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Why is the square root of negative one an important relation for imaginary numbers and quaternions, etc.?

I mean, if we are going to have imaginary numbers, couldn't they be imaginary without needing to be defined in terms of the square root of negative one?

I.e. why couldn't we have imaginary numbers without them having any definition in terms of a relation to the real numbers?

Seems to me that you could say imaginary numbers are based on the square root of x, where x is some number that's not on the real number line (but not necessarily square root of negative one—maybe instead, 1/0).

CommaToast
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  • With regards to division by zero, whenever you expand upon the number system to include more you might inadvertently cause some nice properties that you previously had to fail in the new system. When moving to complex numbers, we lost the property that the set was totally ordered. When moving to quaternions, we lost the property that multiplication is commutative. If you were to include $1/0$ far worse things happen. See here. – JMoravitz Jan 19 '21 at 18:56
  • With regards to other extensions to $\Bbb R$ which don't rely on $\sqrt{-1}$, see things like the Split Complex Numbers and Hypercomplex Numbers in general. – JMoravitz Jan 19 '21 at 18:59
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    This thread might be of use to you. – Joe Jan 19 '21 at 19:10

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Because historically, the question that needed to be addressed was not "What definition can we give to a second continuum of numbers?" with the answer "Let's say they're multiples of $\sqrt{-1}$." Rather, it was "What can we do when multiples of $\sqrt{-1}$ emerge from algebraic computations?" with the answer "Let's treat them as a second continuum of numbers."

user10478
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    I second this. The problems we want to solve generally come first. New definitions, operations, techniques etc, all come later to solve the desired question. – JMoravitz Jan 19 '21 at 19:04
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    It occurred to me, the story is quite different for quaternions. Hamilton was looking to upgrade math's RAM capacity to hold four values instead of two, inside a single complex number. Then later octonions doubled math's RAM again, letting us have 8 values. – CommaToast Jul 15 '21 at 08:05
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The modern approach to complex numbers doesn't use square roots at all. The modern approach is to say $i^2=-1$ rather than $i=\sqrt{-1}$. It is more indirect, but actually makes a lot more sense once you get used to working with complex numbers.

How do you work with this? Essentially, complex numbers are polynomials in the variable $i$. They behave like any other polynomials we know and love. With one exception: whenever you see $i^2$, you're allowed to substitute it with $-1$ (this includes cases like $i^3=i^2\cdot i$ and $i^{1000}=(i^2)^{500}$).

That's it. That's what the complex numbers are.

Note that $i^2=-1$ is essential to make complex numbers something other than pure polynomials. So powers of $i$ must be related to real numbers. (Many other quadratic relations can be used instead if you wish, like $i^2=i-1$; basically, any relation that doesn't have a real solution if you interpret it as an equation will give you the complex numbers as a result, but they won't look exactly like the standard complex numbers.)

My gut says it's a better way to teach complex numbers as well. But I haven't tried that enough to actually say for certain.

Arthur
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  • Not sure what you mean by "the modern approach is i^2 = - 1" vs. "i = sqrt(-1)". In my school it was always the latter. Anyway they are the exact same statement. I suppose "i^2 = -1" parlays more nicely into "i^2 = -1 = j^2 = k^2 = ijk" but I'd hesitate to call quaternions "modern" although they have enjoyed a bit of a rennaissance thanks to computer games and, to a lesser extent, quantum physics. Just curious why you would say that. – CommaToast Jul 14 '21 at 02:15
  • @CommaToast Unfortunately, the modern approach to introducing complex numbers is still $i=\sqrt{-1}$. And I assure you, they are not the exact same statement. For instance, one of the statements uses a square root, which in the context of complex numbers is an abomination that introductory students should never have to lay eyes on. – Arthur Jul 14 '21 at 05:22
  • OK I was wrong that sqrt(-1) = i, and i^2 = -1, are exactly the same. What I meant is that, to my mind at least, each of these statements being true obviously means the other is also true. I'm not a math teacher but I was a math student (and always will be, I guess). To me, as a student, I liked sqrt(-1) = i as a first intro because we learned squares and square roots at the same time, and we were told that -1 has no real square root, but unlike division by zero, it doesn't break math—rather, it leads to a whole new realm "complex numbers, which we'll cover later." – CommaToast Jul 14 '21 at 19:07
  • At which point I asked my Pre-Algebra teacher, "OK but what does a complex number look like?" (I was always asking questions two years early.) And Mr. Sullivan said, "You swap out sqrt(-1) with i. And likewise multiplying two complex numbers always gives a real because i^2 = -1." I think it would have made leas sense if he said it the other way around, but that's just me. And yes, I have an eidetic memory, at least for this kind of stuff (sadly not peoples' names though.) – CommaToast Jul 14 '21 at 19:10
  • @CommaToast I am still not completely decided on whether $\sqrt{-1}=i$ or $i^2=-1$ is the best to teach. Don't know if I ever will get the chance to test. But I do know that implicitly telling students that square roots are allowed gives a lot of problems, like this post. We get a variation on that question at least once a month. Mixing square roots and complex numbers require care, and I believe it is better to just avoid. I know the more "abstract" $i^2=-1$ takes time getting used to, but to me it is far superior. – Arthur Jul 14 '21 at 19:38
  • Well, you don't know if that guy was told sqrt(-1) = i first. It also seems that his algebra teacher did not do a great job of teaching about how the rules of algebra work, especially between things inside of sqrt(...) and things outside of it. My teachers always recommended multiplying both sides of the equation by the stuff inside the sqrt(...) to get rid of it before proceeding. Which I think is the main basic point you are trying to make. – CommaToast Jul 15 '21 at 07:58