Prove: If $a\in\mathbb Z$ and $|a| > 1$, then $1/a \notin \mathbb Z$.

Question

Prove: If $a$ is an integer and $|a| > 1$, then $1/a$ is not an integer.

Hi, I need help proving this either by contradiction or contrapositive. I'm not sure where to begin

Answer 1

Suppose $\frac{1}{a} = b$ an integer. I multiply both sides of this equation by a, giving $ab = 1.$ This implies $a = b = \pm 1$ since the only integers who multiply out to $1$ are $\pm 1$ (since they are the units in the ring of integers) . But $|a|>1$. This is a contradiction.

Answer 2

If $|a|>1$ then $$\quad0<\frac{1}{|a|}\text{ and } \frac{1}{|a|}<1.$$ So, $\frac{1}{|a|}$ is a number between $0$ and $1$. Since there is no integer between $0$ and $1$, we conclude that $\frac{1}{|a|}\notin\mathbb Z$.

EDIT: Proof by contradiction, as you want:

Suppose $\frac{1}{|a|}$ is an integer. Since $|a|>1$, we conclude that $$\quad0<\frac{1}{|a|}\text{ and } \frac{1}{|a|}<1.$$ So, $\frac{1}{|a|}$ is an integer between $0$ and $1$. But it's a contradiction, because there is no integer between $0$ and $1$. It follows that $\frac{1}{|a|}\notin\mathbb Z$ and thus $\frac{1}{a}\notin\mathbb Z$ (because $\frac{1}{a}$ is an integer if and only if $\frac{1}{|a|}$ is an integer)

EDIT 2: (to make my answer valid as pointed out by @TomOldfield)

There is no integer between 0 and 1.
Proof: Suppose that there exists a number $p\in\mathbb Z\cap (0,1)$. Let $(x_n)$ be a sequence defined by $x_n=p^n$. Notice that $x_n\in\mathbb Z\cap (0,1)$ for all $n\in \mathbb N$. Since $(x_n)$ is strictly decreasing, it follows that $\mathbb Z\cap (0,1)$ does not have a least element. But it contradicts the well-ordering principle. So, $\mathbb Z\cap (0,1)=\varnothing$.