In the last post we used the multiplication by $n!$ trick to prove that $e$ is not a quadratic irrationality. In this post we will use same technique albeit in a direct fashion to show that $e^{2}$ is irrational.

From the above we can see that LHS is an integer and $S$ is also an integer, thereby implying that the second sum on RHS ($b\cdot R$) is an integer. However this second sum is not bounded (because of large powers of $2$ involved). In fact we can see that \begin{align}R &= \frac{2^{n + 1}}{n + 1} + \frac{2^{n + 2}}{(n + 1)(n + 2)} + \cdots\notag\\ &= 2^{n + 1}\left(\frac{1}{n + 1} + \frac{2}{(n + 1)(n + 2)} + \cdots\right)\notag\\ &< 2^{n + 1}\left(\frac{1}{n + 1} + \frac{2}{(n + 1)^{2}} + \frac{2^{2}}{(n + 1)^{3}} + \cdots\right)\notag\\ &= 2^{n + 1}\cdot\dfrac{\dfrac{1}{n + 1}}{1 - \dfrac{2}{n + 1}} = \frac{2^{n + 1}}{n - 1}\notag\end{align} The above derivation is valid if $n > 1$ (because in summing an infinite GP we need common ratio $2/(n + 1) < 1$). We thus have $$\frac{2^{n + 1}}{n + 1} < R < \frac{2^{n + 1}}{n - 1}$$ Now we need to analyze the sum $S$. We have $$S = n!\left(1 + \frac{2}{1!} + \frac{2^{2}}{2!} + \cdots + \frac{2^{n}}{n!}\right) = \sum_{k = 0}^{k = n}2^{k}\cdot\frac{n!}{k!}$$ We need to analyze the highest power of $2$ in each term $2^{k}\cdot n! / k!$. Clearly the highest power of $2$ in $n!$ is given by $$\left[\frac{n}{2}\right] + \left[\frac{n}{2^{2}}\right] + \cdots \leq \frac{n}{2} + \frac{n}{2^{2}} + \cdots = n$$ Hence the highest power of $2$ in $n!$ is at most $n$. In fact the highest such power occurs when $n$ itself is a power of $2$ and each of $n/2^{r}$ is integral. Let us then take $n = 2^{m}$ and then highest power of $2$ in $n! = (2^{m})!$ is given by \begin{align}\left[\frac{2^{m}}{2}\right] + \left[\frac{2^{m}}{2^{2}}\right] + \cdots &= 2^{m - 1} + 2^{m - 2} + \cdots + 2 +1\notag\\ &= 2^{m} - 1 = n - 1\notag\end{align} From now on we keep $n$ always as power of $2$ i.e. $n = 2^{m}$. Then the highest power of $2$ in term $2^{k}\cdot n!/k!$ is at least $k + n - 1 - k = n - 1$. It follows now that sum $S$ is an integer divisible by $2^{n - 1}$. Similarly $n!a$ is also divisible by $2^{n - 1}$. Thus dividing equation $(1)$ by $2^{n - 1}$ we get \begin{align}\frac{n!a}{2^{n - 1}} &= b\cdot \frac{S}{2^{n - 1}} + b\cdot\frac{R}{2^{n - 1}}\notag\\ A &= b.S' + bR'\tag{2}\end{align} where $A$ and $S'$ are integers and $R' = R/(2^{n - 1})$ and $bR'$ is also an integer because of the above equation.

We have earlier proved that $$\frac{2^{n + 1}}{n + 1} < R < \frac{2^{n + 1}}{n - 1}$$ and therefore \begin{align} 0 &< \frac{4b}{n + 1} < b\frac{R}{2^{n - 1}} < \frac{4b}{n - 1}\notag\\ \Rightarrow 0 &< bR' < \frac{4b}{n - 1} < 1\notag\end{align} provided $n > 4b + 1$ and $n$ a power of $2$ say $n = 2^{m}$. It now follows that we can choose a suitable value of $n$ such that $0 < bR' < 1$ and hence $bR'$ is not an integer. This is the contradiction we needed to achieve and this proves that $e^{2}$ is irrational.

