Math Memes 10

[@Ianweis]

sqrt(-1)/8 = I over ate 😂.

[@user-qc9cd5iz3l]

I say that like I upon eight

[@gapgamer2246]

@@user-qc9cd5iz3l yes

[@AdarshHari708]

I say i by eight

[@shototodoroki7341]

Thanks man I was thinking "1 by 8 imaginary unit" and questioning my entire life.

[@microscopicallysmall]

@@shototodoroki7341 i was thinking “imaginary number divided by 8”

[@justinkianaalfredo6843]

"equations can't have 2 answers" X¹ and X²: allow us to introduce ourselves

[@WojtekXD-bx7jb]

Basically the entire second grade of highschool which is mostly spent on quadratic functions is riddled with X1 and X2

[@meme-bo6hq]

we do not know whether the square root is plus or minus until we check it against a variable collapsing the superposition

[@user-iy6dt4xp5o]

Should be subscripts, not superscripts because it looks like x^1 and x^2. Try using “x1, x2” or “x_1, x_2”

[@Tekdruid]

8:00 Somehow, I was expecting a ...different graphic for Rule 34.

[@Sedyon]

Me too 🤣🤣

[@barnabasszoke6499]

I can say with confidence that rule 34 is false

[@A_literal_cube]

6:30 Jan Misali would like to introduce you to: 6

[@Rangadus]

when you use your calculator to find the digits of pi but it just shows pi:💀

[@GabriTell]

6:29 - Mathematicians after finally convincing their psychologists cups and donuts are the same (there start to appear white vans).

[@Gordy-io8sb]

2:50 Quadratic equations: Am I a joke to you?

[@Pi_guy_3.14]

3:43 Cartesian VS Polar coordinate systems 😂

[@accuset]

3:13 i did solve it, with the help of Wolfram Alpha. It's 3. I'm not even memeing around, it's actually just 3. Granted, this is the solution of the arbitrary, non-zero case where Tuturu = 1, Kurisu-chan therefore = 2, Glasses-chan = 4, and Nyannyan = 0. There may be other solutions, but I don't care about them.

[@mismis3153]

It only has one solution. Also wolfram alpha doesn't seem to be able to compute the closed forms of the integral, but they do exist and they do indeed simplify to exactly 3.

[@accuset]

@@mismis3153 Not sure how your Wolfram input went, but I computed only the ratio of integrals using the math input, with the upper and lower bounds as constants. It solved this for me, but again, I didn't do a system of equations. Just the division. Also, there is another solution with Tuturu, Kurisu, and Glasses = 0 and Nyannyan = 7. Not sure how the integral ratio solves like that, maybe it's undefined, but it's still technically a possibility.

[@mismis3153]

@@accuset your other solution doesn't work, you have an = 6 condition. You can plug the integrals in wolfram alpha but it gives a decimal approximation, which is really close to 3, but it isn't capable of saying whether it is 3 or something really close. If you calculate their closed forms, it becomes obvious that it is 3, but this is like master degree level maths, if you want I can link you a solution for the closed forms. This is way out of my league, but it is possible to do.

[@accuset]

@@mismis3153 On the = 6 bit, I forgot about that one, you're right. On the wolfram bit, I'm an engineer. 3.0000005 = 3, because machines doing calculations have some amount of error due to truncation, data type limitations or conversions, etc. If it's stupidly close to an integer like that, unless I know why it can't be exact, I will assume there is error present and will take the integer approximation. I don't see any 1/(x-3), so this function can be exactly 3. However, with that caveat mentioned, Pi = 3.14 on good days, 3.1415926 on bad days. Bad because I need that level of precision, and precision takes effort. Also gravity = 9.81 m/s^2, and anyone who gives fewer digits is a psychopath. I don't mind 9.80665 and 9.807, but 9.8 is sloppy, and 10 is sinful.

[@watermagle]

You can solve it by transforming both integrals into incomplete Beta functions (complex factors will remove each other out) and then representing them in terms of hypergeometric functions

[@darcash1738]

1:03, use Chio’s reduction twice and then you can do the 3x3 determinant as normal

[@ianweckhorst3200]

Your only option is to be a prime base to do p-adics, 11 or 7 is pretty good in that regard

[@THE_GREMLINZ_OFFICIAL]

-(1/2)🙂 -1/2 🙃 1/-2👽

[@bothenumberblockslogoedito1339]

6:48 That was made by me :)

[@rodrigoqteixeira]

I exploded after seeing "pi halves" 🤣

[@dudono1744]

5:35 seems troll face is analytical

[@KPoWasTaken]

8:50 that video actually talks about the zero ring Dividing non-zero by zero is undefined because it gives no numbers. Dividing zero by zero is indeterminant because it gives every number.

