Exercise 7.1: Problem 16 Transcript | Exercise 7.1: Problem 28 Transcript | Exercise 7.1: Problem 33 Transcript | Exercise 7.1: Problem 52 Transcript

Exercise 7.1: Problem 16 Transcript

Hello. We're going to do an example problem here. This is from section 7.1. This is problem number 16 from the homework. This is a true/false question. We're going to look at a set statement and determine whether it's true or false.

The way to do these problems is to work with the sets, determine two final sets, and, in this case, we're going to determine whether one is a subset of the other.

So the question says, one, two, three, union, three, four, five, and is a subset of three, four, five. So we're going to determine whether this union is actually going to end up being a subset of three, four, five.

The way to do the problem is to perform the union, come up with the final result there-- of course the union takes the two sets and puts them together and makes one bigger set-- and then look at the final result and see if it's a subset of three, four, five.

So the union of one, two, three, and three, four, five, is one, two, three, four, five. And the question is, is that a subset of three, four, five?

Now, a subset would mean that all the elements in this set are also elements in that set, and that obviously, is not true. So this is a false statement. And that would be your answer.

$\begin{array}{l}\text{16) True or False}\\ \left\{1,2,3\right\}\cup \left\{3,4,5\right\}\subseteq \left\{3,4,5\right\}\\ \left\{1,2,3,4,5\right\}\subseteq \left\{3,4,5\right\}\\ \text{False}\end{array}$

The way to do the problems, when you're asked whether things are true or false, is to work them out, determine from your final results whether that's true or false. All right.

Exercise 7.1: Problem 28 Transcript

We're going to do problem number 28 from the homework from 7.1. I'm going to show you how to work through this problem. There's a variety of different things we're going to be doing. There's six different parts to this problem. This is a pretty good example of one that does a whole bunch of different things.

The sets that we're working with, the universal set is going to be a, b, c, d, e, f. The set A is going to be b, c, and the set B is going to be c, d, e.

$U=\left\{a,b,c,d,e,f\right\}A=\left\{b,c\right\}B\text{=}\left\{c,d,e\right\}$

Part a. We're going to take A union B. So we take the sets A and B, and we put them together to make one big set. That's going to be b, c, d, e. That takes all of the elements from both of the sets and make them one big set.

$\text{a) }A{{\displaystyle \cup }}^{\text{}}B=\left\{b,c,d,e\right\}$

Part b. A intersect B. The intersection of two sets is the overlap. It's the elements that are both in A and in B. And that, of course, is just the element c.

$\text{b) A}{{\displaystyle \cap }}^{\text{}}\text{B=}\left\{c\right\}$

Part c A bar. Those are all the elements that are in the universal set, but not in A. So if you look at the universal set, start going through it, and just include the ones that are not in A. So a, small a of course, lower case is not in A. b and c are in A, but d is not. e, f, and that's A bar.

$\text{c) }\bar{A}=\left\{a,d,e,f\right\}$

Part d. B bar. All of the elements that are in the universal set, but not in B. That would be a, b, and f. So those are all of the things in B bar. $$

$\text{d) }\bar{B}=\left\{a,b,f\right\}$

Part e. A intersect B. Bar over the whole thing. You have to pay particular attention because order of operations becomes important on a problem like this. And this problem, you want to do A intersect B first. And then complement that. It's not a matter of taking A complement and B complement, and then intersection. So order of operations is really important with these. So we're going to find A intersect B. I'll do that down here, I guess. A intersect B, we already figured out was c. I guess I didn't have to do that. And then A intersect B complement was a, b, not c, d, e, and f. So pay particular attention to those, and which order things are done in.

$\text{e) }\overline{A{{\displaystyle \cap }}^{\text{}}B}$
$\overline{A{{\displaystyle \cap }}^{\text{}}B}=\left\{a,b,d,e,f\right\}$
$A{{\displaystyle \cap }}^{\text{}}B=\left\{c\right\}$

Part f. On the other hand, A union B complement.  Once again, when the bar's over the whole thing as opposed to barring each A and B separately then unioning that, you find A union B first. And then complement. So, we already know that A union B is b, c, d, e. So A union B complement is a, b, of course is a, f.  That's all the elements that are in the universal set, but not in A or A union B. OK.

$\text{f) }\overline{A{{\displaystyle \cup }}^{\text{}}B}$
$\overline{A{{\displaystyle \cup }}^{\text{}}B}=\left\{a,f\right\}$

Exercise 7.1: Problem 33 Transcript

This time we're going to do another problem from 7.1, question number 33. This is a Venn diagram question. There's a bunch of Venn diagram questions from this section. Some are two set, some are three set. This is a three set problem.

We're being asked to illustrate the Venn diagram for A union B, intersect, A union C.

$\left(A{{\displaystyle \cup }}^{\text{}}B\right){{\displaystyle \cap }}^{\text{}}\left(A{{\displaystyle \cup }}^{\text{}}C\right)$

It's a three set Venn diagram because you have the set A, B, and the set C. We're going to do this in two pieces. When they become complicated enough, a lot of times it's good to do the set up on one Venn diagram, and then the final result on another one.

Over here, to the left, we're going to do our set up Venn diagram. We'll have three circles, A, B, C. And we'll illustrate the two parts, A union B, and A union C. We'll do those in different colors so that you can see the different contrast. For A union B, we'll do forward slashes in yellow. So A union B, will be these. And then for A union C, we'll do backslashes in orange.

So A union C is everything inside the A circle together with everything inside the C circle. Of course it's this up here also. And this a little bit down here. Now, the intersection of those two sets is the overlap.

If I were doing to union here I would include everything that's shaded in, but since this is an intersection it's just going to be where those two sets overlap. And if we just outline this right here, we can see they're overlapping in here, and they're overlapping in here, and they're overlapping in the A circle. So our final result is going to be this. The A circle with this little knob sticking off to the side. So maybe we'll just do that. We probably don't need to at this point, but just to make things a little bit clearer.

A union B, intersect, A union C, it's going to be, there's your A circle, there's your B circle, there's your C circle. And the final result will be everything on the inside of the A circle, and this is a little bit right here. So that is your answer. OK.

Exercise 7.1: Problem 52 Transcript

We're going to look at question number 52 from section 7.1. This is a translation question. We have a bunch of sets. The universal set is all college students. The set M is male college students. Set S is smokers—smoking college students. College students who smoke would be better, but we'll just go with that. And F is freshmen&mdsah;college freshmen, of course.

U=all college students
M=male college students
S=smoking college students
F=freshman

$F\cup S\cup M$

And we're supposed to take the statement F union S union M and translate that using these statements. So you have to be a little bit careful here. There's a temptation to want to say freshmen or smokers or males.

But that's a little ambiguous. We want to be clear about what's going on. We want to make sure that we're being clear that we want to include college students that have not just all those properties together, but any college student that has any of those properties or any combination of those properties. OK?

So we're going to go with the set of all students who are freshmen or male or smokers or any combination of these. So it could be freshmen smokers, male smokers, freshman male smokers, or any combination of those. OK.