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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

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...?

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...?