Here are some more problems for whenever you want to tackle them.
Problem 3: We say that two sets are disjoint if they have no elements in common (or more than two sets can all be mutually disjoint if each set is disjoint with every other). What is a way of saying that sets A and B are disjoint by using the language of unions and/or intersections?
Problem 4*: The difference set, A - B, can be defined as the set of all elements in A that are not also in B. Is it possible to define the difference set in terms of unions and/or intersections of A and B?
Problem 5: The symmetric difference, A Δ B, is defined as everything in A not also in B plus everything in B not also in A. In other words, A Δ B = (A - B) ∪ (B - A). What is another way that we could define the symmetric difference using the union, intersection, and difference operators? (I'm looking for a definition that would probably come to mind directly from a Venn diagram representation of a symmetric difference. By the way, Venn diagrams are sometimes helpful to use when thinking about sets, but I think they are more helpful when you produce them on your own.)
Problem 6: Let A, B, and C be sets such that A and B are both subsets of C. Let A' denote the set of all elements of C not in A, let B' denote the set of all elements of C not in B, and so on for all subsets of C. Show that (A ∩ B)' = A' ∪ B' and that (A ∪ B)' = A' ∩ B'.
*I'm not really sure about the answer. It wasn't a question in the book or anything but just something that I wondered. I think that I know the answer. So you think about it, and then think about it again after problem 6. And then we can talk.
(Prob 3) Sets A and B are disjoint if A ∩ B is {}.
ReplyDelete(Prob 4) Do you mean is it possible to define the difference set A - B ONLY in terms of intersections and unions (ie, without using the difference operator)? If so, my first thought is to define it as A - (A ∩ B). Also, is it ok to use the negation sign? If so, we could say A - B = A ∩ ~B, I think.
that's " { } " on my answer to #3.
ReplyDeleteI have a lot to say about #5.
(Prob 5)
ReplyDeleteYou could also define the symmetric difference as (A ∪ B) - (A ∩ B)
Question: a sentence in problem 5 implies that "union" and "intersection" are operators. Are they really operators in the same sense that addition and subtraction are operators in grade school arithmetic? Then again, it seems like "union" means much the same thing as "add"...but not quite, partly because of the property of sets we were discussing in the previous problem. (More broadly, maybe it's because numbers are a set but not all sets are numbers, so the way that numbers add together is really just a property specific to that set -- the set of numbers -- and therefore saying 2+3=5 is really just expressing this property of performing a union on two members of this set? No idea if this is remotely valid.)
What is an operator, exactly? And another question -- if we do think of unions and intersections as operators, what is the order of operations? In problem 5, for example, is (A ∪ B) - (A ∩ B) the same as A ∪ B - A ∩ B ?
Teaching observation: I agree about the Venn, but that may be because I've seen a lot of Venn diagrams and also because I personally tend towards visual learning. Translating something to a Venn diagram feels natural to me, but it may not be the case for other people...including people who are more skilled at logic and abstract thinking than I am. To be honest I'm not sure if I believe the “different learning types” thing is bullshit or not (visual vs auditory vs haptic or whatever), but I do believe that you bring your own biases into teaching and it’s worth always asking yourself whether an approach that seems straightforward to you is actually very difficult for some of your students.
(Prob 6)
A’ = C – A, meaning “everything that’s in C but not in A”. Likewise, B’ = C – B, meaning “everything that’s in C but not in B”. So, the union of A’ and B’ includes things that are in B but not in A, as well as things that are in A but not in B. The only things in C that are excluded from A’ ∪ B’ are things that are in both A and B… in other words, (A ∩ B). So, (A ∩ B)’ must be the same as A’ ∪ B’
In the same vein, (A ∪ B)’ means “everything in C that is neither in A nor B”. This is the same as saying A' ∩ B'
Problem 3:
ReplyDeleteyep.
Problem 4:
Let's hold off on this for a sec.
Problem 5:
yes to your answer.
did you like how i casually tossed in the word 'operators' or did you find it reckless? i debated doing it. regardless, it's out there now so let's talk about it. i think that you're working too hard to mentally tie the idea of a union to the idea of adding two numbers. they're certainly similar (and earlier, i did use the word 'plus' to talk about a union), but they are also very different. They act on different types of objects. Numbers are not sets. They can belong to sets, but it would be meaningless to add two sets together even if each did only contain a single number.
By 'operator,' I'm really only talking about something that takes two objects and produces a third. So addition (or subtraction or multiplication, etc) takes two numbers and puts out a third based on the rules of arithmetic that we are all familiar with. Unions and intersections (and set differences, etc) take two sets and spit out another based on the rules we are establishing here. In linear algebra (for those familiar), we have the cross product (and vector addition) that takes two vectors and sends them to another. But it wouldn't make sense to take the cross product of two sets or to take the union of two vectors because that would be an attempt at using an operation that has no meaning for the objects that we're working with.
Just to open this up a little more, we can have operators that act on two different types of objects at once. For instance, in linear algebra, we have scalar multiplication, where a vector is multiplied by a number and results in another vector. We also have dot multiplication, where two vectors are combined to make a number. Familiarity with what these objects and operators are isn't important - I'm just trying to demonstrate that operators can be many things, but they are defined only for the objects they are supposed to act on.
Note: We're actually talking about binary operations here. It is conceivable (hopefully) that we could define an operator that acts on 3 or 4 or more objects. Similarly, we could define an operator that acts on one object - maybe this is a familiar concept?
