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The consequence of divisibility definition in integer 
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#1
Feb1514, 07:22 PM

P: 255

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 nono 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
Feb1614, 06:11 AM

P: 965

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. 


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