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The hardest math problem

Anonymous Sun 03 Apr 04:42:03 2022 No.14359963 View Reply Original Report

Quoted By: >>14381439 >>14384346 >>14385864 >>14386711 >>14390751

Try to come up with the hardest math problem that you possibly can. But it has to be original, not Riemann Hypothesis for example. I will start.

>what is the exact average distance between two random points inside of a dodecahedron with a unit side length

Anonymous

Anonymous Sat 09 Apr 2022 00:34:01 No.14377475 Report

Quoted By: >>14378336

>>14377258
Can someone explain this puzzle?
What's being asked for? Is it asking for, some function on a single variable, using multiplication/division, addition/subtraction such that if any number is put in, a number closer to sqrt{2} + sqrt{3} is produced?

Anonymous

Anonymous Sat 09 Apr 2022 04:58:29 No.14377946 Report

Quoted By:

>>14374249
So for 4 pairs you have -2 combinations? :)

Anonymous

Anonymous Sat 09 Apr 2022 05:51:03 No.14378043 Report

Quoted By: >>14382820 >>14383144

>>14362069
0.8 < n
0.16 < 0.2n
n^2 + 0.8n + 0.16 < n^2 + n
n^2 + 0.8n + 0.16 < n^2 + n < n^2 + n + 0.25
(n + 0.4)^2 < n^2 + n < (n + 0.5)^2
n + 0.4 < sqrt(n^2 + n) < n + 0.5

Anonymous

Anonymous Sat 09 Apr 2022 06:07:20 No.14378071 Report

Quoted By:

>>14372099
x = 312308143150995148325518962163387345312703063965080.0000000000005697341796376073246846870137498089580935892157142792693904936291323736624182032609215605084791100716509644735552003653821774941809960278488858787106880810199794338346754783619924595859766987138155339415454223159609658949301287013416304242928934273091109203249732728228145058087378065112009771268506851933333771464567776671855002767603346777962788355854739619540075485725606239480871941858216571870737340295091117357981997464382649836157467291758794177726746045317088520119894662015217091520563969431715824262400443375185223490858266700991012066826098389609484854380637663334967450593036844344584912095170463254596318703359507262699689126023247316158006265724669338042370002391729347606971743170555289692169380453132061744895821328460753992173336782179349052581575422023632836125677423950392558260512212092307543865096106648832640659738231897148630346495857546597943814194873913892507120924687712336234838524666210443529584297637321612496024332098903866620777451987493472531056487189448829176034971669881160684354351760144894770030371001849880721603793069347218789606618201809134144241789324288229181399364355606876155744479156610643526714476773678129696519835608173669060707591509618075076762979318052350689450315907370632159388209614057599285263732348953047810970686825750207740595566178698842326302366478454953388140346366673699834663929277634989643313505810528676807641987240587690459445080155144035167567976896864958251988068759633734686004761052448794137425160396653629898471670500661481334396172254745873231975509249210926663768358522182855671631500298827510687341306135081160759383907143987214597316646224872555058728214421924559951894091544246572681954811593536426389375591806493764852019915443294096125600651912177472073682341116458072445163862450275007713560980310391240538107994601924897812864622730891090542692153973577705269563701552381486659679687689024631184920921411301358744330934090772785810994344572187

Anonymous

Anonymous Sat 09 Apr 2022 09:58:23 No.14378336 Report

Quoted By: >>14378796

>>14377475
>>14377258
Maybe I should have been a bit more specific
sqrt2:={q in Q+: q^2<2}
sqrt3:={k in Q+: k^2<3}
Dedekind cuts can be interpreted as formulas that take a positive rational number and answer whether or not it is in the set or not.
Example. Is 1.4 in sqrt2? Yes
Is 1.6 in sqrt2?No
Is 1.6 in sqrt3?Yes

My question is.How can the real number that is produced by the addition of sqrt2 with sqrt3 be expressed as a Dedekind cut?

I am asking this question because I have never seen in any real analysis book I have read an example of cut addition

Anonymous

Anonymous Sat 09 Apr 2022 13:27:46 No.14378779 Report

Quoted By: >>14378787

>>14377320
five 2 one

Anonymous

Anonymous Sat 09 Apr 2022 13:30:29 No.14378787 Report

Quoted By: >>14378792

>>14378779
err ten 2 one

Anonymous

Anonymous Sat 09 Apr 2022 13:31:51 No.14378792 Report

Quoted By:

>>14378787
err actually its 128 to 1 final answer.

