help-4

[gilded summit]

it wont be a field extention of Z_3?

[slim summit]

no, no, i explained myself badly

what i mean is that given the multiplicative unit 1+I

if by adding multiple copies of 1+I i can only get the elements 0+I, 1+I and 2+I

then we have an extension (though it could fail to be a field) of Z_3

[gilded summit]

if its in Z_3 thats garenteed right

[slim summit]

uh, yeah, because that's basically equivalent to asking the projection morphism is injective

Anyway, going back to taking the quotientby <p(x)>

given what we've said so far

[gilded summit]

ok

[slim summit]

What might
$$
\mathbb{R}[x]/\langle x^2 + 1\rangle ;
$$
be isomorphic to?

[gilded summit]

wait

one sec

im still trying to understand why there is a solution

actually ill solve this

ok so we have (x^2+i)

and all the remainders are

[rocky lotus]

Nanigov

[gilded summit]

les then deg 2

[slim summit]

+1! not +i

typo 😅

[gilded summit]

ok i got scared

coooooolll

ok so

uhhh

yeah

les then deg 2

so all the remainders are coificents right

no

its x+r or r

[slim summit]

indeed

[gilded summit]

not hte last part

[slim summit]

but r is basically 0*x + r

[gilded summit]

so we have

$\mathbb{R}/ \langle a^2 + 1 \rangle = {0, x+r, r}$

and there are infinately many r

[rocky lotus]

BOSS

[gilded summit]

because it just cant be a multiple of 1 right

so there are like infinate cosets?

0.1 + I and 0.2 + I would both be seperate cosets

[slim summit]

There are infinitely many cosets but consider the polynomial

ax + b

it's class isn't x+r

[gilded summit]

im tempted to just use the first iso therem and evaluation homomorpism to show its iso to R

oh nvm

[slim summit]

so you're missing a couple remainder classes

[gilded summit]

-1 isnt in R

ok continue

[gilded summit]

ok

oh

[slim summit]

you are right all remainders have to be lineal or constant

but in general lineal polynomials don't have to be monic

[gilded summit]

$\mathbb{R} [x]/ \langle x^2 + 1 \rangle = {0, ax+b \mid a,b \in \mathbb{R}}$

[slim summit]

odd notation but yes

[gilded summit]

ah

[slim summit]

for the brackets to show up write a \ before them

[gilded summit]

wahts the proper way

[gilded summit]

i forgor one sec

[slim summit]

just ax+b with a, b in R

since you can just take a = 0 for the constant polynomials

[gilded summit]

oh ur right

c is just a=0

ok cool

interesting

wait do u have another example

ill solve it correctly

this is helpful loll

[slim summit]

wait wait, you've not gotten to the punchline here

so all your elements look like ax+b

or well their equivalence classes

[gilded summit]

yes

[slim summit]

now i ask you

what is special about x (or its class rather) in this quotient

what must x + I satisfy?

[gilded summit]

hmm

it must be in the same coset as all (ax+b)+i?

[slim summit]

nope

[rocky lotus]

BOSS

[slim summit]

remember, each ax+b is a different coset

the coset of x+1 is different from the coset of 2x + 3

[gilded summit]

oh there has to be a way to get to 0

[slim summit]

you know that [x^2+ 1] = [0] since by definition x^2 + 1 is in <x^2 + 1>

[gilded summit]

actually idk that was just a guess

[gilded summit]

hmm

o

oh yes

yeah

but what does taht say about ax+b

[slim summit]

[x]^2 + [1] = [0]

solve the equation for [x]

[gilded summit]

[x]^2 + [1] = [0] = [x^2+1] = [0]

uhhh

[x]^2 = [0] - [1]

[slim summit]

uh huh

[gilded summit]

idk

[slim summit]

[0]-[1] = [0-1] = [-1]

so

[x]^2 = [-1]

[gilded summit]

ok

[slim summit]

your cosets look like a[x] + b where [x]^2 = [-1]

what might this ring be?

[gilded summit]

a[x] + b where [x]^2 = [-1]

uhh

hmm

im gonna be so honest im really lost

[slim summit]

fair fair

[gilded summit]

[slim summit]

let's retrace your steps, because you did have the right idea at the start

[gilded summit]

ok

[slim summit]

So we know the quotient R[x]/<x^2+1> is going to consist of (the classes of) all remainders of polynomial division by x^2 + 1

[gilded summit]

yes

and all remainders will look like

ax+b

through the division algo as R[x] is a field so its just all elements less then deg 2

[slim summit]

indeed

[gilded summit]

wait also

these are all possible remainders right

doesnt mean it is a remainder

oh

nvm

[slim summit]

yeah, but it's not too hard to come up with a polynomial that gives it as a remainder

but good catch

[gilded summit]

so does every ax+b exist?

as a remainder

the division algo garentees that there is a r(x)

[slim summit]

yeah, through the polynomial x^2 + ax + (b+1)

[gilded summit]

so if we mod it out would every possible r(x) be an equivilance class

[slim summit]

since it's equal to 1*(x^2 + 1) + (ax+b)

[gilded summit]

ok sure

so in genral every possivble r(x) would be its own equivilance class

making up the quotent ring

ok continue

[slim summit]

yes

Okay now just because of how the operations on the equivalence classes are defined

[ax+b] = [a][x] + [b]

[gilded summit]

So we know the quotient R[x]/<x^2+1> is going to consist of (the classes of) all remainders of polynomial division by x^2 + 1
with remainders ax+b which make up the equivilance classes

[gilded summit]

ok slow down so

ax+b means a is its own coset

same with x

same with b

cool

(ax+b)+ I = (a+I)(x+I)+(b+I)

[slim summit]

yup

Now to make things a bit nicer, consider the fact that [a] = [b] if and only if a = b (assuming a and b are real numbers), because

[a]-[b] = [0] if and only if a-b is a multiple of x^2+1

but the only constant multiple of x^2 + 1 is 0

[gilded summit]

Ok so

[0] is the class that contains our ideal which is all multiples of x^2+1

[slim summit]

[0] is our ideal!

