e.g. $2-\sqrt{2}$ and $2+\sqrt{2}$

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 $\sqrt{2}$ and $\sqrt{2+\sqrt{2}}$ 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 $k$,

${\int }_0^{\infty }{x}^k{e}^{-x}\mathrm{dx}=k!$,

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

${\int }_0^{\infty }{x}^{k}p(x)e^{-x}\mathrm{dx}\equiv k!p(0)(\mathrm{mod}(k+1)!)$

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

$a_0+a_{1}e+a_2{e}^2+...+a_n{e}^n=0$,

and, without loss of generality, $a_0\ne 0$.

Let (for $n$ as above)

$I(\beta ,\alpha )={\int }_{\alpha }^{\beta }{x}^m[(x-1)...(x-n{)]}^{m+1}{e}^{-x}\mathrm{dx}$

and define

$S=a_{0}I(\infty ,0)+a_{1}eI(\infty ,1)+...+a_n{e}^{n}I(\infty ,n)$

and

$T=a_{1}eI(1,0)+...+a_n{e}^{n}I(n,0)$.

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 $I$,

$a_k{e}^{k}I(\infty ,k)=a_k{\int }_0^{\infty }{y}^q{p}_k(y)e^{-y}\mathrm{dy}$

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 $p_0(0)=(-1{)}^{n(m+1)}n{!}^{m+1}$, which, by Fermat's theorem, is congruent to $(-1{)}^{n(m+1)}n!(\mathrm{mod}(m+1))$ when $m+1$ is prime, and since $n!$ isn't divisible by an infinitude of primes, there are an infinitude of $m$ for which $m!p_0(0)$ isn't divisible by $(m+1)!$, and thus an infinitude of $m$ for which $S/m!$ is a nonzero integer.

Lemma 2: $\underset{m\to \infty }{lim}T/m!=0$.

Proof: Let

$M=\underset{0\le x\le n}{max}|(x-1)(x-2)...(x-n)|(x+1).$

It is relatively easy to show that

$|a_{k}I(k,0)|\le k|a_k|M^{m+1}$,

from which it follows

$|T|/m!=O(M^{m+1}/m!)\to 0$.

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

Brad