 Hello there again and we're looking at more methods of proof again. This time we're going to focus our attention on how to prove a bi-conditional statement. So a little bit of review here. What is a bi-conditional statement? We use the double headed arrow here for the symbol and typically we read this P if and only if Q. Bi-conditional statement is a relationship between two statements that basically says they are logically the same. They have the same truth values in all possible situations. One of the things we learn back in the section on logical equivalencies is that P if and only if Q is equivalent to a conjunction. P if and only if Q is equivalent to if P then Q and if Q then P. That is if you make a truth table for P if and only if Q and you make a truth table for this large compound statement on the right, you will get the same truth values in all cases. Now what that means for us when it comes to proofs is that if I wanted to prove a statement of the form P if and only if Q, I need to prove two things. I need to prove first of all if P then Q and then I have to prove if Q then P because of the AND symbol here. P implies Q and Q implies P is equivalent to P if and only if Q. And so proof of a bi-conditional statement is actually a double proof. It's a two stage proof. I have to first of all prove the forward direction P implies Q and then I have to turn and bracket back around and make another proof this time of the converse. Not the contrapositive, but the converse. P implies Q and then prove Q implies P. So the most important thing to remember there is that P if and only if Q proofs involve two mini proofs inside them. So let's look at a simple example here. For any integer K, K is even if and only if K plus 2 is even. So here is the P, if you will, and here is the Q. Okay, P if and only if Q is, so I have to prove two things. First of all, I have to assume or I have to prove that if K is even then K plus 2 is even and then I need to prove the converse. I have to prove if K plus 2 is even then K is even. Once both of those things have been established, I've proved the if and only if statement. So let's get to it. We're going to start by, we're going to do a direct proof of each of these parts here. And I notice each half of a if and only if proof is a proof of a conditional statement. So we could do that either by a direct proof or in some cases we might do this by proof by contraposition. We're going to do both of these if then proofs directly this time. So let's do the first one. So let's do the forward direction and that means I'm going to prove that if K is even then K plus 2 is even. And I'm going to do that directly. I mean I'm not going to form the contrapositive. I'm just going to assume the hypothesis and proceed forward. So let's make a little no-show table for that. There's my step, my no column, and finally my reason column. I think this might be a fairly short proof. The main thing here is to understand the structure of the if and only if. So let's start by assuming the hypothesis. Let's assume K is even, a direct proof. So we're just assuming the hypothesis and that's the reason. The last step of the proof is going to be the conclusion that K plus 2 is even. Okay. So now the flow of this no-show table ought to be pretty familiar to us. Since K is even, there exists an integer, let's say Q such that K is equal to 2Q. And that's the definition of even. Now I want to say something about K plus 2. So let's take line P2 and just add 2 to both sides of this equation. So 2Q plus 2. That's a, I'll call it algebra. What I did was add 2 to both sides of an equation. Now this in my next forward step, I can factor out the 2. Okay, that's factoring. Also algebra. And now you can kind of begin to feel we're coming in for a landing here. Let's say Z equal to Q plus 1. And that is an integer by closure. Okay, because we knew that Q is an integer. It's said so right here in line P1. And so adding 1 to it gives me another integer. And so now I have written K plus 2 equals to 2 times an integer. Okay. By setting Z equal to Q plus 1. And so that gets me to the end. So therefore K plus 2 is even by the definition of even. Okay. Now that completes the first half of the proof, right? I've proven the forward direction. Now I've got to prove the converse direction. I need to assume for any natural number K that if I happen to know that K plus 2 is even, then K is even. Again, another conditional statement, but the hypothesis and conclusion have switched places now. So I have to kind of flip it and prove this direction too. So let's do this with a direct proof as well as to keep it simple. We'll do our step in this line, our no in this column, and our reason in this last column here. This will look very, very similar to the first direction of the proof. So again, let's let P be the hypothesis. We're going to assume that K plus 2 is even. Okay, and that's by the hypothesis assumption. The last line is going to be that K is even. And let's see, so if K plus 2 is even, then there exists another integer. I called a Q in the forward direction. Let's call it R in this direction, an integer R such that K plus 2 equals 2R. Okay, and again, this is the same reason for a definition of even. Okay, now I have something I know about K plus 2. I'm looking to say something about K. So perhaps the right thing to do will be to subtract 2 from both sides. Okay, to solve for K. So I get 2R minus 2. That's about as simple as the alters you can get, right? And now I see that I can factor a 2 off of this side. So I have 2 times R minus 1. That's factoring. And now pretty similar again to the forward direction. I can say that let's call it, let's call it W equals R minus 1 is an integer by closure. R is an integer because of line P1. Said so right there. So R minus 1 by closure is also an integer. And so now what I've done is I've written K equal to 2 times W where W is an integer. Okay, by setting W equal to R minus 1. And that gets me to the end K is now even by definition. Okay, so what I've done here is I've proven, let's see if I can page back to the original statement. I've proven using two complementary direct proofs that if K is even, then K plus 2 is even. Check. And then I turn around and prove the converse. And that is enough to prove the if and only if statement.