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The consequence of divisibility definition in integer

  1. Feb 15, 2014 #1
    So I think I've just proven a preposition, where ##0## is divisible by every integer. I prove it from the accepted result that ##a \cdot 0 = 0## for every ##a \in \mathbb{Z}##. From then, we can just multiply the result by the inverse of ##a##, to show that the statement holds for ##0##. That is to say, there exist an integer ##0##, such that ##a^{-1} \cdot 0 = 0##.

    But then there's another preposition, if ##a \in \mathbb{Z}## and ##a \neq 0##, then ##a## is not divisible by ##0##. Okay we can also use the fact that ##a \cdot 0 = 0##. So far so good. But then I realize that the preposition seems to imply that if ##a=0## then ##a## is divisible by ##0##. The first preposition where ##0## is divisible by every integer also points to the same result because ##0 \in \mathbb{Z}##.

    But we know isn't it, that we cannot divide any number by ##0##, any operation that involves division by ##0## is automatically a no-no in math. It just doesn't sound right. (The preposition comes from a book and I don't propose that myself) Does it mean that technically (according to the definition of divisibility) ##0## is also divisible by ##0##, but it's not a legal operation in cancellation, say when, ##a \cdot 0## = ##b \cdot 0##. We cannot cancel the ##0## in this case. But still again, ##0## is divisible ##0##.
     
  2. jcsd
  3. Feb 16, 2014 #2

    jbriggs444

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    What definition of "is divisible by" are you and your book using? Is it that "a is divisible by b iff a/b is an integer"? Or is it that "a is divisible by b iff there exists an integer c such that a = bc"?

    If it is the former, then "zero is divisible by zero" is neither true nor false -- it is meaningless. If it is the latter then zero is divisible by zero and no contradiction ensues since the definition does not involve division by zero.
     
  4. Feb 16, 2014 #3
    The book uses the latter version, a is divisible by b iff there exists an integer c such that a = bc.
     
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