### Introduction

There are numerous proofs commonly available online for the fact that the Euler's number $e$ is irrational. Then going further we find that $e$ is also a transcendental number which means that it can not be the root of a polynomial equation with integral coefficients and thereby transcends the powers of algebra in a sense. Again the proof that $e$ is transcendental is also available on various places online.In this post I am going to present the proof that $e$ is not a quadratic irrationality. This is based on the paper "Sur l'irrationnalitÃ© du nombre e = 2.718..." by Joseph Liouville.

### An Elementary Proof that $e$ is not a Quadratic Irrationality

What I intend to prove here is the fact that $e$ is not a quadratic irrationality. Or in other words, $e$ is not the root of a quadratic equation with integral coefficients. This follows simply from the earlier mentioned fact that $e$ is transcendental. But here I would like to present a very elementary proof that $e$ is not a quadratic irrationality. This proof is quite simple and easy to grasp compared to proof of transcendence of $e$ (which looks very high-brow).I found this proof long back during my college years in some book by C. L. Siegel, but I am not able to find any references online for the same.

To start with, we assume on the contrary that $e$ is a quadratic irrationality i.e. there exist integers $a, b, c$ not all zero such that $$ae^{2} + be + c = 0\tag{1}$$ Using the fact that $e$ is irrational we can see that we must have $a \neq 0$ and $c \neq 0$. We can rewrite the above equation as follows: $$ae + ce^{-1} = -b\tag{2}$$ For all positive integers $n$ let us define \begin{align}\alpha_{n} &= \frac{1}{n + 1} + \frac{1}{(n + 1)(n + 2)} + \cdots\notag\\ \beta_{n} &= (-1)^{n + 1}\left(\frac{1}{n + 1} - \frac{1}{(n + 1)(n + 2)} + \cdots\right)\notag\end{align} Since we have \begin{align}e &= 1 + \frac{1}{1!} + \frac{1}{2!} + \cdots + \frac{1}{n!} + \frac{1}{(n + 1)!} + \cdots\notag\\ e^{-1} &= 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \cdots + (-1)^{n}\frac{1}{n!} + (-1)^{n + 1}\frac{1}{(n + 1)!} + \cdots\notag\end{align} it follows that \begin{align}n!e &= n!\left(1 + \frac{1}{1!} + \frac{1}{2!} + \cdots + \frac{1}{n!}\right) + n!\left(\frac{1}{(n + 1)!} + \frac{1}{(n + 2)!} + \cdots \right)\notag\\ &= A + \alpha_{n}\notag\end{align} where $A$ is some integer dependent on $n$. Similarly \begin{align}n!e^{-1} &= n!\left(1 - \frac{1}{1!} + \frac{1}{2!} + \cdots + (-1)^{n}\frac{1}{n!}\right)\notag\\ &\,\,\,\,\,\,\,\, + n!(-1)^{n + 1}\left(\frac{1}{(n + 1)!} - \frac{1}{(n + 2)!} + \cdots \right)\notag\\ &= B + \beta_{n}\notag\end{align} where $B$ is some integer dependent on $n$.

If we multiply the equation $(2)$ by $n!$ we can see that \begin{align}a(n!e) + c(n!e^{-1}) &= -b \cdot n!\notag\\ \Rightarrow a(A + \alpha_{n}) + c(B + \beta_{n}) &= -b\cdot n!\notag\\ \Rightarrow (aA + cB) + (a\alpha_{n} + c\beta_{n}) &= -b\cdot n!\notag\\ \Rightarrow S + R_{n} &= -b\cdot n!\tag{3}\end{align} where $S = aA + cB$ is some integer dependent on $n$ and $R_{n} = a\alpha_{n} + c\beta_{n}$. From the equation $(3)$ above it now follows that $R_{n}$ is also an integer for all values of $n$. We will achieve the contradiction by showing that there are infinitely many values of $n$ for which $R_{n}$ is not an integer. This is bit difficult and tricky but still very elementary.

First of all we can note that \begin{align}|\alpha_{n}| &< \frac{1}{n + 1} + \frac{1}{(n + 1)^{2}} + \frac{1}{(n + 1)^{3}}+ \cdots\notag\\ &= \dfrac{\dfrac{1}{n + 1}}{1 - \dfrac{1}{n + 1}} = \frac{1}{n}\notag\end{align} and $$|\beta_{n}| < 1/(n + 1) < 1/n$$ so that $$ |R_{n}| \leq \frac{|a| + |c|}{n} < 1$$ if $n$ is sufficiently large.

In order to prove that $R_{n}$ is not an integer we now only need to prove that $R_{n} \neq 0$ for infinitely many values of $n$.

We need to observe that \begin{align}nR_{n - 1} - R_{n} &= n(a\alpha_{n - 1} + c\beta_{n - 1}) - a\alpha_{n} - c\beta_{n}\notag\\ &= a(n\alpha_{n - 1} - \alpha_{n}) + c(n\beta_{n - 1} - \beta_{n})\notag\\ &= a + (-1)^{n}c\notag\end{align} Now it follows that all the three numbers $R_{n - 1}, R_{n}, R_{n + 1}$ can not be zero simultaneously. For, if it were so we would have $$nR_{n - 1} - R_{n} = 0 \text{ and } (n + 1)R_{n} - R_{n + 1} = 0$$ i.e. $$a + (-1)^{n}c = 0 \text{ and } a + (-1)^{n + 1}c = 0$$ i.e. $c = 0$ which is not the case.

Hence it follows that there are infinitely many values of $n$ for which $R_{n}$ is non-zero and therefore there will be a value of $n$ for which $0 < |R_{n}| < 1$ so that $R_{n}$ is not an integer for some value of $n$. This is the contradiction we needed to achieve and thus it follows that there can not be integers $a, b, c$ such that $ae^{2} + be + c = 0$.

There is an alternative and simpler way to obtain the desired contradiction. Since we have $|R_{n}| < 1$ for all $n > |a| + |c|$ it follows that the only way for $R_{n}$ to be an integer is that $R_{n} = 0$ for all values of $n > |a| + |c|$. This means that \begin{align} a\alpha_{n} + c\beta_{n} &= 0\notag\\ \Rightarrow \frac{\alpha_{n}}{\beta_{n}} &= -\frac{c}{a}\notag\end{align} Now the RHS of the above equation is a constant and hence is of constant sign. But on the LHS, $\alpha_{n}$ is positive and $\beta_{n}$ is alternating sign as $n$ increases one by one. Hence we arrive at a contradiction.

The above also establishes that $e^{2}$ is irrational. In the next post we provide another elementary proof of this fact.

**Note:**The proof above was originally given by Liouville and perhaps independently discovered by C. L. Siegel who included it in his book

*"Transcendental Numbers".*

A simpler proof can be provided if we consider the continued fraction of $e$ given by $$e = 2 + \frac{1}{1+}\frac{1}{2+}\frac{1}{1+}\frac{1}{1+}\frac{1}{4+}\frac{1}{1+}\frac{1}{1+}\frac{1}{6+}\frac{1}{1+}\cdots$$ Since the above continued fraction is not periodic it follows that $e$ is not a quadratic irrational. The main difficulty in this proof is establishing the above continued fraction expansion of $e$ (see here, this proof is based on the continued fraction expansion of $\tan(x)$ presented here).We can also use this continued fraction which is not periodic: $$\frac{e - 1}{e + 1} = \frac{1}{2+}\frac{1}{6+}\frac{1}{10+}\frac{1}{14+}\cdots$$

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