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I require some guidance with the following question:

Consider the following subsets of all integers. $$\begin{align*} A&=\{2n+1\mid n\text{ is an element of all integers}\}\\ B&=\{3n\mid n\text{ is an element of all integers}\}\\ C&=\{3n+2\mid n\text{ is an element of all integers}\} \end{align*}$$ Find each of the following sets, and express it in set-builder notation.

  1. $A-B$.
  2. $B\cap C$.
  3. $C\cap B^c$
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    Please don't yell. (All caps are interpreted as yelling).2011-03-31
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    This is the first time I see that terminology "set builder notation".2011-03-31
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    @Adrian: Unfortunately common in (IMHO bad) books (of which there are far too many).2011-03-31
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    @Ryan P: It's called the "complement of $B$", not the "inverse of $B$".2011-03-31
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    Yes, I am sorry, I meant the complement of B.2011-03-31
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    Yelling has been fixed. :)2011-03-31

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Can you describe in words what $A-B$, $B\cap C$, and $C\cap(B^c)$ are?

For example, for an integer to be in $B$, it must be a multiple of $3$. To be in $C$, it must be an even number plus $2$ (that is, it must be an even number). So to be in $B\cap C$, it must be both even and a multiple of $3$. Can you describe what numbers are both even and multiples of $3$? If so, then you can put that description into the "set-builder notation".

So, start by figuring out what is in each of the three sets (with words). We can then go on from there.

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    @Arturo Magidin: Well, for the first question 'a', in words I see it as "All elements of sort Two times N plus one, which are not multiples of 3".2011-03-31
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    @Ryan P: Can you say "of the sort two times N plus 1" with a single word, commonly used to describe certain kinds of integers?2011-03-31
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    I guess it could be said that it is 'positive'. As we are adding 1 to each possible subset of A.2011-03-31
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    @Ryan: No, it cannot be said "positive". Take $N=-50$. How much is "two times N plus one"? And no, we are not adding 1 to each possible subset of A, we are adding one to the result of multiplying an integer by 2.2011-03-31
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    Hmm, well in any case 2n+1 would result in an odd integer.2011-03-31
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    @Ryan Yes. And when is an odd integer *not* a multiple of $3$? (HINT: what happens if you divide an odd integer by $6$? What happens if you divide a multiple of $3$ by $6$?)2011-03-31
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    An odd integer is not a multiple of 3 if you're looking at the result set formed from A-B. Multiples of 3 can be found in the intersection of both sets or just B.2011-03-31
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    @Ryan P: you missed the point. You are trying to find a way of describing "odd and not a multiple of $3$" so you can describe $A-B$ **without** having to reference $A$ or $B$ (i.e., in "set-builder notation"). You have now figured out exactly *who* is in $A-B$, now you need to find a way of describing them *without* having to say "they are in $A$ but not in $B$". That is: you already figured exactly who lives at this address ($A-B$). Now you are trying to find a way of describing them so you can recognize them if you meet them "on the street".2011-03-31
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    Could it be expressed as simply as this?: {n|n is an odd integer, n cannot be a multiple of 3}2011-03-31
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    OR: {2n+1|n is an element of all integers, n is not a multiple of 3}2011-03-31
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    @Ryan P: I don't know what your book/teacher/professor/grader considers suitable "set builder notation", but that would be one way. But it does not match well with the ways in which $A$, $B$, and $C$ are described, does it? Shouldn't you be trying to come up with something like {`some expression in n` | n is an element of all integer}?2011-03-31
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    No, it is not the way I want it to look, but I cannot figure out how to do it without words...2011-03-31
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    @Ryan P: I repeat: if you take an odd number, and you divide it by $6$, what can happen (as in, what remainders can you get)? What if you take a *multiple of* $3$ and you divide it by $6$?2011-03-31
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    If you take an odd number (say 13) and divide it by 6, you get a decimal number/fraction (2.166 repeating), if you take a multiple of 3 (say 12) and divide it by 6, you get a whole number or 2.2011-03-31
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    @Ryan P: See where I said **remainders** ? If you divide $13$ by $6$, you get a remainder of... If you divide $9$ by $6$, you get a remainder of... If you divide an odd number by $6$, what remainders can you get? And what remainders do you get when that odd number is a multiple of $3$? When it is *not* a multiple of $3$?2011-03-31
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    You get irrational numbers as remainders if you divide an odd number by 6.2011-03-31
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    @Ryan P. No, you don't. First, you cannot get irrational numbers if you divide integers by integers. Second, I said **remainders.** http://en.wikipedia.org/wiki/Remainder You never did division with remainder in grade school?2011-03-31
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    @Ryan P: I have to go, so I can't afford the time to keep pulling teeth: If you divide an integer by 6, the remainders can be $0$, $1$, $2$, $3$, $4$, of $5$. If you divide an *odd* integer by $6$, the remainder will have to be either $1$, $3$, or $5$. But if the number leaves a remainder of $3$ when divided by $6$, then it can be written as $6k+3 = 3(2k+1)$, so it is a multiple of $3$. So an odd number that is *not* a multiple of $6$ will necessarily leave a remainder of either $1$ or $5$, so it can be written as a multiple of $6$ plus or minus $1$, $6k\pm 1$.2011-03-31