Thoughts on 1/7 and other rational numbers (Part 5)

Students are quite accustomed to obtaining the decimal expansion of a fraction by using a calculator. Here’s an (uncommonly, I think) taught technique for converting certain fractions into a decimal expansion without using long division and without using a calculator. I’ve taught this technique to college students who want to be future high school teachers for several years, and it never fails to surprise.

First off, it’s easy to divide any number by a power of $10$ , or $10^k$ . For example,

$\displaystyle \frac{4312}{1000} = 4.312$ and $\displaystyle \frac{71}{10000} = 0.00071$

What’s less commonly known is that it’s also easy to divide by $10^k - 1$ , or $99\dots 9$ , a numeral with $k$ consecutive $9$ s. (This number can be used to prove the divisibility rules for 3 and 9 and is also the subject of one of my best math jokes.) The rule can be illustrated with a calculator:

In other words, if $M < 10^k - 1$ , then the decimal expansion of $\displaystyle \frac{M}{10^k-1}$ is a repeating block of $k$ digits containing the numeral $M$ , possibly adding enough zeroes to fill all $k$ digits.

To prove that this actually works, we notice that

$\displaystyle \frac{M}{10^k - 1} = M \times \frac{ \displaystyle \frac{1}{10^k}}{\quad \displaystyle 1 - \frac{1}{10^k} \quad}$

$\displaystyle \frac{M}{10^k - 1} = M \times \left(\displaystyle \frac{1}{10^k} + \frac{1}{10^{2k}} + \frac{1}{10^{3k}} + \dots \right)$

$\displaystyle \frac{M}{10^k-1} = M \times 0.\overline{00\dots01}$

The first line is obtained by multiplying the numerator and denominator by $\displaystyle \frac{1}{10^k}$ . The second line is obtained by using the formula for an infinite geometric series in reverse, so that the first term is $\displaystyle \frac{1}{10^k}$ and the common ratio is also $\displaystyle \frac{1}{10^k}$ . The third line is obtained by converting the series — including only powers of $10$ — into a decimal expansion.

If $M > 10^k - 1$ , then the division algorithm must be used to get a numerator that is less than $10^k-1$ . Fortunately, dividing big numbers by $10^k-1$ is quite easy and can be done without a calculator. For example, let’s find the decimal expansion of $\displaystyle \frac{123456}{9999}$ without a calculator. First,

$123456 = 12(10000) + 3456$

$123456 = 12(9999 + 1) + 3456$

$123456 = 12(9999) + 12(1) + 3456$

$123456 = 12(9999) + 3468$

Therefore,

$\displaystyle \frac{123456}{9999} = \displaystyle \frac{12(9999) + 3468}{9999}$

$\displaystyle \frac{123456}{9999} = \displaystyle \frac{12(9999)}{9999} + \frac{3468}{9999}$

$\displaystyle \frac{123456}{9999} = \displaystyle 12 + \frac{3468}{9999}$

$\displaystyle \frac{123456}{9999} = \displaystyle 12.\overline{3468}$

This can be confirmed with a calculator. Notice that the repeating block doesn’t quite match the digits of the numerator because of the intermediate step of applying the division algorithm.

In the same vein, it’s also straightforward to find the decimal expansion of fractions of the form $\displaystyle \frac{M}{10^d (10^k-1)}$ , so that the denominator has the form $99\dots9900\dots00$ . This is especially easy if $M < 10^k -1$ . For example,

$\displaystyle \frac{123}{99900} = \frac{1}{100} \times \frac{123}{999} = \frac{1}{100} \times 0.\overline{123} = 0.00\overline{123}$

On the other hand, if $M > 10^k-1$ , then the division algorithm must be applied as before. For example, let’s find the decimal expansion of $\displaystyle \frac{51237}{99000}$ . To begin, we need to divide the numerator by $99$ , as before. Notice that, for this example, an extra iteration of the division algorithm is needed to get a remainder less than $99$ .

$51237 = 512(100) + 37$

$51237 = 512(99 + 1) + 37$

$51237 = 512(99) + 512 + 37$

$51237 = 512(99) + 549$

$51237= 512(99) + 5(100) + 49$

$51237 = 512(99) + 5(99 + 1) + 49$

$51237 = 512(99) + 5(99) + 5 + 49$

$51237 = 517(99) + 54$

Therefore,

$\displaystyle \frac{51237}{99000} = \displaystyle \frac{517(99) + 54}{99000}$

$\displaystyle \frac{51237}{99000} = \displaystyle \frac{517(99)}{99000} + \frac{54}{99000}$

$\displaystyle \frac{51237}{99000} = \displaystyle \frac{517}{1000} + \frac{54}{99000}$

$\displaystyle \frac{51237}{99000} = 0.517 + 0.000\overline{54}$

$\displaystyle \frac{51237}{99000} = 0.517\overline{54}$

In particular, notice that the three $0$ s in the denominator correspond to a delay of length 3 (the digits $517$ ), while the $99 = 10^2 - 1$ in the denominator corresponds to the repeating block of length $2$ .

These can be confirmed for students who may be reluctant to believe that decimal expansions can be found without a calculator.

Thoughts on 1/7 and other rational numbers (Part 5)

Published by John Quintanilla

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Published by John Quintanilla

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