Representation of Sets

Transcript

0:00Introduction

Hello everyone! First, one small puzzle before we start. How many letters are there in the word school? S, C, H, O, O, L: six letters, right? Now tell me, if I make a set of the letters of school, how many elements will it have? Six? Hmm, not so fast. And look, this matters. In JEE, a question often gives you a set in one form and quietly expects you to read it in the other. Today we learn both forms, roster form and set-builder form. So, let's begin.

0:37Roster form: list every element

So, the first way to write a set is roster form, also called tabular form. You simply list all the elements, separated by commas, inside curly braces, the two curly brackets you see on screen, okay? For example: two, three, five, seven, written one by one, with commas, inside braces. That's it, that is roster form. Really simple, right?

1:02Set-builder form: describe the property

Now the second way: set-builder form. Here we don't list the elements. We describe the common property they share. Read this line as: the set of all x, such that x has property P, okay? The colon in the middle is read as such that. Alright, now think about our set: two, three, five, seven. What is common to these four numbers? Yes, exactly! They are the prime numbers less than ten. So, in set-builder form: the set of all x, such that x is a prime number less than ten. Same set, different way of writing. Clear? One small thing. The colon may also be written as a vertical bar, the straight line on screen. Both mean such that. So if you see a bar in a JEE paper, don't get confused, okay?

1:55Rules of roster form

Now, three rules for roster form. Rule one: order does not matter. One, two, three is the same set as three, one, two. Only the elements matter, not the order. Rule two. Look at this set: one, one, two. How many elements does it have? Three? No, no, wait! Repetition is ignored. One, one, two is the same set as one, two. Just two elements. Writing it twice does not count it twice, okay? And now, our puzzle from the start! The letters of school are S, C, H, O, O, L, six of them. But the letter O comes twice, and in a set we write it only once. So the set is S, C, H, O, L: five elements, not six! Mystery solved! Rule three: the three dots. We use them only when the pattern is absolutely clear. Like one, three, five, dot dot dot: everyone can see that seven and nine come next, right? If the pattern is not clear, don't use dots. Describe the property instead.

3:09Topper's trick: roster to set-builder

Now, a trick that toppers use. How do you change roster form into set-builder form? Look at the pattern of the n-th term. First term, second term, third term: how does each one depend on n? Try this one: one, four, nine, sixteen, and so on. What is the n-th term? Yes, n squared! One squared, two squared, three squared, four squared. Perfect. So this set is: n squared, such that n belongs to the natural numbers. The symbol that looks like a rounded E is read as belongs to, and the double-lined N stands for the natural numbers, okay? One more. One by two, two by three, three by four, and so on. What is the n-th term now? Look, the top is n, and the bottom is one more than n. So the set is n over n plus one, such that n belongs to the natural numbers. Simple as that, done! Makes sense?

4:11Set-builder to roster: solve, then filter

Now the other direction: set-builder to roster. Step one: solve the condition, the equation or the inequality, just like you always do. Easy, right? Step two, and this is where marks are lost. Keep only the values from the stated universe. On screen, N is the natural numbers, Z is the integers, and R is the real numbers. Most mistakes happen when students forget this last filter. So remember, solve, then filter. Got it?

4:43Example 1: integers, divisors of 36, letters of MATHEMATICS

Now, Example one. Write three sets in roster form. First, A: the set of all x, such that x is an integer and x squared is less than ten. Second, B: the natural numbers x that divide thirty six. Third, the set of letters of the word mathematics. Pause the video, and try it first. Done? Okay, let's check. Part one. Solve first: x squared less than ten means x lies between minus root ten and plus root ten. The arrow on screen means implies. That is, roughly between minus three point one six and three point one six. Now filter. The question says x is an integer. So keep only minus three, minus two, minus one, zero, one, two and three. Seven elements. Did you forget the negative integers? Many students do! Part two. Which natural numbers divide thirty six? Think in pairs: one and thirty six, two and eighteen, three and twelve, four and nine, and six with itself. So B is one, two, three, four, six, nine, twelve, eighteen and thirty six. And part three, just like our school puzzle. Mathematics has eleven letters, but M, A and T each come twice. Ignoring repeats: M, A, T, H, E, I, C and S, so eight elements, not eleven. With me?

6:22Example 2: roster form with sin, modulus and inequality

Now a harder one, JEE level, again in roster form. First: sine of n pi by two, where n is an integer. Second: integers x such that mod of x minus three is less than five. The two vertical bars on screen mean mod, okay? Third: real x such that x squared, minus three mod x, plus two, is less than zero, and then intersection with Z, the integers. Pause and try. Hint: solve, then filter. Done? Okay, let's check. Part one. As n runs through the integers, sine of n pi by two takes the values zero, one, zero, minus one, zero, one, again and again. So the set has only three elements: minus one, zero and one. Repetition is ignored, remember? Part two. Mod of x minus three less than five means x minus three lies between minus five and five. Add three everywhere: x lies between minus two and eight. Now filter: integers only, so minus one, zero, one, all the way up to seven. Nine elements. Careful, minus two and eight are not included; the inequality is strict. Alright, part three, the real JEE trap. There is mod x inside, so put t equal to mod x. The inequality becomes t squared minus three t plus two, less than zero. That factorises as t minus one, times t minus two, less than zero. So t lies between one and two, which means mod x lies strictly between one and two. Over the real numbers, that is two open intervals: minus two to minus one, and one to two, joined by the cup symbol for union. So the answer is these two intervals, right? No, no, careful! Read the question again. There is an intersection with Z at the end, the cap symbol. We forgot the filter! So, which integers have mod x strictly between one and two? None at all! So the answer is the empty set, the circle with a slash you see on screen. Solve, then filter. That one last step decides your marks.

8:47Quick revision: roster vs set-builder form

So, quick revision. Roster form: list the elements, and remember, order and repeats don't matter. Set-builder form: describe the common property of the elements. And to turn roster into set-builder, spot the pattern of the n-th term. And from set-builder to roster: solve first, then filter by the universe. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!