Most people know that π's digits go on forever, but so do some fractional numbers
such as 1/3 = 0.333….
Fractional numbers are more commonly called rational numbers in mathematical literature
and it is well known that their decimal expansions either terminate or go on a pattern that
repeats indefinitely.
π is different: it's digits have no such pattern.
We know of other numbers that have digits that go on forever and have no such patterns,
such as √2.
Could π be in the same class of numbers as √2?
Let's first say what we mean by that class of numbers.
Specifically, √2 is a number that is a solution to a polynomial equation
with integer coefficients:
$$x^2 - 2 = 0$$
Such numbers are called algebraic numbers.
Observe that rational numbers are also algebraic numbers, since rational numbers are
of the form a/b where a and b are integers,
and therefore they are solutions to the polynomial bx - a = 0.
Thus algebraic numbers are in a bigger class than rational numbers but they include
all rational numbers.
So naturally the question becomes: is π an algebraic number?
The answer turns out to be no, π is in an even bigger class of numbers that we
call transcendental numbers.
These are real numbers that satisfy no such polynomial equation, which means that
numbers like π live in a class that numbers like √2 are not in.
It took mathematicians well over a century to prove that such numbers exist, and longer
to prove that π is one of them.
History
There is a fascinating thread of mathematical history about the dsicovery of different classes of numbers. As one might guess, it starts with ancient Greece.
In the 6th century BC, Pythagoras of Samos had believed that every number
could be described as integers and ratios of integers.
The history is questionable here, but it is often said that that came into challenge
by Hippasus who showed that the length of the diagonal of a 1x1 square, which
is √2, cannot be expressed as such a ratio.
The term incommensurable was used to describe such numbers, which was the
first historical evidence that there is something beyond rationals.
The word "transcendental" first appeared in mathematical literature in
1682 when Gottfried Leibniz was studying the sine function.
Leibniz proved that y = sin(x) does not have a polynomial
relationship.
This is a statement about a function, not about a particular number, but
it did bring about the concept that certain functions "transcend"
algebraic methods.
The following century brought real progress in understanding classes of numbers,
starting with Leonhard Euler who proved that
e was not rational in 1737.
Later, Johann Heinrich Lambert used continued fractions to
prove that π was also not rational and he beloieved that both e and π were
transcendental, but could not come up with a proof.
Although there was doubt, the possibility remained that even though these numbers are not rational,
maybe they could be algebraic?
In 1844, the separation between algebraic numbers and transcendental numbers was finally settled.
French mathematician Joseph Liouville not only showed that there were such numbers
that were not algebraic, but he actually showed how to construct one such example.
Liouville's example had a 1 at every factorial decimal digit (i.e., 1st decimal
digit, 2nd decimal digit, 6th decimal digit, 24th decimal digit, etc....) and zeros elsewhere.
That is
$$0.110001000000000000000001000\ldots$$
is transcendental and not algebraic.
So now all doubt was removed on the existence of such numbers, but the questions remained
open on what classes e and π lived in.
The number e was settled in 1873 by Charles Hermite who
proved that it was transcendental.
Hermite also believed that π was transcendental but wrote that it would be really hard to prove.
The following year, Georg Cantor proved that if you chose a real number completely
at random, then almost certainly it would be transcendental.
He did this by showing that algebraic numbers are "countable" but real numbers are not,
which was a way of showing that some infinite collections are larger than others.
Although a beautiful idea, it was not constructive and it did not get us any closer to
proving that π was transcendental.
The final breakthrough came in 1882, two hundred years after the concept of transcedence was introduced.
Ferdinand von Lindemann found a way to adapt Hermite's method on
the transcedence of e to prove that π is also transcendental.
Finally, we understood the classification of the most historically fascinating number in mathematics!
π being transcendental also solved a 2,000 year old mathematical problem known
as "squaring the circle."
The challenge for that problem was to find a square of area π using only an unmarked straightedge and a compass.
Such a square would have a length of √π.
Previous work on the problem had shown that with such instruments, it is only possible to build
lengths that are algebraic numbers.
Since π is not algebraic, √π is also not algebraic and therefore it is impossible
to square the circle.
Another interesting follow-up was in 1900 when David Hilbert's
famous list of 23 unsolved problems.
The seventh on the list ws whether algebraic numbers raised to an irrational, algebraic
power (such as 2√2) are always transcendental.
This was independently proved to be indeed true by Alexander Gelfond and
Theodor Schneider in 1934.
Such proofs give ways to construct an infinite number (though countable) of transcedental numbers.
Show the math
Proving π is transcendental is not easy, and is beyond what we can capture here.
Instead, we show the classic proof that √2 is not rational, or
"incommensurable" as the ancient Greeks would say.
The proof is by contradiction. Suppose √2 were rational.
Then it could be written as a fraction a/b where
a and b are integers with no common factor:
$$\sqrt{2} = \frac{a}{b}, \quad \gcd(a, b) = 1$$
Squaring both sides:
$$2 = \frac{a^2}{b^2} \quad\Longrightarrow\quad a^2 = 2b^2$$
Since a² is 2 times an integer, it must be even.
But that also means that a must be even, because if it were not, then
it is odd and an odd number squared is always odd.
So we write a = 2k and substitute back into the equation:
$$(2k)^2 = 2b^2 \quad\Longrightarrow\quad 4k^2 = 2b^2 \quad\Longrightarrow\quad b^2 = 2k^2$$
This implies that b² is even, so b must also be even by the same reasoning above.
But that means both a and b are even, which contradicts our initial
assumption that they have no common factor.
Therefore the initial assumption must be false, hence:
$$\sqrt{2} \text{ is irrational.}$$
That's a fairly simple argument, but maybe some readers want more. For those interested learners, we direct them to Mathologer who is among the best for explaining really complex problems in a way that is accessible to a wider audience: