Hello /pol/, I've been looking at the fact that 0.999...

try to find a number between .999... and 1
you cant, so they're the same number

>>8503000
It's obvious bait you tard

1/3 = 0.33..
if 1/3 = 1/3 and 3*1/3 = 1
given 0.33.. * 3 = 0.99..
then 0.99.. = 1
there aren't tricks here

>>8503000
>you cant
But why? What is the logical argument for this?

>>8502978
I'll give you a hint in case it's not bait. First write out an explicit definition of .999... maybe as a geometric series .9+.09+.009... or as a limit of a sequence [.9, .99, .999...].

File: 1=0,999999.png (54KB, 953x613px) Image search: [Google]

54KB, 953x613px

>>8502978
Food for discussion

>>8503000
>you cant, so they're the same number
what number is between 1 and 2 in the naturals?
1 = 2 confirmed?

>>8503487
we are talking about real numbers here kiddo
thanks for trying tho
kisses

File: gorach.jpg (74KB, 800x600px) Image search: [Google]

74KB, 800x600px

>>8502978
0.9999... = 1 - (1/infinity). Think about it. Stop mental masturbating you fucking retards.

Actually, there is a very important reason for this. 1 - 0.999...999 is not zero, but it is an infinitely small decimal meaning that it make an infinitely small difference on the number. TO show this algebraically. The thing about infinite numbers in algebra, if you assign any finite, infinite, or surreal operator to it then you still get an amount that is infinite except for the case of complex (complex infinity) and zero (that's a whole different argument), the inverse is true for the inverse of infinity. If you use any finite operator on it, the inverse of infinity (infinitesimal), will not affect the number, meaning when we take 0.999...999 away from one it isn't zero, but the amount it is is infinitely negligible. Therefore, zero point 9 repeating will always equal one.

>>8503652
You are wrong. Look up the definition of real numbers before talking crap. Your calculus level understanding isn't rigorous enough.

>>8503514
then 0.33333....=1/3 - 1/3inf
or, to put it other way, 0.33333.... =/= 1/3

>>8502978
You're not going to have a formal proof unless you have a meaningful definition for a decimal expansion. The standard definition is,
[math]
b.a_1a_2a_3a_4\dotsc := b + \sum_{n=1}^\infty \frac{a_n}{10^n}
[/math]
So then in particular
[math]
0.999... = \sum_{i=n}^\infty \frac{9}{10^n}.
[/math]
From this, it's trivial.

Suppose you reject this definition because you really want infinitesimal numbers to exist. Then (1) that's stupid, because nothing is actually wrong with the above definition, and (2) it's on that person to come up with a meaningful definition for .999... . The most common definition is "the number infinitely close to, but not equal to, 1". We can prove such a number does not exist. To do this, suppose x had this property. Then x < (x+1)/2 < 1, a contradiction.

>>8503081
try doing what he says before you say shit.

0.999999999999999... and 1
find a number between it.
go on try.
you cannot because the 9s go on forever and so you cannot fit any number in between.
thereby 0.999... = 1

>>8503487
who said anything about natural numbers, stop baitin

>>8502978
>arithmetic tricks
L0Lno fgt pls

Here is a proof by contradiction. First, let's prove that if a < b, then a < (a+b)/2.
Assume that a >= (a+b)/2
2a >= a+b
a >= b, which is false. Thus, if a<b, then a<(a+b)/2.

Now, assume that 0.999...<1.
Using our previous fact, we know that
0.999... < (0.999...+1)/2
Using ordinary division, we find that
0.999...<1.999.../2
0.999... < 0.999...

This is a contradiction. Thus 0.999...>=1. Because it cannot be greater than 1, it must be equal to one.

1=1

>>8503514
And 1/infinity = 0. Your point?

>>8502978

Denote a = 0,99999999....

Hence a > 0, and |a| = |1| then a =1.