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1. Rational And Irrational Numbers 
Of course a rational number m/n satisfies the equation:
nx m = 0,
can be converted to a polynomial equation with integer
coefficients by multiplying both sides by the least common multiple of
the denominators b_{i}'s
and simplify to get the equivalent (having same solution(s)) equation:
c_{n}x^{n} + c_{n}_{1}x^{n}^{1} + c_{n}_{2}x^{n}^{2} + ... + c_{1}x + c_{0}_{ } = 0,
where the c_{i}'s
are integers. So there's no loss of generality that encompasses the rational
coefficients to consider equations
with integer coefficients instead of equations with rational coefficients.
Definition 1.1 Rational And Irrational Numbers

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2. Algebraic And Transcendental Numbers 
Definition 2.1 Algebraic And Transcendental Numbers

Remarks 2.1
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3. Algebraic And Transcendental Functions 
The function:
is constructed by using an operation that doesn't apply
directly or explicitly to polynomials in x
finitely many times. It can be
shown that:
which involves the additions of infinitely many terms containing increasing odd
powers of x
with alternating signs. It's
demonstrated in Section
15.3 Part 3. Note that sin
x of course depends on x, that's why it's a function of x, but not directly in a
finite manner, instead in an infinite manner. An approximate value of sin
x can be calculated by
using the sum of the first k
terms in Eq. [3.3], where k
is a positive integer. The larger the integer k, the better the approximation.
A function f (x) is said to be identically 0 if
it's always 0, ie, if f (x) = 0 for all x in dom( f ). Thus a polynomial p(x) is
identically 0 if all of its coefficients are 0. We see that a function f (x) isn't identically 0 if there's at
least some x
where f (x)
isn't 0. Thus a polynomial p(x) isn't identically 0
if at least one of its coefficients isn't 0.
An algebraic function of x is one that can be constructed by using
finitely many algebraic operations of addition,
subtraction,
multiplication, division, or raising to constant powers applied to polynomials
in x
(the nth
root of an expression is that
expression raised to the constant power of 1/n, so extraction of roots is included in
raising to constant powers). It's a
polynomial in x
or a finite combination of algebraic operations applied to polynomials in x. Now, as suggested
by the fact that
function [3.1] satisfies Eq. [3.2], a function f (x)
can be constructed by using finitely
many algebraic operations iff it satisfies a
polynomial equation in function y
with polynomial coefficients and nonidentically 0 leading coefficient. The
term "algebraic"
means finitely many normal
algebraic operations, in the sense that the values of the function may be calculated
algebraically in
a finite manner for all the values of the variable.
A
function f (x) is said to be a transcendental
function if it's not algebraic, ie, if it can't
be constructed by using finitely many
algebraic operations of addition, subtraction, multiplication, division, or
raising to constant powers applied to polynomials in x,
ie, if it's a function that doesn't satisfy any polynomial equation in function
y with polynomial
coefficients and nonidentically 0
leading coefficient. The term "transcendental" doesn't imply anything
deep or mysterious. It simply means transcending the
normal algebraic operations, in the sense that the values of the function may
not be calculated algebraically in a finite manner
for some or all of the values of the variable.
Definition 3.1 Algebraic And Transcendental Functions

Remarks 3.1
2. Remark that
Eq. [3.4] is similar to Eq. [1.1]: the variable x is replaced by the
function y
and the constants a_{i}'s
are replaced
by the polynomials p_{i}(x)'s. As for the
function sin x,
now we see it's transcendental.
3. Some other examples of algebraic functions:
4. The righthand
side of Eq. [3.3] is the infinite expansion of the transcendental function sin x.
Many algebraic functions also
each have an infinite expansion. So
if a function has an infinite expansion, that doesn't mean that it has to be
transcendental.
The definition of algebraic numbers is analogous to that of
algebraic functions: algebraic numbers are those that satisfy
polynomial equations in variable x
with integer coefficients, algebraic functions are those that satisfy
polynomial equations in
function y
with polynomial coefficients.
The definition of transcendental numbers is analogous to that
of transcendental functions: transcendental numbers are those
that aren't algebraic numbers, transcendental functions are those that aren't
algebraic functions.
A rational number is a number that is or can be written as a
ratio m/n, where m and n are integers and n > 0. Recall that
a
rational function is a function that is or can be written as a ratio p(x)/q(x), where p(x) and q(x)
are polynomials and q(x) isn't
identically 0.
The status of the polynomials in the set of all functions is analogous to that of the integers in the set of all numbers.
Prove that the function f (x) = x is algebraic.
Solution
Let y =
x. Then:
y^{2} = x^{2} = x^{2},
y^{2} x^{2} = 0.
The function f (x) = x satisfies the algebraic equation y^{2} x^{2} = 0, and so is algebraic.
EOS
Theorem 3.1 Transcendency Of Non0 Constant Multiples Of Transcendental Functions
For any non0 constant c, a function f (x) is transcendental iff c f (x) is transcendental. 
Note
Of course 0 f (x) = 0 = 0x for all x in dom( f ), so 0f (x) is an algebraic function, whether f (x) is algebraic or transcendental.
Proof
EOP
Problems & Solutions 
Let:
2. State whether the following functions are algebraic or transcendental.
Solution
a. Algebraic.
b. Transcendental.
c. Algebraic.
3. Prove that the function:
is algebraic.
Solution
Let:
The function f (x) satisfies the last algebraic equation with polynomial coefficients, and so is algebraic.
4. Consider the function f (x)
= sgn x,
where sgn stands for
the Latin word signum and is pronounced siknem. The Latin
word signum means sign. The
function is defined as follows:
Prove that its algebraic by finding an algebraic equation that it satisfies.
Solution
Let y = sgn x.
If x
> 0 then y =
1, so y^{3} = 1 = y, thus y^{3} y = 0,
if x =
0 then y =
0, so y^{3} = 0 = y, thus y^{3} y = 0,
if x
< 0 then y =
1, so y^{3} = 1 = y, thus y^{3} y = 0.
Consequently the function f (x) = sgn x satisfies the algebraic equation y^{3} y = 0, and so is algebraic.
5. Prove that if a function f (x) is transcendental, then 1 + f (x) is transcendental.
Solution
Substitute these expansions into Eq. [1]. Factor all the like powers of f (x). Then we get an equation of the form:
q_{n}(x) f ^{n}(x) + q_{n}_{1}(x) f ^{n}^{1}(x) + q_{n}_{2}(x) f ^{n}^{2}(x) + + q_{1}(x) f (x) + q_{0}(x) = 0,
where the q_{k}(x)'s are polynomials and q_{n}(x) = p_{n}(x) for all x and thus q_{n}(x) is nonidentically 0,
that's satisfied by all x
in
dom( f
). This shows that f (x)
is algebraic, a contradiction. Consequently, 1 + f (x) must be
transcendental.
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