[0] = 0+I = I

[gilded summit]

oh yeah

sorry typo

[slim summit]

but yeah, it is all multiples of x^2 + 1

[gilded summit]

ok so

[a] = [b] if and only if a = b (mod x^2+1)

sure

so

oh

in ur example they have to be the same mod the ideal generator

u asked me a while ago

but sorry continue

[slim summit]

ye

[gilded summit]

we are here

[a] = [b] if and only if a = b (mod x^2+1)

[slim summit]

substract both sides by b to get

a-b = 0 (mod x^2 + 1)

[gilded summit]

yes

[slim summit]

which is equivalent to saying a-b is a multiple of x^2 + 1

[gilded summit]

yes

[slim summit]

but the only constant multiple of x^2 + 1 is the constant polynomial 0

so a-b = 0 so a = b (in R)

[gilded summit]

wait

x^2 + 1 is the constant polynomial 0

yeah

a-b = 0 so a = b (in R)

oh

woah

[slim summit]

yup

[gilded summit]

ok what does that mean for us

[slim summit]

this is just a small sidenote to say that the only constant representative of the class [a] is a itself

[gilded summit]

ok

[slim summit]

so if i were to write (and this would just be notation)
[a][x]+[b] = a[x] + b

there wouldn't be any ambiguity

[gilded summit]

ok so what ur saying is

$(a+I) = (b + I) \iff a=b$

[rocky lotus]

BOSS

[gilded summit]

and the only constant representation of a+I is a, or a is the only constant in a+I

[slim summit]

so we might as well write (b + I) as simply "b" since that's the only constant that would represent it

[gilded summit]

cool

[slim summit]

this is just so the notation looks better

[gilded summit]

yes

ok i agree

[slim summit]

okay now all our elements look like
[ax+b] = [a][x] + [b] = a[x] + b

[gilded summit]

so we are legit jsut working with constants of r

[slim summit]

basically yeah

[gilded summit]

so this is an extention field of R

[slim summit]

it's a long winded way of saying that R is a subring of this quotient

ye exactly

[gilded summit]

as it contains all r+I = r, and then more

[gilded summit]

subring or base field

[slim summit]

well both mean the same thing

[gilded summit]

oh?

[slim summit]

since a field is a ring where everything just so happens to be invertible

[gilded summit]

ok

ill use base field for rn because im trying to get it engraned into my brain

[slim summit]

fair fair

okay now here's the coup de grace or however you spell it

we know that, by definition

[gilded summit]

bet

[slim summit]

[x^2 + 1] = [0]

[gilded summit]

sure

yes

[slim summit]

so,
[x]^2 + [1] = [0]

but we agreed to write [a] as just a so

[x]^2 + 1 = 0

[gilded summit]

ok

[slim summit]

so [x]^2 = -1

[gilded summit]

ok

[slim summit]

now i suggest renaming

[x] to i

[gilded summit]

hm?

[slim summit]

and we have a ring of elements of the form
a[x]+b = ai + b

where i^2 = [x]^2 = -1

[gilded summit]

oh

ok

[slim summit]

so what is our ring?

[gilded summit]

what does this mean

we are in R[i]?

[slim summit]

yes

[gilded summit]

OHHH

al'jnfoajsdg

woahhhhhhhhhh

[slim summit]

:3

and what is R[i] if not

[gilded summit]

my brain hurts but

thats fire

wait

u can do this for

any plinomail

[slim summit]

$$
\mathbb{R}[x]/\langle x^2+1\rangle \cong \mathbb{C}
$$

[rocky lotus]

Nanigov

[gilded summit]

wait

so

for any

F[x] / \langle x^2 + n \rangle

its isomorphic to F[\sqrt{-n}]

[slim summit]

Indeed

[gilded summit]

oObhisabgpfhb

goated

holy shit better explanations then my prof bro

gjgj

[slim summit]

:D

[gilded summit]

ok wait wait wait

i got a

good application for this

warm up problems i didnt get before

but now

look at A3 lmao

i was lost but now i understand

[slim summit]

mhm

[gilded summit]

ok so really quickly

the explanation goes something like

lets say we have $\mathbb{Q}[x]/<x^2-2>$

[rocky lotus]

BOSS

[gilded summit]

then we know that every remainder is under deg 2, so again ax+b where a,b \in Q

then we know that if

a -b = 0

then a = b as the only constant in [0] is 0

so we know a = b or thats the only reepresenation of the coset

so then with that

we jnow

[x]^2 - 2 = 0

x^2 = 2

x = sqrt2

which means this is iomorphic to Q[sqrt2]

@slim summit

[slim summit]

correct

[gilded summit]

im a little iffy on jsut plugging sqrt2 as x

but i think its starting to click

[slim summit]

and the isomorphism is given by [a][x]+[b] -> a*sqrt(2) + b

the whole business with the only constant representative of [b] being b itself guarantees that mapping is injective

and it's obviously surjective

[gilded summit]

oh this works because thats the only way x^2 = 2 as 2 is the only element in the equivilance class

oh

so all they proved here

is that there is an extention field

where we have a zero \alpha?

