For $a,n \in \mathbb{Z}$, where $n>a>0$, prove that $\nexists x \in \mathbb{Z}$ s.t. $nx=a$ Suppose $a,n \in \mathbb{Z}$, and $n>a>0$. How do I prove that $\nexists x \in \mathbb{Z}$ s.t. $nx = a$ ? I'm really not sure where to start on this one. I'd be happy if someone could give me a hint.
Edit: I've solved this by contradiction, but I will not be 'accepting' an answer from below because I did not use any one of them in a significant way to solve the problem.
 A: Assume there is such an $x$. Since $nx = a$, then $0 < nx$ and $nx < n$. Can you now prove that $0 < n$ and $n < 1$? And can you prove that that is a contradiction?
Edit: changed $p < q < r$ statements to $p < q$ and $q < r$ statements. Because hypothetically, I forgot what $<$ means.
Edit 2: Electric Boogaloo
So one way of defining $\mathbb{Z}$ is that it is a commutative ring with a subset $\mathbb{N}$ such that:


*

*Non-trivialty: $\mathbb{N} \ne \emptyset$.

*Closure: for all $a,b \in \mathbb{N}$, $ab \in \mathbb{N}$.

*Trichotomy: For all $x \in \mathbb{Z}$, exactly one of the following is true: $x \in \mathbb{N}$, $x = 0$, $-x \in \mathbb{N}$.

*Well-ordering principle: If $S \subseteq \mathbb{N}$ and is non-empty, there is an $x \in S$ such that for all $y \in S$, $x \le y$.


Now we can define $<$. We say $a < b$ if $b - a \in \mathbb{N}$.
So if you want to prove your statement from the ground up, you should prove:


*

*$a < b$ and $b < c \implies a < c$ (after this point you can write $a < b < c$)

*$ab > 0$ and $b > 0 \implies a > 0$ (useful for part 3)

*$ac < bc$ and $c > 0 \implies a < b$ (division not allowed)

*$0 < 1$, or equivalently $1 \in \mathbb{N}$ (master troll)

*$\not\exists x \in \mathbb{Z} \ 0 < x < 1$ (this one uses well-ordering)

A: Hint: Note that $|nx| = |n| |x|$, and consider the cases $x = 0$ and $|x| \ge1$ separately. Start by noting that $a < n$, so how are $|n| |x|$ and $a$ related?
