# A fraction V in decimal form is an infinite string that comprises the non-repeat string v prefixing infinitely repeating period P of n digits. If the msd (the first digit) in P is m^(th) decimal digit, prove that V = v + 10^(-m) P/(1-10^(-n))?

Oct 19, 2016

In the periodic part, with period P, the place value of the lsd (least

significant digit) =#10^(-m).

So, the value of the periodic part is

$\left(P\right) X {10}^{- m} \left(1 + {10}^{- n} + {10}^{- 2 n} + {10}^{- 3 n} + \ldots\right)$

$= {10}^{- m} \frac{P}{1 - {10}^{- n}}$, using 1+x +x^2+x^3+...ad infinitum = 1/(1-x),

when $- 1 \le x < 1$ Here $x = {10}^{- n} < 1$.

It follows that $V = v + {10}^{- m} \frac{P}{1 - {10}^{- n}}$

Of course, this had been used in related problems that appeared

later.

Elucidation:

Let V = 2.1047 62705 62705 62705...

Here v= 2.1047, P= 62705, m = 9 and n = 5.

So, $V = 2.1047 + {10}^{- 9} / \left(1 - {10}^{- 5}\right) 62703$

$= 2.1047 62705 62705 62705. . .$

This form is quite helpful in storing such values in the computer

memory, with zero truncation error.

After making an elusive correction 'least' for 'most', in this edition I

am confident now in suggesting this memory-oriented application.

What are to be stored are v, m, n and P, for no-loss,due to

truncation. In my opinion, this would sorely enhance precision in

computations involving simple fractions like

17/7 = ,2.428571 428571 428571...

The 10-sd value is 2.428571 429.

The exact value could be stored as

{v, m, n, P} = { 2, 6, 6, 428571 }