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So I was wondering why the only imaginary number we define is $\sqrt{-1}$ why do we not consider stuff like $\log{0}$ or $\frac{1} {0} $ or even $-1!$ as "imaginary"?

I don't understand why we picked $\sqrt{-1} $, is there something special about it?

Thomas Andrews
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Sasha Ivanov
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  • How stringent are you in terms of defining $i$? – imranfat Aug 22 '21 at 02:29
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    $\sqrt{-1}$ obeys the usual rules of arithmetic. There is no way to define those other things that doesn't break arithmetic. – Gerry Myerson Aug 22 '21 at 02:32
  • For the first two examples, $\pm \infty$ are not imaginary. For the last one, that is $-1!$ it is purely real, equals to $-1$. If you meant $(-1)!$, then take a look at the gamma function. – Hussain-Alqatari Aug 22 '21 at 02:35
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    @Hussain, if you go to the Gamma function, you'll still find $(-1)!$ is undefined. – Gerry Myerson Aug 22 '21 at 02:37
  • Firstly, $i$ has several applications in real life, and so it is NOT imaginary, in the literal sense. Secondly, as pointed out above, $-1!$ or $(-1)!$ actually has a value (see Gamma function), and lastly, defining $1/0$ to be something would not be consistent with the existing rules of maths. – ultralegend5385 Aug 22 '21 at 02:39
  • @gerryMyerson Yes I know, "undefined", still not "imaginary". – Hussain-Alqatari Aug 22 '21 at 02:42
  • Because $\sqrt{-1}$ is useful. $\log(0)$ and $\frac{1}{0}$ are not. – morrowmh Aug 22 '21 at 02:46
  • @ultra, $(-1)!$ does not have a (finite, real) value. The Gamma function has a pole where $(-1)!$ would be. – Gerry Myerson Aug 22 '21 at 02:52
  • Fundamentally, because it is useful, while it is not useful - in fact, usually detrimental, to define $\log 0$ or $1/0.$ There are subtle reasons why we can see the complex numbers “should exist” for some definition of “should,” but really too long for a comment. – Thomas Andrews Aug 22 '21 at 02:59
  • Also, yes, “imaginary” is a historical name, not an accurate description. Irrational numbers are not insane, and imaginary numbers are no more fictional than other numbers. – Thomas Andrews Aug 22 '21 at 03:03
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    @Gerry: I didn't realise that while typing (I always forget if $\Gamma(n)=(n+1)!$ or $\Gamma(n+1)=n!$), thanks for correcting :) – ultralegend5385 Aug 22 '21 at 03:37

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The numbers of the form $a+b\sqrt{-1}$, with $a$ and $b$ real, form a field. That means you can add, subtract, multiply, and divide them (except you can't divide by zero); the addition and multiplication are associative and commutative; there are identity elements for both operations; every element has an additive inverse and (excepting zero) a multiplicative inverse; and the distributive laws hold.

As for $1\over0$ (and $\log0$, and $(-1)!$), consider the usual formulas for adding fractions, $${a\over b}+{c\over d}={ad+bc\over bd},\qquad{r\over t}+{s\over t}={r+s\over t}$$ This would give us both $${1\over0}+{1\over0}={0\over0}\qquad{\rm\ and\ }\qquad{1\over0}+{1\over0}={2\over0}$$ So $${0\over0}={2\over0}$$ Now multiply both sides by zero, and use ${m\over n}\times n=m$ to get $0=2$. Well, that didn't work so well. Basically, there's no way to incorporate $1\over0$ in a field.

Joe
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Gerry Myerson
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  • fine, but who said anything about fields? On the "projectively extended real line", which has a nice Wikipedia page, dividing finite quantities by zero is straightforward. – David Aug 22 '21 at 03:15
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    "who said anything about fields?" @David, I did. I'm contrasting extending the reals by $\sqrt{-1}$ with extending the reals by $1/0$, and pointing out that in the former case the usual rules of arithmetic apply, while in the latter case they collapse in a heap. You can take the projective line to be the ordered pairs of reals, $(a,b)$, modulo the equivalence relation, $(a,b)\equiv(c,d)$ iff $ad=bc$, and then $(1,0)$ is sorta kinda like $1/0$. And of course the projective line is useful and important in some parts of math. But doing arithmetic there is problematic, to say the least. – Gerry Myerson Aug 22 '21 at 03:22
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    Right, I knew that you said something about fields, sorry if that wasn't clear. What I meant was that somewhat contrary to the original poster's claim that "we do not consider stuff like... 1/0" mathematicians do indeed consider stuff like 1/0 and the results are rigorous and notable even though, as you said, they can't yield a field. – David Aug 22 '21 at 03:26
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Let me leave aside why we picked this one (which is a historical question) and answer why $\sqrt{-1}$ is the proper unit for an imaginary number.

With imaginary numbers, numbers are no longer on a line, they are on a plane. That is, a given value $x$ might not just be to the left or right of zero, it can also be above and below it! Imaginary numbers extends the concept of number so that numbers don't just fall in a line.

The imaginary unit, $i$, is precisely one unit "above" zero. That is, it is an orthogonal unit to the ordinary unit. It has the same magnitude, but is pointed 90 degrees offset. This means that no amount of movement along the "real" number line will affect the imaginary part, and no amount of movement along the imaginary part will affect the "real" part.

So, that's why $\sqrt{-1}$ is so important, it is literally an orthogonal unit to the standard real value $1$.

johnnyb
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$\imath$ is not defined in the sense of made up. You are making up collections of symbols then defining them to be something useful when they are not.

