# No Divisors Between an Integer and Twice it

1. Nov 4, 2016

### Bashyboy

1. The problem statement, all variables and given/known data
Suppose that $x \in \mathbb{N}$ is such that $n < x < 2n$, where $n$ is another natural number. I want to conclude that it is impossible for $x$ to divide $2n$. Also, is there a way to naturally generalize this to all integers?
2. Relevant equations

3. The attempt at a solution
Since $x \in (n,2n)$< then $x = n + \ell$ for some natural number $\ell \in (0,n)$. Now, suppose the contrary, that $x$ divides $2n$. Then $2n = k(n + \ell)$ for some $k \in \mathbb{Z}$. Immediately we can say that $k$ must be a positive integer. Rearranging the equation gives

$n(2-k) = k(n+\ell)$

If $k$ is either $1$ or $2$, then we get $n = \ell$ or $2 \ell = 0$, both of which are contradictions. If $k > 2$, then the LHS will be negative but the RHS will be positive, a contradiction.

How do I make this a little more rigorous?

Last edited: Nov 4, 2016
2. Nov 4, 2016

### Staff: Mentor

$2n = k(n + \ell) \nRightarrow n(2-k) = k(n+\ell)$ and I don't see why you may assume $k \in \{1,2\}$.
However, I don't think $\ell$ is needed at all. What do you get by simply assuming $xk=2n$?

3. Nov 4, 2016

### Bashyboy

Whoops! Rearranging the equation should actually give $n(2-k) = k \ell$. And I am not not assuming that $k$ always takes the values $1$ and $2$: I was just considering special cases.

To answer your question, I don't know what $xk=2n$ implies.

4. Nov 4, 2016

### Staff: Mentor

I meant, if you substituted $x$ by $\frac{2n}{k}$ in your assumption $n<x<2n$, you could divide the whole thing by $n$ and see where it get's you to.