Discuss

Discuss.

It is.

/thread

it's not 0 either

Technically, it is

It is as close to one as you can possibly get.

no it is exactly 1

Yup.
NEXT

>It is as close to one as you can possibly get without being one.
One is closer to one. So FTFY.

1/3 =.3 repeating
.3 repeating x3 = .9 repeating
.9 repeating = 1

There's nothing to discuss here except "you're wrong."

x = 0.999...
10x = 9.999...
10x - x = 9.999... - 0.999...
9x = 9
x = 1

0.999... = 1 now fuck off

X =.999
10x = 9.999
10x = 9 + x
9x = 9
X = 1
Science yo.

>10x - x = 9.999
>9x = 9
nigga wut?
it would be 9x = 9.999...
dumbfuck

If you want anyone to take you seriously, you have to define what "1" is, then build up to defining the real numbers and their decimal representations. And if you do all that (which I have), then you will figure out that .999...=1.

False citation.

X= .99
10x= 9.9
10x-x=9.9-.9
9x=9
X=1

So.9 =1 too?

>Purposely misquoting his post

wowee

Well, it can be said at least that the geometric progression:
9/10 + 9/100 + 9/1000 + ...

Converges to 1

must not let flat earthers take over the maths too!

>9.9-0.99=9

1 is 1.0. Not more and not less. So 0,99999.... is not 1. If it would be 1 than 99 would be 100, which it is obviously not

Here's the proof:
The sum of a infinite series of the form
1 + x + x^2 + x^3 + ...
equals 1/(1-x) as long as |x|

I'm guessing you're one of those Philosophy majors...or a political science major.

You know; the studies that don't matter.

.9 repeating is a geometric series equivalent to (sigma) =.9 * (1/10)^n

In a geometric series with R > 1, the sum can be found using the formula A/(1-r). In this case, a =.9. 1 - 0.1 = 0.9, which makes the summation equation a/(1-r) equal to 0.9/0.9,

Therefore, according to calculus, 0.99 repeating = 1.

In any mathematical system where 0.999... =1 means that 0.999...98=.999...97=0.999.96=0.999...95, etc, etc., ad infinitum

The entire number system breaks down because all numbers exactly equal each other.

Proof?
>no proof.

I wrote down one, just scroll up

If you think I'm going to spend the time explaining how an infinite geometric series works over Sup Forums, you must be clinically retarded.

If you're legitimately curious about it and want to know more, there are a shitload of good websites you can check out. I think there's even a khan academy video about it. It's actually pretty interesting if you take the time to learn.

There is no last number.

0.999... is an increasing number that gets closer and closer to, but never reaches, 1.

.099... is a real number that is the same as the real number 1, since it is greater than every rational number strictly less than 1. (This follows from the definition of a dedekind cut.) Another way to put this is, there is no x a subset of 1 such that x is not also a subset of .999... therefore 1 is not strictly greater than it.

However, it is not equal to the natural number 1, which is just {{}}; this would violate extensionality.

OK, I'll bite.
Now prove that the sum of an infinite series of the form 1+x+x^2+x^3+... equals 1/(1-x) as long as |x|>1.

>0.099... is greater than every rational number less than 1
It's not even greater than 0.5 kek

try again

thoumst mom is a gay

Missed a decimal place on mobile. eat a log of shit out of my ass.

Ok gimme a minute

0.99999999 is infinitely far away from 1.0

>makes basic errors in his attempt at a troll post
>immediately rages when called out

try harder.

Literally it is not it holds its own numerical value but in a everyday sense, it's 1.

And then prove that it is not also the case between 0.999....99 and .0999...98.

Because both are of "equal distance" on any number line. Meaning if 0.999... = 1 then all other numbers equally distant from each other ALSO equal each other. Which then collapses the entire numbering system as all numbers equal each other.
Go ahead, prove me wrong.

It's not? if 1/3 = 0.333... => 3/3 = 0.999... since 3/3=1 => 1=0.999...

It's not

That only works if you pass to limit

Look morons...
.9999...=x. Multiply by 10
9.9999...=10x subtract from each side
9= 9x divide by 9
1= x ergo
1=.9999... By definition of terms.

58 yr old Boomer here. You fuckers didn't learn shit in school, got participation trophies, are addicted to technology, have short attention spans. Just kill yourselves already.

You should learn the definition of a decimal representation before posting idiotic shit like this.

>Converges

Ok you did succeed in rustling me slightly, but my post is not a troll post, it's grounded in accepted principles of classical set theory

DOES not = x.

Otherwise move along nothing to see here.

Here:
Suppose that x is a number that could be either complex or real

Consider the finite series

1 + x + x^2 + x^3 + ... + x^n

Store the entire sum into a number, namely s_n:

s_n = 1 + x + x^2 + x^3 + ... + x^n (*)

Multiply both sides by the same x:

x(s_n) = x + x^2 + x^3 + ... + x^n + x^(n+1) (**)

Subtract (*) from (**):

(1-x)s_n = 1 + x + x^2 + x^3 + ... + x^n - (x + x^2 + x^3 + ... + x^n + x^(n+1))

A lot of terms will cancel out in this sum. In fact, only 1 and x^(n+1) will survive because the appear in either (*) or (**) once. Hence:

(1-x)s_n = 1 - x^(n+1)

Now, since dividing by zero makes no sense, we have to admit that x is not equal to 1. If that is the case then:

s_n = ( 1 - x^(n-1) ) / (1-x)

Now, that is the sum of a finite geometric progression. The sum of a infinite series however, is defined as the LIMIT of such sum as n tends to infinity. In this case the limit is:

lim s_n = lim(1 - x^(n-1)) / lim(1-x)

I used the fact that the limit of a quotient is the quotient of limits as long as the denominator isn't zero

Now,

lim s_n = (1 - lim(x^(n+1))) / (1-x) (***)

(Subtraction of a limit is the limit of a subtraction and 1-x is continuous on x not being 1)

From (***), you can already see that the limit will only exist if x^(n+1) gets smaller as n gets big, which only happens for |x|

you would have to subtract 9x not 9 in order to get just x.

