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> I think the point is not defining 0.9... to be 1, the point is that “...” means an infinite number of 9s. If you shift the decimal point by 1, then nothing ch

> I think the point is not defining 0.9... to be 1, the point is that “...” means an infinite number of 9s. If you shift the decimal point by 1, then nothing changes, there are still an infinite number of 9s. If you shift the decimal point by 5 places, there are still an infinite number of 9s to the right. And here is the logical (induction) step: if you shift the decimal point by an infinite number of places, then there are still an infinite number of 9s to the right. This works for any repeating fraction, in groups of more than 1 repeating digit. I've heard this argument many times. I understand the intuitive reasoning. I just don't find it a proof. I mean with reasoning like this why can't you have an infinite amount of 9s and then just put a 7 after that? What's keeping you from doing that? It's just a hand-wavy argument with no rules of any kind of what are allowed. > Do you mean you disagree with the result, or that you agree with the result but don’t believe the proof is really a proof? I've never considered that specific "proof" a proof. When 0.9... is given a proper definition of limits and considered within the real numbers, then sure of course it's true and the proof is legitimate.

dahart 7y ago
> why can’t you have an infinite amount of 9s and then just put a 7 after that? You can. The proof still works if you do that. What you’re refusing to accept here is the definition of infinity.
- CogitoCogito 7y ago
  > You can. The proof still works if you do that. > What you’re refusing to accept here is the definition of infinity. I'm refusing to accept the definition of infinity? I have no idea what you mean by that. Would you make the same statement were you aware that I do in fact have a PhD in mathematics in the field of analysis? That I have in fact studied logic? Just as a hypothetical scenario.
  - dahart 7y ago
    The proof hinges on the definition of infinity. 0.9bar7 is a completely nonsensical number precisely because of the definition of infinity. Yes, I think you’re failing to accept the definition of infinity. You’re rejecting a proof that many other PhDs in math accept along with some notable mathematicians like Euler. I reject your appeal to authority; having an advanced degree doesn’t mean you’re somehow automatically right, lots of PhDs in math have failed the Monty Hall problem, lots of people with advanced degrees are wrong all the time. I myself am a walking example on occasion. I haven’t heard a reason yet why the logic of the proof doesn’t work, I’ve only heard that you don’t accept it. What is the reason it doesn’t work?
    - CogitoCogito 7y ago
      > The proof hinges on the definition of infinity. 0.9bar7 is a completely nonsensical number precisely because of the definition of infinity. Yes, I think you’re failing to accept the definition of infinity. You’re rejecting a proof that many other PhDs in math accept along with some notable mathematicians like Euler. I reject your appeal to authority; having an advanced degree doesn’t mean you’re somehow automatically right, lots of PhDs in math have failed the Monty Hall problem, lots of people with advanced degrees are wrong all the time. I myself am a walking example on occasion. I wasn't appealing to authority. I never said that implied that I was correct. (Had I wanted to do that, I would have brought that up much earlier in this comment thread.) I was really just curious if you still believed that I was "failing to accept the definition of infinity" even if you knew that about me. Apparently you do. > I haven’t heard a reason yet why the logic of the proof doesn’t work, I’ve only heard that you don’t accept it. What is the reason it doesn’t work? https://news.ycombinator.com/item?id=23007600 https://news.ycombinator.com/item?id=23007600
