The Universe of Disco

Imaginary units
Yesterday I had a phenomenally annoying discussion with the pedants on the IRC #math channel. Someone was talking about square roots, and for some reason I needed to point out that when you are considering square roots of negative numbers, it is important not to forget that there are two square roots.

I should back up and discuss square roots in more detail. The square root of x, written √x, is defined to be the number y such that y² = x. Well, no, that actually contains a subtle error. The error is in the use of the word "the". When we say "the number y such that...", we imply that there is only one. But every number (except zero) has two square roots. For example, the square roots of 16 are 4 and -4. Both of these are numbers y with the property that y² = 16.

In many contexts, we can forget about one of the square roots. For example, in geometry problems, all quantities are positive. (I'm using "positive" here to mean "≥ 0".) When we consider a right triangle whose legs have lengths a and b, we say simply that the hypotenuse has length √(a² + b²), and we don't have to think about the fact that there are actually two square roots, because one of them is negative, and is nonsensical when discussing hypotenuses. In such cases we can talk about the square root function, sqrt(x), which is defined to be the positive number y such that y² = x. There the use of "the" is justified, because there is only one such number. But pinning down which square root we mean has a price: the square root function applies only to positive arguments. We cannot ask for sqrt(-1), because there is no positive number y such that y² = -1. For negative arguments, this simplification is not available, and we must fall back to using √ in its full generality.

In high school algebra, we all learn about a number called i, which is defined to be the square root of -1. But again, the use of the word "the" here is misleading, because "the" square root is not unique; -1, like every other number (except 0) has two square roots. We cannot avail ourselves of the trick of taking the positive one, because neither root is positive. And in fact there is no other trick we can use to distinguish the two roots; they are mathematically indistinguishable.

The annoying discussion was whether it was correct to say that the two roots are mathematically indistinguishable. It was annoying because it's so obviously true. The number i is, by definition, a number such that i² = -1. This is its one and only defining property. Since there is another number which shares this single defining property, it stands to reason that this other root is completely interchangeable with i—mathematically indistinguishable from it, in other words.

This other square root is usually written "-i", which suggests that it's somehow secondary to i. But this is not the case. Every numerical property possessed by i is possessed by -i as well. For example, i³ = -i. But we can replace i with -i and get (-i)³ = -(-i), which is just as true. Euler's famous formula says that e^ix = cos x + i sin x. But replacing i with -i here we get e^-ix = cos x + -i sin x, which is also true.

Well, one of them is i, and the other is -i, so can't you distinguish them that way? No; those are only expressions that denote the numbers, not the numbers themselves. There is no way to know which of the numbers is denoted by which expression, and, in fact, it does not even make much sense to ask which number is denoted by which expression, since the two numbers are entirely interchangeable. One is i, and one is -i, sure, but this is just saying that one is the negative of the other. But so too is the other the negative of the one.

One of the #math people pointed out that there is a well-known Im() function, the "imaginary part" function, such that Im(i) = 1, but Im(-i) = -1, and suggested, rather forcefully, that they could be distinguished that way. This, of course, is hopeless. Because in order to define the "imaginary part" function in the first place, you must start by making an entirely arbitrary choice of which square root of -1 you are using as the unit, and then define Im() in terms of this choice. For example, one often defines Im(z) as !!z - \bar{z} \over 2i!!. But in order to make this definition, you have to select one of the imaginary units and designate it as i and use it in the denominator, thus begging the question. Had you defined Im() with -i in place of i, then Im(i) would have been -1, and vice versa.

Similarly, one #math inhabitant suggested that if one were to define the complex numbers as pairs of reals (a, b), such that (a, b) + (c, d) = (a + c, b + d), (a, b) × (c, d) = (ac - bd, ad + bc), then i is defined as (0,1), not (0,-1). This is even more clearly begging the question, since the definition of i here is solely a traditional and conventional one; defining i as (0, -1) instead of (0,1) works exactly as well; we still have i² = -1 and all the other important properties.

As IRC discussions do, this one then started to move downwards into straw man attacks. The #math folks then argued that i ≠ -i, and so the two numbers are indeed distinguishable. This would have been a fine counterargument to the assertion that i = -i, but since I was not suggesting anything so silly, it was just stupid. When I said that the numbers were indistinguishable, I did not mean to say that they were numerically equal. If they were, then -1 would have only one square root. Of course, it does not; it has two unequal, but entirely interchangeable, square roots.

The that the square roots of -1 are indistinguishable has real content. 1 has two square roots that are not interchangeable in this way. Suppose someone tells you that a and b are different square roots of 1, and you have to figure out which is which. You can do that, because among the two equations a² = a, b² = b, only one will be true. If it's the former, then a=1 and b=-1; if the latter, then it's the other way around. The point about the square roots of -1 is that there is no corresponding criterion for distinguishing the two roots. This is a theorem. But the result is completely obvious if you just recall that i is merely defined to be a square root of -1, no more and no less, and that -1 has two square roots.

Oh well, it's IRC. There's no solution other than to just leave. [ Addenda: Part 2 Part 3 Part 4 Part 5 ]

[Other articles in category /math] permanent link