/sci/ - Science & Math

[Anonymous No.12232230]

0.99999... != 1
lemme explain. suppose you have 1 infinitely large program, so an infinite number of lines of code. you are doing a file transfer from 1 infinitely large storage to another but 1 conditional statement the program relies on fails to transfer so you have 0.99999... of the program transferred. the program therefore does not operate.
0.99999... != 1
qed

[Anonymous No.12232246]

>>12232230
But 0.999... of the program is the whole program.

[Anonymous No.12232254]

>>12232246
except it's missing 0.0000...1 of it and doesn't run.

[Anonymous No.12232263]

>>12232230
>thinking ∞-1 is a meaningful statement

infinity is bigger than you think

[Anonymous No.12232273]

>>12232254
>0.0000...1
No such thing

[Anonymous No.12232280]

∞ - 1 = ∞
So you have equal amounts of code in both programs. Into the trash your proof goes.

[Anonymous No.12232291]

>>12232273
yes there is.
1-0.99999... = 0.0000...1

>>12232280
clearly that is wrong as this thought experiment shows. math has been wrong for all these years until my proof by shitpost on this laotian finger painting forum.

[Anonymous No.12232300]

>>12232291
>yes there is.
>1-0.99999... = 0.0000...1
Wrong.

[Anonymous No.12232337]

>>12232300
nope, infinitely large program missing 1 line of code is that.

[Anonymous No.12232346]

>>12232337
Wrong

[Anonymous No.12232348]

>>12232291
Two proofs 0.0000...1 = 0:

Via Cauchy sequences:

0.000...1 = (0.1, 0.01, 0.001,...) = lim as n-> inf of 1/10^n = 0

Via Dedekind cuts:

Assume to contradiction that x is a rational number such that

0 < x < 0.000...1

0.000...1 is less than any finite string of 0s followed by a 1 so for any natural number n

0 < x < 0.000...1 < 1/10^n

x < 1/10^n

10^n < 1/x

n < log(1/x)

Let n = ceiling(log(1/x))+1

ceiling(log(1/x))+1 < log(1/x)

This is a contradiction, so x does not exist and 0.000...1 = 0.

[Anonymous No.12232364]

>>12232348
>lim as n-> inf of 1/10^n = 0
that is a limit so it approaches 0 but isn't true 0.

>Assume to contradiction that x is a rational number such that
>0 < x < 0.000...1
this x dne to begin with. 0.000...1 is the smallest possible number.

[Anonymous No.12232367]

IT IS 1 BY DEFINITION YOU IMBECILES! THERE IS NO 0.9999...

[Anonymous No.12232371]

>>12232364
*before 0.

[Anonymous No.12232388]

>>12232364
>that is a limit so it approaches 0 but isn't true 0.
No, the limit is 0. What isn't 0 is 1/10^n for any natural number n. Welcome to elementary school math.

>this x dne to begin with.
Then 0.001... = 0 by construction of the reals.

>0.000...1 is the smallest possible number.
Yes, 0.

[Anonymous No.12232413]

>>12232230
Idk man I ate .999.. bagels today

[Anonymous No.12232566]

>>12232348
>Via Cauchy sequences
All this shows is that 000...1 approaches 0, not that it is actually equal to 0. You may say that it is *effectively* equal to 0, but when talking with exact precision as mathematicians should, 0.000...1 merely rounds down to 0 in physical circumstances. It does not actually exactly equal 0.
>Via Dedekind cuts
There is no fraction between 0.000...1 and 0. Sure. 0.000...1 is still not equal to 0, onefag. Doesn't prove anything.

Now to thoroughly BTFO you drooling retard onefags, I'll provide two proofs that 0.999... does not equal 1.

Limiting behavior
[eqn]0.9<1\\0.99<1\\0.999<1\\\ldots[/eqn]Now by examining the behavior of this sequence as it extends to infinity, we see a uniform pattern here. That they are NEVER equal, and that 0.999... is ALWAYS less than 1. Hence, [math]0.\overline{9}\neq1[/math].

Proof by contradiction
[eqn]1-0.\overline{6}=0.\overline{4}\\0.\overline{9}-0.\overline{6}=0.\overline{3}\\0.\overline{4}\neq0.\overline{3}\\0.\overline{9}\neq1[/eqn]

[Anonymous No.12232586]

If 0.999...99 = 1, then what's stopping 0.999...98 = 0.999...99?
You may as well just say 1 = 0 and be done with it.

[Anonymous No.12232590]

>>12232586
>0.999...98 = 0.999...98?
In truth, I was the retard.

[Anonymous No.12232597]

>>12232566
>All this shows is that 000...1 approaches 0
0.000...1 is not a function, it doesn't "approach" anything.

>not that it is actually equal to 0
It shows that it's exactly equal to 0, look up the construction of the reals via Cauchy sequences. Do you have an actual argument?

>There is no fraction between 0.000...1 and 0.
Then they are equal. Thanks for playing.

>Now by examining the behavior of this sequence as it extends to infinity, we see a uniform pattern here.
Yes, a pattern for finite amounts of 9s, not a proof of anything. This tells us nothing about infinite 9s. And I can easily show that 0.999... is not less than 1 by a proof similar to the one you failed to counter.

Assume to contradiction that x is a rational number such that

0.999... < x < 1

0.999... is greater than any finite string of 9s so for any natural number n

1-1/10^n < 0.999... < x < 1

1-x < 1/10^n

10^n < 1/(1-x)

n < log(1/(1-x))

Let n = ceiling(log(1/(1-x)))+1

ceiling(log(1/(1-x)))+1 < log(1/(1-x))

This is a contradiction, so x does not exist and 0.999... = 1.

[Anonymous No.12232601]

>>12232586
>If 0.999...99 = 1,
0.999...99 is not a thing at all, and thus doesn't equal anything. Only 0.999... is.

