6 ms·

Both are false. 0.99999... is not less than 1. It is the same as 1.

by OskarS 7y ago

amelius 7y ago
Yes, because someone defined it that way. It is because the "limit" in 0.999... = lim[eps->0] 1-eps is implicit and defined as being applied before anything else. But you might as well define that implicit limit as applying over the entire expression. UPDATE: So instead of interpreting the expression as: (lim[eps->0] 1-eps) < 1 which is indeed false, you can also interpret the expression as: lim[eps->0] ((1-eps) < 1) which is true (assuming that -> denotes a limit from above). Note that here the "lim" has been taken out and acts over the entire expression.
- gspr 7y ago
  > But you might as well define that implicit limit as applying over the entire expression. What is this supposed to even mean?
- OskarS 7y ago
  It's still not true. No no sense is it true. It is true that 0.999... = lim[eps->0] (1 - eps), but it is ALSO true that lim[eps->0] (1 - eps) = 1. That's how limits work. If you accept both of those two (which you should, because they're correct), then since equality is transitive, 0.999... = 1. Therefore it is not less than 1, it is equal to 1.
  - amelius 7y ago
    I think you should look more closely at what I did with the order of operations, and the fact that "<" now is part of the expression acted over by the limit.
    - danharaj 7y ago
      It is not reasonable to interpret the notation that way. For one, variable binding becomes incoherent.
      - amelius 7y ago
        I think it is perfectly reasonable for two reasons. One is that (when it works) the notation becomes more intuitive, as in 0.9999... < 1 And the second reason is that mixing implicit limits (or series or infinity) and numbers is a mathematical hack, and when the notation doesn't work it is a clear indication that something is wrong.
    - OskarS 7y ago
      Limits are defined for functions and "1 - eps < 1" is not a function, it's an inequality. But even if we pretended that limits were defined for inequalities, there only reasonable interpretation of it is that: lim[eps->0]( 1 - eps < 1) is equal to 1 - 0 < 1 which is obviously false. Again though, limits are defined for functions, not inequalities. If you have the limit lim[eps->0] (x - eps) Then it is equal to x - 0 which is equal to x. You're simply wrong here. You can read that wikipedia article if you don't trust me, or you can watch any of the thousands of youtube videos of mathematicians explaining this to you. This is one of those things that is hard to do. Once you have an idea in your head that you're sure is right, it is incredibly difficult to dislodge it. It takes an enormous amount of humility and intellectual flexibility. But it is healthy and good to do it, every time you do it, you become a better person for it.
      - GolDDranks 7y ago
        You know, inequalities can be interpreted as boolean functions. The function under the limit produces consistently true as you approach the limit. (Of course one-sided one; the case where epsilon is getting smaller). It isn't a valid argument for 0.999... != 1 but an interesting one nevertheless.
- gspr 7y ago
  > which is indeed false, you can also interpret the expression as: > lim[eps->0] ((1-eps) < 1) Assuming you don't mean some special notion of limit, I would guess that by `(1-eps) < 1` you mean the function from the reals to `Y = {false, true}` that is defined as sending `x` to `true` if `1-x` is strictly less than `1`, and `false` otherwise. Let's call this function `f`. I assume you're endowing `Y` with the discrete metric? If so, `f` does indeed have a limit from above that is `false` and a limit from below that is `true`. Where do you wanna go from here? Edit: Corrected stupid wrong assertion about limit from below, d'oh.
  - amelius 7y ago
    Have a look here: https://www.wolframalpha.com/input/?i=Limit%5BSign%5B1-x-1%5D%2C+x-%3E0%5D https://www.wolframalpha.com/input/?i=Limit%5BSign%5B1-x-1%5... This shows that the limit does exist from both sides (but is different from both sides).
    - gspr 7y ago
      Oops, sorry, that was a stupid blunder. Thanks for the correction.
  - amelius 7y ago
    > If so, `f` does indeed have a limit from above that is `false` and a limit from below that is `true`. Where do you wanna go from here? That's the other way around. From above you get true, and from below you get false. Note that 0.999... represents the limit from above for 1-eps. Hence the result is true.
    - gspr 7y ago
      > That's the other way around. From above you get true, and from below you get false. Yeah, my bad. > Note that 0.999... represents the limit from above for 1-eps. Hence the result is true. I mean you can define 0.999… like that if you want. You'll get 1. So what? Your detour via `f` provided nothing.

