# Irrationality of ${\pi}$ and $e^{x/y}$

I attempt to prove that $${\pi}$$ is an irrational number. For this, I use the beautiful continued fraction given by Brouncker who rewrote Wallis' formula as a continued fraction, which Wallis and later Euler (1775) proved to be equivalent.

The continued fraction given by Brouncker is $$\frac{4}{\pi} = 1 + \cfrac{1}{3 + \cfrac{4}{5 + \cfrac{9}{7 + \dots}}}$$

Now we know that every infinite continued fraction is irrational. Does it mean that since RHS has an infinite continued fraction the LHS $$\frac{4}{\pi}$$ must also be irrational? And hence $${\pi}$$ must be an irrational number?

EDIT 1: From this post General Continued Fractions and Irrationality If, in the continued fraction $$\cfrac{b_1}{a_1 + \cfrac{b_2}{a_2 + \cfrac{b_3}{a_3 + \dots}}}$$ the values $$a_{i}, b_{i}$$ are all positive integers, and if we have $$a_i \geq b_i$$ for all $$i$$ greater than some $$n$$, then the value of the continued fraction is irrational(Is this a necessary and sufficient condition? I am in doubt since the golder relation $$\phi$$ has $$a_i \geq b_i=1$$ for all $$n\geq1$$). As the condition is not satisfied the proof is actually wrong.

EDIT 2: How to prove the following continued fraction of $$e^{x/y}$$

$${\displaystyle e^{x/y}=1+{\cfrac {2x}{2y-x+{\cfrac {x^{2}}{6y+{\cfrac {x^{2}}{10y+{\cfrac {x^{2}}{14y+{\cfrac {x^{2}}{18y+\ddots }}}}}}}}}}}$$

Since $$a_i \geq b_i$$ for all $$i \geq 1$$. By the condition of irrationality of generalized continued fraction, proving this directly proves that $$e^{x/y}$$ will be an irrational number!

• Proving $\pi$ is irrational number is much harder than it seems... Commented Feb 4, 2021 at 10:03
• math.stackexchange.com/questions/895611/…
– user65203
Commented Feb 4, 2021 at 10:07
• That's the idea behind Lambert's proof of the irrationality of $\pi$. See this and this. Commented Feb 4, 2021 at 10:07
• Your right-hand side is not really a continued fraction but a generalised continued fraction, with numerators greater than the first term of the denominators, so it does not quite work Commented Feb 4, 2021 at 10:08
• @Henry Does the generalized infinite continued fraction is not an irrational number? Commented Feb 4, 2021 at 10:10

To see that not every generalized infinite continued fraction is irrational, note that if $$x=1 + \frac{2}{1 + \frac{2}{1 + \frac{2}{1 + ...}}}$$ then $$x=1+{\large{\frac{2}{x}}}$$, so $$x=2$$.

Of course that assumes that the continued fraction converges.

To prove $$x=2$$ more rigorously, we can argue as follows . . .

By definition, $$x$$ is the limit, if it exists, of the infinite sequence $$x_0,x_1,x_2,...$$ defined recursively by $$x_0=1$$ and $$x_n=1+\frac{2}{x_{n-1}}$$ for $$n\ge 1$$.

Clearly we have $$x_n \ge 1$$ for all $$n$$.

For $$n\ge 1$$ we have $$x_n=1+\frac{2}{x_{n-1}}\le 1+\frac{2}{1}=3$$ hence, since $$x_0=1$$, we have $$x_n\le 3$$ for all $$n$$.

Then for all $$n\ge 1$$ we get $$x_n=1+\frac{2}{x_{n-1}}\ge 1+\frac{2}{3}=\frac{5}{3}$$ From the recursion we get $$x_n-2=\frac{2-x_{n-1}}{x_{n-1}}$$ for all $$n\ge 1$$, hence for all $$n\ge 2$$ we get $$\left|x_{n}-2\right|\le\frac{\left|x_{n-1}-2\right|}{\left({\large{\frac{5}{3}}}\right)}$$ and then an easy induction yields $$\left|x_n-2\right|\le\Bigl(\frac{3}{5}\Bigr)^{\large{n-1}}$$ for all $$n\ge 2$$.

