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Definition of Divisibility

  1. Sep 26, 2015 #1

    S.R

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    Question:
    If x | y, (is true), then x ≤ y and x ≠ 0.

    For instance, if x > y, then there are no integer solutions to equation kx = y and thus, x does not divide y.
    Is this a correct proposition?
     
  2. jcsd
  3. Sep 26, 2015 #2

    mfb

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    2016 Award

    Staff: Mentor

    What do you think?
     
  4. Sep 26, 2015 #3

    WWGD

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    Gold Member

    Up to sign, if you extend divisibility to the whole of ## \mathbb Z ##.
     
  5. Sep 26, 2015 #4

    S.R

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    I suppose if x and y are negative, then the converse is true.
    For instance, -2 | -4 is true, but -4 | -2 is false.
     
    Last edited: Sep 26, 2015
  6. Sep 27, 2015 #5

    Mark44

    Staff: Mentor

    Why aren't you using the usual definition of divisibility; i.e., For integers x and y, x | y iff y = kx for some integer k.
     
  7. Sep 27, 2015 #6
    Happens to be correct, but merely a proposition unless you also prove it.
    If [itex]x|y[/itex] and [itex]x>y[/itex] then there exists ##c\in\mathbb{Z}\setminus\{0\}## such that ##cx = y##. Figure out how this contradicts and you are home free :) Also, ##\mathbb{Z}## does not form a group under multiplication, so you can't "divide" both sides by some non zero scalar.
     
  8. Sep 27, 2015 #7

    S.R

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    Assuming x, y ∈ N:
    If x | y and x > y, then there exists an integer k such that kx = y or k = y/x.
    However, since x > y, the expression y/x is not an integer.
    Therefore, we can conclude x does not divide y, since no integer k exists such that kx = y.
     
    Last edited: Sep 27, 2015
  9. Sep 27, 2015 #8
    Were you operating with real numbers (in a complete ordered field) the statement ##k = \frac{y}{x}## is meaningful, but you can't do that if you are strictly operating within ##\mathbb{Z}##.
    The conclusion is correct that there is no such integer ##k## that produces the desired result.
    Alternatively we could use the given that ##x>y##. Let's look for a candidate ##k\in\mathbb{Z}## such that ##kx=y##
    Since ##k## is an integer it can be expressed as a sum ##k :=\pm ( 1+1+...+1)## (there is actually a reason why this is true, but I don't want to burst the bubble)

    Looking at positive integers first
    Then ##kx = y## becomes ##(1+1+...+1)x = y##. Due to the distributive property the previous can also be written as ##x + x + ... + x = y##
    But we already have that ##x>y## therefore ##x+x+...+x = y## can never be true.

    And if we look in the negative integers then ##-x -x ..-x = y##. Under addition, ##\mathbb{Z}## does form a group, so we can say there exists ##-(-x) = x## such that ##x+(-x) = 0## and what would follow is ##y+x+x+...+x = 0##, which can only be true if ##y>x##, violating one of our established criteria.
    The initial assumption that such an integer exists is therefore incorrect. And ##x|y## only if ##0\neq |x|\leq |y|##

    In actuality, such divisibility is defined in a ring with no zero terms. A ring is an algebraic structure that forms an abelian group under addition, multiplication and addition are connected via the distributive property. A ring with no zero terms (can't be too sure of the English terminology here) is a ring where no two ring elements' product is 0.

    We don't know anything about "dividing both sides by.." for this problem. There is no group under multiplication.


    EDIT: oop, You assumed x,y to be in N, then the negative integer part is unnecessary.
    Actually, this proof contains inaccuracies, as well :/
     
    Last edited: Sep 27, 2015
  10. Sep 27, 2015 #9

    Mark44

    Staff: Mentor

    If x | y, then it's not possible for x to be larger than y. "x divides y" means that x is a factor of y.
    I don't get what you're trying to do here.
     
  11. Sep 27, 2015 #10
    Challenge the definition :D I agree, it's fruitless.
     
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