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I am still new to the proof game so please be kind! This is my third time attempting a proof. Any feedback would be greatly appreciated. Thank you in advance.

Claim: For all integers $n$: if $7n+4$ is even, then $5n+6$ is even.

Proof: Assume $5n+6$ is odd. If $5n+6$ is odd, then $5n+6=2k+7 \Rightarrow 5n=2k+1$ for some $n,k \in \mathbb{Z}$. If $7n+4$ is even, then $7n+4=2k$ for some $k \in \mathbb{Z}$. Therefore, $5n=7n+4+1 \Rightarrow 2n=-5$ for some $n \in \mathbb{Z}$. Clearly, $2n \neq -5$, which contradicts our assumption that $5n+6$ is odd.

I feel like I completely lost myself, but do not know where to go. Please be kind! Any words of wisdom and insight would be great. Thanks.

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What I see is if $7n+4$ is even then $7n$ is also even, and so $n$ is even too.

Then $5n$ is even and thus $5n+6$ is even.

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  • $\begingroup$ So, $n$ is even. Since $n$ is even, $5n$ is even also. Thus, $5n+6$ is even. Am I understanding that correctly? $\endgroup$
    – Ryan
    Jul 27, 2018 at 19:36
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    $\begingroup$ yep! hope this helped you $\endgroup$ Jul 27, 2018 at 19:40
  • $\begingroup$ It has. Thank you for your feedback! $\endgroup$
    – Ryan
    Jul 27, 2018 at 19:41
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Walking through your attempt:

Claim: For all integers $n$: if $7n+4$ is even, then $5n+6$ is even.

Proof: Assume $5n+6$ is odd.

Normally proving $P_a\Rightarrow P_b$ you would start by assuming $P_a$. Here you're assuming $\lnot P_b$, suggesting you want to prove the contrapositive, $\ \lnot P_b\Rightarrow \lnot P_a,$ which is equivalent to $P_a\Rightarrow P_b$. But you need to understand where you are trying to get to.

If $5n+6$ is odd, then $5n+6=2k+7 \ldots$

This is unusual - it would be more usual to say $5n+6=2k+1 $

$\Rightarrow 5n=2k+1$ for some $n,k \in \mathbb{Z}$.

Which would then be $\Rightarrow 5n=2k-5$.

If $7n+4$ is even,

Having started in on the contrapositive, you are now switched over... perhaps for the purpose of contradiction, but it is starting to get hard to follow.

then $7n+4=2k$ for some $k \in \mathbb{Z}$.

You shouldn't reuse $k$. We'll say $7n+4=2m$ for some $m\in \Bbb Z$. This was your main mistake.

Therefore, $5n=7n+4+1$

Not true. This is the result of re-using $k$ inappropriately.

$\Rightarrow 2n=-5$ for some $n \in \mathbb{Z}$. Clearly, $2n \neq -5$, which contradicts our assumption that $5n+6$ is odd.

Actually - if this were valid - you should have it contradict your later statement that $7n+4$ is even, putting you on track to prove the contrapositive.

You don't need to come into the proof backwards on this occasion. It's easy enough to start with the the premise:

Taking $\mathit{ 7n+4}$ as even, we have $\mathit{ 7n+4=2m}$ for some $\mathit{ m\in \Bbb Z}$. Then $\mathit{ 5n+6 = (7n-2n+4+2) = 2m-2n+2 = 2(m-n+1)}$ and since $\mathit{ (m-n+1)\in \Bbb Z}$ we have ${\mathit 5n+6}$ is even.

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    $\begingroup$ Thank you so much for walking me through it! I followed it perfectly step by step. I took a lot away from your solution. Thanks again. $\endgroup$
    – Ryan
    Jul 27, 2018 at 21:07
  • $\begingroup$ How would I approach this by method of contradiction? That is, what should I be setting out to prove? $\endgroup$
    – Ryan
    Jul 27, 2018 at 21:20
  • $\begingroup$ I only ask because the instructions for this specific problem was to approach it using the contradiction method. Thanks again. $\endgroup$
    – Ryan
    Jul 27, 2018 at 21:22
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    $\begingroup$ No problem. If this is your preferred answer, you can mark it with the $\checkmark$. To prove this claim by contradiction you would take $7n+4$ as even (as given), and then assume $5n+6$ is odd and work towards some impossible statement. Exactly what you work towards depends on what you have previously proven (or are allowed to assumed true). $\endgroup$
    – Joffan
    Jul 27, 2018 at 21:33
  • $\begingroup$ Great. Thank you again. I marked this as my preferred answer. Sorry, a little new to math exchange so wasn't aware of these things. Thank you for your help! $\endgroup$
    – Ryan
    Jul 27, 2018 at 21:39
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Here is another approach. To make $5n+6$ appear in $7n+4$, consider their difference: $$ (7n+4)-(5n+6)=2n-2=2(n-1) $$ Therefore, $$ 5n+6=(7n+4)-(2n-2) $$ is even, being the difference of two even numbers.

