APPENDIX C

Two Properties of Continued Fractions

In chapter 3, Lambert’s proof of the irrationality of π required two particular properties of continued fractions, which we establish here and, for convenience, restate.

For the continued fraction

1. If {λ₁, λ₂, λ₃, . . .} is an infinite sequence of non-zero numbers, the continued fraction

has the same convergents as y and, if there is convergence, converges to y.

2. Now suppose that b₀ = 0.

If |a_i| < |b_i| for all i 1, then:

(a) |y| 1;

(b) writing

if, for some n and above, it is never the case that |y_n| = 1, then y is irrational.

Both results are susceptible to proof by induction, which we will undertake, but in an informal manner. Unfortunately the formalization of each has something of an unwelcoming shape and we leave it to the reader to manufacture the mould, if this is desired.

Proof of 1

We will simply list the first three cases, which expose the pattern.

Stage 1:

Stage 2:

Stage 3:

We hope that, with these, the mystery of the result is laid bare.

Proof of 2

(a) Since

and so

which means that

From this we have, with both sides being positive,

Using the alternative form of the Triangle Inequality, |α − β| |α| − |β|, we then have

Now, the a_i and b_i are integers and so it must be that, with |a_i| < |b_i|, |b_i| − 1 |a_i| and this means that

and so

We continue the same argument, working towards the start of the continued fraction.

We have

so

so

and

which means

Continue the process to the start of the continued fraction and we have that the modulus of its expansion to its i + 1th term is always less than 1 for all i: this must mean that, in the limit, |y| 1.

(b) Suppose that y is rational and so

It must be that |p₁/p₀| 1 and so |p₁| |p₀|. Now write y = p₁/p₀ = a₁/(b₁ + y₂). Then y₂ must be the rational number

and |y₂| 1 implies that |p₂| |p₁|. Continue the process indefinitely until we reach n so that |y_n| < 1 and therefore |p_n+1| < |p_n| and we generate the decreasing infinite sequence of positive integers:

|p₀| |p₁| |p₂| · · · |p_n| > |p_n+1| > · · ·,

which is clearly impossible. The only reconciliation is that y is irrational.