From: wyniijj5@gmail.com   
      
   On Sun, 2026-01-25 at 18:52 +0000, Lawrence D’Oliveiro wrote:   
   > 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.   
      
   Not different from the popular 0.999...=1 magic, but 'rewrote' in more ugly   
   form (I just thought real experts like simple proof, semi-experts like to   
   complicated proof and believe it, even though they don't really understand   
   it!).   
             
       X * 10 ** m - a₁a₂...aₘ = 0.b₁b₂...bₙb₁b₂...bₙ...   
           = (b₁b₂...bₙ ÷ 10 ** n) + (b₁b₂...bₙ ÷ 10 ** n²) + ...   
           = (b₁b₂...bₙ ÷ (10 ** n - 1))  // what rule makes this valid   
   derivation   
                                          // from the above line?    
             
   https://sourceforge.net/projects/cscall/files/MisFiles/RealNumbe   
   2-en.txt/download        
   ...[cut]    
          (1) x= 0.999...   
          (2{ 10x= 9.999... // may have implicitly defined 0.999... is 1   
          (3) 10x= 9+x   
          (4) 9x=9    
          (5) x=1    
          Answer: There is no axiom or theorem to prove that (1) <=> (3).   
          (3) is one of the infinite interpretations of (1), or (3) is the   
          'introduction' definition of 0.999..., etc. In short, there is no   
          necessary relationship between (3) and (1), or it still needs to be   
          proved. For example, 0.999... formed by 1/2+1/4+1/8+... does not have   
   the   
          property of (3).   
   ----------        
   You might be using infinite series, see the link.    
        
   Anyway, whatever the specific proof is, you still need to refute the proof    
   shown in the previous post, Prop1,Prop2,Prop3. You cannot pretending blind to   
   escape it.   
      
   > > > 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?   
      
   No, I don't believe 'infinite-precision integer' is representable.   
   See the link.   
      
   --- SoupGate-Win32 v1.05   
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