# How to define the following sets, given the provided formulas?

I have this problem: Define the following sets: $$a) A=\left\{x\in \mathbb{Z} \mid \frac{6x}{2x+1} \in \mathbb{Z}\right\},$$

$$a) B \cap \mathbb{N}, \text{where } B = \left\{ \frac{105}{2}, \frac{106}{3}, \frac{107}{4}, \cdots, \frac{n+104}{n+1}, \cdots \right\}, n \in \mathbb{N}$$

I don't really understand what "define" means in this context, they, especially the first one seem well defined. If I write the second one as $B=\left\{x=\frac{n + 104}{n + 1} \in \mathbb{N} | n \in \mathbb{N} \right\}$, it's pretty much the same thing. Maybe "defining" means listing all the elements, if there's a finite number of them, but I'm not sure, and that would require a proof. How would you interpret it?

Edit: People have suggested that both of these sets have a finite number of elements, so i'm looking for a thought process to list the elements of these sets respectively.

• Yes, they are defined as they are. Doesn't the problem have a second part asking to do something more substantial with these sets? – Mirko Dec 2 '15 at 22:55
• The word "define" does strike me as strange in this context. But it is true that both of those sets are finite, so your "list" interpretation may be what's intended. – Ravi Fernando Dec 2 '15 at 22:56
• Hmm. The first is well defined but it isn't "well described" without actively thinking about it, I don't know if it has any elements, finite number of elements, what elements are or are not in it. So, I'm guessing, "dfining" it means providing a clearer definition. But you are right. It is currently "defined" – fleablood Dec 2 '15 at 22:58

If $\frac {6x}{2x + 1} = k \in \mathbb Z; x \in \mathbb Z$ then either $x = 0$ or $6 = k*(2 + 1/x)$ so $2 + 1/x \in \mathbb Z$ so $x = \pm 1; k = 2, 6$. So A = {0, 1, -1}. (It'd be hard to argue that that isn't a clearer definition.)
I'll point out that in your post $B\ne\left\{x=\frac{n + 104}{n + 1} \in \mathbb{N} | n \in \mathbb{N} \right\}$ but $C = B \cap \mathbb N =\left\{x=\frac{n + 104}{n + 1} \in \mathbb{N} | n \in \mathbb{N} \right\}$.
$\frac {n + 104}{n + 1} = 1 + \frac {103} {n+1}$. As 103 is prime, n =0 or n= 102. Well, it depends on whether you define 0 to be a natural number but C = {104, 2} if 0 is natural, or {2} if 0 is not.