$-\xi = \langle -a_0-1,1,a_1-1,a_2,a_3,\cdots\rangle\ \ if \ \ \ a_1>1. $

Question

Let $\xi$ be irrational number, $\xi = \langle a_0,a_1,a_2,\cdots\rangle$. To verify that $$-\xi = \langle -a_0-1,1,a_1-1,a_2,a_3,\cdots\rangle\ \ if \ \ \ a_1>1. $$ and $$-\xi \langle -a_0-1,a_2+1,a_3,a_4,\cdots\rangle\ \ if \ \ \ a_1=1.$$

Can any one give some hints which will help to solve the problem?

Thank You.

Answer 1

Here $$\xi = a_0 + \frac{1}{a_1+\frac{1}{a_2+\frac{1}{a_3}+\ \cdots}}$$

Let us assume $\frac{1}{a_2+\frac{1}{a_3}+\ \cdots} = b.$

So, $\xi = a_0 + \frac{1}{a_1+b}$

Now, $$-\xi = \langle -a_0-1,1,a_1-1,a_2,a_3,\cdots\rangle \\ \implies -\xi = (-a_0-1) + \frac{1}{1+\frac{1}{a_1-1 +\ b}} \\ \implies \xi = a_0 + 1 - \frac{a_1-1 +\ b}{a_1 +\ b} \\ \implies \xi = a_0 +1 -1 + \frac{1}{a_1 +\ b} \\ \implies \xi = a_0 + \frac{1}{a_1+b}$$

which is true.