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

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

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

iheartducks Group TitleBest ResponseYou've already chosen the best response.0
means that there is an integer c such that b=a*c =====>>>> a  b
 2 years ago

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

KingGeorge Group TitleBest ResponseYou've already chosen the best response.0
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: \(c>0\)
 2 years ago

iheartducks Group TitleBest ResponseYou've already chosen the best response.0
contradiction when a>b?
 2 years ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.0
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.
 2 years ago

iheartducks Group TitleBest ResponseYou've already chosen the best response.0
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?
 2 years ago

iheartducks Group TitleBest ResponseYou've already chosen the best response.0
therefore making a smaller than b?
 2 years ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.0
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.
 2 years ago

iheartducks Group TitleBest ResponseYou've already chosen the best response.0
ohhh ok. Gotcha. Sorry for the bother
 2 years ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.0
no problem.
 2 years ago
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