/sci/ - Science & Math

>>9390164
wtf is 0.(9)?

>>9390164
>>9390182
[math]
\displaystyle \lim_{n\to \infty} \sum_{k=1}^n \frac{9}{10^k} = 1
[/math]
btw

>>9390185
You just wrote "0.(9) = 1".

>>9390185
Not equals, only converges to one.

>>9390194
Then: "0.(9) -> 1"

>>9390185
n can not reach infinity

>>9390164
[math] \displaystyle
1 = \frac {3}{3} = 3 \cdot \frac {1}{3} = 3 \cdot 0. \bar{3} = 0. \bar{9}
[/math]

>>9390202
so not converges to one

>>9390205
the problem with this is that 3 * 0.333... =/= 0.999...

>>9390216
No, the problem is 1/3 =/= 0.333...

>>9390216
problem with the 1/3 = 0.(3)

because 1/3 ≠ 0.(3)
0.(3) -> 1/3

>>9390222
I mean 1/3 -> 0.(3) woops

>>9390221
[math]
\begin{align*}
&0.33... \\
3 & \overline{)1_0 \;\;\;\;\;} \\
& \;\;\;\underline{9} \\
& \;\;\;1_0 \\
& \;\;\;\;\; \underline{9} \\
& \;\;\;\;\; 1 \;\; etc
\end{align*}
[/math]

>>9390205

>>9390232
0.1 = 1/10^1
0.01 = 1/10^2
0.0...1 = 1/10^inf = 0

>>9390221
prove it

>>9390164
https://en.wikipedia.org/wiki/Least-upper-bound_property

>>9390232
>if 0.00......1 is not a real number
It is

>>9390164
>>9390194
>>9390202
>>9390207
https://en.wikipedia.org/wiki/0.999......
Just read an article you brainlets

>>9390245
Prove it

>>9390237
0.999...9 + 0.000...1 = 1
0.999...9 = 1 ->
-> 0.000...1 = 0
Infinitesimal = 0

>>9390245
I have already read everything and many times

>>9390224
>1/3 -> 0.(3)
so
1/3 -> 0.(3) |x3
3/3 -> 0.(9)
1 -> 0.(9) hmmm
but 0.(9) < or = 1.
so 0.(3) x3 ≠ 0.(9) or/and 1/3 <- 0.(3)

>>9390249
If 0.(9) != 1
It must be whether 0.(9) > 1 or 0.(9) < 1
But 0.(9) > 1 is wrong because integer part of 0.(9) , which is 0, is smaller than 1

Also, since both are real number there must be 1 another real number between 0.(9) and 1

So let's say 0.(9) < a < 1 and a exists
Since a < 1, a = 0.b1b2b3b4.....bn
If b1 < 9, 0.(9) <a Is wrong, so b1 = 9
If b2 < 9, 0.(9) <a is wrong, so b2 = 9
If we keep doing this, we can find bn=9
However, if bn=9, it means a = 0.(9) so 0.(9) < a doesn't make any sense

So there's no real number between 0.(9) and 1

By construction of real number (https://en.m.wikipedia.org/wiki/Construction_of_the_real_numbers)), 0.(9) = 1

>>9390246
>>9390253

>>9390252
just 0.(3) -> 1/3 ok

>>9390247
>0.999...9 + 0.000...1 = 1

wow so 1+0=1
ty mathsuperman

>>9390253

1) If 0.(9) = 1, then infinitely small = 0.
But this is absurd.

2) If 0.(0)1 does not belong to the set of real numbers, then and 0.(9) does not belong to the set of real numbers.
They simultaneously cease to belong to the set of real numbers.

>>9390260
>1) If 0.(9) = 1, then infinitely small = 0.
But this is absurd.
Prove it then

>2) If 0.(0)1 does not belong to the set of real numbers, then and 0.(9) does not belong to the set of real numbers.
They simultaneously cease to belong to the set of real numbers.

0.(0)1 is fucking real number you retard

Do you know why is sqrt(2) a real number?

>>9390260
Above all, you didn't falsify my argument; you just sperg out another autistic claiming

>>9390262
1) school-level mathematics (Wikipedia+Google please)
2) https://en.wikipedia.org/wiki/Talk:0.999......

dealing with repeating decimals isnt well defined
3*1/3 is defined
3*0.(3) isnt defined
try to avoid repeating decimals

>>9390262
>>9390260
0.(0)1 is just 1-0.(9) so

Let's say (1-(0.(9)) != (1-1)

Then (1-(0.(9)) must be bigger than (1-1) or smaller than it

So let's say (1-(0.(9)) < (1-1)
Then must be 0.(9) > 1 but we already figured out it can't be, so 0.000....1 can't be smaller than 0

What if (1-0.(9)) > (1-1)?
Then there must be one number between both
Let's say 1-(0.(9)) > a > 1-1 and a exists
Hmm.... then 0.(9) < 1-a < 1 and 1-a exists
1-a = 0.b1b2....bn « bn € {0,1,2,3,4,5,6,7,8,9} »

But we already know bn = 9 so 1-a = 0.(9)
a = (1-(0.(9)) so

1-(0.(9)) > a doesn't make sense

So 1-(0.(9)) = 1-1 -» 0.(0)1 = 0

>>9390278
Maths have been mistaken for centuries.
Is it possible?

>>9390282
I see no reason to continue to communicate with you, I'm sorry.

>>9390274
>school-level mathematics
>just look up wikipedia and google
Not an argument.

You also learn 0.999999.... = 1 in school tho

>0,000...01
oh for fuck's sake

>>9390285
Because you can't prove 0.(9) != 1.

You are just try hard edge brat teenager thinking you are somehow better than most of mathematicians.

>>9390283
well the problem with repeating decimal is they arent complete you are repeating a task over and over again because you cant complete it
you always end up with a reminder

0.999...9 + 0.000...1 = 1
0.999...9 < 0.999...9 + 0.000...1/2 < 1
0.999...9 < 1 - 0.000...1/2 < 1

>>9390303

I mean
0.999...9 + 0.000...1 = 1
0.999...9 < 0.999...9 + 0.(0)05 < 1

>>9390303
You must prove 0.(0)1 != (0.(0)1)/2 and 0.(0)1 > (0.(0)1)/2 first tho

>>9390313

0.(0)05 x 2 = 0.(0)1
proved

>>9390253
>HAND WAVING INTENSIFIES

>>9390291
>you always end up with a reminder
just reminding you are faggot

>>9390194

That's the same thing.

>>9390325
whatever cuntface

>>9390319
0.(0)05 x 2 is 0.(0)10, not 0.(0)1 you retard

"0.999... = 1" is what's keeping us from interstelar travel

>>9390332
is 0.(0)100 or 0.(0)10000 or 0.(0)1(0)
1 = 1.000000000
1.1 = 1.1000000000

>>9390332
>0 x 2 is 0, not 0 you retard
kek

>>9390356
ban

>>9390278
Of course it's well defined. You should know this from your Calculus course.

>>9390356
>>9390334
>>9390332
>>9390319
>>9390313
[eqn]0.\dot01=a=0.c_1c_2c_3...c_n (c_k \in \{0,1,2,3,4,5,6,7,8,9\})[/eqn]
[eqn]
0=0.\dot0=b=0.d_1d_2d_3...d_n (d_k \in \{0,1,2,3,4,5,6,7,8,9\})
[/eqn][eqn]
\text{if }a \ne b \text{ is true}
[/eqn][eqn]
\text{then }c_final \ne d_final\text{ is true}
[/eqn][eqn]
0=c_1=c_2=c_3=...=c_\infty=d_final\text{ but }c_final = 1,\text{ thus } c_\infty \ne c_final
[/eqn][eqn]
\text{Let's say }c_final = c_{\infty +1}
[/eqn][eqn]
\mathbb{C}=\in\{c_1,c_2,c_3,...,c_{\infty + 1}\}, \mathbb{D}=\in\{d_1,d_2,d_3,...,d_\infty\}
[/eqn][eqn]
n(\mathbb{C})=\infty +1, n(\mathbb{D})=\infty,
[/eqn][eqn]
\text{and } n(\mathbb{C}) \ne n(\mathbb{D})
[/eqn][eqn]
\text{so } \infty \ne \infty +1 [/eqn]
But it isn't(https://en.wikipedia.org/wiki/Hilbert's_paradox_of_the_Grand_Hotel)),
[eqn]
\text{thus }\infty=\infty+1[/eqn][eqn]
n(\mathbb{C})=n(\mathbb{D}), c_final=c_{\infty+1}=c_\infty
[/eqn][eqn]
\text{ but }c_\infty=0\text{ so }c_final \ne d_final\text{ is not true}[/eqn]
[eqn]\text{thus }a \ne b \text{ is also not true}[/eqn]
[eqn]\therefore 0.\dot01 = 0[/eqn]

>>9390194
that limit is equal to 1 you silly man

>>9390432
>=∈
*=

>>9390432
>delirious

>>9390432
>Get this triggered

>>9390232
By that logic
(1÷3)3=0.(9)? How? An operation should work same forward as backward

>>9390164

Let x be any number that is smaller than 1
Clearly x <= 0.(9) is true for all x, also it's clear that 0.(9) <= 1 is true.

So we have x <= 0.(9) <= 1, where x can be any number smaller than 1.

If you agree that 0.(9) is a real number, then it has to be equal to 1.

>>9390232
0.000...0, 0.999...9, 0.000...1 are not the same as 0.000..., 0.999..., 1*10^(-n). If you write a number as the former, you imply the number terminates after some infinite amount of decimal places back. By definition, a number extending to infinity NEVER terminates. Thus the proof of convergence for 0.999 = 1 holds.

We need a base 18 for decimals to et past this logic problem of infinite repeating single digits like 0.333•••
I've proposed this a couple times but just keep sharing it
01a2b3c4d5e6f7g8h9

1/3 = 0.c, where each lower letter represents and infinite single decimal. Its important to make this distinction cause 0.3••• is as different frm 0.3 as 0.3 is from 0.4, as to say that 0.3••• is a unique number to perform arithmetic on, thus would make more sense to have its own symbol like 0.c

0.3 × 4 = 1.2
0.33 × 4 = 1.32
0.333 × 4 = 1.332
0.3••• × 4 = 1.3•••
Logic would dictate there would be a 2 at the end of that repeating sequence of 3's but there never is.
a = 0.111•••
b = 0.222•••
c = 0.333•••
...
h = 0.888•••

So thaht 0.c × 3 = 1.0 easily, and provable by counting on the number line of 0abcdefgh. 0.c as the analog for the repeating "3" gives us a base
0.c×3 = 3×3 on the decimal number line or 9 counts
abc def gh0 <- 0, we hit 0 and count up a whole = 1.0
0.c×4 = 4×3 or 12 counts
abc def gh0 abc <- c, we hit 0 = 1.c

>>9390993
1/3 = [math]0.1_3[/math]

>>9391040
Thanks for the pointless post.

