#math posting again

[worldwithoutmiracles]

actually pissed off that this is clearly Google Sheets, and the pun is even better that way

[smileyobrien]

The B in Beckett stands for... [insp.]

[charcubed]

do I think that Dean Winchester knows he's bisexual? Yes. Do I think he knows he wants to fuck Cas? Also yes.

[xxplastic-cubexx]

hm.

[catgrandpa]

I'm listening to ABBA again and that never means anything good for anyone

Imagine the Batfam all doing their own Bat Things in the cave, just existing in the same space. The kids convinced Bruce to play music over the speakers, he agreed only because he's doing work on the bat computer and therefore has ultimate control over the music.

Chiquitita comes on and Bruce swiftly moves his hand to change the song.

Dick: Ah, actually B, could we keep this song on? ABBA always makes me feel closer to my mom, and this in particular was one of her favorite songs.

Jason: Weird Dickie, ABBA was my mom's favorite band too.

Bruce: Hn. They were my mothers favorite too, that's why I was changing it.

Jason: Of course, Old Man, cant expect you to let yourself feel an emotion, after all, can you?

Tim: Not to interrupt your regularly scheduled make-fun-of-Batman enrichment time Jay, but ABBA was my mom's favorite too. She would play Money, Money, Money every time she got ready for any big business deal.

Duke: Okay, that's really weird? My parents danced to I Do, I Do, I Do, I Do, I Do at their wedding. 5 of our moms having the same favorite band is a wild ass coincidence.

Steph: 6 actually.... *narrows her eyes and looks around the room* Well, at least we don't have to worry about The Daughter of the Demon liking ABBA.

Damian: And to no one's surprise, Brown is confidently wrong once again. *dodges a thrown book* Mother never found ABBA as frivolous as other western music, for whatever reason.

Cass: I don't believe Lady Shiva has ever listened to music before so I think I'm out of this one.

Tim: Ahh, sorry Cass but Shiva did actually play music during our training at times, and ABBA was on heavy rotation.

Cass: :(

Bruce: Hnnnn

*music cuts out*

Barbara, over the speakers: Before any of you pull out some ridiculous conspiracy theory revolving around 80's Swedish Pop sensation ABBA, let me remind you that they are literally peak Mom Music. Moms love ABBA, it's not weird. It's hardly even a coincidence. I would be more surprised if your mothers DIDN'T listen to ABBA when you were growing up.

(Bruce is still compiling a file of possible ways ABBA could be used for evil.)

[spirkbitch]

(Didn’t include tas since it’s pretty much just tos but animated)

Put in the tags what you think your job/role would be too (because I think it would be fun)

[physicallyimprobable]

what's the 3-dimensional number thing?

Well I'm glad you asked! For those confused, this is referring to my claim that "my favorite multiplication equation is 3 × 5 = 15 because it's the reason you can't make a three-dimensional number system" from back in this post. Now, this is gonna be a bit of a journey, so buckle up.

Part One: Numbers in Space

First of all, what do I mean by a three-dimensional number system? We say that the complex numbers are two-dimensional, and that the quaternions are four-dimensional, but what do we mean by these things? There's a few potential answers to this question, but for our purposes we'll take the following narrative:

Complex numbers can be written in the form (a+bi), where a and b are real numbers. For the variable-averse, this just means we have things like (3+6i) and (5-2i) and (-8+3i). Some amount of "units" (that is, ones), and some amount of i's.

Most people are happy to stop here and say "well, there's two numbers that you're using, so that's two dimensions, ho hum". I think that's underselling it, though, since there's something nontrivial and super cool happening here. See, each complex number has an "absolute value", which is its distance from zero. If you imagine "3+6i" to mean "three meters East and six meters North", then the distance to that point will be 6.708 meters. We say the absolute value of (3+6i), which is written like |3+6i|, is equal to 6.708. Similarly, interpreting "5-2i" to mean "five meters East and two meters South" we get that |5-2i| = 5.385.

The neat thing about this is that absolute values multiply really nicely. For example, the two numbers above multiply to give (3+6i) × (5-2i) = (27+24i) which has a length of 36.124. What's impressive is that this length is the product of our original lengths: 36.124 = 6.708 × 5.385. (Okay technically this is not true due to rounding but for the full values it is true.)

This is what we're going to say is necessary to for a number system to accurately represent a space. You need the numbers to have lengths corresponding to actual lengths in space, and you need those lengths to be "multiplicative", which just means it does the thing we just saw. (That is, when you multiply two numbers, their lengths are multiplied as well.)

There's still of course the question of what "actual lengths in space" means, but we can just use the usual Euclidean method of measurement. So, |3+6i| = √(3²+6²) and |5-2i| = √(5²+2²). This extends directly to the quaternions, which are written as (a+bi+cj+dk) for real numbers a, b, c, d. (Don't worry about what j and k mean if you don't know; it turns out not to really matter here.) The length of the quaternion 4+3i-7j+4k can be calculated like |4+3i-7j+4k| = √(4²+3²+7²+4²) = 9.486 and similarly for other points in "four-dimensional space". These are the kinds of number systems we're looking for.

