# Must a binary relation on a set $X$ be defined for all elements in $X$?

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..)

EDIT:

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.)

• 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? – Gus Kenny May 23 at 3:54
• 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 $\frac{1}{x-y} \land x \neq y$? – Gus Kenny May 23 at 4:02
• by "well-defined" above, I meant in the sense of "well-travelled", hehe – Gus Kenny May 23 at 4:09
• @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 $\frac1{x-y}$. – Brian M. Scott May 23 at 4:24
• @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 $\frac12$. – Brian M. Scott May 23 at 5:07