There is one loose end related to exponentials that I want to tie up here. It’s the issue of multiple roots (or multiple-valuedness as it’s called in the context of inverse functions).

**Introduction**

You’ll remember that, for integer exponents n, we had two *inverse* operations for a^{n}:

- The logarithm: the instruction here is to find n (i.e. the exponent) given the value a
^{n}and given*a*(i.e. the base). - The ‘n
^{th }root’ function: the instruction here is find a (i.e. the base) given the value a^{n}and given*n*(i.e. the exponent).

We have two inverse operations because the exponentiation operation is not commutative: while a + b = b + a (and, therefore, a×b = b×a, so multiplication is commutative as well), a^{n }is surely *not* the same as n^{a} (except if a = n, of course).

Having *two *inverse operations is somewhat confusing, of course. However, when we expand the domain of the exponential function to also include *rational *exponents, the ‘n^{th }root’ function becomes an exponential function itself: a^{1/n}. That’s nice, because it tidies things up. We only have one inverse operation now: the logarithm.

Now, my kids understand exponentials, but they find logarithms weird. There are two reasons for that. The most important one is that we don’t learn about the logarithm function when we learn about the exponential function. We only learn logarithms later – much later. Therefore, we are not as familiar with them as we should be. There is no good reason for that but that’s what it is. [I guess I am like Euler here: I’d suggest logarithms and complex numbers should be taught earlier in life. Then we would have less trouble understanding them.]

The second one is notation, I think. Indeed, x = log_{a}(y) looks much more frightening than y = a^{x }because… Well… Too many letters. It would be more logical to apply the same economy of symbols. We could just write x = _{a}y instead of log_{a}(y), for example, using a *subscript* in front of the variable–as opposed to a *superscript* behind the variable, as we do for the exponential function. Or, else, we could be equally verbose for the exponential function and write y = exp_{a}(x) instead of y = a^{x}. In fact, you’ll find such more explicit expressions in spreadsheets and other software, because these don’t take subscripts or superscripts.

In any case, that’s not the point here. I will come back to the logarithmic function later. The point that I want to discuss here is that, while we sort of merged our ‘n^{th }root function’ with our exponential function as we allowed for rational exponents as well (as opposed to integers only), we’re actually still taking roots, so to say, and then we note another problem: the square root function yields not one but *two* numbers when the base (a) is real and positive: ± a^{1/2}.

In fact, that’s a more general problem.

**Odd and even rational exponents **

You’ll remember the following rules for exponentiation:

**1.** For a positive real number a, we have always have two real n^{th }roots when n is even: a^{1/n}: ± a^{1/n}. That’s obviously a consequence of having two real *square *roots ± a^{1/2}, because the definition of even parity is that n can be written as n = 2k with k any integer, i.e. k ∈ **Z** (so k can be negative). Hence, a^{1/n }can then be written as a^{1/2k }= a^{1/2k }= (a^{1/k})^{1/2}. Hence, whatever the value of a^{1/k }(if k is even, then we have two k^{th }roots once again, but that doesn’t matter), we will have two real roots: *plus* (a^{1/k})^{1/2 }and *minus* (a^{1/k})^{1/2}

**2.** If n is uneven (or *odd* I should say), so n ∈ {2k+1: k ∈ **Z**}, we have only one real root a^{1/2k+1}: that root is positive when a is positive and negative when a is negative.

**3.** For the sake of completeness, let me add the third case: a is negative and n is even. We know there’s no real n^{th }root of a in that case. That’s why mathematicians invented *i*: we’ll associate an even root of a negative real number with two complex-valued roots: a^{1/n}: ± *i*a^{1/n}.

The first and second case are illustrated below for n = 2 and n = 3 respectively. The complex roots of the third case cannot be visualized because y is a real axis. Of course, we could *imagine *the complex-roots ± *i*a^{1/n }if we would flip or mirror the blue and red graph (i.e. the graphs for n = 2) along the vertical axis and re-label that axis as the *i*y-axis, i.e. the imaginary axis. But so I’ll leave that to your imagination indeed.* *

How does this parity business turn out for *rational* exponents?

