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\huge{DETECTING IRRATIONALITY\\From 
\(\sqrt{2}\)  to $\zeta (3)$}
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{\bf Introduction}.  In this article we give a 
survey of some methods used to show 
irrationality of numbers.Our style is 
expository. We do not entend to give a 
proof of every theorem we state; rather 
we refer the reader to the suitable 
reference.

{\bf Sec1. The Fundamental Theorem of 
Arithmetic.} This theorem simply states 
that
``Every natural number can be written as 
a product of primes".This marvelous
theorem was the first tool used to 
detect irrationality. One can go one 
step
farther and say it is the one that 
brought irratoinal numbers into the 
realm
of mathematics.The story is very 
short.Using our modern terminology, the
Phythagoreans thought that all numbers 
are rational.They were shocked when they 
found that the length of the diagonal of 
the unit square is not rational. That
is to say, $\sqrt{2}$ is not 
rational.For if  $\sqrt{2}  = \frac{a}{b}$
for some relatively prime integers a and 
b, then we would have $a^{2}=2b^{2}$,
this says a square is divisible by an 
odd power of 2,which contradicts the 
fundamental theorem of arithmatic, since 
a square (if divisible by 2)
is divisible by an even power of 2. Thus 
 \(\sqrt{2}\)  is irrational. The same 
argument can be used to show that
 \(\sqrt[n]{m}\), where m is not a power 
of n,is also irrational. An almost 
similar argument 
can show that $\log[m] n $ is irrational 
if (i) n is not a power of m and (ii) 
there is a prime
that divides n or m, but not both. 

{\bf sec2. The Rational zeros Theorem.}  In the 
theory of equations, we have the 
following

{\bf Theorem:}Let 
$a_{n}x^{n}+a_{n-1}x^{n-1}+...+a_{1}x
+a_{0}=0$, where $a_{i}$ are integers, 
$a_{n}\neq0$.If $\frac {p}{q}$,  where 
(p,q)=1, is a zero of the given 
equation, then p divides $a_{0}$ and q 
divides $a_{n}$.

So this theorem determines all rational 
zeros of equations with integer 
coefficients. Now let $\alpha$ be a real 
number.If we can show that $\alpha$ is a 
root of some polynomial equation with 
integer coefficients and with no 
rational roots,then $\alpha$ is 
necessarily irrational. For example,let 
$\alpha$=\(\sqrt{3}\)+\(\sqrt{5}\). Then 
$\alpha$ is a root of 
$x^{4}-16x^{2}+4=0$.The possible 
rational zeros are: +1, -1, +2, -2, +4, 
-4. None of which is a zero. So $\alpha$ 
is irrational. 

{\bf sec3. Decimal Expansions.}  In the theory 
of decimal expansions, there is a fact 
which states that ``A real number is 
rational if and only if its decimal 
expansion terminates or recurs." So this 
gives a way to detect irrationality. 
But, How practical is this method? One, 
in fact, can use decimal expansions to 
construct irrational numbers. For 
example, the numbers
    $\alpha =.a_{1}a_{2}a_{3}...$, where $a_{i}=1$ 
if i is prime and 0 otherwise,
and $\beta = .p_{1}p_{2}p_{3}...$, where $p_{i}$ 
is the sequence of primes given in 
increasing order,
are irrational. But the proofs are not 
trivial, particularly for 
$\beta$.See[1,sec 9.4]. The main point 
to observe here is that there is a rule 
defining the digits in $\alpha$ and 
$\beta$. Then using the theory of 
primes, one can show their 
irrationality. But suppose one 
constructed a decimal number by giving 
the digits in a random way. This number 
is most likely to be irrational; but, 
how to prove it? No clue.

{\bf Sec4. Lindemann Theorem.} Equally 
difficult is the question of determining 
the transcendence of numbers. The 
transcndence of $\pi$ was proved by 
Hermite in 1873 and that of e was proved 
by Lindemann im 1882. Lidemann gave a 
general theorem that proves the 
transcendence of a class of numbers 
including $\pi$ and e. Before we state 
the theorem and its corollary, we remind 
the reader that every transcendental 
number is irrational as can be trivially 
shown.

