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I know that the question doesn't really match the level of this site, but I will be very grateful if someone showed me the proof:

$\forall x,y,z \in \mathbb{Z}$ if $xz \hspace{4 pt} \vdots \hspace{4 pt} (z - y)$ then $xy \hspace{4 pt} \vdots \hspace{4 pt} (z - y)$,

When $x \hspace{4 pt} \vdots \hspace{4 pt} (z - y)$, it's obvious that for any $y \in \mathbb{Z}$ this holds. But how do I prove for all the other cases?

Thank you!

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    I think that means x is divisible by y. In other words, there exists $c \in \mathbb{Z}$, such that $x = y\cdot c$.2012-04-07
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    Your question is not stated quite right. $\forall x,y,z \in \Bbb{Z}$ must not be written after the implication, but before.2012-04-07
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    Sorry, I didn't know that was important. I corrected it.2012-04-07
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    @Beni: that’s not really an issue — in formal syntax, a quantifier is indeed usually put before the formular it quantifies, but in prose, it’s often put afterward. The thing that it’s really important to be unambiguous about is the extent of the formula being quantified — but in this question, that’s unambiguous anyway, so it’s not a problem.2012-04-07
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    Your question *does* match the level of this site!2012-04-07

2 Answers 2

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In the notations I know, $x \vdots y$ is equivalent to $y |x$. I'm not sure what you mean, but if this is it, then the proof is obvious:

$y-z$ divides $xz=xy-(y-z)x$

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    LaTeXspeak for $|$ is `\mid`, which gets the spacing correct.2012-04-07
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    I see now: $xz = xy - (y - z)x \hspace{2 pt} \vdots \hspace{2 pt}(y - z) \implies \exists c \in\mathbb{Z} : xy - (y - z)x \hspace{2 pt} = \hspace{2 pt}(y - z)c \implies xy = (y - z)c + (y-z)x = (y-z)(c+x) = -(z - y)(c+x) \implies xy \hspace{2 pt} \vdots \hspace{2 pt} (z-y)$. Thank you! You helped me so much!. Definitely accepted!2012-04-07
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Hint $\rm\ \ x\:(z-y)\: =\: xz - xy\:\ \Rightarrow\:\ z-y\ |\ xz\!\iff\! z-y\ |\ xy$

i.e. $\rm\ mod\ z-y\!:\:\ z\equiv y\ \: \Rightarrow\:\ xz\equiv xy\:\ \Rightarrow\:\ xz\equiv 0\!\iff\! xy\equiv 0$