1 - 0.9999... = 0.000...1
0.9999... =/= 1
prove me wrong
protip:you cant
This is bait.
>>7781261
it doesnt matter how many times u roll a 10 faced dice, there is one case such as 77777.... you cant deny its existence.
That is, y = (10^x - 1)/10^x, if y can be 1, then 10^x = 10^x - 1, then 1 = 0.
0.999... is just not 1, give it up
>>7781275
The probability of it being 777777... is 0. Note that probability 0 [math]\neq[/math] impossible.
Sage because bait.
https://en.wikipedia.org/wiki/0.999...
>>7781287
it is infinitesimally small, not 0.
If u divide by 0, 1/0, it is indefinite, bc any number fit, but if u divide by a infinitesimally small number, it is infinity and nothing else.
>>7781288
i read that but can u prove me wrong?
protip: you cant
>>7781249
fuck I can't take it anymore.
Is this just a shitty /sci/ meme to provoke newfags like me?
How the fuck can anyone in this board not know 0.9999... == 1?
>>7781331
>still cant prove me wrong
>shitty /sci/ meme
0.000000...1/10
Tarded bait.
>1 - 0.9999... = 0.000...1
>0.9999... =/= 1
Pretty much because you're cutting out information.
A-B=C
A=C+B
Get your shit together.
>>7781249
Nothing goes on for infinity, OP.
Stop thinking abstract ideas are real, you're worse than the people who think a negative number of things can physically exist.
>>7781249
Taking 0.(9) as 1 is just to make things easier.
That's how we solved Zeno's paradox with Achilles and tortoise.
>>7781302
the probability of 777777... is (1/10)*(1/10)*..., which is zero. If it would be infinitesimally small, you would get a rather strange probability space, in which the universe event has more mass than 1.
>>7781249
That depends, are you counting infinitesimals as irrationals or integers that have been multiplied by infinity...
... if you're clever you might notice there's a flaw in the difference there teehee
1-0.99999... = 0.1-0.099999... = 0.01 - 0.0099999...
that is, for some n,
1-0.9999... = .1^n - .1^n*0.99999...
= .1^n(1-0.99999...)
Take the limit as n->infinity
1-0.9999... = lim .1n ( 1-0.99999...)
= (1-0.9999...) * lim 0.1^n
Since lim 0.1^n = 0, 1-0.9999... = 0
If you are too stupid to use infinities, then you are too stupid to learn math.
>>7781359
So if you have five cows and you add a negative 5 cows, those negative cows magically made the real cows disappear, right?
5 cows + -5 cows = (poof) no cows
Is that what you are saying?
>>7781249
0.000...1 = 0
prove me wrong
protip: you can't
>>7781249
Okay, I'll try to take this on. Consider this: the uniqueness of a number is determined by the ability to find a number in between it and any other given number. You posit that there is a non-zero (though infinitesimal difference) between .999.... and 1. So I ask you this:
is .999....9 "bigger" than .999.....? How can appending a single 9 to an infinite string of 9's increase the value? Is .000...01 less than .000...1? The answer to all of these is no. The 1 simply has no place; the string of 0's is countably infinite, and does not terminate. Adding an additional 0 does not change the value, making it a degenerate case. Adding an additional 9 does not change the value of .999...., making it a degenerate case. There is no number between 1 and .999...., meaning that by definition, .999... = 1.
1/9 = 0.111...
*9
= 9/9 = 0.999...
<=> 9/9 = 1 <=> 0.999... = 1
prove ME wrong
>>7781384
well if they were meticulously and clearly defined that there are a variety of infinities then people might actually give a nod to putting science over religion, just in the nature that one preceeds and has given meaning to the later as something that it is something that can be relied on doesn't give the later enough credence, there has to be a middle ground... such as 'god' being 'infinite' 'in all things' makes for the sense that there would be a variety of inf's but nooo the community has to be pussy little bitches about defining zero instead of just calling it a null set of accountingment but for the coherence of shape, which is what it is! Just an arbitrary middle ground that can be reduced to anywhere in the number line with other arbitrary middle grounds, which if you have ever studied those sand folks who invented calculus, you'll see that such is the basis of such!
>>7781400
You're not wrong
The point is each: fraction is subject to the very point of the radius which is defined by pi and by such has means for lacking wholeness
>>7781400
>mfw 1 doesn't exist
>every x is actually x-0.000...1
For those who don't believe in infinites in math, perhaps a career in accounting would be what you need. They don't have to deal with infinities at all. For the most part, two decimal places is more than enough.
