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Can you explain what you mean with "real" in that sentence? Because in the context of maths, a real number is "a number in ℝ", which this absolutely qualifies f

Can you explain what you mean with "real" in that sentence? Because in the context of maths, a real number is "a number in ℝ", which this absolutely qualifies for. Whereas in plain English the term doesn't really have a clear definition. You might consider "real" numbers (in plain English) to mean physically measurable quantities, but there are plenty of numbers that we can write out because they're infinitely long, but trivially "made" (such as π, which just requires grabbing a compass and drawing a circle) The bit you should be wondering about is why 0.666...6 and 0.666...7 are the same number: infinities cause digits written on paper (or a computer screen) to look like a kind of number that they're not. The two fractions (numbers in ℚ) 0.6666 and 0.66667 are 0.00001 apart, but the two reals (numbers in ℝ) 0.666...6 and 0.666...7 are 0.000...1 apart. That looks like a tiny tiny fraction, but it's not a fraction, it's an infinite number of zeroes, and thanks to that, this number, while it looks like a fraction, is just a silly way to write zero. So thanks to infinities, the most-definitely-not-a-fraction number that we write as 0.666... is the same as the most-definitely-not-a-fraction number 0.666...(some numbers here). The difference between the two is zero. Infinities are fun. And difficult. But also fun.

hilbertseries 7y ago
Infinities are difficult and your explanation is wrong. This notation .666...7 is meaningless. The notation .666... indicates a decimal representation that does not end. The representation has infinitely many sixes and does not end. If there is a 7 in the representation then it must be at a specific decimal. If it was at the end of the representation then it would be a finite decimal representation. The notation is meaningless with respect to the real numbers.
- enedil 7y ago
  Yes, exactly, .666...7 means a number that has 6 at each decimal position and 7 for index k, where k is greater than any natural number. This exactly is .666... (just as parent commenter explained).
  - gspr 7y ago
    > Yes, exactly, .666...7 means a number that has 6 at each decimal position and 7 for index k, where k is greater than any natural number. This exactly is .666... (just as parent commenter explained). Ill-defined. Try again.
- ping_pong 7y ago
  It's as much nonsense as someone saying "infinity plus one is bigger than infinity!"
- exmadscientist 7y ago
  I don't agree with you, or with the sibling comment that claims that "infinity plus one" is "nonsense". It is in fact precisely well-defined, to a mathematician working in a framework that admits it and makes it worth discussing. The blog post that (I think) started off this chain of posts, at https://blog.plover.com/misc/half-baked.html https://blog.plover.com/misc/half-baked.html , has the most concise and lucid explanation I've found, so I'm just going to straight-up copy it: 0.666...7 is an “an object... said to ‘have order type ω+1’, and is completely legitimate.” It's not very useful – it's exactly equal to 0.666... ! – but it's legitimate and well-defined. My absolute favorite construction of objects like this is Conway's surreal numbers. These things appear perfectly naturally in the surreal numbers, and are completely well-defined, if (again) not very useful.
- TheRealPomax 7y ago
  This reasoning is understandable, but also incorrect, and we can point to where it breaks down: the idea that the 7 is "at a specific decimal" doesn't hold true, due to that pesky ellipsis. The crazy bit about the infinite repetition is that the 7 in 0.666...7 is not at a specific decimal. It's not even at "the last decimal" because there is no last. So, let's show this via a proof by contradiction: --- 1. we assert that 0.666...7 is a sequence of digits. 2. we assert that each digit in a sequence can be assigned an integer index corresponding to its position in the sequence. I.e. we can defined each digit's index as "the number of digits that precede this digit". 3. from (1) and (2) it follows that the index for 7 must be an integer. 4. the ellipses represents an infinite number of digits (infinitely repeating the repeated digit pattern preceding it). 5. from (2) and (4) it follows that the index for 7 must be the integer value "infinity", because it has an infinite number of digits preceding it. 6. (3) and (5) cannot both be true, because infinity is not an integer. 7. from (6) it follows that (1), and/or (2), and/or (4) must be false 8. (4) is, by definition, true. 9. from (7) and (8) it follows that (1) and/or (2) must be false. 10. (1) is our fundamental assertion. If (1) was false then there is wouldn't even be a sequence of digits for us to reason about. So (1) is true. 11. from (6), (8), and (10) it follows that (2) must be false for there to be no contradiction. QED --- Now, certainly, for finite length numbers the assumption that each digit in a sequence has an integer index holds true, but it turns out we have mathematical notation that lets us write down numbers for which that property does not hold. Infinities are fun. And difficult. But also fun.
  - gspr 7y ago
    > 1. we assert that 0.666...7 is a sequence of digits. A sequence in the mathematical sense is a function whose domain is the natural numbers. Please define that function for the creature you're working with here. Otherwise you're trying to prove things about an object with no definition. You will end up in trouble.
    - TheRealPomax 7y ago
      Why? https://news.ycombinator.com/item?id=23009160 https://news.ycombinator.com/item?id=23009160 posited that we were in a situation where (1) holds, so that's where we're starting. If we assume (1) holds, then (2) cannot hold. We can abandon (1) but then we're no longer replying to that specific comment, now we're trying to prove something else.

