/sci/ - Science & Math

If you claim "1 != 0.999...", then no-one gets baited, because most of the retards here actually think that it's true.

On the other hand, if you claim "1 = 0.999...", then you will get lots of angry replies.

t. A guy with MSc in baitology

>>9323707

>>9323608
[math] \displaystyle
1 = \frac {3}{3} = 3 \cdot \frac {1}{3} = 3 \cdot 0. \bar{3} = 0. \bar{9}
[/math]

>>9323880
prove that [math]\frac{1}{3} = 0.\bar{3}[/math]

>>9323608
okay you cuck

let' say that you have a variable x, where
x = 0.999....
then
10x = 9.999...
so what happens if you do
10x - x = 9.999... - 0.999...
you get
9x = 9
x = 1
but x = 0.999...
QED
>>9323891

nice bait fucker

>>9323891
[math] \displaystyle
\begin{align*}
\frac{1}{3} = \left (\frac{3}{10} + \frac{1}{30} \right )
&= 0.3 + \frac{1}{30} \\
= 0.3 + \left ( \frac{3}{100} + \frac{1}{300} \right )
&= 0.33 + \frac{1}{300}\\
= 0.33 + \left ( \frac{3}{1000} + \frac{1}{3000} \right )
&= 0.333 + \frac{1}{3000} \\
= 0.333 +\left ( \frac{3}{10000} + \frac{1}{30000} \right )
&= 0. \underset{n}{ \underbrace{3333}} + \frac{1}{3 \underset{n}{ \underbrace{0000}}} \\
&\vdots
\end{align*}
\\ \displaystyle
\Rightarrow 0.\overline{3} = \frac{1}{3}
[/math]

>>9323891
3 thirds = 1 whole
0.3¯ is not 1/3

>>9323891
[math]
\begin{align*}
&0.33... \\
3 & \overline{)1_0 \;\;\;\;\;} \\
& \;\;\;\underline{9} \\
& \;\;\;1_0 \\
& \;\;\;\;\; \underline{9} \\
& \;\;\;\;\; 1 \;\; etc
\end{align*}
[/math]

>>9323608
>t. I never took analysis

brainless morons on /sci/ don't realize that all induction arguments are flawed because they use infinity which isn't a number.

>>9324854
pls reply to my bait

>>9324355
the simplest proofs are the best i like this one

>>9324854
infinity is made up of numbers and only numbers. therefor its a numbers. stop being retarded

>>9323608
Wrong by definition.

>>9324865
>infinity is made up of numbers
wrong.
> and only numbers
Also wrong.

>>9325315
prove your statement, because its absurd.

>>9323608
Take analysis brainlet.
[math]0.\overline{9} \equiv \sum_{k=1}^{\infty} 9 \cdot \left( \frac{1}{10} \right)^k[\math]
[math]0.\overline{9} = \sum_{k=0}^{\infty} 9 \cdot \left( \frac{1}{10} \right)^k - 9[\math]
[math]0.\overline{9} = 9 \left( \frac{1}{1-\frac{1}{10}} \right) - 9[\math]
[math]0.\overline{9} = 1[\math]

>>9323608
>>9326308
Crap I did it again
>[math]0.\overline{9} \equiv \sum_{k=1}^{\infty} 9 \cdot \left( \frac{1}{10} \right)^k[/math]
>[math]0.\overline{9} = \sum_{k=0}^{\infty} 9 \cdot \left( \frac{1}{10} \right)^k - 9[/math]
>[math]0.\overline{9} = 9 \left( \frac{1}{1-\frac{1}{10}} \right) - 9[/math]
>[math]0.\overline{9} = 1[/math]

>>9323608
I will never understand how anyone can not see how 1 != 0.999...

>>9323608
we can prove that a 0.9999.. part of a line contains all points within that line. relaying on the fact that 0.999... > 1 - 1/(10^n) for all positive n.

>>9326313
[math]\sum_{k=0}^{\infty} 9 \cdot \left( \frac{1}{10} \right)^k = 9 \left( \frac{1}{1-\frac{1}{10}} \right)[/math]

Wrong. This is an approximation. Just like [math]0.\overline{9} = 1[/math] is an approximation and not actually true.

