A real number is called *algebraic* if it is the root of a polynomial with integer coefficients. Examples of algebraic numbers are (it is the root of ), (), the golden ratio (), and the single real root of the quintic polynomial (which cannot be expressed with radicals).

A real number that is not algebraic is called *transcendental*. It is easy to see that every transcendental number is irrational, but, as we see above, not every irrational number is transcendental. In 1874 Georg Cantor proved that there are only countably many algebraic numbers. Thus there must be uncountably many transcendental numbers! The vast, vast majority of real numbers are transcendental.

Despite the fact that almost every real number is transcendental, it is very difficult to prove that a given number is transcendental. Joseph Liouville discovered the first transcendental number in 1844:

In 1873 Charles Hermite proved that is transcendental and nine years later Ferdinand von Lindemann proved that is transcendental. Some other transcendental numbers are: (we have not yet proved that is transcendental), , , , and (yes, this is a real number: ).

In this sequence of three blog posts I will prove that is transcendental. At a glance the proof looks long and complicated, but the proof is really quite straightforward (this post is the longest of the three). Most of the proof is nothing more than algebraic manipulations and divisibility arguments with integers. There is some differential calculus, but nothing beyond Calculus I: the mean value theorem, the derivative of the product rule, and the limit . The proof also uses the infinitude of primes.

This proof is based on Adolf Hurwitz‘s 1893 simplification of Hermite’s proof. (To be specific, I used Herstein’s *Topics in Algebra* as a source for the details of Hurwitz’s proof.)

**Three-step proof that is transcendental
Step 1**

Step 2

Step 3

**Theorem.** is transcendental.

This is a proof by contradiction. Suppose is algebraic. That is, is the root of a polynomial with integer coefficients. In other words,

We may assume that . (If is zero divide by , or a sufficiently large power of ; if , multiply by .)

* Step 1.*

In this first post we will prove the following lemma. We use the familiar notation that is the th derivative of (and ).

**Lemma 1.** Suppose is a root of the polynomial . Let be a polynomial and . Then there exist such that , where .

Suppose is a polynomial of degree with real coefficients. Then . In particular, the function is a finite sum:

Notice that the derivative of has the following simple form

Define a new function . The derivative of also has a compact form. Using the product rule we obtain

At this point we must invoke the mean value theorem.

**Theorem.** [Mean value theorem] Suppose is a differentiable function and . Then there exists such that .

Since the number is between and , we can write it as where . Thus we may restate the theorem as follows.

**Theorem.** [Mean value theorem (vers. 2)] Suppose is a differentiable function and . Then there exists such that .

The function is differentiable, so we may apply the mean value theorem to the interval (, ). It guarantees such that

Using the definition of and our expression for this equality becomes

A little algebra yields the following equality (we will call this common value ).

We now repeat this for all of the intervals for , where is the degree of the polynomial for which . Each application of the mean value theorem yields its own and its own corresponding :

Multiply the left and right sides of this equality by (the th coefficient of ) to obtain

Summing these equation for we obtain

But, since , we know that . So the equality simplifies to

This proves the lemma.

In the next step we will choose a specific polynomial for the lemma.

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