[slim summit]

yes

[gilded summit]

humf

[slim summit]

Basic, but it does allow you to like, all of modern Galois theory

[gilded summit]

idk how it works in thsi example ngl

but

could just be me

ill work it out again

[slim summit]

o7

[gilded summit]

using the example we just solved

[slim summit]

notice that the important part of [a] = [b] iff a = b depended on the fact that you can't multiply (x^2 + 1) by anything other than 0 to get its degree down

[gilded summit]

ok

[slim summit]

which depends on teh fact Z_2 is a field

i mean R

because it has no zero divisors

[gilded summit]

wait does this work only in fields

[slim summit]

^^

[gilded summit]

like F[x]/<p(x)>

[gilded summit]

so integral domains too?

D[x]/<p(x)>

[slim summit]

yeah

[gilded summit]

hmm

ok

that makes sense

because we have

[slim summit]

Z[x]/<x^2+1> is also Z[i]

[gilded summit]

cancelation law

and no other things can be 0

ok wait

let look at this spesific example

[slim summit]

also important to note F[x]/<p(x)> won't always be a field since p might not be irreducible

[gilded summit]

yeah

[slim summit]

but you'd still get an isomorphism to F[\alpha] with \alpha some root of the polynomial, it just won't be a field

[gilded summit]

its a PID and integral domain

but not a field

wait no

thats F[x]

my bad

ok wait so

for

$\mathbb{Z}_2 [x] / \langle x^2 + x + 1 \rangle$

[rocky lotus]

BOSS

[gilded summit]

We basicly know this element is irreduceable

so

this is a field

which is why this applies

because no zero devisors

so for any prime and maximal ideals this would work?

because then its at least an integral domain

[slim summit]

yeah

in the sense that you'll get an extension of the rign with an added root

[gilded summit]

ok so let me just speedrun this example really quickly

[gilded summit]

i still dont completelly get what this means

but

oh

we are getting an extention of Z_2 tho not Z_2 x right

[slim summit]

yes

[gilded summit]

ok so

$\mathbb{Z}_2 [x] / \langle x^2 + x + 1 \rangle$ has all elements r(x)+I such that r(x) has deg less then 2

[rocky lotus]

BOSS

[gilded summit]

this is precicely ax+b, but we are in Z_2 so we get

0, 1,x+1,x

so

[slim summit]

Yes, though remember stricktly speaking you have to prove the other inclusion

[gilded summit]

$\mathbb{Z}_2 [x] / \langle x^2 + x + 1 \rangle = {0, 1,x+1,x}$

[rocky lotus]

BOSS

[slim summit]

but that's easy just make up a polynomial that obviously has that remainder

[gilded summit]

doesnt the division algo garentee existance?

we can just add 0,1,x+1,and x to x^2+x+1 to show it tho lmao

[slim summit]

yeah basically

[gilded summit]

now as 1+1 is 0, we know this containz Z_2

and is an extention of Z_2

[slim summit]

yes but it would be better to phrase it as the projection morphism is injective

since the only element of Z_2 that maps to 0 + I is well, 0

[gilded summit]

ok the goal here is to construct a field extension of Z_2 containing an element α such that p(α) = 0.

so

we are solving for x^2+x+1 = 0

right

[slim summit]

yeah

[gilded summit]

ok so

[x^2]+[x]+1 = 0

[x^2]+[x] = 1

as 1 is the only constant in 1+I

uhh hmm

[slim summit]

yes

[gilded summit]

should i lwkt just factor this out

[slim summit]

you don't need to solve it tho actually

[gilded summit]

lmao

oh

[slim summit]

like just the fact
[x^2]+[x]+[1] = [0]

already tells you that this is a solution

by definition

[slim summit]

now you might not be able to "solve for [x]" but you know it satisfies the equation [x]^2 + [x] + [1] = [0]

so it's already a solution automatically

[gilded summit]

i domnt notice that alpha ^2 + alpha +1 = 0

[slim summit]

that's because p(alpha) = 0

so like by definition alpha^2 + alpha + 1 = p(alpha) = 0

are you familiar with presentations of groups by any chance?

[gilded summit]

uhhh

[slim summit]

if not it doesn't matter i just thought tha tmight've helped a lil lol

[gilded summit]

probably but i forgot the topic name because its been a minute since group theory

[slim summit]

so you know when you write the dihedrals as like
<r, s | r^3 = s2 = e, rs = sr^-1>
?

[gilded summit]

yes

[slim summit]

the equations on the rhs are referred to as "relations" which is to say equations these elements satisfy by definition

here the element alpha by definition satisfies the relation p(alpha) = 0

[gilded summit]

rotations + idenitity and then that multiplied with the reflection up front

ok yeah i dont understand that

id have to review lmao

ok wait

[slim summit]

yeah dw abt it

[gilded summit]

This is where my logic kinda gets lost

[slim summit]

it was just in case

[gilded summit]

I understand what E is and what its an extention

i dont understand why E has a zero

or like has to have a zero

based on what we did we proved taht certain quotent rings have zeros

but here let me set

x^2+x+1 =0

and we have elements {0, 1,x+1,x}

uh

hmm

[slim summit]

so you know {0, 1, x+1, x} is a field (because the polynomial you took the quotient by was irreducible), and you'd like to know what the operations on that set actually look like

[gilded summit]

yes

it would be addition and multiplication right

[slim summit]

yes

[gilded summit]

jsut normal quotent ring opperations

ok

x^2+x+1 =0 = [x]^2 + [x]+[1]

right

[slim summit]

yes

[gilded summit]

ok so

we know that 1 is the only constant in 1+I because lets say if a-b=0, then as the only constant in I is 0 we have it so a = b

with the equivilance classes

so we have it so

[x]^2 + [x] = -1

[slim summit]

yes

[gilded summit]

ok so then with this where do we go from here?

