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Here is a question,

A = {1,2,3,4,6} = B, $aRb$ iff $a$ is a multiplier of $b$ .

Now I think the whole cartesian product of AxB should be the relation as every number is somehow a multiplier of another. Please help me out by sharing your review. Thanks

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    How does $3$ and $4$ stand in the relation? Or $4$ and $6$?2011-11-03
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    $3*(1/3)$ would give you $1$.2011-11-03
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    multiplier or multiple ? Integer multiple ?2011-11-03
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    multiplier. Its not mentioned about real or integer.2011-11-03
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    Seems like a translation error or the like. As you say, every number can be *used* as a multiplier for another number; this is not a static relation in which the numbers stand independent of context. One can only guess that the intended meaning was "multiple" in the sense of "integer multiple".2011-11-03
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    @joriki: The exercise is from a very popular book by Kolman,Busby and Ross. Any idea?2011-11-03
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    @Akito: Any idea about what? Why don't you tell us the title of the book? If you're referring to the book *Discrete Mathematical Structures*, a Google Books search of that book gives [zero hits for "multiplier"](http://books.google.com/books?id=7ccZAQAAIAAJ&dq=%22discrete+mathematical+structures&q=multiplier#search_anchor) and [thirty hits for "multiple"](http://books.google.com/books?id=7ccZAQAAIAAJ&dq=%22discrete+mathematical+structures&q=multiple#search_anchor).2011-11-03

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I have Kolman and Busby, Discrete Mathematical Strucxtures in Computer Science. Problem 9 on page 100 is

$A=\lbrace\,1,2,3,4,6\,\rbrace=B$; $a\,R\,b$ if and only if $a$ is a multiple of $b$.

I'm sure that what is intended is $a$ is an integer multiple of $b$. So for example $(6,3)$ will be in the relation, but $(3,6)$ won't be.