If α and β are two real numbers, we say that β is equivalent to α if there are integers a, b, c, and d such that ad−bc=±1 and β=aα+bcα+d.
How can I show that two irrational numbers α and β are equivalent if the tails of their simple continued fractions agree, that is, if
α=[a0;a1,a2,…,aj,c1,c2,c3,⋯]β=[b0;b1,b2,…,bk,c1,c2,c3,…]
where ai, i=0,1,2,…,j; bi, i=0,1,2,…,k; and c1, i=1,2,3,… are integers, all positive except perhaps a0 and b0?
This is what I tried: I assumed the above and tried to show that there exist a, b, c, and d such that ad−bc=±1 and
β=aα+bcα+d⟹b0+1b1+1b2+⋯=a(a0+1a1+1a2+⋯)+bc(a0+1a1+1a2+⋯)+d
However, the operations on the fractions are confusing, and I have the feeling that algebraic manipulations will not help or that there may be an easier way to tackle this problem.
Answer
Hint This - like many properties of continued fractions - is clearer when continued fractions are viewed as sequences of Möbius maps. For an introduction to the Möbius viewpoint see the following exposition: A. Beardon, Continued Fractions, Discrete Groups and Complex
Dynamics.
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