help-49

[grand pond]

ImOakley

[mystic condor]

Which has no surd

So the denominator is rationalised

Sorry a-b^2

Not a^2

[midnight plank]

@maiden bridge Has your question been resolved?

[midnight plank]

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[tawdry kraken]

First I wrote this Lemma to help me

\begin{lemma}
\label{dominotile}
  Dominos can be used to tile any $m \cross 2n$ rectangle.
\end{lemma}
\begin{proof}
  Place the dominoes horizontally across the first row. Each covers two units, so with $n$ dominos, a $1 \cross 2n$ rectangle is tiled.
  Repeat this process for rows $2$ through $m$.
\end{proof}

[grand pond]

Coolempire2026

[junior flower]

[tawdry kraken]

These proofs were all pretty straightforward, just looking for a quick lookover.

\begin{proof}[Proof of \textbf{41}]
  Rotate the board so that the two missing corners are at the top of the board.
  Note that between the two removed corners, a $1 \cross 6$ rectangle is formed.
  The rest of the board forms a $7 \cross 8$ rectangle.
  By Lemma \ref{dominotile}, both of these regions can be tiled.
\end{proof}
\begin{proof}[Proof of \textbf{42}]
  Note that between the two missing corners at top and bottom, $1 \cross 6$ rectangles are formed.
  In the interior of the board, a $6 \cross 8$ rectangle is formed.
  By Lemma \ref{dominotile}, all three of these regions can be tiled.
\end{proof}
\begin{proof}[Proof of \textbf{43}]
  Recall that the area of a rectangle is given by $A = \ell \cdot w$.
  Therefore, if the area is even, then $2 | \ell$ or $2 | w$.
  If $2 | w$, we have by Lemma \ref{dominotile} that the board can be tiled.
  If $2 | \ell$, rotate the board 90 degrees. Now we have a $w \cross \ell$-sized board, which can thus be tiled by Lemma \ref{dominotile}.
\end{proof}

[grand pond]

Coolempire2026

[tawdry kraken]

Then this one I have a strong suspicion is false but I have to figure out how to go about

\begin{proof}[Disproof of \textbf{44}]
  :)
\end{proof}

[grand pond]

Coolempire2026

[tawdry kraken]

👋 😄

[junior flower]

for that, find an invariant that applies when you put dominos down

related to the colors

[tawdry kraken]

(I was really smart for splitting the lemma into a different message, look how it turned my references)

[tawdry kraken]

Well I did also want to make sure that they seem valid/rigourous

Since I've never done proofs with physical objects before

Enumerations are my thing not constructions :P

[junior flower]

the proofs for 41, 42, 43 look fine to me

[tawdry kraken]

Ah wait it may be simple then

[junior flower]

||whenever you place a domino, you cover 1 black and 1 white tile. so the number of black and white tiles must be the same to have any chance at tiling||

[junior flower]

this will not be the case in problem 44

[tawdry kraken]

\begin{proof}[Disproof of \textbf{44}]
  Note that a domino covers both a white and a black square of the checkerboard when placed. 
  Therefore, for the checkerboard to be tileable with dominos, there must be the same number of white squares as black squares.
  In a $5 \cross 5$ checkerboard, all of the corners have the same color, WLOG call them white.
  Then in total there are 13 white squares in the checkerboard and 12 black squares.
  Removing three of the white squares (the three corners), then, leaves the board with 10 white squares and 12 black squares, meaning tiling is impossible.
\end{proof}

[grand pond]

Coolempire2026

[junior flower]

true facts

[tawdry kraken]

This is where things start getting hard

[junior flower]

proof by exhaustion...

[tawdry kraken]

Maybe because I'm no expert in symmetry like I should be 😂

[junior flower]

you can just use this same idea

what do they mean proof by exhaustion

[tawdry kraken]

As in demonstrate that every possible attempt results in a contradiction

[junior flower]

so is this allowed or not lol

[tawdry kraken]

Hint: number the squares of the

[junior flower]

[tawdry kraken]

No, because they already did that proof

XD

[junior flower]

oh ok lol

[tawdry kraken]

Regardless this feels nasty

[junior flower]

agreed

it's literally telling you to do it a nasty way

X112 5442 5633 ?6?X

if you cover tile 2 horizontally (with domino 11)

then you need to cover tile 4 vertically (with domino 22)

and there are more decisions forced on you by tiles that can only be tiled one way and you end up there

which can't be completed

i wrote them in order. so like 33 is the next one that is forced when you place 11 and 22

the case where you tile tile 2 (not domino 22) vertically is similar. you have tiles you can only tile 1 way

so i'll say it might not be THAT bad depending on how much explanation is expected

[tawdry kraken]

\begin{proof}
  Number the checkerboard rowwise 1 through 16 and remove two corners, 1 and 16.
  Begin, WLOG, by placing a tile on square 2. 
  This must be either vertical or horizontal.
  Throughout this proof, we will refer to ``forced" placings.
  If a square only has one (untiled) square adjacent to it, a domino is forced to cover those two squares for a complete tiling.

  Now, if the tile placed on square two is vertically oriented, i.e., covering tiles 2 and 6, then the placement of the domino covering tiles 5 and 9 is forced, which forces one to be placed covering 13 and 14, which forces one to be placed covering 10 and 11, which leaves no untiled squares adjacent to 15, meaning that tiling in this manner is impossible. The last unforced tile placement was that covering square 2, so it cannot be placed vertically.

  Therefore, the domino covering square two must be placed horizontally. This placement forces a domino covering tiles 4 and 8, which forces a domino covering tiles 12 and 11, which forces a domino covering tiles 15 and 14, which forces a domino covering tiles 13 and 9, which forces a tile covering dominos 5 and 6, leaving squares 10 and 7 with no adjacent squares, and thus they are untileable.
  The last unforced placement was that covering square two, so it cannot be placed horizontally.

  Because no domino can be placed such that square 2 is covered, this board cannot be tiled.
\end{proof}

[grand pond]

Coolempire2026

[junior flower]

i feel like a picture says a million words here

[tawdry kraken]

Almost certainly so

Yes now come the hard ones

Proving that you can instead of that you can't

[junior flower]

what's the top left tile color?

just for reference

[tawdry kraken]

Actually I might be able to do it

[junior flower]

your lemma from earlier should be helpful

[tawdry kraken]

My thoughts exactly

[midnight plank]

@tawdry kraken Has your question been resolved?

[junior flower]

this is my hint for the problem i think

ok there are different places you could go with that hint but i think i have a really good proof

[tawdry kraken]

Almost done (halfway)

But once I finish this tbh I could actually just remove the first parts

And condense 😂

Yeah this is too long I'm going straight for the big money (bad idea)

Here's what I had so far, just so you could see

[junior flower]

i'm gonna write mine up i just need to make the pictures

[tawdry kraken]

\begin{proof}
  We go by considering a series of smaller checkerboards.
  
