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 INTERMEDIATE ALGEBRA
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 Section 12
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 REVIEW OF A FIRST COURSE IN LINEAR ALGEBRA
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 Some Proofs of the Existence of Irrational Numbers
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 Complex Numbers,Complex Functions and Contour Integrals
 APPLICATIONS OF LINEAR EQUATIONS
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 Fractions
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 Intermediate Algebra
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 ON SOLUTIONS OF LINEAR EQUATIONS
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 Fractions
 Fabulous Fractions
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 linear Equations
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 PreCalculus 1
 Compositions and Inverses of Functions
Functions
Equal Functions
Two functions f and g are equal if they have the same domain D, and f(d) = g(d) for all d in D. 
Inverse Functions
Let f be a onetoone correspondence from the set
A to the set B. The inverse function of f is the function that assigns to an element b belonging to B the unique element a in A such that f(a) = b. The inverse function of f is denoted f ^{1}. Thus, f ^{1}(b) = a when f(a) = b, so f ^{1}(f(a)) = a. 
As we observed, the function f(x) = x + 1 from the set of
integers to the set of integers is a
bijection. To reverse the correspondence, let y = f(x) = x + 1. Thus, we know
that x = y – 1.
Consequently, f ^{1}(y) = y – 1.
Note that not all functions have an inverse. A onetoone correspondence is
called invertible since
we can define an inverse of this function. However, a function is not
invertible
if it is not a oneto
one correspondence, since the inverse of such a function does not exist. For
example, the
function f(x) = x^{2 }from the set of integers to the set of integers is
not invertible since it is not oneto
one. Since f(1) = f(1) = 1, the inverse of f would have to assign two elements
to 1, which
violates the definition of a function. Hence, an inverse of f cannot exist in
this case.
Composition of Functions
Consider two functions: successor as applied to integers, and a sqr (“square”)
function. If we take
the output of the successor function and use it as input to the sqr function,
the two functions are
operating together. They take as input an integer n, add one to it and then
square n+1. We call this
a composition of two functions.
Let g be a function from the set A to the set B
and let f be a function from the set D to the set C (where B is a subset of D). The composition of f and g, denoted by (f o g), is a function defined by: (f o g)(a) = f(g(a)) where a is an element of A. f o g may also be read "f circle g". 
Here is a diagram to illustrate (in this case B = D):
So (f o g)(1) = y, (f o g)(2) = y, and (f o g)(3) = z.
always true that g(f(a)) = f(g(a)). Indeed, for a simple
example of this, consider the case where g
has the set A as a domain and g has the set D as a domain, and the sets A and D
contain no
elements in common. Then, while f(g(a)) may be defined, g(f(a)) is undefined,
since f does not
operate on element a. In the diagram above, we cannot form g o f, because the
codomain of f is
not a subset of the domain of g. f always produces a result in the set {x, y,
z}, and g does not
operate on these values.
Notice also that in the successorsqr example, the composition is not
commutative since (n + 1)^{2} ≠
n^{2} + 1 (except for n = 0).
As an additional example, let f and g be functions from the real numbers to the
real numbers
defined as follows:
f(x) = 2x + 3
g(x) = 3x + 2
The composition of f and g is: (f o g)(x) = f(g(x)) = f(3x
+ 2) = 2(3x + 2) + 3 = 6x + 7
The composition of g and f is: (g o f)(x) = g(f(x)) = g(2x + 3) = 3(2x + 3) + 2
= 6x + 11
Thus both compositions are defined, but (f o g)(x) ≠ (g o f)(x).
A Few Important Functions
Now that we have all the definitions out of the way, there are a few functions
that appear on
occasion in computer science that are worth specifically identifying.
Identity Function
Let I be a function from the set A to set A
defined by I(a) = a for all a in A. The function I is called the identity function. 
Floor Function
The floor function assigns to the real
number x the largest integer that is less than or equal to x. The floor function applied to x is denoted . 