>>9391040
The extra numbers of abcdefgh would only be used for the decimal point, so really it's just an extension of base-10, the one everyone uses. Its nice that 0.333••• is 0.1 in base-3 but no one uses base-3. The problem isn't the base, the problem is how to interpret repeating decimals in a base.

0.3 with a hundred, a thousand, even a million 3's after it, is a literal different number than 0.3•••, as the finite limit of 3's in the non-infinite-repeating case eventually allows base10 arithmetic to be performed.
For example:
0.3 × 12 = 3.6
0.33 × 12 = 3.96
0.333 × 12 = 3.996
0.3333 × 12 = 3.9996
0.33333 × 12 = 3.99996
0.333333 × 12 = 3.999996
which leads to the faulty logic of
0.333••• × 12 = 3.999•••6
where base-10 arithmetic fails the repetiton.

>>9391050

>>9390282
>>9390432
>>9390993
>>9391092

Thanks for the pointless posts

>>9390232
>some retard went through the effort to make this
That's not how our number system works

>>9391100
Kill yourself f a m

>>9390669
1/3 ≠ 0.(3)
I will not repeat anymore

>>9391143
1/3 = 0.333•••
This is not very difficult to understand.s

>>9390672
what kind of nonsense are you talking about?

>>9390691
I do not understand your logic
just a statement without proof

>>9390993
>>9391092
stop talking nonsense

>>9391163
It isn't nonsense.
0.999••• = 1 is nonsense. It cannot be proved by itself and requires getting it as a sum or using division from it, such that 1/3 = 0.333••• so 3/3 = 0.999•••, while we know N/N = 1, or that 0.999••• ÷ 3 = 0.333•••, and 1.0 ÷ 3 = 0.333•••

Thats why the new decimal number line would have these repeating decimal steps. For numbers 1-8, where 0.999••• doesnt exist because it can be proven that 0.c × 3 = 1.0 by itself.

>>9391153

Try and contradict it.

>>9391147

0.3 -> 1/3
0.3333 -> 1/3
0.3333333... -> 1/3
0.(3) - Infinity in this model is unreachable.
1/3 - here infinity is already achieved.

1/3 - "numerical machine" and not the final product.

Incompatibility of theoretical models.

>>9391181
no
0.(0)1

>>9391187
>0.(3) - Infinity in this model is unreachable.

moron, infinity is in that syntax's definition

>>9391187
>>9391192
I dont know what the hell you're talking about. What the fuck does 0.(n) mean?

>>9391187
0.3 is not equal to 1/3 you mongol.
try dividing 1 by 3. Do you even know what long division is. Did you even kow fractions are writte decimally by doing the division?

fucking street shitting pajeet go throw your dirty trousers on someone else's yard.

>>9391187
so we have two "0.(9)"
0.(9) -> 1 --- real 0.(9)
0.(9) = 1 --- Imaginary 0.(9)

>>9390194
>Not equals, only converges to one.
yes equals, because 0.(9) is the notation for the limit, not the sequence of partial sums.

>>9390164
*sobs*

>>9391192
Infinity is NaN and so is the infinitesimal also NaN. These are concepts, not numbers. Pi is a number, infinity is not a number. Pi has a seemingly infinite amount of non-repeating decimal places, but "infinity" itself is not a property of pi, because infinity is a concept and not a number, no more than than a Ford F150 is a number, or a plastic bag is a number. You do arithmetic on numbers, and using infinity in arithmetic leads to useless results, because infinity is not a number but instead a vague, undefined amount of something, much the same as trying yo determine the numerical value of "some" or "a lot".
1÷3 = 0.3••• this is true
3÷3 = 1.0 this is true
0.3••• × 3 = 1.0 this is true
0.3••• × 3 = 0.9••• this is not true

the idea that 0.3••• × 3 = 0.9••• extends from the FLAW of thinking "0.3•••" is related to "3" for the obvious reason that both numbers contain a 3 and we assume 33×3 = 99, 3×3 = 9, 0.3 × 3 = 0.9, so 0.3••• × 3 must also have a 9 in the answer.

0.333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333 is NOT the same number as 0.3•••, and it is as truthfully as different as 1 is different from 2, meaning 0.3••• is a UNIQUE number that is confusingly written using established numbers.
It would be like instead of a number existing for 8, you were meant to write colon-three instead ":3" so that the number line would be
0 1 2 3 4 5 6 7 :3 9, and instead of ":3" being a combined character on your keyboard where 8 would be, there is no such combination and you're expected to manually write ":" and "3" individually everytime you wanted to use the number eight. Its dumb, and as dumb as writing 0.n•••, or 0.n..., or 0.(n) or whatever the fuck you do, which is where the extended decimal base-18 system comes into play to solve by ease of interpretation.

>>9391253
You only know how to use someone else's theories and stupidly prove them every time by any means.
But you do not want to think with your head.
You can not create something new yourself.

Our conversation is over.

>>9391253
In short 0.999••• doesnt equal 1 because (1÷3)×3 does not actually equal 0.999•••, but instead the equation is "([a unique number])×3 so that that the answer is 1"

>>9391270
I dont know what you're talking about retard. This is my theory one, and we weren't having a conversation either. So far as I could tell you were just shitting out brain diarrhea which no, that does not pass for information conversion much less conversation. Stay a forever brainlet.

0.9
0.99
0.999
0.999...9

Look, there are only nines here.
There is no one here.
0.999...9 - absolute limit.
The largest number of model.
After him there is nothing.
And this number itself is unattainable.

But you say that there is something after infinity.
You violate the laws of the model.
You are violating our old arrangements.

1,2,3,... -> infinity -> another reality
What kind of absurdity?
It is impossible to combine these models together.

An infinite number of points in a segment.
There are two incompatible realities.
0.(9) = 1 only in the world of segments, but not in the world of points, they do not even know about the existence of 1.

But you prove that they know about the existence of 1 and that their world must live according to the laws of your world.

>>9391303
>you ever smoked so much weed you became a foreigner?
honestly my guy, it is very difficult to understand you. What is your native language?

>>9391361
To reiterate, "infinity" is NaN. Not a number.
It is as useful as replacing with "some" or "a lot". You cannot do an infinite amount of something because you don't know how much to do, just as the same as doing a thing "some" amount of times, "many" amount of times, or "a lot" of times. "Infinity" has the same numerical meaning as some, many, a lot, an undefined "x" variable in a equation not equal to anything, whatever. It's not a number, and therefore you can't do arithmetic with it that results in a number.

For example:
>5x+4x =
We can evaluate it to be
>5x+4x = 9x
but X still has no value and is therefore not an actual arithmetic equation presenting a single number solution. Just as valid would be the answer
>5x+4x = 8x-x
And you can see that all of this has different interpretations and is meaningless towards producing some kind of intelligble single unit answer.

You could also rewrite it
>5[math]\infty[/math]+4
and its still just the same. Infinity is not absolutely not number, and can't be used in any arithmetic to get a number that isn't already afflicted with infinity or a property of infinity, with the latter extending to 0.999•••, but 0.999••• is also not a number that is sanely represented in readable or writeable arithmetic. Ergo, it makes sense to give the repeating single digit numbers 0.1 through 0.8 a unique identifying symbol or character that understands the problem of infinite repeating singles being a nothing more than the product of using Base-10.

To reiterate, "infinity" is NaN. Not a number.
It is as useful as replacing with "some" or "a lot". You cannot do an infinite amount of something because you don't know how much to do, just as the same as doing a thing "some" amount of times, "many" amount of times, or "a lot" of times. "Infinity" has the same numerical meaning as some, many, a lot, an undefined "x" variable in a equation not equal to anything, whatever. It's not a number, and therefore you can't do arithmetic with it that results in a number.

For example:
>5x+4x =
We can evaluate it to be
>5x+4x = 9x
but X still has no value and is therefore not an actual arithmetic equation presenting a single number solution. Just as valid would be the answer
>5x+4x = 8x-x
And you can see that all of this has different interpretations and is meaningless towards producing some kind of intelligble single unit answer.

You could also rewrite it
>5∞+4∞
and its still just the same. Infinity is absolutely not number, and can't be used in any arithmetic to get a number that isn't already afflicted with infinity or a with property of infinity, with the latter extending to 0.999•••, but 0.999••• is also not a number that is sanely represented in readable or writeable arithmetic. Ergo, it makes sense to give the repeating single digit numbers 0.1 through 0.8 a unique identifying symbol or character that understands the problem of infinite repeating singles being nothing more than the product of using Base-10.

>>9391314
google translater
What sentence?

>>9391394
>"Infinity" has the same numerical meaning as some, many, a lot,
nope, infinity means, surpassing all bounds. Whatever upper bound you make, infinity is larger
and if you say that infinity does not exist then you also say that [math]0.\bar{9}[/math] does not exist.

Sum to infinity means that you let your index grow without bounds and look what happens to the total

>>9391401
Its not just Base10 that has repeating numbers but you get the idea. Base-16 would need a similar layout like
0123456789ABCDEF
abcdefghjklmn where
j = 0.AAA•••
k = 0.BBB•••
l = 0.CCC•••
m = 0.DDD•••
n = 0.EEE•••

Now this is all very fresh and I'm open to changing the abcdefgh stuff to different symbols if not creating new symbols or some kind of lowercase number or strikethru that wouldn't conflict with subscript numbers or the broken idea of overline numbers, neither of which are easily reproducable on a keyboard, but i feel using symbols that look like the established base10 arabics would just continue to lead to this logic fuckup of misinterpretation for normies and brainlets.

