1

I've got struck by curiosity: Are there axioms or theorems about the decimal termination of numbers? For example:

$$\frac{1}{3}=0.3333333333333333\ldots$$

And

$$\frac{1}{78}=0.01282051282051282\ldots$$

The only thing I've heard was that real numbers have non-repeating infinite decimal terminations and rational numbers have finite decimal terminations or infinite repeating decimal terminations. Is there a mathematics field of study that concers with properties in the decimal terminations?

Michael Hardy
  • 1
Red Banana
  • 23,956
  • 20
  • 91
  • 192
  • 1
    Here's http://math.stackexchange.com/questions/198810/proof-that-every-repeating-decimal-is-rational/847020#847020 Michael Hardy's argument presented in a different way. –  Jun 25 '14 at 12:59

3 Answers3

3

Theorems, yeah, axioms, not really. If you expand $0.a_1a_2a_3\ldots$ as $0 + \frac{a_1}{10^1} + \frac{a_2}{10^2} + \frac{a_3}{10^3} + \cdots$, then you can analyze it more easily. If you want a way to prove it, try this (I don't know any better/more general ways):

  1. If we have a decimal $0.\overline{a_1a_2\ldots a_n}$, prove it's the same as $[a_1a_2\cdots a_n]/(10^n - 1)$, where the brackets mean just stick the digits together, not multiplication. Clearly this is rational.

  2. Prove that a decimal like $0.b_1b_2\ldots b_m\overline{a_1a_2\ldots a_n}$ is also rational. This proves all repeating expansions are rational.

  3. Prove that all rational numbers have a repeating or terminating decimal expansion. (Hint, get a $10^n$ or $10^n - 1$ in the denominator.

EDIT: After looking at some other questions, you may find these links useful.

Community
  • 1
Henry Swanson
  • 12,972
2

A rational number will terminate if the only prime factors of the denominator (when it is in lowest terms) are $2$ and/or $5$. In that case, you can multiply numerator and denominator by something to make the denominator $10^n$. Otherwise it will have an infinite repeat. You can take a terminating one and make it repeating by decrementing the last digit by $1$ and appending $999999\dots $ at the end.

Ross Millikan
  • 374,822
2

It is gross malpractice that things like this are presented dogmatically rather than honestly in schools. It results from laws requiring everyone to be taught mathematics whether they want to or not. Consequently only that which everyone can be forced to do is taught.

The pigeonhole principle is involved here. That says if you put seven objects into six boxes, at least one box gets more than one object. Consider the rational number $7/26$. Let's find its decimal expansion by long division: $$ \begin{array}{cccccccccccccccccccc} & & 0 & . & 2 & 6 & 9 & 2 & 3 & 0 & 7 & \\ \\ 26 & ) & 7 & . & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ & & 5 & & 2 \\ \\ & & 1 & & 8 & 0 & & & & & & & \longleftarrow\\ & & 1 & & 5 & 6 \\ \\ & & & & 2 & 4 & 0 \\ & & & & 2 & 3 & 4 \\ \\ & & & & & & 6 & 0 \\ & & & & & & 5 & 2 \\ \\ & & & & & & & 8 & 0 \\ & & & & & & & 7 & 8 \\ \\ & & & & & & & & 2 & 0 \\ & & & & & & & & & 0 \\ \\ & & & & & & & & 2 & 0 & 0 \\ & & & & & & & & 1 & 8 & 2 \\ \\ & & & & & & & & & 1 & 8 & 0 & \longleftarrow & \text{repetition} \end{array} $$ The remainder $18$ has been repeated.

There are just $26$ possible remainders: $0,1,2,3,4,\ldots,25$.

If $0$ occurs as a remainder, the decimal expansion terminates.

If $0$ never occurs then there are $25$ possibilities. By the pigeonhole principle, the process cannot go $26$ steps without a repitition. Once a repetition occurs, we're just doing the same problem again, and must get the same answer.

Thus every rational number has a terminating or repeating expansion.

There is also an algorithm for converting this repeating decimal back to the fraction it came from, $7/26$.

Michael Hardy
  • 1