 Mathematical symbols fall into three major categories. Operators, symbols that indicate an action, like plus or times or divide. Operands, what these operators act on, symbols like five or x or pi. And relational symbols, symbols that indicate a relationship between two expressions, equals greater than approximately. So let's take a look at those relational symbols. So to begin with, let's define an equivalence relation. Let R be a relational symbol for operands in some set F. We say R is an equivalence relation if, for any x, y, and z in our set, it is reflexive, so that for any x, x is related to itself. Symmetric, if x is related to y, then y is related to x. And transitive, if x is related to y and y is related to z, then x is related to z. It's important to keep in mind there are only so many symbols. And so while we define equivalence relation in terms of this symbol R, R could be any symbol we wish to use to represent a relation. The most familiar symbol is this one, and let's see if this is an equivalence relation for the elements of the set of real numbers. Definitions are the whole of mathematics. All else is commentary. Our definition of an equivalence relation can be viewed as a set of requirements for entry into a very exclusive club. And what we want to do is we want to check the requirements at the door. If a relation meets these requirements, then it gets in, and if not, it's rejected. So let's suppose x, y, and z are real numbers. And we'll check our three requirements. So first we see that equals is reflexive since x is equal to x, which means our first requirement equals is symmetric since whenever x is equal to y, y is equal to x, which means our second requirement, and finally equals is transitive since whenever x is equal to y and y is equal to z, we have x equal to z. And that meets our third requirement, which means that equals is an equivalence relation, and it can be let into the club. How about greater than or equal to? Well, again, we'll check our requirements for any x, y, and z in R. Well, greater than or equal to does satisfy reflexivity, x is greater than or equal to x, but it fails symmetry. It is not symmetric. If x is greater than or equal to y, it will not generally be true that y is greater than or equal to x. And the important thing to recognize here is that as soon as we fail any of the entrance requirements to the club, then we don't get in the club. And so as soon as we see that greater than or equal to is not symmetric, we know that it is not an equivalence relation. Again, that relation can be anything we want. So let x and y be integers, and let x be related to y if x minus y is a multiple of five. Is let's call this squiggle an equivalence relation. So again, to show that squiggle is an equivalence relation, we need to show that for any x, y, and c that are integers, x squiggle x. If x squiggle y, then y squiggle x. And if x squiggles y and y squiggles z, then x squiggles z. Now, definitions are the whole of mathematics. And remember, we want to rewrite definitions as two conditionals. And while we know our definition for equivalence relation, the other thing to recognize here is that we have a definition for squiggle. And our definition says that if x squiggle y, then x minus y is a multiple of five. And also, if x minus y is a multiple of five, then x squiggle y. So here's another useful proof strategy. You can always write down the consequence of a conditional. So we want to prove that x squiggle x. So let's write that down as our conclusion. But remember, this has to be the last line of our proof. And so everything we write has to go above this line. Definitions are the whole of mathematics. All else is commentary. If I want to conclude that x squiggle x, then I have to have that x minus x is five times something. Well, I know that x minus x is equal to zero, so I can write that as the preceding line. And I can fill in our something. x minus x is five times zero. And so I conclude that x squiggle x. Again, we can always assume the antecedent of a conditional. We want to prove that squiggle is symmetric. So suppose x squiggle y. We want to conclude that y squiggle x. Definitions are the whole of mathematics. All else is commentary. In order to conclude that y squiggle x, then y minus x must be a multiple of five. It's got to be five times something. Since we know that x squiggle y, we know that x minus y is five times something. And we might notice that we can convert x minus y into a y minus x by multiplying through by negative one. And a little algebra says that we have a minus k inside those parentheses. And so y squiggle x. And our last requirement is transitivity. So again, we can always assume the antecedent of a conditional. Suppose x squiggle y and y squiggle z. We want to conclude x squiggle z. So we can conclude x squiggle z if x minus z is five times something. Since we know x squiggle y and y squiggle z, we know that x minus y is five times something, and y minus z is five times something. Now since we want to end up with x minus z, we might notice that we can just add the two equations. And that tells us that x minus c is in fact five times something. And so x squiggle z. And now let's put our pieces together. Since squiggle is reflexive, we prove that x squiggle x. Symmetric, we prove that if x squiggle y, then y squiggle x. And transitive, we prove that if x squiggle y and y squiggle z, then x squiggle z. It follows that squiggle is an equivalence relation.