  Begin with a checkerboard of size $2 \cross 2$.
  When both a white and a black square are removed, they must be adjacent, leaving an adjacent white and black square which can be tiled by a single domino.

  Now consider a checkerboard of size $2 \cross 2n$.
  There are a few cases by which a white and a black square may be removed.
  First consider when they are adjacent.
  If they are in the same row, say at positions $i$ and $i+1$, then we have a $2 \cross 2$ rectangle (formed with positions $i$ and $i+1$ of the other row), which can be tiled.
  Furthermore, if $i$ is even, then the first $i-2$ columns of the two rows form a size $2 \cross 2m$ rectangle, which can be tiled by Lemma 1, and the $i-1$ column can be tiled across both rows with one domino.
  A similar procedure can be applied to the latter $2n-(i+1)$ columns, because if $i$ is even, then $2n-(i+1)$ is odd.
  Otherwise, if $i$ is odd, then the first $i-1$ columns and latter $2n-(i+1)$ columns of the two rows form a size $2 \cross 2m$ rectangle, which can be tiled by Lemma 1.
  Now we consider the same case, but when the missing tiles are not adjacent.
  Because the removed tiles are of different colors, they must be an odd L1 distance away from each other.
  This is important, because we can treat them as the edges of 

  Finally, consider a checkerboard of size $2m \cross 2n$.
\end{proof}

[grand pond]

Coolempire2026

[junior flower]

lemma 2: an m x n checkerboard with m even, odd where opposite corners are removed can be tiled

proof: not right now

now back to the 8x8, let b,w be black and white tiles. consider the lime green rectangles drawn below around b and w (i'm just using a picture rn to "define" them with an example. i want this to be as intuitive as possible, not a formal proof yet). the portions of the checkerboard outside the green can be tiled by lemma 1 since they are made of rectangles of length or width 8. now notice that one of these green rectangles is an 8 x even rectangle (this must be so, because if the horizontal and vertical distance between b and w is the same parity, they are the same color). that means the region outside the red rectangle can be tiled. furthermore, the red rectangle can be tiled by lemma 2

oops wrong images

that's the other one

for proof of lemma 2, just tile the right and left borders vertically then use lemma 1 for the remaining (m-2) x n size board

[tawdry kraken]

\begin{proof}
  Suppose that the two removed tiles are located at positions $(a,b)$ and $(c,d)$ (row-column counting from 1 from the top left, so (1,1) is the white top left square).
  First, note that we can cut off groups of two columns/rows (starting from the edges of the checkerboard) outside of the rectangle formed with corners $(a,b)$ and $(c,d)$, because these can be tiled by Lemma 1.

  Therefore, we work with a size $2m \cross 2n$ checkerboard with $a,c \in \{1,2,2m-1,2m\}$ and $b,d \in \{1,2,2n-1,2n\}$.
  Note that the case where $m = n = 1$ is trivial.
  Now, we can break our $2m \cross 2n$ checkerboard into smaller further into $2 \cross 2$ boxes.
  We assume that the removed tiles are in different boxes (since otherwise, $m = n = 1$), and we have by definition that both boxes are on the edge and at corners of our smaller checkerboard.
  We can tile the box containing one of the removed squares with two dominos, such that one domino sticks out into an adjacent square.
  We now treat the square that the domino sticks into like the `removed square', redrawing the smaller checkerboard until the last domino sticks into the box with the other removed square.
  Now, one may note (by observation) that the domino which sticks out must cover a square of the same color as that of the removed square whose box it is exiting.
  Therefore, the `last domino' which sticks into the box with the other removed square must cover a square adjacent to it, morphing into the case of $m = n = 1$, and we are done.
\end{proof}

[grand pond]

Coolempire2026

[tawdry kraken]

Turns out the (a,b) (c,d) stuff wasn't necessary but I think that's it

Also our power went out jfyi

[junior flower]

wow did we have the same proof

(i haven't finished reading yet)

hm no we did not

i can't follow your proof

[tawdry kraken]

Oh why not

Give me two squares to remove I can draw it

Also maybe it's clearer if you skip the coordinate stuff at the top

[tawdry kraken]

Actually even the making the checkerboard smaller isn't necessary

[raw vector]

What's the current problem? @tawdry kraken

[tawdry kraken]

It's just saying that the two points are in the first two (or last two) rows and columns

But it turned out unnecessary

[raw vector]

I see the pins got 4 problems

[junior flower]

and it's none of those

[tawdry kraken]

(Because if they were further in the interior then we would have cut off more rows/columns)

[tawdry kraken]

I'll repin

[tawdry kraken]

Last pinned q

[raw vector]

46?

[tawdry kraken]

Yes

[junior flower]

what is the 2m x 2n checkerboard?

[tawdry kraken]

But yeah I can simplify that a little bit (since it turned out unnecessary

[junior flower]

i need a drawing

[tawdry kraken]

That's why I said give me two squares and I can draw it for you

[tawdry kraken]

A smaller checkerboard within the 8x8 one (the minimum size necessary to fit both of the removed squares)

[junior flower]

these, since i had this image on hand

[tawdry kraken]

By only cutting of groups of 2 rows/2 columns from the 8x8 checkerboard

Okay, now I have to figure out how to do this with no power

Ah I can preview the image on the computer

Here is the 2mx2n checkerboard (smallest checkerboard I can make by cutting off groups of 2 rows and columns from the 8x8)

Now I split this checkerboard into 2x2 boxes

[raw vector]

Well you can tile a 8x8 checkerboard with all vertical dominoes then remove the domino on the removed white square (which creates an empty black square). By rotating the domino orthogonally adjacent to that empty square 90°, you can make the empty slot moves diagonally until it's the same column as the removed black square. Then by shifting the domino of that column only, you can get a tiling

[tawdry kraken]

I would've liked a hint to figure that out instead of the solution but I'll consider it after I finish explaining this and then reading Layla's idea

[raw vector]

I'm not sure if my solution is correct lol

[tawdry kraken]

Oh the bottom 3 boxes can go

Turns out I can remove 4 rows from the bottom

Regardless the idea is thus

[raw vector]

My last part of shifting the dominoes only in the column is only true if none of the dominoes in the column between the empty slot and the remived black square was rotated in the process

[tawdry kraken]

I can tile the box with the removed square

And one domino will stick out into another box

It will have the same color as the removed square

So we treat it like a new 'removed square'

And then repeat

Say I tiled like this

Now I have to do it again, in this case it is forced

Again

Finally they both lie within a single box

But since they do, and the domino that's sticking out into the box has the same color as the other removed square, it must be adjacent to the removed square in that box

Thus we can tile the remaining area

And the rest of the board is 2x2 boxes

Meaning we can tile by Lem. 1

[raw vector]

||What does a square being white and a square being black tell us about the rectangle with them as opposite corners||

[junior flower]

that's the idea used in my proof

[tawdry kraken]

Although I think the proof can go even quicker if you prove that no 'stick-out' tiling can cause a losing scenario

And then you're done in like 2 paragraphs instead of 5

[junior flower]

the spoilered message

[tawdry kraken]

Oo nice

Omg i'm almost as good as layla

😂

[junior flower]

[tawdry kraken]

What does it tell us 👀

[junior flower]

||it is an even x odd rectangle||

[tawdry kraken]

XD proof not right now

Yeah I thought about the MBR

But I though that the even MBR might be eaiser

[junior flower]

MBR?

