Question about repeating decimal?

Question

For simple fraction, we can easily convert it to repeating decimal by calculator. Ex. $\frac 1 3 = 0.33333\ldots$, $\frac 1 7=0.(142857),\ldots$ But some fraction fraction like $10/29, 1/97,...$ The repeating part of them are too long, so it can't fully show on the calculator. So is there any algorithm to find the repeating part for that fraction?

Answer 1

Yes. Here is one I once wrote in the R language. It uses the 'gmp' (GNU Multiple Precision) package.

#########################################################
### Functions for converting back and forth between   ###
### rational numbers and decimals. Repeating decimals ###
### are denoted by brackets around the digit block    ###
### repeats. For example, 11/6 is 1.8[3] in decimal.  ###
#########################################################

#############################
### Author: Roman Chokler ###
#############################
library(gmp)

bigq.to.dec <- function(q)
{
  sgn <- sign(q)
  num <- abs(numerator(q))
  den <- denominator(q)
  d <- den
  c2 <- 0
  c5 <- 0
  while(d %% 2==0)
  {
    d <- as.bigz(d/2)
    c2 <- c2 + 1
  }
  while(d %% 5==0)
  {
    d <- as.bigz(d/5)
    c5 <- c5 + 1
  }
  transient <- max(c2,c5)
  dec <- as.character(as.bigz(num/den))
  rem <- num %% den
  if (rem != 0)
  {
    dec <- paste0(dec,".")
    if (transient>0)
    {
      for(i in 1:transient)
      {
        rem <- rem * 10
        dec <- paste0(dec,as.bigz(rem/den))
        rem <- rem %% den
      }
    }
    if (rem != 0)
    {
      dec <- paste0(dec,"[")
      r <- rem
      rem <- rem * 10
      dec <- paste0(dec,as.bigz(rem/den))
      rem <- rem %% den
      while (rem != r)
      {
        rem <- rem * 10
        dec <- paste0(dec,as.bigz(rem/den))
        rem <- rem %% den
      }
      dec <- paste0(dec,"]")
    }
  }
  if (sgn == -1)
  {
    dec <- paste0("-",dec)
  }
  return(dec)
}

dec.to.bigq <- function(dec)
{
  bq <- 0
  sgn <- 1
  n <- nchar(dec)
  if (regexpr("-",dec)[1]==1)
  {
    sgn <- -1
    dec <- substr(dec,2,n)
    n <- n - 1
  }
  pnt <- regexpr("\\.",dec)[1]
  rstart <- regexpr("\\[",dec)[1]
  rend <- regexpr("\\]",dec)[1]
  if (pnt == -1)
  {
    return(as.bigq(dec) * sgn)
  }
  if (n==pnt)
  {
    return((as.bigq(substr(dec,1,pnt-1))) * sgn)
  }
  bq <- as.bigq(substr(dec,1,pnt-1))
  if (rstart == -1)
  {
    transient <- substr(dec,pnt+1,n)
    den <- pow.bigz(10,nchar(transient))
    transient <- sub("^0+","",transient)
    if (transient == "")
    {
      transient <- "0"
    }
    return((bq + div.bigq(transient,den)) * sgn)
  }
  den <- as.bigz(1)
  if (rstart > pnt + 1)
  {
    transient <- substr(dec,pnt+1,rstart-1)
    den <- pow.bigz(10,nchar(transient))
    transient <- sub("^0+","",transient)
    if (transient == "")
    {
      transient <- "0"
    }
    bq <- bq + div.bigq(transient,den)
  }
  if ((rend == -1) || (rstart >= rend - 1) || n > rend)
  {
    return(as.bigq(NA))
  }
  rep <- substr(dec,rstart+1,rend-1)
  den <- den * (pow.bigz(10,nchar(rep)) - 1)
  rep <- sub("^0+","",rep)
  if (rep == "")
  {
    rep <- "0"
  }
  return((bq + div.bigq(rep,den)) * sgn)
}

Answer 2

For, say, $\large \frac{1}{97}$, start by running the Python program

R = []

numer = 1
denom = 97

for i in range(0,denom):
    r = numer * 10**i % denom
    try:
        ndx = R.index(r)
    except:
        ndx = -1
    if ndx >= 0:
        print(i,ndx)
        break
    R.append(r)

The output of the program is

96 0

The period length is

i - ndx

and is therefore equal to $96 - 0 = 96$.

The offset (to the right of the decimal) is $0$ so the period begins immediately after the decimal.

So we need to put a bar over the first $96$ digits.

To get those decimal digits you can conveniently use wolfram, and compute

integerPart((1 + 1/97) * 10^96)

The output is

1010309278350515463917525773195876288659793814432989690721649484536082474226804123711340206185567

and you have to drop the leading $1$ (used to 'light up' any zeroes right after the decimal point) to get the final answer (with period broken down into blocks of $25$),

$\quad \large \frac{1}{97} =$

$0.\overline{0103092783505154639175257}$
$\; \; \, \overline{7319587628865979381443298}$
$\; \; \, \overline{9690721649484536082474226}$
$\; \; \, \overline{804123711340206185567}$