Enjoy.
>>7781249
Ok, but first you need to prove that 1 - 0.9999... = 0.000...1
>>7781415
>Input interpretation
>0.(0) (repeating decimal) x 1
>0 x 1
You just proved yourself wrong haha, it's saying 1-0.999... is 0.
>>7781377
the limit of 0.999... is 1 there is no doubt about it.
>>7781367
> If it would be infinitesimally small, get a rather strange probability space
call me a newfag, what is the probability of the 10 faced dice (of 1 to 10) rolling eleven(s)? 0. It is not infinitesimally small, it is 0, nothing.
But P(7777....) is infinitesimally small, it means arbitrarily close to 0 but not 0, if it would be 0, what makes 777.. and rolling eleven different? Do you really think that they have the same probability?
>in which the universe event has more mass than 1
no, infinitesimally small event integrate to 1, ε(1/ε) = 1.
>>7781395
>is .999....9 "bigger" than .999.....? How can appending a single 9 to an infinite string of 9's increase the value? Is .000...01 less than .000...1?
I am searching through infinity to answer ur question.
>>7781395
> the uniqueness of a number is determined by the ability to find a number in between it and any other given number.
Quite right, by far i think you are winrar.
x=0.999...
10x=9.9....
10x-x=9.999...- 0.999...
9x=9
x=1
>>7781563
Second step is illegal.
>>7781629
how?
Proof by python here you go.
>>7781249
Fuck this meme board.
X=0.99999...
10x=9.99999999...
9x=9
X=9/9
0.999999...=1
MEMES MEMES MEMES
>>7781747
:^)
>>7781629
Only in the minds of those who can't do math.
>>7781249
There are an infinite amount of zeros in 0.000...1, so the 1 is never reached. Therefore, 0.000...1=0.
>>7781249
image is wrong, lol
Wouldn't this depend on the context?
I'm assuming this comes from the fact that 1/3 * 3 = 1. It makes sense in this case, but simply arguing 0.999... = 1 seems illogical.
What I mean is fractions should be treated as a special case. 1/3 "represented" in decimal, say, 0.333...(frac) is different from 0.333...(non-frac)
So with my shitty arbitrary notation
0.333...(frac) + 0.333...(frac) + 0.333...(frac) = 1
But:
0.333...(non-frac) + 0.333...(non-frac) + 0.333...(non-frac) != 1
0.333...(frac) + 0.333...(frac) + 0.333...(non-frac) != 1
You get the idea.
Waiting for mathfags to point out any flaws, don't go too hard plox I'm just trying to learn.
>Using 0.9999... =/= 1 to troll autistic math nerds on 4chan for nearly a decade
>People still falling for it
>>7781302
Let j = 0.000...0001
Let k = j/2
if k = j: j is zero
if k < j: j is not infinitesimal, contradicting the definition of j
>>7782411
>0.333...(frac) + 0.333...(frac) + 0.333...(frac) = 1
>0.333...(non-frac) + 0.333...(non-frac) + 0.333...(non-frac) != 1
>1/3 "represented" in decimal, say, 0.333...(frac) is different from 0.333...(non-frac)
Now what, what is the difference between the two?
>>7782417
>Using 0.9999... =/= 1 to troll autistic math nerds on 4chan for nearly a decade
*trollface* *evil smile*
>>7782084
>There are an infinite amount of zeros in 0.000...1, so the 1 is never reached.
>so the 1 is never reached.
for a continuous object, a solid, there are infinite amount of points from its center to its boundary.
The boundary is behind that infinite amount of points.
>>7781389
(+ -5) represents minus 5 cows, not plus negative 5 cows.
>>7781389
Think about it like your bank account anon.
You have $5. You additionally decide to take away $5. See what I did there? :D
>>7781249
Well, since plenty of others will take the bait, might as well, too.
0.999... goes on forever, there is no last 9. There is no end.
0.000...1 has an end. There is a finite amount of zeros between the decimal point and that 1. So, your first statement is just wrong.
Infinitely repeating numbers like 1/3 are just an,"error," that base 10 can't express properly. Consider doudecimal. 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B.
