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I want to give a direct proof of a conditionnal $$\forall\, n\in \Bbb Z : \left[ P(n) \Rightarrow \lnot Q(n) \right]$$ such that $$P(n)= n>2$$ and$$ Q(n)= \exists\ m \in \Bbb Z: (\,m+n=mn \wedge n|m\,)$$ hence, I want to prove that $$\forall\, n\in \Bbb Z : \left[\,n > 2 \Rightarrow \lnot \exists\ m \in \Bbb Z: (\,m+n=mn \wedge n|m\,) \,\right] $$ is true.

I think that I proved the statement:

I take the converse, so $Q(n) \Rightarrow \lnot P(n)$ and assume $Q(n)$ hence, $n|m$.

It follows that $$m=nk: k\in\Bbb Z $$ Thus, $$k=\frac 1{n-1} \Rightarrow \left[m=\frac n{n-1}\in \Bbb Z \iff n=2\right] \Rightarrow \lnot P(n)$$

Therefore, $$\left[ Q(n) \Rightarrow \lnot P(n) \right]\Rightarrow \left[P(n) \Rightarrow \lnot Q(n)\right]$$ is true, by contrapositive.

What would be a direct proof of this statement? Is the contrapositive considered a form of direct proof?

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    @Matthew Leingang Thank you I edited my question, I think it is better now – 2017-01-25
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    Yes, that helps. Note that statement $P$ is false, so $P\implies Q$ is true for all $Q$. – 2017-01-25
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    $Q$ has a free variable, namely $n$. – 2017-01-25
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    As noted, the statement $\forall n \in \Bbb Z : n > 2$ is false. Instead I think you mean to say, Let $n \in \Bbb Z$ with $n > 2$. – 2017-01-25
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    I suspect No Thought-No Concept is trying to prove $$\forall n\in \Bbb Z:\Big[ n>2 \implies \neg\exists m\in\Bbb Z:\big( m+n=mn ~\wedge~n\mid m\big)\Big]$$ – 2017-01-25
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    @ajotatxe Is the free variable "hiding" in the fact that if $Q$ is true $n|m$? @ Graham Kemp Yes! that is what I want to prove thank you I shall edit my post! – 2017-01-25
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    If @GrahamKemp is right, then you can't split it up into $P$ and $Q$ that way. Instead, the statement is of the form $\forall n \in \mathbb{Z} : [P(n) \implies \neg Q(n)]$, where $P(n)$ is $n>2$ and $Q(n)$ is $\exists m\in \mathbb{Z} : (m+n = mn \wedge n | m)$. – 2017-01-25
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    Could it be that a proof by contrapositive is the solution? – 2017-01-25
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    @NoThought-NoConcept I proved it. $n+m=nm$ only has two integer solutions. Simple examination of both finds that $m$ doesn't exist. – 2017-01-25
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    @user4894 Perhaps you've forgotten that, as Graham shows, $\forall n\in \mathbb Z\Big[n\gt 2 \rightarrow \lnot \exists m \in \mathbb Z((m+n = mn) \land (n\mid m))\Big]$. Which means for every n that's an integer, **IF** $(n\gt 2)$, **then**.... The entire proposition says nothing about those $n$ such that $n\leq$. – 2017-01-25

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You wish to prove the following statement.

For all integers $n$ such that $n > 2$, there does not exist an integer $m$ such that $m + n = mn$ and $n$ is divisible by $m$. We will prove this directly.

We first note that $mn = n + (m-1)n$. Therefore, $(m-1)n=m$ and $n = \frac {m}{m-1}$. I now cite some previous knowledge you may or may not know of. Two adjacent integers $a$ and $a+1$ do not share any prime factors. Therefore, $n = \frac {m}{m-1}$ is only true when $m-1$ has no prime factors to prevent the division from resulting in a whole number integer $n$. the only numbers with that quality are $-1$ and $1$. Therefore, $m$ can only be one of two values $m = 0$ or $m = 2$. This results in two values for $n$, which are $2$ and $0$. Note that the only time $m$ exists satisfying the first condition is when the value for $n$ does not satisfy the given statement. Therefore, $m$ does not exist and the statement is true.

Comments/Note: I didn't use the divisibility portion of the statement. That's because I didn't need it. $n$ and $m$ are always equal by the equation and equal to either $2$ or $0$. I can only presume that the mention of divisibility is a red herring meant to throw off the op and make them think more critically about what is and is not useful in a proof.

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    I was thinking that I proved it by contrapositive if I assumed $Q$ and so if $n|m$ then, $m=nk\,:\,k \in \Bbb Z$ so we have $nk \,+\, n=n^2k$ which means $k=\frac 1{n-1}$ hence, $m=\frac n{n-1} \, \in \Bbb Z \iff n=2$ which means $\lnot P$. Therefore, since $Q \Rightarrow \lnot P$ we have $P \Rightarrow \lnot Q$ by contrapositive. I think your direct proof is more beautiful because of the argument you used though. – 2017-01-25
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    @NoThought-NoConcept thanks. I tried to do as much as I could without referring to the divisibility. Most things tend to get very bloated when that is thrown in. Plus, it seemed pretty strange to have $n > 2$. Looking at the equation, it seemed like only $2$ satisfied it which tied back to the inequality. The fact that $0$ did as well was actually a happy accidental find. At that point, there was no need to think of divisibility. Basically, I proved a stronger statement that was simpler and "cleaner" than yours. – 2017-01-25
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    Should I always try to avoid referring to divisibility and simplify that kind of statement before I try to prove it or does that means taking the risk of loosing meaningful information? – 2017-01-25
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    @NoThought-NoConcept Think of it this way. It's true because the other statement is true. However, if this is a homework, don't bother reducing it. Just ignore the unneeded restriction.If it's a proof for a paper or something else, maybe mention that the restriction is unneeded in hindsight like I did. However, for a simple proofing class, the general premise of a proof is to be efficient, clear, and concise.Needlessly saying the restriction is unneeded just serves as wasteful dialogue. Think of it like a post office package. You get charged by the pound (character). Don't waste characters. ;) – 2017-01-25
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    @NoThought-NoConcept honestly, proofs are a combination of english writing and mathematical reasoning. So, my proof is sufficient in terms of reasoning so long as you accept my previous knowledge. As for style and whatnot.. that's really a matter of personal preference and knowledge of good english writing. Sadly, I am not good at editing. So, I'd default to not mentioning that as it breaks the flow but feel free to include it if you feel it helps in readability. – 2017-01-25
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    @NoThought-NoConcept oh. I read that all wrong. slightly ignore the last two comments. I would not remove divisibility. I didn't do that either. I just wrote the statement in English words to make it sound pleasant to human readers and then just never used that restriction to solve regarding $m$. Pure coincidence it occurred. Proofs are done however you choose. Just think through it, but NEVER change the original statement. Rewording it slightly is fine,but changing it won't remove information. It just means you are proving an equivalent statement and never stating you proved the original. – 2017-01-25