Proof by contrapositive: if $a$ and $b$ are consecutive integers then the sum $a + b$ is odd if $a$ and $b$ are consecutive integers then the sum $a + b$ is odd
Proof by contrapositive
Contrapositive form:
if the sum of $a$ and $b$ is not odd then $a$ and $b$ are not consecutive integers
I am struck here, so if $a + b$ is not odd means $a + b$ are even
$a + b = 2p$, where $p\in\mathbb Z$.
What are the next steps to show $a$ and $b$ are not consecutive?
 A: $a+b$ is not odd, therefore it is even, so $a+b=2k$.
Now, two integers $a,b$ are consequtive if $a-b=\pm 1$. In your case, you have $a-b = a+b-2b = 2k - 2b = 2(k-b)\neq \pm 1$ (because $1$ is not even, $2(k-b)$ is even, so $a,b$ are not consequtive.
A: You know that $a\neq b$ and you can assume, without loss of generality, that $a<b$. Then $b-a=1$ and $a+b=2p$. Therefore$$2b=a+b+b-a=2p+1,$$which is impossible, because $2b$ is even and $2p+1$ is odd.
A: We have $$a+b=p+p$$
Let $i = a - p$ and $j = b-p$. Then we have 
$$
i + p + j + p = p + p
$$
$$
i + j = 0
$$
$$
j = -i
$$
Thus, $a = p + i$ and $b = p - i$ for some integer $i$. It's impossible for $a$ and $b$ to be consecutive since either they are both equal to $p$ (if $i=0$), or $p$ is an integer somewhere between $a$ and $b$ (for $i \neq 0$).
A: A direct proof is so much clearer:
$b=a\pm1$ implies $a+b= 2a\pm1$, which is odd.
But if you must use contrapositive:
Let $b=a+d$. Then $a+b=2a+d$ is even iff $d$ is even. Therefore, $|a-b|=|d|$ is even and so is never $1$.
A: WLOG $a=k$ and $b=k+1$ for some natural $k$.
Hence the sum 
$$a+b=2k+1$$
Since $2k$ is always even $a+b$ is always odd
A: $a+b=2n  \implies a-b=2n-2b=2m$ where $m\in \mathbb Z \implies   a-b\not =\pm1 $
A: If $a + b$ is not odd, then $a + b = 2k$ for some integer $k$.
Then $k = (a + b)/2$.
Then $2$ divides $a + b$.
Then $a + b$ is even.
Then $a$ and $b$ are both odd or both even.
Therefore they aren't consecutive.
If $a$ and $b$ were consecutive one would be even, one would be odd.
