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 9s. (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:

TI999

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.

TI9999

green line

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 0s 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.

TI99000

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2 Comments

  1. Formula for an infinite geometric series (Part 11) | Mean Green Math
  2. Thoughts on 1/7 and Other Rational Numbers: Index | Mean Green Math

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