Why do we miss 8 in the decimal expansion of 1/81, and 98 in the decimal expansion of 1/9801?

Why do we miss 8 in the decimal expansion of 1/81, and 98 in the decimal expansion of 1/9801? I’ve seen this happen that when you divide in a fraction using the square of any number with only nines in the denominator. Like in

\frac{1}{9^2}=\frac{1}{81} =
0.01234567\!\underset{\uparrow}{}\!9
01234567\!\underset{\uparrow}{}\!9
01234567\dots\,,

and in

\frac{1}{99^2}=\frac{1}{9801} =
0.0001020304050607080910111213 \dots 9697\!\underset{\uparrow}{}\!99000102 \dots\;\,,

the decimals go on predictably when suddenly in the first one you miss 8, and in the second you miss 98 and it keeps going on forever. How does this happen? Why do we miss numbers like 8 in the decimal representation of \frac{1}{9^2}, or like 98 in the decimal of \frac{1}{99^2}, or 998 in \frac{1}{999^2}, or 9998 in \frac{1}{9999^2}\;?

Answer

For \frac{1}{81}, there was an 8, but it got bumped up. We can write \frac{1}{81} as this sum:


\large{\frac{1}{81}}=\;\;\;\small
\begin{align}
&0.0
\\+\;&0.01
\\+\;&0.002
\\+\;&0.0003
\\+\;&0.00004
\\+\;&0.000005
\\+\;&0.0000006
\\+\;&0.00000007
\\+\;&0.000000008
\\+\;&0.0000000009
\\+\;&0.00000000010
\\+\;&0.000000000011
\\+\;&\underline{\quad\vdots\quad\quad\quad\quad\quad}
\\ &0.01234567901 \dots
\end{align}

This kind of effect of “carrying the 1” when the nine digit flips to a ten is the thing that is causing the behavior in all of the fractions you are describing.


To follow-up, this should provide a bit more insight as to why interesting patterns appear in the decimal representations of fractions with a power of 9 or 11 as the denominator, and see why we can write those numbers like \frac{1}{81} as that sum. First note that

\frac{1}{9} = \left( \frac{1}{10}+ \frac{1}{100}+ \frac{1}{1000}+ \dotsb \right)

so if we were to consider \frac{1}{81} like before, we would have

\frac{1}{81} = \left(\frac{1}{9}\right)^2
= \left( \frac{1}{10}+ \frac{1}{100}+ \frac{1}{1000}+ \dotsb \right)^2

Then if we were to want to know the value of the ten-thousandth’s decimal place of \frac{1}{81}, we would just have to find the numerator of the term in the expansion of this square with a denominator of 10\,000, which we can readily see is
\begin{align}
\frac{1}{81} = \Big( \dotsb + \Bigg(\Big(\frac{1}{10}\Big)\Big(\frac{1}{1000}\Big)+\Big(\frac{1}{100}\Big)&\Big(\frac{1}{100}\Big)+\Big(\frac{1}{1000}\Big)\Big(\frac{1}{10}\Big)\Bigg) +\dotsb \Big) \\
= \Big( \dotsb + \Bigg(\frac{3}{10000}&\Bigg) +\dotsb \Big)
\end{align}

So it looks like these evident patterns that appear in the decimal expansions of fraction with a multiple of 9 in the denominator is due at least partly to the fact that this infinite sum representation of \frac{1}{9} consists entirely of terms with a numerator of 1, so multiplying this sum into things may result in “predictable” behavior that results in a pattern.

As for why having a multiple of 11 in the denominator makes similar patterns, note that
\begin{align}
\frac{1}{11} &= .090909090909090909 \dots \\
&= \left(\frac{9}{100}+\frac{9}{10000}+\dotsb\right) \\
&= \left(\frac{10-1}{100}+\frac{10-1}{10000}+\dotsb\right) \\
&= \left(\frac{1}{10}-\frac{1}{100}+\frac{1}{1000}-\frac{1}{10000}+\dotsb\right)
\end{align}

So again we have an infinite sum of terms each with a numerator of 1 (just alternating sign this time) that will result in certain “predictable” patterns when multiplied by things.

Attribution
Source : Link , Question Author : Mathster , Answer Author : Mike Pierce

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