>>9391430
but anon, in base-16 0.FFFFFF(...) is 1.000000(...)
and in base-3 0.222222(...) is 1.000000(...)
and so on

simply because infinitesimals don't exist in the real numbers.
In hyperreal numbers? Sure
In surreal numbers? Sure
but not in real numbers

>>9391414
You cannot take a measurement of the total without concatenating and deciding at what number you've cut off to take the measurement, which is no different than saying 0.33333 × 3 = 1 which is, for a fact, false because 0.33333 × 3 = 0.99999
You can't say that 0.333••• × 3 = 0.999••• = 1, because you are basically saying 0.3x × 3 = 0.9x = 1, which is also false.
0.9x is a number even if we don't know what x is and is therefore only equal to itself without arithmetic. 0.9x = 0.9x.
You can try to solve for x in 0.9x = 1 all you want but you wont get very far with this understanding.
>divide both sides by x
Ok 0.9 = [math]\fract{1}{x}[/math]
x = 1.111•••
ah, but we are defining infinite repitition as a variable, so really we get
x = 1.1y
For the equation
>0.9(1.1y) = 1
Which has literally solved nothing and only made the equation more complex, and will continue to do so for every attempt at defining the variable.

It is DISGUSTING that infinity is used in "higher" math as if its suppose to be an intelligble number. You cannot set a limit to infinity cause that is just saying no limit, which then precludes and invalidates writing a limit into the equation outright.

>>9391469
brainlet

>>9391469
>You can't say that 0.333••• × 3 = 0.999••• = 1, because you are basically saying 0.3x × 3 = 0.9x = 1, which is also false.
I have no idea what the fuck you're trying to say.
how do you get from [math]0.\bar{3} \cdot 3 \eq 0.\bar{9} \eq 1[/math] to [math]0.3 \cdot x \eq 0.9 \cdot x \eq 1[/math]?

>>9391495
fuck it, just replace all \eq with =

>>9391439
The point of this thing that i'm just going to call intermantissa is to get rid of writing single digit repeating decimals, and writr them in a way that is distinct from the single number they're composed of to get around logic problems.
0.3 × 3 = 0.9 we can do the math and know this
0.4 × 3 = 1.2 we can do the math and know this
the reason we know to do the math this way is cause we understand 0.4 has a unique and different value and written symbol than 0.3
If 3 wasn't even a number and we were meant to write three as "-22-", two 2's surrounded by hyphens, this would get very old and cumbersome very fast, and arithmetic with "-22-" would be confusing where 2 × 4 = 8 but -22- × 4 = 12 rather than what might logically be seen as "-88-". Its a battle of logic reminding yourself that "-22-" is not "two 2's" but instead "three"; much the same as 0.333••• or any single digit repeating decimal has no logical meaning when used in arithmetic. "0.333•••" isn't "zero point three three.. three repeating", cause 3 isn't "hyphen two two hyphen". Where the logic problem of -22-×4 = 12 instead of -88-, we get a similar logic problem where 0.333••• × 4 = 1.333••• instead of 1.333•••2

Really, just use the abcdefgh decimal system for a couple problems and it becomes easy to understand.

>>9391502
>If 3 wasn't even a number and we were meant to write three as "-22-", two 2's surrounded by hyphens, this would get very old and cumbersome very fast, and arithmetic with "-22-" would be confusing where 2 × 4 = 8 but -22- × 4 = 12 rather than what might logically be seen as "-88-". Its a battle of logic reminding yourself that "-22-" is not "two 2's" but instead "three"; much the same as 0.333••• or any single digit repeating decimal has no logical meaning when used in arithmetic. "0.333•••" isn't "zero point three three.. three repeating", cause 3 isn't "hyphen two two hyphen". Where the logic problem of -22-×4 = 12 instead of -88-, we get a similar logic problem where 0.333••• × 4 = 1.333••• instead of 1.333•••2

>>9391506

>>9391507
I've read that like three times and still have no Idea what you're trying to say
please condense it

In "9...9 world" you can determine the location of the point of 1.

In "0.9...9 world" 0.(9) is infinity, but infinity is not a point.

You can not find the real beginning and real end of the little point world in your segment (-inf, +inf).

"Infinity of nines after zero is point one" - is a absurd.

You fucking idiots.

The notation 0.999... is DEFINED to be the limit. The ellipsis INDICATES that one should take this limit. Hence 0.999... = 1.

Here's a document to quickly explain intermantissa and it's usage, as well as some possible symbols for the intermantissa numbers.

Here's a quick explanation of intermantissa and how to use it. Had some typos so reuploading it.

>>9391668
kek forgot something, use this instead

>>9391514
Say we lived in a world where we had no symbol for the number three, that "3" sideways butt or catface icon didn't exist, but we wrote three as "-22-" or "hyphen two two hyphen", and more than that we did not have typing mechanisms or keyboards with the single button/key "-22-" so we would always have to type three in as I'm doing now, "hyphen two two hyphen", four distinct button inputs. This can be fairly fatiguing and annoying that there isn't a single convenient button on our keyboards in this alternate world to write the numeral three. When we do arithmetic with -22-, like -22- × 4, we know this is "three times four", therefore -22- × 4 = 12. It might frequently trip up people though, since they see two 2's where 2 still has the same meaning as a number. They might accidentally evaluate -22- × 4 = -88-, cause they know 2 × 4 = 8. We have a logic problem based on a confusing written interpretation of three.

Now back to this reality, we have the problem 0.333••• × 4.
We know:
33 × 4 = 132
3 × 4 = 12
0.3 × 4 = 1.2
0.33 × 4 = 1.32
0.333 × 4 = 1.332
0.3333 × 4 = 1.3332
And we see a pattern that multiplying 3 and 4 in this way should always result in the last number of the answer being "2"
Yet 0.3••• × 4 does not equal {1.3•••2} and instead equals {1.3•••} without the 2. This is a logic problem, and the problem extends from 0.3••• actually being a unique number unrelated to 3.

So in our other alternate world where there were accidents of evaluation because three was written using already established 2's, in this reality we see accident of evaluating numbers with repeating decimals because they are written using already established 3's.

If we have a different symbol to represent the decimal representation of 1/3, as prescribed in >>9391507 , it becomes easy to avoid the evaluation problem of trying to add a 2 onto the end result of 0.3••• × 4 = {1.3•••} to get {1.3•••2}

>>9391549
?
What does the notion of location as you are using it mean? In traditional constructions of the real numbers, like dedekind cuts, it's very trivial to show that .999... = 1. If we accept that .999... specifies a real number, then it is also quite easy to show that it must equal 1.

As for your notion of infinity being a point, it is again standard in lots of undergrad courses to show that the set of infinite decimals under the standard expected operations is isomorphic to any complete ordered field

>>9391709
no

>9 x 10[math]^{-∞}[/math] = 1

>>9391574
Limits can't be unlimited. That is the exact opposite of what limit means. Infinity is not a limit, and nothing can be limited to infinity. The sun will consume earth before a final integer of pi is calculated, and whatever the last integer counted the moment before the world ends will be a definite, countable number of digits to pi - not infinite, and neither will another number come after that last decimal digit.
Infinity isn't even a number anyway, it's just an abstract concept.

>>9391796
>Infinity isn't even a number anyway, it's just an abstract concept.
anon, all numbers are abstract concepts
two, zero and 5000 don't exist in reality

>>9392218
You can count two and 5000 of things though. You can't count an infinite amount of things, because you'd never get a total. Similarly you can't count "some" or "many" things cause these are just abstract concepts. Tending to infinity is not a valid solution to a problem, as once you concatenate the result as
>m-muh infinity
then you can no longer rework the problem from the reverse to get an inverse equation.

This planet is filled with brainlets because calculus is retard tier garbage completely seperates from practical engineering, be it mechanical or biological. You may as well be attending daydreaming class if your math courses deal with infinity. Shit is meaningless as a number or a number of things.

>>9391673
1 2 3 4 5 6 7 8
ㅗㄷד ב フסノo
just gotta get the unicode consortium to dissassociate the hebrew letters from force right-to-left and we're good.

>>9391673
How is someone supposed to say those numbers? Do they have names?

>>9392449
inter-one, inter-two, inter-three
idk

>>9391673
So
0.999--- = 1
but only cause 0.999--- doesn't actually exist as finite numbers with which to do maths?
And 0.111--- or 0.222--- or 0.333--- or 0.nnn--- also dont actually exist, so there would be no reasonable way of evaluating 0.999--- as a sum?
So 1/3 = null point inter-three
null point inter-three × 3 = 1?
8/9 = null point inter-eight
1/9 = null point inter-one
null point inter-one + null point inter-eight = 1

What about a number like 1/11?
Would that be 0.0inter-nine or what? That evaluates to 0.090909--- and you don't have an inter number for either 0 or 9.

You didn't think this through

Don't know if already someone posted legit proof, but here's mine:
Let x = 0.(9) , therefore
10*x = 9.(9), so
10*x - x = 9, so 9x = 9, x = 1

>>9392491
Hm

If [math]\bar{0}[/math] isn't a thing by itself and 0.[math]\bar{9}[/math] assumes to equal 1 precluding writing [math]\bar{9}[/math] with the methodology behind intermantissa simply being getting rid of evaluation logic errors, maybe overline could be kept for examples like 0.[math]\bar{09}[/math]?

lets try some math idk
0.9 × 4 = 3.6
0.09 × 4 = 0.36
0.0909••• × 4 = 0.3636••• || 0.בס
i dont see any logic error of wanting to finalize the significand with a different number than 3 or 6 here.

1/30 = 0.0[math]\bar{3}[/math] which could be written 0.0ב

If you find an example where overline for 0 and 9 wouldn't work by loosely evaluating a desired number as the final significand that's different than the preceeding repetition then I'll consider retracting this idea of intermantissa cause the extra symbols for 0 and 9 would break the easy proofs like 3 × 0.ב
9 countable places on the intermantissa line up to 1.0 where an internine would just give the result internine and no therefore different than the current repeating 9 model.

>>9392520
x = 0.9•••
10x = 9.9•••
10x - x = 9.9••• - x
9x = 9.9••• - x
x = 1.1••• - x
2x = 1.1•••
x = 0.0555•••

>>9392545
>>9392520

x = 0.9•••
10x = 9.9•••
10x - x = 9.9••• - x
9x = 9.9••• - x
x = 1.1••• - [math]\frac{x}{9}[/math]
x+[math]\frac{x}{9}[/math] = 1.1•••

>>9392545
>>9392553
I dont get the point

yall seriously need to take a course on analysis

>>9392530
Oh thats kinda neat with how you could write successive alternating repeating digits just with successive internumbers but without the overline. I guess that makes sense if intermantissa are supposed to be the final digit(s) of a decimal.

