Very simple proof help about integers

Question

I was wondering if anybody could help me with proving something very simple: that $9n\ne6$ when n is any integer. It seems extremely intuitive but I don't know how to make it into part of a rigorous proof.

It's useful to prove the contrapositive. Namely, prove that if $9n=6$ for some real number $n$, then $n$ is not an integer. The contrapositive is logically equivalent to the statement you're trying to prove. — Shawn O'Hare, Jul 15 '14 at 18:20

score 2 · Accepted Answer · answered Jul 15 '14 at 18:18

2

$9n=6$ so $n=6/9=2/3$ which is not an integer.

Alternatively you might say $9=3^2$ and $6=2\cdot 3$ and the result follows from the fundamental theorem of arithmetic.

answered Jul 15 '14 at 18:18

lemon

3,548

oh my gosh I can't believe that I missed that. Thanks! – hyperdo Jul 15 '14 at 18:20
This shows that $,9n = 6,\Rightarrow,3n = 2.\ \ $ That the latter has no integer solution $,n,$ still requires proof. – Bill Dubuque Jul 15 '14 at 18:43
If $3n=2$ then it follows from the FT of arithmetic that $3|2$. Done. – lemon Jul 15 '14 at 19:07

score 1 · Answer 2 · answered Jul 15 '14 at 18:27

If $9n=6$ then we consider two cases for n:

1) n is odd. Then 9n is odd too. But $2|6$. A contradiction. Therefore $9n\neq 6$.

2) n is even. Let $n=2k$ for some integer k. So $9\times 2k =6$ or $3k=1$. In the latter, 3 divides the LHS but does not divide RHS. This is a contradiction.

score 0 · Answer 3 · answered Jul 15 '14 at 18:22

0

Hint: If $n$ is an integer, then either $n\geq 1$ or $n \leq 0$. Would either case allow $9n = 6$?

answered Jul 15 '14 at 18:22

Semiclassical

15,842

Very simple proof help about integers

3 Answers3