If an integer $n$ has the form $3k+1$, then $n$ does not have the form $9l+5$

Question

I am trying to do the following proof by contradiction and need verification if the proof is correct:

If an integer $n$ has the form $3k+1$, then $n$ does not have the form $9l+5$

This is my proof:

Let $n$ be an integer of the form $3k+1$ where $k$ is some integer.

We want to prove that $n$ does not have the form $9l+5$ where $l$ is some integer.

Suppose, on the contrary, $n$ does have the form $9l+5$

It follows that :

$3k +1 = 9l +5$

$3k -9l = 5-1$

$3(k-3l) = 4$

$k - 3l = \frac{4}{3}$

Since $k -3l$ is an integer by the closure properties of integers $k-3l$ can't equal $\frac{4}{3}$, which is a rational number. Therefore the assumption that $n$ does have the form $9l+5$ is false and the original statement holds.

I am new to proofs. Is this proof correct? Is it stylistically correct as well?

Answer 1

Yes, this works as a proof of it. Nicely done!

I would use that only $9l, 9l+3$ and $9l+6$ are multiples of $3$ for $l\in\Bbb Z$, so only $9l+1, 9l+4, 9l+7$ are of the form $3k+1$.

Answer 2

Apart from a typo in the second line ($4k$ should be $3k$), the proof is good.

You don't need to use $4/3$: the third line implies $4$ is a multiple of $3$, which it isn't.