e.g. 2-Ö2 and 2+Ö2

Proof that e's irrational isn't that horrible, is it? I remember it being on an old STEP..

Sarah

and it's easy to check the left hand side is a polynomial with integer coefficients.

Thus it follows that numbers such as Ö2 and

Ö

2+Ö2

are algebraic.

There is however another proof of e's transcendence, one which in my opinion is a good deal simpler. It's due to Hilbert.

Recall that for an integer k,

ò₀^¥ x^k e^-x dx=k!,

and thus for an integral polynomial p (with integer coefficients, that is)

ò₀^¥ x^k p(x) e^-x dx º k! p(0) (mod (k+1)!)

Suppose then that e is algebraic. There exist integers a_i such that

a₀+a₁ e+a₂ e² + ... + a_n eⁿ=0,

and, without loss of generality, a₀ ¹ 0.

Let (for n as above)

I(b, a)=ò_a^bx^m[(x-1)...(x-n)] ^m+1 e^-x dx

and define

S=a₀ I(¥, 0)+ a₁ e I(¥, 1)+...+a_n eⁿ I( ¥, n)

and

T=a₁ e I(1,0)+...+a_n eⁿ I(n,0).

Note that S+T=0.

Now, we require two lemmas:

Lemma 1: For infinitely many m, S/m! is a nonzero integer.

Proof: Substituting y=x-k into the integral for I,

a_k e^k I(¥, k)=a_kò₀^¥ y^q p_k(y) e^-y dy

where q=m or m+1 as k=0 or not, and the p_k are polynomials with integer coefficients. Using (1), we have that the first term of S is divisible by m!, and that the rest are divisible by (m+1)!. But p₀(0) = (-1)^n(m+1)n!^m+1, which, by Fermat's theorem, is congruent to (-1)^n(m+1)n! (mod (m+1)) when m+1 is prime, and since n! isn't divisible by an infinitude of primes, there are an infinitude of m for which m! p₀(0) isn't divisible by (m+1)!, and thus an infinitude of m for which S/m! is a nonzero integer.

Lemma 2:

lim m®¥

T/m!=0

.

Proof: Let

M=

max 0 £ x £ n

|(x-1)(x-2)...(x-n)|(x+1).

It is relatively easy to show that

|a_kI(k,0)| £ k|a_k| M^m+1,

from which it follows

|T|/m!=O(M^m+1/m!)® 0.

But, since S+T=0, Lemma 1 and Lemma 2 contradict each other, and thus e cannot be algebraic.

Brad