Anyway what about 1/1.3?
That evaluates to 0.769230---
0.769230--- × 7 = 5.384615---
which should "logically" calculate to
5.340510--- if we ignore the repeating sign like your 0.3--- × 4 examples imply.

>>9391187
you have legitimate schizophrenia

>>9390432
>so∞≠∞+1
>hasn't heard of ordinal arithmetic
jesus christ what a fucking retard

>>9390672
>>9391185
you're a retard, but you're on the right track
you need to show that any [math]x[/math] STRICTLY less than 1 is also strictly less than 0.999...

>>9391143
>>9391187
>i will not repeat any more
>keeps making his schizophrenic posts
i wish you would just die

>>9392583
Copy pasting the hebrew and greek letter representations of the numbers is really broken so i'm just going to write the intermantissa as abcdefgh
A rule for this to work is when using 9 within the intermantissa, count it after 0 and before a

abc def gh0 [9]
this way, 0.0999••• + 0.0333••• evaluates as [9]->a->b->c = 0.1333•••
[9] + 1count = a
[9] + 2counts = b
[9] + 3counts = c
this should be added to the intermantissa number line so that 9 and 0 occupy the same visual space.
Anyway, to solve this problem with intermantissa:

>1 ÷ 1.3 = 0.769230•••
>1 ÷ 1.3 = 0.gf9bc0
>0.769230••• × 7 = 5.384615•••
>0.gf9bc0 × 7 = ?

0.0 × 7 = 0
>0
0.g0 × 7 = 4.9 (7×7 inters = 49 counts)
>0 + 4.9 = 4.9
0.0f0 × 7 = 0.db (7×6 inters = 42 counts)
>4.9 + 0.db = 5.cb
0.00[9]0 × 7 = 0.0fc (7×9 inters = 63 counts)
>5.cb + 0.0fc = 5.chc
0.000b × 7 = 0.00ad (7×2 inters = 14 counts)
>5.chc + 0.00ad = 5.chdd
0.0000c × 7 = 0.000ba (7×3 inters = 21 counts)
>5.chdd + 0.000ba = 5.chdfa
Goofy left-to-right maths aside, up to this point it looks like we have the correct answer. 5.chdfa = 5.38461, but our next step is to multiply 0.00000[0] by 7 which would give us 0 instead of the required addition of 0.000005
Based on knowing only repeating numbers come in the significand and not the integer whole before the decimal point, we should be able to evaluate that e should come next as that is the intermantissa representation of the 5 integer ahead of the decimal. Would this work as a rule of thumb that if the final decimal was 0 preceeded by intermantissa, this inter-zero would instead become the first number in the answer up to this point in order to get that 5?
>5.chdfa0 ~> 5.chdfae = 5.384615•••

>>9392699
>0.09--- + 0.03--- = 0.1333---
>0.09--- + 0.0c = 0.13 = 0.ac
Well how does this work then? 0.ac would imply 0.13131313--- which isn't the answer.

>>9392699
>Would this work as a rule of thumb that if the final decimal was 0 preceeded by intermantissa, this inter-zero would instead become the first number in the answer

Well why did 1 / 1.3 = 0.gf9bc0 instead of 0.gf9bcg?

>>9392752
Because the answer starts with a 0?
>0.gf9bc0
>^-----------^

>>9392728
Gotta create rules for this so its really not as dead simple as I'd hoped, but since 0.09 and 0.0c add together into the null space decimal after 0 of 0.0, that place should evaluate using a regular number instead of intermantissa. 9+c = 13, the 1 is in null space
>0.09
>0.0c
>0.1c
so it stays 1, and the 3 becomes c as it is in the same decimal place as non-zero digits in the equation

>>9392774
>0.09
>0.0c
>0.1c

>0.09 + 0.0c = 0.1c
>0.09 + 0.0b = 0.1b
>0.09 + 0.0a = 0.1a
>0.09 + 0.00 = 0.10
>0.09 * 10 = 0.1 * 10
>0.9 = 1.0

m8

>>9392789
Truthfully that was supposed to be a repeating 9 which would have instead naturally evaluated to 0.1 instead of 0.09•••

>>9392699
>db = 42
>fc = 63
These double intermantissa values aren't right. 42 should be 40, d, count 2, ->e->f, so df= 42.
63 should be 60, f, count 3, ->g->h->0, so f0 = 63

Intermantissa are single digits though, so the result isnt "42 forty-two", its "4 & 2 four and two" so 4 counts to d from 0, and 2 counts to b from 0.

>>9392311
so from that rant I assume that you're not a fan of imaginary and complex numbers?

>>9393011
Set a limit to i and tell me how that goes.

>>9390287
this

.000...01 only exists in the hyper-reals, which are non-standard for a reason, these threads specifically.

Is this entire thread seriously debating the .999... = 1 shit. It isn't hard to prove but this thread seems to have the world's most disgusting way of thinking about this and the worst notation to explain it.

>>9390202
>n can not reach infinity

>>9390323
>>HAND WAVING INTENSIFIES
No, anon gave a perfectly legitimate explanation.

>>9390202

>>9393133
>this thread seems to have the world's most disgusting way of thinking about this and the worst notation to explain it.
It's kind of beautiful; I've never seen mathematical notation this terrible before.

>>9393441
Well it can be proven that 0.9••• doesn't equal 1 if your really want to go there.

in base 16, 0.F••• would be closer to 1.0 by a factor of 6 than base 10's 0.9•••, since base 10's 0.9••• is still written as 0.9••• in base 16.

0.9••• can't equal 1 since it's 60% less than 0.F•••

the real trick is realizing single digit repeating numbers aren't actually useful "numbers" in their own right, not arguing about how to make it equal to 1 because of a flaw in the base system.

>>9393528
>in base 16, 0.F••• would be closer to 1.0 by a factor of 6 than base 10's 0.9•••, since base 10's 0.9••• is still written as 0.9••• in base 16.
See, that's the kind of thing I was talking about. This is to mathematics what one of those Markov chain-bots is to language - All the right symbols are there and they're arranged in a way that looks coherent, but there's no actual meaning in any of it.

>>9393528

>>9393579
Thanks for proving my point. As you can plainly see, base 16 reachs it's 0.999999999 equivalent before base 10 does. 7 steps for base 17, 9 steps for base 10. Even taken at the first step, 15/16 = 0.9375 which is clearly larger than 9/10 = 0.9

For every step, even uncountably infinite as many steps, 0.F••• will always be greater than 0.9•••

What is sorely misleading about your image is concatenating and rounding after so many steps to equal 1, but regardless it still serves to prove the point.

>>9393583
proves nothing

>>9393586
>proves nothing
[math]\frac{15}{16}[/math] > [math]\frac{9}{10}[/math]
[math]\frac{15}{16} × n[/math] > [math]\frac{9}{10} × n[/math]

let d be the real number having decimal expansion .999999999999...
then d is equal to the sum 9/(10^i) where i ranges from 1 to infinity
it is clear that the limit of the partial sums approaches 1
then d = 1

>>9393595
[math]\frac{15}{16} * \infty[/math] = [math]\infty[/math]

[math]\frac{9}{10} * \infty[/math] = [math]\infty[/math]

[math]\frac{15}{16} * \infty[/math] = [math]\frac{9}{10} * \infty[/math] = [math]\infty[/math]

>>9393611
[math]\frac{1}{10} × \infty[/math] = [math]\frac{9}{10} × \infty[/math] = [math]\infty[/math]

>>9393608
sry, clear that the sequence of partial sums approaches 1

>>9393608
>let

So what is the definition of infinity suppose to be in terms of limits or repeating decimals?

0.9 ≠ 1
0.99 ≠ 1
0.999 ≠ 1
0.9999 ≠ 1
0.99999 ≠ 1
0.999999 ≠ 1
0.9999999 ≠ 1
0.99999999 ≠ 1
0.999999999 ≠ 1
0.9999999999 ≠ 1
it looks like every step of the way, this still doesn't equal 1. So why would an infinite amount of steps allow it to be equal to 1 rather than just an infinite amount of 9's ?

inf x 999 = inf
inf x 555 = inf
555 = 999
0.(9) = 1

>>9393595
idiot

1/x and 1/x2 have a different slope and that says absolutely nothing about having a different limit+ at zero

0.(9) x inf = inf
1 x inf = inf
0.(9) = 1
Yep

>>9393683
1 = 9/10 + 1/10 = 0.9 + 9/100 + 1/100
= 0.99 + 9/1000 + 1/1000 = ...

it looks like every step of the way, this still equals 1. So why would an infinite amount of steps allow it to not equal to 1 with an infinite amount of 9's ?

>>9393699
You do realize this equation clearly says that 1 = [math]0.\bar{9} + \frac{1}{1\bar{0}} + \frac{9}{1\bar{0}}[/math], right?

>>9393725
it says 1 = 9/10 + 1/10
it also says
9/10 + 1/10 = 0.9 + 9/100 + 1/100

don't know wtf you are saying

>>9393734
You said an infinite amount of steps. 0.9->0.99->0.999... -> [math]0.\bar{9}[/math]

Your equation says [math]1 = 0.\bar{9} + \frac{1}{\bar{1}}[/math].

So [math]0.\bar{9}[/math] can't equal 1 cause it requires adding something to it to sum 1, and no number divided by another number = 0 so you're not simply adding 0 to it.

>>9393749
so where exactly does it deviate from 1 ?

hard mode: no hand waving

Maybe 0,(9) is not 1 if 0,(0)1 exists. But then, 0,(9)9 is 1 right?

>>9393752
the moment a perfectly round circle completes a single perfectly measured roll across a perfectly flat surface.
idk what you're asking for. Tending towards infinity is not tending towards a number, it's just tending towards infinity. Thats the answer you give and thats the answer your professor accepts.

>>9393762

>>9393767
okay enjoy being a gay fag. How about you provide a proof for [math]0.\bar{9}[/math] = 1 that can't be easily contradicted.

In b4
>[math]\frac{1}{3} × 3 = 0.\bar{9}[/math]
cause it ain't. 1/3 × 3 = 3/3 = 1.
[math]0.\bar{3}[/math] × 3 = 1, not [math]0.\bar{9}[/math]

>>9393775
If there is no deviation from 1, then that is the proof.

