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.999~ = 1
ok, bear in mind you need to comprehend at least 8th grade math to understnd this(stolen from LUE on gamefaqs.com quite some time ago, but good for conversation)
1/3 =.333~ 2/3 =.666~ .333~+.666~=.999~ 1/3 + 2/3 = 1 .999~=1 or .999~=1 x = .999 10x= 9.999~ 9x = 9 x=1 1=.999~ discuss. |
Re: .999~ = 1
1/3 as .333 is an estimate and since it goes on forever u cant say it equals 1 exactly
but i no what u mean and its clever |
Re: .999~ = 1
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So, in reality ~1/3 + ~2/3 = ~1. (~ is the sign for approximation [I think]). ;) |
Re: .999~ = 1
He's using the ~ to represent a repeating a repeating decimal, I think.
1/9 = .111~ 2/9 = .222~ 7/9 = .777~ 9/9 = .999~ = 1 This can also be done with 1/11 = .0909~ 2/11 = .1818~ 5/11 = .4545~ 10/11 = .9090~ 11/11 = .9999~ So, yes. .999~ = 1 I <3 math. |
Re: .999~ = 1
while we're bending the rules of mathematics, how about this one?
1 = 1 -1 = -1 -1/1 = -1/1 -1/1 = -1/1(-1/-1) -1/1 = 1/-1 (-1/1)^.5 = (1/-1)^.5 ((-1)^.5) / (1^.5) = (1^.5) / ((-1)^.5) cross multiply, and you get ((-1)^.5) * ((-1)^.5) = (1^.5) * (1^.5) -1 = 1 discuss that one why don't ya |
Re: .999~ = 1
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1 = 1 -1 = -1 -1/1 = -1/1 -1/1 = -1/1(-1/-1) -1/1 = 1/-1 (-1/1)^.5 = (1/-1)^.5 ((-1)^.5) / (1^.5) = (1^.5) / ((-1)^.5) cross multiply, and you get Did you forget order of operations? :p Exponents first :) (1/-1) / (1/1) = (1/1) / (1/-1) (-1) / (1) = (1) / (-1) -1 != 1 -1 = -1 Atleast I'm pretty sure that's right... :confused: |
Re: .999~ = 1
What do you mean flawed math? A lot of people don't like this..
.999~ = 1 (~ means "going on forever") ...but its true. He already gave this proof, so I don't know why I'm showing it again. x = 0.9~ - 10x = 9.9~ ___________ (subtract and the .9~'s cancel out -9x = -9 x = 1 Therefore: 1 = -.9~ Like it or not, its true. |
Re: .999~ = 1
There are acctually slight errors that can be found in all of those. Try this one, in calculus, you'll deal with "conditionally convergent series". which, if summed to infinity, can equal any number.
(Infinity) (Sigma) (-1)^n * (1/n) n=1 or in other words, plug 1 in for N, plug 2 in for N and add it, plus 3 in..... to infinity. And the number you get? Whatever you want, it can be 5, it can be 0, it can be 450005.4343. Crazy, huh? |
Re: .999~ = 1
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If you check the math on the second one, though, you end up with 9 * .999~ = 8.999999999999999999999999999991 (give or take .000000000000000000000000000001) but it never equals 9. I could graph it and show where the line never intersects, but I'm wayyyy lazy to do that. :p In calculus they always tell you there's more than one calculus for every problem, and think these are definitely proofs of that atleast ;) |
Re: .999~ = 1
How about this proof:
given: e^(i * pi) = -1 e^(3 * i * pi) = e^(i * pi) * e^(i * pi) * e^(i * pi) = -1 * -1 * -1 = -1 therefore: e^(i * pi) = e^(3 * i * pi) ln e^(i * pi) = ln e^(3 * i * pi) (i * pi) ln e = (3 * i * pi) ln e i * pi = 3 * i * pi 1 = 3 |
Re: .999~ = 1
Back when I was 229 Team Leader, I got a random email from some mathematician out to prove that dividing something by zero actually equaled zero.
http://members.lycos.co.uk/zerobyzero/ It's pretty weird. It was definitely strange to get an email from him, apparently he just searched the net randomly for people to hear his theory, and our team name ("Division by Zero") led him to me. |
Re: .999~ = 1
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Some of the theories on there are cool...but most of them are just kind of strange. He really likes the number zero... :( His samples really don't add up to me, though. Ex: (1/2) / (0/2) = (1/2) * (2/0) = 0 You can't divide something into portions of zero! I know he's using the basis of cross multiplication but yeah...it's still wrong. Back to the topic at hand, the last proposed proof (the e^x one) has officially blown my mind. I think it's time for a cookie. |
Re: .999~ = 1
Mathematicians would agree that .999... and 1 represent the same real number. In fact, from my real analysis book, "Elementary Analysis" by Kenneth Ross, it says:
[short proof, similar to above] "Thus 0.9999... and 1.0000... are different decimal expansions that represent the same real number!" Later on the book proves that the only case where this can occur is when the two expansions for the same real number end in an infinite series of nines and infinite series of zeros. |
Re: .999~ = 1
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let a=1, b=1 a = a a^2 = ab a^2 - b^2 = ab - b^2 (a + b)(a - b) = b(a - b) a + b = b 1 + 1 = 1 have fun =D |
Re: .999~ = 1
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x = 3, y = -3 x^2 = y^2 sqrt(x^2) = sqrt(y^2) sqrt = square root of however, x != y even though sqrt(x^2) = x and sqrt(y^2) = y and sqrt(x^2) = sqrt(y^2). |
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