Anonymous

Anonymous Sat 09 Apr 2022 13:32:50 No.14378796 Report

Quoted By:

>>14378336
sqrt2 + sqrt3 = { q+k in Q+ : q^2 < 2 & k^2 < 3 }

Anonymous

Anonymous Sat 09 Apr 2022 14:55:41 No.14379047 Report

Quoted By:

>>14366264
>>14366217
No he's quite right, that's true.

Anonymous

Anonymous Sat 09 Apr 2022 15:58:36 No.14379250 Report

Quoted By: >>14379333 >>14379364 >>14380511 >>14380530 >>14383416

I thought of a probability problem recently
It seems pretty easy to define but I haven't been able to find a problem like it
You count down once a second from a number n, each time you count down you have a p chance of having to start over from n.
What is the expected value of how many seconds it takes you to count down to 0?
I got this idea from this Aqua Teen Hunger Force Clip:https://www.youtube.com/watch?v=eAgUoSIG3Jo
Where Meatwad starts singing the "bottles of beer on the wall" song at 3 million, but sometimes loses count and has to start over from 3 million

Anonymous

Anonymous Sat 09 Apr 2022 16:33:28 No.14379333 Report

Quoted By:

>>14379250
I thought about it again and its actually equivalent to Expected Number of Trials to get N Consecutive Heads (with heads having probability 1-p)
So its something like [1/(1-p)] ^ {n+1} - 1/(1-p)

Anonymous

Anonymous Sat 09 Apr 2022 16:48:52 No.14379364 Report

Quoted By:

>>14379250
Sounds like you could set up a discrete valued markov chain to solve this. I don't exactly remember how to do that though, but I think the answer should just be the sum of expected amount of time of reaching 0 from each state. So 1/(1-p)^n + 1/(1-p)^(n-1) + .... + 1/(1-p) Not sure though

Anonymous

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Anonymous Sat 09 Apr 2022 17:07:59 No.14379397 Report

Quoted By:

>>14360159

Anonymous

Anonymous Sat 09 Apr 2022 22:01:56 No.14380511 Report

Quoted By: >>14380530

>>14379250
$\frac{n p^n + 1 - p^n + p\frac{1-p^n-n(1-p)p^{n-1}}{1-p}}{p^n}$

Anonymous

Anonymous Sat 09 Apr 2022 22:07:23 No.14380530 Report

Quoted By:

>>14380511
>>14379250
In my answer p is the probability of decreasing a step, so you would have to substitute 1-p for my p.

Anonymous

Anonymous Sat 09 Apr 2022 22:17:08 No.14380551 Report

Quoted By: >>14380559

>>14372230
$\floor{\frac{n!}{e} + \frac12}$

Anonymous

Anonymous Sat 09 Apr 2022 22:18:59 No.14380559 Report

Quoted By:

>>14380551
$\left \lfloor{\frac{n!}{e} + \frac12}\right \rfloor$

Anonymous

Anonymous Sat 09 Apr 2022 22:29:10 No.14380592 Report

Quoted By:

>>14362920
>>14363805
>>14364515
This all assumes the money in the other envelope is some changing quantity depending on the amount you see in your first choice, which makes no sense.

There are two envelopes, say one contains K dollars, and one contains 2K dollars. Flip a coin and pick one envelope. With probability 1/2, it contains 2K, and switching will lose you K$. With probability 1/2, it instead contains K$, and switching will gain you K$. So, switching has no expected benefit.

Anonymous

Anonymous Sun 10 Apr 2022 04:57:24 No.14381436 Report

Quoted By:

>>14360159
Die

Anonymous

Anonymous Sun 10 Apr 2022 05:03:17 No.14381439 Report

Quoted By:

>>14359963
12 x 18 - 3

Anonymous

Anonymous Sun 10 Apr 2022 05:44:01 No.14381488 Report

Quoted By:

>>14372230
f(n) = (n-1) * (f(n-1) + f(n-2))
where f(1) = 0, and f(2) = 1

0,1,2,9,44,265, etc. for 1,2,3,4,5,6 couples

Anonymous

Anonymous Sun 10 Apr 2022 05:45:18 No.14381492 Report

Quoted By: >>14382533

The value is bounded by (n-1)! and (n!)