>>9326534
This does not prove anything.

>>9325312
What definition? Because such a definition contradicts the fundamental axioms of mathematics.

>>9324355
>>9324341
These are false. Once again you use approximations.

>>9323608
(you)

>>9326693
Arguments in this post: 0

>>9326697
good job, fast and funny reply, here's one more
(you)

>>9326703
Arguments in this post: 0

>>9324336
10 * .999... equals slightly less than 9.999...

All of your "proofs" are circular.

>>9323608
https://en.wikipedia.org/wiki/0.999......

>>9326732
>wikipedia
Not an argument.

>>9326732
kek

>>9326658
Incorrect. You don't understand how infinite sums are defined.
[math]\sum_{k=0}^{\infty} a_k \equiv \lim_{n\to\infty} \sum_{k=0}^{n} a_k [/math]
Now, because I'm assuming your next argument will be something along the lines of "b-but limits to infinity are approximations!!!", I will share with you the definition of a limit of a sequence.
[math]\lim_{x\to\infty} x_n = x[/math] means that for each real number [math]\epsilon >0[/math], there exists a natural number [math]N[/math] such that, for every natural number [math]n\geq N[/math], we have [math]|x_{n}-x|<\epsilon[/math].

>>9326729
Lol prove it fag since everyone else needs to prove obvious shit to you. You literally can't.

>>9324865
This. Cantor and transfinitists can get wrecked.

0.99999 =/= 1 for the same reason that almost-surely doesn't mean the same as certain in probability.

0.99999... is the probability of a random variable with a continuous uniform distribution between 0 and 1 being rolled once and having a value that is transcendental.
0.000000...1 is the probability of that random variable being rolled and having a value that is non-transcendental

>>9324854
This. Cantor and transfinitists can get wrecked.

>>9326671
>What definition?
The definition of the Real numbers.
1=0.999..., inside ZFC with the standard definition of the reals, by definition.

>>9327510
You are uneducated about measure theory, learn what the Lebesgue measure is.
Stop giving your opinions about things you do not understand.

>0.99999 =/= 1 for the same reason that almost-surely doesn't mean the same as certain in probability.
Almost surely means that the set has measure ZERO, aka Probability ZERO.
Educate yourself.

>>932485>>9325838
If you argue outside of ZFC good for your. Whatever you say is worthless.

>>9326477
Because these people have seen a university from inside, which you clearly haven't.

>>9327510
actually anons, this will convince me if it had prof in it

>>9327883
There is no proof, since it is non sense.
If you are interested read about the Lebesgue Measure.

What he is trying to articulate is that the Interval [0,1] has the same "size" as the Interval [0,1) (size has a very specific meaning here, see Lebesgue Measure).
If one had no clue about analysis one might wrongly suspect that the difference in size between [0,1] and [0,1) is not zeros in other words he assumes that a point has width.
From that he is concluding that 0.99...=1.
The flaw in the logic is obvious and there is no serious mathematics behind it.

He puts all that in the Language or Probability theory (Which is exactly the same as my more analytic explanation above) and uses terms such as "almost certain", which are rigorously defined but easily misunderstood if you have no clue about measure Theory.

>>9326679
>approximations
Where in >>9324341 is an approximation?
Every line is 1/3, exactly.

>>9326679
>Division is an approximation

>>9330306
>>9330749
It's not my fault that you are too stupid to see the obvious.

The endless sequence never reaches its target and thus is an approximation.

Yes, some rational numbers cannot be expressed precisely in systems with certain bases. Didn't you learn this in middle school?

>>9331297
>approximation
please point out the line that is an approximation
perhaps #1?
#88?
#986867447?

>>9331309
Hey stop shitting up multiple threads.

1/3 = (3/10 + 1/30) = 0.3 +
(3/100 + 1/300) = 0.33 +
(3/1000 + 1/1000) = 0.333 +
... = 0.333....