Aigner and Ziegler carry on this idea a little forward in their book

$$b(n!e^{2}) = a(n!e^{-2})\tag{3}$$ As shown earlier $n!e^{2}$ can be expressed $n!e^{2} = S_{1} + R_{1}$ where $S_{1}$ is an integer divisible by $2^{n - 1}$ and \begin{align} R_{1} &= \frac{2^{n + 1}}{n + 1} + \frac{2^{n + 2}}{(n + 1)(n + 2)} + \cdots\notag\\ &< \frac{2^{n + 1}}{n + 1} + \frac{2^{n + 2}}{(n + 1)^{2}} + \cdots\notag\\ &= \frac{2^{n + 1}}{n - 1}\notag\end{align} Similarly we can write $n!e^{-2} = S_{2} + R_{2}$ where $S_{2}$ is an integer divisible by $2^{n - 1}$ and $$R_{2} = (-1)^{n + 1}\left(\frac{2^{n + 1}}{n + 1} - \frac{2^{n + 2}}{(n + 1)(n + 2)} + \cdots\right)$$ Since $n$ is even it follows that $R_{2} < 0$ and $$|R_{2}| < \frac{2^{n + 1}}{n + 1}$$ We can now divide equation $(3)$ by $2^{n - 1}$ to arrive at: \begin{align}b\left(\frac{S_{1}}{2^{n - 1}} + \frac{R_{1}}{2^{n - 1}}\right) &= a\left(\frac{S_{2}}{2^{n - 1}} + \frac{R_{2}}{2^{n - 1}}\right)\notag\\ \Rightarrow bB + \frac{bR_{1}}{2^{n - 1}} &= aA + \frac{aR_{2}}{2^{n - 1}}\notag\\ \Rightarrow bB + B' &= aA + A'\tag{4}\end{align} where $0 < B' < 4b/(n - 1)$ and $-4a/(n + 1) < A' < 0$ and $A, B$ are integers. For large values of $n = 2^{m}$ we see an obvious contradiction as the LHS of the last equation $(4)$ is slightly larger than an integer and RHS of the equation is slightly less than an integer.

The same argument can be used to prove that $e^{2}$ is not a quadratic irrational. Since this post has grown considerably in length, we postpone this proof to the next post.

### Proof that $e^{2}$ is Irrational

We know that $$e^{2} = 1 + \frac{2^{1}}{1!} + \frac{2^2}{2!} + \cdots + \frac{2^{n}}{n!} + \frac{2^{n + 1}}{(n + 1)!} + \cdots$$ and hence if we assume that $e^{2}$ is rational say $a/b$ where $a, b$ are positive integers, then we see that \begin{align} n!a = n!be^{2} &= n!b\left(1 + \frac{2^{1}}{1!} + \frac{2^2}{2!} + \cdots + \frac{2^{n}}{n!}\right)\notag\\ &\,\,\,\,\,\,\,\,+ b\left(\frac{2^{n + 1}}{n + 1} + \frac{2^{n + 2}}{(n + 1)(n + 2)} + \cdots\right)\notag\\ \Rightarrow n!a &= b\cdot S + b\cdot R\tag{1}\end{align} where $S$ and $R$ denote first and seconds sums respectively in brackets on the RHS.From the above we can see that LHS is an integer and $S$ is also an integer, thereby implying that the second sum on RHS ($b\cdot R$) is an integer. However this second sum is not bounded (because of large powers of $2$ involved). In fact we can see that \begin{align}R &= \frac{2^{n + 1}}{n + 1} + \frac{2^{n + 2}}{(n + 1)(n + 2)} + \cdots\notag\\ &= 2^{n + 1}\left(\frac{1}{n + 1} + \frac{2}{(n + 1)(n + 2)} + \cdots\right)\notag\\ &< 2^{n + 1}\left(\frac{1}{n + 1} + \frac{2}{(n + 1)^{2}} + \frac{2^{2}}{(n + 1)^{3}} + \cdots\right)\notag\\ &= 2^{n + 1}\cdot\dfrac{\dfrac{1}{n + 1}}{1 - \dfrac{2}{n + 1}} = \frac{2^{n + 1}}{n - 1}\notag\end{align} The above derivation is valid if $n > 1$ (because in summing an infinite GP we need common ratio $2/(n + 1) < 1$). We thus have $$\frac{2^{n + 1}}{n + 1} < R < \frac{2^{n + 1}}{n - 1}$$ Now we need to analyze the sum $S$. We have $$S = n!\left(1 + \frac{2}{1!} + \frac{2^{2}}{2!} + \cdots + \frac{2^{n}}{n!}\right) = \sum_{k = 0}^{k = n}2^{k}\cdot\frac{n!}{k!}$$ We need to analyze the highest power of $2$ in each term $2^{k}\cdot n! / k!$. Clearly the highest power of $2$ in $n!$ is given by $$\left[\frac{n}{2}\right] + \left[\frac{n}{2^{2}}\right] + \cdots \leq \frac{n}{2} + \frac{n}{2^{2}} + \cdots = n$$ Hence the highest power of $2$ in $n!$ is at most $n$. In fact the highest such power occurs when $n$ itself is a power of $2$ and each of $n/2^{r}$ is integral. Let us then take $n = 2^{m}$ and then highest power of $2$ in $n! = (2^{m})!$ is given by \begin{align}\left[\frac{2^{m}}{2}\right] + \left[\frac{2^{m}}{2^{2}}\right] + \cdots &= 2^{m - 1} + 2^{m - 2} + \cdots + 2 +1\notag\\ &= 2^{m} - 1 = n - 1\notag\end{align} From now on we keep $n$ always as power of $2$ i.e. $n = 2^{m}$. Then the highest power of $2$ in term $2^{k}\cdot n!/k!$ is at least $k + n - 1 - k = n - 1$. It follows now that sum $S$ is an integer divisible by $2^{n - 1}$. Similarly $n!a$ is also divisible by $2^{n - 1}$. Thus dividing equation $(1)$ by $2^{n - 1}$ we get \begin{align}\frac{n!a}{2^{n - 1}} &= b\cdot \frac{S}{2^{n - 1}} + b\cdot\frac{R}{2^{n - 1}}\notag\\ A &= b.S' + bR'\tag{2}\end{align} where $A$ and $S'$ are integers and $R' = R/(2^{n - 1})$ and $bR'$ is also an integer because of the above equation.

