5 ms·

I'm actually curious what impact it would have on various proofs if 0.999... wasn't accepted as 1. What gets broken? What consequences do we hit?

saagarjha 6y ago
Arithmetic breaks, as multiplication is no longer the inverse of division. (For example, 1/3 * 3 = 0.999… would no longer work.)
- tomtomtom1 6y ago
  why? 1/3 * 3 could still be equal to one. but 1/3 != 0.33333... that is, 1/3 is not representable in base 10. Which makes way more sense. I wonder if taking 0.9999.. != 1, that is 0.0000...1 exists would allow us to reslove, the fact that some possible events have probability 0?
  - baddox 6y ago
    Limits would break, which might break or at least cause problems with calculus and other branches of math.
  - atq2119 6y ago
    The crux of the matter is that you have to define what things like "0.3333..." mean in the first place. Any reasonable definition of it as a representation of a real number is going to lead to it being equal to 1/3. If you want to redefine it explicitly as not a real number, you can do that, and maybe even get to some amusing math that way, but you're no longer talking the same language as the rest of the world.
    - tomtomtom1 6y ago
      >"but you're no longer talking the same language as the rest of the world" yes, in the standard real numbers 1 = 0.999.., but people have dealt with numbers like "pi" and "sqrt(2)" before the standard real numbers were defined. Hence the question, if we define such a system such as 0.333... != 1/3. what are the consequences? by 0.3333... I mean a countably infinite sequence of 3s.
      - atq2119 6y ago
        I think an important distinction is that in those "old days", people were largely working in what we now know to be subsets of real numbers, and the same conclusion applies there. If you want to go to supersets of real numbers, you may be interested in https://en.wikipedia.org/wiki/Surreal_number https://en.wikipedia.org/wiki/Surreal_number
      - Scarblac 6y ago
        If 0.333... != 1/3, then they are different numbers and the expression 1/3 - 0.333... must have some value different from zero. What is that value?
  - saagarjha 6y ago
    > but 1/3 != 0.33333... But the issue is that this is easy to verify experimentally via (in this case infinitely) long division that you can do by hand. So it’s hard to convince people of this.
    - tomtomtom1 6y ago
      but the long division algorithm never terminates. why would it terminate at countable infinity?
      - recursive 6y ago
        You can show that the long division algorithm is looping. Further, you can show that it will continue, for these inputs, to produce `3`s forever with no change in state. How could `...` be defined such that it wasn't 0.333...?
AlEinstein 6y ago
You can multiply 1/3 by 3 and not get 1.
- rs23296008n1 6y ago
  I have a pizza. I divide it into three parts. You'd be asserting that if I eat the three parts I have not eaten the whole pizza. I'm unconvinced.
  - kps 6y ago
    You lost (1 − 0.999…)th of a pizza to crumbs when you divided it.
    - rs23296008n1 6y ago
      Get a sharper knife.
ltbarcly3 6y ago
So, lets take .9, .99, .999 and so on. If a sequence of rational numbers converges, it converges to a real number. What number does .9, .99, .999, .9999 (and so on) converge to? Which is to say, is there a number that it gets closer and closer to at every step? Clearly it gets closer and closer to 1 at every step, so the sequence converges to 1. This is one of the many (equivalent) ways the real numbers are defined to begin with, https://en.wikipedia.org/wiki/Construction_of_the_real_numbers#Construction_from_Cauchy_sequences https://en.wikipedia.org/wiki/Construction_of_the_real_numbe... There are lots of other ways to define sets with operations, but they won't be anything at all like the normal numbers you are used to.
- saagarjha 6y ago
  I don’t think this reasoning is correct, because the limit of an expression does not have to be the same as the value at the point the limit is being taken. For example, if your sequence is “sin(1/x) * x“, then it’ll slowly appear to converge to one as x approaches infinity, but it cannot reach it. So there’s really no relevant conclusion you can make.
  - ltbarcly3 6y ago
    You're taking a map from R -> R, I'm taking a sequence of discrete values. It's not the same thing. However, I was a bit fast and loose, the sequence .9, .99, .999, .9999 also gets closer and closer to 2 but it doesn't converge to 2, I should have said if you have a metric || and some number X such that for any d, there exists an N such that |X -An| < d for all n > N, then the sequence A converges to X. But I wasn't trying to write a proof. https://en.wikipedia.org/wiki/(%CE%B5,_%CE%B4)-definition_of_limit https://en.wikipedia.org/wiki/(%CE%B5,_%CE%B4)-definition_of...