The place values behind the decimal in that system would therefore be x/10^y where one zero is equal to 12 in base 10. So, for the decimal fractions of the first place value behind the decimal would be x/10^1, or in base 10, x/12. 1/3 ( 4/12 ) in duodecimal would be expressed as 4/10 ( 4/12 or 1/3 ), or .4, 8/10 is .8 ( 8/12 or 2/3 ), and 10/10 ( 12/12 or 3/3 ) is 1.
>>7781249
>prove me wrong
When you wrote "0.000...1",
you proved yourself wrong.
How many digits are in the dots?
>2016
>this thread still exists and /sci/ is still giving it attention
Feels good to be smarter than all of you ne/sci/ents.
[math] \displaystyle
1 = \frac{3}{3} = 3 \cdot \frac{1}{3} = 3 \cdot 0. \overline{3} = 0. \overline{9}
[/math]
>>7781249
Op = faggot
prove me wrong
protip: you cant
>>7782849
0.333... never equal to 1/3
Best proof incoming.
For all reals, if a < b, then there exists such rational r for which a < r < b.
Proof:
Consider natural n which satisfies
[math]\frac{1}{b-a} < n[/math]
Then
[math]1+an<bn \\ a+\frac{1}{n}<b[/math]
And define r as [eqn] \frac{[an]+1}{n}[/eqn]
Then:
[math]a = \frac{an}{n} \leqslant \frac{[an]+1}{n}= r \leqslant \frac{an+1}{n} = a + \frac{1}{n} < b[/math]
Now infinitesimals aren't actually real numbers, because they're infinitelly small enough for you not to be able to stick a number between 0 and 0.000.....1. But 0 is a real number, so that means 0.000...1 is some kind of weird ass fake number (or it's actually equal to zero and our condition that a < b wasn't even satisfied to begin with)
Don't mind me, I'm just preparing for proof-based Calculus exam.
>>7782893
prove it
>>7782893
per definitionem
>>7782895
there is always a remainder(aka difference) if you divide any power of 10 by 3
can we pease just sticky a 0.999... = 1 thread?
so that way retards won't
post
the
same
shit
every
day
0.999... does not equal 1
why would we even invent two ways of writing the exact same number?
>>7782927
1 = 1.000...
>>7781249
> 1 in 3D space
> 0.999... in 4D space
time needs to be a factor in order to say 0.999 = 1
>>7782894
i take [an] mean the floor of an, that mean ([an]+1)/n > an/n, not >=
>infinitesimals aren't actually real numbers, because they're infinitelly small enough for you not to be able to stick a number between 0 and 0.000.....1. But 0 is a real number, so that means 0.000...1 is some kind of weird ass fake number (or it's actually equal to zero and our condition that a < b wasn't even satisfied to begin with)
0.000...1 > 0
0.000...1 > (0.000...1)/10 > 0
prove me wrong
protip: you cant
>>7782927
>why would we even invent two ways of writing the exact same number?
Because binary allows us to represent on/off states and true/false conditions. Hexadecimal allows us to represent fairly large numbers with less digits, and binary, octal, and hexadecimal are all fairly easy to convert among each other.
>>7781249
That .999...=1 is an immediate corollary of the nested sphere theorem (assuming of course that .999... is a real number and not the hyperreal defined by 1-h where h is the infinitesimal element, in which case clearly we would have .999...=/=1)
Proof: Simply recognize .999...≤1 is an upper bound to every interval [0,a), 0<a<1 and apply the result to the intervals [a,1].
>>7781249
>>7782918
prove it
>>7783138
prove me wrong or go home
>>7783172
>i can't prove the shit I say, i'm just shitposting
>>7783172
change the base to 3
>>7783138
>prove it
It's not that hard to prove.
Just start doing long division by hand.
The pattern is very simple, and completely repetitive.
Give up guys.
0.999... is obviously not 1.
If u are too scared to look at these obviously different numbers and only feel safe when supplied with terminologies, there is an infinitesimally small difference between 0.999.. and 1. Meaning 0.999... is not equal to 1.
>>7783183
ts kiddo, a mathematician would get it by now
>>7783188
> totally clueless about infinity
STOP. FEEDING. THE. TROLLS.
>>7781249
0.99999.... = x
10 * (0.99999....) = 10 * (x)
9.9999999...... = 10x
9.99999... -x = 10x - x
9.9999... - (0.99999) = 10(x) - (x)
9 = 9x
1 = x
>>7782847
>Feels good to be delusional
enjoy your Dunning-Kruger Effect, fgt pls
>>7783222
>9.9999...
>0.99999..