ok actually

wait

[slim summit]

well now it would be nice as i said before to know what the operations of the field look like

like yes
x*(1+x) = x + x^2

[gilded summit]

ok

[slim summit]

but that's not an element of {0, 1, x+1, x}

so we need to find out which of those three it's actually equal to

[gilded summit]

wait what is this

oh

nvm u just factored out an x

ok continue

x*(1+x) = x + x^2 = -1

[slim summit]

yes

and -1 = 1 because Z_2

[gilded summit]

ah

[slim summit]

so we've found that
x(x+1) = 1

[gilded summit]

yeah

x(x+1) = 1

[slim summit]

so that's one product, ideally we'd like to complete the whole multiplication table like this

same for addition

[gilded summit]

my book did it

but wait

why does

there have to be a root

like how do we know for sure it contians a root

[slim summit]

because [x^2 + x +1] = [0]

that implies that [x]^2 + [x] + [1] = 0

[gilded summit]

OHH

ok

yeah we did that

ok cool

ok then

h

thats the addition tables

lmfao

[slim summit]

mhm

[gilded summit]

ok if it has a zero

we call that zero alppha

we know that

[slim summit]

yeah

[gilded summit]

$\mathbb{Z}_2 [x] / \langle x^2 + x + 1 \rangle \cong \mathbb{Z}_2(\alpha)$

based on the logic we used before

[rocky lotus]

BOSS

[slim summit]

mhm

and the isomorphism is given by sending [x] to alpha

[gilded summit]

ok

[slim summit]

and [1] to 1

[gilded summit]

yeah

[slim summit]

and extending that linearly

[gilded summit]

Kronecker

goat

ok i think i understand now

[gilded summit]

just to go over why this holds

its what we did before

with R[x]\x^2+1

[slim summit]

yes

[gilded summit]

okie

thank you so much

[slim summit]

np

[gilded summit]

this was incredibally helpful lmao

ill read over that again but yeah explanations make sense

[slim summit]

it feels nice to actually show some understanding after i've spent the whole day banging my head against (funnily enough) Galois theory lol

well Kummer (lmfao) theory in particular

[gilded summit]

lmfaoooo

i think we do like

a smige of galious

to link everything together

my book as an interesting order of things

[slim summit]

Galois is really cool tbh, it's just working out particular examples that can get a bit annoying, since it can involve knowing alot of trivia about finite groups basically

[slim summit]

hmm

[gilded summit]

luckly we just had like 10 mins on it but i do wanna take a class on it later

[slim summit]

that Artin?

[gilded summit]

no clue what ever my prof uses

one sec

if u want it its like

[slim summit]

ANTONIO BEHN

[gilded summit]

10 mb lmao

[slim summit]

THAT'S A PROFESSOR OF MINE WHAT

[gilded summit]

bro what college doyou go to

are you in spain

[slim summit]

Chile

[gilded summit]

ah

makes sense

yeah goat lmfao

[gilded summit]

thats insane

actually insane

[slim summit]

YEAH???

[gilded summit]

lmfaoo

[slim summit]

i mean i knew he was really well connected because i've seen his CV but DAMN

[gilded summit]

he wrote my textbook

so thats something

[slim summit]

anyway he was my Linear Algebra I professor some years ago

[gilded summit]

this was published in 2022

soooooo

you may have had him the same time lmao

ok im gonna go back to learning this to prepare

tysm

[slim summit]

[gilded summit]

.close

[vale dock]

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Closed by @gilded summit

Use .reopen if this was a mistake.

[gilded summit]

.reopen

waiy

[vale dock]

✅

[gilded summit]

@slim summit

sorry

[slim summit]

LMAO

[gilded summit]

are you still here

[slim summit]

yeah?

[gilded summit]

LSGJNA

we arnt doing ideal opperations anymore?

fuck my cafe is closing

one sec

ill brb

[slim summit]

o7

i'll just leave the reply written

[gilded summit]

wahts o7

[slim summit]

a salute

[gilded summit]

ohhh

feel like my parents

bet

[slim summit]

yeah we are (ill write x for alpha)

(1+x)(1+x) = 1^2 + 2x + x^2

we know 2x = 0 since we are in characteristic 2 so

(1+x)(1+x) = 1 + x^2

but since x^2 + x + 1 = 0 (because it is a root of that polynomial) we know that 1+x^2 = -x

but again characteristic 2 means -x is just x

so (1+x)(1+x) = x

[vale dock]

@gilded summit Has your question been resolved?

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

[oak hollow]

for B I used IVT instead

and took the derivative of h

does this still work

someone help

pls

I have ap exam in like 10 hours

[verbal badger]

,rccw

[verbal badger]

lgtm

Organize your work better though

[vale dock]

@oak hollow Has your question been resolved?

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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.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[dense jewel]

Is it correct?

[dense jewel]

? Sorri for writing xd

Is it correct?

@anyone?

Alo?

!showwork

[vale dock]

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

[dense jewel]

@distant pulsar sorri ping really one

@river shale

[river shale]

Why me?

[dense jewel]

Because you are helpful thats what your role says

[river shale]

What’s the original question?

I need it to check if your calculation meets the requirement of the question

[dense jewel]

Like, need to do without expansion

It is determinant

@river shale

Anyone?

.close

[vale dock]

Channel closed

Closed by @dense jewel

Use .reopen if this was a mistake.

