 One of the terms we throw around a lot are strong and weak, and in the context of mathematics it means a following. In the conditional if a then b, we say that a is stronger than b, and that b is weaker than a. Intuitively strong statements claim many things. For example, a strong statement. A number is an even perfect square divisible by 37. And this claims three distinct things. First, a number is even. Second, a number is square. And third, a number is divisible by 37. Now strong and weak have a rather paradoxical relationship in mathematics, and that comes about as follows. Let's consider the conditional if both a and b then a. So let's see if this conditional is true. So remember the definitions for when a compound statement is true, and remember that for conditionals we only care if the antecedent is true. So if both a and b is true, then both a and b have to be true, which means that a has to be true, and so this conditional is true. And this means that the conjunction a and b is stronger than a. And this leads to the following idea. We can produce a stronger premise by assuming more things to be true. But let's follow this a little bit further. Suppose the conditional if a then c is true. What about the conditional if a and b then c? Again, the only case that matters is when a and b is true. Now if a and b is true, then a must be true. But if a is true and a conditional if a then c is true, then c is true. And so we can join premise to conclusion and get a proof of if a and b then c. But again, we can join premise to conclusion and get. And finally, we have our definition of stronger and weaker. The antecedent is stronger than the consequence. And what this means is that if we strengthen the antecedent, we actually weaken the conditional. Now this might not be too bad because many conditionals are too hard to prove. But we might be able to prove a weaker version by strengthening the antecedent. And this leads to what's known as a proof by cases. So if I want to prove the conditional if a then b, I'll make some additional assumptions c. And then I'll try to prove if a and c then b and lather, rinse, repeat. And if the additional assumptions include every possibility for a, we have completed a proof by cases. For example, say we want to prove the sum of consecutive squares is one more than a multiple of four. So let's rewrite this as a conditional. If n is the sum of consecutive squares, then n is one more than a multiple of four. We can always assume the antecedent. So suppose n is the sum of consecutive squares. We want to prove that n is one more than a multiple of four. So let's set that as our end with some distance between the two statements. Definitions are the whole of mathematics. All else is commentary. If we want to say that n is one more than a multiple of four, then we might say that n is a four times something plus one. And if n is the sum of consecutive squares, then n is k squared plus k plus one squared. We can do some algebra. And now we seem stuck. There's no obvious way to join the two n's together. So let's make some additional assumptions about the numbers. So again, suppose n is the sum of consecutive squares. And again, we want to conclude that n is one more than a multiple of four. So n is four times something plus one. But this time we might say something about the numbers. The first number is either even or odd. Well, suppose it's even, then n as the sum of consecutive squares looks like the square of an even number plus the square of the number after it. So that's the square of two k plus the square of two k plus one. And now let's see if we can build our bridge. So doing a little algebra. And now we can write n as four times something plus one. And let's summarize. And here's the important idea. Never promise more than you deliver. So our conditional, the antecedent, is n is the sum of consecutive squares where the first is the square of an even number. And in that case we can conclude n is one more than a multiple of four. Now if this was a pencil's down situation and we have to turn in our work, we've at least proved something. It's not what we wanted to prove, but it's close to it. We can complete the proof by considering the other possibility. So our roadmap remains the same, but instead of assuming that the first number is even, we'll assume it's odd. Then n is the sum of the square of an odd number plus the square of the next number. And we can build our bridge. Let's expand our squares and do some algebra. And we'd like this to be a plus one. So let's split that five into a four plus one. And here we can rewrite it four times something plus one. And again join premise to conclusion if n is the sum of consecutive squares where the first is the square of an odd number, then n is one more than a multiple of four. And here's the important thing to note. The first number is either even or odd. And so are two cases, namely the first number being even or the first number being odd. This includes every possibility. And so we can conclude if n is the sum of consecutive squares, then it is one more than a multiple of four.