If r is a rational number r = m/n, we’ll have to express it as an irreducible fraction first, so the numerator m and denominator n have no other common divisors than 1, or –1 when considering negative numbers. But let’s look at positive numbers first. If we write r as an irreducible fraction m/n, then m and n cannot *both *be even. Why not? Because m and n can then both be divided by 2 and m/n is not an irreducible fraction in that case. Let’s assume m is even. Hence, *n must be odd* in that case. We can then write a^{2k/n }as (a^{k/n})^{2}. This number will always be positive, because we are squaring something. So it doesn’t matter if a^{k/n} has one or two roots: we’ll square them and so the result will always be positive.

Now let’s assume the second possibility: m is odd. We can then write a^{m/n }as (a^{1/n})^{m}. So now it will depend on whether or not n is even. If n is even, we have two real roots, if n is uneven, then we have only one. Let’s work a few examples:

- 8
^{2/3 }= (8^{1/3})^{2 }= 2^{2 }= 4 - 4
^{3/2 }= (4^{1/2})^{3 }= (±2)^{3 }=±2^{3 }=±2^{3 }= ±8 - 16
^{1/4 }= (16^{1/2})^{1/2 }= (±4)^{1/2 }=±4^{1/2 }=±2^{ }= ±2 - (–8)
^{5/3 }= (–8^{1/3})^{5 }= (–2)^{5 }= 32

So we have two roots if m is odd and n is even, and only one root in all other cases. However, we said that m and n cannot both be even, hence, if n is even, m must be odd. In short, we can say that a rational exponent m/n is even (i.e. there will be two roots), if n is even. Does that work for complex roots as well? Let’s work that out with an example:

(–4)^{3/2 }= (–4^{1/2})^{3 }= (±2*i*)^{3 }=(±2)^{3}*i*^{3 }=±8*i*

So, yes! It works for complex roots as well. 🙂

OK. But let’s ask the obvious question now: **where are these even numbers on the real line?**

Well… They are everywhere: we can start from 1/2 and then change the numerator: 3/2, 5/2, etcetera. It’s all fine, as long as we use an odd number. However, we can also go down and change the denominator: 1/4, 1/6, 1/8 etcetera. And then we can, of course, take *odd *multiples of these fractions once again, such as 1025/1024 = 1.0009765625, for example, or on the other side, 1023/1024 = 0.9990234375. So we have two even numbers here *right next* to the odd number 1. We may increase the precision: we could take ± 1/3588 for example. 🙂

Of course, you may have noticed something here. The first thing, of course, is that we’ve defined these two even numbers 1.0009765625 and 0.9990234375 with a precision of 10 digits behind the decimal point, i.e. 1/1024 = 1/2^{10 }= 0.0009765625. The second point to note is that the last digit of these two rational coefficients, when expressed as a decimal, was 5. Now, you may think that should always be the case because of that 1/2 factor. But it’s not true: 1/6, for example, is a rational number that, written in decimal form, will yield 0.166666… This is an expression with a recurring decimal. And 1/10, of course, just yields 0.1. So there’s no easy rule here. You need to look at the fraction itself, and rational numbers are *either *as a finite decimal *or* an infinite repeating decimal. Of course, there are rules for that, but this is not a post on number theory, so I won’t write anything more on this: you can *Google *some more stuff yourself if you’re interested in this.

**Irrational exponents**

How does the business of parity work for irrational exponents? The gist of the rather long story above can be summarized easily. We can write a^{m/n} as a^{m/n} = a^{m·(1/n)} =(a^{1/n})^{m} = a^{1/n}·a^{1/n}·a^{1/n}·a^{1/n} =·… (m times) and so whether or not we have multiple roots (two instead of one) depends on whether or not n is even. Indeed, remember – once again – that exponentiation is *repeated* multiplication, and so for the *sign *of the result, what matters is whether or not the number of times that we do that multiplication is even or odd, not only for integer but for rational exponents as well.