[raw vector]

This

[tawdry kraken]

Minimum bounding rectangle

[junior flower]

ah

[tawdry kraken]

Sorry 😅 CS major

[raw vector]

And yeah this problem definitely also works for any 2nx2n board (obvious, but i just wanted to add that)

[junior flower]

the green rectangles are there just to assist in tiling outside the MBR

[tawdry kraken]

Right

[junior flower]

you can ignore them, this is really the important part

that you can tile the MBR

well that's what lemma 2 says

[tawdry kraken]

AH right

[junior flower]

so you can tile the MBR, and then the rest is easy to tile following one of the green rectangles

[tawdry kraken]

That makes sense

[junior flower]

whichever one is 8 x even

[unkempt warren]

hi guys whats the question

[tawdry kraken]

Last pinned q (46)

[unkempt warren]

also @junior flower ur help channel is open is ur doubt solved its open since last night 😭

[junior flower]

my doubt is UNSOLVED

[tawdry kraken]

It's been open for like 2 days now

It's so unsolved they are even using computational methods to attack it

[unkempt warren]

Oh I'll check that too tho im dumb myself

[junior flower]

it is a tuff one

[unkempt warren]

I'll check this first

[tawdry kraken]

Well techically this one is solved

I'm just rewriting my proof now

And then I'm going to try to write NerdyAsianGuy's idea into a proof

And layla's idea is flawless as usual

[unkempt warren]

this question is tougher for me cz I never played checkers 😭

[tawdry kraken]

See the last pin for a picture of a checkerboard

[unkempt warren]

oh its same as chess board

[tawdry kraken]

I'm realizing how ingenious it is to use the corners, now I feel stupid 😂

[junior flower]

everyone's been playing checkers but i've been playing chess for years

[unkempt warren]

elo?

[junior flower]

800

[tawdry kraken]

Dragon sicilian

Najdorf >>

No actually hyperaccelerated dragon 😂

[unkempt warren]

ah do lmk if u want a game but lets not talk about it here coolempire might feel bad I can understand what if feels to like to get stuck on a problem 😭

[junior flower]

i have not actually been playing chess for years

i have only played a few times in my life

i don't even know the rules

and my elo is probably lower than 800

[tawdry kraken]

That's okay I'll play with you another time if layla doesn't want to

💃 💃

[unkempt warren]

sure dude

also

did u search the solution on google yet?

[tawdry kraken]

No, because I have three working solutions here 😆

[unkempt warren]

oh

[tawdry kraken]

It's not like a homework problem or anything

I'm working on my combinatorics basics

[unkempt warren]

maybe u can ask ur prof and show him that u could solve this amount of question but not rest?

[unkempt warren]

ohh

[junior flower]

this is for the love of combinatorics mr WBJKR

[tawdry kraken]

Basically just grabbing all of the hardest (* and **) problems from the book and skipping the rest

[unkempt warren]

I just typed anything while making my username now its wbjkr

u can call me amay hahaha

[raw vector]

That problem makes me wonder one thing

[unkempt warren]

thats not basics...

[raw vector]

How many ways are there to tile a 2nx2n board with 1x2 tiles?

[tawdry kraken]

That I don't know, although I have an idea on how to start the problem

[unkempt warren]

in my love for permutation and combination or binomial theorem I wont solve toughest I'd go from easiet to hardest

[junior flower]

sounds like a good dynamic programming problem

[tawdry kraken]

I haven't actually executed it to learn yet

[tawdry kraken]

Good idea

[raw vector]

Induction?

[junior flower]

dynamic programming is my second favorite thing. second only to generating functions

[tawdry kraken]

I didn't say I had a technique in mind just an idea 😂

[unkempt warren]

start it. Write it down on paper maybe open the book or smth and look for other info u can use. I might sound dumb cz im not helping and uk that already 😭 but sometimes it helps

[tawdry kraken]

Maybe that's a combinatorics thing then

All my friends hate DP

[unkempt warren]

u cant always solve a question in ur mind

[tawdry kraken]

But it's my personal favorite

[junior flower]

i love DP

[tawdry kraken]

And the Master Theorem makes analysis easy

[unkempt warren]

whats dp?

[raw vector]

Dynamic programming

[junior flower]

i'm so good at self control today

[unkempt warren]

we have 4 chess noobs in chat?

[tawdry kraken]

literally

I thought you were gonna let it out

XD

[unkempt warren]

read ur status I dont think so 😭

[raw vector]

Noobs? Your pronouns are Mag/nus

[unkempt warren]

Jk

[unkempt warren]

Cz I love magnus

loml

nah this aint general chat

lets help cool empire

[raw vector]

What does he need help with rn lol

[tawdry kraken]

Nothing to help rn while I'm writing

XD

[raw vector]

We chilling

[junior flower]

we chilling

[tawdry kraken]

https://tenor.com/view/bingchilling-gif-10067072887923246131

[unkempt warren]

ding chilling

[raw vector]

早上好中国

[junior flower]

your cheekiness will be your downfall

[stiff bison]

uhhhhh so are you all gonna get back on task or leave the channel

[raw vector]

We 're getting more people somehow

[unkempt warren]

ah I am sorry I meant it as a joke but honestly saying I dont think I'll have a downfall I am already dumb

[raw vector]

We're waiting for the guy to write the proof out

[junior flower]

we are on task

[stiff bison]

fair enough

[junior flower]

no rush though coolemplud

[raw vector]

Take as much time as you need, i'm enjoying this chat

Uhm... Do y'all like geo?

[unkempt warren]

lmfao

I suck at geo

I love algebra

[silent dock]

[unkempt warren]

in fact they are teaching us geo rn

which is ur fav topic in math...?

[raw vector]

Eh... I love i equalities so i guess algebra? But generally speaking i like combinatorics

[unkempt warren]

what I feel I dont have combinatorics

or

[junior flower]

[unkempt warren]

is it just permutation and combinations...?

[junior flower]

combinatorics is the best

[raw vector]

True

[unkempt warren]

wait

[raw vector]

Ew

Bad opinion

[raw vector]

Do you like graph theory?

[junior flower]

i'm not very knowledgeable in it but i would say yes

[unkempt warren]

just searched pnc is a part of combinatorics

[junior flower]

coolemplud you need the

[raw vector]

"a part" key word

[unkempt warren]

yea its a fundamental part

anyways if I finish

my syllabus

[junior flower]

[junior flower]

that's my hand

[unkempt warren]

I'll def go deep in combinatorics

[unkempt warren]

hand reveal

actually fingers

[junior flower]

my hand was literally already revealed in my pfp

[unkempt warren]

okay nah I thought that was from pin

[junior flower]

[silent dock]

[unkempt warren]

:AA_Sus: :( 💔

WOW

NO EMOJI FOR ME

[raw vector]

[unkempt warren]

oh nvm

[tawdry kraken]

Oh see pins for the q's btw

[junior flower]

[unkempt warren]

its cz of nitro

[raw vector]

Y'all know much about dumpty points?

