/sci/ - Science & Math

[Anonymous No.8502978]

Hello /pol/, I've been looking at the fact that 0.999... = 1 and the people who deny it and I think the problem is the proofs themselves.

They are really weird. They use arithmetic tricks that don't look right. So I tried to come up with a proof by contradiction which would use inequalities and show why 0.999... cannot be smaller nor bigger than 1... but I haven't been able to do it.

Maybe someone here who knows more about the structure of rational numbers could try to come up with such a proof?

Thanks in advance, /x/.

[Anonymous No.8503000]

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

[Anonymous No.8503009]

>>8503000
It's obvious bait you tard

[Anonymous No.8503014]

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

[Anonymous No.8503081]

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

[Anonymous No.8503241]

>>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...].

[Anonymous No.8503472]

>>8502978
Food for discussion

[Anonymous No.8503487]

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

[Anonymous No.8503499]

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

[Anonymous No.8503514]

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

[Anonymous No.8503652]

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.

[Anonymous No.8503705]

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

[Anonymous No.8503831]

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

[Anonymous No.8503888]

>>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.

[Anonymous No.8504018]

>>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

[Anonymous No.8504019]

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

[Anonymous No.8504023]

>>8502978
>arithmetic tricks
L0Lno fgt pls

[Anonymous No.8504336]

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.

[Anonymous No.8504340]

1=1

[Anonymous No.8504373]

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

[Anonymous No.8504375]

>>8502978

Denote a = 0,99999999....

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