Definitions of algebraic and transcendental numbers
Motivation
of the definition of the term
algebraic number
. The numbers
,
, and
are, in one sense, quite different:

the first is a rational number

the second is a real irrational number, and

the third is (clearly) a complex number (with nonzero 'imaginary' part)
>
evalc(1/2 + sqrt(1  sqrt(2)));
>
I^2; # in Maple 'I' is sqrt(1)
>
But in another sense they are not so different: all three are solutions of equations of the
same kind
:

is a solution of the equation
, where
a
and
b
are integers:

is a solution of the equation
, where
a
,
b
,
c
and
d
are integers:

is a solution of the equation
, where
a
,
b, c
,
d
and
e
are integers:
>
solve(8*x^4  16*x^3 + 8*x^2  1 = 0);
>
evalc(1/2 + sqrt(1sqrt(2))/2);
>
DEFINITION
. A real or complex number
is said to be
algebraic
('over the
integers
') if
is a solution of
some
polynomial equation
, where the
are integers,
.
REMARK
on that
DEFINITION.
You can sometimes see an algebraic number defined as follows: a real or complex number
is said to be
algebraic
('over the
rationals
') if
is a solution of
some
polynomial equation
, where the
are rationals,
.
One should appreciate that it doesn't matter which of these definitions is adopted as the basis for defining the term
algebraic
, for  trivially  a number that is algebraic according to one of these definitions is automatically algebraic according to the other definition, and vice versa. (For example,
x
=
is a solution of the equation
, and so
is algebraic according to the latter definition. But clearly that same
is a solution of the equation
(the earlier one multiplied by the product of the denominators of the rational coefficients) :
>
solve(3/4*x^2  4*x  5/3 = 0);
>
solve(9*x^2  48*x  20 = 0);
>
DEFINITION
. An algebraic number
is said to be of
degree
n
if
is a solution of
some
polynomial equation
, where the
are integers (
), but
is not a solution of any polynomial equation (with integer coefficients, not all zero) of smaller degree.
Example 1.
is clearly an algebraic number of degree 1.
Example 2.
is clearly an algebraic number (set
, then
, and thus
,
, and so
is a solution of the equation
.
Also,
, being irrational, cannot be of the first degree, and thus is of degree 2.
Example 3.
is clearly an algebraic number (but do not jump to the hasty conclusion that it is of degree 2 on the flimsy grounds that
and
are both algebraic, and of the second degree): set
, then
=
,
= 24, giving
, and finally
. Thus
is a solution of the fourth degree equation
.
However, it is not that
is of degree 4 simply because of the fact that the equation is of degree 4. After all, if one set
, then
= 25, giving
, and so
is a solution of the fourth degree equation
. But, the polynomial
happens to factor as the product of two polynomials with integer coefficients, of smaller degree:
, and it's the
that is the fundamental polynomial associated with
.
Without going any further, let me simply remark that
doesn't factor as the product of two polynomials in a similar fashion (and thus
is of degree 4):
>
factor(x^4  25);
>
factor(x^4  10*x^2 + 1);
>
Elementary properties of algebraic numbers (which are easily proved).
1. If
and
are algebraic numbers then so too are
and
(
)
2. If
is algebraic then so also is every one of
__________
Concerning Hardy & Wright
's
It is not immediately obvious that there are any transcendental numbers
...
Why was it natural at one time to wonder about the possibility that
every
real (or complex) number is algebraic? (Of course one has to put out of one's mind the world view after Cantor, with his clarification that there are only a
countable
number of algebraic numbers, but an
uncountable
number of nonalgebraic (=transcendental) ones. The interested reader might wish to consult at my web site my 2nd year BA summary course notes on
The Real Numbers and Cantorian Set Theory
.)
So, imagine starting with a completely empty complex plane, and imagine it being filled up, bit by bit, with every kind of number one can think of... every time one thinks of a new number or numbers (especially an infinite number of numbers) a light comes on at each of those numbers. The plane begins to be lit up with those tiny point specks of light... slowly at first... The integers, even all of them, hardly make any impression, but certainly the rationals fill up a lot of the real line (itself, of course, swamped by the vastness of the entire complex plane), though not, of course, the entire real line: there are all those real irrational numbers... all those ones that are algebraic of the second degree, but they don't fill the entire real line, there are all those of the third, fourth, fifth, sixth, ... degrees (Do they  perhaps  fill the entire real line?). And then there are all those complex numbers with nonzero imaginary parts... all those ones of the form (
), where
and
