Proof by contrapositive help for if $15n$ is even, $9n$ is even

Question

I wanted to check if this was a valid proof for the following:

Let $n \in\mathbb Z$. Prove that if $15n$ is even, then $9n$ is even.

What I have is the following: $$\forall n\in\mathbb Z,p\left( n\right) \rightarrow q\left( n\right) $$ $${\sim q}\rightarrow {\sim p}$$

$p(x)$: $15n$ is even

$q(x)$: $9n$ is even

proof: let $9n=2k+1 \Longrightarrow n=\dfrac {2k+1}{9}$ $$15\left( \dfrac {2k+1}{9}\right) =\dfrac {5}{3}\left( 2k+1\right) $$ $$\dfrac {10}{3}k+\dfrac {5}{3}$$ $$\dfrac {10}{3}K+\dfrac {2}{3}+1=2\left( \dfrac {5k}{3}+\dfrac {1}{3}\right) +1$$

Since $\left( \dfrac {5K}{3}+\dfrac {1}{3}\right) \in\mathbb Z$ and is odd, $15n$ is even.

Help would be appreciated, thank you

How do you know that $\displaystyle{5k\over 3}+{1\over3}\in\mathbb Z$? — Christopher Carl Heckman, Apr 30 '16 at 07:24
An easier trick would be if $9n$ is odd, then $n$ cannot be even, so $n=2k+1$ for some $k\in\mathbb{Z}$. — Levent, Apr 30 '16 at 07:26
An easier trick would be using $$\text{(even)}+\text{(even)}=\text{(even)}$$ — S.C.B., Apr 30 '16 at 07:33
@ChristopherCarlHeckman Well, by the equality and the condition of the original statement right ? Since $2k+1 \in z$ and $n = \frac{2k+1}{9} \in Z$, and $9n \in Z$, so after the simplification, $(\frac{5k}{3}+\frac{1}{3})\in Z$ right ? — PutsandCalls, Apr 30 '16 at 07:34
Or by writing $15n = 9n+6n = 2 \cdot (3n + \cdots) + 1$ ... I suspect this is a homework problem, so the proof must be contrapositive. — Christopher Carl Heckman, Apr 30 '16 at 07:34
@Levent Yes, that was originally given by one of the solutions by introducing a lemma, but would you guys agree that I have shown it or was my proof flawed ? — PutsandCalls, Apr 30 '16 at 07:36
@ChristopherCarlHeckman Yes I wanted to use contrapositive for practice here — PutsandCalls, Apr 30 '16 at 07:36

score 6 · Accepted Answer · answered Apr 30 '16 at 07:32

$\neg q(n)\implies$

$9n\not\equiv0\pmod2\implies$

$9n\equiv1\pmod2\implies$

$\exists{k\in\mathbb{Z}}:9n=2k+1\implies$

$\exists{k\in\mathbb{Z}}:15n=6n+2k+1\implies$

$\exists{k\in\mathbb{Z}}:15n=2(3n+k)+1\implies$

$15n\equiv1\pmod2\implies$

$15n\not\equiv0\pmod2\implies$

$\neg p(n)$

Proof by contrapositive help for if $15n$ is even, $9n$ is even

1 Answers1