[vale dock]

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[boreal solstice]

Let G, A, and I be events.
Let ¬A be A's complement.
Let P(A) be A's probability.

Premise 1: P(G | A) - P(G | ¬A) > P(I | A) - P(I | ¬A)
Premise 2: Something stating that G and I are highly symmetrical events.
Conclusion: ∀x(P(g_x | A) - P(g_x | ¬A) > P(i_x | A) - P(i_x | ¬A)), where g_x is an element of G and i_x an element of I such that g_x and i_x are symmetrical in the relevant sense.

Note that ∀x(P(g_x | A) - P(g_x | ¬A) > P(i_x | A) - P(i_x | ¬A)) doesn't need to be true for all ways of dividing G and I into subevents g_x and i_x. But, I just need one way of dividing G and I into subevents g_x and i_x such that ∀x(P(g_x | A) - P(g_x | ¬A) > P(i_x | A) - P(i_x | ¬A)).

Question: What should premise 2 be? In what sense should G and I be symmetrical for the conclusion to follow?

[vale dock]

@boreal solstice Has your question been resolved?

[vale dock]

@boreal solstice Has your question been resolved?

[boreal solstice]

<@&286206848099549185> anyone whose good with probability theory around? Curious to hear your thoughts :) <3

[midnight pier]

Hi my name is Eshan and I am doing my bachelors in statistics

[midnight pier]

I can help you 🙂

[boreal solstice]

cool!!

thanks

:)

[boreal solstice]

Curious to hear your thoughts on this problem @midnight pier !!

[midnight pier]

It sounds like advance set theory problem

Also, I don't have enough knowledge on symmetrical events so I cant help much in this

[midnight pier]

Try using the properties of conditional probability

Use multiplication theorem and set theory @boreal solstice

[vale dock]

@boreal solstice Has your question been resolved?

[turbid moon]

I don't think this is well formulated, because there's the trivial (and related subdivisions) where you just take the division to be the event itself

further, if you divide into events independent of A, such as C and C's complement, then you automatically get that the right side is zero.

You can take this even further by taking I = A^C, and then you have a negative number

what's the purpose of this?

I've also never heard about "symmetrical" events as a formal definition

[vale dock]

@boreal solstice Has your question been resolved?

[boreal solstice]

sry

let me think of a way to clarif y

[solid mountain]

HOL UP WASSUP MAN

[waxen berry]

what do you mean symmetric in the relavant sense?

also P(G|A) - P(G| -A) looks pretty weird, what are you trying to do with this?

[vale dock]

@boreal solstice Has your question been resolved?

[wet loom]

can anyone confirm this is 16.71m?

[ripe gazelle]

yup got the same thing for the perimeter.

[wet loom]

perfect thanks 🙂

[vale dock]

@boreal solstice Has your question been resolved?

[boreal solstice]

@turbid moon apologies for the unclarity. Here is a clearer version of at least the first part of my question. Would be curious to hear your thoughts!

Let s be an event. \
Let i be an event and $i_j \subset i$. \
Let x be a number. \
Let P(i) be i's probability. \

Premise 1: $P(i) = x P(i_j)$ \
Premise 2: ???? \
Conclusion: $P(i \mid s) = x P(i_j \mid s)$ \

What does premise 2 need to be to make this inference valid?

[rocky lotus]

! If in Oxford, see my about me

[boreal solstice]

Ye, the question was kind in what sense G and I have to be symmetric in order for the inference to be valid, but apologies for not making that clear. I hope this version of the first part of my question is clearer

[boreal solstice]

the intersection between s and i has to be isomorphic to the intersection between s and i_j, but how does one put that formally and how does one prove the conclusion from that?

[rocky lotus]

! If in Oxford, see my about me

[boreal solstice]

Please ignore the below

[rocky lotus]

! If in Oxford, see my about me

[boreal solstice]

ignore the below

[rocky lotus]

! If in Oxford, see my about me

[boreal solstice]

So, I have events i and d, which are partitioned into jointly exhaustive mutually exclusive events i_j and d_j (we chose our event space such that i_j and d_j are also events for all j).
Now, I want to say that the relation between i and {i_1, i_2, ..., i_n} and between d and {d_1, d_2, ..., d_n} is isomorphic, i.e., i is related to {i_1, i_2, ..., i_n}, like d is related to {d_1, d_2, ..., d_n}. How does one say that formally?

[boreal solstice]

Improved version:
So, how can we just that to infer \ $\forall j \exists w_j (P_e(i_j \mid s) = w_jP_e(i \mid s) \wedge P_e(i_j \mid \neg s) = w_jP_e(i \mid \neg s) \wedge P_e(d_j \mid s) = w_jP_e(d \mid s) \wedge P_e(d_j \mid \neg s) = w_jP_e(d \mid \neg s))$ from \

$\forall j \exists w_j (P_e(i_j ) = w_jP_e(i) \wedge P_e(i_j ) = w_jP_e(i ) \wedge P_e(d_j ) = w_jP_e(d ) \wedge P_e(d_j ) = w_jP_e(d ))$

(please ignore the subscript e)

[rocky lotus]

! If in Oxford, see my about me

[waxen berry]

this has to be true for all events i_j which are a subset of i?

[boreal solstice]

yes

[boreal solstice]

this is I think the clearest formulation of my question

[boreal solstice]

but, then I basically need to know that the intersection of i and s, d and s, i and ¬s, and d and ¬s is 'random.'

Like doesn't favour i_j's with high j over those with low j

[boreal solstice]

how does one express that formally?

