Question

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

Answer 1

What does it mean that a|b?

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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: \(c>0\)

Answer 6

contradiction when a>b?

Answer 7

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.

Answer 8

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?

Answer 9

therefore making a smaller than b?

Answer 10

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.

Answer 11

ohhh ok. Gotcha. Sorry for the bother

Answer 12

no problem.