2

So, this is actually 2 questions in 1. I apologize if that is bad practice, but I didn't want to write 2 questions when they're a word different. So, I have

  1. Prove or disprove that if $a|(sb+tc), \forall s,t \in\mathbb{Z}$, then $a|b$, and $a|c$.

and then,

  1. Prove or disprove that if $a|(sb+tc)$ for some $s,t \in\mathbb{Z}$, then $a|b$, and $a|c$.

I know how to prove if $a|b$ and $a|c$, then $a|(sb+tc), \forall s,t \in\mathbb{Z}$, but I'm certain I can't just write the proof backwards, and show that it works. So, how would I tackle these two problems, especially the second one?

Valent
  • 3,228
  • 1
  • 15
  • 28
JCMcRae
  • 843
  • Note that the second proposition implies the first. So a priori, we know it's not possible for the second to be true and the first to be false. – Zubin Mukerjee Oct 13 '14 at 03:18
  • The "for all" in the first statement is a very strong condition; you can choose whichever $s$ and $t$ you want in order to somehow derive $a \mid b$ and $a \mid c$. You might notice that $sb+tc=b$ when $(s,t) = (1,0)$ and that $sb+tc=c$ when $(s,t) = (0,1)$. – Zubin Mukerjee Oct 13 '14 at 03:21
  • The "for some" in the second statement is a very weak condition, so we should intuitively expect it to be false. A good way to approach "prove or disprove" problems in which you think the statement is false is to first look for a counterexample. With a bit of thought you should be able to find an easy counterexample. In general, if it's difficult to find a counterexample you might want to think about whether it actually is false. – Zubin Mukerjee Oct 13 '14 at 03:23

2 Answers2

1

  1. What happens when one of $s,t$ is one and the other one is zero?

  2. $2014| 2014 \cdot 1 + 2014 \cdot 3$ but $2014$ doesn't divide $1$ or $3$...

N. S.
  • 132,525
0
  1. True.

Choose $s = 0, t = 1$, then $a|c$, and choose $s = 1, t = 0$, then $a|b$

  1. False.

Let $a = 2, b = 3, c = 5$, and $s = t = 2$.

DeepSea
  • 77,651