They dont have the same number of digits
>9.9999... - (0.99999) = 10(x) - (x)
and that's where you omit 0.000...9, divide by 9, which is 0.000...1
It comes down to the fact that 0.000..1 is non-zero and you omit it.
/sci/ - taking .999... [math]\neq[/math] 1 bait since 2010.
>>7783255
>Knowing what the Dunning-Kruger effect is
>Not capitalizing your post except for the name of this irrelevant effect
Nice try, psychologist. But I am 100% straight and don’t want to fuck my mom so I have no credit to give to your retarded pseudoscience.
>>7783331
>I am 100% straight
keep lying to yourself, fgt pls
1 - 0.99999.... =/= 0.000...1
Proved you wrong faggot
>>7781249
i like this one
[math]\frac{1}{9} = 0.11111.....
9x\frac{1}{9} = 9 x 0.11111.....
1 = 0.99999....[/math]
>>7783466
what makes u think 1/9 = 0.111..
>>7783554
Its essential to note that not all infinitesimals are descending into zero, but that some are actually summing a number, hence 0.999.... or there as 0.1111.....
Though you could regard the relevance of the later a little more critical, but at a thousands, as engineering points out, you're close enough
>>7783554
Let x=0.111...
Then 10x = 1.111...
10x-x=1.111... - 0.111...
9x = 1
x = 1/9
[0,1] =/= [0,1[
Therefore 0.9999... =/= 1
>>7781331
>Is this just a shitty /sci/ meme
Damn you see me through.
much love, OP
PS:I like to suck cock and 0.999... is obviously equal to 1 (only retards can't understand this).
>>7781359
retard
>>7783172
>prove me wrong
Read up on Russells teapot retard
>>7781249
[eqn]\begin{align} 0.999\ldots &= 0.9 + 0.09 + 0.009 + 0.0009 + \cdots \\ &= 9(0.1 + 0.01 + 0.001 + 0.0001 + \cdots)\\ &=9\left[{1 \over 10} + {1 \over 100} + {1 \over 1000} + {1 \over 10000} + \cdots\right] \\ &={9 \over 10}\left[1 + {1 \over 10} + {1 \over 100} + {1 \over 1000} + \cdots\right]\\ &={9 \over 10} \sum_{n=0} ^\infty \left({1 \over 10}\right)^n \\&={9 \over 10} \left({1 \over 1 - {1 \over 10}}\right) \tag{Geometric series with $|r| < 1$} \\ &={9 \over 10}\left({1 \over {9 \over 10}}\right) \\ &={9 \over 10} \left({10 \over 9}\right) \\ & = 1 \end{align}[/eqn] So [math]0.999\ldots = 1[/math]
>>7783786
This is correct. 0.999... is clearly in the set [0,1[ so it has to be smaller than 1.
>>7784053
Actually, not.
If 0.999... is not equal to 1, then there must be some point between 0.999... and 1. Actually, u ncoutably many such points.
With uncoutably many such points between 0.999... and 1 if 0.999... < 1, then it ought to be easy to identify such a point. Please do so.
Quantifying infinitesimals = 0
>>7781249
MEMEMATICIANS BTFO
0.999 = 1 is a rounding error
>>7781391
0.0000...001=lim_x->Infinity(10^-x)=lim_x->Inf (1/10^x)=0
>>7786102
taking limit doesnt prove equality.
>0.000...1
OK and exactly how many 0's are there?
>>7781249
Isn't this the same as saying the continuum hypothesis is false? That there is a set between the set of reals and the integers. The set of infinitesimals?
>>7787403
infinitely many 0s
x=0.99999 10x=9.99999 10x-x=9 9x/9=1. Get rekt retard
>>7787487
they are not the same. 10x has one less 9 than x after decimal point.
prove me wrong
Protip: you cant
>>7787413
No.
The set of infinitesimal real numbers is {0}
>>7787504
is it Dedekind's cut definition of real?
>>7787480
If there are infinitely many zeros, then the trailing 1 cannot exist if the zeros truly go on forever.
>>7787501
>prove me wrong
there's an infinite number of 9s
infinity minus 1 is still infinity
rekt
e
k
t
>>7787509
>the trailing 1 cannot exist if the zeros truly go on forever
>0.00...001
>you cant imagine such a thing
>dont look at it
0.000...0001/10
>>7787511
>infinity minus 1 is still infinity
>a number minus 1 is still a number
>implying they are equal
meh
told ya yo cant
>>7787513
If the decimal terminates with a 1, it is not infinite, if it does not terminate, the 1 can not exist.