>>9393778
>>Tending towards infinity is not tending towards a number, it's just tending towards infinity.
Tou are literally not even ironically rounding by invoking "i-if it doesn't deviate"

you know what [math]0.\bar{9}[/math] ISN'T deviating from? MORE 9's

Implying that 0.999.... =/= 1, implies that there exists some number 0.000...1 such that 0.999... + 0.000...1 = 1

But this is naive. 0.000...1 cannot exist because, by definition, the 0s go on forever without terminating. There cannot be a trailing 1 because there is no end. Its very existence is a contradiction. It is just zero.

>>9393784
If it isn't 1 at any point (like infinity) then a genius like you should be able to whip up a proof.

Stop hand waving and do some actual math.

Oh boy, it's THIS thread again...

>>9393785
[math]0.\bar{9}[/math] doesn't equal 1 because [math]0.\bar{9}[/math] is not a number, no more than [math]0.\bar{0}1[/math] is a number.

Try to write an equation where the sum is [math]0.\bar{9}[/math].

>>9393792
1 + 0 = 1

>>9393799
Just control-f intermantissa and free your brain from the chains of brainletism

>>9393792

I'll write several.

[math]x = 0.9999...
10x = 9.999...
10x - x = 9.999... - 0.999...
9x = 9
x = 1[/math]

Or
[math]1/3 * 3 = 1 = 3 * 0.333... = 0.9999...[/math]
Or
[math]1=0.999...=\sum_{n=1}^{\infty}\frac{9}{10^n}=\sum_{n=1}^{\infty}\frac{9*(1/10)^n}{1-10^n}[/math]

Fuck, it doesn't even make sense that 0.999... =/= 1, because if you claim that there they aren't equal then there must exist some number which n such that [math]0.\bar{9} < n < 1[/math], which, obviously, cannot exist either. You're making an amateur mistake.

>>9390232
fucking brainlet. A number infinitely close to zero, 0.000...1, IS ZERO. It doesn't exist. It's infinitely small. Imagine a head with an infinitely small amount of gray matter - it shouldn't be hard for you. Because the amount of brain isn't just small, it's infinitely small, there isn't one at all. Pic related.

>>9393792
> 0.9¯ is not a number, no more than 0.0¯1 is a number.

You can't arbitrarily pose claims like that and demand everyone believe it without any proof on a math board. [math]0.\bar 9[/math] is a number and that number is [math]1[/math]. [math]0.0\bar 01[/math] is nonsensical because it implies that the non-terminating 0s terminate. A contradiction.

>>9393792

>>9390205

>>9393789
>[math]9.\bar{9} - 0.\bar{9} = 9.0[/math]

[math]0.\bar{9}[/math] - [math]0.\bar{9}[/math] does not equal 0.
0 = [math]0.\bar{9} - 0.9 - 0.09 - 0.009 - 0.0009 ... - 0.\bar{0}9[/math]
[math]0.\bar{0}9[/math] is not a number

>>9393824
x-x =/= 0

your nobel prize is in the mail

>>9393824

What nonsense are you on about? [math]0.\bar 0 9[/math] is not a number (in fact, it's nonsensical for the same reasons I stated earlier), but [math]0.\bar 9[/math] is. You're making the mistake that any number which can be expressed as an infinite sum is somehow not a number. But again, it's an amateur mistake. Here's a simple proof to show you: [math](\sum_{n=1}^{\infty}\frac{k}{10^n}\forall k \in \mathbb{N})\subset \mathbb{R}[/math]

[math]\frac{1}{3} = 0.\bar 3[/math], for instance. Or are you going to say fractions aren't real numbers either?

>>9393824
> −0.009−0.0009...−0.0¯9
> ...−0.0¯9
> ...
Found your mistake. It never gets to the 'not a number' part. The '...' goes on forever, remember?

>>9393809
Not a single one of these is valid
First
>[math]9.\bar{9} - 0.\bar{9}[/math] = 9.0
can't prove this true.
>[math]0.\bar{9} - 0\bar{9}[/math]
>0 = [math]0.\bar{9}[/math] - 0.9 - 0.09 - 0.009... - [math]0.\bar{0}9[/math] <~ not a number
>[math]0.\bar{9} - 0.\bar{9} \neq 0[/math]

second
[math]\frac{1}{3} × 3 = 1[/math]
[math]\frac{1}{3} = 0.\bar{3}[/math]
[math]0.\bar{3} × 3 = 1[/math]

third
[math] 0.\bar{9} [/math]
>literally just a statement not even an evaluation

Fourth
[math]0.\bar{9}_{10} < 0.\bar{F}_{16} < 1[/math]

>>9393830
>0.0¯9 is not a number

0.9 = 9/10^1
0.09 = 9/10^2
0.0...9 = 9/10^inf = 0

>>9393135
>>9393146
He's correct

>>9393837
>9.9¯−0.9¯ = 9.0

= 9 + 0.9¯ - 0.9¯ = 9 + 0 = 9

>>9393843
Can't divide by infinity.

>>9393846
oh it's done erry day you sweet summer child

>>9393837
Actually, everything I said was true. For the first part, see: >>9393830 and >>9393836
You're trying to terminate a non-terminating

Second, [math]\frac{1}{3}=\frac{0.\bar 9}{3}=0.\bar 3[/math].

Third, you never addressed the fact that there cannot exist a number k such that [math]0.\bar 9 < k < 1[/math], which is a property of subtraction that must exist for every number.

Fourth,
[math]0.\bar 9[/math] is not a number, not a statement or an evaluation. Specifically, it's [math]1[/math] for precisely the last property I just outlined.

Fifth,
[math]0.\bar k_{z}[/math] is nonsense.

Sixth, you're STILL ignoring the fact that you're trying to terminate a non-terminating evaluation.

In YOUR words, you said:
> [math]0.\bar 9 = 0.9 + 0.09 + 0.009 + ... + 0.0\bar 0 9[/math]

But you're forgetting the fact that "last" "non-number", the [math]...+ 0.\bar 0 9[/math], cannot exist in your evaluation.
It just goes
[math]+0.0009 + 0.00009 + 0.000009 + .....[/math]. That's it.
There is no more than that. There is no [math]0.\bar 0 9[/math]. You're trying to terminate something that doesn't terminate, and getting something nonsensical as a result. Of course [math]0.\bar 0 9[/math] isn't a number. But [math]0.\bar 9[/math] is. And it's equal to exactly [math]1[/math] by virtue of that [math]1-0.\bar 9 = 0.000...[/math]

>>9393847
not really. And if you are then you're dumb. Infinity is the biggest brainlet trap math ever became conflicted with, and violates every field of science including the only practical application of higher maths: computer science.
even fourier transforms have limits.
The full concept of infinity is only useful as a generalization for an amount, no different than "lots", "some" or "many". It doesnt define a specific number or limit and therefore can't produce a specific answer that could said to be unattainable without thinking with infinity.

>>9393847

You're a brainlet. There are infinitesimals, which indeed are not numbers (they're concepts), but you trying to treat them as if they numbers are in statements like [math]1 = 0.\bar 9 + 0.\bar 0 1[/math] just shows how little you know what you're talking about.

>>9393852
No, YOU are trying to terminate a non-terminating you hypoxic shitlord. What do you think [math]0.\bar{9} - 0.\bar{9}[/math] implies, fuckstain? Terminating the repetition and returning a 0.

Holy jesus breathe harder my nigger you are doing terrible things to your brain. Fuck reading anything else you gotta say cause it'll just be more dumb bullshit.

>>9393852
fourth was wrong. I meant to say it *is* a number, not a statement or evaluation.

I'm going to sleep. I think I've been trolled, and I'm getting tired.

>>9393857
bless your heart

run along now, adults are talking, not autists

>>9393859
>What do you think 0.9¯−0.9¯ implies, fuckstain?

0. Simple. Because [math]x-x=0[/math] for any real number [math]x[/math].

Can't wait for this meme to die.

>>9393864
Repeating decimal BaseN-1 is not a real number. You may as well be telling me about the last 100 digits of pi.
We're not there, we're never getting there and you only seem retarded by trying to use it in honest to goodness actual arithmetic. [math]0.\bar{9}[/math] doesnt exist. Numbers you mistakingly believe add up to it actually add up to 1 with rral arithmetic or [math]0.\bar{9}[/math] with imaginary arithmetic, but never does a single equation equal both simultaneously.

>>9393871
prove it

>>9393872
>>9391673
Feel free to use intermantissa, a system where repeating decimals [math]0.\bar{1}
- 0.\bar{8} [/math] have distinct characters. It evaluates identicallly to standard arabic base 10 all without ever dealing with [math]0.\bar{9}[/math]

>>9393871
>Repeating decimal BaseN-1 is not a real number. You may as well be telling me about the last 100 digits of pi.

Where's your PROOF, you faggot? Stop making claims you can't back up. Why don't you go bring your 'discovery' to the elites? I'm sure all you'll get a noble prize in hand-waving.

>>9393876

You don't understand the base 10 number system. The repeating decimals is just an artifact, an illusion. Something that doesn't manifest when you deal with decimals, hence why [math](\frac{1}{3}) \times 3=1[/math] but [math]0.\bar 3 \times 3[/math] appears to be [math]0.\bar 9[/math], just as [math]\frac{1}{9} \times 9[/math] also appears to be [math]0.\bar 9[/math] but is in fact [math]1[/math].[math]1[/math] and [math]0.\bar 9[/math] appears to be the same thing because they *ARE*. There is *zero* difference. Saying that they must be different because they appear different is like saying that [math]100[/math] and [math]50+50[/math] must be different. No. They're just different ways of expressing the same value.

0.999... = 1

1 - .999... > 0

You can only pick one.

>>9393877
Okay.
So [math]0.\bar{0}1[/math] doesn't exist, right? why though?
>cause you're trying to terminate, its just a bunch of 0's neverending
Okay
So [math]0.\bar{9}[/math] doesn't exist, right? Why though?
>cause you're trying to terminate, its just a bunch of 9's neverending
Heres the fuckin shitdickbrained stuff we're dealing with.

