 Welcome back to another example of proof using cases. Here's the proposition we're going to study this time. It's a basic property of the absolute value operation. For all real numbers a and b, the absolute value of a times b equals the absolute value of a times the absolute value of b. You learned this property many years ago, but we don't take it as an axiom. It's something we actually need to prove in order to use. So if you're approaching this proposition with an intent to prove it, and you already believe that it's true because you've played with the problem and tried out a bunch of examples first, then before we prove anything, we have to come to terms with three questions. First, what are the precise definitions of the terms in the problem? Second, why should we try a proof using cases this time as opposed to something else? And third, if we use a proof by cases, what are the cases? First question first, what is the precise definition of the main term in the proposition and that's the absolute value function? Now, we might have a vague informal idea or several vague informal ideas about what absolute value means, but in order to prove something about absolute value, we need to have a precise mathematical definition of absolute value that we all agree upon. And here is the most commonly accepted one for absolute value given as a piecewise function. So the absolute value of a real number x is defined to be x itself if x is greater than or equal to zero, or we say non-negative. And the absolute value of x is equal to minus x, or negative x, if x itself is negative. Keep in mind that if x itself is negative, then minus x is actually a positive number. And so this gives what you would expect. For example, minus negative two is positive two. So this is the definition we will use for this problem, and we will keep all other informal interpretations of the absolute value that we might have accumulated over our experiences with math off to the sidelines and out of the proof. Second, why should we use cases on this? I think the best way to see this is just to consider the alternatives. If we try this with a direct proof, what we'd be assuming is that a and b are real numbers, and that's all. And that's not much information. Going from a contrapositive would involve assuming that absolute value of ab does not equal absolute value of a times absolute value of b, and then trying to prove that x and b are not real numbers, and that seems extremely complicated. Proof by contradiction would involve assuming that there exist real numbers a and b such that the equality doesn't hold. And that seems pretty complicated too. So we're going to default to proof by cases here. Less pragmatically, if you look back at the definition of absolute value, that definition itself involves cases. So that's why we have two pieces to the definition. So since the absolute value itself is defined by cases, it seems natural that proofs involving absolute value should be done by cases as well. Third, and finally, what are the cases going to be? And here's where I'm going to hand this off as a concept check. So think about the cases that would be involved with this proof. And the question I want to ask you is, what is the minimum number of cases that we'll need to prove this proposition? Is it two cases, four cases, six cases, or eight cases? Think about that while you pause the video. So we're going to say here that the correct answer is four cases. So what are those four basic cases? First of all, you might have said two because we used two cases in the last proof for even and odd. But remember here that a and b are not integers. And so we can't break the cases into even and odd as we did in the first video. Because obviously some real numbers are not integers, and therefore neither even nor odd. So we have to use cases that actually apply to real numbers and that make sense for the problem. Since we're dealing with absolute values, probably the cases that make the most sense are to think about whether the numbers that we're looking at are either greater than or equal to zero or less than zero. Those are the cases that define absolute value in the first place. So let's just be natural about it. So certainly every real number is either negative or non-negative. And those two cases don't overlap and they cover all possibilities for a single real number. Since we have two real numbers though, we're going to have four cases. And those four cases are the following. Case one, we're going to consider where a and b are both non-negative. Case two, we're going to consider where a is non-negative but b is negative. Case three, we're going to switch that around and assume that a is negative and b is non-negative. And then case four, we're going to look at what happens if a and b are both negative. Now these four cases cover all the possibilities for real numbers a and b. No choice of a and b will fall into two of these at the same time. And if we can prove the result in these four cases, then we've truly proven the result for all pairs of real numbers. So let's dive into it. Let's start with case one. And we're going to assume here additionally that a and b are non-negative. Again, using cases gives us an additional assumption to play with and that's what's so attractive about this method. We want to show that the absolute value of a, b equals the absolute value of a times the absolute value of b. Now remember, since we're proving that an equation holds, the one thing that I can't do with the outset is to write down that equation. That will be assuming what I want to prove. But I can work with each side of this equation individually in isolation from each other. And that's what I'm going to do here. The strategy I'm going to employ is to show that the left side of this equation, absolute value of a, b, and the right side, absolute value of a times absolute value of b, are both equal to a common third expression. And if I can show that both the left and the right independently are equal to the same thing, that will make them equal to each other. So I'm not going to assume that the equation holds, I'm going to prove it by working with both sides independently from each other. So look at the left side first. Now, since a and b are non-negative, I know from the basic axioms of arithmetic that a times b is non-negative. In that case, the definition of absolute value gives me that absolute value of a, b is just a, b. Now look at the right-hand side. Again, since a and b are both non-negative, the definition of absolute value gives absolute value of a equals a and the absolute value of b equals b. Substituting those in, I get the absolute value of a times the absolute value of b is just a, b. And now since the left-hand side and the right-hand side both equal a, b, that means that those two sides actually do equal each other. So that completes the proof of case one. Now let's move on to case two. Let's assume that a is non-negative but that b is less than zero. So again, we're going to look at the two sides of the equation. We want to prove separately from each other. On the left, we know that since a is non-negative but b is negative, then a times b is negative. Therefore, the absolute value of a, b should be minus a, b according to the definition. On the right-hand side of my proposed equation, again, I know that a is non-negative but b is negative. So the absolute value of a would just be a, but the absolute value of b would be minus b. And so if I take the absolute value of a times the absolute value of b, that's a times negative b, and again by axioms of arithmetic, that's negative a, b. So the left and right sides again equal each other and the result is proof for case two. I'm just going to show case three up here because it's very, very similar to case two. And in fact, you could argue that you don't even need a case three. You could just say that case three is really similar to case two. But I think when you're learning in first writing proofs by cases, it's best not to leave stuff out but just to go ahead and write down the case. So here it is. I won't talk you through it. Just let you look at it. We're assuming here that a is negative and b is non-negative and all the things that happen in case two again happen to case three with the letter switched. Finally, let's look at case four. This is where a and b are both negative. So again, by basic axioms of arithmetic, if a and b are both negative, then a times b is positive. Therefore, the absolute value of a times b is just a, b. And since a and b are negative, then the absolute value of a is minus a, and the absolute value of b is minus b. And so absolute value of a times absolute value of b is minus a times minus b. And once again, arithmetic gives us that's equal to a, b. So once again, absolute value of a, b and absolute value of a times absolute value of b are equal to the same third common thing, and so they're both equal to each other. And that proves the result for case four. Now, finally, since we've proved all the result in all four cases and those cases cover all the possible choices of a and b, the proposition is proved for all real numbers a and b, and that ends the proof. So as you can see here, proof by cases can lead to rather lengthy proofs, very workman-like proofs. But in each case, you get extra structure to work with, and so this is kind of an appetizing way to go sometimes. So I think this proof is a good example also of being careful how you choose your cases and very especially making sure you understand the real definitions of the terms that are involved in the problem that you're solving. Thank you for watching.