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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?

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    $\begingroup$ $a+b$ is even implies $a$ and $b$ are both even or both odd. And any two odd integers or any two even integers cannot be consecutive. $\endgroup$ – Bijesh K.S Jan 26 '18 at 14:03
  • $\begingroup$ if { (a + b is not odd) and (a and b are consecutive) } then { a + (a + 1) is not odd } then { 2a + 1 is not odd } then { 2a is odd } which it is not. Therefore { (a + b is odd) or (a and b are not consecutive) } $\endgroup$ – Wouter Jan 26 '18 at 21:53
  • $\begingroup$ If you want to do more work than necessary, don't use contrapositive but prove (1) even + odd = odd and (2) if a, b are consecutive then one of them is even and the other is odd. $\endgroup$ – CompuChip Jan 26 '18 at 22:27
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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 $1$.

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  • $\begingroup$ ya, I know, just want to use different methods to prove this statement as practice $\endgroup$ – skyzhuzhu Jan 26 '18 at 13:59
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$a+b$ 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.

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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.

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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 $i \neq 0$).

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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

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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.

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$a+b=2n \implies a-b=2n-2b=2m$ where $m\in \mathbb Z \implies a-b\not =\pm1 $

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  • $\begingroup$ It should be $a+b=2n$ at the beginning, no? (of course, $a+b$ is even if and only if $a-b$ too). $\endgroup$ – Taladris Jan 27 '18 at 3:30
  • $\begingroup$ @Taladris Yes. That's true: they differ by $2b$... $\endgroup$ – user403337 Jan 27 '18 at 6:21

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