Question

How can I show that \(x_1=[a_0;a_1,a_2,\dots]\) is equivalent to \(x_2=[a_1;a_2,a_3,\dots]\)?

Answer 1

This is elementary number theory: continued fractions.

Answer 2

With equivalence, I mean that there are integers \(a\), \(b\), \(c\), and \(d\) such that \(ad-bc=\pm1\) and \(x_1=\frac{ax_2+b}{cx_2+d}\).

Answer 3

By the way, the notation here means\[x_1=a_0+\frac{1}{a_1+\frac{1}{\begin{matrix}
a_2+ & & \\
& \ddots & \\
& &
\end{matrix}}}\]

Answer 4

seems like x1 = a0 + 1/x2

Answer 5

@experimentX gave you a good clue:\[x_1=a_0+\frac{1}{x_2}=\frac{a_0\cdot x_2+1}{1\cdot x_2+0}.\]This implies that\[ad-bc=a_0\cdot0-1\cdot1=-1,\]and thus \(x_1\) is equivalent to \(x_2\).

Answer 6

Oh! that wasn't a clue at all ... just an observation.