My All 2020 Mathematics A to Z: Extraneous Solutions

Iva Sallay, the kind author of the Find the Factors recreational mathematics puzzle, suggested this topic for the letter X. It’s a fun chance to look at some of the basics of (high school) algebra again.

Extraneous Solutions.

When developing general relativity, Albert Einstein created a convention. He’s not unique in that. All mathematicians create conventions. They use shorthand for an idea that’s complicated or common. Relatively unique is that other people adopted his convention, because it expressed an idea compactly. This was in working with tensors, which look somewhat like matrixes and have a lot of indexes. In the equations of general relativity you need to take sums over many combinations of values of these indexes. What indexes there are are the same in most every problem. The possible values of the indexes is constant, problem to problem, too.

So Einstein saved himself writing, and his publishers from typesetting, a lot of redundant writing. This by writing out the conditions which implied “take the sums over these indexes on this range”. This is good for people doing general relativity, and certain kinds of geometry. It’s a problem only when an expression escapes its context. When it’s shown to a student or someone who doesn’t know this is a differential-geometry problem. Then the problem becomes confusing, and they can’t work on it.

This is not to fault the Einstein Summation Convention. It puts common necessary scaffolding out of the way and highlighting the interesting unique parts of a problem. Most conventions aim for that. We have the hazard, though, that we may not notice something breaking the convention.

And this is how we create extraneous solutions. And, as a bonus, to have missing solutions. We encounter them with the start of (high school) algebra, when we get used to manipulating equations. When we solve an equation what we always want is something clear, like

$x = 2$

But it never starts that way. It always starts with something like

$x^3 - 8x^2 + 24x - 32 + 22\frac{1}{x} = \frac{6}{x}$

or worse. We learn how to handle this. We know that we can do six things that do not alter the truth of an equation. We can regroup terms in the equation. We can add the same number to both sides of the equation. We can multiply both sides of the equation by some number besides zero. We can add zero to one side of the equation. We can multiply one side of the equation by 1. We can replace one quantity with another that has the same value. That doesn’t sound like a lot. It covers more than it seems. Multiplying by 1, for example, is the same as multiplying by $\frac{x}{x}$. If x isn’t zero, then we can multiply both sides of the equation by that x. And x can’t be zero, or else $\frac{x}{x}$ would not be 1.

So with my example there, start off by multiplying the right side by 1, in the guise $\frac{x}{x}$. Then multiply both sides by that same non-zero x. At this point the right-hand side simplifies to being 6. Add a -6 to both sides. And then with a lot of shuffling around you work out that the equation is the same as

$(x - 2)^4 = 0$

And that can only be true when x equals 2.

It should be easy to catch spurious solutions creeping in. They must result from breaking a rule. The obvious problem is multiplying — or dividing — by zero. We expect those to be trouble. Wikipedia has a fine example:

$\frac{1}{x - 2} = \frac{3}{x + 2} - \frac{6x}{(x - 2)(x + 2)}$

The obvious step is to multiply this whole mess by $(x - 2)(x + 2)$, which turns our work into a linear equation. Very soon we find the solution must be $x = -2$. Which would make at least two of the denominators in the original equation zero. We know not to want that.

The problems can be subtler, though. Consider:

$x - 12 = \sqrt{x}$

That’s not hard to solve. Multiply both sides by $x - 12$. Although, before working out $\sqrt{x}\cdot(x - 12)$ substitute that $x - 12$ with something equal to it. We know one thing is equal to it, $\sqrt{x}$. Then we have

$(x - 12)^2 = x$

It’s a quadratic equation. A little bit of work shows the roots are 9 and 16. One of those answers is correct and the other spurious. At no point did we divide anything, by zero or anything else.

So what is happening and what is the necessary rhetorical link to the Einstein Summation Convention?

There are many ways to look at equations. One that’s common is to look at them as functions. This is so common that we’ll elide between an equation and a function representation. This confuses the prealgebra student who wants to know why sometimes we look at

$x^2 - 25x + 144 = 0$

and sometimes we look at

$f(x) = x^2 - 25x + 144$

and sometimes at

$f(x) = x^2 - 25x + 144 = 0$

The advantage of looking at the function which shadows any equation is we have different tools for studying functions. Sometimes that makes solving the equation easier. In this form, we’re looking for what in the domain matches with something particular in the range.

And now we’ve reached the convention. When we write down something lke $x^2 - 25x + 144$ we’re implicitly defining a function. A function has three pieces. It has a set called the domain, from which we draw the independent variable. It has a set called the range. It has a rule matching elements in the domain to an element in the range. We’ve only given the rule. What are the domain and what’s the range for $f(x) = x^2 - 25x + 144$?

And here are the conventions. If we haven’t said otherwise, the domain and range are usually either the real numbers or the complex numbers. If we used x or y or t as the independent variable, we mean the real numbers. If we used z as the independent variable, and haven’t already put x and y in, we mean the complex numbers. Sometimes we call in s or w or another letter; never mind that. The range can be the whole set of real or complex numbers. It does us no harm to have too large a range.

