/sci/ - Science & Math

[Anonymous No.6924982]

Jacob Barnett officially released the proof of 0.9999... = 1

How can mankind thank him? Nobel prize? Fields medal?

[Anonymous No.6924997]

>>6924982
hopefully with enough money for him to go be autistic somewhere else....

>Also having as much attention and pressure as he has been having from such a young age can't be good for his emotional development

[Anonymous No.6925001]

he will never achieve anything like the rest of the savants.

The true achievers are those who have near-savant-like processing power with normal person creativity and outside of the box thinking abilities

PS:

1/3 = .33
2/3 = .66
3/3 = .99

[Anonymous No.6925013]

>>6925001
Nope, he used tripple intergrals to solve the problem

[Anonymous No.6925072]

Does that mean I need to skip 100 when I count now?

[Anonymous No.6925499]

>>6925072
yes

[Anonymous No.6925529]

0.9... = 1
1+1=2

but

0.9 + 0.9 = 1.8
0.99 + 0.99 = 1.98
0.999 + 0.999 = 1.998

0.9... + 0.9... = 1.9...8

1.9...8 != 2

>MFW

[Anonymous No.6925557]

>>6925529
This is probably bait but,

1.99...8 doesn't exist. it's just 1.99...
which equals 2

[Anonymous No.6925572]

x = 0.9999...
10x = 9.999...
9+x = 9.999...
->
10x = 9+x
->
9x = 9
->
x = 1
->
0.999... = 1

[Anonymous No.6925578]

>>6925557
But I proved it does

>0.9 + 0.9 = 1.8
>0.99 + 0.99 = 1.98
>0.999 + 0.999 = 1.998
>0.9... + 0.9... = 1.9...8

[Anonymous No.6925582]

>>6925572
>x = 0.9999...
>10x = 9.999...
>9+x = 9.999...

The problem with this is that if you say that x = 0.999... AND x = 1 then 9+x becomes (0.9...+0.9...+0.9... +0.9...+0.9...+0.9...+0.9...+0.9...+0.9...)+0.9...

And thats 8.9...1 and not 9.

[Anonymous No.6925591]

>>6925578
No, you didn't. You didn't even really prove anything, you demonstrated a pattern. And that pattern holds for finitely long series of nines after the decimal. .99... is not a finitely long string of nines. It's infinitely long.

.999...8 doesn't make sense for the same reason that .999...0 doesn't make sense. You can't have an infinite string of nines ending with an 8, because then your nines wouldn't be very infinite, would they? The "..." after the nine doesn't mean "a lot more", it means infinite.

[Anonymous No.6925604]

>>6925591
Just because you cant ever reach the 8 doesnt mean it isnt there.

When I cant have an 8 at "point inf" you also cant have inf points (nines) before 8. To me it seems like you are just preferring one side of the argument even if both are perfectly logical.

[Anonymous No.6925610]

>>6925591
There's omega 9s and one 8. Don't you believe in infinity?

[Anonymous No.6925614]

>>6925582
>9+x becomes (0.9...+0.9...+0.9... +0.9...+0.9...+0.9...+0.9...+0.9...+0.9...)+0.9...
WAT

[Anonymous No.6925618]

>>6925604
>Just because you cant ever reach the 8 doesnt mean it isn't there.
That's exactly what it means. The problem with people who argue .999...=(1-.000...1) is that they think .999...=.999...0.

But that's wrong. There's no zero there. It's just nines forever. The reason you get an 8 at the end of your sequence is because that's what you get when you add that last nine to itself. But in .999..., there is no "last nine". It's nines all the way down, which means the number you get when you multiply by two is also nines all the way down.

[Anonymous No.6925623]

>>6925614
Are you stupid?

9 = 1+1+1+1+1+1+1+1+1
1 = 0.9...
therfore
9 = (0.9...+0.9...+0.9... +0.9...+0.9...+0.9...+0.9...+0.9...+0.9...)

So when x = 0.9... then
>9+x
(0.9...+0.9...+0.9... +0.9...+0.9...+0.9...+0.9...+0.9...+0.9...)+0.9...

[Anonymous No.6925629]

>plebs still using base 10

[Anonymous No.6925644]

>>6925618
Thats wrong, Ill write it up in another way, maybe you can understand then:

n->inf
8/(10^n) = 0.0..8
(10^n-1)/(10^n) = 0.9...

8/(10^n) + (10^n-1)/(10^n)

We can CLEARLY see that there is an 0.9... AND an 0.00..8

[Anonymous No.6925649]

>>6925644
except one thing. [n->inf]8/10^n=0

[Anonymous No.6925654]

>>6925649
[n->inf]8/10^n is 0 because one can argue that the number 0.0..8 gets infinitly small and therefore converges in 0 but we are talking about if EXISTS or not, we are NOT talking about its size (or convergence).