>>7787501
multiplying by 10 only moves the decimal point though. It doesn't change the number of 9s.
9.99 * 10 = 99.9
see? there are no less 9s.
>>7787514
>>a number minus 1
>implying infinity is a number
>>7787517
pretty much this.
You cannot have it both terminate with a 1 and be infinite. It cannot terminate if it's infinite.
>>7787524
Now you're just moving goalposts.
First you said the number of 9s changes. Now you're saying it changes after the decimal place moves. Well, duh, because the decimal place moves, but that doesn't change the number of 9s. If it's infinite 1 - infinity is still infinity anyway. Don't you know how infinity works? retard.
>>7787522
it is the same as saying two distant points, you cant put infinitely many points in-between, but clearly it is not true.
>>7787528
Nice false analogy.
The distance between digits is uniform.
>>7787532
Bzzzt, try again.
>>7787504
Right but if we were to follow OPs logic here these infinitesimals would "fill the gaps" between the reals and the integers. Would they not? And I thought the continuum hypothesis is based there being no gaps between the set of reals and the integers.
[math]1=\frac{1}{9}\times 9=0.111...\times 9 = 0.9999[/math]
[math]\therefore 1 = 0.999...[/math]
The infinite 9s is just an artifact due to the limitations of the decimal representation of fractions like 1/9. Do the long division yourself and you'll see that no matter how many 9s you have, that there's always a remainder.
So when you multiply the 9 back, you don't get what you had originally. It seems to be missing that 0.000...1
Add in that remainder and you get exactly 1. The infitesimal 0.00...1 doesn't exist.
>>7787527
>If it's infinite 1 - infinity is still infinity anyway.
They are infinities of different size.
>>7787530
>The distance between digits is uniform.
OK, if u drop a painted inelastic ball (like a ping pong) vertically, record it and plot a time graph.
With t=0 its starting point and t=when it halts as ending points, bc it is inelastic the ground absorb its hit and it lands quicker and quicker until it halts.
Each time it lands is like 0.9, 0.09, 0.009..., and the ending point is not fixed, it is dependent on its convergence.
I CAN, I JUST CAN, imagine the ball halts but only with infinite amount of time. If you say the finally 1 doesnt exist, it means the ball doesnt halt.
>>7787536
>the continuum hypothesis is based there being no gaps between the set of reals and the integers.
You are confusing two different definitions of "gaps".
There is in fact a number system with infinities and infinitesimals that fill those "gaps". Maybe you have seen infinity and infinitesimals occasionally been written as "w" and epsilon.
>https://en.wikipedia.org/wiki/Hyperreal_number
Those numbers have the same cardinality as the real numbers and are therefore not "between" the cardinality of the real and natural numbers.
>>7787539
that mean 1/9 =/= 0.111...
We all know 0.999... is equal to 1, but did you know that ...999 is equal to -1?
>>7787541
implying that [math]0.9999... \ne 1[/math] implies that there are an infinite number of real points between them. So, what is that real number? There isn't. And the reason is because they are exactly the same.
You probably haven't gone to a real university to learn this stuff formally, so I'll give you something easy to understand to get you started.
https://www.youtube.com/watch?v=TINfzxSnnIE
>>7787550
Also, this means that ...999.999... = 0
>>7787554
>implying that 0.9999...≠1 implies that there are an infinite number of real points between them. So, what is that real number?
