Contradiction proof question

Question

We had to prove this statement on a test today, and I kind of got lost when writing it out. I started by saying something along the lines of:

"Suppose 3 does divide 3n+2 for all integers n. Then, 3n+2=3a for some a in Z."

After that, I decided to do an odd case and an even case, but I couldn't find a contradiction (or just didn't register if one was there). Was I right to proceed by cases here?

Answer 1

You've already gotten several answers, addressing the (rather obvious) statement you were asked to show, but it seems no one has actually answered the question you've actually asked.

No! It makes no sense to do an odd and an even case.

If it has to be done by contradiction:

Assume that $3n+2$ is divisble by 3, that means a integer $k$ exists such that $3n+2=3k$, we can rewrite that to $2=3(k-n)$. $n$ and $k$ were both integers, so the difference is, and we can see that $3$ divides the right hand side, so we get that $3$ divides $2$, but that's absurd.

Answer 2

This is obvious. We have that $$3n+2\equiv 2\pmod 3.$$

Answer 3

Why use contradiction here?

Simply consider the fact that all multiples of three differ by a multiple of three. It is obvious that $3n$ is divisible by $3$ and that $2$ is not a multiple of $3$, and so $3n + 2$ cannot be divisible by $3$. This may not seem like a ‘proper’ mathematical proof, but it is perfectly rigorous.

Answer 4

$$n+1=\frac{3n+3}3>\frac{3n+2}3>\frac{3n}n=n$$ and there is no integer between $n$ and $n+1$.

Answer 5

I’m not sure exactly what assumptions you’re allowed to make but I’ll assume some basic truths about odd and even numbers. If you want to use cases...

Suppose $3 | 3n + 2$ for some $n \in \mathbb{N}$. Then $3n + 2 = 3k, k \in \mathbb{Z}$.

If $n$ is even, then $3n + 2$ is even because it is the sum of two even numbers. But $3k$ is not even, so one side of the equation is even and the other is not. Contradiction.

If $n$ is odd, then $3n + 2 = 2m + 1$ for some $m \in \mathbb{Z}$. So $2n + 1 = 2m$. One side of this equation is even and the other is not. Contradiction.

By the way, the first statement I wrote here is that there exists an $n$ such that $3n + 2$ is divisible by $3$. This is the opposite proposition to ‘$3n + 2$ is not divisible by $3$ for all $n$’ which is what you aim to disprove. What you wrote is that ‘$3n + 2$ is divisible by $3$ for all $n$’, which is not quite the logical negation. If it is not true that ‘for all $A$, $B$ is true’, then intuitively it must be true that ‘there exists an $A$ such that $B$ is not true’.