## iheartducks 3 years ago Can anyone prove the following? If each of a and b is in ℤ+ and a I b then a <= b

1. KingGeorge

What does it mean that a|b?

2. KingGeorge

I.e., what's the definition of a|b?

3. iheartducks

means that there is an integer c such that b=a*c =====>>>> a | b

4. KingGeorge

Precisely, and both a,b are in $$\mathbb{Z}^+$$. Now, assume that $$a>b$$, and work towards a contradiction.

5. KingGeorge

I would then break it up into two cases, and show why these are both impossible. (also, why is $$c\neq 0$$?) Case 1: $$c<0$$ Case 2: $$c>0$$

6. iheartducks

7. KingGeorge

Right. So assume a>b, and then divide into the two cases where c<0 and c>0. Reach a contradiction with both cases, so a<=b.

8. iheartducks

i'm sorry, i've never been good with contradictions. So like if c>0, then a would HAVE to be < b since (b/c) would result in a fraction?

9. iheartducks

therefore making a smaller than b?

10. KingGeorge

Let's suppose $$c>0$$. Then, $$c\ge 1$$ since $$c\in\mathbb{Z}$$. Hence, $ac=b\ge a$Since we assumed $$b<a$$, this is a contradiction, so $$a\le b$$. Now you just need to work out $$c\le 0$$. That case should be easier.

11. iheartducks

ohhh ok. Gotcha. Sorry for the bother

12. KingGeorge

no problem.