Ceiling Function
The ceiling function assigns to the real
number x the smallest integer that is greater than or equal to x. The floor function applied to x is denoted . 
Examples:
Proofs with Functions
Here are some examples of proofs with functions.
Problem: Prove: if f is a function from set A to set B that is a
onetoonecorrespondence, then the
inverse function f ^{1} is onetoone and onto (i.e., it is also a
onetoone correspondence).
Proof: First we show f ^{1} is onto. If a is in A, then since f
is a function, there is a b in B such that
f(a) = b. Hence, by definition of inverse function, f ^{1} (b) = a,
where b is in the domain of f ^{1} and a
is in the codomain of f ^{1}. Note that for any element a in A, f ^{
1} (b) = a must hold for some b in B
in order for f ^{1} to be the inverse function of f. As a result, f ^{
1} is onto (the set A).
Now, we show f ^{1} is onetoone. Assume that some b' and b exist in B
such that f ^{1} (b) = f ^{1} (b') =
a. If this is true, then f(a) = b and f(a) = b'. But since f is a function, it
implies that b = b'
(otherwise, f would violate the definition of a function). Since b = b' in all
such cases, it follows
that f ^{1} (b) = f ^{1} (b') implies that b = b'. Thus, f ^{
1} is onetoone (by the definition of onetoone).
QED.
Problem: Let g be an invertible function from A to
B and f be an invertible function from B to C.
Prove or disprove: (f o g) ^{1} = g ^{1} o f ^{1}.
Proof: First, note that g and f are bijections, that (f o g) is a
bijection from A to C, and thus
(f o g) ^{1} exists and is a bijection from C to A. (Proofs of these
facts are left to the reader.)
Consequently, by the definition of a function, for any c in C, there must be
some a in A such that
(f o g) ^{1}(c) = a. We now use this as a basis to prove that (f o g) ^{
1}
= g ^{1} o f ^{1}
.
1. (f o g) ^{1}(c) = a implies (f o g)(a) = c  Definition of inverse function applied to (f o g) 
2. (f o g)(a) = c implies f(g(a)) = c, meaning
that there exists b in B such that g(a) = b and f(b) = c 
Definition of composition 
3. g(a) = b implies g ^{1}(b) = a and f(b) = c implies f ^{1}(c) = b 
Definition of inverse function applied to g and f 
Definition of composition  
Substituting f ^{1}(c) = b and g ^{1}(b) = a from step 3 

6. Thus,  Combining steps 1, 4 and 5 
Hence, we have shown that (f o g) ^{1}(c) = g ^{
1} o f ^{1}(c) for any arbitrary c in C, where C is the
domain of (f o g) ^{1}. Thus, it follows that this equality must hold
generally for all elements of the
domain of (f o g) ^{1}, so (f o g) ^{1} = g ^{1} o f ^{
1}.
QED.
Problem: Prove or disprove that
Proof: This statement is actually false. Consider the counterexample
where x = ½ and y = ½.
Here, , whereas
Bibliography
For additional information on functions, consult Discrete Math textbooks or
Calculus
textbooks. The following are excellent references:
S. Epp, Discrete Mathematics with Applications, Belmont, CA: Wadsworth, 1990.
R. McEliece, R. Ash, C. Ash, Introduction to Discrete Mathematics, New York:
Random
House, 1989.
K. Rosen, Discrete Mathematics and its Applications 6th Ed., New York:
McGrawHill, 2007.
Historical Notes
The term "function" and the form (x, f(x)) is attributed to Leibniz who used it
to refer to
several of his mathematical formulas, but the concept of a function was first
presented by René
Descartes (15961650) in his Geometry of 1637. In this book, he presents the
concept of
defining an equation in terms of one variable (i.e., as a function of...), and
how to graph
equations in the Cartesian plane. The formal definition of a function (as a
mapping from
domain to target) was first formulated for sets of numbers by Lejeune Dirichlet
(18051859) in
1837.