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> There is no such thing as 0.9... Yes, there is. Ask yourself: What's the decimal expansion of 1/3, or pi?

by cgmg 9y ago

wyldfire 9y ago
I'm not a math pro but I don't think those are good evidence to refute that claim. I have no idea whether the claim is valid or not, though I vaguely recall this discussion from school. From your examples -- one's a ratio and the other is irrational. Almost as if you suggest "The set of real numbers exists, ergo 0.999999... exists"?
- umanwizard 9y ago
  "0.999... = 1" isn't really a "claim", it's a definition. Well, it's technically a claim, but one that is obviously true once we agree on what 0.999... means. Anyone who doesn't agree that 0.999... = 1 has a different definition of 0.999... from the one used in mainstream math notation.
- yorwba 9y ago
  The battle over the question whether or not something exists in the mathematical sense has essentially been won by the formalists, who will allow any definition to claim the existence of something so long as it is consistent with the rest of mathematics. Then, given that 0.99999... exists, what useful properties could it possibly have, if it is to be treated as a number. 0.99999... = 0.9999... * 1 = 0.99999... * 9/9 = 0.99999... * (10 - 1)/9 = (0.99999... * 10 - 0.99999... * 1)/9 = (9.99999... - 0.99999...)/9 = 9.00000.../9 = 9/9 = 1 So if 0.99999... exists and has the same arithmetic properties as decimal expansions of rational numbers, its value must be 1. Now whether you think it should have these properties is an entirely different question, but most mathematicians seem to like it this way.
- cgmg 9y ago
  No, I'm simply saying those are other examples of numbers with infinite decimal expansions. In fact, most numbers don't have a finite decimal expansion. And the claim is correct. It is the statement that the limit of the partial sums of the inverse powers of 9 is 1, which can be rigorously proven with the epsilon-delta definition of a limit.
umanwizard 9y ago
As much as I believe that 0.999... exists and is equal to 1, the thing where you write 1/3 as 0.3... and multiply both sides by 3 is a really bad proof -- it's pretty much circular reasoning. For someone who doesn't understand what 0.999... means, why should we expect them to understand what 0.333... means and believe that it exists?
- cgmg 9y ago
  Why do you think it's circular reasoning?
  - umanwizard 9y ago
    Because 0.33... means 1/3 for exactly the same reason that 0.99.. means 1. So if you believe one then you should already believe the other. All the intuitive (wrong) arguments against 0.99...=1 apply to 0.33...=1/3 just as well.
    - cgmg 9y ago
      Your first sentence is probably true, but the point is that most people find the former more intuitive. It's a valid demonstration technique. Just because A entails B in the mathematical/logical sense doesn't mean everyone who believes A also believes B. No one is logically omniscient.
      - umanwizard 9y ago
        The point of mathematical proofs isn't really to convince people psychologically that something is true -- it's to show why something is true. Do you really know "why" 0.99.. = 1 by writing 0.99... = 3 * (0.33...) = 3 * 1/3 = 1 ? I'd argue that you just kicked the can down the road.
        cgmg 9y ago
        Did I say it was a mathematical proof?
pacala 9y ago
There is no decimal expansion of 1/3. Or pi for that matter. There are only approximations thereof, with an error margin inverse proportional to the energy we throw at generating said expansion. If you happen to somehow built a decimal expansion of 1/3, please share it with us!
- cgmg 9y ago
  > There is no decimal expansion of 1/3. Or pi for that matter. This is incorrect. A decimal expansion is just a type of Cauchy sequence, which by definition is infinite (a function from the natural numbers to the reals). You are confusing the decimal expansion with its truncated approximations. > If you happen to somehow built a decimal expansion of 1/3, please share it with us! Sure thing! Here it is: def oneThird(place): if place >= 0: return 0 else: return 3 Convert to your favorite programming language/Turing machine/other abstract machine of choice.
  - pacala 9y ago
    oneThird doesn't return a decimal expansion, just one digit in a number. Still waiting for the 1 / 3 decimal expansion you claim it exists. The best I can do is: def oneThird(place): if place >= 0: return 0 else: return 3 def printOneThird(): for x in iter(int, 1): print(oneThird(-x)) Unfortunately, it didn't finish before I edited this response. Perhaps you have more patience, please let me know when you have a link to the decimal expansion of 1 / 3.
    - cgmg 9y ago
      pacala, It seems you don't understand what a sequence is. Formally, a sequence is a function whose domain is the natural numbers (or integers). A decimal expansion is simply a sequence whose codomain is the set of digits {0,1,2,3,4,5,6,7,8,9}. I gave you just such a sequence. Are you really going to continue disputing a basic mathematical fact?
      - pacala 9y ago

> There is no such thing as 0.9... Yes, there is. Ask yourself: What's the decimal expansion of 1/3, or pi?

by cgmg 9y ago

> There is no such thing as 0.9... Yes, there is. Ask yourself: What's the decimal expansion of 1/3, or pi?

