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.
$\endgroup$ 08 Answers
$\begingroup$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.
$\endgroup$ 3 $\begingroup$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.
$\endgroup$ 5 $\begingroup$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.
$\endgroup$ 3 $\begingroup$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.
$\endgroup$ $\begingroup$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......
$\endgroup$ 3 $\begingroup$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.]
$\endgroup$ 3 $\begingroup$Modulo $2$ we have by assumption $$ 0\equiv 7n+4\equiv 1\cdot n+0=n, $$ so that $$ 5n+6\equiv 5\cdot 0+0\equiv 0. $$
$\endgroup$ $\begingroup$$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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