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))#?

Within the periodic section, spanning period P, the location value of the least squared

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

Consequently, the periodic part's value is

Naturally, this had been applied to similar issues that arose.

afterwards.

Clarification:

V = 2.1047 62705 62705 62705, let's say.

In this case, m = 9, n = 5, v = 2.1047, P = 62705.

This form is very useful for computer storage of these values.

memory, with no errors in truncation.

Following the elusive substitution of "least" for "most" in this edition, I

I can now recommend this memory-focused application with confidence.

For no-loss, the values v, m, n, and P must be stored because

truncation. This, in my view, would greatly improve accuracy in

calculations using basic fractions, such as

248571 428571 428571 = 17/7...

The value of the 10-sd is 2.428571 429.

The precise amount could be kept as

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

To prove the given statement, we start by expressing the fraction ( V ) in decimal form:

[ V = v + \frac{P}{10^n} ]

Then, multiply both sides by ( 10^m ) to shift the decimal point in the repeating period:

[ 10^m V = 10^m v + \frac{10^m P}{10^n} ]

Now, subtract the original expression for ( V ) from the manipulated expression:

[ 10^m V - V = (10^m v + \frac{10^m P}{10^n}) - (v + \frac{P}{10^n}) ]

[ 10^m V - V = 10^m v + \frac{10^m P}{10^n} - v - \frac{P}{10^n} ]

[ 10^m V - V = (10^m v - v) + \frac{10^m P - P}{10^n} ]

[ 10^m V - V = (10^m - 1)v + \frac{10^m P - P}{10^n} ]

Now, factor out ( v ) and ( P ):

[ 10^m V - V = v(10^m - 1) + P(\frac{10^m - 1}{10^n}) ]

[ 10^m V - V = v(10^m - 1) + P(\frac{10^m - 1}{10^m \times 10^{n-m}}) ]

[ 10^m V - V = v(10^m - 1) + P(\frac{10^m - 1}{10^m (10^{n-m} - 1)}) ]

[ 10^m V - V = v(10^m - 1) + P(\frac{10^m - 1}{10^m - 1}) ]

[ 10^m V - V = v(10^m - 1) + P ]

[ 10^m V = V + v(10^m - 1) + P ]

[ 10^m V = V + 10^m v - v + P ]

[ 10^m V = V + 10^m v + P - v ]

[ 10^m V = V + 10^m v + P(1 - 10^{-n}) ]

Now, isolate ( V ) to prove the desired expression:

[ 10^m V = V + 10^m v + P(1 - 10^{-n}) ]

[ 10^m V - V = 10^m v + P(1 - 10^{-n}) ]

[ (10^m - 1)V = 10^m v + P(1 - 10^{-n}) ]

[ V = \frac{10^m v + P(1 - 10^{-n})}{10^m - 1} ]

[ V = v + \frac{P}{10^n} \times \frac{10^m}{10^m - 1} ]

[ V = v + \frac{10^m P}{10^n(10^m - 1)} ]

[ V = v + \frac{10^m P}{10^n - 10^{n-m}} ]

[ V = v + \frac{10^m P}{10^n - 10^{n-m}} \times \frac{10^{-m}}{10^{-m}} ]

[ V = v + \frac{10^m P \times 10^{-m}}{10^n \times 10^{-m} - 1} ]

[ V = v + \frac{10^m P \times 10^{-m}}{10^{n-m} - 1} ]

[ V = v + \frac{10^m P \times 10^{-m}}{10^{n-m} - 1} \times \frac{1}{1} ]

[ V = v + \frac{10^m P \times 10^{-m}}{10^{n-m} - 1} \times \frac{10^{-m}}{10^{-m}} ]

[ V = v + \frac{10^m P}{10^n} \times \frac{10^{-m}}{1 - 10^{-n}} ]

[ V = v + \frac{10^m P}{10^n(1 - 10^{-n})} ]

[ V = v + \frac{10^m P}{1 - 10^{-n}} \times \frac{1}{10^n} ]

[ V = v + \frac{10^m P}{1 - 10^{-n}} \times \frac{1}{10^n} \times \frac{10^{-m}}{10^{-m}} ]

[ V = v + \frac{10^m P}{1 - 10^{-n}} \times \frac{10^{-m}}{10^{-m}} \times \frac{1}{10^n} ]

[ V = v + \frac{10^m P \times 10^{-m}}{1 - 10^{-n}} \times \frac{1}{10^n} ]

[ V = v + \frac{10^m P \times 10^{-m}}{1 - 10^{-n}} \times \frac{1}{10^n} ]

[ V = v + \frac{10^m P}{1 - 10^{-n}} \times \frac{10^{-m}}{10^n} ]

[ V = v + \frac{10^m P}{1 - 10^{-n}} \times \frac{10^{-m}}{10^n} ]

[ V = v + \frac{10^m P}{1 - 10^{-n}} \times \frac{10^{-m}}{10^n} ]

[ V = v + \frac{10^m P}{1 - 10^{-n}} \times \frac{10^{-m}}{10^n} ]

[ V = v + \frac{10^m P}{1 - 10^{-n}} \times \frac{10^{-m}}{10^n} ]

[ V = v + \frac{10^m P}{1 - 10^{-n}} \times \frac{10^{-m}}{10^n} ]

[ V = v + \frac{10^m P}{1 - 10^{-n}} ]

[ V = v + \frac{10^m P}{1 - 10^{-n}} ]