Talk:Binary function

I intend to split up the article into sperate sections; examples, definition, generalisations. The aim is to make it easier to read.

Any thoughts? TM-86 (talk) 03:17, 16 December 2011 (UTC)Reply

Changes to structure made: sections made: definition (non labelled), example - division, alternative definitions, restrictions to ordinary functions, generalisations, ternary and beyond, Category Theory.

I've also added a note to explain the meaning of Z^Y to readers under -restrictions to ordinary functions/currying.

I hope that these changes will make it more accessible for beginners. TM-86 (talk) 14:12, 27 December 2011 (UTC)Reply

"For example, if Z is the set of integers, N⁺ is the set of natural numbers (except for zero), and Q is the set of rational numbers, then division is a binary function from Z and N⁺ to Q." I contend that this is not strictly true - (or at least not helpful). The problem I see is that division can be though of as a function working on rational numbers (or even real numbers).

If we consider division d to be a function ${\displaystyle \,d\colon R\times R\rightarrow R}$  and then restrict the domain to ${\displaystyle Z\times N^{+}}$ , then we discover that the image of the function is the rational numbers. But that isn't a very helpful example to someone learning.

I would suggest that we change this example to something along the lines of as 'division of whole numbers can be thought of a function from z and N⁺ to Q.

I may be being overly pedantic, but it seems important to me. — Preceding unsigned comment added by TM-86 (talk • contribs) 02:35, 16 December 2011 (UTC)Reply

Is the following statement true?

A binary function always is a ternary relation.

(I would say it is.) --Abdull (talk) 09:22, 5 September 2010 (UTC)Reply

I'm not an expert but I suspect it's a Binary relation. My reasoning is that if it's mapping an ordered pair (the inputs) to some output. Then the set representing the relation should be a set of ordered pairs each containing an ordered pair and an "output" element: ((a,b), c). Paulmiko (talk) 16:08, 27 June 2011 (UTC)Reply

yes.

In one sense, it (and every function) is a binary relation - or rather functions can be seen as an ordered pair (u,v) within the space UXV, where the function g is g:U->V and U,V are arbitrary sets. So a binary function f could be seen as a collection of ordered pairs, where each pair is ((a,b),f(a,b)). Hence a binary relation.

BUT because one of the parts of this ordered pair is itself an ordered pair (namely the (a,b)) there is an obvious equivalence to the ordered tripple of (a, b, f(a,b)). So it can also be considered as a ternary relation.

What I am trying to say is, every set of ordered pairs (i.e. binary relation) where one 'part' is an ordered pair can be split up into an ordered tripple. This (binary function) is one such occasion when an binary relation can be thought of as a ternary relation.

It is not the case that every binary relation is a ternary relation (should be obvious). However, every ternary relation (or ordered tripple (a,b,c)) can be thought of as a binary relation (e.g. as ((a,b),c) or (a,(b,c)).

Because of this it is more helpful to normally consider any relation as the higher order relation available (in this case ternary).

It may be worth noting that in some cases you may want to think of a binary function as a function and thus a binary relation. Also in some cases it may be possible to consider it as a n-relation.

Hope that helps. TM-86 (talk) 02:10, 16 December 2011 (UTC)Reply