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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?

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Do you include $0\in\mathbb{N}$? (Some people do) –  Zev Chonoles Oct 9 '11 at 18:26
    
Nope, N is without 0, and N0 is with 0. –  jco Oct 9 '11 at 18:30
    
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? –  Zev Chonoles Oct 9 '11 at 18:32
    
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. –  jco Oct 9 '11 at 18:39
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2 Answers 2

up vote 1 down vote accepted

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 $B$.

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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})? –  jco Oct 9 '11 at 18:48
    
@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 $3=2\cdot 1 + 1$. –  Zev Chonoles Oct 9 '11 at 18:58
    
Oh, sorry, I wasn't thinking... –  jco Oct 9 '11 at 20:14
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$C$ 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...

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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... –  user864940 Oct 9 '11 at 18:40
    
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 :) –  jco Oct 9 '11 at 18:50
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