For irrational exponents, we also have repeated multiplication, but now we have an *infinite *expression, not a finite one:

a^{r }= a^{r(1/Δ + 1/Δ + 1/Δ + 1/Δ +…) }= a^{r/Δ}·a^{r/Δ}·a^{r/Δ}·a^{r/Δ}…

I explained this expression in my previous post: 1/Δ is an infinitesimally small fraction. In fact, I calculated rational powers of *e *using the fraction 1/Δ = 1/1024 = 1/2^{10}. I used that fraction because I had started backwards, taking successive square roots of *e*, so *e*^{1/2}, and then *e*^{1/4}, *e*^{1/8}, *e*^{1/16}, etcetera.

However, as I mentioned when I started doing that, there was no compelling reason to cut things up by dividing them in 2. We could use 1/3 as the fraction to start with and, then, of course, or fraction 1/Δ would have been equal to 1/3^{10 }= 1/59049, so we have an odd number in the denominator here. So that’s one problem: we cannot say if Δ is even or odd. And the the second problem, of course, is that it’s an infinite expression and, hence, we cannot say if we multiplied 1/Δ an even or an odd number of times.

That leads to the third problem: we cannot say if r itself is even or uneven, which is basically what we were looking at: can we define irrational exponents as even or odd?

In short, the answer is no. In practice, that means that we will associate a^{r }with one ‘r^{th }root’ only.

Hmm… That obviously makes a lot of sense but how do we ‘justify’ it from a more formal point of view? Where do these *negative* roots (for even powers) go? I am not sure. I guess there must be some more formal argument but I’ll leave that to you to look it up. I am fairly happy with what *Wikipedia *writes on that:

“[Real] Powers of a positive real number are always positive real numbers. […] If the definition of exponentiation of real numbers is extended to allow negative results then the result is no longer *well behaved*.”

In fact, the article actually does give a somewhat more formal argument, as it writes:

- Neither the logarithm method nor the rational exponent method can be used to define
*b*^{r}as a real number for a negative real number*b*and an arbitrary real number*r*. Indeed,*e*^{r}is positive for every real number*r*, so ln(*b*) is not defined as a real number for*b*≤ 0. - As for the rational exponent method, that cannot be used for negative values of
*b*because it relies on continuity. The function*f*(*r*) =*b*^{r}has a unique continuous extension from the rational numbers to the real numbers for each*b*> 0. But when*b*< 0, the function*f*is not even continuous on the set of rational numbers*r*for which it is defined.

I am not quite sure I fully understand the last line, but I guess this refers to what I pointed out above: all these even and odd numbers that are so close to each other. When we go from rational to irrational exponents, we can no longer define odd or even.

**The bottom line**

The bottom line is that, in practice, we will only work with positive real bases. Hence, if b is negative, then we will define b^{r} as –(–b)* ^{r}*.

*Huh?*Yes. Think about it. If b is negative, we’ll just multiply it with –1 to ensure that the base is a positive real number. And then we just put a minus in front to get a graph such as, for example, that x^{1/3 }function *for the negative side of the x-axis as well*.

You should also note that most applications, like the one I use to draw simple graphs like the ones above (rechneronline.de/function-graphs) are not capable of showing you both roots. They do *check* whether the exponent is even or odd though, because it plots the function x^{1/3 }on both sides of the zero point, and the x^{1/2 }graph on the positive side only: it’s just not capable to associate more than one y value with one x value indeed. [In case you’re curious to see what it does with an irrational exponent, go and check it yourself: you can put in x^pi or x^e. Will it give function values for negative values of x as well? What’s your guess? :-)]

You’ll wonder why I am emphasizing this point. Well… I just wanted to note that we should be aware of the fact that, as we go from rational to irrational exponents, we sort of deliberately ‘forget’ about the second (negative) root. The point to note is that the issue of multiple-valued functions – such as discussed in the context of, for example, Riemann surfaces – is not necessarily related to complex-valued functions. We have it here (double roots), and we also have it, in general, for periodic functions.