0.000...0001
prove me wrong
protip: you cant
>>7781249
>0.000...1
An infinite terminating decimal? kek
>>7787747
but there is
>>7783564
>close enough
>engineering
models collapse over and over again because of this thinking
>>7782936
>>7783084
>>7781249
Conjecture: [math] 1 - 0.9999... = 0.000...1 = \varepsilon \neq 0 [/math]
[math] \implies |0.9999...| < 1 [/math]
[math] \varepsilon = 1 - 0.9999... [/math]
[math] \lim_{x \to \infty } \varepsilon^{1/x} = \lim_{x \to \infty } ( 1 - 0.9999...)^{1/x} = 1 + (0.9999...) + (0.9999...)^2 + (0.9999...)^3 + ... (0.9999...)^\infty [/math]
[math] (|0.9999...| < 1) \implies (0.9999...)^\infty = 0 [/math]
[math] \lim_{x \to \infty } \varepsilon^{1/x} = 1 + (0.9999...9) + (0.9999...8) + (0.9999...7) + ... + 0 [/math]
[math] [/math]
[math] \lim_{x \to \infty } \varepsilon^{1/x} = 1 + (1 - \varepsilon) + (1 - 2 \varepsilon ) + (1 - 3 \varepsilon) + ... + (1 - \infty \varepsilon ) [/math]
[math] \implies \infty \varepsilon = 1 [/math]
[math] \lim_{x \to \infty } \varepsilon^{1/x} = \sum^{\infty}_{k=1} (1) - \varepsilon \sum^{\infty}_{k=1} (k) = \infty + \frac{\varepsilon}{12} = \infty [/math]
[math] \lim_{x \to \infty } \varepsilon^{\frac{1}{x} + 1 } = \varepsilon = 1 [/math]
[math] \lim_{x \to \infty } \varepsilon^{\frac{1}{x} + 1 } = \lim_{x \to \infty } \varepsilon^{\frac{1 + x}{x}} = \lim_{x \to \infty } \varepsilon^{\frac{x}{x}} = \varepsilon [/math]
hence
[math] \varepsilon = 1 = 0.000...1 [/math]
therefore the conjecture is wrong
>>7781249
the third to last line was meant to be
[math] \lim_{x \to \infty } \varepsilon^{\frac{1}{x} + 1 } = \infty \varepsilon = 1 [/math]
>>7787842
>summation 1 to infinity = -1/12
>stop reading there
nice try 0/10
>>7787768
>infinite
>terminating
Pick one bro.
There can be a series with an infinite number of values, each with more zeros in the middle than the last, but no single one of those values can be both infinite and terminating.
>>7787876
>There can be a series with an infinite number of values
>infinite
>each with more zeros in the middle than the last
>more zeros in the middle
>terminating
>but no single one of those values can be both infinite and terminating
>sadfrog.jpg
There are an infinite number of distinct proofs that 1=2.
Recall the classic result from elementary algebra for the base case and proceed by induction: let there be n distinct proofs that 1=2. Then n=n+1-1=n+2-1=n+1 such proofs exist. The result follows.
>>7787640
That is not a point that lies between 1 and 0.999...
>>7781249
I only see 4 significant figures. Therefore, you have to round to 4 significant figures. Therefore it equals 1.000. Get fucked mathfags.
>>7788009
it is.
>mfw I see a insignificant math question
$$\aleph_0$$
Is it in latex?¿? How do you guys write here in latex?¿?
Can you prove that $$\aleph_0 < 2^{\aleph_0}$$
?????
PROTIP: In ZFC you CAN'T
>>7781249
>0.000...1
It's infinite you fucktard, infinite 0s, it never ends, you can't just add a 1 at the end of an infinite sequence of numbers
>>7788167
>you can't
>you can't just
i can, i just did, i did that exactly
i shall do it again
0.000...0001
>>7788183
I meant how do you make the latex code to work in the comments! Not how to write latex lol
>>7788239
square bracket "math"
>>7788167
let n be any integer you want.
According to the axiom of choice, there exists p>n sur as the p-th decimal is a 0 and there is still a 1 after.
[ $$ \sum_{i \in yo\,\,moma}^{haha}\huge{Holy shit}$$]
[ "math" $$ \sum_{i \in yo\,\,moma}^{haha}\huge{Holy shit}$$]
>>7788257
you lied to me :(
>>7788264
>sur
?
\[ \sum_{i \in yo\,\,moma}^{haha}\huge{Holy shit}\]
>>7787513
That equals 0 fool
>>7788941
[math] \sum_{i \in yo\,\,moma}^{haha}\huge{Holy shit} [/math] i think i got it
>>7788941
[math] \huge{HAHA!! \quad YES!!!![/math]
OMG let's test
[math]\underline{Definition:}[/math] Yo moma is so fat.
Yo moma is so fat if [math] \displaystlyle \forall \,\, i \in \text{yo moma} \not\exists i : \lambda(i)<M \,\, \forall\,\, M\leq \omega_1 [/math]
>>7788941
Oh it doesn't get all the commands :(
I think I finally got this (i hope)
[math]\mathcal{Definition:}[/math] Yo moma is so fat.
Yo moma is so fat if [math] $$\forall \,\, i \in \text{yo moma} \not\exists i : \lambda(i)<M \,\, \forall\,\, M\leq \omega_1$$ [/math]
>>7788955
Thats not 0 you idiot.
