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)

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 ).

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

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.

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

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.

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