 In previous videos of this lecture series, we've talked a lot about proving conditional statements, things like P implies Q. We have the method of direct proof where you assume the hypothesis and then you derive the conclusion through logical inference. You can also prove this using indirect proof methods like contra positives or proof by contradiction. What I wanna do now is then develop a technique to prove a biconditional statement. So if you wanna prove P is equivalent to Q, how do you do that? Well, the thing to remember about biconditional statements is that they're logically equivalent to the conjunction of two conditional statements. So the biconditional statement P is equivalent to Q is the same thing as the conditional statement P implies Q and its converse Q implies P, right? Which I intentionally drew this one backwards so that P and Q are in the same order here. Now we've talked about before, how do you prove a conjunction? A conjunction is just two statements. We have to both be true for the conjunction be true. So you first prove this one to be true, then you prove that one to be true. Now the first one is a conditional statement P implies Q. So you can prove that by any of the methods we just talked about, direct proof, contra positive, contradiction, et cetera. Then you have to also prove that the converse is true. Now in general, given a conditional statement, its converse might not be true, but if the biconditional statement is true, then the conditional P implies Q and its converse must both be true. But this converse is itself a conditional statement. So you can prove that with direct proof or indirect proof. You can mix it up. This one you can prove with direct proof. This one maybe you prove using contra positives or contradiction, that's a possibility. So these biconditional statements, the way we often read them is we often read it as P if and only if Q. So this is what we mean by biconditional. And this if and only if statement is often abbreviated in mathematics as IFF. So if you ever see an IFF that's not me misspelling if, that's actually an abbreviation for if and only if. So if and only if, you have the last F there, squeezed it together. So if and only if is a biconditional statement. You have to prove that P and Q are logically equivalent to each other. Now because of this observation here, proving if and only if statements is actually two statements that you have to prove. There's the if part. The if part means you will then prove that P implies Q. This is often referred to as the sufficiency of the problem or that P is a sufficient condition for Q. And the reason for that is, if you know that P is true, that's sufficient to tell us that Q is true. Now there could be other reasons why Q is true, but if P is true, that's sufficient to show that Q is true. So that's the if part. The only if part is then the converse. It goes the other way around. Q implies P. This is often referred to as the necessity of the equivalence there, the necessity of the proof or the necessary condition. Which means that if P is true, that means Q is true as well. So for P to be true, it's necessary that P has to be true as well. So be aware that the sufficiency goes in the direction, the first direction, necessity goes in the backwards direction there. And a pneumonic device that logicians sometimes like to use to help you remember which one's sufficiency, which one's necessity. The idea is that, let me see if I can remember it correctly. It's about Socrates. Let's see, it is, oh yes. One can say that knowing that someone is called Socrates is sufficient to know that someone has a name, all right? So knowing that there is Socrates knowing means that people know there's a name. Sufficiency comes before necessity S and N. Knowing Socrates exist implies that names exist because Socrates has a name. So sufficiency comes first, the necessity. But I confess, I forget which one's which all the time. You can survive without the vocabulary, but it's important. It's important I introduce it in this video, right? So what's the general if and only if proof template? Well, if you wanna prove that P is equivalent to Q, you first prove P implies Q, again using direct proof or indirect proof. Then you prove the converse, which is not written correctly here. You prove the converse that Q implies P. So you prove conditional and it's converse and that's what it is. So an if and only if proof comes in two parts always. The necessity comes typically second, but you could do it first if you want to. The sufficiency P implies Q, typically do that first. But again, it doesn't really matter because if P is, if you prove P implies or P is equivalent to Q, that's actually equal to Q is equivalent to P. So this is what I meant earlier. If you mix up the sufficiency and the necessity, it doesn't make much of a difference. You just need to prove that P implies Q and Q implies P. Let's see some examples of that. So let N be an integer. We're gonna prove that six divides N if and only if two divides N and three divides N. So I'm gonna first prove sufficiency. So if six divides N, I didn't wanna prove that two divides N and three divides N. That's the easier direction. So in the first direction, you might say that four sufficiency or something like that. But so the first direction, suppose that six divides N. Well, what does it mean to divide six divides N? Divisibility tells us there exists some integer K such that N is equal to six times K. Now, working with this equation, if N equals six K, that means that N equals two times three K. And since three K is an integer, this shows that two divides N. Now, if I take the equation N equals six and factor it slightly differently, this shows us that N is equal to three times two K. And since two K is also an integer, this shows that three divides N. So we've now shown that if six divides N, two divides N and three divides N, this shows sufficiency, the first direction. Now, for the other direction, for which that's actually how I phrase it here, you should usually give the reader some clue that you're doing the other one. Since it's an if