When defining a binary relation $R \subseteq X^2$, does there have to be a definite "true" or "false" value for a pair $(x,y) \in R$, or does it only have to be "true" to be included, and excluded otherwise?

For example, can I define the relation on $\mathbb{Z}$, $(x,y)\in R$ if and only if $x|y$? Even though $(0,y)$ is undefined for any value $y \in \mathbb{Z}$ (except $0$, depending on which text book you read..)


changed a typo from "$(x,0)$ is undefined..." to "$(0,y)$ is undefined..."


Either $x$ divides $y$, or it does not; whether $x\mid y$ (and hence whether $\langle x,y\rangle\in R$) is not undefined for any $x,y\in\Bbb Z$, though whether $\langle 0,0\rangle\in R$ may depend on your conventions. In particular, $\langle n,0\rangle\in R$ for every $n$ except possibly $0$, and $\langle 0,n\rangle\notin R$ for any $n$.

More generally, a binary relation $R$ on $X$ is simply a well-defined subset of $X\times X$. We may not always be able to tell whether some particular pair $\langle x,y\rangle$ is in $R$, but it either is or is not; there is no fuzzy in-between state. For instance, let

$$R=\{\langle n,d\rangle:\text{the }n\text{-th digit of the decimal expansion of }\pi\text{ is }d\}\;.$$

For sufficiently large values of $n$ we don’t know whether $\langle n,1\rangle\in R$, but we do know that this question has a definite answer.

(I suppose that I should mention that there actually is a notion of fuzzy sets with a corresponding notion of degree of membership, but I assume that we are talking here about the usual notion of binary relation.)

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  • $\begingroup$ Yep, just the usual notion. So because $n|0$ is not defined for any $n \in \mathbb{Z}$, then $(n,0) \notin R$, because it definitely cannot be in $R$, it doesnt matter whether it can be computed or not? Is it considered sloppy to have cases like this? or is it perfectly fine? $\endgroup$ – Gus Kenny May 23 at 3:54
  • $\begingroup$ If we use a less well-defined relation than $x|y$, say $(x,y)\in R$ iff $\frac{1}{x-y}$. Can we still apply this as a relation over $\mathbb{Z}$? or do we have to say $(x,y)\in R$ iff $\frac{1}{x-y} \land x \neq y$? $\endgroup$ – Gus Kenny May 23 at 4:02
  • $\begingroup$ by "well-defined" above, I meant in the sense of "well-travelled", hehe $\endgroup$ – Gus Kenny May 23 at 4:09
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    $\begingroup$ @guskenny83: No, $n\mid 0$ is true for every non-zero integer $n$, and $0\mid n$ is false for every non-zero integer $n$. Whether $0\mid 0$ is a matter of convention: some declare it true, and others declare it false. But whichever convention is used, there is no doubt about which pairs $\langle m,n\rangle$ are in the ‘divides’ relation. As for $\frac1{x-y}$, that isn’t a statement about anything, so it’s meaningless to say that $\langle x,y\rangle\in R$ iff $\frac1{x-y}$. $\endgroup$ – Brian M. Scott May 23 at 4:24
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    $\begingroup$ @guskenny83: Sure; you simply have then that for all integers $x$, $\langle x,x\rangle\notin R$. When $\frac1{x-y}$ is not defined, then it’s certainly not less than $\frac12$. $\endgroup$ – Brian M. Scott May 23 at 5:07

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