help-39

[waxen agate]

No, she cannot.

[autumn trellis]

The only way I can see applying it is by dividing numerator and denominator by x, then apply mvt on $e^{-x}$ to get $$\lim_{x\rightarrow 0} \frac{\frac{e^{-x}-1}{x}+1}{x}=\lim_{x\rightarrow 0}\frac{-e^{-t}+1}{x}$$ where $t\in (0,x)$.

[cinder flower]

[waxen agate]

That is how you’d do it

[glacial charm]

does the MVT come with the question?

[jolly parrot]

what

[sand swallow]

yeah i think we'd use it twice

[sand swallow]

nope, just a suggestion

[autumn trellis]

Well the easiest way will be to write e^{-x} = 1-x+x^2/2+O(x^3) near 0

This immediately gives you the answer

[glacial charm]

or just use L'Hopital

[glacial charm]

is this the series expansion?

[sand swallow]

i don't have the right to use the little o notation

[autumn trellis]

I mean it's just the series expansion. I'm sure you know about it

[sand swallow]

yeah but i'm not allowed to use that either

[cinder flower]

the government has gone too far this time

[oak quiver]

then revolt

lmao

[cinder flower]

i have a meme for this

[sand swallow]

i do but it's beyond the curriculum

[autumn trellis]

How did you define e?

[sand swallow]

ikr 😭

[oak quiver]

pretty sure euler defined it

[sand swallow]

you mean the exponential function

[glacial charm]

[sand swallow]

lmao fr 😭

[bronze heath]

cwazy

[sand swallow]

khay shnu taf o shnu mvt

[vale forge]

nfs chi hh

[sand swallow]

hmmm ana dertha ela f(x)=e^(-x)

att njerb hadi

[sand swallow]

ewa shti lol

[vale forge]

ok

[pearl pond]

@sand swallow Has your question been resolved?

[twin forum]

(1+1/n)**n

As n -> infinity

Or (1+n)**(1/n) as n goes to 0

[sand swallow]

o men bead? baqa F.I

[vale forge]

really?

[pearl pond]

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[pearl pond]

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[sharp smelt]

I suppose I use this?

[shut elm]

augh, my favorite guy is studying stats and not abstract

Meanwhile all the stats people are in #help-6

XD

[sharp smelt]

I really need to do well in stats 🙏

[shut elm]

Interesting

Well good luck 🙂 I can't help because I only did applied version

[ivory basin]

I take offense to being called a stats person lmfao

[ivory basin]

No lol

[sharp smelt]

yea, that's why I struck it out

[ivory basin]

There are no squares here

[shut elm]

Yes pseudomathematician is more palatable good point 🙂‍↕️

[sharp smelt]

mhm, and computing the joint pdf each time by hand is painful

like X_1+2X_2

Then (X_1+2X_2)-X_3

[ivory basin]

I will stab you

[sharp smelt]

and so on

[ivory basin]

Shouldn't you be able to do it all at once

[sharp smelt]

lemme see if I've forgotten something

okay, got it

I just forgot some properties 😔

just need to use this

Or more generally, this

[shut elm]

Oh

I got caught up in the words and didn't actually read the problem

I'm actually familiar with that 😂

[sharp smelt]

$X_1+2X_2-X_3+X_4+X_5 \sim N(0,\left(2+4-2+2+2\right) \cdot 2)= N(0,16)$

[jolly parrot]

wai

[sharp smelt]

oops, shouldn't be -2

should be +2

so N(0,24)

[errant cedar]

just a wider normal

[errant cedar]

also times 2 or times 4, you have std deviation?

[sharp smelt]

2

the var is √2

oops

yea

[errant cedar]

maybe it’s late for me but isn’t standard deviation the square root of variance

[sharp smelt]

shoudl be 4

that just makes it wider

N(0,48)

[errant cedar]

yeah the main bit was this.

[sharp smelt]

got it

thanks

The first one is just $\chi^2(3)$ right

[jolly parrot]

wai

[sharp smelt]

For the next one I have $(X_1+X_2)(X_1-X_2)$

[jolly parrot]

wai

[sharp smelt]

$X_1+X_2 \sim N(0,2);X_1-X2 \sim N(0,2)$

[jolly parrot]

wai

[sharp smelt]

As the vars are indepdendent and Identically dist, we have $N(0,2) \cdot N(0,2)$

[jolly parrot]

wai

[errant cedar]

what’s before is valid, but i disagree with this last statement as X_1+X_2 and X_1-X_2 are dependant of each other even though they are normal

[sharp smelt]

mhm

makes sense

We can still find the joint pdf , it will just be more painful now I guess

[errant cedar]

tbh idk about this one, what i can say is that it can be negative

so it can’t be straight up a chi or smt with only positive outputs

[sharp smelt]

Uh, I tried looking this up

https://math.stackexchange.com/a/4017126/879009

[errant cedar]

smt we usually did in my prob course to figure out joint distribution is to start with P(X_1^2+X_2^2 < x)

but here idk if that’s computable

[sharp smelt]

so You're suggesting $P(0<x^2-y^2<x)$?

[jolly parrot]

wai

[errant cedar]

no I mean usually computing the joint CDF is simpler

[sharp smelt]

mhm, yea

But here I don't really see a way to do that

[errant cedar]

oh one other dubious move i can see would to « condition »on the value of X_2, but i’m not 100% sure it’s legit

[sharp smelt]

Hmm, in that case I suppose it's not really computible, but then I have to determine why

lemme see if I've missed something

oh wait

why are we trying to compute the pdf of the prdouct

instead of the sum

[errant cedar]

for sure in the stat litterature it has been studied and google gemini slop says it’s a non-central chi^2

[sharp smelt]

MSE says the same

This is the best I could find https://mathoverflow.net/questions/89779/sum-of-squares-of-normal-distributions

A comment suggets it's related to Satterthwaite

[errant cedar]

tbh if you haven’t seen non-central chi in your class i would gamble on the fact the expected answer is to justify why the dist is hard to compute

[sharp smelt]

I'm self studying for next sem, so It's fine I suppose

this is for my stats course 😭

[errant cedar]

oh ok so we can have some fun then cool

[sharp smelt]

yup

[errant cedar]

I persisted with my idea of earlier

however the chi CDF is discribed in term of a incomplete gamma function

and that’s ass honestly

[sharp smelt]

