# MSC #29. Recurring decimals.

Our **29 ^{th} MATH-Inic Special for Christmas** is about a “fun, fast and easy” way to convert a fraction with a prime number denominator into a recurring decimal. In our example here, instead of dividing by 19, we will use its

**,**

*Ekadhika***2**. The

**is also discussed in Chapters 1 and 15 of the “30 Master Strategies in Computing”**

*Ekadhika*The ** Ekadhika **of 19 is

**2**, which is “one more than the one before” 9 in 19. So we start with 0 and a decimal point

**1/19 = 0.**

Next, we divide 1 by 2 getting a quotient of 0 and a remainder of 1. We place the remainder **before** the answer figure: **1/19 = 0. _{1}0**

We then divide our next dividend, 10 by 2 to get the next answer figure, 5 **1/19 = 0. _{1}0 5.**

Then we divide 5 by 2 getting an answer of 2 and remainder of 1. Again we place the remainder if front of the answer. **1/19 = 0. _{1}0 5 _{1}2**

Repeat the steps, dividing the last answer figure (including the prefixed remainder) by the Ekadhika, 2, and putting down the result as the next answer figure. Place any remainder before that answer figure.

Repeat the steps, dividing the last answer figure (including the prefixed remainder) by the Ekadhika, 2, and putting down the result as the next answer figure, until we get an answer figure of 1. We stop here, since we started with a numerator of 1 and the sequence of the answer figure will just repeat in a cycle.

**1/19 = 0. _{1}0 5 _{1}2 6 3 _{1}1 _{1}5 _{1}7 _{1}8 9 _{1}4 7 _{1}3 _{1}6 8 4 2 1 **

We can also start
computing from the right. Notice that the last answer figure is 1. Now instead
of dividing by two, we will multiply by 2.
From 1, we have 2, 4, 8 and _{1}6. The next answer figure will
be 6 x 2 + 1 or _{1}3. Then we have 3 x 2 + 1 or 7 and so on. This is
not possible using the conventional method.

There will be 18 digits in the recurring decimal. This is logical because when dividing by 19 there are only 18 possible remainders. To generalize, the recurring decimal equivalent of 1/n will have at most (n-1) digits.

Once we have completed computing the recurring decimal equivalent of 1/19, we can now use the result to determine the decimal form of any fraction with 19 as the denominator. For example, if we want to know what the value of 12/19 is, we just look at the result of 1/19 and we see that the third digit is _{1}2 which is actually 12 and the answer is composed of the digits following it: **0.631578947368421052**.

Now we have purposely divided the answer into two lines of 9 digits each. This is to show that there is a further short cut to this method. Take the difference between the denominator and the numerator, 19 – 1= 18.

Once we see an _{1}8
in the answer line, which is the 9^{th} digit, we can stop computing
and just use “all from 9” to get the next answer figure. You can see the
figures in the first and second lines add up to 9 and if we consider the
remainders the total is 19.

This can be easily be explained by the fact that 1/19 +18/19 = 1 and their decimal equivalents will add up to 0.9999…

More examples for this method is discussed in Chapter 29 of “30 Master Strategies in Computing”