk about analysis and topology (since this construction only deals with algebraic not analytical properties), but this is not a particularly great obstacle - this definition can be refined in various ways to preserve those kinds of structures too.

$\endgroup$
2
  • 1
    $\begingroup$ this doesn't answer the question, here you just throw the usual definition of $\mathbb{C}$ but you are not showing why starting from $i:=\sqrt{-1}$ is problematic, that is what the OP asked $\endgroup$
    – Masacroso
    Jan 12 at 19:30
  • $\begingroup$ @Masacroso The purpose of my answer is to say that starting from $i=\sqrt{-1}$ is basically fine and to show how one common definition of $\mathbb C$ captures that idea. I interpret the question to be asking about how to define $\mathbb C$ without any arbitrary rules for multiplication using the idea that $i$ should be the square root of $-1$. Yes, there is also some valid question about whether one can define things to be "the square root of" rather than "a square root of" which is not answered here, but the context in the question leads me to believe that this is not what is being asked. $\endgroup$ Jan 12 at 19:56
5
$\begingroup$

Here is an answer which addresses the underlying logical issues that your question raises.

I could define a jackalope to be be a rabbit with antelope horns, but then the trouble arises whether any rabbit with antelope horns exists, because if not then my definition is vacuous nonsense.

Similarly, you could define $i$ to be $\sqrt{-1}$, but then you would have to contend with the problem of whether $\sqrt{-1}$ exists. In the real number system it is a theorem that $\sqrt{-1}$ does not exist, or to be more precise there is no real number solution to the equation $x^2=-1$.

So, until someone produces a reasonable number system in which $x^2=-1$ has a solution, the definition $i=\sqrt{-1}$ is vacuous nonsense.

Fortunately, someone has done exactly that, namely produced a number system (in fact a field) in which $x^2=-1$ exists. Namely, those ordered pairs of real numbers with that strange looking multiplication formula that you mentioned in your post.

$\endgroup$
2
  • 2
    $\begingroup$ I think that this is not the point of the question, as you can assume from the beginning that $i$ is a symbol to be added to the set of real numbers, as we add the infinity symbol to the extended real line, so existence is not the point. $\endgroup$
    – Masacroso
    Jan 12 at 3:18
  • 1
    $\begingroup$ My answer is not about non-existence of the symbol $i$. It is about lack of meaning of the expression on the other side of the equation involved in the definition of $i$, namely the expression $\sqrt{-1}$. $\endgroup$
    – Lee Mosher
    Jan 12 at 14:29
1
$\begingroup$

As I see the problem is that, after you extend the square root function from $[0,\infty )$ to $\{-1\}\cup [0,\infty )$ adding a symbol $i:=\sqrt{-1}$ to $\mathbb{R}$, you still need to construct a minimal field (or an algebra) that contains $\mathbb{R}\cup \{i\}$, and you still would need to define what is the value of $i^2$.

One approach could be to set a multiplication operation in the new space such that

$$ \sqrt{a}\cdot \sqrt{b}=\sqrt{a\cdot b}\quad \text{ for all }\,a,b\in \mathbb{R} $$

and this would imply that $i^2=1$, what would give an algebra on $\mathbb{F}:=\{a+bi: a,b\in \mathbb{R}\}$ (the Clifford algebra $\operatorname{Cl}_{0,1}(\mathbb{R})$) but not a field. Thus, by itself, setting the value of $i$ doesn't lead immediately to a value for $i^{2}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.