But that’s for a next post. And there we’ll use our ‘natural’ exponential *e*^{x}, and its inverse function, ln(x), *an awful lot*. So I’ll just conclude here with their graphs, noting, as Wikipedia does, that, nowadays, the term ‘exponential function’ is *almost exclusively* used as a shortcut for describing the **natural exponential function** *e*^{x}. But, to my kids, I say: it’s good that you know where it comes from. 🙂

**Post scriptum**:

When thinking about such minor things, it’s always to good to think about *why* we are manipulating all these symbols. Exponentiation is repeated multiplication. What does it mean to multiply something with a *negative* number? A minus sign is an instruction to reverse direction, to turn around, 180 degrees. So we multiply the magnitudes of both numbers a and b, but we change the direction: if we’re walking down the positive real axis, then now we’re walking down the negative axis.

So *repeated* multiplication with a negative real number means we’re switching back and forth, wildly jumping from the positive to the negative side of the zero point and then back again. You’ll admit you would appreciate being told in advance how many times we need to do the multiplication if the multiplier is negative: if n is even, then we’ll end up going in the same direction: (–1)^{n }= 1. No sign reversal. If n is uneven, then we know that, besides the ‘booster’ effect (i.e. the exponentiation operation), we’re expected to speed in the opposite direction: (–1)^{n }= –1.

Hence, if b would happen to be a negative real number, then defining b^{r} as –(–b)* ^{r}*, or assuming that, in general, our base will be a

*positive*real number makes sense. Of course, the math has to keep track of the

*theoretical*possibility that, if the exponent would happen to be even,

*b*might be a negative number, but you can see it’s more of a theoretical possibility indeed. Not something we’d associate with something happening in

*real life.*

In that sense, I should note that multiplication with a *complex *multiplier is much more ‘real-life’, so to say. Multiplying something with a complex number does the same to the magnitude of both numbers as real multiplication: it multiplies the magnitudes, thereby changing the *scale. S*o the product of a vector that’s 2 units long and a vector that’s 3 units long will still be 6 units long. However, complex numbers also allow for a more gradual change of direction. Instead of just a gear to move forward and backward, we also get a steering wheel so to say: multiplying two complex numbers also adds their angles (as measured from some kind of zero direction obviously), besides multiplying their magnitudes. For example, suppose that the zero direction is east, and we have a vector pointing east indeed (that means its *imaginary *part is zero) that we need to multiply with a vector pointing north (so that’s a vector with a zero *real* part, along the imaginary axis), then the final vector will be pointing north.

However, with that subtlety comes complexity as well. With real numbers, you can go in the same direction by reversing direction two times, and so that’s why we have *two *2^{nd}* ^{ }*roots (i.e. two

*square*roots) of 1: (a) +1, so then we just stay where we are, and (b) –1, so then we rotate two times a full 180 degrees around the zero point: indeed, (–1)(–1) corresponds to two successive rotations by 180 degrees (or π in radians)–clockwise or counterclockwise, it doesn’t matter: one full loop around the zero point will get us back to square one, or

*point*1, I should say. 🙂

With complex numbers, it all depends. The 3^{rd }root (i.e. the *cube* root) of 1 was only 1 in the real space but, **in the complex space, we have three 3 cube roots of unity**. The first one (W

^{1 }= W

^{3}) is the root we’re used to: unity itself, so the angle here is zero, i.e. straight ahead. In fact, with 1, we just stay where we are: 1×1×1 = 1

^{3 }= 1 indeed. But that’s not the only way. The illustration below shows two other ways to end up where we are (i.e. at point 1):