The same reasons [math]0.\bar{9}[/math] shouldn't evaluate ever to 1 are the same reasons [math]0.\bar{0}1[/math] doesn't exist
Reason 1: "you can't terminate a repeating decimal"
Well, how about that. Thanks.
Reason 2: [math]0.\bar{9}[/math] is an infinitesimally small distance away from 1, just as [math]0.\bar{0}1[/math] is an infinitesimally small distance away from 0

The other number repeaters for 1-8 got room to BREATHE, they're all [math]0.\bar{1}[/math] away from each other, but [math]0.\bar{0}1[/math] is crammed down there buried neck deep in 0's asshole and [math]0.\bar{9}[/math] is way up there literally pretending to be 1.

Neither of these numbers are necessary.

Lets try this anyway:
[math]0.\bar{9}[/math] + [math]0.\bar{1}[/math] = what? It equals [math]1.\bar{1}[/math] - [math]0.\bar{0}1[/math]

Whats [math]0.\bar{0}1[/math] + [math]1.\bar{1}[/math]?
It's [math]1.\bar{1}2[/math]

Just toss this useless shit, no one needs it. [math]0.\bar{9}[/math] is the inverse and analog of [math]0.\bar{0}1[/math] just getting by as a wolf in sheeps clothing.

>>9393894
I said FEEL FREE TO USE IT

>>9393894
I'm not sure any modern calculator would allow you to sum [math]0.\bar{9}[/math]

It would just spit out 1 every time. And its not cause [math]0.\bar{9}[/math] = 1, its cause [math]0.\bar{9}[/math] doesnt exist in real arithmetic.
0 to [math]0.\bar{1}[/math] to [math]0.\bar{2}[/math] to [math]0.\bar{3}[/math] to [math]0.\bar{4}[/math] to [math]0.\bar{5}[/math] to [math]0.\bar{6}[/math] to [math]0.\bar{7}[/math] to [math]0.\bar{8}[/math] to 0.

Computer science already basically adheres by this. Intermantissa are just the logic programming for students to emulate and mesh better with how to calculate numbers that calculators will display

>>9393859
>What do you think 0.9¯−0.9¯ implies, fuckstain? Terminating the repetition and returning a 0.
That's... not what "terminate" means.

0.999...9 < 1 < 1.000...01

>>9393844
THATS WHY WE TAKE THE LIMIT, lim x>inf does not mean to set x = inf you fucking retard

>>9394403
[math]n = 1;Lim ->\infty; \frac{9}{10n}++[/math] is not intelligibly different than just stating [math]0.\bar{9}[/math]

>>9394275
You need a proof

>>9393897
This bullshit only works in the hyper reals.
>Reason 2: 0.9¯0.9¯ is an infinitesimally small distance away from 1, just as 0.0¯10.0¯1 is an infinitesimally small distance away from 0
A number being so close to another number that there isn't a number inbetween them means that they are equal you fucking moron

>So 0.0¯10.0¯1 doesn't exist, right? why though?
>>cause you're trying to terminate, its just a bunch of 0's neverending
>Okay

>Lets try this anyway:
>0.9¯0.9¯ + 0.1¯0.1¯ = what? It equals 1.1¯1.1¯ - 0.0¯10.0¯1
>Whats 0.0¯10.0¯1 + 1.1¯1.1¯?
>It's 1.1¯2

fucking brainlet admits it doesn't exist and tries to use it anyway.

>>9394419
What the fuck kind of notation is that

No they arent different thats the fucking point, the point is that .9 repeating equals 1

>>9394429
Mongolian alert, everyone in the thread put your hazard suites on.

[math]0.\bar{9}[/math] can only exist if you accept that [math]0.\bar{0}1[/math] exists.
if you deny [math]0.\bar{0}1[/math] exists, you /must/ also deny [math]0.\bar{9}[/math] to exist.
it can be easily proven [math]0.\bar{8} +
\bar{1} = 1 [/math] with or without your choice to believe in either case, thus it is not a debate or a vote to choose to believe in [math]0.\bar{9} || 0.\bar{0}1[/math], rather it is merely a fact you must accept that no math requires these two numbers.

>>9394444
[math]0.\bar{8} + 0.\bar{1} = 1[/math]

>>9394429
Theres no closer number to 3 than 2 and 4 so that does mean 3 = 2 = 4?

>>9394444
They don't require the other to exist you fucking brainlet. 1 minus .9 repeating equals zero because the nines never terminate.

Writing .0 repeating and then having the fucking gall to put a one after an INFINITE amount of zeros just shows how fucking little thought you've put into this.

.9 repeating exists
.0 repeating with a 1 after doesn't exist.

t. math major

>>9394456

2 < [math] frac{3+2}{2} [/math] < 3

what the fuck is wrong with you. we aren't talking about the naturals that definition of equality doesn't apply to them since the naturals have gaps, we're talking about the reals.

>>9394458
You are a brainless retard.
[math]0.\bar{9}[/math] is a string of never ending 9's. Equating it to 1.0 is terminating the nines.

H o l y
S h i t
F a m

The same exact methodology behind [math]0.\bar{0}1[/math] being unreal, THE SAME EXACT METHODOLOGY, is why [math]0.\bar{9}[/math] is also unreal.

fucking permanent marker huffing motherfucker.

>>9394463
2 < [math] \frac{3+2}{2} [/math] < 3

>>9394465
That isnt what terminating the nines means. Saying it equals one is finding a separate representation of the number, but that representation only equals to a non-terminating sequence of 9's. If you terminate any of the nines, it is no longer equal to 1.

>>9394467
[math]0.\bar{9} < 0.\bar{F} < 1.0[/math]

Just cause you don't know how to write larger numbers doesn't mean they don't exist.

>>9394475
.9 repeating equals .F repeating fucking retard. For the exact same reason that [math] \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ... = \frac{9}{10} + \frac{9}{100} + ... [/math]

>>9394475
Apparently you didn't realize that when I said
>A number being so close to another number that there isn't a number inbetween them means that they are equal you fucking moron

means that there needs to be a number, NOT EQUAL to the other numbers inbetween, didn't think I need to break it down so far for you fucking mouth breathers

>>9394474
Nah, see >>9393871
You cannot naturally get [math]0.\bar{9}[/math] as a sum. Any sum trying to achieve it just ends up becoming 1 in real maths or only attempting to state [math]0.\bar{9}[/math] exists in unreal math, and no equation posits both realities simultaneously. Not because [math]0.\bar{9} = 1[/math], rather instead simply because the answer is 1 for normal number math. For example, [math]\frac{1}{3} × 3 = 0.\bar{3} × 3 = 1[/math]. Evaluating [math]0.\bar{9}[/math] from this is incorrect, and just as incorrect as evaluating [math]0.\bar{3} × 4 = 1.\bar{3}2[/math]

Or how about this common fallacy:
[math]x = 0.\bar{9}[/math]
[math]10x = 9.\bar{9}[/math]
[math]10x-x = 9.\bar{9}-0.\bar{9}[/math]
[math]9x = 9.0[/math]
[math]x = 1[/math]
This doesn't prove anything because it implies [math]0.\bar{9}[/math] is a number you can just come across, and arbitrarily sets it to x to perform arithmetic on it, opposed to the above example this is an unreal invocation.

>>9394511
This is the dumbest post in the thread, congrats
(You)

>>9394511
...so you're trying to say that [math]\frac{9}{10}+\frac{9}{100}+...!= .999...[/math] is that right?

>in real maths
>in unreal math
>normal number math.
meaningless

>>9394525
You can claim that equates to [math]0.\bar{9}[/math] by invoking an infinite repition of actions += [math]\frac{9}{10n}[/math], but because infinity is not a real number neither is the result.

>>9394534
>because infinity is not a real number neither is the result.

...so now you're saying that ANY infinite series can't equate to a real number?

I think you need to back up and relearn the difference between the rationals and the reals

>>9394534
Sorry, should be:
+= [math]\frac{9}{10^n}[/math]
not
> += [math]\frac{9}{10n}[/math]

>>9394542
No, just that BaseN-1's number can't be a repeating decimal. Base10's-1 = 9 cant be used as a repeating decimal. Base16's-1 = F can't be used as a repeating decimal. Base2's-1 = 1 can't be used as a repeating decimal. This is just a basic rule, and this [math]0.\overline{BaseN-1} = 1[/math] brainlet trap presents the probem that
[math]n = 2; Lim -> \infty; 0.\bar{n}_{base n+1} > 0.\bar{n-1}_{base n}[/math], defining an infinite amount of numbers larger than [math]0.\bar{9}[/math] closer to 1.

>>9394575
Fucking christ Latex is retarded.

n = 2;
Limit to [math]\infty[/math]
[math]0.\bar{n}_{N+1} > 0.\overline{n-1}_{N}[/math]
where subscript Big N is the base numeral system

>>9394575
>This is just a basic rule
Then why did you need to make it up and pull it out of your ass?
>We can use all the numbers in a base in a repeating decimal except for the biggest one because I said so

Argue about bases all you want, but by your own logic [math]\frac{9}{10} + \frac{9}{100} + ... = .999... [/math] in every base bigger than ten.
But changing bases fundamentally doesn't change the value of what you're writing bc it's just different ways to write the same thing.

So it ought to work for base 10 as well if the values don't change.

(1/inf) =/= 0
=>
1 - (1/inf) =/= 1
0.999999... =/= 1

>>9394593
just think about this logicall ok
Think about the normal integers placed on the outside edge of a circle
1-2-3-4-5-6-7-8-9-0~1.
What can be said about them.
They are equidistant? They are each 1 unit away from their nearest neighbor?
ok.
Think about 0.5 decimals now
1.5 - 2.5 - 3.5 - 4.5 - 5.5 - 6.5 - 7.5 - 8.5 - 9.5 - 0.5 ~ 1.5
These too are equidistant from each nearest neighbor

Now lets think in repeating decimals
[math]1.\bar{1} =\frac{1}{9}[math] -
[math]2.\bar{2} =\frac{2}{9}[math] -
[math]3.\bar{3} =\frac{3}{9}[math] -
[math]4.\bar{4} =\frac{4}{9}[math] -
[math]5.\bar{5} =\frac{5}{9}[math] -
[math]6.\bar{6} =\frac{6}{9}[math] -
[math]7.\bar{7} =\frac{7}{9}[math] -
[math]8.\bar{8} =\frac{8}{9}[math] -
[math]0 =\frac{9/0}{9}[math] ~
[math]1.\bar{1} =\frac{1}{9}[math]
These too are equidistant from each nearest neighbor.

