Please, please tell me no one on Veeky Forums actually fell for the "0.999999... = 1" meme

here's the definitive proof op. unless you think the meme spread to wikipedia...

en.wikipedia.org/wiki/0.999...

>he thinks Wikipedia is an objective source
>he thinks Wikipedia isn't pushing an agenda
How fucking stupid can one person get?

>is 1.111... equal to 1.2

no, because 1.12 comes between the two. there are an infinite number of numbers between 1.1111... and 2.

Assume .9999... and 1 are two different numbers op. what's 1 - .9999... ?

>Is 1.111... equal to 1.2?

No because 1.1112 is between those two numbers. And infinitely many more are aswell.

0.000...001

You're comparing two universes
>an inifinite number of paper slips with 0 written on them plus one paper slip with 1 written on it
>an inifitine number of paper slips with 9 written on then plus one paper slip with 0 written on it
They are not the same because they're written differently.

[math]\mathbb{R}-{0}[/math] is a group under multiplication.

[math]0.999... \times 1 = 0.999...[/math]

[math]0.999... \times 0.999... = 0.999...[/math]

Substituting,

[math]0.999... \times 1 = 0.999... \times 0.999...[/math]

Left cancelling,

[math]0.999... = 1[/math]

>no, because 1.12 comes between the two. there are an infinite number of numbers between 1.1111... and 2.
Yeah, I just noticed I made a typo--meant 1.12.
>Assume .9999... and 1 are two different numbers op. what's 1 - .9999... ?
0.00000...1? Is that supposed to be a trick question? Because the answer is very clear. It's certainly not 0, it's an asymptote forever approaching 0. Nowhere near the same thing.

What is the pentium bug for $100, Alex.

>.99999....≠1

i'm wondering if anyone has actually argued this or seen someone argue this in a university math class. if you guys have any stories, please share.

1 - 0.999... = 0.00...1
1 - 1 = 0
0.00...1 != 0

>Yeah, I just noticed I made a typo--meant 1.12.

no i made the typo. there are an infinite number of numbers between 1.1111... and 1.12. there are no numbers between .9999... and 1.

>0.00000...1? Is that supposed to be a trick question? Because the answer is very clear. It's certainly not 0, it's an asymptote forever approaching 0. Nowhere near the same thing.

asymptotes DO hit 0 after an infinite amount of time. That's an extremely important thing. calculus wouldn't work otherwise.

the answer to the question is 1 - .9999... = 0. you can't have .00000...1. how can there be an infinite number of 0s and THEN 1? if you're putting a 1 at the end of your 0s then you don't have infinite 0s.

here's a cambridge mathematician explaining it. has the meme spread to research mathematicians now?

youtube.com/watch?v=G_gUE74YVos

If you admit that there is a number 1 at the """end""" of the number 0.000...1, then the number 0.999... does not expand forever.
Go learn about infinity again.

We have this paradox because real numbers are imperfect. It's just a quantifying approximation of continuous line

But it's a very easy proof and you can Google it.

op is either a troll or he's just started high school maths and thinks his limited knowledge of quadratic equations makes him knowledgable in maths.

[math](0.999...)^\infty = 0[/math]

[math]1^\infty = 1[/math]

retards

This is a notational problem. Notational problems, like all problems of language, are resolved by appeal to consensus.

When most mathematicians talk about "0.999 repeating", they are referring to a particular construct, a particular definition, where it is in fact equal to 1.

Further, one of the properties of the Real Numbers is that two real numbers are inequal if and only if there is a third Real number in between them. By appealing to informal arguments, "0.999 repeating" names a Real number, and there is no Real number between "0.999 repeating" and 1, and therefore they name the same Real number.

>how can there be an infinite number of 0s and THEN 1?
[math]1+x+x^{2}+... = 1/(1-x)[/math] right retard? how can you have an infinite series equally a single term?

1/3 = .3333333333333333333....
So 3 * .333333333333333333...
=1

>1/3 = .3333333333333333333....
no, 1/3 = .333333... plus the 0.00...00000 and 1/3 at the end.

Welcome to calculus.

[math] \displaystyle
1 = \frac{3}{3} = 3 \cdot \frac{1}{3} = 3 \cdot 0. \overline{3} = 0. \overline{9}
[/math]

1/3 =/= 0.3333...

yes it does yuo fucknard

but it literally is

The only story I have was intro to analysis where the lecturer seemed kind of disappointed nobody objected to his claim that (0,1) had no maximum.

