5 ms·

Sorry for my naivety, but why one couldn't prove by induction that adding 9s never close the gap, or let's say, that by definition the operation is such, that i

by vfinn 6y ago

Sorry for my naivety, but why one couldn't prove by induction that adding 9s never close the gap, or let's say, that by definition the operation is such, that it never closes the gap. If you can always halve the pie, then you can continue eating forever. To me it would be much easier to accept that (1/3)*3 is not 1.

saagarjha 6y ago
Induction can be non-intuitive to people who are troubled by this problem.
- jhanschoo 6y ago
  Induction has nothing to do with this. It's about a taking the limit of a process.
jhanschoo 6y ago
It's because the rigorous mathematical definition is more subtle. It's defined as the smallest real number such that repeating this process (putting more 9's in the end) can't result in a number that's greater than it. So as long as you have a number that's smaller than 1, there's a gap there, and repeating the process of adding 9's will eventually give you a number that lies in that gap.
- vfinn 6y ago
  Yeah, it makes sense if the whole foundation of mathematics is laid out in such way that the above has to be true (kinda circular, but still), and this would mean that what I'm talking about is actually not mathematics but something else :).
mFixman 6y ago
Let's try a naïve inductive proof of 0.999… ≠ 1. Base case: Given 𝑎⁰ = 0, 𝑎⁰ ≠ 1. Inductive case: Given 𝑎ⁿ⁺¹ = 1 - (1 - 𝑎ⁿ) / 10, 𝑎ⁿ ≠ 1 ⇒ 𝑎ⁿ⁺¹ ≠ 1 This proves that for every 𝑎ⁿ = 0.999…9, there's an 𝑎ⁿ⁺¹ that's a 9 larger and still different than 1, which is similar to your halving the pie example. However, you can see that it always happens that 𝑎ⁿ⁺¹ > 𝑎ⁿ, so the "last" infinite 0.999… is not part of the sequence of the inductive case. My intuitive way to see this is that infinitely repeating decimals are an abomination that breaks the nice property of decimal notation where each number contains a single representation (without zeroes at the beginning or at the end of the decimal). Fractions are the one true way to represent rational numbers.
topaz0 6y ago
By induction, you can prove that no _finite length_ sequence of 9s after the decimal reach one. It has nothing to say about the infinite limit, though.
kosievdmerwe 6y ago
Other people have pointed out that induction never makes the jump from a finite number of 9s to an infinite number of 9s. I feel the easiest "proof" is a proof by contradiction. First hopefully we can agree that if we have two real numbers x and y that are not equal then we have a number z, such that x < z < y. The easiest example is z = (x + y)/2. If 0.999...!= 1 then there must exist a number A, such that 0.999... < A < 1. Now since 0 < A < 1 (as 0 < 0.999...) it should be easy to see that A's decimal expansion is of the form 0.abcdef... . Since, A != 0.999... one of the digits in the decimal expansion of A has to be something other than a 9. For instance, we might have A = 0.99998999... . However, this would mean A < 0.999... as all digits other than 9 are smaller than 9 [1], but this contradicts our initial assumption that 0.999... < A and since we're dealing with strict inequalities both can't be true at the same time! Thus no such A should exists and thus 0.999... = 1. Now this isn't a rigorous proof, the thing that makes me most uncomfortable is the bit that I state A < 0.999..., but I'm uncomfortable because A might have multiple decimal expansions and I don't know how the algorithm in [1] interacts with that, however, if someone quibbles about that bit of the proof for that reason I feel they should have already accepted that 0.999... = 1 via another rigorous proof. [1] If this is not clear think about how you would compare two decimal expansions to see if one is smaller than the other. You go through every digit until you find one that is different between the two numbers and then you compare those.
- vfinn 6y ago
  "If 0.999...!= 1 then there must exist a number A, such that 0.999... < A < 1." Thanks for the reply, but I don't know how you can make the above claim, because to me the question of what we mean by 0.999... is intertwined with the claim itself. I mean to me it seems to be a matter of how you interpret the approach to infinity. I don't see why there has to be A in between, if you interpret 0.999... as the "biggest possible number below 1", as then there would be also a "difference of the smallest possible amount" between those numbers approaching 0, but not quite getting there. But then again, if it's by some fundamental definition (limit) that 0.999... = 1, then ok. I'm slightly embarrassed I don't know more mathematics, but I'm trying to learn some more...
  - 6y ago

by vfinn 6y ago

saagarjha 6y ago

Induction can be non-intuitive to people who are troubled by this problem.

