/sci/ - Science & Math

>>5659152
1+1 = 2

x = 0.999... | equation times 10
10x = 9.999... | equation - x (x = 0.999...)
10x - x = 9.999 - 0.999...
9x = 9 | divide equation by 9
x = 1

>>5659167
I get that, but how can 0.999 = 1?

>>5659152
That would be a contradiction. That means something in the process is wrong.
In the third step 10x-x != 9x.
There are an infinite number of nines so subtraction would never terminate.
This could be made true by saying 9x is approximately 9 which would make x approximately 1.
Which IS true.

>>5659177
Here how

you can also think of it as 1-0.999... (which is continuous by the way in case you didn't realize) and 1-0.999... would give you 0.000... where there is always another zero, making the answer zero. 1-0=1 therefore 0.999...=1

It is a proof that .9999999.... (infinite 9's) = 1.

It is correct and not a troll. There is a Wikipedia page devoted to this.

>>5659152
Lets stop this shit.
It's not fun. It's not interesting. It's not intellectual.
It's been done over and over and we achieve nothing by arguing aboutit.

What I'm confused about is: how do you not understand. Or is it a trolling thing?

0.9.... is different from 0.9 followed by an arbitrary amount of 9's.

Its confusing to look at, but its no different then comprehending how 3 is not the same as 5.

>>5659187
I agree with this.
0.999... probably isn't considered a rational number because it just keeps going. But if there was a way to express this as a fraction then you could try proving it.

>>5659202
>using geometric series to shut .999... =/= 1 retards up

why didn't i think of this

>>5659251
Geometry series converging is depended on the reals having the least upper bound property
least upper bound property in the decimal construction is proved BY ASSUMING 1=.999...

Therefore the proof is assume 1=.999..., therefore 1=.999... which is not a real proof.

You can't prove 1=.9999.... if your real numbers systems is the infinite decimal construction.

>>5659251
because somebody stupid enough not to believe .999...=1 would probably have no clue what they were looking at

here you go

>>5659216
That's assuming that 0.000... = 0

Which it does. It is an infinite series, so there is never a 1 at the end like you might think just more zeros.

>>5659337
why would 0.0... = 0
what...

>>5659152

x = 0.999.....

*algebra*

x = 1

therefore, 0.999..=1

OR

there is no such number as 0.999....

>>5659222
Don't fight it. It's never going away. This board is a way to waste time so just continue wasting.

>>5659216
I don't think a binary operation can be considered continuous.

>>5659202
The green box is not induction.

.9 is that one? Nope.
.99 is that one? Nope.
.999 is that one? Nope.
Fuck your 'proof.'

if .9999...=1 then
.99999...8=.9999...=1
so eventually 0=1

>>5659363
Actually, it can; this just means that it's continuous in the product topology. Here's a simple example: addition is a function <span class="math">+: \mathbb{R} \times \mathbb{R} = \mathbb{R}^2 \to \mathbb{R}[/spoiler], and graphing f(x, y) = x + y in Euclidean 3-space, you can easily see that addition is a continuous binary operation.

ITT idiots use estimates to "prove" .9999=1

>>5659648

Wow. I used to think .999... = 1 but your post has shown me the light.

>ask junior math teacher if .999=1
No
>entire class laughs at me and calls me stupid
>accept it and go back to finishing all of the class work within 10 minutes
Looking back, I wish I had taken a higher level class

>>5659648

lol strong retard/10

the 0.9999... never ends. The 9s continue indefinitely. Yet to have a number like 0.9...98 implies that 8 is the last digit, which in turn implies it is of finite length and therefore =/= 0.99...999

Do you even math?! So many idiots on /sci/ today...

>>5659674

ITT people who don't understand mathematics. No "estimates" have been used you cunt face.

look at this phaggot >>5659202

>>5659712
This shit wouldn't happen where I live.

ITT: trolls trolling.

I have no Idea what I'm doing but...
Assume
x = 0.9990

10x = 9.990
10x - x = 8.991 <<-- Would need the extra 0.009
9x = 8.991
8.991/9 = x
0.999 = x

So do you need and extra 0.000...9 for this to be true 0.999... =1?
0.999...+000...9

Infinity plus something = infinity ?

>These threads every time

It's abstract nonsense, with no real-world value. Like most of math.

