From: wyniijj5@gmail.com   
      
   On Mon, 2026-01-26 at 21:18 -0500, James Kuyper wrote:   
   > On 26/01/2026 16:51, wij wrote:   
   > ...   
   > > https://sourceforge.net/projects/cscall/files/MisFiles/RealN   
   mber2-en.txt/download   
   > >           3. 1/3 = 0.333... + non-zero-remainder (True identity.   
   How to deny?)   
   > > How would you deny it, and call the cut-off 'equation' identity?   
   >    
   > I deny it easily - the remainder is exactly 0.   
   >    
   > > You cut off non-zero-remainder to stop repeating, so yes, you see the part   
   you   
   > > want to see, i.e. the front part without "...", and forgot the definition   
   > > "infinitely repeat" is invalidated.   
   >    
   > There's no non-zero-remainder to cut off, nor is there any need to stop   
   > repeating. It repeats endlessly, and it is only because of the endless   
   > repetition that the remainder is 0. If it ever ended, the remainder   
   > would be non-zero, as you claim.   
      
   Don't you see that you made a number of assertions beyond your standard allows.   
   "... can have smaller numerators and denominators..." ... How? Prove it, not   
   asserting it.   
      
   > The flaw is in your property 2, which claims that an infinite sum of   
   > rational numbers is not a rational number. That's unambiguously not the   
   > case in the standard real number system. Your proof of that claim is   
   > based upon asserting that the numerator and denominator of each step in   
   > the infinite series has a larger number of digits (which isn't   
   > necessarily true - but that's unimportant), but ignores the fact that   
   > the limit of an infinite sequence can have smaller numerators and   
   > denominators than the terms that make up the sequence.   
      
   Accordingly, I suppose you deny this passage:   
      
            // ... [cut]   
            1. If 0.999... = 1 holds, then it can be proved that the unique prime   
               factorization theorem of positive integers does not hold:   
               0.999... = 999.../1000... = 9*(111...)/(5*2)*... =1   
               <=> 3*3*(111...) = (5*2)*...   
               The integer to the left of the equal sign contains the prime number   
               3, but the integer to the right cannot contain 3... Prime   
               factorization is not unique.   
      
   > ...   
   > >         Prop 2= Repeating Q+Q infinitely does not yield rational   
   number.   
   > >                 (precisely, positive rational number)   
   >    
   > I dispute the validity of the proof of Prop 2.   
      
   No problem, you just need to prove it yourself.   
      
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