
# 3.1: Binary Representations


Suppose $$\left\{a_{n}\right\}_{n=1}^{\infty}$$ is a sequence such that, for each $$n=1,2,3, \ldots,$$ either $$a_{n}=0$$ or $$a_{n}=1$$ and, for any integer $$N,$$ there exists an integer $$n>N$$ such that $$a_{n}=0 .$$ Then

$0 \leq \frac{a_{n}}{2^{n}} \leq \frac{1}{2^{n}}$

for $$n=1,2,3, \dots,$$ so the infinite series

$\sum_{n=1}^{\infty} \frac{a_{n}}{2^{n}}$

converges to some real number $$x$$ by the comparison test. Moreover,

$0 \leq x<\sum_{n=1}^{\infty} \frac{1}{2^{n}}=1.$

We call the sequence $$\left\{a_{n}\right\}_{n=1}^{\infty}$$ the binary representation for $$x,$$ and write

$x=.a_{1} a_{2} a_{3} a_{4} \dots.$

## Exercise $$\PageIndex{1}$$

Suppose $$\left\{a_{n}\right\}_{n=1}^{\infty}$$ and $$\left\{b_{n}\right\}_{n=1}^{\infty}$$ are both binary representations for $$x .$$ Show that $$a_{n}=b_{n}$$ for $$n=1,2,3, \ldots$$.

Now suppose $$x \in \mathbb{R}$$ with $$0 \leq x<1$$. Construct a sequence $$\left\{a_{n}\right\}_{n=1}^{\infty}$$ as follows: If $$0 \leq x<\frac{1}{2},$$ let $$a_{1}=0 ;$$ otherwise, let $$a_{1}=1 .$$ For $$n=1,2,3, \ldots,$$ let

$s_{n}=\sum_{i=1}^{n} \frac{a_{i}}{2^{i}}$

and set $$a_{n+1}=1$$ if

$s_{n}+\frac{1}{2^{n+1}} \leq x$

and $$a_{n+1}=0$$ otherwise.

## lemma $$\PageIndex{1}$$

With the notation as above,

$s_{n} \leq x<s_{n}+\frac{1}{2^{n}}$

for $$n=1,2,3, \ldots$$.

Proof

Since

$s_{1}=\left\{\begin{array}{ll}{0,} & {\text { if } 0 \leq x<\frac{1}{2}} \\ {\frac{1}{2},} & {\text { if } \frac{1}{2} \leq x<1}\end{array}\right.$

it is clear that $$s_{1} \leq x<s_{1}+\frac{1}{2} .$$ So suppose $$n>1$$ and $$s_{n-1} \leq x<s_{n-1}+\frac{1}{2^{n-1}}$$. If $$s_{n-1}+\frac{1}{2 n} \leq x,$$ then $$a_{n}=1$$ and

$s_{n}=s_{n-1}+\frac{1}{2^{n}} \leq x<s_{n-1}+\frac{1}{2^{n-1}}=s_{n-1}+\frac{1}{2^{n}}+\frac{1}{2^{n}}=s_{n}+\frac{1}{2^{n}}.$

If $$x<s_{n-1}+\frac{1}{2^{n}},$$ then $$a_{n}=0$$ and

$s_{n}=s_{n-1} \leq x<s_{n-1}+\frac{1}{2^{n}}=s_{n}+\frac{1}{2^{n}}.$

Q.E.D.

## Proposition $$\PageIndex{2}$$

With the notation as above,

$x=\sum_{n=1}^{\infty} \frac{a_{n}}{2^{n}}.$

Proof

Given $$\epsilon>0,$$ choose an integer $$N$$ such that $$\frac{1}{2^{n}}<\epsilon .$$ Then, for any $$n>N,$$ it follows from the lemma that

$\left|s_{n}-x\right|<\frac{1}{2^{n}}<\frac{1}{2^{N}}<\epsilon .$

Hence

$x=\lim _{n \rightarrow \infty} s_{n}=\sum_{n=1}^{\infty} \frac{a_{n}}{2^{n}}.$

Q.E.D.

## lemma $$\PageIndex{3}$$

With the notation as above, given any integer $$N$$ there exists an integer $$n>N$$ such that $$a_{n}=0$$.

Proof

If $$a_{n}=1$$ for $$n=1,2,3, \dots,$$ then

$x=\sum_{n=1}^{\infty} \frac{1}{2^{n}}=1,$

contradicting the assumption that $$0 \leq x<1 .$$ Now suppose there exists an integer $$N$$ such that $$a_{N}=0$$ but $$a_{n}=1$$ for every $$n>N .$$ Then

$x=s_{N}+\sum_{n=N+1}^{\infty} \frac{1}{2^{n}}=s_{N-1}+\sum_{n=N+1}^{\infty} \frac{1}{2^{n}}=s_{N-1}+\frac{1}{2^{N}},$

implying that $$a_{N}=1,$$ and thus contradicting the assumption that $$a_{N}=0$$. $$\quad$$ Q.E.D.

Combining the previous lemma with the previous proposition yields the following result.

## Proposition $$\PageIndex{4}$$

With the notation as above, $$x=. a_{1} a_{2} a_{3} a_{4} \ldots$$.

The next theorem now follows from Exercise 3.1.1 and the previous proposition.

## Theorem $$\PageIndex{5}$$

Every real number $$0 \leq x<1$$ has a unique binary representation.