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I was working on a problem asking me whether on not the relation is symmetric, reflexive, or transitive? The defined relation R on N by aRb if a/b is an element of N. I found that it is reflexive. Not transitive and not symmetric. The counter example for symmetry was easy but I am getting caught off by finding numbers to prove a/b, b/c does not equal a/c some further explanation on why I can't find a counter example would help.

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    if it helps I already know my definitions of what it means for a relation to be reflexive transitive and symmetric. – 2017-02-27
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    So, $a/b\in \Bbb N$ is another way of saying $b\mid a$ is another way of saying $a=bk$ for some $k\in\Bbb Z$. So... if $a\mathcal{R}b$ and $b\mathcal{R}c$ then there exists an integer $k$ such that $a=bk$ and there exists an integer $l$ such that $b=cl$ which means... – 2017-02-27
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    I can work that around and say that a=c*k*l which is an integer, since k*l is an integer. oh my gosh since they are integers they can not be part of the natural numbers for sure. Oh my gosh thank you so much. I was defining a,b,c to be elements of the natural numbers when I should've defined them to be apart of Z. – 2017-02-27
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    Well... the natural numbers are a subset of the integers... all natural numbers are integers too. Since $a$ and $b$ are both natural numbers, i.e. positive integers (*unless you count zero too in which case what do you do about $0\mathcal{R}0$?*) you can note that a positive number divided by a positive number must also be positive, so for this specific case you can note that since $\frac{a}{b}\in \Bbb N$ that $a=bk$ for some *natural number* $k$. – 2017-02-27
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    $k$ is not only an integer, it is a natural number as well. For example the number $5$ is all of the following: a natural number, an integer, a rational number, a real number, a complex number, a quaternion, etc... – 2017-02-27
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    so then my relation is in fact transitive then. – 2017-02-27
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    but only for the way it is currently defined. If the definition for the relation had changed then it would be difficult to show transitivity. – 2017-02-27
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    "*if the definition had changed*"... well yes of course if we don't know how the relation is defined we can't say anything about the properties of the relation. If you are worried about how I seemingly reworded the relation you are working with, thats okay since they are equivalent definitions – 2017-02-27
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    thank you again this is useful insight I was considering the relation to fail both transitivity and symmetry but it works out. The way explained it. – 2017-02-27
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    yeah I mean just reading it to see a is congruent to 0modb then it becomes a bit different correct? – 2017-02-27
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    $a\equiv 0\pmod{b}$ is equivalent to $b\mid a$ is equivalent to $\frac{a}{b}\in\Bbb N$ is equivalent to... so no, it is exactly the same problem – 2017-02-27
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    I meant to say a is congruent to bmod0. had a dyslexic moment. – 2017-02-27
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    $a\equiv 0\pmod{b}$ is equivalent to $0\equiv a\pmod{b}$ – 2017-02-27

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