[junior flower]

no

[raw vector]

Sadge

[junior flower]

miss me with that geometry

[silent dock]

coolemplud i'll have a look if this remains unresolved for long enough

can't be bothered with all the chats

[raw vector]

If this is still open after the weekend I'm gonna take a serious look cuz it looks interesting enough

[tawdry kraken]

It's solved dw

[silent dock]

I'm currently investigating an important matter for the history of mathscord

[tawdry kraken]

At least until I start 47-50

[unkempt warren]

do u guys use disc the most..?

I quit instagram 🙏

[tawdry kraken]

\begin{proof}[Proof of \textbf{46+}]
  Take a size $2n \cross 2m$ checkerboard and divide it into size $2 \cross 2$ boxes.
If the removed squares are in the same box, then, because they are of different colors, they are adjacent, and we can tile the remaining two adjacent squares with one domino.
Tiling $2 \cross 2$ rectangles is trivial, so the remaining $nm-1$ boxes can also be tiled, and we are done.

  If the removed squares are not in the same box, then WLOG consider a box containing one of the two removed squares.
We can tile this box using two dominos, such that one of them extends into an adjacent box, covering a same color square as the removed square from the original box in consideration.
If the adjacent box contains the other removed square, then the domino extending into it must be adjacent to it (since they cover different colored squares), and we can tile the board from here as if the removed squares were in the same box originally.

  Finally, we note that the induction here is complete (with treating the tile extending into the adjacent box as a removed square and repeating the original procedure) because there is never a forced losing scenario.
That is to say, one can tile the whole board by this procedure (left as an exercise to the reader), meaning that it is always possible to, by a series of box tilings, tile into the box with the other removed square.
\end{proof}

[grand pond]

Coolempire2026

[tawdry kraken]

Might be just as hard to follow

But even if I can't formalize it I got the idea

[junior flower]

i'm not gonna lie i don't understand exactly what the construction of the 2n x 2m rectangle is. but i agree that you can tile a rectangle of 2x2 squares containing the punctured tiles with your method except for one issue

let me draw a picture

it kinda just reads like you start with one of the punctured 2x2 blocks pick any adjacent 2x2 block to go through and it works. but you can't e.g. tile like this

so you should pick a path that leads to the other punctured block

ok like the last paragraph may or may not have this in mind, i can't tell

[tawdry kraken]

Yeah that was the intention with it

[junior flower]

is that the exercise that was left to me

[tawdry kraken]

Yes 😂

[junior flower]

lol

[unkempt warren]

this is so tough dude Thank you so much whenever I see a tough problem I see it in my dream

[tawdry kraken]

And this is why your method is a lot more sensible

[unkempt warren]

u have already given me a dream for tonight

[tawdry kraken]

Okay great

Now that I'm done helping

Simple, remove the two squares adjacent to the corner square

[tawdry kraken]

Now it's starting to look like combi 🥹

[junior flower]

triominoes

hm there are a lot of squares

[unkempt warren]

I need to study rational point on elliptical curves seems interesting saw it on reddit

[junior flower]

oh yea i used to have a banner about that

[unkempt warren]

when? WAIT U HAVE THAT

[junior flower]

i have no more comments to make regarding this

[unkempt warren]

why lmfao

bad memories?

[tawdry kraken]

Get your elliptic curves here come one come all

[junior flower]

also it's 'elliptic' curve. this is what an elliptical is

[unkempt warren]

mb...

💔

[junior flower]

no

look i have no further comment

[unkempt warren]

oh ohkies was gonna ask 😭 its alr

[tawdry kraken]

At least it's just squares and not rects

Then it'd really be combi

Well you can form rects by adding to squares anyway

And subtracting that is

But no no squares squares

So first the simple version of the coloring argument

Which is numbering

The resulting board must have a multiple of 3 size

So I need a difference of squares equalling to a multiple of 3

[junior flower]

yea

[tawdry kraken]

8^2 - x^2 = 0 (mod 3)
(8 + x)(8 - x) = 0 (mod 3)
8 + x = 0 (mod 3)
or
8 - x = 0 (mod 3)

We have x = 1, 2, 4, 5, 8

Oh I could have simplified further

\begin{align*}
x &= \pm 8 \pmod{3} \\
x &= {-1,2} \pmod{3}
\end{align*}

[grand pond]

Coolempire2026

[tawdry kraken]

Ah yes -1 and 2 like those aren't different numbers

Should be 1,-1

1,2,4,5,7,8

Wait 7 counts

ah 7 + 8 = 15

Missed it the first time

[junior flower]

also another slight simplification. we can just think about squares whose bottom left corner is in the purple region. every possible square looks like one of those but rotated

[tawdry kraken]

idk why my scrren keeps jumping upwards every time I send a msg

[tawdry kraken]

I was originally planning to move the squares all around but I realized they meant subtract not add so then I was like o,o

[tawdry kraken]

Okay right then the rotation argument which of course you have pointed out

And then we multiply by 4 for the rots

[junior flower]

not quite

[tawdry kraken]

Ah

You're right

The ones centered in the center don't rnotate

Now to think about if there's a way to do this without placing down each one

Let me pull out the visual aid

Ah yes of course there must be the same number of every color

Currently there are 21 blue, 21 black, 22 white

Which they used to make an ingenious proof that you can't just remove the corner

Because although it gives you 63 tiles total

If you rotate it, now the corner is white

And now you have a problem

Actually I think that invalidates all squares of size 1

Because you can rotate so that it's a different color

And by rotate I mean flip the board really

[tawdry kraken]

That is to say it must start on a white

Ah wait there's one white that can't be flipped into a black or blue

This guy right here

Every other white square can undergo a symmetry transformation so that it's no longer white

[junior flower]

makes sense

[tawdry kraken]

Ah and it's fairly easy to tile with triominoes for that one

Okay so we've got 4 squares so far

Next are the 2x2's

Which I expect to have more candidates

[junior flower]

hm a lot of them will remove 2 blue and 1 black

or other way around

[tawdry kraken]

I feel like this could be automated with the dihedral group in a computer but I digress

[junior flower]

all the 2x2 squares with 2 white tiles are candidates?

