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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 $6$? –  Asaf Karagila Nov 3 '11 at 17:46
    
$3*(1/3)$ would give you $1$. –  Fahad Uddin Nov 3 '11 at 17:49
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multiplier or multiple ? Integer multiple ? –  Henry Nov 3 '11 at 17:58
    
multiplier. Its not mentioned about real or integer. –  Fahad Uddin Nov 3 '11 at 18:10
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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" and thirty hits for "multiple". –  joriki Nov 3 '11 at 21:34

1 Answer 1

up vote 1 down vote accepted

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.

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