Part D: Assertion–Reason and Concept Checks

Transcript

0:00Introduction

Hello everyone! Okay, let me start with a small proof. Take a relation R that is symmetric and transitive. If x is related to y, then by symmetry, y is related to x. And then, by transitivity, x is related to itself. So R is reflexive! Two lines, nice and clean, right? But here is the shock: this statement is false. Somewhere in those two lines there is a hole. Can you find it? Today we play the true or false game, the way JEE plays it, in assertion reason and statement questions. One wrong tick costs you marks, so let's learn to catch the trap. Let's get started.

0:45The rule of the game

So, here is the game. In each item you get a statement, and you decide: true or false. And not just a tick, okay? You also give a one-line reason, or a counterexample. That is what makes it JEE level. Now tell me. You try one example, and the statement works. So it is true, right? Hmm, think again. No, no, careful! This is where students lose marks. To say true, your reason must work in every case: every set, every relation, every function. One nice example proves nothing. Clear? But for false, the story flips. Just one counterexample is enough. One single case where it fails, and the whole statement is dead. That's all it takes! Let's try one. Claim: if A intersection B equals A intersection C, then B equals C, okay? The double arrow on screen means implies. It looks like we can just cancel A from both sides, right? Hmm, let's test it. Take A as the set with just one, B as the set one, two, and C as the set one, three. Now, A intersection B is the set with one. And A intersection C is also the set with one. Equal! But B and C are clearly different, and the equals sign with a slash says not equal. One counterexample, and the claim is false. You cannot cancel A, okay?

2:15Topper's trick: tiny sets and the empty set

Now a trick that toppers use. Where do you look for a counterexample? Not in big, scary sets. In the smallest ones: a set with one element, or two elements, just like the ones we used, okay? And the sneakiest one of all: the empty set, the circle with a slash on screen. It has no elements at all. So before you tick true, always ask: what happens if one of the sets is empty? Try this claim. If A cross B equals B cross A, then A equals B, okay? The cross means the Cartesian product, the set of all ordered pairs. Sounds true, right? Can you break it? Take A as the empty set, and B as the set with just one. There is nothing in A to make a pair with, so A cross B is empty. And B cross A is empty too! Equal products, but A is not equal to B, okay? False! But look, if A and B are both non-empty, the statement becomes true. So in the exam, read every word. One small condition, like non-empty, changes the answer. Got it?

3:27Example 1: true or false on symmetric, transitive relations

Now for example one, and our puzzle from the start is hiding inside it. Four statements. One: every symmetric and transitive relation is reflexive. Two: the union of two transitive relations is transitive. Three: if R is an equivalence relation, so is R inverse. And four: a relation can be both symmetric and antisymmetric. True or false, with a reason. Pause the video, and try it first. Done? Okay, let's check. One is false. On the set one, two, take R with just the pair one, one. Symmetric? Flip one, one, and you get one, one. Yes. Transitive? Yes, nothing breaks. But reflexive needs the pair two, two as well, and the crossed belongs to sign says it is not in R, okay? So where was the hole in our proof? We assumed every x is related to some y! Here, two is related to nothing, so symmetry and transitivity never even start. Mystery solved! Two is also false. The relation with only the pair one, two is transitive, and so is the relation with only the pair two, three, because neither has a chain to complete. But take their union. Now one goes to two, and two goes to three. Transitivity demands the pair one, three, and it is not there. False! With me? Three is true. R inverse, the R with a little minus one on top, means flip every pair, okay? But an equivalence relation is symmetric, so flipping gives back the same pairs. In fact, R inverse equals R, so of course it is an equivalence relation. And four is true, which surprises many students. Symmetric and antisymmetric sound like opposites, right? But take the identity relation, where each element is related only to itself. Flip any pair and you get the same pair, so it is symmetric. And antisymmetric only demands that if x is related to y, and y to x, then x equals y, which is always the case here. Even the empty relation works. So, true.

5:42Example 2: one-one and onto, composites, finite vs infinite

Now a harder one, on composite functions. Read it with me. One: if g composed with f is one-one, then g is one-one. Two: if g composed with f is onto, then g is onto. Three: any function from a finite set to itself that is one-one must be onto. Four: any function from the natural numbers to the natural numbers that is one-one must be onto. Pause the video and try. Hint: think small. Done? Okay, let's check. The small circle on screen means composition: first apply f, then apply g, okay? Now, if the whole thing is one-one, who must be one-one, f or g? Most students say g, because g acts last. Hmm, careful! This is the JEE trap. Let's break it with tiny sets. Let f go from the set with just one to the set a, b, with f of one equal to a, okay? The arrow means from this set, to that set. And let g send both a and b to zero. Now, g composed with f has only one input, so it is automatically one-one. But g sends two different elements, a and b, to zero, so g is not one-one. So statement one is false. It is f, the first function, that must be one-one. Remember that! Statement two is true. Take any c in the final set. Since g composed with f is onto, c equals g of f of a, for some a, okay? So c has a pre-image under g, namely f of a. Every c gets one, so g is onto. Got it? Statement three is true. Say the set has n elements. One-one means n different inputs give n different images. And n distinct images inside a set of n elements fill the whole set. Nothing is left out, so the function is onto. Now, statement four looks exactly the same, doesn't it? The double struck N in the question means the natural numbers. So one-one must be onto here too, right? No! Take f of n equals n plus one. Different inputs give different outputs, so it is one-one. But which natural number is never an output? One! Nothing lands on one, so f is not onto, and the statement is false. See, the only difference was one word: finite. Read every word, okay?

8:24Quick revision: counterexamples, tiny sets, the empty set

So, quick revision. To say true, give a reason that works in every case. To say false, one counterexample is enough, and try tiny sets and the empty set first. And read every word: finite, non-empty, and which function comes first. Now go and try the practice problems on this page. Four more rounds of true or false are waiting for you. All the best!