{\bf Lindemann Theorem:} Given any distinct 
algebraic numbers 
$\alpha_{1},\ldots,\alpha_{n}$, the 
values $e^{\alpha_{1}},\ldots, e^{\alpha_{n}}$ are linearly 
independent over the field of algebraic 
numbers.[2,ch.9]

{\bf Corollary:}The following numbers are 
transcendental:
e, $\pi$, $e^{\alpha}$, $sin\alpha$, $cos\alpha$, 
$tan\alpha$, $sinh\alpha$, $cosh\alpha$, $tanh\alpha$, 
($\alpha\neq$0, algebraic)
$log\beta$, $arcsin\beta$, $arccos\beta$, $arctan\beta$,
 $arcsinh\beta$, $arccosh\beta$, $arctanh\beta$,
  ($\beta\neq$0,1, algebraic)
 
Before we leave this section, we point out that the 
irrationality of e and $\pi$ can be proved independetly 
of being transcendental.
See[1].

{\bf Sec5. Gelfand-Schneider Theorem.}  This theorem, proved 
independently by Gelfand and Schneider, gives another
class of transcendental and hence irrational numbers.

{\bf Theorem:} If $\alpha$ and $\beta$ are algebraic numbers
with $\alpha\neq$0, 1 and $\beta$ is not a real rational
number, then every value of $\alpha^{\beta}$ is
transcendental.
See [2] for a proof.
For example, the numbers $2^{i}, 3^{\sqrt{5}}, \sqrt{2}^{\sqrt{5}},i^{-2i}$
are all transcendental. Since $e^{\pi}$ is one value of $i^{-2i}$, then
$e^{\pi}$ is also transcendental.

{\bf Sec6. Irrationality of $\zeta(3)$.}  The series $\zeta(k)=\sum_{n=1}^{\infty}
\frac{1}{n^{k}}$

converges for $k>1$. In fact we know more; $\zeta(2k)=\frac{(-1)^{k}(2\pi)^{2k}
B_{2k}}{2(2k)!}$
and so $\zeta(2k)$ is irrational for $k\geq1$. But What about $\zeta(2k+1)$ 
for $k\geq1$?. Recently, Apery showed the irrationality of $\zeta(3)$. The other
values, $\zeta(2k+1)$, $k>1$, are still unknown. How did Apery prove it?.

For practical purposes, irrational numbers are always approximated by rationals.
In this direction, a whole theory was developed. One of its beautiful theorems
is the following characterization of irrationality:

{\bf Theorem:} If there is a $\delta>0$ and a sequence $\frac{a_{n}}{b_{n}}$ of 
rational numbers such that $\frac{a_{n}}{b_{n}} \neq \alpha$ and $|\alpha-
\frac{a_{n}}{b_{n}}|<\frac{1}{q_{n}^{1+\delta}}, n=1, 2, 3,...$,
then $\alpha$ is irrational.

For $\zeta(3)$, Apery was able find a $\delta$ and a sequence 
$\frac{a_{n}}{b_{n}}$ of rational numbers satisfying the inequality of the 
above theorem. Namely, given the recursion
\[ n^{3}u_{n}+(n-1)^{3}u_{n-2}=(34n^{3}-51n^{2}+27n-5)u_{n-1}\]
define  $ u_{0}=a_{0}=0, u_{1}=a_{1}=6, a_{n}=u_{n}$ for $n \geq 2$,
and  $u_{0}=b_{0}=1, u_{1}=b_{1}=5, b_{n}=u_{n}$ for $n \geq 2$.
Further, take $\delta =0.080 529 ... >0$. For more details, the reader is 
encouraged to read [4].

{\bf Conclusion.}  As the reader observed, there is no one method that works for all.
Rather, different methods take care of different classes of numbers. But, still,
the story is not complete. There are some numbers whose irrationality or 
transcendency are not yet known. Examples include Euler's constant $\gamma, 
\pi^{\sqrt{2}}, \pi^{e}, e+\pi, 2^{e}, 2^{\pi}$.

{\bf References:}

1. G. H. Hardy and E. M. Wright. An Introduction to the Theory of Numbers. Fifth
   edition, 1979, Oxford Science Publications.
   
2. Ivan Niven. Irrational Numbers. 1956. The Mathematical Association of America.


3. H. E. Rose. A Course in Number Theory. Second edition,1994, Oxford Science 
   Publications.
   
4. Alfred van der Poorten. A Proof that Euler Missed...The Mathematical 
   Intelligencer,Vol 1,pp.195-203,1979.
   

