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 bi's and simplify to get the equivalent (having same solution(s)) equation:
cnxn + cn1xn1 + cn2xn2 + ... + c1x + c0 = 0,
where the ci'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
2. Algebraic And Transcendental Numbers
Definition 2.1 Algebraic And Transcendental Numbers
3. Algebraic And Transcendental Functions
is constructed by using an operation that doesn't apply
directly or explicitly to polynomials in x
finitely many times. It can be
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,
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 non-identically 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.
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 non-identically 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
2. Remark that
Eq. [3.4] is similar to Eq. [1.1]: the variable x is replaced by the
and the constants ai's
by the polynomials pi(x)'s. As for the function sin x, now we see it's transcendental.
3. Some other examples of algebraic functions:
4. The right-hand
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
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
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
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.
Let y = |x|. Then:
y2 = |x|2 = x2,
y2 x2 = 0.
The function f (x) = |x| satisfies the algebraic equation y2 x2 = 0, and so is algebraic.
Theorem 3.1 Transcendency Of Non-0 Constant Multiples Of Transcendental Functions
For any non-0 constant c, a function f (x) is transcendental iff c f (x) is transcendental.
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.
Problems & Solutions
2. State whether the following functions are algebraic or transcendental.
3. Prove that the function:
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 sik-nem. The Latin
word signum means sign. The function is defined as follows:
Prove that its algebraic by finding an algebraic equation that it satisfies.
Let y = sgn x.
> 0 then y =
1, so y3 = 1 = y, thus y3 y = 0,
if x = 0 then y = 0, so y3 = 0 = y, thus y3 y = 0,
if x < 0 then y = 1, so y3 = 1 = y, thus y3 y = 0.
Consequently the function f (x) = sgn x satisfies the algebraic equation y3 y = 0, and so is algebraic.
5. Prove that if a function f (x) is transcendental, then 1 + f (x) is transcendental.
Substitute these expansions into Eq. . Factor all the like powers of f (x). Then we get an equation of the form:
qn(x) f n(x) + qn1(x) f n1(x) + qn2(x) f n2(x) + + q1(x) f (x) + q0(x) = 0,
where the qk(x)'s are polynomials and qn(x) = pn(x) for all x and thus qn(x) is non-identically 0,
that's satisfied by all x
dom( f ). This shows that f (x) is algebraic, a contradiction. Consequently, 1 + f (x) must be transcendental.