I don't know what the order of operations are or if there are any firmly established for set operators. Order of operations are just rules (that I think we spend way too much time on in elementary school or wherever) that allow us to short-hand equations. We'll be using parentheses and asking for clarification if one of us writes something that could be interpreted multiple ways like in your example.
Let's keep the discussion about operators going because I'm pretty sure I didn't explain this in the best way. So let me know what you're thinking about them now.
As for the venn diagram stuff - are you suggesting that I include a few?
Problem 6:
Comment coming soon!
Problem 6:
ReplyDeleteOk, you're pretty much there, and I like the way you've reasoned out through description that the first two must be the same. However, your initial reasoning led to some confusion (at least for me) - "the union of A’ and B’ includes things that are in B but not in A, as well as things that are in A but not in B." It does include these things, you're right, but it also has stuff in neither (which you kind of addressed in the next sentence). So it's clear to me that you understand why these things are equal conceptually, but let me show you a method of showing it in a more step-by-step way:
(A ∩ B)' = A' ∪ B'
Let x be an element of the set A' ∪ B'. This implies that x is either in A' or B' or both. If x is in A', then x is not in A (in the overall context of C). If x is in B', then x is not in B. In either case, x is not in the intersection of A and B. Therefore, x is in (A ∩ B)'. Therefore, A' ∪ B' must be a subset of (A ∩ B)' because we have shown that an element of the first set is always an element of the second and not the other way around.
In the other direction, we can assume that x is an element of (A ∩ B)' which means that x is not an element of A ∩ B. There are three cases in which x is not in the intersection of A and B - x is not in A, x is not in B, or x is not in either. Thus, either x is in A', x is in B', or both. So x is in the union of A' and B'. Hence, (A ∩ B)' is a subset of A' ∪ B'. Because each is a subset of the other, then they must be equal.
I know that writing things out in this way kills a little bit of the joy of the problem when it is intuitive to you in another way, but sometimes it is necessary to keep everything straight. Would you like to try the second part of the problem using this technique?
Also, notice where I used the word "not" repeatedly to mean "not in A (or B or whatever), but still in C." This is helpful where there is a superset overriding the sets you are dealing with that is understood - a universe it's sometimes called (I think). A' is called the complement of A in C or alternatively can be called not A when C is understood. This relates to your answer to problem 4.
First, I'll try the second part of the problem using the method you used:
ReplyDeleteLet x be an element of the set A'∩ B'. Then, x is not in A and x is also not in B. Since x is not in the union of A and B, x is in (A ∪ B)' by definition. Therefore, A'∩ B' must be a subset of (A ∪ B)'. In the other direction, let x be an element of (A ∪ B)'. Since A is a part of A ∪ B, x must not be in A and must therefore be in A'.
Likewise, since B is a part of A ∪ B, x must not be in B and must therefore be in B'. So, x is in the intersection of A' and B' and (A ∪ B)' must be a subset of A'∩ B'. These sets are equal because they're subsets of each other.
Second: regarding the joy-killing thing -- no, I don't think it does. While I basically agreed the Lockhart essay, I kept thinking about something after I read it. For me, part of why I like math is that my mind doesn't naturally work in a rigorous and systematic way; it works in an irregular, creative way. I enjoy math because it allows me to focus my thoughts with more rigor and method than I'm normally accustomed to. I agree that creativity is an essential part of problem solving and that math devoid of creativity is empty (which is how I practiced it growing up). However, I think a big part of the enjoyment for me is the fact that its logic constrains thought in a very deterministic structure. So, definitely show and teach method.
ReplyDeleteI agree with you. It's fun to me because it demands creativity, but never lets you get goofy like other disciplines. I'm pretty sure we're all still on the same page with this (you, me, and Paul Lockhart - the big 3), but I wish he was here to add his two cents from time to time about what I'm doing.
ReplyDeleteYour answer to number 6 is very nice. These two equations are called De Morgan's Laws if anyone cares, and they have an alternate form that is very popular with regard to truth statements - replace the complement signs with the "not" sign and replace unions and intersections with ORs and ANDs.
This relates to what you said in your response to problem 4. If we use the idea of "not" or "complement," then yes, we can define the set difference operation with unions and intersections, but otherwise, I think the answer is that we can't. However, we always have to be careful when using the complement because the concept requires that there be some superset that the complement is with respect to. Does that make sense?
Also, how did you feel about what I said about operators? We can shelve that discussion for now if you'd rather, but I'm curious as to whether what I said made any sense at all.
Yes, that all makes sense. And I do remember DeMorgan's Laws in regards to truth statements, though I didn't make the connection here. I also get what you're saying about the complement and "not".
ReplyDeleteAbout the operators -- I think I mostly get it, but I'll have to see it in action before I can say. What's confusing to me is how this fits with the language and concept of functions, because I think of a function as "something that takes one object and returns another object". In contrast, I tend to think of operators as rather inert and uncomplicated component parts of a function (or whatever relation/expression), sort of like how a piston or a valve would relate to the "machine" of the function. A function takes an object and does something to it. But an operator doesn't seem to act with such...agency.
So I understand what you're saying about operators being things that takes two objects and return a third, and that the operators must be defined for the type of object they receive. I also see how an operator can be defined for more than two objects, or just one (would taking a power or a square root of a number be an example of this?). What I'm not as clear about is where this broader, more abstract definition of "operator" is headed and how it fits into my current conceptual understanding of math.
Yes, exactly, taking a power is an operator as are all of the trig functions. We will be visiting functions soon, and I have an example saved for that that will hopefully bring it all together. It seems like you understand it pretty well anyway, but I'll keep hitting the idea where I can.
ReplyDelete