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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

As Ive noted, a function can be formally defined as an ordered triple consisting of its domain, codomain, and graph: http://www.mathsisfun.com/forum/viewtop … 465#p74465. In fact, functions are special cases of the more general notion of binary relations.

A binary relation from *X* to *Y* is an ordered triple (*X*,*Y*,*R*), where *R* is a subset of *X*×*Y*. *X*, *Y* and *R* are called the domain, codomain and graph of the binary relation respectively although if it is clear what *X* and *Y* are, it is much more usual to refer to the relation as just *R* rather than (*X*,*Y*,*R*). Also, for any *x* ∈ *X*, *y* ∈ *Y*, if (*x*,*y*) ∈ *R*, it is much more usual to write *x*R*y* instead of (*x*,*y*) ∈ *R*.

Now consider the following four conditions which the binary relation *R* may satisfy:

(i) for any *x* ∈ *X*, there exists *y* ∈ *Y* such that *x*R*y*

(ii) for any *y* ∈ *Y*, there exists *x* ∈ *X* such that *x*R*y*

(iii) if *x*R*y* and *x*′R*y*′, *x* = *x*′ ⇒ *y* = *y*′

(iv) if *x*R*y* and *x*′R*y*′, *y* = *y*′ ⇒ *x* = *x*′

It can be seen that a function is a binary relation that satisfies (i) and (iii). If a binary relation satisfies (i), (ii) and (iii), it is a surjective function, while a binary relation that satisfies (i), (iii) and (iv) is an injective function. A bijection is a binary relation that satisfies all four conditions above.

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

**Inverse binary relation**

Let *R* be a binary relation from *X* to *Y*. The inverse of *R*, denoted *R*[sup]−1[/sup], is the binary relation from *Y* to *X* such that for any *x* ∈ *X* and *y* ∈ *Y*, *y*R[sup]−1[/sup]*x* if and only if *x*R*y*.

It is obvious that

(a) *R* satisfies condition (i) above if and only if *R*[sup]−1[/sup] satisfies (ii)

(b) *R* satisfies (ii) if and only if *R*[sup]−1[/sup] satisfies (i)

(c) *R* satisfies (iii) if and only if *R*[sup]−1[/sup] satisfies (iv)

(d) *R* satisfies (iv) if and only if *R*[sup]−1[/sup] satisfies (iii)

(e) (*R*[sup]−1[/sup])[sup]−1[/sup] = *R*

In general, if is a function, the inverse [sup]−1[/sup] is not a function, only a binary relation. [sup]−1[/sup] is a function if and only if is bijective (in which case [sup]−1[/sup] itself is also bijective).

*Last edited by JaneFairfax (2007-06-22 07:39:12)*

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Something which I personally find fun is to try to invent sets and binary operators and then investigate to see if they have any cool properties. Here is one I worked on a little, and always meant to get back to, but never have:

Rationals mod n.

Let a and b be integers, b non-zero. Note that it need not be (a, b) = 1. Define (a, b) + (c, d) as:

(ad + cb (mod n), bd (mod n) ) which would be the equivalent of:

And multiplication similarly. See if you can find any properties such as abelian, commutative, identity, heck, even well defined.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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