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

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

KingGeorge Group TitleBest ResponseYou've already chosen the best response.0
I.e., what's the definition of ab?
 one year 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
 one year 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.
 one year 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: \(c>0\)
 one year ago

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

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

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

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