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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

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  1. KingGeorge
    • 3 years ago
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    What does it mean that a|b?

  2. KingGeorge
    • 3 years ago
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    I.e., what's the definition of a|b?

  3. iheartducks
    • 3 years ago
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    means that there is an integer c such that b=a*c =====>>>> a | b

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

  5. KingGeorge
    • 3 years ago
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    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
    • 3 years ago
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    contradiction when a>b?

  7. KingGeorge
    • 3 years ago
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    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
    • 3 years ago
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    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
    • 3 years ago
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    therefore making a smaller than b?

  10. KingGeorge
    • 3 years ago
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    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
    • 3 years ago
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    ohhh ok. Gotcha. Sorry for the bother

  12. KingGeorge
    • 3 years ago
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    no problem.

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