3/3 = (9/10 + 1/10) = 1 +
(9/100 + 1/100) = 1.1 +
(9/1000 + 1/1000) = 1.11 +
... = 1.111...

3/3 = 0.999... = 1 = 1.111... > 1

>>9331337
> i have nothing useful to say

>3/3 = 1.111...
top kek

>>9323608
1 = 10/10 = 9/10 + 1/10 = 9/10 + 9/100 + 1/100 = 9/10 + 9/100 + 9/1000 + 1/1000 + ... = 0.999...

>>9331309
I do not see that many lines. Here is where the approximation happens.

>>9331427
how exactly?
since at every line before that the value is exact? No drifting of the sum value.

show your
[math]
\displaystyle
lim_{n \rightarrow \infty}
[/math]

>>9331427
Any infinite repeating single digit decimal should be undefined. 1/3 in a decimal based calculator should spit out "ERR" if not just [math]\frac{1}{3}[/math]

>>9331462
how about[math] 0.1_3 [/math]

>>9323608
>1 + 1 = 2

>>9331457
The approximation used is:
[math]\underset{n \to \infty }{lim}\frac{1}{3\underset{n}{\underbrace{0}}}=0[/math]

[math]\underset{n \to \infty }{lim}\frac{1}{n}\neq 0[/math] because [math]\underset{n \to \infty }{lim}\frac{n}{n}=1\neq 0[/math] however [math]\underset{n \to \infty }{lim}0\cdot n=0[/math].

Why do you people not see the obvious and are aware that limes is an approximation?

>>9331744
I want to mention that I do not use lim properly in this post. The lim I use in the bottom is not actually the lim but the true value to showcase why limes is an approximation.

Because after some other posts ITT I am not sure if everyone here is intelligent enough to understand my post if I do not point this out.

>>9331744
[math]
\lim_{n \to \infty } \frac{1}{n} \neq 0
[/math]
[citation needed]

>>9331744
https://www.wolframalpha.com/input/?i=lim+1%2Fn+as+n-%3Einfinity

>>9331772
>>9331812
Illiterate idiots. See >>9331751

>>9331470
[math]\frac{1}{9}

>>9331470
[math]\frac{1}{9}[/math]

>>9331751
>desperate handwaving

>>9323707
look what you've done

Found a near calculator app.

Funnily enough it stores results only up to 15 decimal places of accuracy, much like most computer numbers when stored only as double.
If i remove " + 0.000 000 000 000 1", the answer becomes 0.999 999 999 999 8[...], or if I add 0.000 000 000 000 01, the answer becomes 1.

if there were no limit by using a custom string function to produce the answer rather than storing it as a double, the answer would be 0.999[...]8 which doesn't even equal 0.999[...]9 much less 1.0

therefore it is an abstract, non-mechanical concept to say 0.999[...]9 = 1

it is like the observation paradox in particle waves. As soon as you observe an end to the repeating 9's, a finite limit no matter how gargantuan, you change the behaviour of the outcome. Unobserved, you can do no practical arithmetic on the value. Observed with a limit, there will be a final 9 where ([math]frac\{9}{10n}[/math]
+ [math]frac\{1}{10n}[/math]) can satisfy as true to return 0 decimal and equate to 1.

When people say 0.999[∞ ] = 1, what they actually are saying is 0.999[n] = 1, where n is any huge finite number too inefficient to describe yet less than infinity. I would say "infinity - 1" but that doesnt make sense as that would too violate the concept of being unable to perform useful arithmetic with infinity.

The true issue lies within the base counting system which makes it incalculable. For example in Base-16, the concept would be written as 0.FFF[∞ ] , but that would not equate to 0.999[∞ ] in Base-10 as there are 5 more orders of accuracy beyond 9 in Base-16, that is to say [math]0.F_16[/math] is closer to 1.0 than [math]0.9_10[/math] is close to 1.0 - meanwhile Base-3, would have 0.222[∞ ] as the representation for that concept, but translated from base-10's [math]\frac{1}{3}[/math]×3 equation, would not equal to base-10's 0.999[∞ ] result, as [math]0.333(∞ )_10[/math] = [math]0.1_3[/math], so that 0.1 + 0.1 + 0.1 = 1.0.

