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From: "Achava Nakhash, the Loving Snake"
Newsgroups: sci.math
Subject: Re: How Do You Prove A Number Is Irrational?
Date: Mon, 28 Sep 1998 17:06:39 -0700
Kurt Foster wrote:
> [query as to methods of proving irrationality deleted]
>
> There aren't all that many methods, and, as you suspect, none is
> comprehensive. The best known are (1) Unique factorization ("fundamental
> theorem of arithmetic"). (2) Irreducibility of polynomials (various
> results, Eisenstein's criterion perhaps being best known. Zeroes of
> irreducible polynomials of degree > 1 are not rational numbers; if the
> zeroes are also real, they are real irrational numbers. (3) Infinite
> series which "converge too fast" can be proven to have sums which are
> irrational, or even transcendental.
> Examples:
>
> Method (1): log_2(3) = a/b --> 2^a = 3^b, violates unique factorization.
>
> Method (2): sqrt(2) satisfies x^2 - 2 = 0, x^2 - 2 irreducible by
> Eisenstein's criterion.
>
> Method (3): zeta(3) = SUM, n >=1, 1/n^3 (irrational - Apery, 1979)
>
> Number with unknown character: Euler's constant "gamma" = .577+
To expand a bit on what was already said,
The Eisentein irreducibility criterion says that a polynomial of the form
x^n + a_(n-1)x^(n-1) + ... + a_1x + a_0
such that, for some prime p, a_0, a_1, ..., a_(n-1) are divisible by p, but
a_0 is not divisible by p^2, is necessarily irreducible over the rational
numbers. Hence any of its roots are irrational, especially if they are
real.
Examples of such polynomials are x^2 - 2, x^3 - 5, x^2 - 7x - 7, and many
more.
Notice that this technique proves that all sqare roots of square-free
integers are
necessarily irrational. Of course, the unique factorization theorem (which
is used in the proof of the Eisenstein irreducibility theorem) can be used
to prove this directly.
To elaborate on Kurt's third technique, let us suppose that x = a/b is a
rational number. Let us try to approximate it by some other irrational
number p/q, where p and q are any integers, not necessarily prime, and
obviously such that q is not equal to 0. Also assume that a/b and p/q are
not actually equal to each other and that both fractions are presented in
lowest terms. Then
|a/b - p/q| = |(aq-bp)/bq| >= 1/bq
by the Fundamental Theorem of Transcendental Number Theory which says that a
non-zero integer must have an absolute value greater than or equal to 1.
In particular, if x is rational, there is a constant c, depending only on x,
such that for sufficiently large q, |x - p/q| > c/q.
Thus if we have an infinite sequence of rational numbers, p_n / q_n and a
real number x such that |x - p_n / q_n| 0 and for all
sufficiently large n, then x must be irrational.
Armed with this load of baggage, it is not too difficult to demonstrate that
e = 1/0! + 1/1! + 1/2! + 1/3! + ...
is irrational by letting the n'th partial sums be our p_n / q_n and
estimating the
difference, 1/(n+1)! + 1/(n+2)! + ... to see that our sequence of rationals
meets
the above sequence criterion. I have been a little sloppy about when things
are in reduced terms and so on, but it is quite easy to make this argument
rigorous, and it can also be found in several books, some of which have been
mentioned in previous messages in this thread.
The approach by finding good rational approximations to x is also behind
many of the transcendance proofs in the literature. Liouville demonstrated
way back when that a real number x that satisfies an irreducible polynomial
with integer coefficients of degree n has a number C associated with it such
that any rational p/q satisfies
|x - p/q| > C / q^n
and then used this to demonstrate that numbers of the form
1/2^(0!) + 1/2^(1!) + 1/2^(2!) + ...
could not satisfy any irreducible polynomial of degree n with integers
coefficients for an n, and hence such numbers must be transcendental.
Many years were to pass before mathematicians found arguments of
extraordinary cleverness to prove that e and pi were transcendental. In
1909, Thue managed to lower than exponent attached to q in Liouville's
theorem, and proved as a consequence of this sharpening that a class of
Diophantine equations could have only a finite number of solutions.
Research continue to this day to improve upon this result and to extend the
significance of this result for the solution of Diophantine equations. I am
not move than superficially familiar with any recent developments, but I
believe they are rather spectacular.
Regards,
Achava