It follows that $$\lim_{n\to\infty}x_n = 2$$ so $$x=2$$.

As to the question you raised in your edit, consider the generalized continued fraction $$1 + \frac{3}{1 + \frac{3}{1 + \frac{3}{1 + ...}}}$$ which can be shown to be equal to $$\frac{1+\sqrt{13}}{2}$$ Thus the condition for irrationality that you referenced is a sufficient condition, but not a necessary condition.

• To be extra rigorous, one would also have to prove that it's convergent. Commented Feb 4, 2021 at 10:24
• Agreed. I'll edit in a proof of that. Commented Feb 4, 2021 at 10:25
• $x=1+\frac{0}{1+\frac{0}{1+\frac{0}{1+\dots}}}$ is rational as well, with arguably more apparent convergence. Commented Feb 4, 2021 at 10:28

Regarding the second part of the question.

I think this might be a solution.

The Continued Fraction Expansion of the hyperbolic tanh function discovered by Gauss is

$$\tanh z = \frac{z}{1 + \frac{z^2}{3 + \frac{z^2}{5 + \frac{z^2}{...}}}} \\$$

We also know that the hyperbolic tanh function is related to the exponential function with the following formula

$$\tanh z=\frac{e^z-e^{-z}}{e^z+e^{-z}}$$

Putting $$\frac{x}{y}$$ in place of $$z$$ in the previous equation we get $$\frac{e^{\frac{x}{y}}-e^{-{\frac{x}{y}}}}{e^{\frac{x}{y}}+e^{-{\frac{x}{y}}}}= \frac{{\frac{x}{y}}}{1 + \frac{{\frac{x}{y}}^2}{3 + \frac{{\frac{x}{y}}^2}{5 + \frac{{\frac{x}{y}}^2}{...}}}} \\$$

This continued fraction can be simplified into

$$\frac{e^{\frac{x}{y}}-e^{-{\frac{x}{y}}}}{e^{\frac{x}{y}}+e^{-{\frac{x}{y}}}}= \frac{x}{y + \frac{x^2}{3y + \frac{x^2}{5y + \frac{x^2}{...}}}} \\$$

This equation can be further be simplified as

$$1+\frac{2}{e^{\frac{2x}{y}}-1}=y + \frac{x^2}{3y + \frac{x^2}{5y + \frac{x^2}{...}}} \\$$

$$\frac{e^{\frac{2x}{y}}-1}{2}=\frac{1}{(\frac{y}{x}-1) + \frac{x}{3y + \frac{x^2}{5y + \frac{x^2}{...}}}}$$

From this equation after some algebraic manipulation, we finally get the continued fraction expansion of $$e^{x/y}$$ as

$${\displaystyle e^{x/y}=1+{\frac {2x}{2y-x+{\frac {x^{2}}{6y+{\frac {x^{2}}{10y+{\frac {x^{2}}{14y+{\frac {x^{2}}{18y+\ddots }}}}}}}}}}}$$

This is an infinite generalized continued fraction. We will now state the necessary and sufficient condition for the continued fraction proved by Lagrange given in Corollary 3, on page 495, in chapter XXXIV on "General Continued Fractions" of Chrystal's Algebra Vol.II to converge into an irrational number.

\begin{theorem}The necessary and sufficient condition that the continued fraction $$\frac{b_1}{a_1 + \frac{b_2}{a_2 + \frac{b_3}{a_3 + \dots}}}$$

is irrational is that the values $$a_{i}, b_{i}$$ are all positive integers, and if we have $$|a_i| > |b_i|$$ for all $$i$$ greater than some $$n$$ \end{theorem}

In the continued fraction of $$e^{x/y}-1$$ as we have derived $$a_{i}, b_{i}$$ are equals to $$2(2i-1),x^2$$ except $$i=1$$.Therefore we have $$|a_i| > |b_i|$$ for all $$i>\frac{\frac{x^2}{2}+1}{2}$$. Hence we have proved that $$e^{x/y}-1$$ is irrational which in turn means $$e^{x/y}$$ is irrational.