# Simple Infinite Continued Fraction is Uniquely Determined by Limit

## Theorem

Let $\sequence {a_n}_{n \mathop \ge 0}$ and $\sequence {b_n}_{n \mathop \ge 0}$ be simple infinite continued fractions in $\R$.

Let $\sequence {a_n}_{n \mathop \ge 0}$ and $\sequence {b_n}_{n \mathop \ge 0}$ have the same limit.

Then they are equal.

## Proof 1

Follows immediately from Continued Fraction Expansion of Limit of Simple Infinite Continued Fraction equals Expansion Itself.

$\blacksquare$

## Proof 2

Recall that by Simple Infinite Continued Fraction Converges, they do indeed have a limit.

The result will be achieved by the Second Principle of Mathematical Induction.

Suppose $\sqbrk {a_0, a_1, a_2, \ldots} = \sqbrk {b_0, b_1, b_2, \ldots}$ have the same value.

First we note that if $\sqbrk {a_0, a_1, a_2, \ldots} = \sqbrk {b_0, b_1, b_2, \ldots}$ then:

$a_0 = b_0$

since both are equal to the integer part of the common value.

This is our basis for the induction.

Now suppose that for some $k \ge 1$, we have:

$a_0 = b_0, a_1 = b_1, \ldots, a_k = b_k$

Then all need to do is show that $a_{k + 1} = b_{k + 1}$.

Now:

$\sqbrk {a_0, a_1, a_2, \ldots} = \sqbrk {a_0, a_1, \ldots, a_k, \sqbrk {a_{k + 1}, a_{k + 2}, \ldots} }$

and similarly:

$\sqbrk {b_0, b_1, b_2, \ldots} = \sqbrk {b_0, b_1, \ldots, b_k, \sqbrk {b_{k + 1}, b_{k + 2}, \ldots} }$

As these have the same value and have the same first $k$ partial denominators, it follows that:

$\sqbrk {a_{k + 1}, a_{k + 2}, \ldots, } = \sqbrk {b_{k + 1}, b_{k + 2}, \ldots}$

But now $a_{k + 1} = b_{k + 1}$ as each is equal to the integer part of the value of this simple infinite continued fraction.

Hence the result.

$\blacksquare$