Anonymous

Anonymous Sun 10 Apr 2022 16:20:15 No.14382525 Report

Quoted By: >>14382628

Ok the exact equation is this:
(n-1)! x (
$1$ +
$\sum_{2}^{n-2}\frac{1}{a}$ +
$\sum_{2}^{n-2}\frac{1}{b} \sum_{2}^{b-2}\frac{1}{a}$ +
$\sum_{2}^{n-2}\frac{1}{c} \sum_{2}^{c-2}\frac{1}{b} \sum_{2}^{b-2}\frac{1}{a}$ + ...)
The sum is done for $\lfloor n/2 \rfloor$ rows.
The number of terms in each row r starting at 0 is ${(n-2)-r}\choose{r}$ , and
the total number of terms being added is the (n-1) $^{th}$ number in the fibonacci sequence starting at 0

dont judge my autism

Anonymous

Anonymous Sun 10 Apr 2022 16:23:50 No.14382533 Report

Quoted By: >>14382628

>>14381492
shit this means it's not upper bounded by n!, wtv kmn

Anonymous

Anonymous Sun 10 Apr 2022 16:49:28 No.14382628 Report

Quoted By: >>14382796

>>14382525
>>14382533
Do you know what a derangement is anon?

Anonymous

Anonymous Sun 10 Apr 2022 17:34:03 No.14382796 Report

Quoted By:

>>14382628
oh yeah that is nicer

Anonymous

Anonymous Sun 10 Apr 2022 17:36:02 No.14382805 Report

Quoted By:

>>14360226
>>14360239
You seem to have convinced yourself. Now think about what it would take to convince someone else and write that.

Anonymous

Anonymous Sun 10 Apr 2022 17:42:13 No.14382820 Report

Quoted By:

>>14378043
Nice. I don't think there's a simpler proof.

Anonymous

Anonymous Sun 10 Apr 2022 19:24:49 No.14383144 Report

Quoted By:

>>14362069
>>14378043
That's fun. The same form of argument works with $\sqrt[3]{n^3 + 3/2 n^2}$ when $n \gt 1$

Anonymous

Anonymous Sun 10 Apr 2022 20:39:21 No.14383416 Report

Quoted By:

>>14379250
Let E(k) be the expected number of seconds to reach 0 from k.

E(0) = 0
E(k+1) = p E(n) + (1-p) E(k) + 1

E(1) = p E(n) + 1

$\frac{E(k+1)}{E(1)} = \frac{p E(n) + (1-p) E(k) + 1}{E(1)} = 1 + (1-p) \frac{E(k)}{E(1)}$

$\frac{E(k)}{E(1)} = \sum_{i=0}^{k-1} (1-p)^i = \frac{1 - (1-p)^k}{p}$

$E(n) = \frac{1 - (1-p)^n}{p} (p E(n) + 1)$

$E(n) = \frac{1}{p (1-p)^n} - \frac{1}{p}$

Anonymous

Anonymous Sun 10 Apr 2022 21:35:23 No.14383581 Report

Quoted By:

>>14377258
I believe that
$\sqrt{2}+\sqrt{3}:=\{q\in\mathbb{Q}^{+} \ | \ q^{4}-10q^{2}<-1 \}$
This is obtained by evaluating the minimal polynomial of $\sqrt{2}+\sqrt{3}$ over the rationals, then expressing such as an inequality for the defining property of the Dedekind cut. Unfortunately this means we work with the already know answer and then convert it into a Dedekind cut, instead of somehow adding the two Dedekind cuts together in a more self-contained manner. I think one would have to know how to describe the minimal polynomial of the sum of two algebraic numbers knowing the minimal polynomials of the two starting algebraic numbers.

Anonymous

Anonymous Sun 10 Apr 2022 23:01:59 No.14383819 Report

Quoted By:

>>14370998
the problem with this, especially with politically charged labels, is that while the bar for being labeled one thing is constantly lowered, the connotations and repercussions remains the same.
for example with current "definitions" making a comment about how it's unfair black people have an easier time getting into contested jobs and college programs will get you labeled a racist, white supremacist etc. which then has the same repercussions for you as if you'd burned a cross on somebody's lawn.
it's a very effective trick but you should be aware of how bullshit it is so you can push back on it

Anonymous

Anonymous Mon 11 Apr 2022 02:27:01 No.14384346 Report

Quoted By: >>14385679

>>14359963
My dad's location

Anonymous

Anonymous Mon 11 Apr 2022 16:11:14 No.14385679 Report

Quoted By:

>>14384346
Under your desk.

