Uniqueness Infinite Continued Fraction
Given any irrational number
\[x\]
determined from \[x\]
by the continued fraction algorithm converges to \[x\]
and no other finite continued fraction converges to \[x\]
.Proof
Using the continued fraction algorithm, the irrational number
\[x\]
produces a sequence of finite continued fractions\[x=x_1, \; [ a_1, \; x_2 ] , \; [ a_1 , \; a_2, \; x_3 ] , \; [ a_1, \; a_2, \; a_3, \; x_4 ] ,...\]
where
\[a_n= int(x_n)\]
and \[x_{n+1}= \frac{1}{x_n-a_n}\]
.To show that the infinite continued fraction has limit
\[x\]
let \[[ a_1, \; a_2,..., \; a_{2n} ] = \frac{p_k}{q_k}=C_k\]
, the \[2n\]
th convergent, then from the properties of the convergents, \[\| x -\frac{p_{2n}}{q_{2n+1}} \| \lt | C_{2n}-C_{2n-1} | = \frac{1}{q_{2n} q_{2n-1}} \lt \frac{1}{2n(2n+1)}\]
hence \[C_{2n} \rightarrow x\]
.We can prove that if
\[[ a_1. \; a_2,... ], \; [b_1, \; b_2,... ]\]
have the same value, then they are the same continued fraction by induction.Let
\[P(k)\]
be the statement that \[a_k=b_k\]
. \[a_1=b_1\]
since both continued fractions have the same integer part, so \[P(1)\]
is true.Suppose then that
\[a_1=b_1, \; a_2=b_2,..., \; a_k=b_k\]
.\[[ a_1, \; a_2,..., a_k [ a_{k+1},... ] ] = [ b_1, \; b_2,..., b_k [ b_{k+1},... ] ]\]
Hence
\[a_{k+1}=b_{k+1}\]
and equality of the continued fractions is true by induction. 
