Proving that the inverse of 2 is not an integer

Question

I have been trying to prove that the inverse of $2$ does not exist in integers, but I did not succeed.

I tried to assume negatively that is an integer, but I have refrained from moving on... I am not sure if I can divide because we live in the integers, etc...

I would be glad to have your help, thanks.

Answer 1

Let $k \in \mathbb{Z}$ s.t. $2k=1$ then, since $k$ is an integer, we have (integers are the free group over the element $1$) $k=\pm(1+1+...+1)$, where the 1 appears k times and $k\neq 0$. But $1=2k=k+k= \pm(1+1+1+1...+1)=1$. Now since there are at least two $1'$s involved in the second to last equation we get $1+1+...+1=0$ which is a contradiction to the assumption

Answer 2

It is hard to answer that question without knowing which properties of $\mathbb Z$ you can use. But a possible answer is: if $n\in\mathbb Z$, then $2n$ is even. Therefore, it cannot be $1$, since $1$ is odd.

Answer 3

I'll try to use the least facts as possible.

If $a$ is a positive integer, $2a=(1+1)a=a+a>a\ge 1$

If $a\le 0$ then $2a\le 0$.

So $2a\neq 1$.