      - dahart 7y ago
        Okay, I understand what you’re trying to say. I accept that the proof does not define what “...” means formally, and that is the problem you have with it. The infinity is understood implicitly to have the property that the 9s never end. In that sense, I think the proof does make use of limits, it just relies on definitions not written as part of the proof. Isn’t that okay, doesn’t that actually happen very often? The proof also doesn’t define what addition, multiplication, and equality mean either, but other kinds of mathematicians might complain on those grounds. How complete is complete, and what is the purpose of a proof if not to demonstrate a truth economically, relying on, rather than restating, the already laid foundation? How could this particular proof be shown to have weakness or fail before adding rigorous definitions of limits? Would it be any more apparently true to a wide variety of mathematicians and students if it had the type of rigor you’re advocating?
gjulianm 7y ago
There is no "after infinity". You can justify the idea by defining a decimal representation of a number x as a vector x_2, x_1, x_0, x_{-1}, x_{-2} ..., with x_n ∈ {0, 1, ..., 9}. Negative indexes are digits after the comma, positive indexes before the comma. You can recover the original number simply using x = \sum_{n=-∞}^{n=∞} 10^n x_{n} (1) (this sum always converges as long as x_n = 0 when n > N, for some big enough N ∈ ℕ). For example, the number three is represented by x_0 = 1, x_n = 0 otherwise. 0.9... is defined as x_n = 0 for n >= 0, 9 for n < 0. Now, by using limits, we could recover the original number using the formula above. However, we can do it without them. For that, we define just enough operations for the proof. 1. If z = x - y and for all x_n, y_n we have that x_n >= y_n, then z_n = x_n - y_n for all n. 2. If z = 10x, then z_n = x_{n-1}. For the first operation, in order to be rigorous, we need to ensure that if z_n = x_n - y_n in the same conditions, then z = x - y. The proof of this part just consists of plugging the recovery formula (1): z = \sum 10^n z_n = \sum 10^n (x_n - y_n) = (\sum 10^n x_n) - (\sum 10^n y_n) = x - y. We can perform all those operations as we are guaranteed that (1) always converges. Now, let y = 0.9... defined as in the example above (y_n = 0 for n >= 0, 9 otherwise), and let x = 10y (therefore x_n = 0 for n > 0, 9 otherwise). Now, define z_n = x_n - y_n, so that z_n = 9 for n = 1, 0 otherwise, which yields z = 9. As we demonstrated above, this implies that z = x - y, therefore 9 = 10y - y => y = 1, so 1 = 0.9... . PS: I don't think one can make a proof without at least using some bits of limits to be able to switch between decimal representation as a vector and the number itself. However I don't think it's a problem, because you need the same bits to be able to talk of "0.9..." as a well-defined number.
michaelcampbell 7y ago
Don't conflate "I don't understand it" with "hand-wavy". The 2 are different things. Another proof by contradiction I've heard of this is that if 0.9... != 1.0... then there exists a number in between the 2. What is it?
- ubercow13 7y ago
  It is hand-wavy, because it is using a vague conception of infinity which doesn't correspond to the real, useful definition of the infinity labelled "..."