hilbertseries 7y ago

Infinities are difficult and your explanation is wrong. This notation .666...7 is meaningless. The notation .666... indicates a decimal representation that does not end. The representation has infinitely many sixes and does not end. If there is a 7 in the representation then it must be at a specific decimal. If it was at the end of the representation then it would be a finite decimal representation. The notation is meaningless with respect to the real numbers.

enedil 7y ago
Yes, exactly, .666...7 means a number that has 6 at each decimal position and 7 for index k, where k is greater than any natural number. This exactly is .666... (just as parent commenter explained).
- gspr 7y ago
  > Yes, exactly, .666...7 means a number that has 6 at each decimal position and 7 for index k, where k is greater than any natural number. This exactly is .666... (just as parent commenter explained). Ill-defined. Try again.
ping_pong 7y ago
It's as much nonsense as someone saying "infinity plus one is bigger than infinity!"
exmadscientist 7y ago
I don't agree with you, or with the sibling comment that claims that "infinity plus one" is "nonsense". It is in fact precisely well-defined, to a mathematician working in a framework that admits it and makes it worth discussing. The blog post that (I think) started off this chain of posts, at https://blog.plover.com/misc/half-baked.html https://blog.plover.com/misc/half-baked.html , has the most concise and lucid explanation I've found, so I'm just going to straight-up copy it: 0.666...7 is an “an object... said to ‘have order type ω+1’, and is completely legitimate.” It's not very useful – it's exactly equal to 0.666... ! – but it's legitimate and well-defined. My absolute favorite construction of objects like this is Conway's surreal numbers. These things appear perfectly naturally in the surreal numbers, and are completely well-defined, if (again) not very useful.
TheRealPomax 7y ago
This reasoning is understandable, but also incorrect, and we can point to where it breaks down: the idea that the 7 is "at a specific decimal" doesn't hold true, due to that pesky ellipsis. The crazy bit about the infinite repetition is that the 7 in 0.666...7 is not at a specific decimal. It's not even at "the last decimal" because there is no last. So, let's show this via a proof by contradiction: --- 1. we assert that 0.666...7 is a sequence of digits. 2. we assert that each digit in a sequence can be assigned an integer index corresponding to its position in the sequence. I.e. we can defined each digit's index as "the number of digits that precede this digit". 3. from (1) and (2) it follows that the index for 7 must be an integer. 4. the ellipses represents an infinite number of digits (infinitely repeating the repeated digit pattern preceding it). 5. from (2) and (4) it follows that the index for 7 must be the integer value "infinity", because it has an infinite number of digits preceding it. 6. (3) and (5) cannot both be true, because infinity is not an integer. 7. from (6) it follows that (1), and/or (2), and/or (4) must be false 8. (4) is, by definition, true. 9. from (7) and (8) it follows that (1) and/or (2) must be false. 10. (1) is our fundamental assertion. If (1) was false then there is wouldn't even be a sequence of digits for us to reason about. So (1) is true. 11. from (6), (8), and (10) it follows that (2) must be false for there to be no contradiction. QED --- Now, certainly, for finite length numbers the assumption that each digit in a sequence has an integer index holds true, but it turns out we have mathematical notation that lets us write down numbers for which that property does not hold. Infinities are fun. And difficult. But also fun.
- gspr 7y ago
  > 1. we assert that 0.666...7 is a sequence of digits. A sequence in the mathematical sense is a function whose domain is the natural numbers. Please define that function for the creature you're working with here. Otherwise you're trying to prove things about an object with no definition. You will end up in trouble.
  - TheRealPomax 7y ago
    Why? https://news.ycombinator.com/item?id=23009160 https://news.ycombinator.com/item?id=23009160 posited that we were in a situation where (1) holds, so that's where we're starting. If we assume (1) holds, then (2) cannot hold. We can abandon (1) but then we're no longer replying to that specific comment, now we're trying to prove something else.