# Show that rational numbers correspond to decimals

• Feb 10th 2011, 06:58 AM
page929
Show that rational numbers correspond to decimals
Show that rational numbers correspond to decimals which are either repeating or terminating.

Hint: If q = m|n, then when dividing m by n to put q into decimal form there are at most n different remainders. Conversley, if d is a repeating decimal, then find s,t such that 10^sd - 10^td is an integer.

I took an example and understand, but don't know how to write it in a general form.

My example:

Part I
2/7
0.2857142...
72.000...
-14
60
-56
40
-35
50
-49
10
- 7
30
-28
20
-14
6....
So the my bold numbers can only be 1-6

Part II
R = 12.34545...
100R = 1234.54545...
100R - R = 1222.2
99R = (12222 / 10)
R = (12222 / 990)

Any help on writing this in a general form would be greatly appreciated.
• Feb 10th 2011, 07:53 AM
HallsofIvy
I would recommend that you interpret "terminating decimal" as a "repeating decimal" where the repeating part happens to be 0. That way, you can simply show that any rational number is a "repeating" decimal.

let m and n be two integers. Then $\displaystyle \frac{m}{n}$ means "m divided by n". If m> n, then there exist q and r ("quotient" and "remainder") such that m= qn+ r with r< n. That is, $\displaystyle \frac{m}{n}= q+ \frac{r}{n}$ so it is sufficient to prove this for $\displaystyle \frac{r}{n}$ where r is less than n. When we "long divide" n into r, we will have, at each decimal place, a "remainder" that is less than n and larger than or equal to 0, then bring down a "0" and continue. The critical point is that there can only be n different remainders, 0 to n-1 as I said. After at most n steps, a remainder must repeat, we bring down a "0" again so we are dividing n into exactly the same number as before and everything repeats.
• Feb 10th 2011, 09:00 PM
CaptainBlack
What Halls has shown is that for any rational there is some repeating decimal which is equal to it. We can also show that any repeating decimal is rational:

Let $\displaystyle $$x be a repeating decimal, we can split it into two parts: \displaystyle x=a+10^{-l} p where \displaystyle$$a$ is the terminating non periodic part of length $\displaystyle $$l digits and \displaystyle p\in[0,1] is the purely periodic. Let the period be \displaystyle$$d$ digits long and the terminating decimal with $\displaystyle $$d digits which are equal to \displaystyle$$p$ terminated after the $\displaystyle $$d -th be \displaystyle$$u$ digit then:

$\displaystyle \displaystyle p=\sum_{n=0}^{\infty} u \times 10^{-n\times d}=u\sum_{n=0}^{\infty} 10^{-n\times d}$

Now the sum on the right is a convergent geometric series and so we can write down its sum, and it is rational, and as $\displaystyle $$u is rational \displaystyle$$p$ is rational. Hence as $\displaystyle $$a is rational \displaystyle$$x$ is rational.

CB
• Feb 11th 2011, 04:50 AM
HallsofIvy
On a different forum, many years ago, a person asked how to show that any rational number could be written as a fraction with integer numerator and denominator. It took me a moment to realize that he must have been taught that the definition of "rational number" is a number that can be written as a terminating or repeating decimal. Of course, it can be done either way.