and only of proof, there's two directions. It's good to put some type of Q to show that, oh, we did the first direction. That's now done, which is what this told us right here. We're done with that direction. Now, for the other direction, we're starting the other direction. So suppose that two divides N and three divides N. We have two assumptions here. We need to show that six divides N. Now, because two divides N, that means there's an integer R, such that N equals two R. And likewise, because three divides N, there's some integer S, such that N is equal to three S. Now, I want you to focus on this part just for a moment. Two R equals three S. Since two clearly divides two R, that means that two must also divide three S. Now, by Euclid's lemma, if two divides three S and two was a prime, that means that two divides three or two divides S. But wait a second, two does not divide three because three is also a prime number and it's only divisors are one and three, not two. Euclid's lemma then gives us that two divides S. Now, if two divides S, there likewise has to be an integer T, such that S is equal to two T. And if I take this equation and substitute it into the equation above, we then discover what we're looking for. N, which equals three S, is also three times two T, three times two is six. And so therefore we now see that six divides N. This then proves the necessity of the problem. And so since we've now proven both directions, we get the equivalence we're looking for. Six dividing N is equivalent to two and three dividing N. So we've now proven this if and only if statement. I wanna sort of make a parallel here that when you wanna prove that two sets are equal to each other, A equals B, we do this by showing that A is less than or equal to B and B is less than or equal to A. There's two directions to that proof that's very similar to this statement of logical equivalence. To show that two statements are logically equivalent, that is if and only if, we show implications in both direction, both the sufficiency and necessity. Now I wanna give a variation of this because with the last proof, we proved two conditionals P implies Q and Q implies P. We did both of those using direct proof. We assume the hypothesis and then derive the conclusion. But you can also prove implications using some type of indirect proof like contra positives. Now, if you apply that to an if and only if statement, you can actually vary it in the following way. If you wanna prove that P is equivalent to Q, you can first do the first direction, the P implies Q direction. You can do that directly. So assume P, then prove Q. To prove the other direction, Q implies P, this is logically equivalent to the inverse. So Q implies P, the converse is logically equivalent to the inverse, not P implies not Q for which as this is also a conditional statement, you can prove that by assuming not P and then proving not Q. And so basically, if you're gonna prove the converse using contra positives, you're actually proving the inverse and you can clean up the language by using this template here. Assume P, then prove not Q. So assume P, then prove Q. And as a second step, assume not P and prove not Q. So this can be a lot simpler than having to introduce contra positives or contradictions when you come to equivalence. So I'm gonna use that technique on the following proof here. The integer N is odd if and only if N squared is odd. So I'm gonna prove sufficiency first. So N being odd implies N squared is odd. So first suppose that N is odd. Well, if N is odd, that means there exists some integer K such that N is equal to 2K plus one. And then I'm gonna plug that into N squared and see that N squared is equal to 2K plus one squared. Which if you foil it out, you get 4K squared plus 4K is equal to plus one. For instance, if you factor that, you can write that as two times 2K squared plus 2K squared plus 2K plus one. And as this is an integer, you get two times an integer plus one that's odd. So this shows that N squared is odd. This gives us the sufficiency. Now to go the other direction, if we try to prove that directly, we would have to assume that N squared is odd, which we've looked at this before, it gets a little bit awkward. We'd rather plug the simpler term in the more complicated term. So when we try to prove that N squared being odd implies N is odd, we usually did that like contrapositor or contradiction. Since we have an equivalence here, I'm just gonna use the inverse. I'm gonna assume that N is not odd. Suppose that N is not odd, that is N is even. That's gonna be a lot easier in this situation. Now be aware, since we're going to the other direction, it's good to offer some type of guidepost to the reader, so conversely, helps me know I'm considering the other case, which admittedly I'm doing the inverse right now. So if you wanna say inversely, you can, but conversely is a more natural thing to say in English there. So suppose that N is not odd, that is it's even. So there exists some integer K, such that N equals 2K. Now if you're wondering, I already used the symbol K, but that wasn't the first half of the proof. Kind of like in a computer programming setting, I have a local variable. K only mattered for that proof. Now that I'm in a different proof in a different locality, that I can reuse that symbol. It's fine, no big deal. If there's not all any confusion here, you can introduce a new symbol, but honestly I think K works just fine. So N equals 2K. Well, if N equals 2K, then N squared equals 2K quantity squared, which is four times K squared, which is the same thing as two times 2K squared. This shows that N squared is even. So this proof template here was really, really nice. To show if and only if, you prove the implication in both directions, or in this case, what you can do is you can assume one, prove the other, then you can negate that same one and then prove the negation of the other, which we ended with of course a not, we ended with an even, which is the same thing as not odd. And this shows us how we can improve if and only if statements, which occur all the time in mathematics.