I don't get this

[errant cedar]

$f_{X_2}(y) = P(X_2=y)$

[jolly parrot]

pola_touche

[sharp smelt]

Like what have you written in green

[errant cedar]

oh I say it’s a normal PDF

[sharp smelt]

mhm

I see

Tysm

[errant cedar]

yeah i apply homeomorphisms to my writing

[errant cedar]

anyhow from here we would need to spot out what distribution this is. That does not seems easy imo

[sharp smelt]

We could just say it's a distbution with this CDF

[errant cedar]

i mean yeah, what would be cool would be to reduce this to the non-central chi cdf and call it a day

[sharp smelt]

I just read about chi squraed yesterday, so I suppose I'll read a bit more about this and see if I get any insights>

[errant cedar]

hum, now that I think about it maybe trying to do this with the moment generating functions would be easier

[sharp smelt]

I forgot that method existed tbh

☠️

[errant cedar]

but that’s just a thought

https://en.wikipedia.org/wiki/Moment-generating_function

[sharp smelt]

mhm, I'll eat and get back to this I guess

Thanks so much!

.close

[pearl pond]

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[pearl pond]

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[sharp smelt]

really silly question, but this isn't the product of the density functions is it

[frozen bluff]

because it’s a random sample the joint density is just the product of the exponential densities

[sharp smelt]

Like $\theta^3 e^{-\theta (x_1+ x_2+x_3)}$

[jolly parrot]

wai

[sharp smelt]

huh, okay

[frozen bluff]

for x₁, x₂, x₃ ≥ 0 (and 0 otherwise)

[sharp smelt]

Thanks!

.close

[pearl pond]

Channel closed

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[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
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[desert patrol]

Why is it not -sqrt(72)?

[desert patrol]

B is the middle point of AC

Asking where the slope on the left intercepts

Also the angle between the slopes in 90

[bronze heath]

6sqrt(2) is the same

[desert patrol]

Yeah but the answer is -6sqrt(3)

[bronze heath]

oh

A would be2sqrt3

nah wair

[desert patrol]

Yeah first I did that but the answer is supposed to be E

Then I tried doing the formula for calculating the distance

That gave me 6sqrt2

[midnight haven]

hi

[desert patrol]

Hey

[midnight haven]

hru

[desert patrol]

I'm fine wbu

[midnight haven]

am i in the wrong channel

[desert patrol]

Yeah

[midnight haven]

im good

[midnight haven]

sorry mb

[desert patrol]

It's okay

[bronze heath]

@desert patrol what was your slope for perpendicular?

[desert patrol]

sqrt3/3 is for the positive one

is for the negative -3/sqrt3

[bronze heath]

you have to do y1-y2/x1-x2

[desert patrol]

I know

oh

[bronze heath]

so should be the reciprocal

[desert patrol]

The opposite then

[bronze heath]

yes

3/sqrt3 = sqrt3

[desert patrol]

-sqrt3/3 for the negative

[bronze heath]

= -1/sqrt3

now 6/x = -1/sqrt3

[desert patrol]

AC is between -6 and 2sqrt3

[bronze heath]

(0,-6) and (2sqrt3,0)

[desert patrol]

Yeah

So the slope is 3/sqrt3

[bronze heath]

yes

[desert patrol]

The opposite is -sqrt3/3

[bronze heath]

yes

[desert patrol]

sqrt3 is -6 so what's 3?

[bronze heath]

wdym?

6/x = -sqrt3/3 ?

[desert patrol]

Oh the oppodite

x/6 = -sqrt3/3?

[bronze heath]

6/x

difference of y coordinate in numerator

[desert patrol]

Oh I used the wrong angle mb

Yeah

Got it thank you

.close

[pearl pond]

Channel closed

Closed by @desert patrol

Use .reopen if this was a mistake.

[bronze heath]

np

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
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[desert patrol]

Help, it's asking for the ratio of the colored triangles

[spiral coyote]

!status

[pearl pond]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[desert patrol]

2

[uneven cypress]

Ic you found the angles are equal

So they are similar triangles

[spiral coyote]

Can u explain wht u did...

[pearl pond]

@desert patrol Has your question been resolved?

[desert patrol]

I found one of the sides also the part where the sloped intersrct is a right angle

[desert patrol]

That would mean the answer is 1

Ih nvm

They are similar

[spiral coyote]

You mean tht lines y= -mx+4a and y=mx r perpendicular?

[desert patrol]

But how Am I supposed to use that

Yeah

[spiral coyote]

But The product of slopes is -m²

Not -1

[desert patrol]

Oh...

[spiral coyote]

I suppose you could just find the coordinates of three vertices and find area using determinant method?

For the triangle whose one side is y axis 0.5 × base × height would work

[desert patrol]

I wasn't taught determinant

[spiral coyote]

Find the vertices then use the distance bw a point and a line to find the height and distance bw 2 points to find the base.

[desert patrol]

(0,4a), (0,0) and (4a,0)

I can maybe find the center of gtavity?

Idk

[spiral coyote]

Idk i would js do this for thr shaded triangle

[desert patrol]

Is it even possible

[spiral coyote]

WdymTT can u not find the vertices?

[desert patrol]

This is all I could find

[spiral coyote]

Ahh

[spiral coyote]

Do u want to try this or no.

[bronze heath]

KARMA TEST

[desert patrol]

I can't find them

[spiral coyote]

Tbh im awful at explaining :') but u have the equations for the lines and the vertices are their intersections.

[desert patrol]

So -mx + 4a = mx to find where thry intersrct?

2a = mx

[spiral coyote]

Yep

[rough python]

because that's the point where they share the same y-coordinate. So they intersect.

[spiral coyote]

You can find every vertex in terms of a and m

[desert patrol]

So its actually 90 degtees?

y = 2a and y = -2a

Yeah I dont think thats right

So the bottom part is 2a

[rough python]

i dislike the fact that the question diagram is misleading and not to scale. but dw

can you find the coordinates of these two verticies?

[rough python]

To find the coordinate of this vertix, you were on the right path when you set both line equations together.

From 2a=mx,
if you solve for x, remember, x is the variable, m and a are just some constant numbers.

x=(2a)/m.
You found the x-coordinate of this vertix (the orange dot)

you can plug in that x coordinate in either y=mx or y=-mx+4a to find the y-coordinate of that orange vertix.