[math]0.\bar{9}[math] doesn't exist on the number line in base 10. Do not confuse this as meaning [math]0.\bar{9}[/math] = 1, because [math]0.\bar{9}[/math] doesnt actually exist in base-10.

>>9394646
>= 1, because 0.9¯0.9¯ doesnt actually exist in base-10.

Again, you just made that up.

.9 repeating only uses digits that occur in base 10, so then why the fuck can't you write a number using them.

Again, remember that there aren't unique decimal representations of numbers, more than one decimal can equate to the same number.

>>9394639
Retard

lim n>inf of 1/n equals 0

lim m>inf of (1 - 1/m) equals 1

this is precalc shit

>>9394646
LaTeX is literally ass oh my god.

just think about this logically ok?
Think about the normal integers placed on the outside edge of a circle
1-2-3-4-5-6-7-8-9-0~1.
What can be said about them.
They are equidistant? They are each 1 unit away from their nearest neighbor?
ok.
Think about 0.5 decimals now
1.5 - 2.5 - 3.5 - 4.5 - 5.5 - 6.5 - 7.5 - 8.5 - 9.5 - 0.5 ~ 1.5
These too are equidistant from each nearest neighbor

Now lets think in repeating decimals
[math]1.\bar{1} ||| \frac{1}{9}[/math] -
[math]2.\bar{2} || \frac{2}{9}[/math] -
[math]3.\bar{3} ||| \frac{3}{9}[/math] -
[math]4.\bar{4} || \frac{4}{9}[/math] -
[math]5.\bar{5} ||| \frac{5}{9}[/math] -
[math]6.\bar{6} || \frac{6}{9}[/math] -
[math]7.\bar{7} ||| \frac{7}{9}[/math] -
[math]8.\bar{8} || \frac{8}{9}[/math] -
[math]0.0 ||| \frac{0|9}{9}[/math] ~
[math]1.\bar{1} || \frac{1}{9}[/math]
These too are equidistant from each nearest neighbor

[math]0.\bar{9}[/math] doesn't exist here. The repeating 9 number is merely a byproduct of poor math.

>>9394664
DUDE, you just keep saying that it doesn't exist.

Justify it without this pseudomath shit

Nothing of what you've posted is a proof, its just "well it looks right"

>>9394659
>muh rules
>muh (-1/12)
enjoy ur shitty dogmas retard

>>9394683
And now you're shitting on Complex Analysis bc you're too fucking brain dead retarded to understand Continuation

>>9394672
Everything I've posted is proof. You are just being stubborn.

The only reason you want [math]0.\bar{9}[/math] to exist is because you want to say [math]0.\bar{9} = 1[/math].
but if [math]0.\bar{9}[/math] actually existed as a real number, [math]0.\bar{0}1[/math] would need to exist as well. Now the problem here isn't that you're trying to say [math]0.\bar{9} =
1 && 0.\bar{0}1 = 0 [/math], you are instead flat out denying that latter exists because it's trying to terminate the repeating sequence and place a 1 at the end. The retarded part about that statement though, is trying to equate [math]0.\bar{9} = 1[/math] does the same exact thing; you're trying to terminate the repeating 9's and make it equal to 1 whole. This is a literal paradox and you're a brainlet if you still don't understand why.
Either both [math]0.\bar{9} & 0.\bar{0}1[/math] exist as real numbers, or neither do, and in either case [math]0.\bar{9} \neq 1[/math], because either [math]0.\bar{9} + 0.\bar{0}1[/math] = 1 or [math]0.\bar{9}[/math] doesn't exist as a real number.

So towards proving that the only reason you want [math]0.\bar{9}[/math] to exist is to set it equal to 1, this post alone proves that [math]0.\bar{9}[/math] does not exist for that purpose.

>>9394695
>The only reason you want 0.9¯0.9¯ to exist is because you want to say 0.9¯=10.9¯=1.
>but if 0.9¯0.9¯ actually existed as a real number, 0.0¯10.0¯1 would need to exist as well.
FALSE YOU DUMB NIGGER

Not a proof, the assumptions are wrong.

>is trying to equate 0.9¯=10.9¯=1 does the same exact thing; you're trying to terminate the repeating 9's and make it equal to 1 whole.
You're genuinely confused as to what the unholy fuck "terminate" means

You're wrong, you haven't put the thought into this bc the definitions you're using are flawed.

>>9394705
Everything I've posted is proof. You are just being stubborn.

>>9394718
You don't know what a proof is. You have not posted any proofs. You are just being stubborn.

t. math major

Real numbers have the property that the bounded infinite sum of positive real numbers is a real number because of completeness.

.999... is a real number bc it can be written as a sum of infinitely many real numbers due to the fact that the reals are complete

>>9394748
Define bounded
Define infinite

ITT: autists confusing a number with its notation

>>9394754
These are well defined terms you insufferable idiot.

the partial sums of 9/10 + 9/100 + ... will never be larger than 1 and will never be less than the previous partial sum. It is bounded

>define infinte
unending, you fucking shitstain

>>9394695
>0.0¯1 would need to exist

but it does
0.0¯1 = 0

[math]0.\bar{1}[/math] is not 0.1 or 0.12
[math]0.\bar{2}[/math] is not 0.2 or 0.23
[math]0.\bar{3}[/math] is not 0.3 or 0.34
[math]0.\bar{4}[/math] is not 0.4 or 0.45
[math]0.\bar{5}[/math] is not 0.5 or 0.56
[math]0.\bar{6}[/math] is not 0.6 or 0.67
[math]0.\bar{7}[/math] is not 0.7 or 0.78
[math]0.\bar{8}[/math] is not 0.8 or 0.89
[math]0.\bar{9}[/math] is not 0.9 or 1.0

please stop shouting kids, i'm try to sleep

>>9390205
>he thinks 0.(3)=1/3
And this was supposed to be the smart board too.

>>9394659
oh god my eyes...

[math] \displaystyle
\lim_{n \rightarrow \infty} \frac{1}{n}=0
\\
\displaystyle
\lim_{m \rightarrow \infty} \left ( 1 - \frac{1}{m} \right ) = 1
[/math]

>>9394819
What does 1/3 = ?

>>9394819

>>9390225

>>9394819
You fucking autist:
http://www.wolframalpha.com/input/?i=0.3+repeating

>>9393868
>there are no integers between 2 and 3
>therefore 2=3

>>9394762
So... you're just concatenating and rounding?

>>9394822
You can't accurately represent 1/3 in base 10 decimal notation.

>>9394832
bs
[math]0.\overline{3}
[/math]
works just fine

>>9394819
>>9394832

it's called an infinite series, they're fucking equal

>>9394830
No, I'm summing the terms, no concatenating.

There is no rounding when you look at the partial sums, as you look at later and later partial sums, you're getting closer and closer to the actual answer.

>>9394841
No it doesn't, hence this thread.

>>9394851
prove it

>>9394851
It do work tho

[math] a = .(3) [/math]
[math] a + 3 = 3.(3) [/math]
[math] a + 3 = 10a [/math]
[math] 3 = 9a [/math]
[math] a = \frac{1}{3} [/math]

>>9394856
>hence this thread

>>9394860
idiots shitting in a corner
don't prove why corners exist

>>9394844
So the actual answer is self evidently [math]0.\bar{9}[/math] by that logic.

Tell me [math]senpai[/math], you got an equation to look at a specific latter section of pi that hasn't been calculated? Say the 999 sextillionth digit?

>>9394859
that's circular reasoning, kys
you can't just go 9*0.(3) = 3 when trying to prove 0.(3) = 1/3

>>9394868
good that
9*0.(3) = 2.(6)
then

p.s.
awesome math skills you got there

>>9394868
Do you have any problem with intermantissa prescribed in this thread, and inter-three?
Cause i agree.
[math]\frac{1}{3} \neq 0.\bar{3}[/math]
[math]\frac{1}{3} = [/math] 0.ב
and 1 = 3 × 0.ב

>>9394868
>circular reasoning
All I did was divide both sides by 9

>>9394875
>i invented a squiggle
>all hail the squiggle

>>9394882
3 is just a squiggle if you think about it.

>>9394878
No faggot, you did this
>a+3=10a
you will first have to prove 0.(3)= 1/3 before i can accept that.

>>9394875
I would just use a fraction. Or in practical applications whatever precision i need for the decimal.

>>9394887
how fucking stupid are you, a is [math] .333... [/math]

its obvious that [math] 3.333... [/math] is [math] 10 * .333... [/math] that is unrelated to a being a third.

>>9394884
you are just a shitpost now that I think about it

>>9394882
>>9394884
Seriously you can't tell me the problem with repeating decimals doesn't stem from the sinple fact that they are written with numbers we already have preconceived notions of how to use with arithmetic.
You can't tell me that you don't think [math]0.\bar{3} × 3 = 0.\bar{9}[/math] because [math]0.3 × 3 = 0.9[/math]

the fact of the matter is you're getting the wrong result from [math]0.\bar{9}[/math] cause you are incorrectly assuming [math]0.\bar{3}[/math] is related to the number 3.
You wouldn't say 4 × 3 = 9, would you?
No? Why not?
Cause four has a different symbol and different value.
If you use intermantissa symbols, you no longer get confused about repeating decimals. You begin to understand that
0.ב is not explicitly related to the number 3, no more than the number 4 is related to the number 3, and you then use inter-three to calculate, resulting in real actual answer 1
with no accident of believing [math]0.\bar{9}[/math] = 1 because [math]0.\bar{9}[/math] was never a factor in finding the solution.

Occams razor

what is easier to believe and work with
A) [math]0.\bar{9}[/math] doesn't exist
B) [math]1.0[/math] doesn't exist

[math] .\bar{3} * 3 = .\bar{9} = 1 [/math]

https://youtu.be/XFDM1ip5HdU

>>9394901
how fucking stupid are YOU?
the contested point is that 1/3 =/= 0.(3)
i'm saying that i think 1/3 > 0.(3)
and you're just going "lol 9*0.(3)=3" in your proof

>>9394945
Im solving for a you fucking asshat

I assumed that a = .(3)
which led me to 3 = 9a
im not assuming that 9* .(3) = 3
But I get there legitimately through algebraic manipulation.

Theres absolutely no circular reasoning

>>9394954
>I get there legitimately through algebraic manipulation
>a+3=3.(3)
>a+3=10a
You can't go from former to the later without proving 1/3=0.(3) first, cunt. The fact that i disagree on 1/3=0.(3) means i disagree on 3.(3)=10a too.