dahart 7y ago

> why can’t you have an infinite amount of 9s and then just put a 7 after that? You can. The proof still works if you do that. What you’re refusing to accept here is the definition of infinity.

CogitoCogito 7y ago
> You can. The proof still works if you do that. > What you’re refusing to accept here is the definition of infinity. I'm refusing to accept the definition of infinity? I have no idea what you mean by that. Would you make the same statement were you aware that I do in fact have a PhD in mathematics in the field of analysis? That I have in fact studied logic? Just as a hypothetical scenario.
- dahart 7y ago
  The proof hinges on the definition of infinity. 0.9bar7 is a completely nonsensical number precisely because of the definition of infinity. Yes, I think you’re failing to accept the definition of infinity. You’re rejecting a proof that many other PhDs in math accept along with some notable mathematicians like Euler. I reject your appeal to authority; having an advanced degree doesn’t mean you’re somehow automatically right, lots of PhDs in math have failed the Monty Hall problem, lots of people with advanced degrees are wrong all the time. I myself am a walking example on occasion. I haven’t heard a reason yet why the logic of the proof doesn’t work, I’ve only heard that you don’t accept it. What is the reason it doesn’t work?
  - CogitoCogito 7y ago
    > The proof hinges on the definition of infinity. 0.9bar7 is a completely nonsensical number precisely because of the definition of infinity. Yes, I think you’re failing to accept the definition of infinity. You’re rejecting a proof that many other PhDs in math accept along with some notable mathematicians like Euler. I reject your appeal to authority; having an advanced degree doesn’t mean you’re somehow automatically right, lots of PhDs in math have failed the Monty Hall problem, lots of people with advanced degrees are wrong all the time. I myself am a walking example on occasion. I wasn't appealing to authority. I never said that implied that I was correct. (Had I wanted to do that, I would have brought that up much earlier in this comment thread.) I was really just curious if you still believed that I was "failing to accept the definition of infinity" even if you knew that about me. Apparently you do. > I haven’t heard a reason yet why the logic of the proof doesn’t work, I’ve only heard that you don’t accept it. What is the reason it doesn’t work? https://news.ycombinator.com/item?id=23007600 https://news.ycombinator.com/item?id=23007600
    - dahart 7y ago
      Okay, I understand what you’re trying to say. I accept that the proof does not define what “...” means formally, and that is the problem you have with it. The infinity is understood implicitly to have the property that the 9s never end. In that sense, I think the proof does make use of limits, it just relies on definitions not written as part of the proof. Isn’t that okay, doesn’t that actually happen very often? The proof also doesn’t define what addition, multiplication, and equality mean either, but other kinds of mathematicians might complain on those grounds. How complete is complete, and what is the purpose of a proof if not to demonstrate a truth economically, relying on, rather than restating, the already laid foundation? How could this particular proof be shown to have weakness or fail before adding rigorous definitions of limits? Would it be any more apparently true to a wide variety of mathematicians and students if it had the type of rigor you’re advocating?

gjulianm 7y ago

There is no "after infinity". You can justify the idea by defining a decimal representation of a number x as a vector x_2, x_1, x_0, x_{-1}, x_{-2} ..., with x_n ∈ {0, 1, ..., 9}. Negative indexes are digits after the comma, positive indexes before the comma. You can recover the original number simply using x = \sum_{n=-∞}^{n=∞} 10^n x_{n} (1) (this sum always converges as long as x_n = 0 when n > N, for some big enough N ∈ ℕ). For example, the number three is represented by x_0 = 1, x_n = 0 otherwise. 0.9... is defined as x_n = 0 for n >= 0, 9 for n < 0. Now, by using limits, we could recover the original number using the formula above. However, we can do it without them. For that, we define just enough operations for the proof. 1. If z = x - y and for all x_n, y_n we have that x_n >= y_n, then z_n = x_n - y_n for all n. 2. If z = 10x, then z_n = x_{n-1}. For the first operation, in order to be rigorous, we need to ensure that if z_n = x_n - y_n in the same conditions, then z = x - y. The proof of this part just consists of plugging the recovery formula (1): z = \sum 10^n z_n = \sum 10^n (x_n - y_n) = (\sum 10^n x_n) - (\sum 10^n y_n) = x - y. We can perform all those operations as we are guaranteed that (1) always converges. Now, let y = 0.9... defined as in the example above (y_n = 0 for n >= 0, 9 otherwise), and let x = 10y (therefore x_n = 0 for n > 0, 9 otherwise). Now, define z_n = x_n - y_n, so that z_n = 9 for n = 1, 0 otherwise, which yields z = 9. As we demonstrated above, this implies that z = x - y, therefore 9 = 10y - y => y = 1, so 1 = 0.9... . PS: I don't think one can make a proof without at least using some bits of limits to be able to switch between decimal representation as a vector and the number itself. However I don't think it's a problem, because you need the same bits to be able to talk of "0.9..." as a well-defined number.

michaelcampbell 7y ago

Don't conflate "I don't understand it" with "hand-wavy". The 2 are different things. Another proof by contradiction I've heard of this is that if 0.9... != 1.0... then there exists a number in between the 2. What is it?

ubercow13 7y ago
It is hand-wavy, because it is using a vague conception of infinity which doesn't correspond to the real, useful definition of the infinity labelled "..."