Found a neat calculator app.

Funny enough it stores results only up to 15 decimal places of accuracy, much like most computer numbers when stored only as double.
If i remove " + 0.000 000 000 000 1", the answer becomes 0.999 999 999 999 8[...], or if I add 0.000 000 000 000 01, the answer becomes 1.

if there were no limit by using a custom string function to produce the answer rather than storing it as a double, the answer would be 0.999[...]8 which doesn't even equal 0.999[...]9 much less 1.0

therefore it is an abstract, non-mechanical concept to say 0.999[...]9 = 1

it is like the observation paradox in particle waves. As soon as you observe an end to the repeating 9's, a finite limit no matter how gargantuan, you change the behaviour of the outcome. Unobserved, you can do no practical arithmetic on the value. Observed with a limit, there will be a final 9 where ([math]frac\{9}{10}_n[/math]
+[math]frac\{1}{10}_n[/math]) can satisfy as true to return 0 decimal and equate to 1.

When people say 0.999[∞ ] = 1, what they are actually saying is 0.999[n] = 1, where n is any huge finite number too inefficient to describe yet less than infinity. I would say "infinity - 1" but that doesnt make sense as that would too violate the concept of being unable to perform useful arithmetic with infinity.

The true issue lies within the base counting system which makes it incalculable. For example in Base-16, the concept would be written as 0.FFF[∞ ] , but that would not equate to 0.999[∞ ] in Base-10 as there are 5 more orders of accuracy beyond 9 in Base-16, that is to say[math]0.F_1_6[/math]is closer to 1.0 than[math]0.9_1_0[/math]is close to 1.0 - meanwhile Base-3, would have 0.222[∞ ] as the representation for that concept, but translated from base-10's[math]\frac{1}{3}[/math]×3 equation, would not equal to base-10's 0.999[∞ ] result, as0.333(∞)[math]_1_0[/math]=0.1[math]_3[/math], so that 0.1 + 0.1 + 0.1 = 1.0.

>>9331915
Fuck me i would love to know why this bullshit isn't formatting properly.

>>9331918
for starters it's not frac\{9}{10}_n
but \frac{9}{10}_n

>>9323608
...999.0 = -1.0

>>9331915
>When people say 0.999[∞ ] = 1, what they are actually saying is 0.999[n] = 1, where n is any huge finite number too inefficient to describe yet less than infinity.

nah, inf is inf, ooohit'sbig doesn't cut it

>>9331915
>>9331918
it is like the observation paradox in particle waves. As soon as you observe an end to the repeating 9's, a finite limit no matter how gargantuan, you change the behaviour of the outcome. Unobserved, you can do no practical arithmetic on the value. Observed with a limit, there will be a final 9 where ( [math]frac\{9}{10}[/math]
+[math]frac\{1}{10}[/math] +...) can satisfy as true to return 0 decimal and equate to 1.

When people say 0.999[∞ ] = 1, what they are actually saying is 0.999[n] = 1, where n is any huge finite number too inefficient to describe yet less than infinity. I would say "infinity - 1" but that doesnt make sense as that would too violate the concept of being unable to perform useful arithmetic with infinity.

The true issue lies within the base counting system which makes it incalculable. For example in Base-16, the concept would be written as 0.FFF[∞ ] , but that would not equate to 0.999[∞ ] in Base-10 as there are 5 more orders of accuracy beyond 9 in Base-16, that is to say[math]0.F_{16}[/math]is closer to 1.0 than[math]0.9_{10}[/math]is close to 1.0 - meanwhile Base-3, would have 0.222[∞ ] as the representation for that concept, but translated from base-10's[math]\frac{1}{3}[/math]×3 equation, would not equal to base-10's 0.999[∞ ] result, as [math]0.333(∞)_{10}[/math]= [math]0.1_3[/math], so that 0.1 + 0.1 + 0.1 = 1.0

So
[math]0.111_2[/math]... ≠
[math]0.222_3[/math]... ≠
[math]0.999_{10}[/math]... ≠
[math]0.FFF_{16}[/math]...
Since
[math]0.1_2[/math] ≠
[math]0.2_3[/math] ≠
[math]0.9_{10}[/math] ≠
[math]0.F_{16}[/math]

And if none of these numbers equal each other, how could they all equal 1?

