0.99999999... = 1 ?

[delmarco]

http://www.linuxforums.org/forum/topic-59751.html


holy shit.....

this linux forums was utterly almost destroyed with a civil war over this problem...

see link...

[delmarco]

lim( m -- > ∞ ) sum ( n = 1)^m (9)/(10^n) = 1
0.9999... = 1

Thus

x = 0.9999...
10x = 9.9999...
10x - x = 9.9999... - 0.9999...
9x = 9
x = 1.

was i wrong again?


http://www.blizzard.com/press/040401.shtml

[Speedinfusion]

e=mc squared


i win

[komodo]

Originally Posted by delmarco

http://www.blizzard.com/press/040401.shtml

They issued a press release on it? Haha!

[donald1476]

The equations 1/3=.333..., 2/3=.666..., 3/3=.999... are all incorrect.

The fraction 1/3 when represented as a decimal does not terminate. That is to say, in order to make the decimal exactly equal to the fraction one needs an infinite number of places in the decimal. Therefore 1/3 does not even equal 0.333333333333333333333333333333333333333333333333 3333333333333333333333333 and so forth as I can never enter an infinite number of 3's. Thus all of the three equations above are incorrect and so "the proof" is not correct.

1/3 is a fraction and 0.333 is called a decimal representation.

[xsperf]

? thats why you say POINT THREE REPEATING

[delmarco]

x = 0.9999...
10x = 9.9999...
10x - x = 9.9999... - 0.9999...
9x = 9
x = 1.

actually this equation set is flawed. Obviously flawed. And wheter or not 1 = 0.9999 this equation does not prove the numbers are equivalent.
I've been thinking of this equation for sometime now and I realize this,
at line a; we say x = 0.999
at line b; we multiply both sides by 10 and get 10x= 9.999 which is true.
at line c; we subtract X from both sides...here is where the flaw comes in...
for 10x - x (we must assume x is not equal to 1 yet so the solution is NOT 9x but something like 9.00000000001x)
and then on the otherside we have 9.999 - x or 9.999 - 0.999 which is 9

I'll stop here because at this point we are assuming the answer for x is 1 before proving the answer that x WILL BE 1. The is also a law broken here because we cannot assign x to 2 different numbers (1 and 0.999) before solving for x. In other words you can't tell a blind man that a red car is blue and when he can see a week later you show him an identical blue car and tell him that is the same car.
i'll continue
in the next line d; we have
9.0000000001x = 9
thus x = 0.9999 and not 1.
9.000000001 is not the same kind of number as 0.99999 because the zero's in 9.0000000001 will go on from 0 to infinity but the final number after the zero will be 1.

This is solvible but you can't do it with a simple algebraic equation you need knowledge of continuing fractions and limitations found in calc2 to prove whether or not 1 is equal to 0.9999

My argument is that this x = 0.999 equation set is the worst way of hustling someone into thinking 1 is not equal to 1 but 0.9999

if that was the case then 1 - 1 will not be 0 and you will have broken all sorts of conservation and euclidean geometry laws.
and the Universe will explode!!!
ha ha ha ha

[voca]

Originally Posted by delmarco

at line c; we subtract X from both sides...here is where the flaw comes in...
for 10x - x (we must assume x is not equal to 1 yet so the solution is NOT 9x but something like 9.1111x)
and then on the otherside we have 9.999 - x or 9.999 - 0.999 which is 9

uh, your not assuming x = 1 yet. the 10x comes from part b.

10 * .999~ = 9.999~ (this is what we conclude is part be.)

so if 10x = 9.999~

then

10x - x = 9

or

9.999~ - .999~ = 9


I am not arguing that .999~ = 1. I, in fact, believe that that's bogus and this whole thing is just a small paradox that doesn't at all effect my life so I couldn't care less.

the above proof was released by Blizzard Entertainment® (PC game developer and publisher of Warcraft, etc.) and if you want to argue with it you really should take it up with Mike Morhaime. good luck

[c1apton]

Originally Posted by Speedinfusion

e=mc squared

In my world it's E=mc sq.(+/- 3dB)

[beem]

i remember my old calc professor showed us something like this.. except he used 2.9999 and 3. I'll try and get it from him and post it sometime next week

[delmarco]

http://ned.ucam.org/~sdh31/maths/pointnine.html

here is a great link that features all the various proofs and critiques them and then critiques the critiques.