[To be explicit, for those who know the words: What we are looking for is a vector algebra over the real numbers with a prescribed basis under which the Euclidean norm is multiplicative and the integer lattice forms a subring.]

Part Two: Sums of Squares

Now for something completely different. Have you ever thought about which numbers are the sum of two perfect squares? Thirteen works, for example, since 13 = 3² + 2². So does thirty-two, since 32 = 4² + 4². The squares themselves also work, since zero exists: 49 = 7² + 0². But there are some numbers, like three and six, which can't be written as a sum of two squares no matter how hard you try. (It's pretty easy to check this yourself; there aren't too many possibilities.)

Are there any patterns to which numbers are a sum of two squares and which are not? Yeah, loads. We're going to look at a particularly interesting one: Let's say a number is "S2" if it's a sum of two squares. (This thing where you just kinda invent new terminology for your situation is common in math. "S2" should be thought of as an adjective, like "orange" or "alphabetical".) Then here's the neat thing: If two numbers are S2 then their product is S2 as well.

Let's see a few small examples. We have 2 = 1² + 1², so we say that 2 is S2. Similarly 4 = 2² + 0² is S2. Then 2 × 4, that is to say, 8, should be S2 as well. Indeed, 8 = 2² + 2².

Another, slightly less trivial example. We've seen that 13 and 32 are both S2. Then their product, 416, should also be S2. Lo and behold, 416 = 20² + 4², so indeed it is S2.

How do we know this will always work? The simplest way, as long as you've already internalized the bit from Part 1 about absolute values, is to think about the norms of complex numbers. A norm is, quite simply, the square of the corresponding distance. (Okay yes it can also mean different things in other contexts, but for our purposes that's what a norm is.) The norm is written with double bars, so ‖3+6i‖ = 45 and ‖5-2i‖ = 29 and ‖4+3i-7j+4k‖ = 90.

One thing to notice is that if your starting numbers are whole numbers then the norm will also be a whole number. In fact, because of how we've defined lengths, the norm is just the sum of the squares of the real-number bits. So, any S2 number can be turned into a norm of a complex number: 13 can be written as ‖3+2i‖, 32 can be written as ‖4+4i‖, and 49 can be written as ‖7+0i‖.

The other thing to notice is that, since the absolute value is multiplicative, the norm is also multiplicative. That is to say, for example, ‖(3+6i) × (5-2i)‖ = ‖3+6i‖ × ‖5-2i‖. It's pretty simple to prove that this will work with any numbers you choose.

But lo, gaze upon what happens when we combine these two facts together! Consider the two S2 values 13 and 32 from before. Because of the first fact, we can write the product 13 × 32 in terms of norms: 13 × 32 = ‖3+2i‖ × ‖4+4i‖. So far so good. Then, using the second fact, we can pull the product into the norms: ‖3+2i‖ × ‖4+4i‖ = ‖(3+2i) × (4+4i)‖. Huzzah! Now, if we write out the multiplication as (3+2i) × (4+4i) = (4+20i), we can get a more natural looking norm equation: ‖3+2i‖ × ‖4+4i‖ = ‖4+20i‖ and finally, all we need to do is evaluate the norms to get our product! (3² + 2²) × (4² + 4²) = (4² + 20²)

The cool thing is that this works no matter what your starting numbers are. 218 = 13² + 7² and 292 = 16² + 6², so we can follow the chain to get 218 × 292 = ‖13+7i‖ × ‖16+6i‖ = ‖(13+7i) × (16+6i)‖ = ‖166+190i‖ = 166² + 190² and indeed you can check that both extremes are equal to 63,656. No matter which two S2 numbers you start with, if you know the squares that make them up, you can use this process to find squares that add to their product. That is to say, the product of two S2 numbers is S2.

Part Four: Why do we skip three?

Now we have all the ingredients we need for our cute little proof soup! First, let's hop to the quaternions and their norm. As you should hopefully remember, quaternions have four terms (some number of units, some number of i's, some number of j's, and some number of k's), so a quaternion norm will be a sum of four squares. For example, ‖4+3i-7j+4k‖ = 90 means 90 = 4² + 3² + 7² + 4².

Since we referred to sums of two squares as S2, let's say the sums of four squares are S4. 90 is S4 because it can be written as we did above. Similarly, 7 is S4 because 7 = 2² + 1² + 1² + 1², and 22 is S4 because 22 = 4² + 2² + 1² + 1². We are of course still allowed to use zeros; 6 = 2² + 1² + 1² + 0² is S4, as is our friend 13 = 3² + 2² + 0² + 0².