>>5659788
>Putting things on the end of nonterminating sequences
>ISHYGDDT

>>5659802
Yup, let's not bother understanding the reals
Get back to me when you have solutions to x^2=2 or <span class="math"> e^ix =i [/spoiler] and if you think they're useless then fine, return to your cave.

x = 0.999....
10*0.999... -0.999... = 9.999 - 0.999
=0
x =0.999...
you guys kept the fucking x after you input it's value
more proof
we have the value of x, (x=0.999), proof^
if the equation op gave us was algebraically correct it would look like this
we already know that x=0.999, we cant find it again because we know it, if we were to find the value of the unknown x in op's equation it is a different value, we can't use x twice.
so we need to change the unknown x into a new character, because we know x

X=known (0.999)
Y=unknown
10xy-x=9.999-0.999
9.999y-0.999=9.999-0.999 (correct version of op's equation)
9y=9
y=1
x=0.999

x and y are different values
x =/= y
0.999 =/= 1

Whoever showed you that pic was tricking you. The second line implicitly uses circular reasoning.

If .999... does not equal 1, then there is a number between .999... and 1. Try to find one

>>5659933
For two distinct reals x and y the number (x+y)/2 is between them.

>>5659933
.9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999
If not that, add another nine. It never terminates fools! Why u no infinite?

chekmate athetits

>>5659936

is that 1/(2*AlephNull)?

Assume for the sake of contradiction that 1 =/= 0.999...

Then 1-0.999... Must equal the smallest non-zero real number. However (1-0.999...)^2 must be smaller yet still non-zero, this is a contradiction as 1-0.999... Must already be the smallest non-zero real number, hence 0.999... Is not distinct from 1, they are therefore equal.

>>5660026
>Then 1-0.999... Must equal the smallest non-zero real number.
Why?

>>5660045
Find me a number between 0.999... And 1. Remembering there is an infinite number of 9's, so you can't just tack something on at the end.

>>5660047
That is not a proof of the statement.

>>5660047

There isnt an infinite amount. 'infinite amount' makes no sense. There can only be a finite amount of 9s, and you'll always be able to general a new number.

>>5660047
There are only countable many numbers that can be represented in any way but there are uncountable many number between any two different reals. Just because you can't write a number doesn't mean it doesn't exist.

>>5660056
Not quite true. There are also a countably infinite number of rationals between any two reals. If you can't write a rational between two reals, it must be because the two reals are uncomputable. But then their digits certainly couldn't be given. But their digits are given: one is given as "1" and the other as "<span class="math">0.\overline{9}[/spoiler]".

>>5660056
>>5660058

The 'real numbers' are a mistake. The whole concept is nonsense. Numbers dont exist until they are calculated. If a number can be calculated at all, it is rational.

>>5660065
Between any two rationals is another rational, trivially. And clearly between 1 and 0.999... would be a rational, if these were distinct numbers.

>>5660069

>Between any two rationals is another rational

Not necessarily. I have no doubt a rational could be constructed between any two rationals, but that doesnt mean there is one before we bother to think or calculate it.

>>5660072
What if someone calculated it once, but never wrote it down?

Thats the entire point of this whole thing. The two numbers are not distinct from each other

>>5660065

So the square root of two has no value

>>5660086

Well let be careful. Numbers dont really 'exist' at all. If someone writes down then the numerals on paper exist. To calculate something we follow rules. When we calculate something we can remember how the calculation goes. Perhaps two people might calculate to the same thing the same way. Thats all there really is to say I guess. Two people can invent the same thing independently.

I dont know, what are you asking?

>>5660101

No, its just misunderstood. Square rooting is a kind of calculation. We can do a calculation to a finite extent. We can derive a 'value' that serves our practical need for that number.

>>5660123
I share your sympathies but just so you're aware I've had two three-day bans for being a finitist. Our mod apparently thinks it is trolling.

>>5660123
>No, its just misunderstood. Square rooting is a kind of calculation. We can do a calculation to a finite extent. We can derive a 'value' that serves our practical need for that number.
But we often define the square root of two as the least upper bound of a set, so if square root of two is a number the rest of the reals defined in a similar way must be numbers.

>>5660140

> define the square root of two as the least upper bound of a set

Well then define it differently. People have been worrying about the square root of 2 for much longer than set theory. Dont ask, 'how can this be defined?' If you ask that you are free to define it any way you want. Ask instead what people are doing, or thinking when they work with the square root of 2. The only answer is calculating.

Indeed calculation is the only way to get any actual answer to the square root of 2.

