Besides these operations, which are all simply special cases of the operations on sets, it is also possible to define some special operations on binary relations which depend on the fact that they are sets of ordered couples. The converse of a relation R
is the relation such that, for all x and y, xRy if and only if yRx. Thus the converse of a relation is obtained simply by reversing the order of all the ordered couples which constitute it. The converse of the relation H which holds between x and y if and only if x is the husband of y, is the relation W which holds between y and x if and only if y is the wife of x. Thus,
Since a function is simply a special kind of relation, we can speak of the converse of a function. The converse of a function is always a relation, but in general it will not be a function. |(Suppes 234-235)
A a second example, suppose f is the function such that for every real number x,f(x)=x2if we consider f as a relation, <x, y>∈ f if and only ifx = y2.Consider, now, x = 4. Thenand also <4, –2>∈ f . Henceis not a function.(Suppes 235)
On the other hand, some functions are such that their converses are also functions. For example, if g is the function such that for every real number x,g(x) = 2x + 4.When a function f is such that its converse relationis also a function, then we say thatis the inverse of f. To mark this special situation we use ‘–1’ as a superscript in place of the more general cup notationThus if the converse of f is a function,We use the notation ‘–1’ only whenis a function.(Suppes 235)
g(x) = 2x + 4.(Suppes 235)
It should be noticed that when the inverse of f exists (i.e., the converse of f is a function), we have the following identities: For every x in the domain of f(I) f–1(f(x)) = x,and for every x in the range of f(II) f(f–1(x)) = x.Principles (I) and (II) are both important in solving for f and f–1 as we shall soon see.
At the end of the last section we spoke of many-one and one-many relations. A relation which is both many-one and one-many is one-one.(Suppes 235)
We also speak of a one-one relation as a one-one function. Consider now the following statements about a relation R:
(1) R is a function.(2) R is a many-one relation.(3) The converse of R is a function.(4) R is a one-many relation.(5) R and its converse are both functions.(6) R is a function and its inverse exists.(7) R is a many-one relation and R is a one-many relation.(8) R is a one-one relation.(9) R is a one-one function.(Suppes 236)
The following equivalences hold:
(1) ↔ (2),
(3) ↔ (4),
(5) ↔ (6) ↔ (7) ↔ (8) ↔ (9).
(1) f(x) = 5x – 4
Since (1) holds for every real number x, we may substitute ‘f–1(x)’ for ‘x’ (application of universal specification) and obtain:(2) f(f–1(x)) = 5f–1(x) – 4.
Rule of Universal Specification: US. If a formula S results from a formula R by substituting a term t for every free occurrence of a variable v in R then S is derivable from (v)R.In the above derivation, the term t is simply ‘x’ in both cases of universal specification, which is why we spoke of dropping quantifiers, but in general | we want to be able to replace the quantified variable by any term. Thus if ‘a’ is the name of John Quincy Adams, we may infer ‘Aa → Ma’, as well as ‘Ay → My’, from ‘(x) (Ax → Mx)’. And in the case of arithmetic from the statement ‘(x)(x + 0 = x)’ we may infer by universal specification ‘(x + y) + 0 = x + y’; in this case the term t is ‘x + y’.(Suppes 59-60)
(1) f(x) = 5x – 4(2) f(f–1(x)) = 5f–1(x) – 4.
(II) f(f–1(x)) = x.
(2) f(f–1(x)) = 5f–1(x) – 4(3) x = 5f–1(x) – 4(Suppes 236)
x = 5f–1(x) – 4
x –5f–1(x) = – 4
–5f–1(x) =–x – 4
5f–1(x) =x + 4
(5) f(x) = x2 – 1(Suppes 237)
f(f–1(x)) = f–1(x)2 – 1
(II) f(f–1(x)) = x.
(6) x = f–1(x)2 – 1
–f–1(x)2 + x = – 1
–f–1(x)2 = –x – 1
f–1(x)2 = x + 1
We thus infer thatf–1(3) = 2 and f–1(3) = –2and hence2 = –2which is absurd.(Suppes 237)
The mistake here was in inferring (6) from (5). The substitution of ‘f–1(x)’ for ‘x’ is permissible only if ‘f–1(x)’ is a proper term. If f is not a function, the expression ‘f–1(x)’ is not a term, since it does not designate a unique entity.(Suppes 237)
If R and S are binary relations, then by the relative product of R and S (in symbols: R/S) we mean the relation which holds between x and y if and only if there exists a z such that R holds between x and z, and S holds between z and y. Symbolically,x R/S y ↔ (∃z) (xRz & zSy).If xPy means that x is a parent of y, and xSy means that x is a sister of y, then x(S/P)y means that there is a z such that x is a sister of z and z is a parent of y, and hence such that x is an aunt of y.(Suppes 226)
The relative product of two functions plays an important role in many discussions and is sometimes given a special name: the composition of the functions. Suppose that f and g are functions, and that x, y, and z are entities such that(7) y = f(z)and(8) z = g(x);then x stands to y in the relation g/f. In other terms, from equations (7) and (8) we see, thaty = f(g(x)).Thus the relative product, or composition, of two functions is a function h such that (whenever x is in the domain of definition of g, and g(x) is in the domain of definition of f)h(x) f(g(x)).This operation of forming the composition of two functions is so extensively used in certain branches of mathematics that various special symbols have been used for it; we shall use a small circle ‘∘’. Thus, (f ∘ g)(x) = f(g(x)). We introduce this special notation instead of using the relative product notation because the order of ‘f’ and ‘g’ in ‘f ∘ g’ is the natural one corre- | sponding to their order in ‘f(g(x))’, whereas this order is reversed in the relative product:f ∘ g = g/f.It is clear that composition of functions is associative but not commutative.(Suppes 238-239)
The problem of finding the inverse of a function is actually a special case of the problem of finding “the” function g, given the functions f and f ∘ g. In finding the inverse, g is f–1, and(f ∘ g)(x) = (f ∘f–1)(x) = f (f–1(x)) = x.(Suppes 239)