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25 Dec 2021Mathchan is launched into public

50 / 7 / 46 / ?

Is thrembo real? It's an integer between 6 and 7.
>>
>>136
Thrembo is a number that consists of two parts: the real part and the schizo part. Complex numbers
C\mathbb{C}
have real and imaginary components while schizo numbers
S\mathbb{S}
have real and schizophrenic components. Unlike imaginary numbers, schizo numbers feel "real" but they are always irrational, therefore
SRQ\mathbb{S}\subset\mathbb{R}\setminus\mathbb{Q}
. When you strap a schizophrenic number
sSs\in\mathbb{S}
to the floor (
s\lfloor s\rfloor
) or ceiling (
s\lceil s\rceil
) it stops being schizophrenic and becomes "whole".

If you add a "medical number" to the "schizo number" it stops being schizo. For example:

+ = 6

Therefore, we can conclude thrembo is when number 6 does not take its meds:

= 6 -

Thrembologists have not yet constructed thrembo therefore we do not yet know what
{\lbrace
}\rbrace
is.

References:
https://en.wikipedia.org/wiki/Schizophrenic_number
https://www.who.int/news-room/fact-sheets/detail/schizophrenia
>>
asdf
>>
Working in ZFC, one can prove that there exists no integer x, such that 6<x<7. This also works for most definitions of the integers.
>>
>>136
You could define thrembo as a nimber (surreal number) I think?
You could have a number incomparable to 6.
I don't think you can have an integer nimber in between 6 and 7 though.
>>
>>161
zfc = glowie spam
>>
>>173
No Nimber is an integer besides the integers themselves
>>
its real
>>
Find k such that f is a valid pmf:
f(x)={kln(x)0xϪ0otherwise f(x) = \begin{cases} k\ln(x) & 0 \leq x \leq Ϫ \\ 0 & otherwise \end{cases}
>>

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Image
>>
i still remember my thremboth birthday party
>>
>>357
what was it like
>>

File: thrembo_birthday_math.jpg ( 1.58 MB , 4032x3024 , 1678933794571.jpg )

Image
>>358
i went to chunkie cheese and had Ճ themed balloons
>>
>>

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>Thrembo is NOT a real number
>You WILL take your meds
>You WILL cease your investigations
>You WONT use Thrembo to time-travel
>Why? Because... It just isn't ok? Fucking chud...
>>
d
>>
>>136
yeah if its between 6 and 7 it should be an element of the reals
>>
>>136
uhm sweaty it's actually not real it's fact checked by independent soycientists
>>

File: Icare.png ( 85.72 KB , 800x789 , 1680361817117.png )

Image
>>372
Erm... In fact Fauci already said thrembo is a valid number and that we should all accept it
>>
thrembo is a hypothetical dark number
>>
>>
>>136
lol
>>
gem
>>
>>137
'listering 'emerald
>>
>>136
'm
>>
test
>>
gemerald
>>
adwadadada
>>
its real
>>
>>136
It's as real as derf.
>>
>>137
>schizo numbers feel "real" but they are always irrational
Irrational numbers are real, retard.
>>
>>
>>136
thrembo is misinformation
>>
>>474

/thread
>>
>>137
The word is arheton in the greek unspeakable in English modernized to irrational, read a book
>>
>>137
It's a real digit/number that makes π a rational number, necessary integral for time traveling, used by Ahnenerbe, Vril and Thule society to create anti-gravitational aircrafts and is the keyed integral that allows to restore the ancient Hyperborea.

No wonder (((they))) shut down any study Thrembonomy and advanced occultist esoteric spiritual practices of Thrembology.
>>
Cool
>>
NSA integer group spam safe sage
>>
>>136
erm
>>
>>136
Ok
>>
>>550
Hi birb
>>
>>

File: e9a.png ( 254.35 KB , 680x670 , 1710785235395.png )

Image
>>

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GPT-4o solved the captcha for me
>>
>>829
To solve this problem, we need to determine the elements of the set
S S
that are also in the set
Q Q
, which contains all rational numbers. Let's analyze each element in
S S
:

1.
(1+i32)2019\left( \frac{-1 + i \sqrt{3}}{2} \right)^{2019}
:
This is a complex number raised to a high power. Notice that
1+i32\frac{-1 + i \sqrt{3}}{2}
is a complex number on the unit circle in the complex plane. Specifically, it represents the cube root of unity,
e2πi/3 e^{2\pi i / 3}
. Raising this to the 2019th power will give another root of unity, but not a rational number.

2.
((1+i2)2019)\Im \left( \left( \frac{1 + i}{\sqrt{2}} \right)^{2019} \right)
:
The imaginary part of a complex number raised to a high power. Since
1+i2\frac{1 + i}{\sqrt{2}}
is on the unit circle, raising it to the 2019th power will rotate it on the unit circle, but its imaginary part will generally not be rational.

3.
13\frac{1}{3}
:
This is a rational number.

4.
π3\frac{\pi}{3}
:
This is not a rational number because
π\pi
is irrational.

5.
sinπ3\sin \frac{\pi}{3}
:
sinπ3=32\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}
, which is not rational.

6.
227\frac{22}{7}
:
While
227\frac{22}{7}
is a common approximation of
π\pi
, it is a rational number.

Now, let's count the rational numbers in
S S
:
-
13\frac{1}{3}

-
227\frac{22}{7}


There are 2 rational numbers in
S S
.

Therefore,
SQ|S \cap Q|
equals 2.

The answer is
2BM3FRT2BM3FRT
.
>>
>>830
It's wrong.
1+i32\frac{-1+i\sqrt{3}}{2}
is an a complex cube root of 1 (as can be easily verified by multiplication)
(1+i32)2019=(1+i32)3×673=((1+i32)3)673=1673=1(\frac{-1+i\sqrt{3}}{2})^2019 = (\frac{-1+i\sqrt{3}}{2})^{3\times673} = ((\frac{-1+i\sqrt{3}}{2})^3)^673 = 1^673 =1
. Thus the answer is 3
>>
>>831
Yes but the final answer was right is all what matters for this captcha
>>
>>832
The final answer is 3 and in >>830 it is 2.
>>
>>833
no the answer is clearly 2
two integers cant represent a complex number
>>
sneed
>>
logsday is the thremboth day of the week