1/3 = 0.3333... + 0.000...000 and 1/3, we just leave off the last part because it's really small and the first part is accurate enough

>the 0.00...00000 and 1/3 at the end.
There is no such Real number. If you want to talk about some new kind of please, please give your definitions of terms.

You don't understand how infinity works

i'm not talking about kinds of numbers, i'm just talking about plain old numbers, the kind we all use, not some arbitrary wishy-washy alternative universe math from the ivory towers

Ok. We're interested in formal analysis, not "it makes sense to me". Might I suggest /x/ or /b/ ? Maybe Veeky Forums ?

i am doing a serious analysis, go to Veeky Forums if you want a safe space from the hard truths of math.

>Comparing infinite sums with natural numbers

No

0.99999... = 1 is true only in the same way 1/0 = infinite

You are not using math correctly

Go home

0.9999... = x
10x = 9.9999... / -x
9x = 9 /:9
x = 1
ez

this, there is no infinity in the really existing numbers, you can't take "limits" without the contradictory framework of calculus on top

1 - 0.999... = 0.000...1

you can't just choose to apply infinity when it suits you faggots

Back at you

>people seriously responded to this thread without saging
?

wtf is /:

If 0.00...001 is a nonzero real number what is its reciprocal?

0.999... literally means infinite 9's... You can't just say something is not infinite when it literally is. Pro tip, decimal expansions are not unique. It is something you have to accept in real analysis.

:/

Do you even realize the argument you have right there proves 0.999... = 1? Write 0.999... as a summation of fractions and literally plug it into that equation right there.

So you're going to tell me straight faced that decimal expansions are unique? Give me the proof for that and then you'll have a basis for dismantling very well established mathematics

>0.000...1
>An infinite sequence with something after it.
Jesus. Are you even trying?

In any case, this seems like it's obvious shit being over-thought.
The sequence 0.9, 0.99, 0.999, 0.9999, ... obviously converges to 1 as the number of nines goes to infinity. 0.xxx... is notation that signifies the number of x's goes to infinity.
Problem solved: 0.999... = 1

>Do you even realize the argument you have right there proves 0.999... = 1? Write 0.999... as a summation of fractions and literally plug it into that equation right there.
geomtric series doesn't work like that

I honestly can't tell if people agreeing with OP are trolling.

don't you think trolls would target more serious people and issues? and not brilliant Veeky Forumsentists who, since they're so quick-minded, would pick up on trolling very quickly?

0.9/1 = 0.1
0.99/1 = 0.01
...
0.999.../1 = 0.000...0001

i don't get what's so hard about this, it's like you're all stupid and can't do math

prove it

Infinity is a concept. We can arbitrarily assign whatever properties we want to the concept of infinity. The property P(x) together with the inference rules of mathematical logic dictates that 0.999... = 1. That is enough for most mathematicians.

We do not know if infinity as such is metaphysical possible, but it does not really matter. Does the concept of infinity outside of mathematics have the property P(x)? Who cares... in mathematics it does (because we have we have defined it that way) and most mathematicians are in agreement that P(x).

1/3 is a fraction, 0.3333... is a decimel. There's no isomorphism between rationals and reals... so therefore there's no way to map. it makes sense because with decimels everything is counting in 10s, but you can count in 3's with fractions. 10/3 is undefined in decimal

>There's no isomorphism between rationals and reals... so therefore there's no way to map.
>all morphisms are isomorphisms
Excellent argument. Go back to school, user.

X=0.999
X10=9.999
X10=9+0.999
X10=9+x
X9=9
X=1
Right?

here, let's do it informally OP.
suppose we have two numbers, x and y. can we agree that if there is no number z such that x < z < y, then x = y? okay, good.
can you find a number inbetween 0.9999... and 1.000...? no? then they are equal.

0.999 * 10 = 9.990 you dumb fuck

that's not what i said.

It is an assumption made to move forward with your argument. Maps can be injective but not surjective. Do you have something useful to say?

You can map 1/3 to 0.3333.. but they aren't the same number.

uh, user...

?

t. Norman Wildberger

never thought i'd have to say this to someone who knew what the word "isomorphism" meant, but i guess you really didn't
math.stackexchange.com/questions/335560/is-1-divided-by-3-equal-to-0-333/335561

Changing the subject to a new argument without conceding to your clear mistake? I don't have time for this, feel free to educate yourself rather than putting full faith into the first idea that you consider without thoroughly testing its viability.

No one responded to this. How is this in any way a dismissible response?

Because the trolls in this thread are ignoring arguments they can't counter without making it too obvious that they are trolling.

