.999~ = 1

[pryoplasm]

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.

[matt111]

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

[Jeremy_Mc]

Originally Posted by pryoplasm

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.

That's a pretty flawed proof since .333~ and .666~ are both decimal approximations of a fraction. 1/3 is really a (theoretically) neverending sequence of .33333~ (you get the idea) and the same is true for 2/3.

So, in reality ~1/3 + ~2/3 = ~1. (~ is the sign for approximation [I think]).

[George1902]

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.

[Solace]

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

[Jeremy_Mc]

Originally Posted by Solace

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

Not quite

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?
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...

[Phil 33]

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.

[Case]

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?

[Jeremy_Mc]

Originally Posted by Phil 33

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.

Fair enough the second proof works, but I was referring to the first one.

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.

In calculus they always tell you there's more than one calculus for every problem, and think these are definitely proofs of that atleast

[dk5sm5luigi]

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

[John V-Neun]

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.

[Jeremy_Mc]

Originally Posted by John V-Neun

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.

Haha that site makes me laugh...

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.

[Grommit]

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.

[n00b]

Originally Posted by dk5sm5luigi

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

major probelm: you cannot take the ln of a negative number. therefore, the term ln e^(i * pi) cannot exist. i have another one of those phony proofs, its kinda fun:

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

[n00b]

Originally Posted by Solace

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

square rooting then squaring again cause a loss of information. consider:
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).