I am confused about a homework problem I have, and don't really know where to begin. The statement is that every negative integer can be written as $2a+3b$ where $a$ and $b$ are either positive or negative integers. I need to prove this. Any idea of where I can start. I am not necessarily looking for a solution, but a place to begin.

Show that every negative integer can be written in the form $2a + 3b$ for some (not necessarily positive) integers $a$ and $b$.

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    $\begingroup$ Have you tried induction? If you can write $-k = 2a + 3b$ for some $a,b$, how would you write $-(k+1)$? $\endgroup$ Oct 13 '12 at 19:33
  • $\begingroup$ did not think to do that. Will try now, thanks. Also thanks for the edit. $\endgroup$ Oct 13 '12 at 19:38

HINT: First find integers $a_0$ and $b_0$ such that $$2a_0+3b_0=-1\;.\tag{1}$$ Then let $n$ be any positive integer, and see what happens when you multiply equation $(1)$ by $n$.

  • $\begingroup$ that would probably have worked as well, but I did it with induction as the comment above suggested. $\endgroup$ Oct 13 '12 at 20:55
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    $\begingroup$ @MZimmerman6: Easy as it is in this case, induction actually takes a little more work. Since $2(1)+3(-1)=-1$, it’s immediate that $-n=2n+3(-n)$ for any positive integer $n$. This also gives you a specific representation of each negative integer. $\endgroup$ Oct 13 '12 at 20:58
  • $\begingroup$ yeah, his is the answer I saw first, so I went that route. Thanks for your response though! $\endgroup$ Oct 13 '12 at 21:02
  • $\begingroup$ @MZimmerman6: You’re welcome. (I wasn’t complaining, by the way; induction’s a perfectly reasonable approach. I just figured that since you’d now solved the problem, I might as well finish off the other solution, since it never hurts to see more than one approach.) $\endgroup$ Oct 13 '12 at 21:05
  • $\begingroup$ I entirely agree! Thanks! $\endgroup$ Oct 13 '12 at 22:11

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