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I need to negate the following sentence: "If for the integers $x, y, z$ we know that $x$ divides $y$ and $y$ divides $z$, then $x$ divides $z$."

In this scenario, what does it mean for $x$ to "divide" $y$, et cetera?

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    $\begingroup$ Welcome to Math.SE! Since we don't know where you got this question from, it is difficult to give you a definite answer. It would therefore be helpful if you provided the origin of your question. Also, please include what you yourself think is the meaning of division here. $\endgroup$ – Hrodelbert May 29 '15 at 8:20
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    $\begingroup$ Hello! And this question is from an assignment for my Discrete Structures class. The assignment can be found here: mathstat.dal.ca/~svenjah/math2112/Assign2.pdf And I understood that something can be divisible by something, but I never figured that this being divided is in the passive. I suppose it does make sense that the thing by which something is divided is the thing that divides that something. $\endgroup$ – ConJoJohn May 29 '15 at 8:23
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If $x$ divides $y$, $x$ is a factor of $y$, or in other words, $y$ is divisible by $x$.

Of course, the prerequisite is that $x,y\in{\Bbb{Z}}$.

For instance, $3$ divides $15$ because $15=3\cdot5$, $3$ is a factor of $15$.

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    $\begingroup$ The assumption should be that $x, y \in \mathbb{Z}$. $\endgroup$ – N. F. Taussig May 29 '15 at 10:09
  • $\begingroup$ Thx I've corrected it $\endgroup$ – Mythomorphic May 29 '15 at 10:25
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Saying that $x$ divides $y$ means that there exists an integer, say $q$ such that $y=q.x$ (e.g. $4$ divides $12$ with $q=3$). Your assertion is then: $y=q_1 . x$ and $z=q_2 . y$ then $z=(q_1.q_2).x$ and $x|z$ (read $x$ divides $z$ with $q=q_1.q_2$). Is that clear?

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  • $\begingroup$ Yes it is. Thanks. $\endgroup$ – ConJoJohn May 29 '15 at 8:21
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Does it really matter what is the meaning of "$x$ divides $y$"?

This is a binary relation, $D(x,y)$. The statement you need to negate is $\forall x\forall y\forall z(D(x,y)\land D(y,z)\rightarrow D(x,z))$, and negating this statement has nothing to do with its actual content.

In any case, for the integers, $x$ divides $y$ if there exists some integer $k$, such that $x\cdot k=y$. So for example $1$ divides every integer, and every integer divides $0$.

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  • $\begingroup$ I suppose you're right in that the actual problem doesn't really involve division. But I have a hard time getting past a failure to understand the components of a problem, even if the content of the components is unrelated to the problem itself. $\endgroup$ – ConJoJohn May 29 '15 at 8:21
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    $\begingroup$ Negating a proposition is a syntactic action. It means taking one string of characters and returning another string of characters. Meaning and semantics often get in the way. I do agree that if you want to use the meaning of "divides" for the negation, then it's important to know what is the meaning of that statement. But writing "does not divide" is equally valid in most cases. $\endgroup$ – Asaf Karagila May 29 '15 at 8:25
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    $\begingroup$ @CoJo It is possible that you were assigned this problem precisely so that you can learn to perform syntactical operations like "negate" without needing to understand what the thing you're trying to negate really means. It should be something you are able to do mechanically, without really thinking. $\endgroup$ – Najib Idrissi May 29 '15 at 10:13
  • $\begingroup$ @NajibIdrissi I suppose that's possible, but I assume that the problem was assigned without any thought as to whether or not we would necessarily understand what it means for something to divide something. Though you make a good point. $\endgroup$ – ConJoJohn May 29 '15 at 10:32
  • $\begingroup$ @SufyanNaeem Well, I think we've just agreed on it having nothing to do with the actual problem of negation. $\endgroup$ – ConJoJohn May 29 '15 at 10:34
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In terms of number theory, let $a,b\in\mathbb{Z}$. Then $a$ divides $b$ if $\exists m\in\mathbb{Z}$ such that $am=b$. This is written $a|b$. For example, $7$ is a divisor of $56$ because $7\times 8=56$. This is the algebraic/proper definition for two elements to divide each other. This satisfies the transitive relation \begin{equation*} a|b~\text{and}~b|c\Rightarrow a|c. \end{equation*} Does this help?

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  • $\begingroup$ Yes it does; thank you. $\endgroup$ – ConJoJohn May 29 '15 at 8:19
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x divides y means there exists an integer n such that nx=y. So 7 divides 28, since 4*7=28, but 8 does not divide 28, even though outside number theory we would happily deal with the number 3.5.

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