Question

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))`?

Answer 1

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

`(P) X 10^(-m)(1+10^(-n)+10^(-2n)+10^(-3n)+...)`

`=10^(-m)(P)/(1-10^(-n))`, using 1+x +x^2+x^3+...ad infinitum = 1/(1-x),

when `-1<=x<1` Here `x = 10^(-n)<1`.

It follows that `V = v+10^(-m)(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)/(1-10^(-5)) 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 }