 In general, a proof begins with some things we claim to be true, the premises, deduces other true statements using the rules of inference, and ends with a true statement, the conclusion. Now, stylistically, it's helpful to also summarize our proof by writing it as a conditional in the form, if, premises, then, conclusion. So this is what we're trying to do, and the first problem is how do we start, and in particular, what should we assume? To answer this question, let's consider, in general, we want to prove a conditional, if, a, then, b. Since the only time a conditional is false is when the antecedent, a, is true, and the consequent, b, is false, this gives us a starting point. When trying to prove a conditional, you may always assume the antecedent true. For example, suppose you want to prove all square matrices have non-zero eigenvalues. Let's rewrite this as a conditional, and let's decide what we can assume. So first, let's find our simple statements. A matrix is square, and a matrix has non-zero eigenvalues. Now we can rewrite our statement as a conditional. If the matrix is square, then it has non-zero eigenvalues. And now that it's in this conditional form, remember that when trying to prove a conditional, you may always assume the antecedent true. And so we can begin by saying, suppose m is a square matrix. Well, that's a starting point, but after we assume the antecedent, what then? To proceed with the proof, we can deduce other true statements using conditionals we already know to be true. And here's the important thing to keep in mind. If a conditional is true, then whenever the antecedent is true, the consequent will also be true. And it may help to think of a conditional as a machine that accepts an input and produces an output. The conditional, if a, then b, will accept the input a and return the output b. It won't accept a different input, and you can't use it backwards by supplying the output. So for example, it's supposed to be no if a, then b, and also if b, then c. Let's prove if a, then c. So we can put the conditionals in a sequence, so the output of one is used as the input for the next. We ought to prove if a, then c, and so we may assume that a is true. But wait, since if a, then b is true, then if a is true, b must also be true. In other words, if we supply our conditional with a, we get output b. And so we know that b is true. But wait, we also know that if b, then c is true, and since this is true, then if b is true, then c must also be true. Again, we have b, so if we drop b into this conditional machine, we get c as output, and so c is true. And so if we put all this together, any time a is true, then c is also true, and so the conditional if a, then c is true. So let's try to prove the sum of two odd numbers is even. So our first step is we'll rewrite our claim as a conditional. And that becomes if a number is the sum of two odd numbers, then it is even. And remember, we can always assume the antecedent a number is the sum of two odd numbers. And so the very first line in our proof should be n is the sum of two odd numbers. Now again, we can think about this as building a bridge. We want to start here, n is the sum of two odd numbers, and we want to end by concluding it is even. And if we think about this as a road, we can write down a desired conclusion. The important thing to keep in mind here is the last line of the proof is what you've proven. And so we want to make sure that our conclusion is going to be the last line, so we should leave some space to include the actual proof. Let's try to avoid the infinite regress. So we'll write our conclusion down here somewhere. So remember, definitions are the whole of mathematics, all else is commentary, and this problem talks about odd and even numbers. So we actually need to know a definition for odd and even numbers. And we might have this definition for even numbers. An even number is a product of two and an integer. Remember the reason that definitions are useful is they give us for free two conditionals. And so this definition gives us the two conditionals. If a number is even, then it is the product of two and an integer, as well as if a number is the product of two and an integer, then it is even. So the important thing to recognize here is you can only use a conditional if you have the antecedent. Since we want our conclusion to be n is even, that's our consequent, which means that we have to start with the number as a product of two and an integer. So we might go back a line. The line before we say n is even is that n is the product of two and some integer, and we'll leave the nature of that integer to be determined. Well, let's see if we can fill out the remaining details. So we know that n is the sum of two odd numbers, and then we do a bunch of stuff, hopefully, and get the n is two times something. Now, since we want to say something about the sum of these two odd numbers, let's write down their sum. And so we may say something like the following. Suppose p and q are the odd numbers, then n is the sum p plus q. Well, definitions are the whole of mathematics. All else is commentary. What's an odd number? And we might bring in our definition. An odd number is one more than the product of two and an integer. And this gives us two conditionals. If a number is odd, then it is one more than the product of two and an integer. Also, if a number is one more than the product of two and an integer, then it's odd. Now, in this case, we know that p and q are odd numbers. And so here we have the antecedent of this first conditional, which means that we know that both of these are one more than the product of two and an integer, which means we know that p is 2k plus 1 and q is 2r plus 1. So remember, equals means replaceable. So I have n equals p plus q, but p is equal to 2k plus 1. q is equal to 2r plus 1. So we'll replace, do a little algebra, and it helps to keep in mind our goal. We want to write n as 2 times something. Well, let's factor out a 2. And there's our proof, and we can summarize by writing a conditional where our first line is our antecedent, and our last line is the consequent. So if n is the sum of two odd numbers, then n is even.