Lemma 6.6.1
If
is not a square number then is irrational.

Proof.
Suppose
, with . Then
.
If then let be a prime dividing . Thus
and
so , which is impossible as . Thus and
.

This lemma can also be stated: ``if with
then
''.

Definition 6.6.2
A real number is algebraic if it satisfies a polynomial equation
with coefficients in
.
Real numbers which are not algebraic are transcendental

For instance and are transcendental (the proof is fairly
hard, especially for ). Most reals are transcendental; this can
be proven using the notion of a countable set and we will do it later
(v.i. Theorem thm transcendental numbers -> uncountable).
If the rational
( with ) satisfies a
polynomial with coefficients in
then

so
and
. In particular if then
, which is stated as ``algebraic integers which are rational are
integers''.
Note that if
and
with
then
and
,
and
.
Major open problems in the area of prime numbers are the Goldbach
conjecture (``every even number greater than two is the sum of two
primes'') and the twin primes conjecture (``there are infinitely many
prime pairs and '').
Next:Modular Arithmetic Up:The classical sets of Previous:Uniqueness of prime factorisationContents
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2002-06-10