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

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KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.0What does it mean that ab?

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.0I.e., what's the definition of ab?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0means that there is an integer c such that b=a*c =====>>>> a  b

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.0Precisely, and both a,b are in \(\mathbb{Z}^+\). Now, assume that \(a>b\), and work towards a contradiction.

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.0I 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: \(c>0\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0contradiction when a>b?

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.0Right. 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i'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?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0therefore making a smaller than b?

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.0Let'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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ohhh ok. Gotcha. Sorry for the bother
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