Types of Relations

Transcript

0:00Introduction

Hello everyone! Let's start with one small puzzle. Rahul is Riya's sibling. Now a simple rule: if Rahul is a sibling of Riya, and Riya is a sibling of someone, then Rahul is a sibling of that someone too. Fair enough? But Riya is a sibling of Rahul. So, Rahul is his own sibling? Wait, that's impossible! So where did the logic break? Today we learn the types of relations, and you will crack this puzzle yourself. And remember, JEE asks this topic again and again. Let's begin.

0:35Void, universal and identity relations

So, a relation on a set A is just a subset of A cross A, okay? The smallest choice: no pairs at all, the void, or empty, relation. The circle with a slash is the empty set, and the sideways cup means is a subset of. For example, on the students of a school: two students differ in height by more than five metres. Now tell me, which two students can do that? Nobody, right? Void! The biggest choice: every pair. R equals A cross A, the universal relation. Like, two students study in the same school. Always true. Universal! Clear? And the identity relation, I with a small A below it: only the pairs a comma a, with the curly E meaning belongs to. Every element is related to itself, and to nothing else. Remember this one!

1:25Reflexive, symmetric, transitive

Now the main properties. Here, each dot is an element, and each arrow is a pair, okay? First, reflexive: a comma a is in R, for every element. In the picture, a little loop at every point. Even one missing loop, and it is not reflexive. Next, symmetric: if a comma b is in R, then b comma a must be there too. The double arrow on screen means implies. So every arrow has a return arrow, like the middle picture: one goes to three, and three comes back to one. Like siblings: if Rahul is Riya's sibling, Riya is Rahul's sibling. Perfectly symmetric. Now the third picture. An arrow from two to one, and from one to three. If the relation is transitive, which arrow must also be there? Yes, two to three, the dashed arrow! Transitive: if a comma b and b comma c are in R, then a comma c must be there too. Every two step path has a short-cut. Got it?

2:25Antisymmetric is not the opposite of symmetric

One more: antisymmetric. If a comma b and b comma a are both in R, then a must equal b. So no two different elements are related both ways. With me? Now tell me. Symmetric and antisymmetric sound like opposites, right? So no relation can be both? Hmm, careful! Look, one example breaks this idea completely. Take the identity relation: only loops. Each loop is its own return arrow, so it is symmetric. And no two different elements are related, so it is antisymmetric too. In fact, it has all four properties! And one comma two, two comma one, one comma three is neither. One and two are related both ways, so not antisymmetric. And one comma three has no return arrow, so not symmetric. Got it?

3:13Topper's trick: testing a list of pairs

Now, how do toppers test a list of pairs fast? Reflexive: tick off the loop of every element. Symmetric: for every pair with different entries, find its reverse. One missing, and it fails. Transitive needs care. For each pair a comma b, look at every pair starting with b, say b comma c, and demand a comma c. Loops never cause trouble, so just skip them. Now suppose a comma b and b comma a are both in R, okay? What does transitivity demand? Yes, a comma a, and b comma b! This hidden demand is the most common reason a relation fails to be transitive. And now our puzzle! Rahul to Riya, and Riya back to Rahul, so transitivity demands Rahul is his own sibling. Impossible! So, is a sibling of, is not transitive. Puzzle solved!

4:05JEE traps: the empty relation and the set

Alright, now two JEE traps. First, the empty relation on a non-empty set. Reflexive? No, there are no loops. But symmetric? Transitive? Yes, and yes! Surprised? To break them, you need a pair without its return, or a two step path without its short-cut. With no pairs at all, nothing breaks. This is called vacuous truth. Don't lose marks here! Second trap: reflexivity depends on the set. One comma one and two comma two give a reflexive relation on the set one, two. But on the set one, two, three, the loop three comma three is missing. Not reflexive!

4:43Counting relations: the grid trick

Now a toppers' counting trick. For the set one, two, three, draw a three by three grid, one cell per ordered pair. Each relation switches every cell on or off. So with n elements: n squared cells, two choices each, two to the power n squared relations. For n equals three, five hundred and twelve. Reflexive: the n diagonal cells, the green ones, must be on. So how many cells are free? n squared minus n, exactly! So two to the power n squared minus n; for three elements, sixty four. Symmetric: the two cells of a mirror pair, same colour, go on or off together. So the n diagonal cells give two to the power n, and the n into n minus one by two mirror pairs give two to that power. Multiply: two to the power n into n plus one by two. For three elements, sixty four again.

5:34Example 1: is R reflexive, symmetric or transitive

Now, example one, an old exam question from two thousand four. Set A is one, two, three, four, and R has five pairs: one comma three, four comma two, two comma four, two comma three, and three comma one. Which is correct: R is reflexive, R is transitive, R is not symmetric, or R is a function? Pause the video, and try it first. Done? Okay, let's check. Is one comma one there? No, and the slashed curly E means does not belong to. So, not reflexive. Look, one comma three and three comma one are both there. The hidden demand! Transitivity needs one comma one, and it is missing. Same story as Rahul and Riya, isn't it? Now symmetric. Two comma four has four comma two, and one comma three has three comma one. So it is symmetric, right? No, no, careful! Two comma three is there, but three comma two is not. One missing return arrow, and symmetry fails. And not a function either: two is related to both four and three. So the answer: R is not symmetric, okay?

6:43Example 2: classify (x - y)(x - 3y) = 0 on N

Now a harder one, from JEE Main twenty nineteen. Here x and y are natural numbers, the N with a double line, and x is related to y when x squared minus four x y plus three y squared equals zero. Is R reflexive, symmetric, transitive? Pause and try. Hint: factorise first. Done? Okay, let's check. Factorise: x minus y, times x minus three y, equals zero. So x is related to y exactly when x equals y, or x equals three y. The double headed arrow means if and only if. Reflexive? Yes, x equals x, always. Nice and easy. Symmetric? Three comma one is in R, as three is three times one. But one comma three? One is not three, and not three times three. So no. Transitive? Nine comma three and three comma one are both in R, okay? But nine comma one? Nine is not one, and not three times one. So no. Final answer: reflexive, but neither symmetric nor transitive. Remember, to disprove a property, one counterexample is enough. That is the JEE way.

7:59Quick revision: arrow tests and the empty relation trap

So, quick revision. Reflexive means a loop at every point, and symmetric means every arrow has a return arrow. Transitive means every two step path has its short-cut, and watch the hidden demand. And the empty relation is not reflexive, but it is symmetric and transitive. Now go and try the practice problems on this page. You will find them easy. All the best!