Find logical errors in the proof

Question

I'm finding it really hard to find the logical errors in this proof. I'm still new at this and trying to learn it.

Claim: for all $n$ is in set $\mathbb{N}$(Natural numbers), if $2n+1$ is a multiple of $3$ then, $n^2+1$ is a multiple of $3$.

Proof: by contrapositive. Assume that $2n+1$ is not a multiple of $3$.

It follows that

• If $n = 3k+1$ for $k \in \mathbb{N}$ then $n^2+1 = 9k^2+6k+2$ is not a multiple of $3$.
• If $n = 3k+2$ for $k \in \mathbb{N}$ then $n^2+1 = 9k^2+12k+5$ is not a multiple of $3$.
• If $n = 3k+3$ for $k \in \mathbb{N}$ then $n^2+1 = 9k^2+18k+10$ is not a multiple of $3$.

Where are the logical errors in the proof? I don't understand.

Any help would be appreciated, thank you!

Answer 1

The contrapositive of the statement "If $A$, then $B$" is the statement "If not $B$, then not $A$".

For the statement

If $2n+1$ is a multiple of $3$, then $n^2+1$ is a multiple of $3$

the contrapositive is

If $n^2+1$ is not a multiple of $3$, then $2n+1$ is not a multiple of $3$.

However, the proof assumed that $2n+1$ is not a multiple of $3$ and then proved that $n^2+1$ is not a multiple of $3$, i.e. the proof proved the statement

If $2n+1$ is not a multiple of $3$, then $n^2+1$ is not a multiple of $3$.

This is the inverse, not the contrapositive of the original statement. Proving the inverse of a statement is not equivalent to proving the original statement. Hence, that proof doesn't prove what it is supposed to prove.