- #1
joypav
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I will state the problem below. I don't quite understand what I am needing to show. Could someone point me in the right direction? I would greatly appreciate it.
Problem:
Let p be a natural number greater than 1, and x a real number, 0<x<1. Show that there is a sequence $(a_n)$ of integers with $0 \leq a_n < p$ for each n such that
$ x = \sum\limits_{n=1}^{\infty} \frac{a_n}{p^n} $
and that this sequence is unique except when x is of the form $ \frac{q}{p^n} $, in which case there are exactly two such sequences.
Show that, conversely, if $(a_n)$ is any sequence of integers with $ 0 \leq a_n < p$, the series
$ \sum\limits_{n=1}^{\infty} \frac{a_n}{p^n} $
converges to a real number x with $ 0 \leq x \leq 1 $. If p=10, this sequence is called the decimal expansion of x. For p=2 it is called the binary expansion of x, and for p=3, the ternary expansion.
Problem:
Let p be a natural number greater than 1, and x a real number, 0<x<1. Show that there is a sequence $(a_n)$ of integers with $0 \leq a_n < p$ for each n such that
$ x = \sum\limits_{n=1}^{\infty} \frac{a_n}{p^n} $
and that this sequence is unique except when x is of the form $ \frac{q}{p^n} $, in which case there are exactly two such sequences.
Show that, conversely, if $(a_n)$ is any sequence of integers with $ 0 \leq a_n < p$, the series
$ \sum\limits_{n=1}^{\infty} \frac{a_n}{p^n} $
converges to a real number x with $ 0 \leq x \leq 1 $. If p=10, this sequence is called the decimal expansion of x. For p=2 it is called the binary expansion of x, and for p=3, the ternary expansion.