00:01
So in this question, we're given our universal set and sets a, b, and c, which are subsets of our universal set.
00:07
Okay? and in part a, what we're asked to do is we're supposed to list elements in a complement, right? which means that it's everything that's not in a, or it's in the empty set, right? actually, no, okay.
00:26
I'll explain it like this.
00:30
We're asked to explain everything that's in a complement or our empty set, right? and so, we know that there's nothing in our empty set, right? so this part is pretty much meaningless.
00:43
So what we're focused on here is just finding the elements that are in a complement.
00:49
So if we're going to copy our universal set over here, we know that when we're trying to find the complement of a set, that means that we're looking at everything that is not in that set, right? so if i were to say that we had a universal set s, and then we had this subset, which we can call x, we know that everything that is in our universal set, but not in our x over here, is going to be our x complement, right? and it's important to see it this way because we know that if we take everything that's in our x complement, and then we add in the stuff that's in our x, what we're going to end up with is our sample space.
01:39
So applying that knowledge here, we know that if we have our universal set over there, and then we have our a over here, we're looking for everything in our universal set that is not an a.
01:54
Okay? so it means that if we see a number or an element that's an a, and it's also in u, then we cross it out.
02:00
Okay? so you have a one here.
02:02
We also have a one here, so we can knock it out.
02:05
We have a three in the a and a three here, so we can knock that out.
02:09
Four here, four there, so we knock it out.
02:12
We do the same thing for five, the same thing for 12.
02:16
We do the same thing for 14, and the same thing for 15.
02:23
So it means that all of our remaining elements, we just wanted to go in and erase all of them, like all the ones that we crossed out.
02:34
What's left in our universal set is going to be our a complement.
02:41
Okay? so if we're going to just move everything over, we know that this over here can be defined as a complement or the empty set.
03:13
Okay? so that's part a.
03:17
Now for part b, mir asked to list the elements that are in sets b, and c, right? so it means that it's in the intersection of sets b and c.
03:29
And a visual representation of this is that if we have set b, which is like this, and then we have set c, which is like this, everything that's in here, sorry, set c, everything that's in here is shared with an a and b, right? so it's just all the elements that are in b that are also in c.
03:50
All right so if you're going to look for all the elements that are shared in b and c let's just take a look we know that b has one and we see that c doesn't have one right so one is not going to be in their intersection if we see two in b we don't see a two and c so that's also not going to be a part of the intersection but we see a seven in both b and c right so it means that seven is an element of their intersection now we see a 13 in b and we don't a 13 in c, and so that means that there's no 13 in our intersection.
04:27
And since that's the last element in b, the intersection of b and c only contains one element, which is 7.
04:38
In part c, we are asked to list the elements in a complement or that are in b.
04:48
So when we're dealing with unions, what it means is that when we're looking for stuff that's either in one set or another set, it means that we're looking for a headcount of all of the elements that are found in both sets.
05:00
Okay? and so we're going to apply that knowledge to our situation.
05:06
We already have our a complement, right? which we know is going to be right here.
05:11
I'll just copy it.
05:22
So you know that this is going to be our a complement and this is going to be rb.
05:36
So now we have to look for all the elements that are in either of them, right? so let's take a look.
05:41
We see that we have a 1 that's in b...