Irrationality of π(PI): Lambert’s Proof Contd.

Irrationality of Continued Fractions

We have the following results about the irrationality of some continued fractions:
If $a_{n}, b_{n}$ are positive integers then:
1) The infinite continued fraction $$\frac{b_{2}}{a_{2} +}\,\frac{b_{3}}{a_{3} +}\,\frac{b_{4}}{a_{4} +}\,\frac{b_{5}}{a_{5} +}\,\cdots$$ converges to an irrational value, provided that $a_{n} \geq b_{n}$ for all values of $n$ starting from a certain value $n = n_{0}$.

2) The infinite continued fraction $$\frac{b_{2}}{a_{2} -}\,\frac{b_{3}}{a_{3} -}\,\frac{b_{4}}{a_{4} -}\,\frac{b_{5}}{a_{5} -}\,\cdots$$ converges to an irrational value, provided that $a_{n} \geq b_{n} + 1$ for all values of $n$ starting from a certain value $n = n_{0}$ and the condition $a_{n} > b_{n} + 1$ must hold for an infinite number of values of $n$.

The idea of the proof is simple (we will explain it for the second case mentioned above) and based on the fact that under the conditions mentioned above the value of the continued fraction is less than unity. To proceed further lets assume that the relation $a_{n} > b_{n} + 1$ holds for all values of $n$. In this case the value of the continued fraction is less than unity. Let's assume that this value is a fraction $A_{2} / A_{1}$ where $A_{1}, A_{2}$ are integers. Then we have $0 < A_{2} < A_{1}$.

Thus we have
$$\frac{A_{2}}{A_{1}} = \frac{b_{2}}{a_{2} -}\,\frac{b_{3}}{a_{3} -}\,\frac{b_{4}}{a_{4} -}\,\frac{b_{5}}{a_{5} -}\,\cdots$$ so that $$\frac{b_{3}}{a_{3} -}\,\frac{b_{4}}{a_{4} -}\,\frac{b_{5}}{a_{5} -}\,\cdots = \frac{A_{2}a_{2} - A_{1}b_{2}}{A_{2}}$$ The infinite continued fraction on the left is less than unity hence the integer $A_{3} = A_{2}a_{2} - A_{1}b_{2} < A_{2}$. Thus we have $0 < A_{3} < A_{2} < A_{1}$. This reasoning can be continued indefinitely thereby obtaining an infinite sequence of distinct integers $A_{i}$ lying between $0$ and $A_{1}$. This is obviously false and we reach a contradiction. Therefore the assumption that the original continued fraction represents a rational number is false. Now we need to understand the reason behind the condition $a_{n} > b_{n} + 1$ to hold infinitely many times. If this condition is replaced by $a_{n} = b_{n} + 1$ then we may at some step have the value of continued fraction as equal to unity and then the integers $A_{i}$ will not all be distinct. By requiring an infinity of relations like $a_{n} > b_{n} + 1$ we ensure that we have an infinity of distinct $A_{i}$ and this part is crucial to obtain a contradiction.

Another point to note here is that the conditions on $a_{n}, b_{n}$ should hold starting from some value of $n$. This is easy to understand if we note that if $x$ is irrational then $b_{n} / (a_{n} - x)$ is also irrational. And hence going backwards the original continued fraction will also be irrational even if the conditions required hold after a certain value of $n$.

The proof for the case of first class of continued fractions follows in the same manner because the conditions imposed on $a_{n}, b_{n}$ imply that the value of the continued fraction is less than unity.

Irrationality of $\pi$

It is now easy to deduce the irrationality of $\pi$. We start with the following result:
If $x$ is a non-zero rational number then $\tan x$ is irrational.

To prove it we can assume that $x > 0$ (because changing sign of $x$ will only change the sign of $\tan x$) and set $x = p / q$ where $p, q$ are positive integers have no common factor.

From an earlier post we know that:
$$\tan x = \frac{x}{1 -}\,\frac{x^{2}}{3 -}\,\frac{x^{2}}{5 -}\,\frac{x^{2}}{7 -}\,\cdots$$ Putting $x = p/q$ and simplifying we get $$\tan x = \frac{p}{q}\,\frac{p^{2}}{3q -}\,\frac{p^{2}}{5q -}\,\frac{p^{2}}{7q -}\,\cdots \frac{p^{2}}{(2n + 1)q -}\,\cdots$$ It is easy to see that after some value of $n$ we will start having $(2n + 1) q > p^{2} + 1$ and hence the above infinite continued fraction represents an irrational number.

The irrationality of $\pi$ follows from the fact that $\tan(\pi / 4) = 1$. In the similar fashion we can prove that $\pi^{2}$ is irrational by using the fact that $\tan(\pi) = 0$.

Using a similar argument and the continued fraction expansion $$\tanh x = \frac{x}{1 +}\,\frac{x^{2}}{3 +}\,\frac{x^{2}}{5 +}\,\frac{x^{2}}{7 +}\,\cdots$$ we can prove that $e^{x}$ is irrational if $x$ is a non-zero rational number.

This is how Johann H. Lambert established the irrationality of $\pi, \pi^{2}$ and $e^{x}$ and I think it is simpler (in terms of obviousness) compared to the Niven's proof.

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