Laws of the Algebra of Sets

Transcript

0:00Introduction

Hello everyone! Let's start with one quick question. In algebra, if x plus y equals x plus z, you cancel the x, and y equals z, right? Now sets. If A union B equals A union C, can I cancel A, and say that B equals C? Most students say yes. Hmm. Hold that thought. Sets have their own algebra, much like the one you know, but with a few surprises. And look, a question on exactly this came in two thousand nine, in the exam we now call JEE Main. So let's learn the laws, and then settle it.

0:39Union as plus, intersection as times

So, the secret is in the heading: union, the cup, is like plus, and intersection, the cap, is like times. First, commutative: A union B equals B union A, and the same for intersection. Order does not matter, okay? Associative: bracket A union B first, or bracket B union C first, the answer is the same, and the same for intersection. So with only one operation, brackets do not matter. Now the important one, distributive: A intersection, bracket B union C, equals A intersection B, union, A intersection C, okay? Just like x times, bracket y plus z, equals x y plus x z. Very familiar, right? But now, a twist. Can union also spread over intersection? In normal algebra, is x plus y z equal to x plus y, times x plus z? No, it isn't, is it? But sets do! So A union, bracket B intersection C, equals A union B, intersection, A union C, okay? Mark this one: students forget it, and it cracks many JEE simplifications.

1:55Laws with the universal set and the empty set

Alright, now two special sets: U, the universal set, and the empty set, the circle with a slash. Identity: A union empty is A, and A intersection U is also A, okay? So empty acts like zero, and U acts like one. Next, domination. Now tell me, what is A union U? U itself, yes! Once you take everything, adding more changes nothing. And A intersection empty is empty, just like times zero, okay? Here sets differ from numbers. In algebra, x plus x is two x. But A union A is just A, and A intersection A is A again, clear? These are the idempotent laws. Last, complement. The little dash, A prime, means everything in U outside A, okay? So A union A prime is U, and A intersection A prime is empty. The complement of A prime is A again, of U is empty, and of empty is U itself. Perfectly logical!

3:03De Morgan's laws

Now the most famous law here, De Morgan's law. The complement of A union B: is it A prime union B prime? Just put a prime on each set, right? No, no, careful! Classic JEE trap. Look at the figure. On the left, A prime has blue lines, B prime has red lines, and they cross only outside both circles: that is A prime intersection B prime. On the right, the shaded part is the complement of A union B, the same region! So it is intersection, not union! And its partner: the complement of A intersection B equals A prime union B prime. So remember: the prime goes inside, and union and intersection swap. Got it?

3:48Absorption laws

Now a law that looks strange in algebra, but is obvious in a picture. The dark middle part, A intersection B, lies completely inside A, right? So what is A union, bracket A intersection B? Just A! Adding a piece of A back to A adds nothing new, okay? We say A absorbs A intersection B, and this is the absorption law. And the partner: A union B is the whole shaded region. Intersect it with A, and only the dark part, A itself, remains. So A intersection, bracket A union B, is just A, okay? Keep this one ready: we will use it twice in Example two.

4:30Topper's trick: three quick ways to test an identity

In the exam they often ask: true or false? Three quick checks. One, Venn regions: in the three set Venn diagram with its eight regions, list the regions each side covers. Same list, true. Different list, false, and an element of a differing region is your counterexample. Two, algebra: whenever you see a minus, write A minus B as A intersection B prime. Then use De Morgan, and simplify just like algebra. This one line is your best friend, okay? And three, a tiny counterexample: each of A, B and C belongs to the little list on screen: the empty set, or the set containing just one. Most false identities fail for such small sets. That's it, done in seconds.

5:19Example 1: simplify four set expressions

Now, let's simplify these four. One: A union B, intersection, A union B prime. Two: A intersection B, union, A intersection B prime. Three: A minus, bracket A minus B, and four: A intersection, bracket A prime union B, with me? Pause the video, and try them first. Done? Okay, let's check. Part one: both brackets start with A union, right? So take A union outside, using our twist law: A union, bracket B intersection B prime. But B intersection B prime is empty, and A union empty is just A, okay? Part two is the mirror image. Take A intersection outside: A intersection, bracket B union B prime, which is A intersection U, again just A, okay? Now part three. In algebra, x minus, bracket x minus y, is just y. So A minus, bracket A minus B, is just B, right? No, no, careful! Sets are not numbers. Algebra method: A minus B is A intersection B prime. So we get A intersection, the complement of A intersection B prime. De Morgan makes that complement A prime union B, since the complement of B prime is B itself. Oh, look! That is exactly part four! So let's finish both together. Distribute: A intersection A prime, union, A intersection B, okay? But A intersection A prime is empty. So parts three and four both give A intersection B, not B, right? Only the part of B inside A survives.

7:06Example 2: same union and intersection, so B = C

Now, our puzzle from the start, the JEE Main question of two thousand nine. If A intersection B equals A intersection C, and A union B equals A union C, prove that B equals C, okay? Pause and try it. Hint: start from B, and use absorption. Done? Okay, let's check. First, our hook: can we cancel A with only the union condition? Take A as the set containing one, B as one and two, and C as just two. Then A union B and A union C are both one and two. Equal! But B and C are different. So one condition alone is not enough. Now the proof. We start from set B, okay? By absorption, B equals B intersection, bracket A union B, right? Why? Because now A enters the picture, and we can use what is given. Replace A union B with A union C, that is given. Then distribute: B intersection A, union, B intersection C, okay? Now, B intersection A, which is A intersection B, equals A intersection C, given. So we have A intersection C, union, B intersection C, with me? Take C common, distributive in reverse: bracket A union B, intersection C, okay? And A union B is A union C again. Finally, bracket A union C, intersection C, is just C, by absorption. So B equals C, proved! See, we used both conditions. With one you cannot cancel, but with both, you can.

8:50Quick revision: De Morgan, minus as complement, cancelling

So, quick revision. De Morgan: the prime goes inside, and union and intersection swap. See a minus? Write A minus B as A intersection B prime, then simplify. And no cancelling unless both the intersections and the unions match. Now try the practice problems on this page. You will find them easy, I promise. All the best!