[waxen berry]

why cant i just take i_j = s intersect i and this is wrong, or s has 0 intersection with i, or s contains i?

[boreal solstice]

What do we need to know s and ¬s such that one can validly infer
\ $\forall j \exists w_j (P_e(i_j \mid s) = w_jP_e(i \mid s) \wedge P_e(i_j \mid \neg s) = w_jP_e(i \mid \neg s) \wedge P_e(d_j \mid s) = w_jP_e(d \mid s) \wedge P_e(d_j \mid \neg s) = w_jP_e(d \mid \neg s))$ \ from \

$\forall j \exists w_j (P_e(i_j ) = w_jP_e(i) \wedge P_e(i_j ) = w_jP_e(i ) \wedge P_e(d_j ) = w_jP_e(d ) \wedge P_e(d_j ) = w_jP_e(d ))$

(please ignore the subscript e)

[boreal solstice]

this doesn't work. s is unrelated to i_j and i.

[waxen berry]

so for what i_j do you need this to be true?

[boreal solstice]

in the context of the problem I am working

this needs to be true for all j

[rocky lotus]

! If in Oxford, see my about me

[waxen berry]

we are given s is independent of i_j and s indepent of i?

[boreal solstice]

no

[waxen berry]

so wdym s is unrelared to i_j and i

[boreal solstice]

I am just saying that you cannot relate i_j and i, like this i_j = s intersect i

[boreal solstice]

sry should have clarified

[waxen berry]

you need P(s|i) = P(s|i_j) and P(s|d) = P(s|d_j)

[boreal solstice]

I cannot say that. i_j is a proper subset of i

What do we need to know s and ¬s such that one can validly infer
\ $\forall j \exists w_j (P_e(i_j \mid s) = w_jP_e(i \mid s) \wedge P_e(i_j \mid \neg s) = w_jP_e(i \mid \neg s) \wedge P_e(d_j \mid s) = w_jP_e(d \mid s) \wedge P_e(d_j \mid \neg s) = w_jP_e(d \mid \neg s))$ \ from \

$\forall j \exists w_j (P_e(i_j ) = w_jP_e(i) \wedge P_e(d_j ) = w_jP_e(d )$

(please ignore the subscript e)

[rocky lotus]

! If in Oxford, see my about me

[waxen berry]

so? why does that mean you cant say that?

[midnight pier]

This is absolutely beyond me 😭

[turbid moon]

this is the necessary condition for what you're asking for; also I'm not sure how the second sentence relates to the first

[boreal solstice]

can you prove that?

[turbid moon]

furthermore, "proper subset" when it comes to events isn't very different from "subset" in the sense that in many probability spaces, one can make a proper subset by simply taking out a few outcomes that have measure zero

[boreal solstice]

That is, you claim that

To validly infer
\ $\forall j \exists w_j (P_e(i_j \mid s) = w_jP_e(i \mid s) \wedge P_e(i_j \mid \neg s) = w_jP_e(i \mid \neg s) \wedge P_e(d_j \mid s) = w_jP_e(d \mid s) \wedge P_e(d_j \mid \neg s) = w_jP_e(d \mid \neg s))$ \ from \

$\forall j \exists w_j (P_e(i_j ) = w_jP_e(i) \wedge P_e(d_j ) = w_jP_e(d )$

It's necessary that it's false that $i_j$ is a proper subset of i?

(please ignore the subscript e)

[turbid moon]

what does the wedge mean with respect to probability measures?

oh nevermind it's a parentheses thing; you're using it as logical and

[rocky lotus]

! If in Oxford, see my about me

[waxen berry]

yes

[turbid moon]

can you simplify the notation and use some more english in it? It's really painful to read this sort of nested quantification, and it's unnecessary

[waxen berry]

i mean the two statements are equivalent

given the bottom statement

[waxen berry]

no?

what does proper sbuset have to do with P(s|i) = P(s|i_j)

and P(s|d) = P(s|d_j)

[turbid moon]

P(i) = P(i | s)P(s) / P(s | i), so rewriting the conclusion that you want [P(i | s) = xP(i_j | s)]

P(i | s)P(s) / P(s | i) = x P(i_j | s)P(s) / P(s | i_j)

This gives that P(s) / P(s | i) = P(s) / P(s | i_j), which simplifies to:
P(s | i) = P(s | i_j)

I'm still completely lost on what you're trying to do

[waxen berry]

you have to worry about a few cases (because of division by 0), but yes

[boreal solstice]

Let's start from a more basic point.

How does one prove

$P_e(i \mid h \cup w) = \frac{P_e(i \mid h)P_e(h) + P_e(i \mid \neg h \cap w)P_e(\neg h \cap w)}{P_e(h) + P_e(\neg h \cap w)}$?

(ignore the subscript e)

[rocky lotus]

! If in Oxford, see my about me

[boreal solstice]

Well, we know that $h \cup w = h \sqcup (\neg h \cap w)$

[rocky lotus]

! If in Oxford, see my about me

[boreal solstice]

$\frac{P_e(i \mid h)P_e(h) + P_e(i \mid \neg h \cap w)P_e(\neg h \cap w)}{P_e(h) + P_e(\neg h \cap w)} = \ \frac{P_e(i \cap h) + P_e(i \cap \neg h \cap w)}{P_e(h) + P_e(\neg h \cap w)}$

[rocky lotus]

! If in Oxford, see my about me

[boreal solstice]

Thus the real question is how does one prove

$P_e(i \mid h \cup w) = \frac{P_e(i \cap h) + P_e(i \cap \neg h \cap w)}{P_e(h) + P_e(\neg h \cap w)}$?