>>5660155
Calculating will never give a square root of 2 any more than it could give you the imaginary unit as the square root of -1. The only way to get a square root to 2 is to assert its existence by extending rationals to pairs <span class="math">a+b\sqrt{2}[/spoiler] and defining the typical operations on them.

<span class="math">S = \sum_{n=1}^{\infty}\frac{9}{10^n}=9(\frac{1}{10}+\frac{1}{100}+\frac{1}{1000}+...+\frac{1}{10^{n-1}}+\frac{1}{10^n})\\
S/10 = \sum_{n=2}^{\infty}\frac{9}{10^n}=9(\frac{1}{100}+\frac{1}{1000}+...+\frac{1}{10^{n-1}}+\frac{1}{10^n})\\
S-\frac{S}{10}=\frac{9}{10}\\
\frac{9S}{10}=\frac{9}{10}\\
S=1[/spoiler]

<span class="math">S = \sum_{n=1}^{\infty}\frac{9}{10^n}=9(\frac{1}{10}+\frac{1}{100}+\frac{1}{1000}+...+\frac{1}{10^{n-1}}+\frac{1}{10^n})[/spoiler]
<span class="math">S/10 = \sum_{n=2}^{\infty}\frac{9}{10^n}=9(\frac{1}{100}+\frac{1}{1000}+...+\frac{1}{10^{n-1}}+\frac{1}{10^n})[/spoiler]
<span class="math">S-\frac{S}{10}=\frac{9}{10}[/spoiler]
<span class="math">S=1[/spoiler]

<span class="math">S = \sum_{n=1}^{\infty}\frac{9}{10^n}=9(\frac{1}{10}+\frac{1}{100}+\frac{1}{1000}+...+\frac{1}{10^{n-1}}+\frac{1}{10^n})\\
S/10 = \sum_{n=2}^{\infty}\frac{9}{10^n}=9(\frac{1}{100}+\frac{1}{1000}+...+\frac{1}{10^{n-1}}+\frac{1}{10^n})\\
S-\frac{S}{10}=\frac{9}{10}\\
\frac{9S}{10}=\frac{9}{10}\\
S=1[/spoiler]

<div class="math">S = \sum_{n=1}^{\infty}\frac{9}{10^n}=9(\frac{1}{10}+\frac{1}{100}+\frac{1}{1000}+...+\frac{1}{10^{n-1}}+\frac{1}{10^n})
<div class="math">S/10 = \sum_{n=2}^{\infty}\frac{9}{10^n}=9(\frac{1}{100}+\frac{1}{1000}+...+\frac{1}{10^{n-1}}+\frac{1}{10^n})
S-\frac{S}{10}=\frac{9}{10}
\frac{9S}{10}=\frac{9}{10}
S=1</div></div>

<span class="math">S = \sum_{n=1}^{\infty}\frac{9}{10^n}=9(\frac{1}{10}+\frac{1}{100}+\frac{1}{1000}+...+\frac{1}{10^{n-1}}+\frac{1}{10^n})
S/10 = \sum_{n=2}^{\infty}\frac{9}{10^n}=9(\frac{1}{100}+\frac{1}{1000}+...+\frac{1}{10^{n-1}}+\frac{1}{10^n})
S-\frac{S}{10}=\frac{9}{10}
\frac{9S}{10}=\frac{9}{10}
S=1[/spoiler]

>>5660047
"Prove me wrong" is not how math works, you drooling retard. Either you provide a proof on your own or you fuck off.

<span class="math">S=\sum_{n=1}^{\infty}\frac{9}{10^n}=9(\frac{1}{10}+\frac{1}{100}+\frac{1}{1000}+...+\frac{1}{10^{n-1}}+\frac{1}{10^n})\\
S/10=\sum_{n=2}^{\infty}\frac{9}{10^n}=9(\frac{1}{100}+\frac{1}{1000}+...+\frac{1}{10^{n-1}}+\frac{1}{10^n})\\
S-\frac{S}{10}=\frac{9}{10}\\
\frac{9S}{10}=\frac{9}{10}\\
S=1[/spoiler]