We have earlier proved that $$\frac{2^{n + 1}}{n + 1} < R < \frac{2^{n + 1}}{n - 1}$$ and therefore \begin{align} 0 &< \frac{4b}{n + 1} < b\frac{R}{2^{n - 1}} < \frac{4b}{n - 1}\notag\\ \Rightarrow 0 &< bR' < \frac{4b}{n - 1} < 1\notag\end{align} provided $n > 4b + 1$ and $n$ a power of $2$ say $n = 2^{m}$. It now follows that we can choose a suitable value of $n$ such that $0 < bR' < 1$ and hence $bR'$ is not an integer. This is the contradiction we needed to achieve and this proves that $e^{2}$ is irrational.

**Note:**This proof has been taken from a paper "New Proofs of Irrationality of $e^{2}$ and $e^{4}$" by John B. Cosgrave. The basic idea of dividing by $2^{n - 1}$ is by Liouville.Aigner and Ziegler carry on this idea a little forward in their book

*"Proofs from the BOOK"*and show that $e^{4}$ is irrational. This we present below.### Proof that $e^{4}$ is Irrational

Let us assume on the contrary that $e^{4} = a/b$ where $a, b$ are positive integers. This means that $be^{2} = ae^{-2}$. Multiplying the equation by $n!$ (where $n$ is some power of $2$, say $n = 2^{m}$) we get:$$b(n!e^{2}) = a(n!e^{-2})\tag{3}$$ As shown earlier $n!e^{2}$ can be expressed $n!e^{2} = S_{1} + R_{1}$ where $S_{1}$ is an integer divisible by $2^{n - 1}$ and \begin{align} R_{1} &= \frac{2^{n + 1}}{n + 1} + \frac{2^{n + 2}}{(n + 1)(n + 2)} + \cdots\notag\\ &< \frac{2^{n + 1}}{n + 1} + \frac{2^{n + 2}}{(n + 1)^{2}} + \cdots\notag\\ &= \frac{2^{n + 1}}{n - 1}\notag\end{align} Similarly we can write $n!e^{-2} = S_{2} + R_{2}$ where $S_{2}$ is an integer divisible by $2^{n - 1}$ and $$R_{2} = (-1)^{n + 1}\left(\frac{2^{n + 1}}{n + 1} - \frac{2^{n + 2}}{(n + 1)(n + 2)} + \cdots\right)$$ Since $n$ is even it follows that $R_{2} < 0$ and $$|R_{2}| < \frac{2^{n + 1}}{n + 1}$$ We can now divide equation $(3)$ by $2^{n - 1}$ to arrive at: \begin{align}b\left(\frac{S_{1}}{2^{n - 1}} + \frac{R_{1}}{2^{n - 1}}\right) &= a\left(\frac{S_{2}}{2^{n - 1}} + \frac{R_{2}}{2^{n - 1}}\right)\notag\\ \Rightarrow bB + \frac{bR_{1}}{2^{n - 1}} &= aA + \frac{aR_{2}}{2^{n - 1}}\notag\\ \Rightarrow bB + B' &= aA + A'\tag{4}\end{align} where $0 < B' < 4b/(n - 1)$ and $-4a/(n + 1) < A' < 0$ and $A, B$ are integers. For large values of $n = 2^{m}$ we see an obvious contradiction as the LHS of the last equation $(4)$ is slightly larger than an integer and RHS of the equation is slightly less than an integer.

The same argument can be used to prove that $e^{2}$ is not a quadratic irrational. Since this post has grown considerably in length, we postpone this proof to the next post.

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