﻿ General principles

# General principles

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10.8
Welcome to the wonderful world of inequalities. Inequalities are somehow less glamorous than equations. It’s always famous equations that get the press, but inequalities are at least as important. They’re present in many modern applications. Solving them, well, it’s rather similar to equations in some ways, but we’ll see that there are some important differences. So let’s begin with some terminology. Here is a general inequality of the type we are going to be looking at. We want to find the values of x, for which f of x is less than g of x. Now the inequality less than can be strict. Or it can be non-strict. As we’ll see, this makes only a minor difference in our approach.
60.8
Now, there is a domain, capital D, associated with the inequality, just like for equations. It may be explicitly given or it may be implicitly determined by the functions f and g that are involved. Of course, if f and g are only defined on the subset omega, then the domain should be contained in omega where the functions are defined. What is our goal with this inequality? It’s to find the solution set. A point x in the domain is a solution if it satisfies the inequality f(x) less than g of x. And we want to find the set of all those points for which this is true. Now as usual, we’re going to perform some operations.
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We’re going to massage the inequality until its solution set becomes evident. And there are some things we can do to the inequality that will preserve equivalence. One of these is the use of the additive principle. The additive principle says that if you have the inequality f less than g, and if you add a term h of x to each side, then you have an equivalent inequality. That is, an inequality having the same solutions. To put it another way, you can cancel out additively the h of x from each side and return to the original, fully equivalent inequality.
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That’s the same as saying if you have the inequality f less than g plus h, then you can take the h over on the left side with a minus sign, and consider the equivalent f minus h less than g. It follows that we could always replace the inequality f less than g by the inequality f minus g less than zero, where the right hand side is zero. But there actually is interest in having potentially non-zero expressions on both sides. As in, for example, the use of the multiplicative principle. Now recall that when you have an equation and if you multiply the equation on both sides by a non-zero factor h of x, then you retain equivalence.
188.1
It’s basically the same equation. That is, you could cancel out the h of x from either side of the equation and return to the fully equivalent original one. Now with inequalities, it’s a little more complicated, because if you multiply across by a non-zero h of x, you may reverse the inequality if the h of x is negative. So if h of x is positive, you retain the sense– less than or equal, in this instance– and if h of x is negative and you multiply across by h, then you reverse the inequality. It becomes greater or equal. We saw that, of course, in week one when we discussed arithmetic.
226.9
So for example, let’s apply these two principles that we’ve now seen to a simple affine or linear inequality mx plus b less than or equal c. We’ll assume the m is different from zero. We transfer the b over to the right hand side. So we get c minus b. And then we divide by m on both sides. If the m is positive, the sense of the inequality is preserved. And if the m is negative, it’s reversed. The product principle. Remember, for equations we used many times the fact that if you have f times g equals zero, then either f or g has to equal zero. Well there’s something a little similar for inequalities. Here’s one instance.
272.6
If the product fg is strictly negative, then either f is strictly negative and g strictly positive, or vice versa. See, there are two natural cases. And we could solve such an inequality by looking separately at the two cases. This in fact will be a recurrent theme. Let’s look at a simple example of this. Suppose we want to solve the inequality product of x minus 4 times x minus 1, strictly positive. Then we deduce rather easily that either both terms are strictly positive or both terms are strictly negative in order for their product to be strictly positive. That boils down to, in the first case, x being greater than 4, or in the second case, x being less than 1.
320.9
And so we’ve found the solution set of our inequality. The product principle also extends to non-strict inequalities, as pointed out here. Another remark concerns complementarity. Good news, whenever you solve one inequality, you’ve actually solved two inequalities. What I mean by that is the following. Suppose we have managed to determine the solution set, capital S, of the inequality f less than or equal g. And now suppose we consider the inequality f strictly greater than g with the same domain. Then we immediately know the solution set of this new inequality. It’s simply the complement of the original solution set S relative to the domain D, that is, it’s the set difference D minus S.
375.6
This is fairly obvious when you think about it. It simply stems from the fact that for a given x in D, either f is less than or equal g, or else f is greater than g, and the two are mutually exclusive. Now let’s look at an example, an example that will have a moral. Perhaps you recall that at one point we solved this equation, and we solved it in cowboy style. We simply squared both sides to get rid of the radical. We simplified the resulting equation. It turned out to be a quadratic one. And we found two possible solutions, x equals 1 or x equals 4.
418.6
Now the cowboy in us has found an extra cow here, you might say, because when you go back and check the x equals 4, you find it’s not a solution. But that didn’t bother us too much, because the x equals 1 was a solution, and so we had found the solution set. The singleton 1. Suppose we adopt the same fast procedure to the inequality now. I replace the equation above by the inequality. I proceed to square both sides to get rid of the radical. I then multiply by the denominator, 2 minus x squared, which is positive. That preserves the inequality. It becomes a quadratic one. We solve it rather easily.
458.7
In fact, it’s the product of two terms, as we saw a moment ago, and we find the solutions are the set of all x’s, which are either greater than 4 or less than 1. So there is the solution set, the union of two intervals. Minus infinity 1 and 4 infinity. The trouble with this procedure and with this answer is that it’s completely false. For example, the point x equals 3 is not in this supposed solution set we just found, and yet it certainly is a solution of our original inequality. You can see that because when you put x equals 3, the left hand side becomes strictly negative, which is certainly less than 1.
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So in this example, the cowboy procedure has given you too few cows. There are some missing cows. And this procedure will never be able to really identify for you which ones are missing. So here we need surgeons, not cowboys, to solve inequalities. Or in more formal language, let me summarize by saying, the moral is, in solving inequalities you must use careful thought along the way, even more than for equations.
Equivalence of inequalities, products, complementarity

#### Precalculus: the Mathematics of Numbers, Functions and Equations  