@desert patrol I wonder if you're still here lol

[desert patrol]

Yeah I was solving the other questions

So ((2a)/m)m = y

[rough python]

good

[desert patrol]

2a = y

But why does that also work for the other slope

Shouldnt it be negative

[rough python]

nope. You see that the intersection occurs on positive x.

[desert patrol]

Oh this is not the slope its the coordinates

[rough python]

If it was supposed to be negative, you should also see an intersection in negative y-axis

nope, just the y-axis. m is the slope, but you realise that m got cancelled out when finding the intersection

[desert patrol]

Yeah

[rough python]

with that, and if you understand that the triangle is symmetric, like a 45-45-90 right angle triangle. Then you'll realise that the blue length is equal to this yellow length.

[desert patrol]

How do you know if jts symmetric

a is the height right?

[rough python]

mx and -mx. Give me any random m.

if m=78 for example. very steep slope but you notice that one equation will be y=78x and the other equation will be y=-78x + 4a

[desert patrol]

Yeah

[rough python]

I think I was wrong when I said it's a "45-45-90" right angle triangle, because angles 45-45-90 only happens if m=1. But either way, at the point of intersection, if you draw a vertical line through that point, the "two parts" of the triangle underneath will appear to be symmetric. I.e. isosceles triangle.

[desert patrol]

Yeah

is the other one also isosceles

since they share the same angles

30 60 90 triangle

?

I mean yeah it is the correct answer

Thx

a^2/2 = 2a^2/2 was what I got

So the ratio can be 1/2

And since 2 is not an answer

[rough python]

yeee!

[desert patrol]

Finally :)

Thanks for being patient with me

Can I also ask another one

[rough python]

😄 no worries, and sure

[desert patrol]

It asks for the area of the colored triangle, AD and BC have the same slope meaning they are parallel right? But I don't know what else I can do

[rough python]

brb 2 minutes, ill use chatgpt to translate and figure this out lol

[desert patrol]

kk

[rough python]

AD and BC are parallel.
Distance between AB is 4 units.
Find area of triangle ADC.

We see D is at height 10.

Fun question.

!status

[pearl pond]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[desert patrol]

1

[rough python]

alright, I'll ask a few questions to get you started.
What do we need to find the area of the shaded triangle?

[desert patrol]

2 of the sides and maybe an angle for the formula

Maybe we can also substract something ?

Tbh I would really enjoy math if it wasnt for the exams

[rough python]

lol bare with me, I'm trying to figure this out too.

Your correct with what we need. My thought process is thinking that we should find the slope of AD, which would be the same as slope BC.

[desert patrol]

-10/a?

Maybe we can try -20/2a for the other one

Idk

[rough python]

yeah, but I'm struggling to understand how we can find where C is.

something with angles maybe

[desert patrol]

Maybe we should draw parallelogram?

I don't know hoe that could help though

Can we say that this is a square?

[rough python]

nah doesn't look like a square. looks like a parallelogram

[desert patrol]

Yeah...

I was thinking maybe the triangles were similar

[rough python]

... yeah

<@&286206848099549185> I'll bring backup. I feel like we're missing something crucial xD

[proper nova]

lol

you summoned me

[proper nova]

this?

[desert patrol]

Yeah

[rough python]

hahah yeee

no4

[rough python]

this

[proper nova]

translation please

[desert patrol]

I should've checked it lmao

[rough python]

translated

[proper nova]

your backup ain't crucial lmao

[rough python]

i say "fun question" but fail to solve it myself

[proper nova]

lol

area of shaded triangle

right?

[rough python]

ya

[desert patrol]

Yup

[slate pilot]

Yoo guys

[bronze heath]

Hm.. so we doing ts?

[rough python]

gang gang

[desert patrol]

Yeah

[rough python]

---ive made an error-

[proper nova]

A and B is not fixed

🙏

[proper nova]

what are you trying to do here

[proper nova]

are you asking the question or giving the solution?

[slate pilot]

I have question

[proper nova]

#❓how-to-get-help

[proper nova]

!occupied

[slate pilot]

Yes

[pearl pond]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[slate pilot]

Ok

[desert patrol]

You should send it in one of the channels that are avalible

The one with the green tick

[proper nova]

A is not fixed and B depends on A so B is not fixed either

so probably let $A$ have the coordinate of $(x_A, 0)$ which gives the coordinate of $B$ as $(x_A + 4, 0)$

[jolly parrot]

1 divided by 0 equals Infinity

[desert patrol]

Yeah

[proper nova]

and $AD // BC$

[jolly parrot]

1 divided by 0 equals Infinity

[desert patrol]

So they have the same slope

[proper nova]

cool

[desert patrol]

Which is negative

[proper nova]

if we assume $x_A > 0$

[jolly parrot]

1 divided by 0 equals Infinity

[proper nova]

which it is lol

[desert patrol]

The shape is on the right though?

[proper nova]

so what's the slope here

[sweet junco]

you can make a line(any line) here that passes through the point (0, 10) and then translate it to the right by 4 units to get a line for BC, after that you can get a length for BC and AD

[rough python]

slope = -10/a was found from earlier, ya

[desert patrol]

Yeah

[proper nova]

i mean

only hope is to find the coordinates

so using the parallel then we can find the formula for $y_C$

[jolly parrot]

1 divided by 0 equals Infinity

[proper nova]

and hopefully $x_C$

[jolly parrot]

1 divided by 0 equals Infinity

[rough python]

there's a formula for y_C? :o

[desert patrol]

isnt yc 0 since it is on the axis

[proper nova]

since when do you say that $C$ is on the x axis

[jolly parrot]

1 divided by 0 equals Infinity

[sweet junco]

it would be less than 4, not 4, because it's not touching the line BC, if it touches the line BC and is parallel to the x axis, then it'd be length 4

[proper nova]

it isn't

[rough python]

nah, by y_C, we mean the y-coordinate for point C.

[desert patrol]

Oh we are talking about that

Mb

[rough python]

oh right! mb

[sweet junco]

AD can be equivalent to y=-x+10
BC can be equivalent to y=-x+14(translated to the right by 4 units)
wait doesn't the area for this vary depending on the slope or how steep it is + there's no conditions for C

is ADC a right angle?

