[pnqk]

[iwillsurvivecollege]

Shout to Witch of Agnesi, my favorite curve 🙏

[uuuuuhhhhhhhhhhhhhhhhhh]

"contrapositives are for cowards"

[eulerseverything]

If I were writing a math textbook I’d have a section in the beginning called “A Note on Notation” where I introduce all the notation Im using. And everyone would want to fuck me sooooo bad

[ms-demeanor]

Got mostly rid of my bad/reactive/aggressive mood by badly hand-poking the alphabet on a banana.

[pnqk]

I like your mathbb font

[pnqk]

imagine our world finally overcoming capitalism, oligarchy, wars, and creating a society where everyone is equal in rights and has access to education and healthcare, and then like in 2 days there begins the apocalypse

[pv1isalsoimportant]

Broke: using co- to talk about the dual version

Woke: using antonyms to talk about the negation, bonus points if it only makes sense in your native language

[hoes-cant-handle-my-swag]

I’m so sick of people constantly telling me im “way behind in math” or just plain stupid for not understanding certain subjects. I’ll have you know that my school is severely underfunded and can barely afford to keep the walls up, let alone have good teachers. So everything im learning is done by my own initiative. I teach myself this, i sit through 12 hour courses on geometry just to understand it, because no one else can provide it for me.

[zzkt]

Why 7/6 and not 1/6? There is no spoon

I used a more convoluted method If f is the homomorphism that sends elements of Z_12 into products of powers of roots of unity, there are 12 different possible selections for what f(1) equals. f(1) = e^(pii/6) may be the most obvious. since 1 generates Z_12 f(1) is all you need to define to define f, but when the group is not cyclic you need to use a more general method to find the homomorphism. first you would have to find the basis of the group by representing it as products of groups of prime power order, in this case represent Z_12 as the direct sum of Z_3 and Z_4. 0 = (0, 0) 1 = (1, 1) 2 = (2, 2) 3 = (0, 3) 4 = (1, 0) .... 11 = (2, 3) then you know that Z_12 as ordered pairs of elements Z_3 and Z_4 can be generated as linear combinations of (1, 0) and (0, 1). So in order to define a function f that sends elements of Z_12 into products of roots of unity, you only need to define f((1, 0)) to be ANY 3rd root of unity and f((0, 1)) to be ANY fourth root of unity. In my case I arbitrarily defined f((1,0)) = e^(2ipi/3) and f((0, 1)) = e^(2ipi/4), which gives f((1, 1)) (the generator of the cyclic group) to be f((1, 0)) * f((0, 1)) = e^(7ipi/6) Since there are 3 third roots of unity and 4 fourth roots of unity, there are 3*4 = 12 possible different functions and hence twelve different representations of Z_12 using the exponential function (when the group is not cyclic, the homomorphisms will not be isomorphisms).

“Typhoid & swans - it all comes from the same place.”

[zzyriphian]

My favorite thing in math is equivalent conditions. Give me a thousand ways to say the same thing please. "Two numbers are coprime if and only if their least common multiple is equal to their product", and "A matrix is invertible iff its kernel has a cardinality of 1" sound so wrong because that's not at all how those two things are defined but they are also provably true.

I got so much extra credit on a linear algebra test once for being able to name almost every equivalent condition of invertability of a matrix we learned over the semester.

[bubbloquacious]

Honestly name a more iconic duo than existence and uniqueness

[kojakaj]

it is really funny in number theory seeing like. 1 million theorems about ODD primes and being like, a complete math newb like "why is it odd primes? aren't all primes odd, except 2" and then ur professor is like yeah 2 fucking sucks and like does not fit any of our proofs so we have to deal with it separately and etc. and they have like such vitriol for. prime number 2.

[pnqk]

one simple idea just came to my mind

so, ℤ/2ℤ is a group {0, 1} with binary operation addition mod 2. this is a group where –1 = 1, and a – b = a + b. in other words, the operation and its inverse operation work identical.

this holds for addition in any ring with characteristic 2. it is quite useful in practice when computers use Galois Fields GF(2ᵐ), where a computer can be programmed the same way for addition and subtraction. and I personally got quite used to it: addition=subtraction.

but all groups of order 2 are isomorphic to ℤ/2ℤ, for example {+1, –1} with operation multiplication. and what we get here is that for elements A, B, C of this group the three statements are equivalent:

A = B ⋅ C

C = A ⋅ B

B = A ⋅ C

now, division=multiplication, and my brain still can't get used to it

there is quadratic reciprocity law, which can be expressed using Legendre symbols:

(p/q) = (q/p) ⋅ (–1)^{(p – 1)(q – 1)/4}

which in some books written this way:

(p/q) ⋅ (q/p) = (–1)^{(p – 1)(q – 1)/4}

sometimes, there's a small 2 seconds gap when my brain rebels against it. but they are equivalent, nothing wrong there

crazy

[hyperoperationfractallisation]

Addition and multiplication are isomorphic in Z/2Z

[pnqk]

what

[pnqk]

one simple idea just came to my mind

so, ℤ/2ℤ is a group {0, 1} with binary operation addition mod 2. this is a group where –1 = 1, and a – b = a + b. in other words, the operation and its inverse operation work identical.

this holds for addition in any ring with characteristic 2. it is quite useful in practice when computers use Galois Fields GF(2ᵐ), where a computer can be programmed the same way for addition and subtraction. and I personally got quite used to it: addition=subtraction.

but all groups of order 2 are isomorphic to ℤ/2ℤ, for example {+1, –1} with operation multiplication. and what we get here is that for elements A, B, C of this group the three statements are equivalent:

A = B ⋅ C

C = A ⋅ B

B = A ⋅ C

now, division=multiplication, and my brain still can't get used to it

there is quadratic reciprocity law, which can be expressed using Legendre symbols:

(p/q) = (q/p) ⋅ (–1)^{(p – 1)(q – 1)/4}

which in some books written this way:

(p/q) ⋅ (q/p) = (–1)^{(p – 1)(q – 1)/4}

sometimes, there's a small 2 seconds gap when my brain rebels against it. but they are equivalent, nothing wrong there

crazy

[msexcelfractal]

Parentheses are great, but have you ever put your expression inside a fortified enclosure?

𓊆x+1𓊇

[michaelrotonal]

cursed number notation. can you figure it out

1 = []

2 = [[]]

3 = [][[]]

4 = [[[]]]

12 = [[[]]][[]]

0 =

-1 = -[]

19 = [][][][][][][][[]]

20 = [[[]]][][[]]

-2 = -[[]]

1/2 = [-[]]

sqrt(2) = [[-[]]]

72^1/6 = [[-[]]][[][-[]]]

5/4 = [-[[]]][][[]]

[marigold-22]

I love mathematical abstraction cause you can be working with funny little arrows in 2d space and some guy will hit you in the head and now your doing work with different colored frogs or something and you have to think about if negative frogs exist and if they do it turns out that frogs are actually the same thing as the funny little arrows, and in fact if the frogs have uncountably many traits that can all be added and scaled and such the frogs are actually analogous to functions from the reals to the reals.