In math, we do not go around arbitrarily defining things. It is not arbitrary taxonomy. We simply name useful things.

$\imath$ is VERY useful because it provides the link to understand many mysteries that come from real numbers. But $\imath$ is just a symbol that represents a concept. The things you are trying to define do not represent anything because they are meaningless(they do not have any meaning, $\imath$ does). You are picking arbitrary things out of the universe and trying to make them be useful, $\imath$ was defined because it was precisely a useful thing to define.

There are many many ways to think about $\imath$:

  1. You can think of it as an operation on a tuple: (a,b), then $\imath$ is equivalent o (-b,a) so it is a swap and negation of the first element.

This comes from extending the algebra on the reals to pairs:

Suppose we have (a,b) and (c,d) (where a,b,c,d are reals) and we want to add and multiply them, how can this work?

Well we want some way to add and multiply: $(a,b) + (c,d) = ? (a,b)*(c,d) = ?$

If we magically define them to be $(a,b) + (c,d) = (a+b, c+d) and (a,b)*(c,d) = (a*c-b*d, a*d+b*c)$

Then it turns out that we can write $(a,b) = a + \sqrt(-1)*b$ and we can recover our magic formulas.

So where does one get the magic formulas? Well, $(a,b)$ looks just like a point on the cartesian plane and so that is how they came about. When you add "points(which you think as arrows) you connect them end to end and that is precisely what $+$ is doing. When you multiply them you are representing a certain structure that is very useful in physics which is the curl and dot product.

So $i$ helps us simplify dealing with these strucutures because it all works out very nicely. It turns out there is only one way to extend R in a way that makes it natural. What is natural?

Suppose we have a real polynomial f(x). Now we have our new structure with our "points" and we want to convert f to this new structure. We can then just change x from real to "complex" and everything works out the same. Our polynomial f(z) behaves exactly in the same way as it does with f(x) as far as the algebra is concerned. So we have a true extension of the reals and do not have to learn some entirely new thing.

Furthermore we can then solve our real polynomial when they are not solvable in the reals(of course we have to go out of the reals to do this but it is in to a nice space).

  1. You can think of it as a type of "modulo" on polynomials. Specifically if you have R[x] then you can "modulo it" much like you can Z to get Z_n. You do this in a sense by "moduloing" by x^2 + 1.

So I will go in to this a little more:

In Z we have our integers. Now, if we equate some integer to 0, e.g., 7 = 0, then we get Z_7 = integers modulo 7 = Z mod 7 = Z/7Z. We are pretending 7 is the same as 0 and this "wraps" everything around on itself.

In math this was well understood because of number theory. It is much easier to think of these "pieces" of Z rather than Z itself in many problems involving integers(and rationals since we can convert between them).

Now, we have $x^2 = -1$ or $x^2 + 1 = 0$ and this is a problem since it "has no solutions"... but does it?

Well, we can have other polynomials like $x^4 + 3x^2 - 3$. Think of it as analogous to some integer like 43.

But now we can modulo the polynomials in $x$ by $x^2 + 1$ which is equating $x^2 + 1 = 0$ just like we can do $7 = 0$ and $43 mod 7 = 1$. $x^4 + 3x^2 - 3 mod x^2 + 1 = (-1)^2 + 3*(-1) - 3 = 1 - 3 - 3 = -5$

So that quartic is equivalent to -5 modulo x^2 + 1. All I did was use x^2 + 1 = 0 and substitute in. If you do this you cannot get anything more than a linear poly(any higher powers of x will be reduced.

But remember, in back of your mind you are thinking of $x^2 + 1 = 0$ as defining $\imath$ but you might not know what it really is or what to call it at first so you have to just use the defining equation.

now, you can write $43 = 7*6 + 1$ and similarly

$$x^4 + 3x^2 - 3 = (x^2 + 2)*(x^2+1) - 5$$

BUT if we label x^2+1 as $\imath$ then we have

$$(x^2 + 2)*(x^2+1) - 5 = (x^2 + 2)*\imath - 5$$

and so it is sort of as if we are moduloing by $\imath$ and it is all exactly analogous to modulo in the integers.

In this case the polynomial $x^4 + 3x^2 - 3$ in R[x] becomes $-5 + \imath$ in R[x]<x^2+1>.

So, you can see that the complex numbers arise from doing modulo on polynomials but what they do is reduce polynomials to two reals(or in this case it will be integers because we have Z[x]).

What this does is allow us to understand complex numbers in terms of things we already understand, which is modulo structures.

So you probably don't understand that $\imath$ is actually a very useful definition because it helps us simplify things greatly in to structures we already understand but it does. If we define ln(0) = e it not only does not simplify anything, it makes things illogical:

ln(0) = e but e = ln(e^e) so this 0 = e^e

But clearly 0 != e^e.

So the arbitrary definition I made created a logical contradiction! It would be destructive to use it.

Now, we do this in math in modulo! 7 = 0! and if 7 = 0 then 14 = 7 + 7 = 0 + 0 = 0 so how is it any different?

Well, isn't not except the whole point of modulo structures is to keep track the contradiction. We know, because we are dealing with modulo 7, that we are conflating 0 and 7 but we do this because it is helpful.

You could do the same with ln(0) = e but you would have to create a structure in which it is valid and keep it distinct from other structures in which it is a logical contradiction in those structures.

Now, that structure better have some use or be logically consistent in and of itself or it would be pointless. Chances are, arbitrary definitions are going to be useless.

Gupta
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