[desert patrol]

It's not given

[sweet junco]

the problem doesn't have a specific solution(I'm guessiing this from what the panda pfp person has sent)

[sweet junco]

does C stop anywhere or is C just on that specific line?

oh wait

no nvm

height doesn't change along with that

but the slope does change the height of the triangle

[bronze heath]

by any chance do you know finding area by determinants?

[desert patrol]

No I wasn't taught and I din't think it's in the curriculum

[bronze heath]

k

[sweet junco]

alr I'll give this q the benefit of the doubt
let y=-x+10 be a line that has line segment AD(let m=-1 for simplicity sake), A would be at (10, 0)
this would form a 45 45 90 right triangle with O where the length between AD is 10*sqrt2
now, consider BC, since BC is parallel with AD, we can consider a 45 45 90 right triangle with AB as the hypotenuse, where the leg would be equivalent to the distance between AD and BC(or the altitude of the triangle ADC with AD as base)
this would then give us 4/sqrt2 or 2sqrt2, whhich you can then multiply to the 10sqrt2 earlier and divide by 2 to give you the final answer of 20
I don't see an algebraic way to solve this that isn't complicated asf

[desert patrol]

Damn they suck at drawing

OA looks much shorter than DO

[rough python]

genius, but slightly annoying q. you choose A to be at (10, 0). That would never cross my mind

[bronze heath]

that doesn't matter diagrams suck

[desert patrol]

This line?

[sweet junco]

yes that line would be the leg of the 45 45 90 right triangle with AB as hypotenuse

[desert patrol]

Thanks

I knew for a fact that could be a square lol

That angle definitely looks like an acute but whatever

.close

[pearl pond]

Channel closed

Closed by @desert patrol

Use .reopen if this was a mistake.

[bronze heath]

.reopen

@desert patrol do you know how to find perpendicular distance from a point to a line?

[desert patrol]

Yeah

[surreal siren]

Ye

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[bronze pike]

Let (A) be a set of positive natural numbers not exceeding (10^{2025}).

For a set (A):

• (A + A = {a+b : a,b \in A}),
• (A - A = {a-b : a,b \in A}),
• (A + A + A = {a+b+c : a,b,c \in A}),
• (|A|) denotes the cardinality of (A).

Determine which of the following statements are true:

A) If (|A + A| \le 5|A|), then (|A - A| \le 50|A|).

B) (|A + A| \le |A - A|).

C) If (|A| < 10^{1000}), then (|A + A| \le 10^{2024}).

D) If (|A + A| \le 5|A|), then (|A + A + A| \le 25|A|).

E) If (|A + A| > 10^{2025}), then (A + A + A) contains a number of the form (2^k - l) for some integer (k > 6000) and (l < 600).

[lilac ocean]

is \le supposed to be \leq or is it just french...

what have you tried

[plucky python]

$\le = \leq$

[jolly parrot]

LY

[lilac ocean]

hohoho

i will remmeber this...

[bronze pike]

tez z polski jestem

[lilac ocean]

:)

[bronze pike]

nie mam pomyslu jak z tym zadaniem ruszyc, jestem w 3 klasie i mam wrazenie ze to co mialem w szkole nijak odnosi sie do tego

tak z inuticji sobie to rozpisalem i wlasnie na tym utknalem

[bronze pike]

anyone want to help?

[naive zinc]

It seems to involve additive combinatorics? There is a combinatorics theory channel below. If no one can help you maybe you can try that channel

[pearl pond]

@bronze pike Has your question been resolved?

[pearl pond]

@bronze pike Has your question been resolved?

[pearl pond]

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[swift lava]

How do i prove this? I don't really get the problem intuitively

[bitter herald]

do you know Wilson's theorem

[swift lava]

No

[bitter herald]

ok ignore that

anyways, here is a hint

||rearrange the product||

hint 2: ||pair elements with their inverses where you could||

[swift lava]

what does that mean again...

[bitter herald]

the group is abelian, yes? you are free to group the elements in teh product $\prod_{g\in G}g$ in any way you like

[swift lava]

add them with their inverses?

[bitter herald]

add?

i mean it is a product is it not

[swift lava]

oh i thought it was like the operation of the group

[bitter herald]

What im basically saying is to group every element in the product next to its inverse in the product. e.g., $g\2g^{-1}$

[bitter herald]

Question now becomes: what does $g\2g^{-1}$ evaluate to?

[swift lava]

the identity element

[bitter herald]

yeah

[swift lava]

<@&268886789983436800>

mods i swear im not pingign you for help

[bitter herald]

LOL

[bitter herald]

but also, this is true, but there is a catch

what is the catch

[swift lava]

uhm

[bitter herald]

in your case, what happens if g = g^{-1}?

[swift lava]

g is the identity element isnt it

[arctic siren]

Yeah this one is kind of just a "find the trick" question

[bitter herald]

g^2 is the identity element

[swift lava]

oh righy

[bitter herald]

but like

this is a set right

[swift lava]

oh so it can only have one g

[bitter herald]

could you really have g*g

[swift lava]

ergo if we multiply everything we end up with identity elemtns * g

[bitter herald]

right. so there is leftover

[swift lava]

ohhh got it

thanks!!

[bitter herald]

yeppie

[swift lava]

.close

[pearl pond]

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[pearl pond]

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[low matrix]

I want to find the strongly connected components of this graph

[low matrix]

idk how to find the bigger ones, i could only find two, a-i and d-e

[bitter herald]

Are the bottom and top rows of vertices strongly connected

[low matrix]

b is connected to h but h is not to b

[pearl pond]

@low matrix Has your question been resolved?

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[pearl pond]

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[coral dew]

i need help regarding lateral surface area, i am in 7th grade.

[warm dagger]

are 7th graders supposed to be on discord?

welp anyways what is ur question?

[coral dew]

Hassan built a fence around a square yard. It took 48 meters squared of lumber to build the fence. The fence is 1.5 meters tall. What is the area of the yard inside the fence?

[warm dagger]

!show

[pearl pond]

Show your work, and if possible, explain where you are stuck.

