Thoughts on 1/7 and other rational numbers (Part 6)

In Part 5 of this series, I showed that fractions of the form \displaystyle \frac{M}{10^d}, \displaystyle \frac{M}{10^k - 1}, and \displaystyle \frac{M}{10^d (10^k-1)} can be converted into their decimal representations without using long division and without using a calculator.

The amazing thing is that every rational number \displaystyle \frac{a}{b} can be written in one of these three forms. Therefore, after this conversion is made, then the decimal expansion can be found without a calculator.

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Case 1. If the denominator b has a prime factorization of the form 2^m 5^n, then \displaystyle \frac{a}{b} can be rewritten in the form \displaystyle \frac{M}{10^d}, where d = \max(m,n).

For example,

\displaystyle \frac{3}{160} = \displaystyle \frac{3}{2^5 \times 5}

\displaystyle \frac{3}{160} = \displaystyle \frac{3}{2^5 \times 5} \times \frac{5^4}{5^4}

\displaystyle \frac{3}{160} = \displaystyle \frac{3 \times 5^4}{2^5 \times 5^5}

\displaystyle \frac{3}{160} = \displaystyle \frac{1875}{10^5}

\displaystyle \frac{3}{160} = 0.01875

The step of multiplying both sides by \displaystyle \frac{5^4}{5^4} is perhaps unusual, since we’re so accustomed to converting fractions into lowest terms and not making the numerators and denominators larger. This particular form of 1 was chosen in order to get a power of 10 in the denominator, thus facilitating the construction of the decimal expansion.

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Case 2. If the denominator b is neither a multiple of 2 nor 5, then  \displaystyle \frac{a}{b} can be rewritten in the form \displaystyle \frac{M}{10^k - 1}.

For example,

\displaystyle \frac{3}{11} = \displaystyle \frac{3}{11} \times \frac{9}{9}

\displaystyle \frac{3}{11} = \displaystyle \frac{27}{99}

\displaystyle \frac{3}{11} = 0.\overline{27}

This example wasn’t too difficult since we knew that 9 \times 11 = 99. However, finding the smallest value of k that works can be a difficult task requiring laborious trial and error.

However, we do have a couple of theorems that can assist in finding k. First, since k is the length of the repeating block, we are guaranteed that k must be less than the denominator b since, using ordinary long division, the length of the repeating block is determined by how many steps are required until we get a remainder that was seen before.

However, we can do even better than that. Using ideas from number theory, it can be proven that k must be a factor of \phi(b), which is the Euler toitent function or the number of integers less than b that are relatively prime with b. In the example above, the denominator was 11, and clearly, if 1 \le n < 11, then \gcd(n,11) = 1. Since there are 10 such numbers, we know that k must be a factor of 10. In other words, k must be either 1, 2, 5, or 10, thus considerably reducing the amount of guessing and checking that has to be done. (Of course, for the example above, k=2 was the least value of k that worked.)

In general, if n = p_1^{a_1} p_2^{a_2} \dots p_r^{a_r} is the prime factorization of n, then

\phi(n) = n \left( \displaystyle 1 - \frac{1}{p_1} \right) \left( \displaystyle 1 - \frac{1}{p_2} \right) \dots \left( \displaystyle 1 - \frac{1}{p_r} \right)

For the example above, since 11 was prime, we have \phi(11) = 11 \left( \displaystyle 1 - \frac{1}{11} \right) = 10.

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Case 3. Suppose the prime factorization of the denominator b both (1) contains 2 and/or 5 and also (2) another prime other than 2 and 5. This is a mixture of Cases 1 and 2, and the fraction \displaystyle \frac{a}{b} can be rewritten in the form \displaystyle \frac{M}{10^d (10^k-1)}.

For example, consider

\displaystyle \frac{11}{74} = \displaystyle \frac{11}{2 \times 37}

Following the rule for Case 1, we should multiply by \displaystyle \frac{5}{5} to get a 10 in the denominator:

\displaystyle \frac{11}{74} = \displaystyle \frac{11}{2 \times 37} \times \frac{5}{5} = \frac{55}{10 \times 37}

Next, we need to multiply 37 by something to get a number of the form 10^k - 1. Since 37 is prime, every number less than 37 is relatively prime with 37, so \phi(37) = 36. Therefore, k must be a factor of 36. So, k must be one of 1, 2, 3, 4, 6, 9, 12, 18, and 36.

(Parenthetically, while we’ve still got some work to do, it’s still pretty impressive that — without doing any real work — we can reduce the choices of k to these nine numbers. In that sense, the use of \phi(n) parallels how the Rational Root Test is used to determine possible roots of polynomials with integer coefficients.)

So let’s try to find the least value of k that works.

  • If k = 1, then 10^1 - 1 = 9, but 9 \div 37 is not an integer.
  • If k = 2, then 10^2 - 1 = 99, but 99 \div 37 is not an integer.
  • If k = 3, then 10^3 - 1 = 999, and it turns out that 999 \div 37 = 27, an integer.


\displaystyle \frac{11}{74} = \frac{55}{10 \times 37} \times \frac{27}{27}

\displaystyle \frac{11}{74} = \frac{1485}{10 \times 999}

\displaystyle \frac{11}{74} = \frac{999 + 486}{10 \times 999}

\displaystyle \frac{11}{74} = \frac{999}{10 \times 999}+ \frac{486}{10 \times 999}

\displaystyle \frac{11}{74} = \frac{1}{10}+ \frac{486}{10 \times 999}

\displaystyle \frac{11}{74} = 0.1 + 0.0\overline{486}

\displaystyle \frac{11}{74} = 0.1\overline{486}

6 thoughts on “Thoughts on 1/7 and other rational numbers (Part 6)

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