A place for everything about math

Random thought on magic squares:

If I view the smallest possible non-trivial magic square

2 7 6
9 5 1
4 3 8

since its rows and diagnoals sum up to 2+5+8 = 2+7+6 = 4+5+6 = 2+9+4 = … = 15

in the article as a 3x3 Matrix, its determinant is Δ = -360 . Its inverse:

-37/360 19/180 23/360
17/90 1/45 -13/90
-7/360 -11/180 53/360

note how this is a magic square, rows and diagonals sum up to 1/15.

https://matrix.reshish.com/inverse.php

Now if you are really bored (I can not do this): proof that for any non trivial magic squares the inverse …

• exists (i.e. every non-trivial magic square has an inverse)
• is a magic square.

Two students who discovered a seemingly impossible proof to the Pythagorean theorem in 2022 have wowed the math community again with nine completely new solutions to the problem.

While still in high school, Ne'Kiya Jackson and Calcea Johnson from Louisiana used trigonometry to prove the 2,000-year-old Pythagorean theorem, which states that the sum of the squares of a right triangle's two shorter sides are equal to the square of the triangle's longest side (the hypotenuse). Mathematicians had long thought that using trigonometry to prove the theorem was unworkable, given that the fundamental formulas for trigonometry are based on the assumption that the theorem is true.

Jackson and Johnson came up with their "impossible" proof in answer to a bonus question in a school math contest. They presented their work at an American Mathematical Society meeting in 2023, but the proof hadn't been thoroughly scrutinized at that point. Now, a new paper published Monday (Oct. 28) in the journal American Mathematical Monthlyshows their solution held up to peer review. Not only that, but the two students also outlined nine more proofs to the Pythagorean theorem using trigonometry.

Calculator: https://www.omnicalculator.com/everyday-life/dilution-ratio

If I type in the dilution ratio and final volume it calculates the concentrate amount and water amount but I don't know how it does that and want to find out how it does that

Then I am stuck. I think the provided answer contains an error. But even if they are right, why does this last step equal f(x,y) + g(y) ????

(For the sake of intuition, 1/√0=0)

I'm thinking re the latest vid of @mindyourdecisions

No need to view his vid. Here's the problem –

Brian has some boxes of paper clips. Some boxes hold 10 clips and some boxes hold 100. He has some paper clips left over. He has 3 more boxes with 100 paper clips than he has boxes with 10 paper clips. He has 2 fewer paper clips left over than he has numbers of boxes with 100 paper clips. What number of paper clips could he have?

• let x1 be the number of boxes with 10 clips
• x2 be the number of boxes with 100 clips
• n be the number of leftover clips

I thought of 100x2 = 10x1 + 300

Is that equation right? Something tells me I shouldn't equate 100x2 to 10x1 plus 300. Something tells me I shouldn't make an equation re number of clips as it isn't explicit in the problem. I'm confused.

Hi,

I found online a nice (and seemed easy) math problem.

Rocket A travel from Mars to Earth in 200 days
Rocket B travel from Earth to Mars in 150 days, but take off 30 days later

When they cross each other, which one is the closet to the earth ?

So they give a "flat" answer, without giving any explanation on how they reach this conclusion.

What would be your simplest Mathematical solution for this ?

Thanks.

So this is bugging me for a while and I'm just do dumb to get how I solve this, but here's the situation:
Given I take a local backup of my system daily and have a retention policy that keeps a backup of the past 7 days each, a backup of the past 4 weeks each and a backup of the past 6 month each. That's either 17 backups or less if you consider some backups being counted as a daily and weekly or as a weekly and monthly. But that's not that important.
The interesting part is, that I also take a remote backup of my local backup daily, which has the same retention policy, so it's cascading. Here there is obviously a huge overlap of backups, but I can't wrap my head around, how I calculate this.
Is anybody willing and/or interested to solve this for and with me?