[coral dew]

So like the lesson I learned is like, solving for rectangular prism's volume

And so I'm stumped at what to do when faced with an area question regarding like

prisms/3d objects

[warm dagger]

well it's not regarding prisms

just treat this like how you usually would with a 2d geometry problem

this isnt a 3d problem at all

[vital estuary]

possible to be 13 during 7th grade

[coral dew]

fyi im 13 and on 7th

i was out of school for a yr

bc a piece of debris struck my left eyebrow area and i had to get it removed

[warm dagger]

damn bro

sorry u had to go through all that

[coral dew]

its alr, im just tryna get thru school...

[coral dew]

wait so i just do like

area divided by the uh

height?

[warm dagger]

no

ur given the perimeter of a square

what can u infer about its side length?

[coral dew]

all sides are equal i guess?

so area divided by 4 = 12?

[warm dagger]

yea

and so how to you find the area

[coral dew]

48m (squared) is area i think

[warm dagger]

what

no the area of a square is its side length squared

so you have 12^2 or 12x12=144

[coral dew]

oh

[warm dagger]

hold on

I think the problem may be asking for something with the 1.5 meters

[coral dew]

by squared do you mean power of 2 or just the label

[warm dagger]

power of 2

[coral dew]

oh alr

[warm dagger]

hold on

it says the heigh is 1.5

so we may need to divide 48 by 1.5 first

then do the rest

the wording is a bit weird

yea I think this is what it means

so it would be 32/4=8

and 8x8=64

but I don't know

lemme get confirmation

[sonic marten]

That’s what I got as well

[warm dagger]

yea that might be the answer then

the question is worded very weirdly

[coral dew]

huh

but like what do i do???

bc i gotta show my work and stuff and i dont wanna keep spamming this with the same questions

[warm dagger]

I think the 1.5 height was there for a reason so just take the 64 answer

[coral dew]

wow its right whew!

[coral dew]

ty!!!!!!

[sonic marten]

The area of a single piece of fence is 12. You divide 12/1.5 to find the length of a single piece of fence. The length of the fence is the same as the length of the yard

Since the yard is square you just do the length squared

[warm dagger]

its sorta weird tho since they gave the height and no width

[warm dagger]

np

[coral dew]

.close

[pearl pond]

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[coral dew]

i completed my reasoning and logic for a question i was trying to solve in a test. i want feedback and to know if my logic is incorrect or not. grade: 7th

[shut elm]

!da2a Just post your question and we can give feedback!

[pearl pond]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question!

[coral dew]

a volume word problem is:
Q: "Jamie wants to know the volume of his gold ring in cubic inches. He gets a rectangular glass with a base 3 in by 2 in and fills the glass 4 in high with water. Jamie drops his gold ring in the glass and measures the new height of the water to be 4.25 in.

What is the volume of Jamie's ring in cubic inches?"
My answer and logic:
to find it, understand that when his ring is placed in the water, it displaces an equal volume of water and pushes it up. so the volume of his ring would be the incremental volume of which the height increased. what is the incremental volume? to find the incremental volume, multiply the length (3 cubic inches) by the width (2 cubic inches) by the height amount increased (0.25). so to find the incremental volume, we do 3 times 2 times 0.25. 0.25 is the same as dividing into fourths, or into 1/4 or one-fourth.
so 3 times 2 is 6, times 0.25 is 1.5, as half of 6 is three, half of three is 1.5. so the incremental value is 1.5 cubic inches.

any feedback on my logic and reasoning?

[shut elm]

Sounds good to me

[coral dew]

any ways i can improve it?

[shut elm]

It's the same as my reasoning but expressed slightly differently

My reasoning was volume at the end must be volume at the beginning + change in volume

The change in volume must be the volume of the ring

Multiplying, volume at the end = 3*2*4.25 = 25.5 cu. in.
Volume at the beginning = 3*2*4 = 24 cu. in.

Therefore volume of the ring = end volume - beginning volume = 25.5 - 24 = 1.5 cubic inches

[coral dew]

Alr, ty!!

.close

[pearl pond]

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[shut elm]

No problem!

[pearl pond]

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[hallow remnant]

hello

omg this is indian question writing style

"has exactly two solution is"

[hexed cargo]

idk

[hallow remnant]

uhh regardless

[hexed cargo]

pls help

[hallow remnant]

uhhh can you graph the function y=|3x+2|+... mentally?

[hexed cargo]

yessss

[hallow remnant]

ok so can u take a pic of a rough sketch of it?

[sharp smelt]

||wouldn't finding the maxima and minima ( local and global) be enoug||. Instead of a sketch

[hallow remnant]

||uhh if certain assumptions exist then yes||

namely ||end behaviors both go towards the same infinity|| and ||convexity||, both of which happen to be true in this, but i feel like a sketch is just more intuitive

[sharp smelt]

Fair.

[hexed cargo]

Na, bruh, because I
am from the web.

[hallow remnant]

bhaiya kya bolre ho

[hexed cargo]

i mean ki main website sai hu

phone sai nhi tho photo nhi bej paunga

[hallow remnant]

okay but does ur sketch have 'two' branches

[hexed cargo]

yeh

[hallow remnant]

so when are there two possible values of x corresponding to y/k?

its going to be, when we can draw a horizontal line y=k

and it passes through the function exactly twice

[hexed cargo]

yess

okk ok

i understand

thank youu

[pearl pond]

@hexed cargo Has your question been resolved?

[pearl pond]

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[crimson nebula]

If f:A->B is surjective , g:B->C is surjective.Show that g o f is surjective.
.

Kindly check this proof .

[analog walrus]

help?

[pearl pond]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[naive zinc]

Looks correct

Though in the final part better remove “any” in “any b in B”, just say there exists a such that…

[dense jasper]

^ You could do this a lot more concisely though:

[jolly parrot]

Civil Service Pigeon

[dense jasper]

Alternatively, if you're fine with image language, you can one-line this:

[jolly parrot]

Civil Service Pigeon

[dense jasper]

Btw, if you put . at the start of your first message, then the bot will assume that you're not claiming the channel (which is bad)

[naive zinc]

Consider (x-1)(Σk x^k). And you should claim a vacant channel

[ivory basin]

This is their channel actually

Cuz the last person added a . Before their message

[naive zinc]

Oh, the first person added .

[ivory basin]

Please repost this in another channel without the dots lol

[crimson nebula]

.close

[kindred glade]

Guys, what is this type of example called? (Need a topic name)

[crimson nebula]

What should I do ...