[tawdry kraken]

So starting from the bottom left we have 2W1A1U

[tawdry kraken]

Until proven otherwise

[signal ibex]

looks like it yeah

[tawdry kraken]

Which it looks like bottom left is good

[junior flower]

well yes just candidates

as in that's a necessary thing

not sufficient

[tawdry kraken]

Ah there are only 3 possible squares in the bottom left quadrant anyway

XD

[junior flower]

aren't there 4?\

[tawdry kraken]

Where is the 4th 👀

Also out of those 3 looks like only the bottom left works

[junior flower]

actually 5

the 5 with these bottom left corners

those all pass the necessary 2 white tiles

[signal ibex]

2 of those are partially in the quadrant

thats where the confusion lies

they do still need to be considered

[tawdry kraken]

Oh right I forgot that

[tawdry kraken]

Also hiiiii

[signal ibex]

and the partials clearly pass the check

[tawdry kraken]

Today I wrote some uglier proofs than last time if you scroll up in chat

Which is funny given last time I had a few drinks and this time I didn't

[junior flower]

wow i don't get a hiiiii

[signal ibex]

i havent shown up for ~25 hours

it is almost 1am here

why do i do mathenatics at this hour? hotel? trivago.

[junior flower]

every hour is a good hour for combinatorics

[last slate]

whats going on

[signal ibex]

math

[junior flower]

you are so pretty

[last slate]

😭

[last slate]

nah 😭

[tawdry kraken]

Us being in the same time zone is the craziest part of this

[junior flower]

it's only 12:44 am

[tawdry kraken]

I think you were on before I got on

But yes

hiiiiii 💃

[junior flower]

thank you

[last slate]

hii!!

[junior flower]

hiiiiii

[signal ibex]

accurate but i am trying to be less of a degen with my sleeping time

the other day i woke up at 14:30 which isnt even the worst

[last slate]

where are u from??

[junior flower]

the usa

[tawdry kraken]

Oh are we all in the same time zone

That's funny

[signal ibex]

canada has 2 time zones to the east of this one

which is crazy

[last slate]

cool I am asian

[signal ibex]

the name "est" barely makes sense here

[last slate]

from ph

anyways whats the question

[junior flower]

oh cool

[junior flower]

this one

[signal ibex]

and now we're doing the removal of 2 by 2 squares

[junior flower]

wait oops 6. also the one on the top right of the red

[tawdry kraken]

Yes still because I'm a bit slow (I accidentally jumped on to help someone while in the middle of the problem)

Yep good notice

[signal ibex]

right, theres that partial too

which clearly passes

[tawdry kraken]

So yeah I agree that these are the squares that satisfy the color test for 2:

[signal ibex]

i tested that one off the dome just didn't clock that it was partially in all 4 quadrants

[tawdry kraken]

Now to count and see if I can tile them

[last slate]

nvm I dont thinK i can help I am dumb sorry coolemp and sleighla

[junior flower]

it's ok

[signal ibex]

all the partials are tileable

[junior flower]

you're still awesome and pretty

[tawdry kraken]

Yes but I have to see that for myself 😆

[signal ibex]

you have the two 3 by 8 sections (tileable because one of the dimensions is 3)

and then you're left with some 2 by 3 sections

[tawdry kraken]

Looks like the top left and bottom right ones are obvious

Oh actually

Both of those only count as 1

One is a rotation of the other

So that adds 4 to our total

Now we're at 8

Next the bottom left

[signal ibex]

the 6 by 2 parts are tileable (6 is divisible by 3)

then the remaining 6 by 6, same idea

[tawdry kraken]

Ah yes the other two are really sensible too

So that's another 4, and 1

Gives us 13 so far

[signal ibex]

did we check the other ones?

or do we only have 9?

[last slate]

thank you. you too!

also should I change pfp cz I have been encountering creeps from other server but I dont think I'll here cz its a math server..

[tawdry kraken]

9 + the 4 of size 1

I have to do 4x4's next

[signal ibex]

oh yeah, if we count the total

3 by 3 has 0

55 is not divisible by 3

[tawdry kraken]

Yes we did out the numbers

[junior flower]

oh you might i get creeps in my dms from here a lot

[tawdry kraken]

So for 4x4

We need 6 whites, 5 blues 5 balcks

[signal ibex]

move straight to 4 by 4

[last slate]

just for pfp or u upload selfies here anyways im closing my dms ig 😭

[junior flower]

i send pics of myself sometimes but probably mostly just my pfp and status

like this one i got earlier

and this one

[last slate]

insane... anyways @tawdry kraken I am sorry we are having this convo here..

[last slate]

wtf...

[tawdry kraken]

No no you're fine 😆

[junior flower]

but yea i get a new creep message like every day

[last slate]

that too on christmas.. bro was waiting for his gifts mfker

[signal ibex]

we only have 4, 5, 7, and 8 left

7 and 8 are extremely easy

[last slate]

(i called him mf not u 😭 )

[last slate]

I am sorry

[junior flower]

if anyone creeps on you you can send it to @shadow scaffold

[signal ibex]

6 and 7 are both 0

8 is 1

[last slate]

sure I will but I wont upload selfies here 😭 other servers where u might find creeps mostly say "ur hot" or smth but this is insanity

[junior flower]

bro really wanted that for christmas

[last slate]

yea lmfaoo Ihope ur fine dont respond to them

the next step they have in mind is sending a vid or smth

wait I'll show u

(NOT VIDEO)

[tawdry kraken]

Yes

Again the delay is much 😅

[last slate]

what i encountered yesterday

[tawdry kraken]

So we need this

[signal ibex]

man 4 is annoying to test

you need 6 whites and 5 of the other two colours

[tawdry kraken]

I see only 2 candidates

[junior flower]

men on discord are unhinged

[tawdry kraken]

Nope now I see 4 candidates

Lt me mark them

[last slate]

feet pics is the most weird thing ever...

[signal ibex]

yeah i was thinking theres more than 2

its wherever white squares would occupy 2 of the corners

[tawdry kraken]

Jesus

[signal ibex]

but some are eliminated because of rotation

[tawdry kraken]

Looks like I just cut it all up

[tawdry kraken]

Yeah I think two of these are out by rotation

[junior flower]

yea usually they ask for these

[tawdry kraken]

Two o these 5

[last slate]

is this the funniest one yet?

[junior flower]

no i have funnier ones

[tawdry kraken]

I lie there's more

[last slate]

Coming from a physics server this is weird.. newton died virgin for these ppl

[tawdry kraken]

But I know that one won't work because it's a corner

I can rotate it so that there's only 5 whites

Same for the bottom left two

[junior flower]

https://cdn.discordapp.com/attachments/1149587891652141116/1279139532570230876/qQFYiVQ.png?ex=69501f85&is=694ece05&hm=23cade597ed2986a4232346dbeaec03117e9980d38ac41626b5f764e0e1ea13e

these are all old but some are hilarious

[tawdry kraken]

Yeah only the middle one works

[last slate]

wait I'll get my magnifying glass..

[junior flower]

loll

if you open it in browser or on phone you should be able to see it

[signal ibex]

so 4 by 4 has 1

5 by 5

[last slate]

the most followers I have on my ig are Indian even tho I am from ph..

