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Is this proof correct?

Theorem: The product of an even integer and an odd integer is even.

Proof: Let $a$ and $b$ be integers. Assume $a$ is even and $b$ is odd, so there exists an integer $p$ so that $a=2p$ and there exists an integer $q$ so that $b=2q+1$. If $a \cdot b$ is even then by definition of even there exists an integer $r$ such that $a \cdot b = 2r$. So we have $a \cdot b = (2p) (2q+1) = 2r$, where $r$ is an integer.

Therefore, $a \cdot b$ is even.

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    $\begingroup$ You're just missing the order. Look, $ab=2k.(2q+1)=2(k(2q+1))=2r$, where $r=k(2q+1)$, so $ab$ is even. $\endgroup$ – Koto Sep 29 '17 at 22:06
  • $\begingroup$ Where you say "by definition of even", technically this permits you to then go on to show that $a\cdot b$ satisfies the definition of an even number, and this be a proof. But in this form it's a little more difficult to follow than if you reorder your proof into a sequence of statements whose truth follows from the preceding statements. If you do that then you have a proof by direct implication which may be more clear to follow. Also the statement "so we have" is perhaps vague. A more precise statement would be that propositions A and B combine to imply proposition C. $\endgroup$ – user334732 Oct 4 '17 at 11:12
  • $\begingroup$ It also helps the reader if you give each proposition its own new line. $\endgroup$ – user334732 Oct 4 '17 at 11:14
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No. You are assuming what you are trying to prove.

Consider this "proof" (the same as yours but I'm replacing the gray text with red text:

Theorem: The product of an even integer and an odd integer is $\color{gray}{\text{even}}$ $\color{red}{\text{odd}}$.

Proof: Let $a$ and $b$ be integers. Assume $a$ is even and $b$ is odd, so there exists an integer $k$ so that $a=2k$ and there exists an integer $q$ so that $b=2q+1$. If $a⋅b$ is $\color{gray}{\text{even}}$ $\color{red}{\text{odd}}$ then by definition of$\color{gray}{\text{even}}$ $\color{red}{\text{odd}}$ there exists an integer $r$ such that $\color{gray}{a*b=2r}$ $\color{red}{a*b=2r+1}$. So we have $a⋅b=(2p)(2q+1)=\color{gray}{2r}$ $\color{red}{2r+1}$, where $r$

is an integer.

Therefore,$ a⋅b$ is $\color{gray}{\text{even}}$ $\color{red}{\text{odd}}$.

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  • $\begingroup$ Good point. Although I also think it depends on the precise meaning of "So we have". If "so we have" means "From which we can directly imply the true statement..." then you couldn't validly make this substitution. $\endgroup$ – user334732 Oct 4 '17 at 11:25
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The basic idea is correct, but it is not presented precisely.

If ab is even then by definition of even there exists an integer r such that ab = 2r. So we have ab=(2p)(2q+1)=2r,where r is an integer. Therefore,ab is even.

Why "if"? What you need is rather the other direction, so: If $ab=2r$ for an integer $r$, then $ab$ is even. We have $ab = 2p(2q+1)= 2(p(2q+1))$. Since $p(2q+1)$ is an integer, the claim follows, setting $r = p(2q+1)$.

As mentioned in a comment, it is not really needed to say $b=2q+1$. It is simpler to work with $b$ directly in this case.

If $ab=2r$ for an integer $r$, then $ab$ is even. We have $ab = 2pb= 2(pb)$. Since $pb$ is an integers, the claim follows, chosing $r = pb$.

An additional point, you use once $p$ and once $k$ for the same thing.

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This proof is not clearly written, it could be interpreted as you proving your statement by assuming it first to be true (however not in the inductive sense. This could be a clearer way:

Let $a=2n$ (ie $a$ is even) and $b=2m+1$ (ie $b$ is odd). Then the product $ab$ is: $$\begin{array}{rcl} ab & = & (2n)(2m+1) \\ & = & 2(n(2m+1)) \end{array} $$ Letting $k=n(2m+1)$, this gives $ab=2k$, which implies $ab$ is even (since $k$ is an integer).

Edit: having seen the updated question, your working is all correct, however it could be made more explicit what your $r$ is.

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  • $\begingroup$ I do not understand the "Edit" Nothing seems to have changed. Your critique still applies. $\endgroup$ – quid Sep 29 '17 at 22:13
  • $\begingroup$ @quid I believe the proof is laid out roughly how mine is (with a bit less clarity), however what its missing now is the step where what $r$ is, is made clear $\endgroup$ – aidangallagher4 Sep 29 '17 at 22:14
  • $\begingroup$ Yes, but "if" and "so" are not used correctly as you say. I will not further press this point, but except for correcting the oversight $p$ for $k$ there just is no update to the substance to begin with, it's just formatting. Your original answer was a good one, I feel it is slightly degraded by the "edit." I still have upvoted it. $\endgroup$ – quid Sep 29 '17 at 22:21
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I think the proof clarifies if you employ the Distributive Property fully: ab = (2n)(2m + 1) = (2n2m) + 2n = 2( 2mn + n ) only ONE 2 pulls out of the first product Since m and n are integers, ab will always have a factor of 2 among its Prime factors.

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$a$ is even $\implies a=2n:n\in\mathbb{Z}$

$b$ is odd $\implies b=2m+1:m\in\mathbb{Z}$

$ab=2nm+2n=2n(m+1)$

$2n$ factors $2n(m+1)\implies ab$ is even.

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