[ivory basin]

Repost in another channel

Without the leading dots

[ivory basin]

!occupied

[pearl pond]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[crimson nebula]

My doubt has been cleared.

[ivory basin]

Ah cool

Leave it be then

[naive zinc]

Consider (x-1)(Σk x^k)

[kindred glade]

Who can tell ???

[pearl pond]

@analog walrus Has your question been resolved?

[amber compass]

poor justin, he came into my channel by accident

and now even after he claimed it he was kicked out by bot lol

[uneven cypress]

its called math, primarily a quadratic equations question ... Also claim a non-occupied channel next time

[pearl pond]

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[random ermine]

what have you tried

[subtle crystal]

let the og polynomial be ax³ + bx² + cx + d Let's assume they get og polynomial by adding so the polynomial to which they added derivative MUST be ax³ + (b-3a)x² + (c -2(b-3a))x + d - (c - 2(b-3a))

But it's previous operation is subtract
So it must be
ax³ + bx² + but for third terms we need (b-3a)as coefficient of x² which cannot be because we have b-3a as as second term so we need b

[naive zinc]

(At the beginning of a class, the teacher wrote a third-degree polynomial on the blackboard. Then the students alternated among themselves, each performing one of the operations with the polynomial written on the board (which was then substituted by a new one):

(a) add to the polynomial its derivative.
(b) subtract from the polynomial its derivative.

At the end of the class the polynomial initially written by the teacher reappeared on the blackboard. Prove that at least one of the students made a mistake.)

[subtle crystal]

That's right

Also this

Is this logic right

Base case
Getting same polynomial by 1 operation - obviously they made mistake

Assume that if we getting poly after n operations, there's an error

Now assume we did it in n+1 trials
And prove that it's not possible

Hence, no matter how many operations u do, it's never possible to get same poly back

[random ermine]

oh do the students have to keep switching operations? so one does a then b then a etc?

[naive zinc]

Consider the matrix $\begin{matrix}0&0&0&0\1&0&0&0\0&2&0&0\0&0&3&0\end{pmatrix}$

[jolly parrot]

Cogwheels of the mind
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[naive zinc]

Prove that the power of this matrix won’t be +/- identity

Using Jordan form I think

Sorry I meant product of multiple I+A , I-A can’t be I

Where A is this matrix

(I+A)^m (I-A)^n doesn’t equal I

Still, Jordan form

[pearl pond]

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[subtle crystal]

Wdym 🤔
Let's say we start with poly f
Then let's say first student chose addition so he will write f+f'
Now consider f+f'=g
So next student MUST subtract because alternate operation they are performing
So he will write g-g'
So on

[random ermine]

yes that's what i meant, thanks

[naive zinc]

Since your operations are linear, corresponding I+A and I-A respectively. Then their canonical form will be I+J and I-J where J is the Jordan form of A

[subtle crystal]

Am sorry but am in high school 😭
So I was expecting a solution that fits into my syllabus
Ik math doesn't work like it, but I would like to solve it with resources I hve

Also, can u check my solution

[subtle crystal]

@naive zinc @random ermine

[naive zinc]

I see no reason why the second operation has to be subtraction. You assumed there are two operations performed?

[random ermine]

they're going backwards i think

[subtle crystal]

Nope
The operations are in alternate manner no.

[subtle crystal]

YEP

[naive zinc]

Exactly. So your reasoning fails

[subtle crystal]

How

[random ermine]

so in the 3rd last polynomial, what's the coefficient of x^2?

[naive zinc]

You said the second operation has to be subtraction

And you just said yourself it doesn’t have to be

[random ermine]

2nd last polynomial according to u is (in the case last operation is addition)
ax³ + (b-3a)x² + (c -2(b-3a))x + d - (c - 2(b-3a))

i haven't checked this btw but i'll assume this is right

[naive zinc]

Anyway I can’t think of another method except using matrix

[random ermine]

i think we're first considering the case where the last operation is addition

[subtle crystal]

We split it into cases

First We assumed that we got the og polynomial by adding derivative to whatever is written there
But since operations are alternating
Second last operation MUST be subtraction

[subtle crystal]

This has to be the Polynomial
Because we want the og polynomial by adding it's derivative to this 2nd last polynomial
So you can check by additing derivative, we do get out og poly back

[random ermine]

yep and u also need to make sure it's the only such polynomial g st g + g' = f

[subtle crystal]

It is it is
Because....

While adding or subtracting derivative, coefficient of x³ remains same
Instead while adding derivative x² is ALWAYS ADDED with 3a so coefficient of x² in second last poly SHOULD be b-3a

By such reasoning we can find coefficient of x and constant too

[random ermine]

alr

[naive zinc]

Anyway, I don’t have to split cases since I+A commute with I-A the canonical form of A is J ((k,k+1) entries being 1 others 0), n addition, m subtraction done gives you n-m, nC2+mC2-nm, nC3-(nC2)m+n(mC2)-mC3=0. Should give you m=n=0
Yeah nC2+nC2-n^2=0 does give you n=0=m

If translating to elementary staff I guess you can prove those two operations commute with each other

[subtle crystal]

Hmm..will look into ur method too
Ig that's only the better solution

[random ermine]

let g = f + f'
g - g' = f + f' - (f' + f'') = f - f''

let h = f - f'
h + h' = f - f' + (f' - f'') = f - f''

so the order of operations doesn't matter

and because addition adds 3a to the coeff of x^2 and subtraction subtracts 3a to the coeff of x^2,
the number of times each operation is done is the same (n=m)

[naive zinc]

Matched

n-m, nC2+mC2-nm, nC3-(nC2)m+n(mC2)-mC3. These three should be 0. First condition matched

[random ermine]

so you're basically doing f --> f - f'' multiple times to get f again

so you can reduce the problem to this 1 operation

this operation keeps subracting 6a from the coeff of x (the coeffs of x^3 and x^2 remain unchanged)

a != 0 so contradiction

@subtle crystal hopefully my solution is comprehensible

[subtle crystal]

so you're basically doing f --> f - f'' multiple times to get f again

• And - goes alternatively

[random ermine]

yes

actually this works even if the + and - ops don't alternate

and can be in any order

[subtle crystal]

Why did u consider such an operation tho?

