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i am having trouble proving the statement below using Proof by Contrapositive. I have negated the statements as required and then i prove that $n$ is odd if and only if $7n+4$ is odd. However, from there i am stuck and am not sure how to prove the 1st part that if $n$ is negative integer, then $n$ is odd. My work is show below.

Original Statement: if n is a positive integer then n is even if and only if is 7n+4 is even.
Contrapositive: If n is negative integer then n is odd if and only if 7n+4 is odd.

Therefore by definition of odd: n = 2k+1
Substitute n:
=7(2k+1)+4
=14k+7+4
=14k+11
=2(7k)+11
Therefore, n is odd and 7n+4 is odd.

Thats as far as i got and i dont even know if what i did above is even right though. Thanks.

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    $\begingroup$ That's not the contrapositive. $\endgroup$ Jan 29, 2015 at 23:06
  • $\begingroup$ then what is? isnt the contrapositive of lets say pVq, not p V not q? $\endgroup$
    – a22asin
    Jan 29, 2015 at 23:07
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    $\begingroup$ If you have a statement "If P then Q" and want the contrapositive you get "If $\lnot Q$ then $\lnot P$. $\endgroup$
    – User
    Jan 29, 2015 at 23:08
  • $\begingroup$ well, wouldn't "if n is not even" be equivalent to "n is odd"? same goes for "if 7n+4 is not even" would be equivalent to "7n+4 is odd"?. $\endgroup$
    – a22asin
    Jan 29, 2015 at 23:10
  • $\begingroup$ @a22asin I would use a direct proof here (as I gave in my answer). You really want to reserve proofs by contradiction, contraposition, etc., for when doing a direct proof is annoying, far more difficult, etc. At least that's my opinion. $\endgroup$ Jan 29, 2015 at 23:31

3 Answers 3

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There really is no need to prove by contrapositive (unless your teacher is requiring you to do that for some reason). The following direct proof works just fine.


Problem: Suppose $n\in\mathbb{Z}$ ($n$ being positive is really not necessary). Then $n$ is even if and only $7n+4$ is even.

Proof.

($\to$): Suppose $n$ is even. Then $n=2\ell$, where $\ell\in\mathbb{Z}$. Thus, we have that $$ 7n+4=7(2\ell)+4=14\ell+4=2(7\ell+2)=2m, $$ where $m=7\ell+2$ and $m\in\mathbb{Z}$. Thus, the forward direction is true.

($\leftarrow$): Suppose $7n+4$ is even. Consider what happens when $n$ is odd or even.

  • $n$ odd: We have $n=2\ell+1$ for some $\ell\in\mathbb{Z}$ and so $$ 7(2\ell+1)+4=14\ell+11=2(7\ell+5)+1=2m+1, $$ where $m=7\ell+5$ and $m\in\mathbb{Z}$.
  • $n$ even: We have $n=2\ell$ for some $\ell\in\mathbb{Z}$ and so $$ 7(2\ell)+4=14\ell+4=2(7\ell+2)=2m, $$ where $m=7\ell+2$ and $m\in\mathbb{Z}$.

The above analysis shows that $n$ is even if and only if $7n+4$ is even, as desired.

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  • $\begingroup$ It should be noted that a proof by contrapositive will be very similar. The problem would be stated as follows when proving by contrapositive: $7n+4$ is odd if and only if $n$ is odd. $\endgroup$ Jan 29, 2015 at 23:34
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I think you want to rephrase the original statement as:

"Let $n$ be a positive integer. Then $n$ is even if and only if $7n+4$ is even."

With "if and only if" proofs, we would call the forward direction of this proof

"If $n$ is even then $7n+4$ is even"

This is equivalent to its contrapositive, "If $7n+4$ is not even then $n$ is not even." As you pointed out, this contrapositive is the same as saying "If $7n+4$ is odd then $n$ is odd"

Next, the backward direction is

"If $7n+4$ is even then $n$ is even"

which is equivalent to the contrapositive statement "If $n$ is odd then $7n+4$ is odd."

So to complete your proof, you have to complete the proof in the forward direction (whether you do it directly or prove the contrapositive) and similarly you have to show the backward direction. Once you have done this, you may conclude that $n$ is even iff $7n+4$ is even.

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Your statement of the contrapositive is correct, except that the first "negative" should be "positive".

The argument you have given proves correctly that if $n$ is odd, then $7n+4$ is odd.

It remains to prove that if $7n+4$ is odd, then $n$ is odd. Here is one possible proof.

Let $n$ be an integer and assume that $7n+4$ is odd. Now $6n+4$ is obviously even, and the difference between an odd and an even number is odd, therefore $(7n+4)-(6n+4)$ is odd. That is, $n$ is odd. This completes the proof.

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  • $\begingroup$ I'm not answering what was originally posted, but what I can see :) $\endgroup$
    – David
    Jan 29, 2015 at 23:35
  • $\begingroup$ Interesting. The OP kept editing his/her post while comments came in at the beginning so the flawed logic doesn't show up in the edit history now. :( What a mess. $\endgroup$ Jan 29, 2015 at 23:38

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