This year's Abel Prize has just been awarded to Michael Talagrand. I didn't knew about his work, but it seems really interesting and he made an effort to make it really accessible both to read and access.

Isn't it just "composite"?

Every arrow in category can be composed, the set(or class or whatnot..) of that is composite.

In trying to figure out the answer to my homework problem, I came across this volume, which I thought the community might find interesting and/or helpful.

I've been knocking out the trig problems in this section with minimal difficulty so far, but I've run straight into a brick wall on this "Algebraic" part. I'm asked to find sin(x)=0 between [0,2π). If I graphed the unit circle this would be a trivial exercise to show sin(θ)=0 when θ=0 or π.

Where I have trouble is- I'm very explicitly being told here that the solution is ALGEBRAIC, and I'm struggling to figure out a way to rearrange sin(x)=0 to come up with the known answer. Further, unit circles are not in this chapter, they wouldn't likely ask me to exercise a skill taught in another chapter. What am I missing?

It's not just 31, either. Looking ahead at eg 37, I can easily show sin(-x) = -sin(x) on a unit circle. I could maybe fuck around with inverse trig ratios but those are in section 3- this is only section 1.

Help me out here, drop a hint, share a link: how do I solve sin(x)=0 on [0,2π), but algebraically? I suspect it's something glaringly obvious and/or very very simple I've overlooked.

It's homework help, but I'm not asking for the solution. The problem only asks for cos, sin, tan, cot, csc given sec. I found those pretty quickly on my own, and confirmed solutions with the back of the book.

Where I run into confusion is when I try to find angle theta on my own. Arccos of found cos gives 2.06, arcsin of found sin gives 1.08, and arctan of found tan gives -1.08. Problem givens exclude possibility of the negative angle found by arctan(-15/8), but the other two are possible and conflicting. And why wouldn't they all be the same? I reattempted because there were so many erase marks from trying to figure this out that it was almost illegible.

Am I wrong? Did the book give me a point not on the unit circle or something, assuming I wouldn't try to find theta on my own? Have I used arcfuncs wrong- I checked the domains against the function definitions? Have I found a hole in math?

We have "triangle" "rectangle" "pentagon"...etc "tetrahedral" "cube" "octahedron" ..etc

Instead of having to say "group" all the time, like "dihedral group" "cyclic group", if we make it into one word it will sound more like an elementary mathematical object.

What would be a nice suffix for group?

Hello.

I am currently inventing a language, and have created a base 4 number system for it. Unfortunately, I am horrible with numbers, even in decimal. So it was a hard slog. But I finally got there.

It would be great if I could know of any practical applications quaternary has (if any), so I can incorporate it into the language and make it more naturalistic. Thanks.

I feel like this has to be a math/logic thing that has a name already and I wanna know what it's called so I can look it up when I'm no longer extremely drunk.

In this phone game the objective is to get all the people on all the same color floors with as few stops at any floor as possible. When the last few moves look like this, you just have to go through in the right order and only stop at each stop once (except the first/last floor).

But sometimes there's different little sub-sets of pairs inside the bigger set of pairs that are self-contained, and for each one of those there's another floor that has to be started and stopped on to complete that loop. That makes the minimum number of moves to solve: the sum of the number of pairs in both sub-sets together plus the number of subsets. (And only counting the number of pairs in both subsets because if one of the pairs is already matched it won't count for the moves).

So like these two are all one big continuous loop: A-E, B-A, C-B, D-C, E-D and A-B, B-E, C-A, D-C, E-D

And this one has one already matched leaving a single complete loop in need of matching: A-B, B-E, C-A, D-D, E-C

These ones, however, have two loops. one loop that's three floors long (four moves) and one that's two floors long (three moves): A-B, B-C, C-A, D-E, E-D and A-D, B-E, C-A, D-C, E-B

And these ones have one already matched pair, and two sub-sets of two that still need to be matched: A-B, B-A, C-C, D-E, E-D and A-D, B-B, C-E, D-A, E-C

What is this called?