[random ermine]

which one

[subtle crystal]

The operation which u considered which only subtracts from coefficient of x

[random ermine]

that's f - f''

[random ermine]

which is because of this

[pearl pond]

@subtle crystal Has your question been resolved?

[subtle crystal]

Umm gimme some time I'll process this

Really sorry, I am a bit slow

[feral sedge]

I'm confused

Define by A the operator f←f+f'

A^-1 • A = f-f''

now define B:=A^-1•A

B^-1•B = f+f'''' = f for cubic f?

x^3
+f' = x^3+3x^2
-f' = x^3-6x
-f' = x^3-3x^2-6x+6
+f' = x^3

[autumn trellis]

x^3-3x^2-6x+6 +(3x^2-6x-6)=x^3-12x not x^3

[pearl pond]

@subtle crystal Has your question been resolved?

[subtle crystal]

Damn dude

Understood it now

😭😭😭😭😭😭😭

[pearl pond]

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[magic ruin]

please teach my...in it we have to find elimination matrix

[toxic lichen]

are you familiar with row operations yes or no.

[magic ruin]

i got it ann

btw long time no see

[toxic lichen]

ok so do you need more help or not

[magic ruin]

no...

[toxic lichen]

ok

[magic ruin]

thank you

.close

[pearl pond]

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[hollow sonnet]

How would one go about solving this

$f(xy)=f(x)f(y)-f(x+y)+1$
If given $f(1)=2$, find $f(10)$.

[jolly parrot]

d1 shitposter

[hollow sonnet]

is it something involving testing x=1 y=1 and x=-1 y=-1

[uneven cypress]

x=y=1 gives you f(2), try find f(2) and f(3) and see if there is a recursion or pattern @hollow sonnet

[hollow sonnet]

right

[naive zinc]

Let y=1

[hollow sonnet]

just moving the things around

f(10) seems to be 11

can someone double check

thanks

[naive zinc]

Correct

[hollow sonnet]

extra question

,, f(\frac{x+1}{x})=2x+3

[jolly parrot]

d1 shitposter

[hollow sonnet]

can I multiply the inside by x to multiply the outside as well

f(x+1)=2x^2+3x type shi

[naive zinc]

Let y=(x+1)/x you see x=1/(y-1)

[hollow sonnet]

a function is a thing that maps values of x to y right

[naive zinc]

f•T=g then f=g•T^-1 actually (on the suitable domain)
T here is a linear rational transformation, your case T(z)=(z+1)/z

Two sets X and Y, a function from X to Y is a subset F of X times Y such that

1. any x in X there exists y in Y such that (x,y) is in F
2. for any (x,y) and (x’, y’) in F, if x=x’ then y=y’
   Given such F, for any x in X, by 1) and 2) there exists a unique y in Y such that (x,y) is in F, we denote this y by f(x)

[hollow sonnet]

X times Y here is a cartesian product or something right

[naive zinc]

Cartesian product

[hollow sonnet]

okay

why so rigorous

🤧

[naive zinc]

Have to

[hollow sonnet]

makes sense

[naive zinc]

Intuition explanation comes later, if necessary. Without definition any explanation is void

[hollow sonnet]

A function from $X \rightarrow Y \subset \subset F \ mid$

[jolly parrot]

d1 shitposter

[hollow sonnet]

wait what

nvm

.close

[pearl pond]

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[eager jewel]

i used IBP in the first step

to get integral nk (1-x^k)^n-1(x^k) from 0 to 1

and now idk what to do

@toxic lichen is it still wrong

[coarse locust]

it is correct now, any ideas?

[eager jewel]

nope

[coarse locust]

well clearly u need an I_(n-1) term

what can u do with the x^k

rewrite it somehow

[eager jewel]

x x^(k-1)?

[coarse locust]

that doesnt really help tho

okay try to split it into 2 terms

one with I_(n-1) and other with In

[eager jewel]

(1-x^k)^n-1 x^k and ur saying use IBP again?

[coarse locust]

no no

[eager jewel]

o

what do i split

[coarse locust]

well u have x^k and u want it gone, now can u write it using 1 and (1-x^k)

[eager jewel]

oh (x^k -1 + 1 ?

[naive zinc]

Integration by part you should have obtained a recursive relation for I_n

[eager jewel]

thats what im trying to obtain..

[coarse locust]

yeah 1-(1-x^k)

[eager jewel]

oh ok

ohh alr alr

got the relation

thank u

[coarse locust]

[eager jewel]

i was not able to see this step that easily

.close

[pearl pond]

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[pearl pond]

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[rough forge]

[still hamlet]

mb

[eager jewel]

kings rule would be a good idea if there was no x^2

[fierce sky]

queens rule works

then convert to double angle and its just arctan integrals

[eager jewel]

uh

(pi-x)^2 = pi^2 + x^2 - 2pix

how to simplify this

[still hamlet]

I'm guessing you take each integral and apply kings again except for the x² one

[eager jewel]

why queens rule?

[still hamlet]

i mean queens

[fierce sky]

well they're all the same

[eager jewel]

whats the same

[fierce sky]

apply kings rule, then expand the quadratic, you will get -I in the resultant integral

so you can move that to the other side and get 2I = stuff, I = stuff/2

[eager jewel]

wont u get +I?

[fierce sky]

?

you get an extra negative factor

[eager jewel]

oh from the cos

alr alr got it

thanks

.close

[pearl pond]

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[fierce sky]

👍

[pearl pond]

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[hollow sonnet]

$xf(x)+2f(-x)=-1$, find $\sum_{i=1}^{100}f(\frac {1}{i})

[jolly parrot]

d1 shitposter
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[autumn fossil]

Do we know anything else about f? And is f defined for all complex numbers?

try plugging in -x maybe

[naive zinc]

You actually can solve the expression of f, try replace x with -x to obtain another equation

[hollow sonnet]

it's not stated but my course content would make me assume f is defined for all R

[hollow sonnet]

brilliant idea

thanks

oh crap I forgot an x

$xf(x)+2xf(-x)=-1$, find $\sum_{i=1}^{100}f(\frac {1}{i})$.

