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3.1: Binary Representations

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    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 \(n=1,2,3, \ldots\).



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

    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

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


    Proposition \(\PageIndex{2}\)

    With the notation as above,

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


    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

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


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


    lemma \(\PageIndex{3}\)

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


    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.

    This page titled 3.1: Binary Representations is shared under a CC BY-NC-SA 1.0 license and was authored, remixed, and/or curated by Dan Sloughter via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.