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

In summary, the conversation discusses the concept of 1 and 0.999... being equal, with some arguing that they are equal based on a proof using fractions and others arguing that they are not equal based on the concept of surreal numbers. The conversation also mentions the possibility of using surreal numbers in game theory and physics, as well as the idea that the number of repetitions of 9 in 0.999... can affect the equality of 1 and 0.999....
  • #1
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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  • #3
Always find these fascinating, even if most of it goes over my head at this point.
 
  • #4
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
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
 
  • #7
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.
 
  • #8
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.
 
  • #9
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.
 
  • #10
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 ##
 

1. What is the definition of "0.999..."?

The notation "0.999..." represents an infinite decimal expansion, where the digit 9 repeats infinitely. It is equivalent to the decimal number 1.

2. Why is the statement "1 = 0.999..." considered a rigorous proof?

This statement is considered a rigorous proof because it is based on the definition of decimal notation and the properties of infinite decimal expansions. It can also be proven using mathematical induction and the concept of limit in calculus.

3. Is there any room for error in the statement "1 = 0.999..."?

No, there is no room for error in this statement. It is a mathematical fact that has been proven and accepted by the mathematical community.

4. Can you provide a visual representation of "1 = 0.999..."?

Yes, a visual representation of this statement can be shown through a number line, where the numbers 1 and 0.999... are infinitesimally close to each other and essentially overlap.

5. How does the concept of infinity play a role in "1 = 0.999..."?

The concept of infinity is crucial in understanding this statement. The decimal expansion of 0.999... is infinite, meaning it never ends or repeats. Therefore, it is equivalent to the infinite number of digits in the decimal expansion of 1, which is also infinite. This infinite nature of both numbers is what makes them equal.

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