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

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Discussion Overview

The discussion revolves around the question of whether there is a rigorous proof that 1 equals 0.999..., exploring various mathematical perspectives and number systems. It includes informal proofs, alternative number systems, and the implications of definitions in different contexts.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • Some participants share informal proofs, such as the relationship between fractions and their decimal representations, specifically noting that 1/9 equals 0.111..., leading to the conclusion that 9/9 equals 0.999... which equals 1.
  • Others introduce the concept of surreal numbers, suggesting that in this number system, 1 may not equal 0.999..., depending on the definitions used for 0.999....
  • It is noted that surreal numbers contain values that lie strictly between 0.999... and 1, raising questions about the nature of equality in this context.
  • Some participants discuss the implications of taking different counts of repetitions of 9 in 0.999..., suggesting that the equality may depend on whether countably many or uncountably many repetitions are considered.
  • A mathematical perspective using the properties of metric spaces is presented, indicating that the distance between 1 and 0.999... can be made arbitrarily small, leading to the conclusion that they are equal under certain conditions.

Areas of Agreement / Disagreement

Participants express a mix of agreement and disagreement, with some supporting the idea that 1 equals 0.999... based on informal proofs, while others argue against this equality in the context of surreal numbers and different definitions. The discussion remains unresolved with multiple competing views.

Contextual Notes

The discussion highlights the dependence on definitions and the implications of different mathematical frameworks, such as surreal numbers and metric spaces, without reaching a consensus on the equality of 1 and 0.999....

micromass
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micromass submitted a new PF Insights post

Is There a Rigorous Proof Of 1 = 0.999...?

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Always find these fascinating, even if most of it goes over my head at this point.
 
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
 
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
 
gill1109 said:
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

Whether ##1=0.9999...## in the surreals depends highly on the definitions for ##0.9999...##. Some definitions make it equal, others don't.
 
micromass said:
Whether ##1=0.9999...## in the surreals depends highly on the definitions for ##0.9999...##. Some definitions make it equal, others don't.
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.
 
gill1109 said:
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.

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.
 
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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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