I'm pretty sure 75% of the people in this thread already have that idiot filtered

Go to bed Norman

> 0.9/1 = 0.1
>Not 0.9
Pic related

Clearly he meant set minus on the dedekind cuts with the parameters reversed, right?

> A sequence of infinite zeros followed by a one
That's not a real number user. When the hell are you going to reach the end of an infinite sequence to put a one at the very end?
Also here's a simple proof that .9999...=1:

.9999... = 9/10+ 9/100 + ... + 9/(10^n) (n is a member of the natural numbers)

Hence, using the sum of an infinite geometric series:

.999... = (9/10)/(1 -(1/10)) = (9/10)/(9/10) = 1

I seriously don't get how hard this is to grasp. There's certainly a plethora of strange shit in the many fields of math but this isn't one of those things.

>Also here's a simple proof that .9999...=1
you cant prove that without first saying which number system you're using.

if you're using the reals, the answer is so blatantly obvious it defies belief anyone could even beg the question in the first place. All we have to do is check whether the sequences 0.9, 0.99, 0.999... and 1, 1, 1, ... are equivalent cauchy sequences of rational numbers, which they fucking obviously are.

However, there are number systems where 0.9999... is not equal to 1. The main point to note here is that to this day I have not heard of a single reason why anyone should care about these extensions of the reals. Obviously we can do calculus just fine without the need for actual infinitesimals, so why bother with these things?
So that fucktards on Veeky Forums my troll other fucktards?
literally the only reason I can think of

True. I was just implying we were in the reals with respect to the proof I gave.

I'm curious, you can't conceive infinity, what's the biggest number you can think?

>>elementaryschool

top kek m8, copied that pasta presto.

(0.9999....∞) + (0.0000...∞1) = 1

:^)

Are we getting invaded by /v/ or something?

tell me a number that fits in between 0.9... and 1 and ill accept that they're not equal

No, I agree there are infinitely many 9s in 0.999.... But when you take 1-0.999..., you get 0.000...1, where you have a 1 in the [math]\omega[/math] decimal point.

1...000, i.e. a 1 followed by infinitely many 0s to the left of the decimal point. Come on, this isn't even difficult.

How about 0.999...5?

0.000...1 is nonsensical. So a infinite expansion of zeros ends with 1? Basically, you are telling us that an infinite decimal expansion is finite. Nice contradiction.

>someone disagrees with me and claims two completely-different numbers are not equal to one another
>h-he must be in high school!
Fuck off, you walking testament to eugenics.

I understand where you're coming from, but one of those numbers is still defined as being higher than the other, regardless of the fact that it's only by a very miniscule amount. If I show you a step on a staircase and then point to the one above it and say, "There are no steps between this step and the one before it, therefore they are at the same elevation," I would be wrong. An asymptote is not the same as the number it is approaching. It will, by its very nature, never actually reach that number, but merely get closer and closer to it. To equate the two is just silly. Yes, the fraction multiplication is a neat trick, but that's all it is--a neat trick. These are still two different numbers with two different values.

>0.999...×1=0.999...×0.999...
How is that possible? Isn't .999×.999 != 1? I'm not that one guy bitching, I'm lurkin and think this is fascinating

Where the flying fuck do you get this shit from?

You're confusing a set being well-ordered with being finite. 0.000...1 has (countably) infinitely many 0s before the 1, and then a 1 afterwards. Are you disagreeing that the decimal points of a number are well-ordered, that it's possible to have an infinite well-ordered set, or that it's possible to have an well-ordered set where one element is larger than infinitely many other elements? In any case, you're wrong: these are well-known results in set theory.

Yeah and you faggots don't understand that 0.000...1 is infinite 0s FOLLOWED by a 1

If you can have infinite 9s followed by a 9 then you can have infinite 0s followed by a 1.

this is a contradiction

so, 0.999...9 != 1

QED

X = 1
1/3 of x * 3 = 0.333... * 3 = 0.999...

X = 9
1/3 of x * 3 = 3 * 3 = 9

See?

That's like saying "Tell me an integer that fits between 1 and 2 and I'll accept that they're not equal".
Why would two "consecutive" (I know the word doesn't really apply to the set of real numbers, but I'm not sure what you'd call this) real numbers have to be equal? Although there are other proofs for 0.999...=1, this one's fairly weak.

If it ends it's not infinite.

actually 0.333... * 3 = 1, not 0.999...

This 0.999... = 1 is one of the most annoying memes in math and it only works if you ignore constraints inherent to infinities.

That's not even remotely true. The interval [0,1] ends at 1, but there are undeniably infinitely many points in between (including, for example, both 0.999... and 1).