I'm not a math pro but I don't think those are good evidence to refute that claim. I have no idea whether the claim is valid or not, though I vaguely recall this discussion from school. From your examples -- one's a ratio and the other is irrational. Almost as if you suggest "The set of real numbers exists, ergo 0.999999... exists"?

umanwizard 9y ago
"0.999... = 1" isn't really a "claim", it's a definition. Well, it's technically a claim, but one that is obviously true once we agree on what 0.999... means. Anyone who doesn't agree that 0.999... = 1 has a different definition of 0.999... from the one used in mainstream math notation.
yorwba 9y ago
The battle over the question whether or not something exists in the mathematical sense has essentially been won by the formalists, who will allow any definition to claim the existence of something so long as it is consistent with the rest of mathematics. Then, given that 0.99999... exists, what useful properties could it possibly have, if it is to be treated as a number. 0.99999... = 0.9999... * 1 = 0.99999... * 9/9 = 0.99999... * (10 - 1)/9 = (0.99999... * 10 - 0.99999... * 1)/9 = (9.99999... - 0.99999...)/9 = 9.00000.../9 = 9/9 = 1 So if 0.99999... exists and has the same arithmetic properties as decimal expansions of rational numbers, its value must be 1. Now whether you think it should have these properties is an entirely different question, but most mathematicians seem to like it this way.
cgmg 9y ago
No, I'm simply saying those are other examples of numbers with infinite decimal expansions. In fact, most numbers don't have a finite decimal expansion. And the claim is correct. It is the statement that the limit of the partial sums of the inverse powers of 9 is 1, which can be rigorously proven with the epsilon-delta definition of a limit.

umanwizard 9y ago

As much as I believe that 0.999... exists and is equal to 1, the thing where you write 1/3 as 0.3... and multiply both sides by 3 is a really bad proof -- it's pretty much circular reasoning. For someone who doesn't understand what 0.999... means, why should we expect them to understand what 0.333... means and believe that it exists?

cgmg 9y ago
Why do you think it's circular reasoning?
- umanwizard 9y ago
  Because 0.33... means 1/3 for exactly the same reason that 0.99.. means 1. So if you believe one then you should already believe the other. All the intuitive (wrong) arguments against 0.99...=1 apply to 0.33...=1/3 just as well.
  - cgmg 9y ago
    Your first sentence is probably true, but the point is that most people find the former more intuitive. It's a valid demonstration technique. Just because A entails B in the mathematical/logical sense doesn't mean everyone who believes A also believes B. No one is logically omniscient.
    - umanwizard 9y ago
      The point of mathematical proofs isn't really to convince people psychologically that something is true -- it's to show why something is true. Do you really know "why" 0.99.. = 1 by writing 0.99... = 3 * (0.33...) = 3 * 1/3 = 1 ? I'd argue that you just kicked the can down the road.
      - cgmg 9y ago
        Did I say it was a mathematical proof?

pacala 9y ago

There is no decimal expansion of 1/3. Or pi for that matter. There are only approximations thereof, with an error margin inverse proportional to the energy we throw at generating said expansion. If you happen to somehow built a decimal expansion of 1/3, please share it with us!

cgmg 9y ago
> There is no decimal expansion of 1/3. Or pi for that matter. This is incorrect. A decimal expansion is just a type of Cauchy sequence, which by definition is infinite (a function from the natural numbers to the reals). You are confusing the decimal expansion with its truncated approximations. > If you happen to somehow built a decimal expansion of 1/3, please share it with us! Sure thing! Here it is: def oneThird(place): if place >= 0: return 0 else: return 3 Convert to your favorite programming language/Turing machine/other abstract machine of choice.
- pacala 9y ago
  oneThird doesn't return a decimal expansion, just one digit in a number. Still waiting for the 1 / 3 decimal expansion you claim it exists. The best I can do is: def oneThird(place): if place >= 0: return 0 else: return 3 def printOneThird(): for x in iter(int, 1): print(oneThird(-x)) Unfortunately, it didn't finish before I edited this response. Perhaps you have more patience, please let me know when you have a link to the decimal expansion of 1 / 3.
  - cgmg 9y ago
    pacala, It seems you don't understand what a sequence is. Formally, a sequence is a function whose domain is the natural numbers (or integers). A decimal expansion is simply a sequence whose codomain is the set of digits {0,1,2,3,4,5,6,7,8,9}. I gave you just such a sequence. Are you really going to continue disputing a basic mathematical fact?
    - pacala 9y ago