Fucking get it right you degenerates.

>.9999999....
white people
>1
niggers

Bertrand Russell (of Russell's teapot fame) disproved Set Theory over a century ago.

I said that already

It is.

Explanation:
Let's imagine what 9.9 really is. It's 10-1/10. What is 9.99? It's 10-1/100. So then what is 9.999repeating? It's 10-1/∞ or in other words 10-0, which equals 10.

So what you're all saying is that 1 - 0.999... = 0.
Therefore n - 0.999... = n.
Thus all numbers equal the two closest numbers themselves. And in an infinite progression, all numbers are equal to all other numbers. All because .999... = 1

Mathematically, yes. Effectively, no.

subtract what? 0.9999? 10-0.9999 is 9.0001 not 9? Also how can you subtract from 10x if there's still x?

>1/∞=0
multiply both sides by ∞ and get 1=0

Nope. 1/∞ is a tiny amount but not 0.

Literally arguing over 0.00001

OK let me make this a bit more logical
1 - .999... = 0
Therefore
1 + (1-.999...) = 1
and
1 - (1-.999...) = 1

Which means
n = n +/- (1-.999...) = n +/- 0
>+/- in this case means both plus AND minus.

Now turn that into an infinite progression and you get all numbers equaling all other numbers.

Not sure if retarded or trolling

This is not really a coherent mathematical statement, but for the benefit of others reading, I'll say that Russell made a great contribution to set theory by describing paradoxical sets that result from unrestricted comprehension.

no they are both exactly 1

check mathologer and he will explain it to you simply.

if you can't grasp it after watching his video then you never will and should not engage in discussions on the matter.

Doesn't matter what formulas you throw at it, .9999 repeating is not equal to 1. Because by definition .999 repeating is not 1, it is .999 repeating.

Sup Forums in a sentence

But actually, we're arguing over .000... the whole point of the ... is that there is no last digit.

effectively and mathematically yes.
draw circle
draw 3 lines from center dividing circle into 3 equal pieces. each piece is 33.3repeating%
erase 1 line
1 piece is 66.repeating % 1 piece is 33.3repeating %
remove last line and 99.9repeaing is left is the remaining circle somehow smaller?
no
99.9repeating% = 100%

But that also makes all numbers exactly all other numbers.

And all you've done by claiming 1=0.999... is constructed a similar nonsensical statement to that which Russell reveled in.

Congratulations. You've dug yourself into a corner that the genius Bertrand Russell constructed over a century ago.

> 1 - 0.999... = 0
correct

>Therefore n - 0.999... = n
incorrect

n - 0.999... = n - 1

The last digit is 1, it just comes after an infinite number of 0s.

Number LINE, not number circle. What you are saying is +infinity = -infinity because they occupy the same point on a circle.

eh?

flat earth = ignorant misunderstanding of blatant evidence and basic physics

.9r=1 = advanced math

how are these things related?

in reality 0.99999 isnt one but effectively it is

X= .99
10x= 9.9
10x-x=9.9-.99
9x=8.91
X=0.99

FTFY

0.9... is literally equal to one

google it faggot

kek

>FTFY
fuck off back to plebbit

1/3=0.33..
x3
3/3=0.99..=1

F(x)= 1/(x-1)

F(.999....) = -1,000,000,000...
F(1) = undefined

If they are not equal what is their difference? The real numbers (can be) defined as an equivalence class of limits of sequences of rational numbers. The elementwise difference between the sequence (1,1,1,1...) and (0, 0.9, 0.99, 0.999,...) converges to 0, so they are equivalent, so they are the same number.

oh shit your right! 2=46 wtf, serious, check it out.

It's not my fault you don't understand set theory and I'm not going to try and explain it to you. All you're doing is saying "Bertrand Russell" but not actually doing any math.

>Oh no, he's right, I should make fun of him.

Sir, you have injured me. I am mortally wounded. Ow! How will I recover? Woe!

of course it is you retard

>technically
The most autistic kind of -ically.

Nigga I ain't even the guy you responded to, now stop acting like an autist

This is a correct explanation, thank you for contributing

Are you claiming that -1,000,000,000... is a real number?

it is because the god said so

>converges
But are never equal.

I understand it better than you do, as you managed to make statements that were disproven over a century ago.

It is not real but it is defined

1/0 cannot be defined, even as infinity

At no matter what place you lead the 9s to, it will always be smaller, than 1.
And only get to equal 1 when you round up

The "sick burns" continue. I may be an autist, but you are truly special needs.

Real numbers are limits. Convergence is the same thing as equality

Whoops, meant to say
If 1 - .999... = 0
And n - 0 = n
then n - (1-.999...) = n - 0 = n
Thus, all numbers are equal to themselves minus (1-.999...) AND all numbers are ALSO equal to themselves PLUS (1-.999...) Thus, all numbers are equal to themselves AND their nearest decimal number themselves.

If this is true of all numbers, then all numbers are equal to all other numbers.

Up is down. Left is right.

It is. The most important part is the "...". Without it, then it is not.

What do you mean when you say defined? What type of number system are you working in? If we were working with the projective reals then 1/0 = infinity, no ifs ands or buts, so the question matters.