saagarjha 6y ago

Arithmetic breaks, as multiplication is no longer the inverse of division. (For example, 1/3 * 3 = 0.999… would no longer work.)

tomtomtom1 6y ago
why? 1/3 * 3 could still be equal to one. but 1/3 != 0.33333... that is, 1/3 is not representable in base 10. Which makes way more sense. I wonder if taking 0.9999.. != 1, that is 0.0000...1 exists would allow us to reslove, the fact that some possible events have probability 0?
- baddox 6y ago
  Limits would break, which might break or at least cause problems with calculus and other branches of math.
- atq2119 6y ago
  The crux of the matter is that you have to define what things like "0.3333..." mean in the first place. Any reasonable definition of it as a representation of a real number is going to lead to it being equal to 1/3. If you want to redefine it explicitly as not a real number, you can do that, and maybe even get to some amusing math that way, but you're no longer talking the same language as the rest of the world.
  - tomtomtom1 6y ago
    >"but you're no longer talking the same language as the rest of the world" yes, in the standard real numbers 1 = 0.999.., but people have dealt with numbers like "pi" and "sqrt(2)" before the standard real numbers were defined. Hence the question, if we define such a system such as 0.333... != 1/3. what are the consequences? by 0.3333... I mean a countably infinite sequence of 3s.
    - atq2119 6y ago
      I think an important distinction is that in those "old days", people were largely working in what we now know to be subsets of real numbers, and the same conclusion applies there. If you want to go to supersets of real numbers, you may be interested in https://en.wikipedia.org/wiki/Surreal_number https://en.wikipedia.org/wiki/Surreal_number
    - Scarblac 6y ago
      If 0.333... != 1/3, then they are different numbers and the expression 1/3 - 0.333... must have some value different from zero. What is that value?
- saagarjha 6y ago
  > but 1/3 != 0.33333... But the issue is that this is easy to verify experimentally via (in this case infinitely) long division that you can do by hand. So it’s hard to convince people of this.
  - tomtomtom1 6y ago
    but the long division algorithm never terminates. why would it terminate at countable infinity?
    - recursive 6y ago
      You can show that the long division algorithm is looping. Further, you can show that it will continue, for these inputs, to produce `3`s forever with no change in state. How could `...` be defined such that it wasn't 0.333...?

AlEinstein 6y ago

You can multiply 1/3 by 3 and not get 1.

rs23296008n1 6y ago
I have a pizza. I divide it into three parts. You'd be asserting that if I eat the three parts I have not eaten the whole pizza. I'm unconvinced.
- kps 6y ago
  You lost (1 − 0.999…)th of a pizza to crumbs when you divided it.
  - rs23296008n1 6y ago
    Get a sharper knife.

ltbarcly3 6y ago

So, lets take .9, .99, .999 and so on. If a sequence of rational numbers converges, it converges to a real number. What number does .9, .99, .999, .9999 (and so on) converge to? Which is to say, is there a number that it gets closer and closer to at every step? Clearly it gets closer and closer to 1 at every step, so the sequence converges to 1. This is one of the many (equivalent) ways the real numbers are defined to begin with, https://en.wikipedia.org/wiki/Construction_of_the_real_numbers#Construction_from_Cauchy_sequences https://en.wikipedia.org/wiki/Construction_of_the_real_numbe... There are lots of other ways to define sets with operations, but they won't be anything at all like the normal numbers you are used to.

saagarjha 6y ago
I don’t think this reasoning is correct, because the limit of an expression does not have to be the same as the value at the point the limit is being taken. For example, if your sequence is “sin(1/x) * x“, then it’ll slowly appear to converge to one as x approaches infinity, but it cannot reach it. So there’s really no relevant conclusion you can make.
- ltbarcly3 6y ago
  You're taking a map from R -> R, I'm taking a sequence of discrete values. It's not the same thing. However, I was a bit fast and loose, the sequence .9, .99, .999, .9999 also gets closer and closer to 2 but it doesn't converge to 2, I should have said if you have a metric || and some number X such that for any d, there exists an N such that |X -An| < d for all n > N, then the sequence A converges to X. But I wasn't trying to write a proof. https://en.wikipedia.org/wiki/(%CE%B5,_%CE%B4)-definition_of_limit https://en.wikipedia.org/wiki/(%CE%B5,_%CE%B4)-definition_of...