(ignore the subscript e)

[rocky lotus]

! If in Oxford, see my about me

[boreal solstice]

Let's rewrite:

$(P_e(h) + P_e(\neg h \cap w)) * P_e(i \mid h \cup w) = P_e(i \cap h) + P_e(i \cap \neg h \cap w)$

$(P_e(h) + P_e(\neg h \cap w)) * P_e(i \cap (h \cup w))/P(h \cup w) = P_e(i \cap h) + P_e(i \cap \neg h \cap w)$

[rocky lotus]

! If in Oxford, see my about me

! If in Oxford, see my about me

[boreal solstice]

<@&286206848099549185> does anyone know how to prove this, using $h \cup w = h \sqcup (\neg h \cap w)$?

[rocky lotus]

! If in Oxford, see my about me

[boreal solstice]

$(P_e(h) + P_e(\neg h \cap w)) * P_e(i \mid h \cup w) = P_e(i \cap h) + P_e(i \cap \neg h \cap w)$

$(P_e(h) + P_e(\neg h \cap w)) * P_e(i \cap (h \cup w))/P(h \cup w) = P_e(i \cap h) + P_e(i \cap \neg h \cap w)$

$(P_e(h) + P_e(\neg h \cap w) - P(h \cap (\neg h \cap w)) * P_e(i \cap (h \cup w))/P(h \cup w) = P_e(i \cap h) + P_e(i \cap \neg h \cap w)$

$P_e(h \cup (\neg h \cap w)) * P_e(i \cap (h \cup w))/P(h \cup w) = P_e(i \cap h) + P_e(i \cap \neg h \cap w)$

$P_e(h \cup w) * P_e(i \cap (h \cup w))/P(h \cup w) = P_e(i \cap h) + P_e(i \cap \neg h \cap w)$

$P_e(i \cap (h \cup w)) = P_e(i \cap h) + P_e(i \cap \neg h \cap w)$

$P_e(i \cap (h \cup w)) = P_e(i \cap h) + P_e(i \cap \neg h \cap w) - P(i \cap h \cap i \cap \neg h \cap w)$

$P_e(i \cap (h \cup w)) = P_e((i \cap h) \cup (i \cap \neg h \cap w))$

$P(i \cap (h \cup w)) = P_e( (i \cup (i \cap \neg h \cap w)) \cap (h \cup (i \cap \neg h \cap w)) )$

$P(i \cap (h \cup w)) = P_e(i \cap (h \cup w)) )$

[rocky lotus]

! If in Oxford, see my about me

[boreal solstice]

fixed

[boreal solstice]

$P_e(i \mid h \cup w) = \
\frac{P_e(i \cap (h \cup w))}{P_e(h \cup w)} = \
\frac{P_e((i \cap h) \cup (i \cap \neg h \cap w))}{P_e(h \cup (\neg h \cap w))} = \
\frac{P_e(i \cap h) + P_e(i \cap \neg h \cap w) - P(i \cap h \cap i \cap \neg h \cap w)}{P_e(h) + P(\neg h \cap w) - P(h \cap (\neg h \cap w))} = \
\frac{P_e(i \mid h)P_e(h) + P_e(i \mid \neg h \cap w)P_e(\neg h \cap w)}{P_e(h) + P(\neg h \cap w)}$

[rocky lotus]

! If in Oxford, see my about me

[hexed sundial]

That's why I quit math, too much sweats

[random nebula]

yeah bro this is crazy wth and this person has undergrad role too

[desert knot]

Premise 2 could be (Premise 1) => (Conclusion)

Logically that's the weakest possible statement under which the argument would be valid

It could also be 1=0, which is the strongest possible statement under which the argument would be valid

are either of those answers satisfactory?

[waxen berry]

the statement i gave was such that
Premise 1 => (Premise 2 <=> Conclusion)

so slightly stronger

[desert knot]

premise 2 can't reference itself, otherwise it's not well defined

[desert knot]

or ok you mean this is your premise 2?

yeah that's probably the best there is, though it's still really bizarre

[waxen berry]

that is not premise 2, premise 2 is a statement such that the statement i gave is true

[wicked relic]

hi guys i will create telegram group for calc2, 3, 4, 5,
is there anyone who wants to join that group

[sleek nebula]

!occupied

[vale dock]

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).

[wicked relic]

Yes

[sleek nebula]

..

dont advertise here?

[wicked relic]

Im not advertise that group for fun

[sleek nebula]

sigh

Its someone else's help channel blud

[vale dock]

@boreal solstice Has your question been resolved?

[turbid moon]

in all fairness, a lot of what's been written so far is imo overcomplicated and ill-defined

so don't get discouraged lol

[turbid moon]

for readability purposes, can you please stick to one notation (sets or logical operators) and not use * for multiplication in LaTeX? It makes what you're writing insanely hard to follow

[turbid moon]

also I have no idea what you're doing here; your equality comes almost immediately from the fact that probability is additive over disjoint sets

I'm not going to write with i, h, w because I honestly can't remember which is which, but you have:

$P(A | B \cup C) = \frac{P(A \cap (B \cup C)}{P(B \cup C)}$ by the definition of conditional probability, and then since

$A \cap (B \cup C) = (A \cap B) \uplus (A \cap B^C \cap C)$, you have by additivity your desired numerator and likewise for the denominator $B \cup C = B \uplus (B^C \cap C)$.

[rocky lotus]

Saccharine

[turbid moon]

I still completely fail to see what you're trying to accomplish here, and it smells iffy

[vale dock]

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Use .reopen if this was a mistake.