- The second cube root is W
^{2}: 120 degrees. You can see we get back at 1 by making three successive turns of 120 degrees indeed, so that’s one full loop around the or<igin. Using complex numbers (in polar notation), we write*e*^{2π/3}×*e*^{2π/3}×*e*^{2π/3}=*e*^{6π/3}=*e*^{2π }=*e*^{0 }= 1. - The third cube
^{ }root is W^{1}: that’s 240 degrees ! Indeed, here we get back at square one by making three successive turns of 4π/3 radians, i.e .by making*two*loops, in total, around the origin:*e*^{4π/3}×*e*^{4π/3}×*e*^{4π/3}=*e*^{12π/3}=*e*^{4π }=*e*^{0 }= 1.

In short, we gain flexibility (of course, we have four 4^{th} roots (with which we make 0, 1, 2 and 3 loops around the origin respectively), 5^{th} roots, and so on), and the great Leonhard Euler was obviously fully right: complex numbers are more ‘natural’ numbers as they allow us to model real-life situations much better.

However, if you think that double roots are a problem… Well… Think again ! With *complex* numbers, the problem of multiple-valuedness is much more *‘real’*, I’d say. 🙂

**P.S**: As mentioned in my previous post, I talk about that problem of multiple-valuedness when talking about Riemann surfaces in my October-November 2013 posts, so I won’t repeat what I wrote there. It’s about time I get back to both Feynman as well Penrose. 🙂

Just one last (philosophical) question to test your understanding. Negative real numbers have no real square root. That includes –1 obviously. Why is that? Why do we have *two* square *real *roots for +1 and *no *(none!) (real) square roots for –1?

[…] No? Come on!

[…] OK. Let me tell you: it’s all a question of definition. What’s implicit here is that we have only one *real *direction: from zero to infinity along the positive axis, and then –1 is nothing but a *reversal* of direction. So it’s an operation really, not a ‘real’ number. In a philosophical sense, of course: negative numbers don’t exist, so to say! Indeed, ask yourself: what is a negative number? It’s an operation: we subtract things when we use the minus sign, and we reverse direction when multiplying numbers with –1. So, if we multiply something with –1 two times in succession, we are back where we are.

Of course, we could say that the negative direction is the ‘real’ direction and, hence, that it’s the positive numbers that don’t ‘really’ exist. Indeed, math doesn’t care about what we say, so let’s say that the negative axis is the ‘real’ one, in a physical sense. What happens then? Well… Let’s see… Let’s do what we did before. We still define –1 as a reversal of direction, or a rotation by 180 degrees and, hence, doing that two times should bring us back where we want to be, so that’s –1 now. OK. So we have (–1)(–1)(–1) = (–1). But so that means that (–1)(–1) = 1, and… How can we write something like that for –1? What number a gives us the result that a×a = –1. Hmm… Only this imaginary number: *i*×*i* = *i*^{2} = –1. So, no matter how hard you try: the way we use symbols is pretty consequent, and so you will find that (–1)(–1) = 1×1 = 1 (so we have two square roots of 1), but we will *not *find that 1×1 = –1. If you would want to do that, you’d have to define +1 as a reversal of direction, so that basically means that the + sign would take the function of the – sign. *Huh?*

🙂 You must think I’ve gone crazy. I don’t think so. The idea I want to convey here is that, no matter how abstract math may seem to be – when everything is said and done – it’s intimately connected to our most basic notions of space, and our motion in that space. We go from here to there, or backwards, we change direction, we count things, we measure lengths or distances,… All that math does is to capture that in a non-ambiguous and consistent way. That also results in terse ‘truths’ such as: 1 has two real square roots, +1 and –1, but the square roots of –1 are only *imaginary*: ± *i*.

However, that terse statement hides another fun ‘truth’: +*i* and −*i *are as real as –1. Indeed, they are a rotation by 90 degrees, counterclockwise (+*i*) or clockwise (−*i*), as opposed to, for example, a rotation by 180 degrees (–1), or a full loop (1). 🙂