[signal ibex]

9 whites and 8 of the other two

[tawdry kraken]

Yap

[last slate]

is this js cuz of pfp 😭

[signal ibex]

all 16 of them can just be checked manually

[tawdry kraken]

Man how annoying to count

[signal ibex]

this is the last one

[tawdry kraken]

Oh right our total is 14 now

[junior flower]

those are all old, i had different profiles back then

[signal ibex]

6 7 and 8 are all degenerate

[tawdry kraken]

Okay 5 in the corner doesn't work

because it's a corner

[last slate]

oh dayum must have been controversial jkjkjk

[tawdry kraken]

Ah wait it does

It's centered around the case with size 1!!!

[signal ibex]

i think the corner ones actually do work this time

[tawdry kraken]

So now we have 18 total

[signal ibex]

are those the only 4?

or are there others?

i think the non-corners are all eliminated, but not 100% sure

[last slate]

two ppl are talking bout creeps two ppl are doing math
😭

[junior flower]

😭

[last slate]

I honestly thought its some edited image of nyc with all buildings

[junior flower]

i wish it was that

[tawdry kraken]

Ah nope this fails because you can flip the board

So yeah I think just the corners as well

[signal ibex]

it doesnt tile

[tawdry kraken]

That too

[signal ibex]

6 by 6 is 0 so we skip that

[last slate]

wait I'll stalk u to find out

[tawdry kraken]

Okay next 7

[signal ibex]

7 by 7 is 0

you have one row of length 8 in all 4 cases

[last slate]

nah im too lazy

[signal ibex]

that can never tile

[tawdry kraken]

True

And also it fails for symmetry

[signal ibex]

8 by 8 is 1

[last slate]

do u guys use ipad, notebook or whatever idk where do u do math

[signal ibex]

blank boards do tile

[last slate]

I hate ipad

[tawdry kraken]

Goodnotes on my ipad

[signal ibex]

total is 19

[tawdry kraken]

But I prefer a whiteboard

A really big whiteboard

[junior flower]

i used to take thirst trap pics of myself with math books to use as banner and that attracted a lot of creeps

[tawdry kraken]

@junior flower verif?

[junior flower]

oh i haven’t been paying any attention i was just happy someone else wanted to help with this one

[signal ibex]

if you verified 18 total up to 5 by 5

then 19 total for all of them

because 6 and 7 are both 0

[tawdry kraken]

I mean that's how calculations always go

[signal ibex]

i think 18 is right though

[junior flower]

i can’t send them here 😭 but it was for attention

[tawdry kraken]

"If you got it right up to the last step, and you did the last step right, then the whole thing is right"

😂

[junior flower]

same with everything about my current profile

[tawdry kraken]

Two more

[last slate]

lmfao 😭 my dms are open jk
anyways im happy ur honest bout it

[tawdry kraken]

Damn problems stretched my brain

[signal ibex]

ok im convinced our 5 by 5 is correct, only 4 there

[signal ibex]

uhhh

aren't there 7?

[tawdry kraken]

Of what size

[signal ibex]

4

tetris has 7 tetrominoes

[tawdry kraken]

Invariant under rotation and symmetry?

[signal ibex]

ok theres that

not invariant under reflection

but invariant under rotation

5 of the 7 tetrominoes tile the board

[tawdry kraken]

Ah wait

I forgot the zigzag one

[signal ibex]

you're missing

--

[junior flower]

i just found a pic in this server of an old banner

[tawdry kraken]

I drew two o the same one

😂

[signal ibex]

and its mirror image

[last slate]

wait so u js like trolling the creeps or u js like the validation
sometimes the msgs from creeps are funny

[signal ibex]

they only reflect, not rotate

[junior flower]

both

[signal ibex]

ok

[tawdry kraken]

There we go

[signal ibex]

4 of these tile the grid

[last slate]

ah dont send here ig I dont think u'd like new msgs from creeps or maybe throw it in #discussion and get dms lmfao

[signal ibex]

find out how

[junior flower]

#bots message 😭

[last slate]

use reddit then they will call u a catfish tho but get verfied if u have instagram u might get 1k requests but mostly Indians

[signal ibex]

try to perfectly tile a 4 by 4 grid (again, 4 of them can do this)

[tawdry kraken]

crazy work

[junior flower]

coolemplud don't look

[last slate]

what what 😭

[tawdry kraken]

The 2x2 and the straight tetromino are obvious

[last slate]

oh wait

[tawdry kraken]

I skip those

[last slate]

dAMN

[junior flower]

when that guy was talking about rational points on elliptic curves earlier...

[last slate]

lmfao

[junior flower]

yea i don't think i could get away with that here these days

[signal ibex]

there are only 4 rational points on a unit squircle

[junior flower]

but yea i used to take lots of pics like that and make them my banner and the dms were insane

[last slate]

damn dm me the msgs u got

[junior flower]

some of them were in that image i sent earlier

[tawdry kraken]

For the L-shaped tetromino you can stick two together to make a 2x4

So that case is trivial too

[signal ibex]

yeah

the T one is the hardest

[tawdry kraken]

With the T-shaped tetromino I can make a 4x4 square, that case is trivial

[signal ibex]

thats all 4

the 5th one doesnt work

[tawdry kraken]

Yes now I have to disprove it

Which I'm going to try their proof by exhaustion technique they taught in 45

(see the pins)

[signal ibex]

you are never able to tile 2 of the corners

[tawdry kraken]

Yes but that sounds hard to prove

[junior flower]

that would work on me

[last slate]

ahh 😭

[last slate]

insane I get these a lot maybe u should start using reddit

[junior flower]

😭

[junior flower]

https://www.youtube.com/watch?v=Pu4W8gH4-5A

[last slate]

😭 girl

[tawdry kraken]

\begin{proof}[Disproof of \textbf{49}]
  Let us attempt to tile the board with the S-shaped tetromino.
  Number the squares of the board 1 through 64 rowwise.
  Consider that there are only two possible orientations of a tetromino in a given corner, and they are symmetric about the diagonal.

  Therefore we, WLOG, place the S-shaped tetromino covering squares 1,2,10,11.
  Then the placement of an S-shaped tetromino covering squares 3,4,12,13 is forced, which forces one covering 5,6,14,15, which leaves square 7 inaccessible.
  Therefore, no tiling is possible of the checkerboard with S-shaped tetrominoes.
\end{proof}

[grand pond]

Coolempire2026

[signal ibex]

i think i have an argument

[signal ibex]

similar to mine

i was going to colour the ends of the S white

and the between as black

then all S tiles tile 2 connected black tiles but 2 disconnected white tiles

and then use the same argument to show that a corner tile is untileable by forcing placement of an S so that it connects a white tile

yours is less clunky though

[tawdry kraken]

Since we had to use it for 45 already most likely

I almost said this is trivial but then I read it again

Looks like I actually have to use brain

I have an interesting argument

[signal ibex]

looks like disproof

[tawdry kraken]

This is interesting yes

[last slate]

so do u usually report em or troll

[junior flower]

troll then report

or just report

trolling usually means picking some emoji or sticker and only sending that repeatedly and nothing else

[signal ibex]

a colouring argument should work

if theres 26 of a colour on the board it is untileable

[midnight plank]

@tawdry kraken Has your question been resolved?