jhanschoo 6y ago
Induction has nothing to do with this. It's about a taking the limit of a process.

jhanschoo 6y ago

It's because the rigorous mathematical definition is more subtle. It's defined as the smallest real number such that repeating this process (putting more 9's in the end) can't result in a number that's greater than it. So as long as you have a number that's smaller than 1, there's a gap there, and repeating the process of adding 9's will eventually give you a number that lies in that gap.

vfinn 6y ago
Yeah, it makes sense if the whole foundation of mathematics is laid out in such way that the above has to be true (kinda circular, but still), and this would mean that what I'm talking about is actually not mathematics but something else :).

mFixman 6y ago

Let's try a naïve inductive proof of 0.999… ≠ 1. Base case: Given 𝑎⁰ = 0, 𝑎⁰ ≠ 1. Inductive case: Given 𝑎ⁿ⁺¹ = 1 - (1 - 𝑎ⁿ) / 10, 𝑎ⁿ ≠ 1 ⇒ 𝑎ⁿ⁺¹ ≠ 1 This proves that for every 𝑎ⁿ = 0.999…9, there's an 𝑎ⁿ⁺¹ that's a 9 larger and still different than 1, which is similar to your halving the pie example. However, you can see that it always happens that 𝑎ⁿ⁺¹ > 𝑎ⁿ, so the "last" infinite 0.999… is not part of the sequence of the inductive case. My intuitive way to see this is that infinitely repeating decimals are an abomination that breaks the nice property of decimal notation where each number contains a single representation (without zeroes at the beginning or at the end of the decimal). Fractions are the one true way to represent rational numbers.

topaz0 6y ago

By induction, you can prove that no _finite length_ sequence of 9s after the decimal reach one. It has nothing to say about the infinite limit, though.

kosievdmerwe 6y ago

Other people have pointed out that induction never makes the jump from a finite number of 9s to an infinite number of 9s. I feel the easiest "proof" is a proof by contradiction. First hopefully we can agree that if we have two real numbers x and y that are not equal then we have a number z, such that x < z < y. The easiest example is z = (x + y)/2. If 0.999...!= 1 then there must exist a number A, such that 0.999... < A < 1. Now since 0 < A < 1 (as 0 < 0.999...) it should be easy to see that A's decimal expansion is of the form 0.abcdef... . Since, A != 0.999... one of the digits in the decimal expansion of A has to be something other than a 9. For instance, we might have A = 0.99998999... . However, this would mean A < 0.999... as all digits other than 9 are smaller than 9 [1], but this contradicts our initial assumption that 0.999... < A and since we're dealing with strict inequalities both can't be true at the same time! Thus no such A should exists and thus 0.999... = 1. Now this isn't a rigorous proof, the thing that makes me most uncomfortable is the bit that I state A < 0.999..., but I'm uncomfortable because A might have multiple decimal expansions and I don't know how the algorithm in [1] interacts with that, however, if someone quibbles about that bit of the proof for that reason I feel they should have already accepted that 0.999... = 1 via another rigorous proof. [1] If this is not clear think about how you would compare two decimal expansions to see if one is smaller than the other. You go through every digit until you find one that is different between the two numbers and then you compare those.

vfinn 6y ago
"If 0.999...!= 1 then there must exist a number A, such that 0.999... < A < 1." Thanks for the reply, but I don't know how you can make the above claim, because to me the question of what we mean by 0.999... is intertwined with the claim itself. I mean to me it seems to be a matter of how you interpret the approach to infinity. I don't see why there has to be A in between, if you interpret 0.999... as the "biggest possible number below 1", as then there would be also a "difference of the smallest possible amount" between those numbers approaching 0, but not quite getting there. But then again, if it's by some fundamental definition (limit) that 0.999... = 1, then ok. I'm slightly embarrassed I don't know more mathematics, but I'm trying to learn some more...
- 6y ago