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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.

Sunny22
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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. – Sunny22 Feb 27 '17 at 03:42
  • 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... – JMoravitz Feb 27 '17 at 03:43
  • I can work that around and say that a=ckl 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. – Sunny22 Feb 27 '17 at 03:52
  • 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 $k$. – JMoravitz Feb 27 '17 at 03:54
  • $k$ 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... – JMoravitz Feb 27 '17 at 03:58
  • so then my relation is in fact transitive then. – Sunny22 Feb 27 '17 at 03:59
  • 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. – Sunny22 Feb 27 '17 at 04:01
  • "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 – JMoravitz Feb 27 '17 at 04:07
  • 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. – Sunny22 Feb 27 '17 at 04:09
  • yeah I mean just reading it to see a is congruent to 0modb then it becomes a bit different correct? – Sunny22 Feb 27 '17 at 04:11
  • $a\equiv 0\pmod{b}$ 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 – JMoravitz Feb 27 '17 at 04:14
  • I meant to say a is congruent to bmod0. had a dyslexic moment. – Sunny22 Feb 27 '17 at 04:17
  • $a\equiv 0\pmod{b}$ is equivalent to $0\equiv a\pmod{b}$ – JMoravitz Feb 27 '17 at 04:17

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