>>9395000
dude what the fuck, you dont need 1/3 = .(3) to show that 3.333... = 10* .3333...

where in the holy fuck do you need to apply 1/3 = .(3)

[math] .\bar{3} = \frac{3}{10} + \frac{3}{100} +... [/math]
[math] .\bar{3} * 10 = 3 + \frac{3}{10} + \frac{3}{100} +... [/math]
[math] .\bar{3} * 10 = 3 + .\bar{3} [/math]

not once did I need to use .(3) = 1/3 you fucking autist

>>9395000
Why in the everloving godfuck do you keep writing 0.(3) holy shit dude use the LaTeX its math \bar{3} /math

>>9395029
[math]0.\bar{3}[/math] = 0.ב

>>9395045
your notation is cancer

it doesn't make any of this easier to explain and it only adds symbols for things we can already write

>>9395049
You can write repeating 3 in any way you want but if you use "3" i its base then you're only keeping yourself confused. The single repeating decimal numbers are not related to their first mantissa. You cannot perform finite arithmetic with them based on the 0-9 number line, else you get the wrong answer.
Stop fucking shitposting anime you clod, open your mind and try using intermantissa. Write them as abcdefgh if that helps you.

>>9395054
>You cannot perform finite arithmetic with them based on the 0-9 number line, else you get the wrong answer.

Wrong, I just fucking used it and got the right answer in >>9395029
>>9395054

>open your mind and try using intermantissa.
you arent fucking enlightened you moron, we've understood how repeating decimals have worked for hundreds of years. There's no fucking reason to upend our notation for the retards that fundamentally can't understand it

>>9395054
>you're only keeping yourself confused.
Im not the one who's confused you brainlet fuck.

>>9392311
>Tending to infinity is not a valid solution to a problem

That's the fucking point. IT DOESN'T TEND TO INFINITY DUMBASS. It tends to 1/3.

>>9392311
>calculus is retard tier garbage
lmao

>>9394851
>cites thread full of idiots as evidence

>>9395067
I dont know what mongoloid problem you were attempting to solve here>>9395029
You just wrote out some shit and didn't evaluate anything

http://m.wolframalpha.com/input/?i=%281+%2F+3%29+×+3
Is the amswer [math]0.\bar{9}[/math] or 1.
What does wolfram alpha say, tell me.

I am not inventing a concept with intermantissa, i'm merely addressing a concept that has existed since the late 70s in a mathematical field that extends far beyond your grasp, being computer science and computer engineering.

Modern calculators will not ever return [math]0.\bar{9}[/math] unless they're designed to handle expressions and abstract concepts like infinite limits, and all you'd be able to do with those is merely prove you can somehow construct the number with irrational, unusably unreal mathematics. You could have just as easily scribbled a dick in your notebook and that would have been just as much as a good use of your time than irrationally forcing something to evaluate as [math]0.\bar{9}[/math]

You are unironically the actual brainlet here.

>>9395099
>intermantissa

>>9394918
oh squiggle love me
squiggle give me power
all wiggle in praise of the Squiggle

>>9395106
>[math]0.\bar{9}[/math] = 1

>>9395098
the whole number part in 1.0... is greater than in 0.99... which means 1>0.99...
QED

>>9395110
Crackpot retard alert

>>9394929
>feel better already
you don't exist

>>9395099
fuck thats why you were being so fucking autistic about this shit ok

yeah calculators wont give you an infinite decimal, because theyre fucking calculators retard. Calculators don't give you the sqrt of 2 either.

>http://m.wolframalpha.com/input/?i=%281+%2F+3%29+×+3
>Is the amswer 0.9¯0.9¯ or 1.
>What does wolfram alpha say, tell me.
Wolfram is not the say all end all in mathematics, it can be wrong sometimes. Stop worshiping computers they aren't fucking perfect and they sure as shit don't decide what numbers exist or not.

the limitations of calculators do not affect numbers themselves.

>far beyond your grasp
says the brainlet nigger who cant understand that numbers exist outside computers

>You are unironically the actual brainlet here.
this

>>9395113
>the whole number part

fuck off with your baby math

You are without a doubt a certified math retard if you believe any of the following
>[math]\infty[/math] is a number
>[math]0.\bar{1}_{base2} \geq 0.\bar{9}_{base10} \geq 0.\bar{F}_{base16}[/math]
>there are more than 2 genders

>>9395128
not an argument

>>9395124
>im better at do maff dan wetodded computor an calculatoooooor
What a time to be alive.

>>9395140
and neither is
>the whole number part in 1.0... is greater than in 0.99... which means 1>0.99...

Its simply not rigorous

I know you retards in CS don't understand rigorous proofs, but realize that "looks close enough" doesn't mean its equal

>>9395134

That second one is very interesting.

They would all have to be equal to each other, but they wouldn't appear like it except logically or except when the infinite sum is completed.

It really isn't surprising though, because it is just approaching the same limit at various rates

>>9395146
You're fucking stupid as shit if you think computers are better at infinite series than calc students

>compyuta can do maff parfect
what time to be alive, kys retard. You are in the wrong, t. math major

>>9395147
>Its simply not rigorous
the brainletest post itt

>>9395179
Whine all you fucking want you little brainlet bitch, it still isn't rigorous

>>9395146
>>9395113
>>9395099
>>9395054

>>9395151
Yes, infinity has connotations of time more than just an hugd amount of a thing. Thats why infinity is not a number, and really shouldn't be used in arithmetic or to define a property of numbers.
When i read infinite limit functions, i laugh that they surmise a result. Its the ultimate copout of lazy bullshit and is the complete opposite of "show your work". I could rewrite these functions in programming, but the diffeference there is when I go and run the prorgram i wont actually get a result back.

for(n = 1; n != 0; n++){
x = 1/(10^n);
}
print x;

this program will never print a result. It enters the for-loop and then knows its crunch time and to dedicate all time and energy to evaluating the for loop.
"For n=1, while n is not 0, increment n"
and then it'll calculate what is between the bracket block of { }
I put my print function outside the block cause i want the answer when it has finished calculating rather than a thousand different answers every second, and this print message never occurs though because the program never leaves the loop block. It will take an infinite amount of time to count an infinite amount of increments of n, which means a finite answer cannot be calculated in finite time with aspects of infinity. You can daydream all you want as a mathematician over how the answer might or should look in your imagination, but the fact is you will one day die and no longer be able to imagine anything, and whatever infinite sequence that made up your neurons and imagination woul have come to a definite end.
So yes, infinity and infinite repititions have connotations of time inherently attached to them, and is why the second example was related to the first.

Why does this always create so much discussion? It will always be smaller than 1. If the amount of decimals are infinite it would be an infinitely small amount away from 1.

I hate this thread.

>>9395191
>but the fact is you will one day die and no longer be able to imagine anything, and whatever infinite sequence that made up your neurons and imagination woul have come to a definite end.

the worst argument, "your life is finite so you cant talk about the infinite"
Math isn't limited by physics.

Mathematics is all about abstraction, why the hell can't you abstract infinity, it's no more controversial than the irrationals and those are perfectly consistent.

>>9395203
>infinitely small amount

= 0

>>9395191
But you have no problem with epsilon-delta definitions of limits?

>>9395207
And at which digit of pi does a perfectly round wheel roll upon to complete a single rotation?

Only some things in math can be abstract, but dont try to hijack the field entirely. Math is meant to produce resal results.

>>9395220
>Only some things in math can be abstract
says fucking who

>dont try to hijack the field entirely
Yeah, follow your own advice and don't apply rules to things that you just made up.

>Math is meant to produce real results.
Irrelevant. uncertainty in QM doesn't mean we can't write numbers smaller than Planck's constant

>>9395214
0 is not infinitely small. Its not even small it's just no thing. If i invited you to a party and you offered to bring over food to feed the whole party but said you had a small car so it might take multiple trips to bring all the food over, only for you to never show up at all because you actually had no car, i'd be upset and probably stop talking to you thinking you must be a retard.

>>9395214
no, itll be always larger than 0

>>9395253
No, it will always be larger than or equal to 0

>>9395256
no, if its infinitely small it will always be larger than 0. 0 is nothing.

>>9395266
infinitely small is equivalent to nothing.

if it is infinitely small then nothing is in between it and 0. and then, by the completeness of the Reals, they are equal.

The only bound on [math] \frac{1}{x} [/math] as x goes to infinity is zero, and for every next x+1 it will only ever get closer.

>>9395252
So if you have a half size car, it would take 2 tips to bring all the food over.
A quarter sized car would be 4 trips
A 1/100th car would be 100 trips
[math]\frac{1}{\infty}[/math] sized car would take an infinite amount of trips
Yet 0 car, no car, would take 0 trips. You could complete some amount countable of trips with an infinitesimally sized car, but you coud never complete even 1 single trip with 1 car.

Therefore [math]0.\bar{0}1 > 0.0[/math] and [math]0.\bar{9} < 1.0[/math]
and [math]0.\bar{9} + 0.\bar{0}1 = 1[/math]

>>9395295
Cabt complete 2 trip with no car

>>9395279
>>if it is infinitely small then nothing is in between it and 0
No, an infinitely small amount is between it and 0.

Infinity will always be infinitely close in approach, but never actually there. Its the point of infinity. It makes no sense in logical terms, its fucking infinity.

>>9395252
>If i invited you to a party
no thanks I'm not gay

>>9395301
>Infinity
>approach
no you dufus, it's already there

>>9395321
No, if its already there why would you call it 0.(9) instead of 1?

>>9395295
you're logic is incomplete and its really hard to show why in this tiny ass reply box.

First, >Yet 0 car, no car, would take 0 trips.
No, it would take an undefined number of trips.

you're asking about the sum of [math] \frac{1}{n}, n [/math] times. As n limits to infinity the sum still equals 1. But you're trying to split up the limit by evaluating [math]\frac{1}{n}[/math] before you add them all together. That can't be done in a limit for the same reason that [math] 1^{\infty} [/math] is indeterminate.

If it infinitely approaches 1 then its also infinitely far away from 1.

>>9395335
Dont get all philosophical m8, us autists dont understand that.

>>9395328
why does a dog lick its balls?

Infinity is not the point.
1 is the point.
0.(9) ≠ 1

>>9396629
They can not be at the same point.