Chris G

• Chris G 16 Jun 2019 13:54 UTC

  1 point

  in reply to: Slider’s comment on: 0.999...=1: Another Ra­tion­al­ity Lit­mus Test

  Still not entirely convinced. If 0.A > 0.9 then surely0.A… > 0.9...?

  Or does the fact this is true only when we halt at an equal number of digits after the point make a difference? 0.A = ¹⁰⁄₁₁ and 0.9 = ⁹⁄₁₀, so 0.A > 0.9, but 0.A < 0.99.

• Chris G 2 Jun 2019 15:29 UTC

  1 point

  in reply to: Slider’s comment on: 0.999...=1: Another Ra­tion­al­ity Lit­mus Test

  So would you say that 0.999...(base10) = 0.AAA...(base11) = 0.111...(base2)= 1?

• Chris G 1 Jun 2019 6:18 UTC

  1 point

  in reply to: Slider’s comment on: 0.999...=1: Another Ra­tion­al­ity Lit­mus Test

  I think I see your first point.

  0.A{base11} = ¹⁰⁄₁₁

  0.9 = ⁹⁄₁₀

  0.A − 0.9 = 0.0_09...

  0.AA = ¹⁰⁄₁₁ + ¹⁰⁄₁₂₁

  0.99 = ⁹⁄₁₀ + ⁹⁄₁₀₀

  0.AA − 0.99 = 0.00_1735537190082644628099...

  Does this mean that because the difference or “lateness” gets smaller tending to zero each time a single identical digit is added to 0.A and 0.9 respectively, then 0.A… = 0.9...?

  (Whereas the difference we get when we do this to say 0.8 and 0.9 gets larger each time so we can’t say 0.8… = 0.9...)

• Chris G 31 May 2019 6:22 UTC

  0 points

  in reply to: rossry’s comment on: 0.999...=1: Another Ra­tion­al­ity Lit­mus Test

  0.9{base10}<0.99{base10} but 0.9...{base10}=0.99...{base10}

  0.9{base10}<0.A{base11} but 0.9...{base10}=0.A...{base11}

  0.8{base10}<0.9{base10} and 0.8...{base10}<0.9...({ase10}

  0.9{base10}<0.A{base11} and 0.9...{base10}<0.A...{base11}

  
  

  I’m not trying to prove “0.999...{base10}=1 “is false, nor that “0.111...(base2)=1” is either—in fact it’s an even more fascinating result.

  Also “not(not(true))=true” is good enough for me as well.

• Chris G 26 May 2019 12:09 UTC

  1 point

  on: 0.999...=1: Another Ra­tion­al­ity Lit­mus Test

  It’s instructive to set out the proof you give for 0.999...=1 in number bases other than ten. For example base eleven, in which the maximum value single digit is conventionally represented as A and amounts to 10 (base ten). 10 (base eleven) amounts to 11 (base ten). So

  Let x = 0.AAA...

  10x = A.AAA...

  10x—x = A

  Ax = A

  x = 1

  0.AAA… = 1

  But 0. A (base eleven) = ¹⁰⁄₁₁ (base ten) which is bigger than 0.9 (base ten) = ⁹⁄₁₀ (base ten). So shouldn’t that inequality apply to 0.AAA… (base eleven) and 0.999… (base ten) as well? (A debatable point maybe). If so, then they can’t both equal 1, unless we say something like 0.999...=1 and 0.AAA...=1 are both valid but base dependent equations, as indeed any such equation would be when using the top valued single digit of its base. This would mean 0.111...=1 in binary.