[tawdry kraken]

I actually failed at the proof

\begin{proof}[Disproof of \textbf{50}]
  Note that a $5 \cross 5$ checkerboard can be tiled with straight tetrominoes such that 1 square is left over, and that a $10 \cross 10$ checkerboard is composed of four $5 \cross 5$ checkerboards.
  Importantly, the top left $5 \cross 5$ must always leave a white square untiled, because there are 13 white tiles and 12 black tiles in this sector of the board.
  After tiling this sector, 
\end{proof}

[grand pond]

Coolempire2026

[tawdry kraken]

This is what I was going for but it wasn't going to work

[signal ibex]

just label the boxes 1 2 3 4 in order

youll have 26 of one of the numbers

which makes it fail

[tawdry kraken]

Ah right coloring

How exactly do I get 26 of a color when 100 is divisible by 4

[signal ibex]

itll be 26 of one and 24 of another

just do the labeling

[tawdry kraken]

If I only do 50 yeah

[signal ibex]

just like the 8 by 8 grid with two opposite corners removed

you will have 30 of one colour and 32 of the other

[tawdry kraken]

Right

[signal ibex]

same idea here except nothing is removed

its just the actual 10 by 10 grid

[raw vector]

Straight tetrominoes are...

1x4 tiles?

[tawdry kraken]

Yes

[raw vector]

Oh just do the coloring thing

Color 26 squares black such that any tetromino can only cover 1 black square

[tawdry kraken]

All colored

[raw vector]

Ew numbers

[tawdry kraken]

Oh I see it can't cover 1313

[raw vector]

I meant only 1 color

[tawdry kraken]

So it has to go 1234

But then 12 is broken

Because it can't cover 1313

But that feels like forcing the tetrominoes to be in one orientation

i.e. I feel like I lose generality

[raw vector]

Yep

[raw vector]

Try this @tawdry kraken

Should be pretty straightforward once you see the pattern

[tawdry kraken]

Oh I think I've seen such a coloring before

[signal ibex]

the 1313 should be another 1234

then below that is 2341

[tawdry kraken]

Ah I guess it would be because we have seen this coloring before

[signal ibex]

yeah

26 reds

[raw vector]

Yeah

[tawdry kraken]

Right and only 25 tetrominoes must be there exactly

[raw vector]

Also i just wanna add, your 2x2 squares idea could work too, i'm surprised you didn't try that

[tawdry kraken]

Okay makes sense

[tawdry kraken]

Sounds harder for 4-long objects

[raw vector]

Nah

10x10 can be perfectly split into 2x2 squares

And you can probably see why it works

[signal ibex]

26 of this one

24 if you chose different squares to colour

[tawdry kraken]

This will be the question for tomorrow (before chapter 2 begins)

[signal ibex]

the other 2 actually have 25 of them

[raw vector]

What's Chomp?

[tawdry kraken]

Right regardless exists not 25

[signal ibex]

yeah

[tawdry kraken]

I'll post the description with it as well

[signal ibex]

if you find one with <25 that is also valid

[tawdry kraken]

But here it is in advance

[raw vector]

In which case each 1x4 tile will have 2 black squares and 2 white squares

And... Done

[signal ibex]

that doesnt do the job

[raw vector]

It does, there would be 52 squares of a color and 48 of the other color

[tawdry kraken]

Okay @junior flower time for bed the question I'll be tackling tomorrow is this one

Which will be harder I presume because they say that it's not really dealt with till chapter 5

I think that's when the structural induction is

[junior flower]

i haven't been paying any attention to the math in this channel since you had that first triominoe problem that was really annoying and i was talking to madi but ok

[tawdry kraken]

Yeah but polyomino tilings are done!

[junior flower]

i appreciate the channel

[tawdry kraken]

I (barely) got through them all 🙂 feels like I learned quite a bit

[junior flower]

good

there is nothing better than learning combinatorics

except finding a bf

[tawdry kraken]

.close

[midnight plank]

Channel closed

Closed by @tawdry kraken

Use .reopen if this was a mistake.

[midnight plank]

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[frozen arch]

tan 2x = -1

[frozen arch]

is 3pi/8 + npi/2, where n belongs to Z

a solution

or

the soluiton

for general solutions

[fathom onyx]

Yes, it's the general solution i.e. all the solutions

[midnight plank]

@frozen arch Has your question been resolved?

[frozen arch]

i meant, is it correct

[fathom onyx]

I mean

,w tan 2x = -1

(that's functionally the same answer)

[grand pond]

Results provided by WolframAlpha

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[fathom onyx]

,w expand (1/8)(4 pi n - pi) + pi/2

why am I getting 5

oh

[grand pond]

Results provided by WolframAlpha

Click here to refine your query online
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Requested by waes_wires

[fathom onyx]

@frozen arch The same answer you've got (bumping it up by the periodicity of the function)

[midnight plank]

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[midnight plank]

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[bright shoal]

I need some help is there any lemma or theorem that limits how many divisors of a number excluding itself can sum up greater than the number for an abundant number

[lusty python]

i don't really think you can limit the amount of divisors of a number

[bright shoal]

look i feel we can

[lusty python]

thus by chance there will be more abundant number (i think also increase in frequency)

[bright shoal]

said cause we are looking for abundant number

any number must have a minimum number of divsiors to be called abundant

understand

we are talking here about s(n) aliqout sum

excluding the number

[short ibex]

isn't the size more important

[bright shoal]

understand

[short ibex]

of the divisors

[bright shoal]

any number with less than 3 proper divisors can be

a prime

or a semi prime

[lusty python]

you mean

minimum of divisors for a number to be an abundant number?

is that what you mean?

[bright shoal]

i am not sure man but for an abundant number there should be atleast 3 cause except one only semi primes have 2 proper and primes have only 1

[bright shoal]

yes

i can prove that 2 is not enough

[lusty python]

12 requires 4 proper divisors

6 requires 3 proper divisors

[bright shoal]

we are talking about any general number and i am asking any limit

for being an abundant number

like minimum number of divisors

:>

:>

[bright shoal]

i feel 3 is minimum

[bright shoal]

6 is a perfect number

:>

not abundant

[soft stone]

3 isn’t enough even 4 proper divisors won’t do it

[bright shoal]

look

[soft stone]

any abundant number needs at least 5 proper divisors (6 total) after that size matters more than count

[bright shoal]

below 3 proper divisors abundance is impossible

eg

a prime

0 divisor excluding 1
of the form p^k
so only proper divisor is prime and its powers so if we calculate sum we would have p^k-1/p-1 which is kess than p^k and if semiprime it can be shown

excluding 1 as a divisor a number cannot be abundant if it has less than 3 proper divisors

eg

12 = 1 + 2 + 3 + 4 + 6
has 4

[bright shoal]

at least 3 proper divisors greater than 1 should exist to be an abundant number

i just want to say below 3 abundance is impossible

oh wait

[soft stone]

..

