From: ldo@nz.invalid   
      
   On Sun, 25 Jan 2026 16:46:43 +0800, wij wrote:   
      
   > On Sun, 2026-01-25 at 08:15 +0000, Lawrence D’Oliveiro wrote:   
   >>   
   >> On Sun, 25 Jan 2026 13:28:31 +0800, wij wrote:   
   >>   
   >>> On Sat, 2026-01-24 at 23:06 -0500, James Kuyper wrote:   
   >>>>   
   >>>> This generalizes to work with any recurring decimal. When the   
   >>>> recurrence is n digits long (in the above example, n=2), just   
   >>>> multiply by 10^n.   
   >>>   
   >>> Such arithmetic is called approximation ...   
   >>   
   >> No, it’s exact.   
   >   
   > Prove it. No one wants blind belief.   
      
   Sure. Consider the general case of a fractional number X which can be   
   represented by a repeating decimal:   
      
       X = 0.a₁a₂...aₘb₁b₂...bₙb₁b₂...bₙ...   
      
   where the a’s and b’s are decimal digits, such that a₁a₂...aₘ   
   represents the initial non-repeating part, consisting of m digits,   
   where m ≥ 0, and b₁b₂...bₙ represents the repeating part, consisting   
   of n digits, where n > 0. First, separate out the non-repeating part:   
      
       X * 10 ** m = a₁a₂...aₘ.b₁b₂...bₙb₁b₂...bₙ...   
      
   from which   
      
       X * 10 ** m - a₁a₂...aₘ = 0.b₁b₂...bₙb₁b₂...bₙ...   
           = (b₁b₂...bₙ ÷ 10 ** n) + (b₁b₂...bₙ ÷ 10 ** n²) + ...   
           = (b₁b₂...bₙ ÷ (10 ** n - 1))   
      
   Notice that’s a closed-form expression: no more indefinitely-repeating   
   parts at all. The right-hand side is a ratio of two integers, which is   
   what makes it “rational”. If it’s not already in its lowest terms, it   
   can be made so, by cancelling out common factors. Since there are only   
   a finite number of integers between those values and 1, those lowest   
   terms exist somewhere along the point between the two, and can always   
   be found in a finite number of steps. QED.   
      
   >> I think your problem is you are trying to think about maths using a   
   >> specific low-level programming language (C or C++) which only   
   >> supports finite-precision integers.   
   >   
   > Your problem is just repeating what you are told to repeat, no   
   > meaning.   
      
   You’re not familiar with languages that support infinite-precision   
   integers, are you?   
      
   --- SoupGate-Win32 v1.05   
    * Origin: you cannot sedate... all the things you hate (1:229/2)