vary in all possible ways over the rational numbers... Every one of those is a solution of a quadratic equation with integer coefficients (and so is of degree 2). They fill up a lot of the complex plane, but not, of course, all of it. But then we imagine switching on lights at all the complex solutions of third, fourth, fifth, sixth, ... degree algebraic numbers. Have we now filled up the entire complex plane; or, at the very least, earlier, the entire real line?
Suppose,
just for the moment
, that we hadn't filled the entire complex plane, and let us return to the above definition of the actual term
algebraic
, with its reference to 'integer' (or 'rational') coefficients. Let's now make the observation that those coefficients are
merely
(with hindsight) algebraic numbers of the first degree, and, in the light of that, let us now
speculate
that we might  perhaps  get our hands on a real or complex number that isn't an algebraic number (according to our earlier definition) by doing attempting
like
this: let's
try
to find a polynomial equation that doesn't have an algebraic solution by choosing some (or possibly all) of its coefficients not to be integers (or rational numbers)...
Note that every such polynomial equation has solutions (as many, in fact, as the degree of the polynomial, thought some may be repeated), every one of which is a complex number: that's because of the socalled
Fundamental Theorem of Algebra
(remarkable itself in its day, and still so...), namely that
every
polynomial equation with
real/complex
coefficients has
exclusively complex solutions
(some of which may be real).
Thus (we wonder) could we find/construct a polynomial, all of whose coefficients are algebraic numbers (but where some/all of them are of degree greater than 1), such that the corresponding equation (all of whose solutions we know from the Fundamental Theorem of Algebra must be complex numbers) possesses at least one solution that is
not
an algebraic number? For example, consider the following equation in which all of the coefficients are algebraic numbers (not all rational):
... (E)
with coefficients
(2nd degree),
(1st degree), and
(2nd degree algebraic).
The solutions of that equation are:
>
solve(sqrt(7)*x^2  9*x/11 + 3/4  sqrt(5/3) = 0);
>
A beginner might not immediately notice, but those two solutions
are
algebraic numbers according to the standard definition of being algebraic '
over the integers
'.
Incidentally, it is an entirely elementary matter to prove  independent of actual calculation  that the two solutions of the above equation are algebraic numbers (my 2nd year BA students do that sort of thing). In fact, it is an entirely elementary standard result (this is not the place in which to prove it) that no matter what polynomial equation one takes of whatever degree (
m
), in which all of the coefficients are algebraic numbers of whatever degrees, then not only does that equation have
m
complex solutions (some of which may be real numbers), possibly with some 'repeated' solutions, but every one of those solutions is an algebraic number (according to the standard definition). In other words, if
... (B)
is a polynomial equation of degree
m
in which every coefficient
is an algebraic number (according to the standard definition, and
) then every solution of (B) is a solution of:
... (A)
in which every coefficient
is an integer, and
. (The standard way of expressing the above succinctly is to say that the field of algebraic numbers is
algebraically closed
.)
In short, any hope of creating/finding/... a nonalgebraic number by merely tinkering with the coefficients (in the manner suggested above) is
doomed to failure
.
The question then, in Euler's time (or possibly earlier than Euler?), was:
are there
any
(real or complex) numbers that are not algebraic?
Euler (or someone earlier?) defined a
transcendental number
to be a (real or complex) number that is not algebraic; it "transcendended the power of the algebraic".
Of course
if
there were any,
then
there would automatically be infinitely many. That's a matter that a novice might like to think about.
I return to Euler in another subsection.
_____________________________
A final word about computer computation/demonstration. I'd like to return briefly to the example (E) above, just in case a novice (using Maple or whatever) attempts to vary the coefficients, and wonders about why sometimes things don't appear to work out nicely... Consider the example:
... (E')
in which I have merely changed in (E) the
to
, but otherwise kept everything else unaltered. Now try 'solving', and you get:
>
solve(sqrt(7)*x^5  9*x/11 + 3/4  sqrt(5/3) = 0);
>
The software appears to fail, and with good mathematical reason... Here a beginner needs to know that one has now come up against a barrier: the classical result (Abel and Galois) concerning the nonexistence of a solution 'using radicals' for general polynomial equations of degree at least 5... But that is
"another day's work"...
You can of course do things like:
>
plot(sqrt(7)*x^5  9*x/11 + 3/4  sqrt(5/3), x = 2..2);
>
plot(sqrt(7)*x^5  9*x/11 + 3/4  sqrt(5/3), x = 1..1);
>
>
fsolve(sqrt(7)*x^5  9*x/11 + 3/4  sqrt(5/3) = 0);
>
Enough of this.