[bright shoal]

you are right even 3 is not possible

4 minimum

[soft stone]

yep as a necessary condition that’s fine <3 proper divisors can’t work

just note it’s not sufficient abundance actually starts at 5 proper divisors (12)

[bright shoal]

excluding 1

lookn @soft stone

using this fact can we say that aleast 3 divisors are needed of an abundant number to sum it

:.

:>

below 3 abundance is not possible

necessary condition so any with greater than must atleast have more than 3 proper divisors but abundance is possible beyond three so minimum 3 divisors can sum up a abundant number

[soft stone]

it is true that below 3 proper divisors (>1) abundance is impossible (necessary condition)

but that does NOT mean 3 divisors can sum to an abundant number

[bright shoal]

any counter example

or is this is a conjecture

[soft stone]

well it’s not a conjecture it can be proved directly

[bright shoal]

i checked counter example

look

carefully

,w 20 divisors

[grand pond]

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[bright shoal]

no 3 divisors can sum up

:<

[soft stone]

it has four proper divisors?

[bright shoal]

no like i was thinking that for any abundant number a k set of any divisors can sum up that abundant number but no not true :>

[soft stone]

yep abundance is about the total sum of proper divisors not whether some subset can add up to the number

[bright shoal]

but yes we can say a number cannot be abundant if it has not atleast you say 4 divisors greater than 1 proper

least example 12

2,3,4,6

[soft stone]

mhm

[bright shoal]

can we claim this apart from some finite numbers minimum 3 factors can never sum up an abundant number

[soft stone]

not just apart from finitely many, never

three proper divisors >1 are never enough for abundance

[bright shoal]

look i feel yes look understand product is always bigger than sum except from the worst case 31/30 lmao

wait

lemme write some stuff

lets see

[soft stone]

alright

[soft stone]

yep but the real reason is divisor bounds, not product vs sum

[bright shoal]

fk

[soft stone]

with only 3 proper divisors you can never reach n so abundance just can’t happen

[bright shoal]

i checked every 30 multiple can be summed up using exactly 3 divisors

check

,w 30 divisors

[grand pond]

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[bright shoal]

15+10+6

lmao

,w divisors 120

[grand pond]

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[bright shoal]

60+40+30

lmao

yes oh fk

so infinitely man numbers can be summed up by 3 divisors lmao

;-;

fk

[soft stone]

uhh subset sum ≠ aliquot sum

[bright shoal]

i know that

i mean that 3 numbers can sum up greater than the number itself

:<

[soft stone]

your examples hit n but abundance needs the entire divisor set to exceed n, which 3 divisors can never do

[bright shoal]

we cannot set a minimum bound also lmao

[soft stone]

some 3 divisors yeah

[bright shoal]

no

30m family cannot

we cannot set a lower bound even lmao

but yes minimum 3 is needed

we can say this not a necessary condition

[soft stone]

okay so for subset sums there’s no lower bound infinitely many numbers (like30m) have 3 divisors summing to ≥n

[bright shoal]

yes no lower bound

[soft stone]

but abundance has nothing to do with subset sums right so those examples don’t contradict the abundance condition

[bright shoal]

but yes atleast min 3 divisors can sum

greater than n

not necessary

[soft stone]

okay, agreed

[bright shoal]

is this is a standard lemma

:>

[soft stone]

not really a standard lemma abundance theory doesn’t study subset sums

that’s a different problem

this is more a combinatorial/divisor structure observation than a named lemma

[midnight plank]

@bright shoal Has your question been resolved?

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[midnight plank]

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[sand flume]

Is my logic sound. We assume that there exists an A such that for all epsilon > 0 |sin(1/x) - A| < epsilon whenever |x| < delta. |sin(1/x)-A) <= |sin(1/x) + |-A| = |sin(1/x) + |A|. Now this needs to be less than epsilon. Say epsilon = 0.5. Next we know that (sin(1/x) <= 1. |sin(1/x) + |A| <= 1 + |A|. If we had 1+|A| < 0.5 then |A| < -0.5 which cannot be true since absolute value is always positive and there is no delta we can find to amend this.

[pure wraith]

There's a problem

$\abs{\sin(1/x) - A} < \epsilon$ is what you have to work with, and you have $\abs{\sin(1/x) - A} \leq \abs{\sin (1/x)} + \abs{A} \leq 1 + \abs{A}$

[grand pond]

jewels!

[pure wraith]

However, just because the upper bound fails the condition, it doesn't mean that the actual value fails it

[lavish venture]

you're supposed to argue that for any delta > 0 there is some x with |x| < delta but |sin(1/x) - A| >= eps

so you need to find some x that would contradict it for a suitable choice of epsilon

[sand flume]

your saying I need to find an x such that I have |1-A|?

[autumn canopy]

Usually the idea is the explicitly give two sequences for which sin(1/x) evaluate to different but constant values, and these two sequences should both go to 0

Let $x_{n}=\frac{1}{2n\pi +\pi /2}$ and $y_{n}=\frac{1}{2n\pi +3\pi /2}$. Now what is $\sin(1/x)$ for both of these

[grand pond]

Kepe

[autumn canopy]

It's clear that both go to 0

[sand flume]

why?

[sand flume]

also why are we bringing in sequences?

[autumn canopy]

Because as n -> infinity, the denominator gets larger and larger

[autumn canopy]

Assume A exists and sin(1/x) converges to A for x -> 0. Then, because these two sequences go to 0, sin(1/x_n) and sin(1/y_n) also go to A

And because

the idea is the explicitly give two sequences for which sin(1/x) evaluate to different but constant values,

We will get the contradiction

That's the idea behind it

[sand flume]

what about using limits?

[autumn canopy]

You can do that. What's your definition of continuity at a point?

Epsilon-delta or that you can interchange function and limit?

[gaunt imp]

Continuity can be perfectly characterized by sequences

[autumn canopy]

Ok, so epsilon-delta

[sand flume]

by the second thing do you mean that the limit is equal to the value at that point

[pure wraith]

The negation of the sequential definition for continuity is useful in showing functions are discontinuous

[sand flume]

havent seen any sequence defintions yet

[autumn canopy]

Yes, for every sequence $(x_n){n \in \mathbb N}$ with $x_n \to x^*$, we get $\lim{n \to \infty } f(x_n) = f(\lim_{n \to \infty} x_n) = f(x^*)$. That's one way to put it. But you use $\varepsilon$-$\delta$ in your book.

[grand pond]

Kepe

[gaunt imp]

They said they hadn't covered this characterization yet, though

[autumn canopy]

Yep

So let's go with epsilon-delta

[gaunt imp]

So they probably want to go with standard epsilon-delta

[autumn canopy]

sin swallows up the 2n pi, so sin(1/x_n) = sin(pi/2) and sin(1/y_n) = sin(3pi/2), right?