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  • $\begingroup$ Thank you for your response. If I wanted to prove this statement by contradiction, what adjustments or corrections should I make in my proof? $\endgroup$
    – Ryan
    Jul 27, 2018 at 19:31
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    $\begingroup$ @Ryan If it were odd it would differ from the given even number by an odd amount, but you get a contradiction by verifying that the difference is even. Incidentally, a lot of beginners unnecessarily create a proof-by-contradiction wrapper for a proof that can just be written as direct. $\endgroup$
    – J.G.
    Jul 27, 2018 at 19:38
  • $\begingroup$ I understand. Direct looks like it would be an easier approach. My practice problem just instructed me to use proof by contradiction, which is why I approached it that way. $\endgroup$
    – Ryan
    Jul 27, 2018 at 19:40
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If $7n+4$ is even, then $7n+4=2k$ for some integer $k$. Then you can write the expression $7n+4$ as $7n+4=(5n+6)+(2n-2)$ which implies that

$5n+6=(7n+4)-(2n-2)=2k-(2n-2)=2(k-n+1)$

This proves directly that $5n+6$ is even.

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In the second statement "if $7n+4$ is even...." when you write $7n+4=2k$, then you are using the same $k$ as mentioned in the first statement. This is incorrect. Instead you should say some thing like, $7n+4=2l$ for some $l \in \mathbb{Z}$.

In fact, you may approach on the following lines:

You already have that $5n=2k+1$, now consider $7n+4=5n+2n+4=(2k+1)+2n+4=2(k+n+2)+1$. So this makes $7n+4$ odd......

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  • $\begingroup$ Ahh, I had a feeling that was incorrect. Thank you for your feedback. $\endgroup$
    – Ryan
    Jul 27, 2018 at 19:26
  • $\begingroup$ Wow, that was extremely clever! I would have never thought of that. Thank you so much! $\endgroup$
    – Ryan
    Jul 27, 2018 at 19:39
  • $\begingroup$ Since $7n+4$ is odd, how do I conclude my proof? Should it be along the lines of: If $5n+6$ is odd, then $7n+4$ must also be odd, which is a contradiction. Would that be plausible? Thank you for your help. $\endgroup$
    – Ryan
    Jul 27, 2018 at 20:04
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Forget about proof by contradiction and just work forwards from what you are given, thinking about what you can deduce from statements like "$a + b$ is even" and "$ab$ is even":

If $a + b$ is even, then $a$ and $b$ are either both even or both odd. As $4$ is even, if $7n + 4$ is even then so is $7n$. If $ab$ is even, then at least one of $a$ and $b$ is even. $7$ is not even so if $7n$ is even, then $n$ must be even. But then, as $n$ is even, so are $5n$ and $5n + 6$ (because $6$ is even).

[Aside: the advice about reasoning "modulo 2" in other answers is a slick way of packaging the above way of thinking.]

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  • $\begingroup$ Thank you for your response. I cannot just "forget" proof by contradiction because the homework instruction specifically states to use the method of contradiction to prove it. $\endgroup$
    – Ryan
    Jul 27, 2018 at 21:08
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    $\begingroup$ OK. Just add "Assume that what we are trying to prove is false" to the beginning of your proof. You don't need to use that assumption. (If you are feeling brave, then challenge your instructor as to why he or she asked you to give a proof by contradiction when it's completely unnecessary.) $\endgroup$
    – Rob Arthan
    Jul 27, 2018 at 22:41
  • $\begingroup$ Will do. Thanks again for your response! $\endgroup$
    – Ryan
    Jul 27, 2018 at 23:51
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Modulo $2$ we have by assumption $$ 0\equiv 7n+4\equiv 1\cdot n+0=n, $$ so that $$ 5n+6\equiv 5\cdot 0+0\equiv 0. $$

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$7n+4$ is even $\Leftrightarrow 7n\equiv 0 \mod 2$ then $n\equiv 0 \mod 2$.

Then, $5n+6\equiv 0 \mod 2$, since $5n+6\equiv5(0)+6\equiv 6\equiv0 \mod 2$

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