4.5

## University of Padova

Skip to 0 minutes and 11 secondsWe focus here on single variable equations. We're going to have some general principles about when equations can be modified in a certain way and still be, sort of, essentially the same. So we're looking at an equation of the form f of x equals g of x, and we're interested for solutions of this equation in a given domain D. The domain might be explicitly given, or quite often, it's implicitly defined, based on the expressions that appear in the equation and the underlying functions. If the functions, f and g, are defined on a domain omega, then naturally, the domain of the equation should be contained in omega.

Skip to 0 minutes and 57 secondsBy the way, that's a capital omega, one of the bonuses of mathematics is you get to know the Greek alphabet. We use these letters, not just because they are beautiful, but because we run a bit short of letters, if we use only the usual alphabet. A solution to our equation means, of course, a number x in D for which f of x is equal g of x, and the goal is to find all solutions, to determine the solution set, capital S. Our first equivalence principle concerns adding something to each side. And it says that if you have the original equation, f equals g, and if you add h of x to each side, then you get an equivalent equation.

Skip to 1 minute and 44 secondsNow of course, the function h would have to be defined on the domain D. That almost goes without saying. Although come to think of it, in mathematics, very little goes without saying. Anyway, informally speaking, we say that the term h can be additively cancelled from each side, and you come back to the original equation, f equals g. That, by the way, is the same as saying that if you have f equals g plus h, then that's the same as f minus h equal to g. That's just adding minus h to both sides of the equation. There's also a multiplicative principle for equivalence.

Skip to 2 minutes and 21 secondsIf you have, again, the original equation, f equals g, you can multiply each side by an expression h of x, provided that h is never 0 in D. Informally speaking, we say that the multiplicative factor h can be cancelled out, and you have essentially the same equation. Now, notice the condition that h never be 0. So in particular, you can multiply or divide an equation on each side by a non-zero constant without really changing it. However, you must learn to be careful when you multiply or divide by an expression involving the variable itself. Here's a simple example of a typical error. Suppose we want to solve this equation.

Skip to 3 minutes and 7 secondsWell, we'd be tempted to divide both sides by x minus 2, because then we obtain a simpler equation, x plus 1 equals 3x. In other words, we've cancelled out the x minus 2. If we then proceed to solve the simpler equation, switching terms left to right, it becomes a minus 2x equals minus 1. Divided by minus 2, we find x equals 1/2, and that's not wrong. x equals 1/2 is a solution of the original equation. The problem is we've lost equivalence. More precisely, we have lost one of our solutions, x equals 2, because you see, x equals 2 is clearly a solution of the original equation. So we performed an operation here in which equivalence was lost.

Skip to 3 minutes and 50 secondsHow did it happen? We divided it by x minus 2, which can be 0 in the domain, namely when x is 2. Here's another example. We want to solve root x over 2 minus x equals 1. We think about the domain. We obviously require x positive, since we have the square root of x. We also want to avoid x equals 2, because that would put 0 in the denominator. So the natural domain might be R+ but minus the point 2. A more insightful choice for D would be possible. We could notice that any solution x of the original equation must be such that 2 minus x is a positive number.

Skip to 4 minutes and 34 secondsAnd that would lead us to take the domain to be the half open interval 0, 2. That would be insightful, indeed. Be that as it may, let's multiply the equation across by 2 minus x, which is never 0 on the domain, and then we'll square both sides. And we arrive at this equation, in which the radical has disappeared. Now, we could proceed to solve this equation, but let's just observe here that x equals 4 is certainly a solution of this new equation. Right? When you plug in x equals 4, it works. However, x equals 4 doesn't solve the original equation, as you can see. So what's going on? Have we made a mistake? Is this a paradox of some kind?

Skip to 5 minutes and 19 secondsNo, it's simply that we did something to the original equation that did not preserve equivalence. We've produced, in fact, what is called an extraneous or surplus solution of the equation. Here is the general fact. When you look at the equation f equals g, and then the equation where you square both sides, f squared equals g squared, they're not generally equivalent. The second equation will, in general, have more solutions than the first. They will be phony solutions, not of the original equation. A very simple example, x equals 1 is not the same equation as x squared equals 1, because minus 1 is a solution of x squared equals 1, but not of the original equation.

Skip to 6 minutes and 5 secondsNow, there's a moral to this story. Whenever you find the solutions of a given equation, you should always go back to the original equation to check that they really are solutions. Now, it's always a good policy to check anyway, as you know. They say that to err is human. So if you're not an Android, do always check your answers.

# Equivalence

Additive and multiplicative equivalence, squaring does not preserve equivalence