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Insights Is there a rigorous proof of 1 = 0.999...? - Comments

  1. Aug 30, 2015 #1

    micromass

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  2. jcsd
  3. Aug 30, 2015 #2

    jedishrfu

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    Great insight, thanks Micro.
     
  4. Aug 30, 2015 #3
    Always find these fascinating, even if most of it goes over my head at this point.
     
  5. Aug 30, 2015 #4

    kreil

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    Nice!

    The informal proof I always share with people is that 1/9=0.111..., 2/9=0.222..., 3/9=0.333..., and so on until 9/9=0.999...=1
     
  6. Aug 30, 2015 #5
  7. Aug 31, 2015 #6

    gill1109

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    Here is a number system in which 1 is not equal to 0.9999.... and it moreover is rather useful in game theory, and some people even imagine it might be useful sometime in the future, in physics: J H Conway's "surreal numbers". https://en.wikipedia.org/wiki/Surreal_number
     
  8. Aug 31, 2015 #7

    micromass

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    Whether ##1=0.9999...## in the surreals depends highly on the definitions for ##0.9999...##. Some definitions make it equal, others don't.
     
  9. Aug 31, 2015 #8

    gill1109

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    Yes you are right. However the surreals do include numbers (lots and lots of them!) which are strictly between all of 0.99999...9 (any number of repetitions) and 1.
     
  10. Aug 31, 2015 #9

    micromass

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    That depends how many repetitions of ##9## you take in ##0.999...##. If you take countably many, then sure. But if you index over all ordinals, then no.
     
  11. Aug 31, 2015 #10

    WWGD

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    You can also use, though not as nice, the perspective of the Reals as a metric space, together with the Archimedean Principle: then d(x,y)=0 iff x=y. Let then ## x=1 , y=0.9999.... ##Then ##d(x,y)=|x-y| ## can be made indefinitely small ( by going farther along the string of 9's ), forcing ## |x-y|=0## , forcing ##x=y ##. More formally, for any ##\epsilon >0 ##, we can make ##|x-y|< \epsilon ##
     
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