Proof by contradiction of rational numbers

Question

I am trying to proof by contradiction the following statement: "Every integer is rational", so far I have done the following:

I can convert the expression to "If $n\in \mathbb{Z}$ then $n\in \mathbb{Q}$" so that I can have it the form $p\rightarrow q$.
For what I know in a contradition proof I should negate q and if I assume that p is true then I should end up with a contradiction in this term.
So I can say that n is an irrational or that $n\in \mathbb{I}$. For this to hold I should assume that n cannot be represented as the ratio of two numbers p and q, but this is a contradiction with
$n\in \mathbb{Z}$ because if n is integer there should be result of a ratio of two integer numbers.

I would like to know is this proof is correct or am I missing something?

Answer 1

Here's how you do it by contradiction:

Let $n \in \Bbb Z$. Suppose $n\not\in \Bbb Q$. That means $n$ can't be written as a quotient of two integers, but $n=n/1$. A contradiction.