[Anonymous No.12232604]

>>12232586
>You may as well just say 1 = 0 and be done with it.
there is no end to .999...
you cant flip the "last 9" to an eight
even if you could, you couldnt ever remove infinitely many 9's before it to reach 0

[Anonymous No.12232686]

>>12232586
>You may as well just say 1 = 0 and be done with it.
that's true though, in mod 1

[Anonymous No.12232756]

>>12232291
>1-0.99999... = 0.0000...1
1-1=0

[Anonymous No.12232761]

>>12232566
>let's pretend finite is infinite

[Anonymous No.12232771]

>>12232586
>0.999...99 = 1, then what's stopping 0.999...98 = 0.999...99
nothing, anything after the ellipsis is just n*10^-inf=0
0.999...98 = 0.999...99 = 0.9... = 1

[Anonymous No.12232822]

>>12232230
It’s literally close enough, don’t be such a fucking sperg about it

[Anonymous No.12232840]

>>12232586
>0.999...98 = 0.999...99

Because those are nonsensical notations that you made up. Show one example of any one ever using repeating digit notation in that way besides you.

[Anonymous No.12232848]

>>12232230
Fuck you, stop shitting up the post with mathematical illiteracy and take a calc 2 course.

[Anonymous No.12233360]

>>12232230
This is a traversal problem like Hercules chasing the tortoise.

[Anonymous No.12233421]

>>12233360
You're an idiot.

[Anonymous No.12233429]

>>12233360
it was Achilles, retard

[Anonymous No.12233482]

>>12232230
No, the cardinality of the string of nines is equal to the infinite number of nines.
.99...=1
>x=.99...
>10x=9.99...
>10x-x=9.99...-.99...
>9x=9
>x=1

[Anonymous No.12233501]

>>12232848
>let me define 0 to be 1,therefore 0 = 1!!!! 1!1!!
Fuck off

[Anonymous No.12233504]

The very fact that this is even an issue, and furthermore there are so many 'proofs' that either support or debunk this, means to me that our implementation of mathematics is *broken*.

[Anonymous No.12233511]

>>12232230
Your face = your butt
lemme explain. suppose you have 1 infinitely gay face, so an infinite homosexual faggotry. you are doing a shit cum bomb from 1 infinitely gay face to another but 1 giant poopy asshole and the face relies on poop to transfer so you have 100% of the poop blasted onto the butt, the butt then has to be the face.
your face is your butt and your butt is your face
qed.

[Anonymous No.12233642]

>>12232230
This explains the situation. Talk given in layman terms by a guy with a degree in Pure Mathematics, and a doctorate in Algebraic Number Theory
https://vimeo.com/24014769
Thank me later.

[Anonymous No.12233652]

>>12233501
0.9...=0!!!

[Anonymous No.12233657]

>>12233504
>means to me that our implementation of mathematics is *broken*.
it actually means that the typical human is incredibly poorly suited for thinking, as your comment shows

[Anonymous No.12233945]

>>12233504
>The very fact that this is even an issue
it's not an issue

[Anonymous No.12234349]

btfo

[Anonymous No.12234363]

>>12232566
Your proof is shit.
You have to show that your pattern is true for any rational number, but you have instead just assumed that it's true until infinity.
You can't assume your conclusion.

[Anonymous No.12234432]

>>12232566
>Now by examining the behavior of this sequence as it extends to infinity, we see a uniform pattern here.
what "uniform pattern" are you talking about? what is the exact relation between 0.999... and the lines 0.9, 0.99, 0.999,... ? can you prove that this relation preserves strict inequalities?

[Anonymous No.12235141]

>>12234349
Argument 3 is nonsensical.

“...” is used to denote repetends in decimals not for infinitely extended operations. The “white space” you put in (2) doesn’t mean anything. You just bastardized the symbol for something it wasn’t meant for, did a nonsensical thing, and got a nonsensical result. It has nothing to do with whether 9.999... - 0.999... = 9 because it’s clear that what it’s referring to, that it’s a form of recurring decimal rotation. What you’re doing with the 1-1+1-1+1-1+... is your made own made up bastardization of recurring decimal notation.

[Anonymous No.12235160]

>>12235141
>>12234349

But let’s say for sake of argument that you can use the symbol that way. “...”

The pattern is:
1 - 1 =0
1 - 1 + 1 - 1 = 0
1 - 1 + 1 - 1 + 1 - 1 = 0

It just always 0’s out. If you added expression (1) and (2) together you’d just get 2x = 0
X = 0

Wow how interesting.

[Anonymous No.12235355]

>>12235160
9.9 - 0.9 = 9.0
9.99 - 0.99 = 9.0
9.999 - 0.999 = 9.0
9.9999 - 0.9999 = 9.0
9.99999 - 0.99999 = 9.0
...
9.999... - 0.999... = 9.0

These are all the same:
[math]9.99... = 9.999... = 9.\overline{9}[/math]

X = 0.333...
X * 10 = 3.333...
10x - 1x = 3.333... - 0.333...
9x = 3.0
X = 3/9
X = 1/3

0.3 - 0.3 = 0
3.3 - 0.3 = 3.0
3.33 - 0.33 = 3.0
3.333 - 0.333 = 3.0
3.3333 - 0.3333 = 3.0
3.33333 - 0.33333 = 3.0
...
3.333... - 0.333... = 3

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

[Anonymous No.12235373]

>>12235355
It’s not “absurd”... it’s not “weird”...

.999... and 1 are just two different notations of the same value.

[Anonymous No.12235376]

>>12234349
>1-1+1-1+1-1+
any proof dependent on the order of adding is bs

[Anonymous No.12235583]

>>12234349
>cant into Divergent series
fucking undergrads