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|    comp.lang.c    |    Meh, in C you gotta define EVERYTHING    |    243,242 messages    |
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|    Message 243,069 of 243,242    |
|    James Kuyper to wij    |
|    Re: Collatz Conjecture proved.    |
|    26 Jan 26 21:18:24    |
      From: jameskuyper@alumni.caltech.edu              On 26/01/2026 16:51, wij wrote:       ...       > https://sourceforge.net/projects/cscall/files/MisFiles/RealNum       er2-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.              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.              ...       > 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.              --- SoupGate-Win32 v1.05        * Origin: you cannot sedate... all the things you hate (1:229/2)    |
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