>>9331942
[math]0.1_2[/math] = [math]\frac{1}{2}_10[/math]
[math]0.2_3[/math] = [math]\frac{2}{3}_10[/math]
[math]0.9_{10}[/math] = [math]\frac{9}{10}_10[/math]
[math]0.F_{16}[/math] = [math]\frac{15}{16}_10[/math]

>>9331942
[math]0.1_2[/math] = [math]\frac{1}{2}_{10}[/math] = 0.50
[math]0.2_3[/math] = [math]\frac{2}{3}_{10}[/math] = 0.666...
[math]0.9_{10}[/math] = [math]\frac{9}{10}_{10}[/math] = 0.90
[math]0.F_{16}[/math] = [math]\frac{15}{16}_{10}[/math] = 0.9375

>>9331942
>how could they all equal 1?

they just do

>>9331942
>how could they all equal 1?

they just do

>>9331942
To be fair, [math]0.222..._3[/math] rounding or whatever to [math]1.0_3[/math], isn't the same as [math]0.999..._{10}[/math] to [math]1.0_{10}[/math]

The value of [math]1_{10}[/math] is [math]\frac{1}{10}[/math] to equal "10", while the value of [math]1_3[/math] is [math]\frac{1}{3}[/math] to equal "10".

While [math]0.2_3[/math] isn't equal to [math]0.9_{10}[/math], an infinite series would put both equally as close to their relative values of 1.
[math]0.222..._3[/math] = [math]1_3[/math] × [math]10_3[/math] = [math]10_3[/math] = [math]3_{10}[/math]

[math]0.999..._{10}[/math] = [math]1_{10}[/math] × [math]10_{10}[/math] = [math]10_{10}[/math] = [math]100_{3}[/math]

Saying that [math]1_{10}[/math] isn't the same number as [math]1_{3}[/math]

>>9331996
>[math]1_{10}[/math] ≠ [math]1_3[/math]

>>9323608
Okay then what is 1.0 - 0.999... ?

[math]0_3[/math] = [math]0_{10}[/math]
[math]0.1_3[/math] = [math]0.333..._10[/math]
[math]0.2_3[/math] = [math]0.666..._10[/math]
[math]1_3[/math] = [math]1_{10}[/math]
[math]2_3[/math] = [math]2_{10}[/math]
[math]10_3[/math] = [math]3_{10}[/math]
[math]11_3[/math] = [math]4_{10}[/math]
[math]12_3[/math] = [math]5_{10}[/math]
[math]20_3[/math] = [math]6_{10}[/math]
[math]21_3[/math] = [math]7_{10}[/math]
[math]22_3[/math] = [math]8_{10}[/math]
[math]30_3[/math] = [math]9_{10}[/math]
[math]31_3[/math] = [math]10_{10}[/math]

>>9332014
1 - 0.999... = ω
yet somehow 0.999... = 1
so ω × ∞ = 0

>>9331996
1 is 1 in any base
wtf are you rambling about

>>9332036
>0.999[∞] is a number
>but ω is just a different way to write 0
Thanks for being gay, mathematicians.

>>9332019
30_3 ?
base 3 only uses digits 0,1,2 retard

>>9332041
Oh fuck my bad.
[math]100_3[/math] = [math]10_
{10}[/math]

>>9332072
Should be 9.
[math]100_b = b^2\\
10_b = b^1\\
1_b = b^0\\
0.1_b = b^-1
[/math]
etc, for any base b

>>9332072

Cant let this mathfacts thread diem there's too lany mathfacts here.

>>9331462
>what even is a Cauchy sequence

>>9323707
obviously you haven't studied enough

⤶day going well, then suddenly: this thread

>>9333157

>>9323608
yall mofos need to take analysis before spouting bullshit like this

>>9332014
0.0...1 obviously
what are you retarded?