Both are false. 0.99999... is not less than 1. It is the same as 1.

by OskarS 7y ago

Both are false. 0.99999... is not less than 1. It is the same as 1.

Yes, because someone defined it that way. It is because the "limit" in 0.999... = lim[eps->0] 1-eps is implicit and defined as being applied before anything else. But you might as well define that implicit limit as applying over the entire expression. UPDATE: So instead of interpreting the expression as: (lim[eps->0] 1-eps) < 1 which is indeed false, you can also interpret the expression as: lim[eps->0] ((1-eps) < 1) which is true (assuming that -> denotes a limit from above). Note that here the "lim" has been taken out and acts over the entire expression.

gspr 7y ago
> But you might as well define that implicit limit as applying over the entire expression. What is this supposed to even mean?
OskarS 7y ago
It's still not true. No no sense is it true. It is true that 0.999... = lim[eps->0] (1 - eps), but it is ALSO true that lim[eps->0] (1 - eps) = 1. That's how limits work. If you accept both of those two (which you should, because they're correct), then since equality is transitive, 0.999... = 1. Therefore it is not less than 1, it is equal to 1.
- amelius 7y ago
  I think you should look more closely at what I did with the order of operations, and the fact that "<" now is part of the expression acted over by the limit.
  - danharaj 7y ago
    It is not reasonable to interpret the notation that way. For one, variable binding becomes incoherent.
    - amelius 7y ago
      I think it is perfectly reasonable for two reasons. One is that (when it works) the notation becomes more intuitive, as in 0.9999... < 1 And the second reason is that mixing implicit limits (or series or infinity) and numbers is a mathematical hack, and when the notation doesn't work it is a clear indication that something is wrong.
  - OskarS 7y ago
    Limits are defined for functions and "1 - eps < 1" is not a function, it's an inequality. But even if we pretended that limits were defined for inequalities, there only reasonable interpretation of it is that: lim[eps->0]( 1 - eps < 1) is equal to 1 - 0 < 1 which is obviously false. Again though, limits are defined for functions, not inequalities. If you have the limit lim[eps->0] (x - eps) Then it is equal to x - 0 which is equal to x. You're simply wrong here. You can read that wikipedia article if you don't trust me, or you can watch any of the thousands of youtube videos of mathematicians explaining this to you. This is one of those things that is hard to do. Once you have an idea in your head that you're sure is right, it is incredibly difficult to dislodge it. It takes an enormous amount of humility and intellectual flexibility. But it is healthy and good to do it, every time you do it, you become a better person for it.
    - GolDDranks 7y ago
      You know, inequalities can be interpreted as boolean functions. The function under the limit produces consistently true as you approach the limit. (Of course one-sided one; the case where epsilon is getting smaller). It isn't a valid argument for 0.999... != 1 but an interesting one nevertheless.
gspr 7y ago
> which is indeed false, you can also interpret the expression as: > lim[eps->0] ((1-eps) < 1) Assuming you don't mean some special notion of limit, I would guess that by `(1-eps) < 1` you mean the function from the reals to `Y = {false, true}` that is defined as sending `x` to `true` if `1-x` is strictly less than `1`, and `false` otherwise. Let's call this function `f`. I assume you're endowing `Y` with the discrete metric? If so, `f` does indeed have a limit from above that is `false` and a limit from below that is `true`. Where do you wanna go from here? Edit: Corrected stupid wrong assertion about limit from below, d'oh.
- amelius 7y ago
  Have a look here: https://www.wolframalpha.com/input/?i=Limit%5BSign%5B1-x-1%5D%2C+x-%3E0%5D https://www.wolframalpha.com/input/?i=Limit%5BSign%5B1-x-1%5... This shows that the limit does exist from both sides (but is different from both sides).
  - gspr 7y ago
    Oops, sorry, that was a stupid blunder. Thanks for the correction.
- amelius 7y ago
  > If so, `f` does indeed have a limit from above that is `false` and a limit from below that is `true`. Where do you wanna go from here? That's the other way around. From above you get true, and from below you get false. Note that 0.999... represents the limit from above for 1-eps. Hence the result is true.
  - gspr 7y ago
    > That's the other way around. From above you get true, and from below you get false. Yeah, my bad. > Note that 0.999... represents the limit from above for 1-eps. Hence the result is true. I mean you can define 0.999… like that if you want. You'll get 1. So what? Your detour via `f` provided nothing.