Just because your penis is infinitly small doesnt mean it doesnt exist. You wouldnt like to be called the guy without a dick just because your penis is inf small.

[Anonymous No.6925668]

>>6925654
But the sequence only ends in 8 if n is an integer. Why is an infinity an integer? Is it so you can count it on your fingers you fucking idiot?

[Anonymous No.6925669]

THIS WHOLE THREAD

[Anonymous No.6925675]

>>6925629
In base 2:
0.111... = 1

In hexadecimal:
0.EEE... = 1

[Anonymous No.6925682]

>>6925668
How can you go from n to inf if inf isnt an integer, idiot?

Its not like n somehow becomes a goat on its way to inf.

[Anonymous No.6925689]

>>6925675
you mean 0.FFFFF...

[Anonymous No.6925691]

>>6925689
No, I meant pentadecimal :^)

(...thanks for pointing it out. My bad.)

[Anonymous No.6925699]

>>6925529
0.9... = 1
1 + 1 = 2
0.9 + 0.9 = 1.8
0.9... + 0.9... = 1.9....8

0.9 + 0.9 + 0.9 + 0.9 + 0.9 + 0.9 = 5.4
0.9.... + 0.9.... + 0.9.... + 0.9.... + 0.9.... + 0.9...= 5.9....4 = 6 ?
?

[Anonymous No.6925700]

>>6925529
<span class="math">\displaystyle
1 = \frac{3}{3} = 3 \cdot \frac{1}{3} = 3 \cdot 0.\overline{3} = 0.\overline{9}
[/spoiler]

[Anonymous No.6925705]

>>6925699

[Anonymous No.6925706]

>>6925557
>"I deem it insignificant, therefore it does not exist"
>maths
>physicists

[Anonymous No.6925712]

>>6925675
This

>>6925682
Because we're talking about limits and those are typically continuous. But fine, even keeping it at integers. As a fun little experiment, let's keep the number system in base ten but the base of your exponential 4. So it's [n->inf]x/4^n. But keeping it base ten just for giggles. make x, let's say, 4. Well,[n->inf]4/4^n. is exactly the same as [n->inf]1/4^n. Why? because 4/4^2=1/4, and infinity+1 is infinity. The number in the numerator is arbitrary. As long as it's a constant, it limits to zero.

Same with 8/10^n. The number in the numerator is arbitrary. It limits to zero. And the thing about infinitely repeating numbers is that they are DEFINED by their limits because that's what infinity fucking means. It doesn't mean "a lot". You can't just bundle it up and set it in the middle of a fucking number. You can't treat it like a quantity. It's infinite.

>>6925706
the reason it doesn't exist is not because it's insignificant. The reason it doesn't exist is because IT ACTUALLY DOES NOT EXIST.

[Anonymous No.6925714]

>>6925712
Infinity doesn't exist.

[Anonymous No.6925724]

>>6925714
as a number no you're right it doesn't, if it did then 0.999...1 or other such bullshit would be valid.

[Anonymous No.6925728]

>>6925712
>IT ACTUALLY DOES NOT EXIST.

So your argument is that 8 cant be there because infinity doesnt exist?

But wouldnt that mean that 8 IS there because, if inf doesnt exist, we would end up with addint "the last 9s"

[Anonymous No.6925729]

>>6925724
>as a number no you're right it doesn't
So if I index each of the terms, we'll never hit infinity. Then we can put an 8 at the end.

[Anonymous No.6925741]

>>6925729
If it never "hits" infinity then it never "hits" 8

>>6925728
No, having a last nine would require infinity to be a number. It would just be "infinity nines, ending with a nine". But the fact that they go on infinitely means that for any nine you select, there's a nine after it. You cant multiply any 9 and get that 8 because there's always another 9 behind it to carry the one

[Anonymous No.6925748]

>>6925741
>No, having a last nine would require infinity to be a number. It would just be "infinity nines, ending with a nine". But the fact that they go on infinitely means that for any nine you select, there's a nine after it. You cant multiply any 9 and get that 8 because there's always another 9 behind it to carry the one

Ah, now I understand what you mean.

You are doing the following: You write up the fist infinite 9s, then the second, and when you try to add them, coming to the conclusion that when you add the last one there is always one more to add so there is never an 8.

What I am doing is: I write up 0.9 and then I add 0.9 becoming 1.8, then I write up 0.99 + 0.99 etc. So I am constructin the 9s step by step.

Since you can never write inf 9s in the first place your argument is shit to begin with and my constructive approach makes way more sense, putting an 8 at the end of inf 9s.

[Anonymous No.6925762]

>>6925748
>Since you can never write inf 9s in the first place your argument is shit to begin with
Your failure to conceptualize an infinite sequence does not mean the logic is not sound.

>putting an 8 at the end of inf 9s.
>at the end
>inf 9s
>end
>inf

Whatever, man. It's like describing the color blue to a blind person.