[jolly parrot]

d1 shitposter

[fierce sky]

I believe it should still work

You still get two linear equations

[hollow sonnet]

yup

3xf(x)=4

f(1/x)=4x/3

then it just becomes the sum $\sum_{n=1}^{100}\frac {n}{3}$ right

[naive zinc]

No

I obtained f(x)=1/x

[hollow sonnet]

shi

seems like I forgot a sign change somewhere

oh

[quartz night]

Guys what 2+2?

Pls help me

[hollow sonnet]

ok got the same answer

mb

[naive zinc]

Good

[jolly parrot]

d1 shitposter

[naive zinc]

No

f(x)=1/x, i have no 3 anywhere

[hollow sonnet]

bah

brain melting brb

f(1/x)=x

[hollow sonnet]

this sum should be uhhh 5050 right

[naive zinc]

Yeah

[hollow sonnet]

thanks

[naive zinc]

Np

[hollow sonnet]

last question of the day

Let f be a function such that

$f(\frac {x+1}{x})=2x+3$ \

For all $x \in \mathbb{R}$, let set A and B be subsets of R, where $D_f=\mathbb{R}-A$, and $R_f=\mathbb{R}-B$. Express $A \cup B$ in roster notation.

[naive zinc]

Can’t understand

[hot timber]

Uh

[hollow sonnet]

forgive my translation

[naive zinc]

Still makes no sense

R_f, D_f has anything to do with f?

[jolly parrot]

d1 shitposter

[naive zinc]

I mean can’t understand mathematically

[hollow sonnet]

the notation here is domain and range of function

[naive zinc]

Oh

Should have said that

[hot timber]

Ouhhh

[hollow sonnet]

my bad

[naive zinc]

So (3x-1)/(x-1)

=(3x-3+2)/(x-1)=3+ 2/(x-1)

Rest is clear

[hollow sonnet]

is the -1 for matrix (1,1),(1,0) supposed to denote a division linear transformation or something

^-1

[naive zinc]

Yeah

[hollow sonnet]

I see

[naive zinc]

T(x)=(x+1)/x corresponds to the matrix (1,1;1,0)

[hollow sonnet]

I'll need to study up on that

[naive zinc]

Don’t have to though, certainly make things smoother

[hollow sonnet]

(Ax+B)/(Cx+D) should the correspond to the matrix (A,B;C,D) right

[naive zinc]

Yeah

[hollow sonnet]

makes sense

[naive zinc]

Matrices / ~ where A~ kA for some constant k

[hollow sonnet]

I don't understand functions enough for this yet

F maps to G by applying a linear transformation to f

which means that f can be obtained by applying the inverse linear transformation on g

[naive zinc]

There is no map that maps mappings to something. All those are just composition

f•T=g
(composition of f and T)=g

composite with T^-1 in the front of both sides

f=(f•T)•T^-1=g•T^-1

(On a suitable domain)

[hollow sonnet]

i see

in this case, the transformation is the inverse of (1,1;1,0)

multiply g by inverse of function to get f(x)

where f(x)=3+2/(x-1)

sheesh

[naive zinc]

Yeah

[hollow sonnet]

okay I think I'll stop here\

thank you for the help

😭 💀💀💀

[naive zinc]

Np

[hollow sonnet]

.close

[pearl pond]

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[low matrix]

@cinder flower lmao

ptsd

[cinder flower]

i have helped with this exact problem too lol

[low matrix]

That's why I pinged, I read the chat...

[cinder flower]

lol

[low matrix]

How many checkerboards are there in total

we have 4 cells, each with either red or blue

So ig its 16

[cinder flower]

yea

[low matrix]

is this it

[cinder flower]

bottom line is wrong

[low matrix]

did i miss one

[cinder flower]

that class should have 4 elements, and also you wrote the same one twice

[low matrix]

only BRRB and RBBR?

i cant find the other two

what operation did u perfor?m

[cinder flower]

oops. i thought the was the left red, right blue square

yea that one just has 2

[low matrix]

so just four classes?

[cinder flower]

what about RBBB etc?

[low matrix]

oh

[low matrix]

there wil be two more [RRRB] and [BBBR]

each will have four elements

[cinder flower]

yes

[low matrix]

now the paritition is complete, all classes sum to total 16 elements

nice

[cinder flower]

yes

[low matrix]

ty

deja vu

.close

[pearl pond]

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[cinder flower]

i’ve just been in this place before

[pearl pond]

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[low matrix]

I am drawing the hasse diagram

[low matrix]

Are both of these acceptable

<@&286206848099549185>

[edgy bane]

Replace the top 12 by 20 and it's acceptable

Just a slight mistake

Aside from that the Hasse diagram on the right is acceptable

[low matrix]

whyy tho

al my arrows are upward

[edgy bane]

10 clearly doesn't divide 12

[low matrix]

10 * 2 = 20

that is why i did not conect 10 and 12

[naive zinc]

I assume the left one is your answer? Remove 5->20

[low matrix]

why? 5 divides 20

[naive zinc]

You already have 5->10->20

[low matrix]

oh

btw how do i know where to place the vertices

the diagrams i draw turn out to be ugly lmao

[pulsar flax]

Also you should have 2 -> 4

[low matrix]

i did

[edgy bane]

This mistake

That's the only mistake in your right diagram.

[low matrix]

yes yes mb

but that is not my diagram hehe

i copied from the solution

mine is the left

[edgy bane]

Oh

[low matrix]

how can i improve the structure

i end up with weird shapes

[naive zinc]

By elements“rank” if your poset has some good property like polytopes. But in general I don’t think we can guarantee to make the heights reflecting their order

[low matrix]

how could i know 2 and 5 are at same level

and that 25 is at the same level as 4 and 10

[naive zinc]

You couldn’t know in advance. Now that you know you can modify it

[low matrix]

can i put 25 at the same level as 12 and 20?

[edgy bane]

Yes since they're all maximal here

[low matrix]

12,20 and 25 are maximal?

[naive zinc]

You just guarantee one thing:

[low matrix]

which is the greatest elemnt?

[naive zinc]

If x and y are adjacent , and x<y, then y must be placed above x

[edgy bane]

There is no unique maximum here. It's a partially ordered set

[low matrix]

yeah cuz the arrows are upward

[low matrix]

does this hasse diagram have a greatest element?

[naive zinc]

Elements having the same height, or one being higher can still be incomparable, it’s allowed