So 0.9999.... == 1 is simply a meme way of saying three thirds equal one?
>>170822923 x 0.333333... is just shy of 1.
>>17082299Most folks don’t mind, but maths types are sticklers for accuracy.
>>17082299cool it with the anti semitism
>>170822920.999..., 1/1 and 1 are representations of the same numerical quantity as an integer, fraction or decimal number.0.333... and 1/3 are also representations of their respective quantity, but they have no integer equivalent because the number they represent is a real/rational, not a natural/integer.
>>17082292>>17082299>>17082580Once you get past the eighth decimal place, the difference doesn't matter to anybody outside of particle physics whatsoever.
>>17082597On computer systems, such small diference get rounded anyway.
>>17082292Depending on how you define the decimal representation 0.999... might not even be a valid number. Either you have it as sum of digits times ten to the power of negative the digits place. Or you say eath digit is chosen such that no larger digit could be chosen in any place. The latter avoids all the nine-repeating patterns and gives every number a uniqe representative. My analyisis professor did it like that.
>>17082292
>>17082292Not quite. It's a statement of dogma.
>>17082292It's an artifact of the decimal notation.Pragmatic proof: you can't add 0.<arbitrary number of zeroes> 1 to 0.999... without overshooting 1.
Claim: [math] 0.999_{\dots} \neq 1[/math]Proof: We use induction. The base case is trivial: [math] 0.9 \neq 1[/math]. Next we introduce the notation that [math]0.9_n = \underbrace{0.9999999}_{n-\text{many nines}}[/math] is the decimal with n-many 9s.Now the inductive step: we assume [math]0.9_n \neq 1[/math]. Then trivially [math]0.9_{n+1} \neq 1 [/math]. It might help to notice that [math] 1 - 0.9_{n+1} \neq 0[/math]. This implies that [math]0.9_n \neq 1 \qquad \forall n\in \mathbb{N}[/math]Finally, we define [math] 0.999_{\dots} := \lim_{n\to\infty} 0.9_n[/math].[math]\therefore 0.999_{\dots} \neq 1 \qquad \square [/math]
Claim: [math] 0.999_{\dots} = 1 [\math] Proof: We use counterexample. Assume >>17083047 is true and that [math] \phi [\math] is a function of two numbers that produces the text of that prove if the first argument is [math] 0.999_{\dots} [\math] and the second is 1. (The first argument has to start repeating at after some digit, otherwise the function is undefined).Notice:- [math] \phi(0.999_{\dots}, 1) \iff \phi(0.888_{\dots}, 0.888_{\dots}) [\math]- But [math] 0.888_{\dots} = 0.888_{\dots} [\math], obviouslyThus [math] \phi(0.999_{\dots}, 1) [\math] is false and thus [math] 0.999_{\dots} = 1 [\math] \qquad \square
>>17083086idk how i fucked that up
>>17083088Your slashes are all ass backwards
>>17083090i see, thank you:Claim: [math] 0.999_{\dots} = 1 [/math]Proof: We use counterexample. Assume >>17083047 is true and that [math] \phi [/math] is a function of two numbers that produces the text of that prove if the first argument is [math] 0.999_{\dots} [/math] and the second is 1. (The first argument has to start repeating at after some digit, otherwise the function is undefined).Notice:- [math] \phi(0.999_{\dots}, 1) \iff \phi(0.888_{\dots}, 0.888_{\dots}) [/math]- But [math] 0.888_{\dots} = 0.888_{\dots} [/math], obviouslyThus [math] \phi(0.999_{\dots}, 1) [/math] is false and thus [math] 0.999_{\dots} = 1 [/math][math] \qquad \square [/math]
>>17083047Congrats on failing first-year analysisShowing that [math] 0.9_n\neq1\ [/math] for every finite [math] n [/math] doesn't imply that its limit isn't 1For that, you'd need to show that there exists an [math] \varepsilon>0\ [/math] such that [math] |1-0.9_n|>\varepsilon [/math] for all [math] n\in\mathbb N [/math]But [math] 1-0.9_n=10^{-n}\to0 [/math] so the limit is precisely 1
>>17082993>Depending on how you define the decimal representation 0.999...The nines are defined the same way the zeros are defined for an integer. It's a background digit. There are as many as you need, to complete an operation (or as many as you can get to, to try). There is never any situation, finite or infinite, where 0.9... ≠ 1.
>>17082292the real numbers defined as the completion of the equivalence classes of cauchy sequences of rational numbers whose difference is equivalent to a null sequence makes the answer to the question crystal clear, but requires a lot of definitions and thought to get there
>>17083902You don't need any of that for 0.9... to be equal to 1
>>17082292I don’t get why this is so hard for people with more than an iota of math knowledge. 0.999… by definition is equal to 1.It’s literally just shorthand for the limit of the summation of (0.9)(10^-n) from n=0 to i as i goes to infinity. Which is 1.
>>17083912Summation doesn't cut it for people who don't like infinity. The nines are there, in every decimal place, for as long as you can carry out an operation, just like the zeros are there for an integer.
>>17083917You can make the symbolic representation, but the representation cannot be made physically real or meaningful. You can write it out to be different, but for all physical intents and purposes, 9.999... is 1.
>>17083917There is no infinity. The infinity in the limit is a placeholder for asking “if I keep increasing this value, will I be able to get arbitrarily close to this other value?”There is no “actually infinite operation” being done.
>>17082292Correct. As per Mr.Nordnothingburger.
>>17083955>arbitrarily closeYou don't even need the concept of "keep increasing" "to get arbitrarily close." If you subtract the decimal representation of 1 (which has a zero wherever you need it) from the decimal representation of 0.9... (which has a nine wherever you need it), you'll never get anything but 0.0..., which is the decimal representation of zero, the additive identity.>>17083921>physically real or meaningful0.9... is no more or less real than 1.0..., in decimal notation.
>>17082292it will never be 1