The domain, though. We do insist that everything in the domain match to something in the range. And, like, $\frac{1}{x - 2}$? That can’t mean anything if x equals 2.

So we take an implicit definition of the domain: it’s all the real numbers for which the function’s rule is meaningful. So, $\frac{1}{x - 2}$ would have a domain “real numbers other than 2”. $\frac{6x}{(x - 2)(x + 2)}$ would have a domain “real numbers other than 2 and -2”.

We create extraneous solutions — or we lose some — when our convention changes the domain. An extraneous solution is one that existed outside the original problem’s domain. A missing solution is one that existed in an excised part of the domain. To go from $x^2 = 4x$ to $x = 4$ by dividing out x is to cut $x = 0$ out of the space of possible solutions.

A complaint you might raise. What is the domain for $x - 12 = \sqrt{x}$? Rewrite that as a function. $f(x) = x - 12 - \sqrt{x}$ would seem to have a domain “x greater than or equal to 0”. The extraneous solution is $x = 9$, a number which rumor has it is greater than or equal to 0. What happened?

We have to take that equation-handling more slowly. We had started out with

$x - 12 = \sqrt{x}$

The domain has to be “x is greater than or equal to 0” here. All right. The next step was multiplying both sides by the same quantity, $x - 12$. So:

$(x - 12)(x - 12) = \sqrt{x}(x - 12)$

The domain is still “x is greater than or equal to 0”. The next step, though, was a substitution. I wanted to replace the $(x - 12)$ on the right with $\sqrt{x}$. We know, from the original equation, that those are equal. At least, they’re equal wherever the original equation $x - 12 = \sqrt{x}$ is true. What happens when $x = 9$, though?

$9 - 12 = \sqrt{9}$

We start to see the catch. 9 – 12 is -3. And while it’s true that -3 squared will be 9, it’s false that -3 is the square root of 9. The equation $x - 12 = \sqrt{x}$ can only be true, for real numbers, if $\sqrt{x}$ is nonnegative. We can make this rigorous with two supplementary functions. Let me call $g(x) = x - 12$ and $h(x) = \sqrt{x}$.

$h(x)$ has an implicit domain of “x greater than or equal to 0”. What’s the domain of $g(x)$? If $g(x) = h(x)$, like we said it does, then they have to agree for every x in either’s domain. So $g(x)$ can’t have in its domain any x for which $h(x)$ isn’t defined. So the domain of $g(x)$ has to be “x for which x – 12 is greater than or equal to 0”. And that’s “x greater than or equal to 12”.

So the domain for the original equation is “x greater than or equal to 12”. When we keep that domain in mind, the extraneous nature of $x = 9$ is clear, and we avoid trouble.

Not all extraneous solutions come from algebraic manipulations. Sometimes there are constraints on the problem, rather than the numbers, that make a solution absurd. There is a betting strategy called the martingale. This amounts to doubling the bet every time one loses. This makes the first win balance out all the losses leading to it. This solution fails because the player has a finite wallet, and after a few losses any player hasn’t got the money to continue.

Or consider a case that may be legend. It concerns the Apollo Guidance Computer. It was designed to take the Lunar Module to a spot at zero altitude above the moon’s surface, with zero velocity. The story is that in early test runs, the computer would not avoid trajectories that dropped to a negative altitude along the way to the surface. One imagines the scene after the first Apollo subway trip. (I have not found a date when such a test run was done, or corrections to the code ordered. If someone knows, I’d appreciate learning specifics.)

The convention, that we trust the domain is “everything which makes sense”, is not to blame here. It’s normally a good convention. Explicitly noting the domain at every step is tedious and, most of the time, unenlightening. It belongs in the background. We also must check our possible solutions, and that they represent things that make sense. We can try to concentrate our thinking on the obvious interesting parts, but must spend some time on the rest also.

I am surprised to be so near the end of the 2020 A-to-Z, and to 2020, I hope. This and all the other glossary essays for the year should be at this link. All the essays from every A-to-Z series should be at this link. Thank you for reading.

What We Mean By x

[ Oh, wow. Yesterday’s entry had way fewer hits than average. I also put an equation out right up front where everyone could see it. I wonder if this might be a test of Stephen Hawking’s dictum about equations and sales. Or maybe I was just boring yesterday. I’d ask, but apparently, nobody found me interesting enough yesterday to know for comparison. ]

It shouldn’t be too hard to translate the the idea “I want to know the population of Charlotte at some particular time” into a polynomial. The polynomial ought to look something like y equals some pile of numbers times x’s raised to powers, and x somehow has to do with the particular time, and y has something to do with the population. And it’s not hard to do that translating, but I want to talk about some deeper issues. It’s probably better explaining them on the simple problem, where we know what we want things to mean, than it would be explaining them for a complicated problem.