|
|
|
![]() |
|
|||||||
|
||||||||
![]() |
| Thread Tools | Rate Thread | Display Modes |
|
#16
|
||||
|
||||
|
Re: .999~ = 1
Quote:
Last edited by dk5sm5luigi : 26-01-2004 at 01:58. |
|
#17
|
|||
|
|||
|
Re: .999~ = 1
A little while ago I wrote a long calculus proof about why .999... = 1 but I can't find it.
But it's true. Think about it this way: 1 - .999... = 1/inf 1/inf = 0 (by definition). For the "proof" that 3 = 1, my trusty TI-89 says that ln[e^(3*i*pi}] = i*pi not 3*i*pi. I don't know much about imaginary numbers, so don't ask me why. For the "proof" that 2=1, that's easy. At one point you divide both sides by (a-b). a = b therefore a-b = 0 and you are dividing by 0. |
|
#18
|
|||||
|
|||||
|
Re: .999~ = 1
Quote:
let a=1, b=1 a = a a^2 = ab a^2 - b^2 = ab - b^2 (a + b)(a - b) = b(a - b) This line gives you a division by zero error when you divide both sides by (a-b), because a-b = 0. Also, e^i*pi is a perfectly valid expression that doesn't take the ln of a negative number. e^i*pi = cos(pi) + i sin(pi) = -1 Last edited by Kris Verdeyen : 26-01-2004 at 11:20. |
|
#19
|
||||
|
||||
|
Re: .999~ = 1
Don't think .999999999... = 1?
Think about it! Between any two numbers, there must needs be another number. For example, between 1 and 3 there is 2; between .5 and .6 there is .55; you can satisfy yourself if you don't buy it. Anyway, try to find a number between .999999... and 1 ... you can't, as they are the same number (just as you probably couldn't find a number between 2 and, well, 2). Anyway, check out the authority on the matter: Dr. Math |
|
#20
|
|||
|
|||
|
Re: .999~ = 1
Quote:
|
|
#21
|
|||||
|
|||||
|
Re: .999~ = 1
Quote:
no. order of operations only matters when you are actually evaluating something. you are not yet evaluating at this point, only at the end. The reason why this little tidbit of confusion can exist is because the laws governing the use of the subdivision of fractions involving roots do not extend into the realm of imaginary numbers. |
|
#22
|
|||||
|
|||||
|
Re: .999~ = 1
Quote:
Therefore, the final step of the proof will state (+-)1=(+-)1, which is true. Here is another interesting one: a proof by induction that everyone who reads Chief Delphi is the same age! First, a little refresher of proof by induction. In an inductive proof, we prove that statement S(n) is true when n=1, and then prove that if S(n) is true, then S(n + 1) is true. If we prove these two things, we have then proven that S(n) is true for all values of n. Statement S(n): In any group of n Chief Delphi readers, everyone in that group has the same age. First I prove that S(1) is true:
Last edited by ahecht : 26-01-2004 at 15:03. |
|
#23
|
|||||
|
|||||
|
Re: .999~ = 1
Also, the infinite decimal expansion 0.99999... is the geometric series .9+.09+.009+...
a = 0.9 (the first term) r = 0.1 (the constant multiple). Geometric series with |r|<1 converge to a/(1-r). In this case, that's 0.9 / (1-0.1) = 0.9/0.9 = 1 So 0.9999... = 1. |
|
#24
|
|||||
|
|||||
|
Re: .999~ = 1
Quote:
That's because you can't have decimals and fractions in a real number...so I guess you could say yes, .999~ and 1 are the same real number. [Edit: Wait...just kidding. Im dumb.] |
|
#25
|
||||
|
||||
|
Re: .999~ = 1
well, for those who still have doubts about the difference between the number. first of all, because i am too lazy to find the alt + some odd number to find a 9 with a repeating symbol above it, yes i did use .999~ to represent an infinite amount of 9's, so that the only numerical difference between .999~ and 1 is .000~1, which is an infinitesimly small amount, and therefore in math, ignored.
also when i said infinitesimly, if i made up a word there, sorry, but it sounds good.... |
|
#26
|
||||
|
||||
|
Re: .999~ = 1
Quote:
|
|
#27
|
|||||
|
|||||
|
Re: .999~ = 1
Quote:
|
|
#28
|
||||
|
||||
|
Re: .999~ = 1
Quote:
|
|
#29
|
||||
|
||||
|
Re: .999~ = 1
Quote:
|
|
#30
|
|||||
|
|||||
|
Re: .999~ = 1
Quote:
![]() His 9x=9 didn't come from multiplying 9*.999.... (Which would indeed give you 8.999...1.) Rather it came from subtracting the two equations above it. [Actually, if you were to multiply out 9*.999..., since the .999... never ends, the 8.999...1 would be 8.999.... You'd never get to the 1. According to the postulate, this would be equal to 9, but we can't rely on that, since it's the postulate that we're trying to prove.] Last edited by Greg Ross : 29-01-2004 at 00:40. Reason: Added an afterthought. |
![]() |
| Thread Tools | |
| Display Modes | Rate This Thread |
|
|