I'm solving some problems for practice, and I've come across a something I don't quite understand... So here's the deal:

$A = \{x \in \mathbb{N}: -1 \leq x < 2\}$

$B = \{x \in \mathbb{Z}: -10 < x \leq 0\}$

$C = \{n \in \mathbb{Z}: n = 2k + 1, k \in \mathbb{Z}\}$

a) $C \setminus(A\cap B)$

b) $(B\cup C)\setminus A$

How am I supposed to solve this when $C$ has infinite members?

  • $\begingroup$ Do you include $0\in\mathbb{N}$? (Some people do) $\endgroup$ – Zev Chonoles Oct 9 '11 at 18:26
  • $\begingroup$ Nope, N is without 0, and N0 is with 0. $\endgroup$ – jcora Oct 9 '11 at 18:30
  • $\begingroup$ Also, I'm not sure what your question is: do you know which numbers are in the sets $C\setminus(A\cap B)$ and $(B\cup C)\setminus A$, and simply you're not sure how to describe the answers to a) and b) in mathematical notation, or is the fact that $C$ is infinite posing a problem for you in solving the problem? $\endgroup$ – Zev Chonoles Oct 9 '11 at 18:32
  • $\begingroup$ I do know which numbers are supposed to be there, but I can't simply write {..., bla, bla, bla, ...}, so yeah writing that down is the problem. $\endgroup$ – jcora Oct 9 '11 at 18:39

Let's come up with some more explicit descriptions of these sets:

$$A=\{1\}$$ $$B=\{-9,-8,-7,-6,-5,-4,-3,-2,-1,0\}$$ $$C=\{\ldots,-3,-1,1,3,5\ldots\}=\{\text{odd numbers}\}$$

First, do you understand why the above statements are true? If not I can explain in more detail.

Can you describe $A\cap B$? Then $C\setminus (A\cap B)$ consists of every element of $C$ that isn't in $A\cap B$ (hint: there is a very easy description of this set using $C$).

The second part is a bit trickier. $B\cup C$ consists of every element of either $B$ or $C$, so it is the odd numbers, together with $-8,-6,-4,-2,0$ (the odd numbers $-9,-7,-5,-3,-1$ are in both $B$ and $C$, but they don't get counted twice or anything). Then, $(B\cup C)\setminus A$ consists of every element of $B\cup C$, just throwing away anything in $A$.

EDIT: We can describe the set explicitly as $$(B\cup C)\setminus A=\{n\in\mathbb{Z}: n=2k+1,k\in(\mathbb{Z}\setminus\{0\})\}\cup\{-8,-6,-4,-2,0\}$$ The first set is just the odd numbers, except for 1 (which we threw out because it was in $A$), and then the second set is the even numbers we have to add in from $B$.

  • $\begingroup$ Thanks, really, I do understand absolutely all the math here, I can list the solutions in my head, the problem was to write them down. Well, you solved both anyway, so thank you. Also, shouldn't it be (Z∖{1}), not (Z∖{0})? $\endgroup$ – jcora Oct 9 '11 at 18:48
  • $\begingroup$ @bane: Glad to help! It is $\mathbb{Z}\setminus\{0\}$, because $1=2\cdot0+1$, while if we used $\mathbb{Z}\setminus\{1\}$, that would get rid of $3=2\cdot 1 + 1$. $\endgroup$ – Zev Chonoles Oct 9 '11 at 18:58
  • $\begingroup$ Oh, sorry, I wasn't thinking... $\endgroup$ – jcora Oct 9 '11 at 20:14

$C$ having infinitely many members should not effect your solution. Assuming that $0\in \mathbb{N}$ it's easy to see that $C$ is simply the set of odd naturals.

You can explicitly write $A = \{-1, 0\}$ and $B = \{-9, -8, ..., 0\}$ if it makes it easier to think about.

Then, essentially, clause a translates to "What is the set of all odd naturals ($C$), excluding (\) those who are both between -1 and 0 ($A$) and ($\cap$) between -9 and 0?".

It's easy to calculate this to be $C$. A shorter version would be to notice that, since $A\cap B \subset A$ and $A\cap C = \emptyset$ you get that result immediately...

This kind of lingual breakdown shouldn't be a part of your mathematical handling after some exercise, but it roughly equates to how you should mentally handle such questions...

  • $\begingroup$ Sorry, didn't notice that $C$ goes over all the odds and not just the natural ones. Though what I said is accurate in essence and could be used as a reference... $\endgroup$ – user864940 Oct 9 '11 at 18:40
  • $\begingroup$ Hey, thanks for the answer, but I do understand the problem here, I just didn't know how to write it down... I'm sorry, I should have been more accurate. I'll give you a vote up, when I can though :) $\endgroup$ – jcora Oct 9 '11 at 18:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.