 Hello there again and welcome to another screencast of More Methods to Proof. This is a second example of proving a bi-conditional statement, an if and only if statement. So this one's a little bit trickier. This is a conjecture actually that shows up a lot in algebra. We're going to prove it. It's not really a conjecture. It's true, but we'll prove it anyway. So for all real numbers x and y, then x squared equals y squared if and only if x equals y or x equals negative y. So we're going to assume a fact from arithmetic. This is actually an axiom from arithmetic that we accept typically without proof that if we take any two real numbers and multiply them together and their product is zero, then either a equals zero or b equals zero. So what we'll just call this our axiom, which is a fancy word for a statement that we accept without proof. Okay, so we're going to prove this. This is an if and only if statement. So we have two things to prove. Remember, we need to first of all prove that if x squared equals y squared, then either x equals y or x equals negative y. And then we've got to prove the converse. The opposite direction would be if x equals y or x equals negative y, then we want to prove that x squared equals y squared. So there's our first part, our second part. We're going to prove that just one step at a time. So let's prove the forward direction first in which we're going to assume that, or if we're going to prove that if x squared equals y squared, then either x equals y or x equals negative y. Put that either. Okay, so let's set them up here in a little bit of another short table. Note it's still a conditional statement. We have a hypothesis and a conclusion. The conclusion has an origin. It's, it's a little strange, but we'll, we'll see if we need to invent anything new when we get there. So here's where I'm going to put my steps. Here's where I'm going to put what I know. In this last column, I didn't leave enough room for this, but I'll write the word reason. That's where we're going to put why we know what we know. So we'll start with the hypothesis. We're going to assume that x and y are real numbers such that x squared equals y squared. That's the hypothesis. Okay? The last line is going to be that x equals y or x equals negative y. It gets a little weird to conclude a disjunction, to conclude that one thing or the other happens. Let's see if it's a, if it causes any problems in the meanwhile. Well, what can we do? Well, the one thing I don't want to necessarily do is take the square root of both sides here for, because there's easier ways to do this. Let me think of something simpler using my axiom. Okay? My axiom tells me that if two real numbers multiplied together, give me zero, then either one of those things is zero or the other is zero. And that looks suspiciously like the thing I'm supposed to conclude here. So I'm going to figure out a way to get the axiom in the game here. To do that, I need to have something equal to zero. So I will, let's do this. Let's subtract y squared from both sides. So x squared minus y squared equals zero. That's just, it's algebra, you know, that's almost too easy to say algebra. We just subtracted y squared from both sides. Now what could I do to push this further along? Well, I see that x squared minus y squared is sitting here on the left hand side. So I believe I could factor that. That's going to be x plus y times x minus y. And that's all equal to zero. It's the difference of two perfect squares. And so that is some real algebra, some factoring I can employ. And now, now's where my axiom can get pulled into play here. I have two things right here and right here. And they're, they're real numbers and they're multiplied together to give me zero. So this is going to tell me that either x plus y equals zero or x minus y equals zero. And the reason for that is my axiom that I was allowed to assume on the previous slide. Okay, so now I think we know how to get, we actually can go straight to here by solving, by doing some algebra, just to solve for x. Solve both of these little equations for x in this line that gets me right to the conclusion. You solve this one for x, you get x equals negative y, or you could solve this one for x and get plus y. Either way we land on the conclusion. So we've done the forward direction of this if and only if proof here. Let's look at the converse direction and see if we can wrap this up. So the converse direction would be the conditional statement if x equals y or x equals negative y, then x squared equals y squared. So let's set this up. Let's go to our step here. And what we know. And over here, our reason for knowing it. Let's try this with a direct proof only because the, a contrapositive proof would be a little awkward. I would have to assume that two things aren't equal and to prove that two things aren't equal to each other. And I like all that negativity. So let's start with, let's do a direct proof here. So we're going to assume that either, I'll put the word either in here just for to emphasize either x equals y or x equals minus y. Okay, that's the hypothesis. Okay, and the conclusion here is going to be that x squared equals y squared. And we don't know again why that is true just yet. So let's think through where we want to head with this. I've assumed that either x is equal to y or x is equal to negative y. Where I'm headed is to say something about x squared. I want to say that x squared equals y squared. So let's start by just writing x squared. Okay. And what could that possibly equal? We want to say y squared. But let's see what we can do. So either x is equal to y or x is equal to minus y. So one of the following two things is true. Either x is equal to y squared or x is equal to negative y squared. And that's really just substitution. Okay, I'm substituting either y or negative y in for x, substitution. That seems a little strange. There's really two possible cases here. We're going to talk about proof by cases in a separate screencast a little later. But I think we can handle this much of it now. If I want to say something about x squared and I know something about x, I can substitute one of two things in for x. And here I'm just doing that kind of simultaneously. And let me work out the right hand sides of those equations simultaneously as well. So x is equal to y squared, okay. Or x is equal to x squared. I'm sorry, I left the squared off. x squared is equal to minus y squared. But we know that's equal to y squared. Okay, so that's algebra. This is carrying out the operation on the right hand sides. Okay, so in either case, I come to the same conclusion that x squared equals y squared. Either way. Okay. What's the reason finally for this? Really, I guess algebra again is the reason for it. Because I did the squaring and it worked out the way that it did, that gives me x squared equals y squared in either case. Okay, so let's recap what we've done. We have proven, go back to the original statement here, I've proven that if you assume that x squared is y squared, then either x equals y or x equals minus y. And that was thanks to our axiom from the tech. And if I assume the opposite, then I, or assume the converse, I should say, if I assume that x equals y or x equals minus y, then x squared equals y squared just amazingly by substitution and pure algebra. Since I've proven both the forward direction and the converse direction, the if and only if statement is now proven.