Transcendental numbers occupy a position in the field of real or complex numbers much like that of insects in the kingdom of animals. Everybody knows they are, by a large margin, the most abundant class, but few know more than one or two of them by name.
Donald R. Newman
It is of course easy to show the existence of transcendental numbers through a countability argument: there are countably many polynomials with integer coefficients, each of which has a finite number of roots, so the set of algebraic numbers is countable. Since the set of real numbers is uncountable, there must exists uncountably many real numbers that are not algebraic. But showing that a particular number is transcendental is much harder.
The first known example of a transcendental number was L = 10-1! + 10-2! + 10-3! + ... = 0.110001000000000000000000100..., which was proved trancendental by Joseph Liouville in 1844. The 1s are in the 1!th, 2!th, ... decimal places.
The proof is by contradiction. Write Lk for the sum of the first k terms in the series of L, 10-1! + 10-2! + ... + 10-k!. We are going to assume that there in fact is a polynomial with degree n such that P(L) = 0, and then show that simultaneously: P(Lk) ≠ 0 for all large enough k, 10n k! P(Lk) is an integer for all k, and 10n k! P(Lk) tends to 0 as k tends to infinity. This is a contradiction, because when k is large enough we will have 0 < P(Lk) 10n k! < 1, but there are no integers between 0 and 1.
So, assume there is a polynomial with integer coefficients that have L as a root, say P(x) = anxn + an-1xn-1 + ... a1x + a0.
Lemma 1. There is a k0 such that P(Lk) ≠ 0 for all k > k0.
Proof. The numbers L1, L2, ... are all distinct. But P is an n-degree polynomial, so it can have at most n distinct roots, so there are at most n indices k such that Lk is a root. So pick k0 to be the maximum of those. ∎
Lemma 2. 10n k! P(Lk) is an integer for all k.
Proof. Lk = 1/101! + 1/102! + ... + 1/10k! = C / 10k! for some integer C. So just substituting it in, we have
P(Lk) = anLkn + an-1Lkn-1 + ... + a1Lk + a0
= anCn/10n k! + anCn-1/10(n-1)k! + ... + anC/10k! + a0,
and multiplying both sides by 10n k!,
10n k! P(Lk) = anCn + anCn-110k! + ... + anC 10(n-1)k! + a010nk!
Now all the numbers involved in the expression on the right hand side are integers, so the result is too. ∎
Lemma 3. 10n k! P(Lk) → 0 as k → ∞
Proof. L and Lk first differ in the (k+1)!th decimal, so |L - Lk| ≤ 2*10-(k+1)!. In other words, Lk converges very quickly towards L. Furthermore, the derivative P'(x) is bounded in the interval [-1, 1]; indeed if |x|≤1, then |P'(x)| = |nanxn-1 + (n-1)an-1xn-2 + ... + a1| ≤ |nan| + |(n-1)an-1| + ... + |a1| = A.
L and Lk are both ≤ 1, so using the mean value theorem,
|P(Lk)| = |P(L) - P(Lk)| = |P'(c)|(L - Lk) ≤ A*2*10-(k+1)!
So 10n k! |P(Lk)| ≤ 10n k! 2A 10-(k+1)!
= 2A 10(n-(k+1))k! → 0 ∎
Thus we have arrived at the required contradiction: pick a k large enough that 0 ≠ 10n k! |P(Lk)| < 1. Then 10n k! |P(Lk)| cannot be an integer.