# .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.

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

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 ). 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.

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… 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.

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?

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 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

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.

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.

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

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).

you just don’t know how to deal with imaginary numbers. plus if you look at how i display it i pull the e and i out of it making it ln e which is a positive number. Although there are other flaws. you just have to find them.

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^(3ipi}] = ipi not 3i*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.

Although, I think it would actually go like this:
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^ipi is a perfectly valid expression that doesn’t take the ln of a negative number. e^ipi = cos(pi) + i sin(pi) = -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

This is related to sines and cosines. To say cos(pi) = cos(3*pi) proves that 1=3 would be incorrect.