Anonymous

Anonymous Mon 11 Apr 2022 17:11:57 No.14385864 Report

Quoted By: >>14386737 >>14386753

>>14359963
find a non-trivial axiomatic system that is both complete and not contradictory.

Anonymous

Anonymous Mon 11 Apr 2022 21:28:38 No.14386711 Report

Quoted By:

>>14359963
0/0

Anonymous

Anonymous Mon 11 Apr 2022 21:37:30 No.14386737 Report

Quoted By:

>>14385864
What do you consider trivial?

Anonymous

Anonymous Mon 11 Apr 2022 21:45:28 No.14386753 Report

Quoted By: >>14386942 >>14388611

>>14385864
Universe U = {0,a}
Axiom. a = 0

Anonymous

Anonymous Mon 11 Apr 2022 22:55:36 No.14386942 Report

Quoted By:

>>14386753
>Universe U != {0,a}

ftfy

Anonymous

Anonymous Tue 12 Apr 2022 15:08:36 No.14388509 Report

Quoted By:

>>14373161
or anything that calls libgmp or a similar library

Anonymous

Anonymous Tue 12 Apr 2022 15:38:56 No.14388611 Report

Quoted By:

>>14386753
Is U = {0}?

Anonymous

Anonymous Tue 12 Apr 2022 15:42:27 No.14388625 Report

Quoted By: >>14389149

What is the biggest finite number you can describe in a reply to this thread without using external links?

Anonymous

Anonymous Tue 12 Apr 2022 18:17:56 No.14389149 Report

Quoted By:

>>14388625
the number you get when you run this program in C on god's computer with infinite computing power/time

#define R { return
#define P P (
#define L L (
#define T S (v, y, c,
#define C ),
#define X x)
#define F );}

int r, a;
P y, X
R y - ~y << x;
}
Z (X
R r = x % 2 ? 0 : 1 + Z (x / 2 F
L X
R x / 2 >> Z (x F
#define U = S(4,13,-4,
T t)
{
int
f = L t C
x = r;
R
f - 2 ?
f > 2 ?
f - v ? t - (f > v) * c : y :
P f, P T L X C
S (v+2, t U y C c, Z (X )))
:
A (T L X C
T Z (X ) F
}
A (y, X
R L y) - 1
? 5 << P y, X
: S (4, x, 4, Z (r) F
#define B (x /= 2) % 2 && (
D (X
{
int
f,
d,
c = 0,
t = 7,
u = 14;
while (x && D (x - 1 C B 1))
d = L L D (X ) C
f = L r C
x = L r C
c - r || (
L u) || L r) - f ||
B u = S (4, d, 4, r C
t = A (t, d) C
f / 2 & B c = P d, c C
t U t C
u U u) )
C
c && B
t = P
~u & 2 | B
u = 1 << P L c C u) C
P L c C t) C
c = r C
u / 2 & B
c = P t, c C
u U t C
t = 9 );
R a = P P t, P u, P x, c)) C
a F
}
main ()
R D (D (D (D (D (99)))) F

Anonymous

Anonymous Wed 13 Apr 2022 04:13:46 No.14390751 Report

Quoted By: >>14391317

>>14359963
The answer is 3
You are unwilling and therefore unable to prove me wrong

Anonymous

Anonymous Wed 13 Apr 2022 11:17:29 No.14391317 Report

Quoted By:

>>14390751
Also unable to prove you right

Anonymous

Anonymous Wed 13 Apr 2022 16:01:15 No.14391977 Report

Quoted By: >>14392267

What number am i thinking of right now?

Anonymous

Anonymous Wed 13 Apr 2022 18:17:45 No.14392267 Report

Quoted By:

>>14391977
-8

Anonymous

Anonymous Wed 13 Apr 2022 18:25:29 No.14392289 Report

Quoted By:

>>14360159
I'll brute force your teeth outta your mouth after i beat you up you fucking codermonkey

Anonymous

Anonymous Thu 14 Apr 2022 10:06:42 No.14394186 Report

Quoted By:

>>14360159
this is the most onions chuuni brainlet reddit faggot post I have seen this year

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