The same fact from the S2 numbers still applies here: since 7 is S4 and 6 is S4, we know that 42 (the product of 7 and 6) is S4. Indeed, after a bit of fiddling I've found that 42 = 6² + 4² + 1² + 1². I don't need to do that fiddling, however, if I happen to be able to calculate quaternions! All I need to do is follow the chain, just like before: 7 × 6 = ‖2+i+j+k‖ × ‖2+i+j‖ = ‖(2+i+j+k) × (2+i+j)‖ = ‖2+3i+5j+2k‖ = 2² + 3² + 5² + 2². This is a different solution than the one I found earlier, but that's fine! As long as there's even one solution, 42 will be S4. Using the same logic, it should be clear that the product of any two S4 numbers is an S4 number.

Now, what goes wrong with three dimensions? Well, as you might have guessed, it has to do with S3 numbers, that is, numbers which can be written as a sum of three squares. If we had any three-dimensional number system, we'd be able to use the strategy we're now familiar with to prove that any product of S3 numbers is an S3 number. This would be fine, except, well…

3 × 5 = 15.

Why is this bad? See, 3 = 1² + 1² + 1² and 5 = 2² + 1² + 0², so both 3 and 5 are S3. However, you can check without too much trouble that 15 is not S3; no matter how hard you try, you can't write 15 as a sum of three squares.

And, well, that's it. The bucket has been kicked, the nails are in the coffin. You cannot make a three-dimensional number system with the kind of nice norm that the complex numbers and quaternions have. Even if someone comes to you excitedly, claiming to have figured it out, you can just toss them through these steps: • First, ask what the basis is. Complex numbers use 1 and i; quaternions use 1, i, j, and k. Let's say they answer with p, q, and r. • Second, ask them to multiply (p+q+r) by (2p+q). • Finally, well. If their system works, the resulting number should give you three numbers whose squares add to 15. Since that can't happen, you've shown that the norm is not actually multiplicative; their system doesn't capture the geometry of three dimensions.

[kayatoastkkat]

guys help random doodles keep appearing in my notebook whenever I try to revise my maths 💔 every time I do like 2 questions this happens how to stop 😔😔

[woodswolf]

a tribute to my favorite post that came out of the pikposting phenomena from august to october of 2023, originally written by @olimar-posting. i have loved this post ever since i first read it and i still think of it very fondly. pure poetry

(Image IDs in alt)

[vinestaff]

whitepine brainrot during math class.

[redge]

You know it's crazy how I am really very convinced that Five is an old man stuck in a teen's body.

Like that Reginald and Five conversation in Season 2 Episode 6? The whole scene I never thought there was something wrong with Five drinking, or Five talking things like "when we were kids" or Five saying he gave Reginald such a hard time as a kid as if it's not a kid's body speaking of it. The scene was simply two old men talking over a drink. Plain and simple.

I remember Aidan answering this question "what message would you give to someone who wants to be an actor" and aside from researching a character, he concluded his answer with "you just really need to make friends with the make believe" and I am very sure Aidan practice what he preach. He loves TUA comics so he knows his character, and he really got me with that make believe that he is indeed an old man in a teenagers body.

In season 4, Aidan is no longer a teenager acting as an adult but he's already an adult acting as an old man but even that was convincing for me. Also the way he played all those Five alternatives? Booth Five is very reminiscent of Five in season 1 and season 2. Drunk Five? We've seen that in season 3. Brisket Five? Oh what a joy to see a smiling-not-living-on-the-edge Five. I mean to separate all those Fives as if they're really different versions of you? What an astounding actor.

I know he is now focusing on his music but I am excited to see him in the next role he will portray just as much as I am excited in the music he is making.

[laser-tripwires]

so he DID cheat. multiple times. and somehow eight years after going no contact he still tried to blame her for the relationship ending because he didn't want to think about his own culpability.

eliot is my favourite character in anything ever, complex characters are the best, and SUCK IT i was right.

[karliahs]

disappointing that we don't know who canonically teaches the gen ed class homerooms because I have to imagine whoever it is wants aizawa's fucking blood

[antlereed]

i feel like crawling into a hole for asking this but: my financial aid from school got routed towards my tuition (an objectively good thing) but that leaves me with no way to cover bills and rent at all so if anyone can toss a little bit my way i would be so appreciative

id be willing to do a little drabble writing for folks who help me out!

[constellation-skirmishes]

disaster siblings again

[lipshits-continuous]

Sorry for the de Rham cohomology posting but another thing I've been thinking about:

I also think it's really cool that we can use tools from algebraic topology to answer "when are closed forms exact?". Like even just the fact we have Mayer-Vietoris means that the problem for certain manifolds becomes a lot easier! You no longer have to worry as much about the differential forms themselves. And then obviously de Rham's theorem gives you an even more powerful tool for this!