[@BRaleatoriedades]

8:07 no, i am a tanplayer

[@RoxanneClimber]

*cotplayer has entered the chat*

[@gasparliboreiro4572]

3:42 that's 0

[@eduardoxenofonte4004]

0 < θ < 2π doesn't even complete the circle, you're missing a point, it should be 0 ≤ θ < 2π

[@psng7416]

So underrated

[@GabriTell]

3:01 - When you spend half the exam time on a proof only to realise the exercise said "refute" :

[@ianweckhorst3200]

The x^5 doesn’t even show up, but the x^4 scares me, even the x^3 is a little scary, but x^2 is easy.

[@williamturner8257]

indeed "Division doesn't exist" is the most correct explanation

[@RoxanneClimber]

13:21 lmao

[@kiti_cat524]

8:08 wait until the tanplayers arrive

[@isaccwewee7868]

was heading to Wolfram Alpha for the first one until I said it out loud... "i over ate" 😆

[@heinrich.hitzinger]

(a+b)²=a²+b² a²+2ab+b²=a²+b² 2ab=0 ab=0 a=0 or b=0 😂😂😂

[@user_sense]

or 2=0

[@Melody_Boi_Piyush]

​@@user_senseas a=0 or b=0 2=0/ab => 2=0/0 Which is indeterminate form, hence 2≠0

[@user_sense]

just choose a and b such that non of them is 0 and nothing will stop you from achiving 2=0@@Melody_Boi_Piyush

[@Melody_Boi_Piyush]

@@user_sense (a+b)²=a²+b² This is not an identity, it's a function, it won't come true if both a and b not equals to zero, you can't just choose numbers, this is not biology, this is maths, and maths is absolute

[@user_sense]

it will be true if both a and b are not equal to zero, if 2=0@@Melody_Boi_Piyush

[@mrblakeboy1420]

4:48 the first two would be swapped for me

[@mrblakeboy1420]

5:48 do they? if that’s true, oh no

[@THE_GREMLINZ_OFFICIAL]

…wait until he learns about multivalued functions. 2:50

[@ploosmaD]

No-one gonna talk about the guy at 0:21 ive been staring at if for 10mins and still dont understand what could lead a person to have what is the maths version of a stroke

[@carlosmanan2152]

Did I understood??? Of course not!!! I also don't understand how I passed math in school after studying so hard for never to use what I learned and forget everything

[@THE_GREMLINZ_OFFICIAL]

5\3=0.6 (the numerator is over the denominator.)

[@Faroshkas]

What is bro talking about

[@THE_GREMLINZ_OFFICIAL]

(an)^2=a^2+2an+n^2 (2+3)^2=2^2+2*3*2+3^2

[@goatgamer001]

(an)² =a²×n²

[@Tcrrxzz]

How do you even solve the first 3 lines of equations? 😢 3:12

[@evank3718]

13:05 Don’t know if this is supposed to be ironic or not…

[@ianweckhorst3200]

Well of course the x squared

[@ChrisMaster2]

The parentheses one has dumb logic: (2+3)^2 = (2^2)+(3^2) = 4 + 9 = 13 IRL is 25

[@thedarkshadow5347]

Ok il becoming back in 4 years when im old enough to understand this stuff someone please comment on this in 4 years

[@jaydentplays7485]

0:11 It’s 25

[@Unname-zq7yr]

sqrt(-1)/8 = 0+0.125i

[@megamaz108]

0:20 I think I know his error he thinks that (2+3)² = 2² + 3² = 4 + 9 = 13

[@Mariofan2943]

(2+3)²=(2+3)×(2+3)=4+6+6+9=25

[@snivydream]

2:35 and 3:09 where tf are these from

[@angelagonzalez8250]

If numbers are real what would imaginary numbers be

[@ENTMusic-cj7wt]

HOLY SHIT I MADE THE MEME AT 1:06 - 1:14

[@Calamahara]

(a+b)^2 is not equal too a^2+b^2 how ever (a*b)^2=(a^2*b^2)

[@ianweckhorst3200]

Aww they’re not reading the communist manifesto, I was hoping they were

[@rexisnox577]

6:29 Be base 6

[@THE_GREMLINZ_OFFICIAL]

|1/0|=♾️

[@rodrigoqteixeira]

i one over eigth 💀

[@mihiguy]

3:34 Theta not zero

[@THE_GREMLINZ_OFFICIAL]

sqrt(a)+sqrt(a)≠2sqrt(a) The square root of any a has multiple values

[@sohamacharya171]

For positive real numbers, the square root is defined to be the positive no whose square yields the number. Thus sqrta+sqrta=2sqrta

[@MrFox-nj3ih]

That's wrong. Sqrt(a) for any real a has only one value. eg, Sqrt 16 cannot be -4, only 4. sqrt(a) + sqrt(a) is always 2sqrt(a).

[@oxidine2968]

0-0=-0