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I'm working on this basic discrete math question and struggling to understand the notation,

We have a relation $R$ on $Z^+$ defined as follows:

$mRn$ if and only if $m|n$

Explain why the relation R is not a function

Let $A$ and $B$ be nonempty sets. Then by definition, a relation $f$ from $A$ to $B$ is a "function" if each element $a\in A$ is related by $f$ to one and only one element of $b$.

Now the question provides the relation $m|n$, which provides us the information "$m$ divides $n$ if $n=km$ for some integer $k$".

How exactly do we related this simple statement to not being a function? I seems that this would be a function as $n=km$ for some integer $k$ appears to have one and only out output for each input.

mathcounterexamples.net
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Bobby B
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1 Answers1

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$2 | 6$ and $2 | 12$. Hence $2R6$ and $2R12$. Meaning that $2$ would have at least two images if $R$ was a function. A contradiction with the definition of a function.

mathcounterexamples.net
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  • So if we expand that relation...$2|6\rightarrow 6=k2\rightarrow k=3$ and $2|12\rightarrow 12=k2\rightarrow k=6$. How is this not a function? If we define $m=2$ then we get a different output $k$ for each $n$. This seems like a function to me as each chosen $n$ integer gives a different output. – Bobby B Dec 29 '20 at 21:35
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    You are not asked here to make any division. Just come back to the definition of what a function is: for any element of the domain you have one and only one image. $2R6$ and $2R12$ mean that $6$ and $12$ would be two different images. – mathcounterexamples.net Dec 29 '20 at 21:44
  • Yes but the question defines the relation $mRn$ as if and only if $m|n$. That conditional is the division of the two chosen integers. How can we simply state $2R6$ and $2R12$ without considering the fact that the relation is the division of two chosen integers? – Bobby B Dec 29 '20 at 21:51
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    Make it simple... Do you agree that $2R6$ and $2R12$? Yes or no? There is no third option... If yes what does it imply regarding the fact that $R$ is a function? – mathcounterexamples.net Dec 29 '20 at 21:54
  • But this isn't simple, because the relation $mRn$ is given a specific condition of existence entirely dependent upon divisibility. We're just supposed to ignore that $m|n$ always provides a one-to-one output? $2$ being capable of dividing two different integers seems exactly like a function to me. I don't understand where the logic is supposed to come from that this isn't a function. – Bobby B Dec 29 '20 at 22:12
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    You haven’t answered to the question I asked, which I think would help you, Therefore I feel of no need to follow on the discussion. – mathcounterexamples.net Dec 30 '20 at 06:28
  • You're completely ignoring my reasonable questions and refusing to offer a literal explanation of yours. I just don't think you have an answer that counters what I'm saying. – Bobby B Dec 30 '20 at 19:13