<div class="math">S=\sum_{n=1}^{\infty}\frac{9}{10^n}=9(\frac{1}{10}+\frac{1}{100}+\frac{1}{1000}+...+\frac{1}{10^{n-1}}+\frac{1}{10^n})</div>
<div class="math">S/10=\sum_{n=2}^{\infty}\frac{9}{10^n}=9(\frac{1}{100}+\frac{1}{1000}+...+\frac{1}{10^{n-1}}+\frac{1}{10^n}</div>
<div class="math">S-\frac{S}{10}=\frac{9}{10}</div>
<div class="math">\frac{9S}{10}=\frac{9}{10}</div>
<div class="math">S=1</div>

>>5660263
<div class="math">S=\sum_{n=1}^{\infty}\frac{9}{10^n} =9(\frac{1}{10}+\frac{1}{100}+\frac {1}{1000}+...+\frac{1}{10^{n-1}}+\f rac{1}{10^n})</div>
random spaces ffs

>>5660263
<div class="math">S=\sum_{n=1}^{\infty}\frac{9}{10^n} =9(\frac{1}{10}+\frac{1}{100}+\frac {1}{1000}+...+\frac{1}{10^{n-1}}+\frac{1}{10^n})</div>

>>5660167

There is no complete number called 'the square root of 2', square rooting is nothing more than a step by step process of producing numbers, like long division.

Its not a question of whether or not calculating produces the square root of two, the square root of two IS a calculation, which produces a number. The output of this calculation is different from the calculation itself.

>>5660314
> square rooting is nothing more than a step by step process of producing numbers, like long division
Please show me this process on two numbers:
a) 2
b) -1

>>5660319
>square root
>of a negative number
troll harder 0/10

>>5659648
How would an infinite chain of digits have a final digit?

>>5659712
Did you say 0.9 recurring or just an arbitrary number of 9s?

>>5660390
Where does it have a final digit?

>>5660394
It wouldn't because an infinite sequence of numbers would have no final digit, that's my point

>>5660401
Then why did you start talking about final digits?

>>5660407
Because the guy I responded to was using them in his "argument"

>>5660053
>>5660054
>>5660056
>>5660253
Guys, he's agreeing that 0.999... = 1

>>5660416
I know, but obviously he has no idea WHY they are equal. He religiously spouts things he was told to remember without knowing anything about mathematical proofs.

It doesn't matter what kind of math you use, as long as it is consistent....

oh wait, Kurt Godel just reminded me..

okay... Finitist math is sure to be consistent...
unless of course some contructivist insists we actually have to have the proof of all the true statements.....

FFFFUUUUUUUUUU


If i can't count it on my fingers and toes... It doesn't real!!!!

>>5660319

Here is how I would calculate the square root of two. I can only do so to a finite extent (as can anyone), but you can see I get pretty close with only a few iterations.

I cant square root a negative number. I think the concept of i was introduced explicitly for the purpose of making this possible. As you can see, we can easily introduce a new rule if we need to to make a calculation work when it otherwise would not.

Our ability to constantly reinvent math to make it work gives an illusion that reality is fundamentally mathematical. If anything its the other way around, mathematics is intentionally designed to apply to the world.

>>5659177
>0.999 = 1
it isn't
0.999...9 is

>>5660419
>>5660419
All he basically is saying is:
1-0.999...<e for all e>0 therefore 1-0.999...=0 This is not so sloppy. Sure the first statement isn't "proved" but it is really obvious (intuitively) it is true. The way to prove this is to represent 0.999... as an infinite sum (this is how infinite decimal expansions are DEFINED) and then sum using the standard formula. But some people do not accept this, because they don't understand limits, infinite sums, the fact that infinitesimals don't exist in our number system and that defining things by their decimal expansion is confusing when we can represent it otherwise. This pseudoproof shows you intuitively that the statement is true, and is more convincing to people who don't understand math than a rigorous one.

>>5660452
you should look at continued fractions of square roots, as they are finite expressions in that they eventually start repeating. For instance, <span class="math">\sqrt{2}[/spoiler] could be written as [1; ,2] -> [1 2 2 2 ...] -> 1+1/(2+ 1/(2 + ...))

>>5660500
ignore the semicolon, or ignore the comma, only one is needed to indicate the start of the repeating portion

>>5660505

Oh interesting. Cool.

>>5660583
You run into a problem if you starting using "computable reals" like this, though, as you cannot take a computable real like <span class="math">\sqrt{2}[/spoiler] and square it to get 2, or subtract it from itself and get 0, etc. Since the inputs to the equation never finish, no algorithm can generally decide what the answer will be, and may not produce any output whatsoever.

0.333... = 1/3
(1/3) * 3 = 1
0.999... = 1

>>5660655