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

Question

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 1

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.

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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.

Answer 2

"Originally" there were 95, 96, 97, 98, 99, 100, 101, 102 ... but since there's only two digit positions for each of them, the one in front of 100, 101, and so forth carries over to the number in front of it and increases that preceding number by 1. This makes 99 into 100, and the one in front of that then carries over to 98 and makes it 99. And that's where the carries stop.

So the reason why exactly 98 is the one that is missing is that 98 is the largest two-digit number that can be increased by one without gaining another digit that would carry.

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It may be instructive to see what happens if we look at some other "original" pattern than 0, 1, 2, 3, ..., 98, 99, 100, 101 ...

For example, $\frac{7}{9801}$ should produce

0, 7, 14, 28, ..., 84, 91, 98, 105, 112, 119 ...

but when carrying out the division the digits we get are:

7/9801 = 0.00071428...849199061320...

because of carries. Here one sees more clearly that 98 is not "missing", it is just increased by one, like the last two digits of the subsequent numbers in the sequence.

We can also look at

9/9801 = 0.00091827...819100091827...

Here, like in $\frac1{9801}$ there's a 99 in the original sequence, but it gets bumped up to 100 by the carry, and then the "original" 90 in turn gets bumped up to 91. Likewise

11/9801 = 0.0011223344556677890011223344...

There's two of every digit except 8 and 9! Again, what really happens is that the original 99 gets bumped to 100, which again bumps 88 to 89.

Answer 3

Here's my answer now: The eight gets bumped to a nine and the nine gets bumped to a zero right here: $0.01+0.002+0.0003+0.00004+0.000005+0.0000006+0.00000007+0.000000008+0.0000000009+0.00000000010=0.012+0.0003+0.00004+0.000005+0.0000006+0.00000007+0.000000008+0.0000000009+0.00000000010=0.0123+0.00004+0.000005+0.0000006+0.00000007+0.000000008+0.0000000009+0.00000000010=0.01234+0.000005+0.0000006+0.00000007+0.000000008+0.0000000009+0.00000000010=0.012345+0.0000006+0.00000007+0.000000008+0.0000000009+0.00000000010=0.0123456+0.00000007+0.000000008+0.0000000009+0.00000000010=0.01234567+0.000000008+0.0000000009+0.00000000010=0.012345678+0.0000000009+0.00000000010=0.0123456789+0.00000000010=0.0123456790$so this is what happens. As you just saw, the 0.00000000010 (0.0000000001) changes the nine into a zero and it carries one into the eight, so it becomes a nine.