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Maybe you will find the proof that the infinite series 0.9999... exactly equals 1 interesting: https://en.wikipedia.org/wiki/0.999 https://en.wikipedia.org/wik

Maybe you will find the proof that the infinite series 0.9999... exactly equals 1 interesting: https://en.wikipedia.org/wiki/0.999 https://en.wikipedia.org/wiki/0.999...

jcul 3y ago
Wow, can't believe I've never realised this. How counterintuitive. The 1/3 * 3 argument, I found the most intuitive.
- quchen 3y ago
  It's a flawed psychological argument though, because it hinges on accepting that 0.333...=1/3, for which the proof is the same as for 0.999...=1. People have less of a problem with 1/3 so they gloss over this - for some reason, nobody ever says "but there is always a ...3 missing to 1/3" or something.
  - Scarblac 3y ago
    The problem is that there are two different ways to write the same number in infinite decimals notation. (0.999... and 1.000...). Thats what's counter intuitive to people, it's not an issue with 1/3. That has just one way to write it as decimals, 0.333...
- thaumasiotes 3y ago
  I like the argument that observes "if you subtract 0.99(9) from 1, you get a number in which every decimal place is zero". The geometric series proof is less fun but more straightforward. As a fun side note, the geometric series proof will also tell you that the sum of every nonnegative power of 2 works out to -1, and this is in fact how we represent -1 in computers.
  - nly 3y ago
    How can the sum of a bunch of positive powers powers of 2 be a negative number? Isn't the sum of any infinite series of positive numbers infinity?
    - defrost 3y ago
      \1 is a good question that deserves an answer. \2 is "not always" .. Consider SumOf 1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 ... an infinite sequence of continuously decreasing numbers, the more you add the smaller the quantity added becomes. It appears to approach but never reach some finite limit. Unless, of course, by "Number" you mean "whole integer" | counting number, etc. It's important to nail down those definitions.
      - thaumasiotes 3y ago
        > \1 is a good question that deserves an answer. The same argument I mentioned above, that subtracting 0.99999... from 1 will give you a number that is equal to zero, will also tell you that binary ...11111 or decimal ...999999 is equal to negative one. If you add one to the value, you will get a number that is equal to zero. You might object that there is an infinite carry bit, but in that case you should also object that there is an infinitesimal residual when you subtract 0.9999... from 1. It works for everything, not just -1. The infinite bit pattern ...(01)010101 is, according to the geometric series formula, equal to -1/3 [1 + 4 + 16 + 64 + ... = 1 / (1-4)]. What happens if you multiply it by 3? ...0101010101 x 11 ------------------- ...0101010101 + ...01010101010 ------------------- ...11111111111 You get -1.
        Dylan16807 3y ago
        But if you look at limits you get "0" and "diverges". And decimal "...999999" is an infinity, which should immediately set off red flags and tell you that you need to be extra careful when analyzing it. In computers your series of 1s is not infinite, there's a modulus that steps in. And this analysis depends on the modulus being an exact power of the base. But you could make a system that's decimal but has a modulus of 999853, for example, and then "-1" would be 999852.
- nly 3y ago
  Another intuition: All the decimals that recur are fractions with a denominator of 9. E.g. 0.1111.... is 1/9 0.7777.... is 7/9 It therefore stands to reason that 0.99999.... is 9/9, which is 1
  - hoosieree 3y ago
    That's a good one! Might replace my current favorite which is: Let x = 0.99... Then 10*x = 9.99... And if we subtract x from both sides, we get: 10x - x = 9.99... - x And since we already defined x=0.99... when we subtract it from 9.99..., we get 9x = 9 So we can finally divide both sides by 9: x = 1

Maybe you will find the proof that the infinite series 0.9999... exactly equals 1 interesting: https://en.wikipedia.org/wiki/0.999 https://en.wikipedia.org/wiki/0.999...

jcul 3y ago

Wow, can't believe I've never realised this. How counterintuitive. The 1/3 * 3 argument, I found the most intuitive.

quchen 3y ago
It's a flawed psychological argument though, because it hinges on accepting that 0.333...=1/3, for which the proof is the same as for 0.999...=1. People have less of a problem with 1/3 so they gloss over this - for some reason, nobody ever says "but there is always a ...3 missing to 1/3" or something.
- Scarblac 3y ago
  The problem is that there are two different ways to write the same number in infinite decimals notation. (0.999... and 1.000...). Thats what's counter intuitive to people, it's not an issue with 1/3. That has just one way to write it as decimals, 0.333...
thaumasiotes 3y ago
I like the argument that observes "if you subtract 0.99(9) from 1, you get a number in which every decimal place is zero". The geometric series proof is less fun but more straightforward. As a fun side note, the geometric series proof will also tell you that the sum of every nonnegative power of 2 works out to -1, and this is in fact how we represent -1 in computers.
- nly 3y ago
  How can the sum of a bunch of positive powers powers of 2 be a negative number? Isn't the sum of any infinite series of positive numbers infinity?
  - defrost 3y ago
    \1 is a good question that deserves an answer. \2 is "not always" .. Consider SumOf 1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 ... an infinite sequence of continuously decreasing numbers, the more you add the smaller the quantity added becomes. It appears to approach but never reach some finite limit. Unless, of course, by "Number" you mean "whole integer" | counting number, etc. It's important to nail down those definitions.
    - thaumasiotes 3y ago
      > \1 is a good question that deserves an answer. The same argument I mentioned above, that subtracting 0.99999... from 1 will give you a number that is equal to zero, will also tell you that binary ...11111 or decimal ...999999 is equal to negative one. If you add one to the value, you will get a number that is equal to zero. You might object that there is an infinite carry bit, but in that case you should also object that there is an infinitesimal residual when you subtract 0.9999... from 1. It works for everything, not just -1. The infinite bit pattern ...(01)010101 is, according to the geometric series formula, equal to -1/3 [1 + 4 + 16 + 64 + ... = 1 / (1-4)]. What happens if you multiply it by 3? ...0101010101 x 11 ------------------- ...0101010101 + ...01010101010 ------------------- ...11111111111 You get -1.
      - Dylan16807 3y ago
        But if you look at limits you get "0" and "diverges". And decimal "...999999" is an infinity, which should immediately set off red flags and tell you that you need to be extra careful when analyzing it. In computers your series of 1s is not infinite, there's a modulus that steps in. And this analysis depends on the modulus being an exact power of the base. But you could make a system that's decimal but has a modulus of 999853, for example, and then "-1" would be 999852.
nly 3y ago
Another intuition: All the decimals that recur are fractions with a denominator of 9. E.g. 0.1111.... is 1/9 0.7777.... is 7/9 It therefore stands to reason that 0.99999.... is 9/9, which is 1
- hoosieree 3y ago
  That's a good one! Might replace my current favorite which is: Let x = 0.99... Then 10*x = 9.99... And if we subtract x from both sides, we get: 10x - x = 9.99... - x And since we already defined x=0.99... when we subtract it from 9.99..., we get 9x = 9 So we can finally divide both sides by 9: x = 1