flog

Not necessarily the most efficient solution:
1) Move the decimal point N places to the right until you have an integer, I.
2) Find the GCM, G, of I and 10^N (Easiest way to do this is get the prime factorization of both and collect like factors).
3) Your FPN is P/Q where P = I/G and Q = (10^N)/G.

Or, of course, you can just use I/(10^N). or (P*k)/(Q*k) for any non-zero integer k.

<i>Don't move
Don't talk out of time
Don't think
Don't worry
Everything's just fine
Just fine </i>

Wow. I'll have to update the population sign post outside of FlogReaderLand. Thanks, Ray.

Ironically, brodo faggins's suggestion turned out to be an order of magnitude faster than RWH's method.

Do you need to get the two smallest integers that form the desired quotient? If not, stop after RWH's step 1. Thus, 1.71285 = 171285 / 100000.

If you do need to reduce this fraction, notice that the denominator has only 2 and 5 as factors. Therefore, repeatedly divide the numerator and denominator by 2 while both results remain integers. Then repeat with 5. In this case, you divide both by a single five, resulting in 34257 / 20000.

Oy. I'm embarrassed at the simplicity of the "don't reduce the fraction" method. I'm ashamed I had to ask...

C:\TEMP>cscript IntRat.vbs
Microsoft (R) Windows Script Host Version 5.6
Copyright (C) Microsoft Corporation 1996-2001. All rights reserved.

1.71285 is 171285 / 100000
C:\TEMP>type IntRat.vbs
Option Explicit

Dim FPN
Dim DecimalPos
Dim N
Dim P
Dim Q

FPN = Inputbox("Enter a number:")
If Not IsNumeric(FPN) Then
  WScript.Echo "I said 'Enter a number:', asshole."
Else

  DecimalPos = Instr(1, Cstr(FPN), ".")
  If (0 = DecimalPos) Then
    WScript.Echo CStr(FPN) & " is " & Cstr(FPN) & " / 1" & vbCrLf
  Else
    N = Len(Cstr(FPN)) - DecimalPos
    Q = 10 ^ N
    P = FPN * Q

    WScript.Echo Cstr(FPN) & " is " & Cstr(P) & " / " & Cstr(Q) & vbCrLf
  End If
End If

WScript.Quit

C:\TEMP>

I forgot why I was toying with this, so I don't know if I need the smallest two integers. I think any one of the infinite number of integer pairs would probably work for my purposes.