[boreal solstice]

apologies.

my question was confused

will make sure to do so next time

[vale dock]

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[modest kettle]

Evaluate: $\lim_{x \to 0} \frac{\sin(x^n)}{(\sin x)^m}$, when $n < m$

[modest kettle]

i got the answer for this a while ago but now I'm getting something with factorials

never mind I'm just dumb

.close

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[quasi valve]

.close

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[burnt forge]

im getting 72 and wanted to see if i was wrong

[burnt forge]

i made pairs of numbers which satisy the conditions that sum of first 2 = sum of last 2 while 7 being the greatest digit present so i got pairs like 74-65 74 being 1st 2 didgit 65 being last 2 you can arrange this is 8 ways and i found 9 such pairs

from there i got 72

[pine prairie]

What are the 9 such pairs you found

[burnt forge]

74-65 73-64 72-63 72-54 71-62 71-53 70-61 70-52 70-43

thats that

[burnt forge]

i dont think theres more

[pine prairie]

Looks good

Now how many ways are there to arrange 4 unique numbers

That’s not right

72 sounds right

[burnt forge]

i get what youre trying to say

its 8

[pine prairie]

2 ways to swap first 2 digits, same for last 2, and 2 ways to arrange the blocks of 2

[burnt forge]

well if only i had done this during my test

thanks

.close

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[lost marlin]

i want to confirm if my proof is correct as it seems a bit circular
prove that gcd(a,b) = gcd(a-kb, b)
let (a,b)=d and (a-kb,b)=e
d|a and d|b thus d|a-kb since (b,a-kb)=e d|e
e|a-kb e|b thus e|a and e|d
=> d=e

[lost marlin]

i am quite new to this

<@&286206848099549185>

[vestal gyro]

What is (a - kb'b)?

[vestal gyro]

!15min

[vale dock]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[shrewd hinge]

idk every prop of gcd but could you send us the proof in a notebook or latex? @lost marlin

[lost marlin]

it was supposed to be a ,

[vestal gyro]

How did you conclude this?
e|a-kb e|b thus e|a and e|d

[lost marlin]

since e|b and e|a-kb we have e|a

now since e|a and b

e|d

[vestal gyro]

Oh.

So e is a common factor of a and b.

And d = any common factor * k.
d = ek.
So e | d.

And d | e.

So e = d

[lost marlin]

yes

[vestal gyro]

Yes, it is correct.

[lost marlin]

thanks

also does this property hold true for lcms too

[vestal gyro]

No.

[5,6] = 30

But [5 + 6, 6] = 11 * 6 because 11 and 6 are coprime.

[lost marlin]

oh

thanks

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[dapper root]

why does it become 3(10.8t+3.6 on the bottom, i wouldve thought it wouldve just been 3?

[stark wedge]

it does not, this is completely wrong

[dapper root]

so would 3 on the bottom be correct?

[stark wedge]

no

this is a very severe kind of mistake

it is not really salvageable even in part

where did you get this?

in general $\int [f(x)]^n \dd{x}$ is very much \textbf{not} equal to $\frac{[f(x)]^{n+1}}{(n+1)f'(x)}$. at all.

[rocky lotus]

Ann

[dapper root]

question is to find volume of revolution from this previous question

[stark wedge]

ok and what

is this part of somebody else's worked solution?

[dapper root]

yeah my lecturers worked example

[stark wedge]

ok so your lecturer is feeding you utter lies then lmao

[stark wedge]

like for real this is pure nonsense.

[dapper root]

what part is wrong?

[stark wedge]

in general $\int [f(x)]^n \dd{x}$ is very much \textbf{not} equal to $\frac{[f(x)]^{n+1}}{(n+1)f'(x)}$. at all.

an exception can be made if $f(x)$ happens to be a linear function --- but in your particular case, it's quadratic, so no dice at all.

[rocky lotus]

Ann

[lone fiber]

is this not chain rule

[stark wedge]

no.

[lone fiber]

if you differentiate it i think it works out

[stark wedge]

there's no such thing as chain rule for integration.

no it doesn't.

[stark wedge]

there is u-substitution but in OP's example there isn't really any workable way to do it.

[dapper root]

i just found a note from chain rule integration - (ax+b)^n+1/a(n+1)

so that covers it

[lone fiber]

so its right?

[stark wedge]

that is the integral of (ax+b)^n.

[dapper root]

its right yeah

[stark wedge]

note that

the stuff inside the power must be a linear function for that to work.

[lone fiber]

no she might be onto something

cos u sub

[stark wedge]

the stuff inside the power must be a linear function for that rule to work.

[lone fiber]

right

the x just dissapears

[stark wedge]

but in your case the stuff inside the power is a quadratic function.

quadratic functions are not linear.

[dapper root]

the note i found answers my question ill leave this to you lot

[lone fiber]

but for a quadratic there remains a t

[stark wedge]

i will reiterate:

[lone fiber]

its to do with u sub right?

[stark wedge]

i guess somewhat.

[jade ivy]

I think the problem is going to be much clearer when the original post doesn't crop out the surrounding page. It is more likely that there is context missing, even though I can't imagine what that would be, than an instructor writing gibberish of that magnitude.

[stark wedge]

it's unimaginably gross incompetence imo.

[lone fiber]

😭

🙏

people make mistakes

[vocal tusk]

they should be fired immediately.

and sentenced to life in prison.

[stark wedge]

i would not go so extreme for just that @vocal tusk

[flat ice]

true

[vocal tusk]

would u agree with this

[stark wedge]

unless i somehow missed that this is one of those signature dead-pan-delivery jokes and/or japes.

[vocal tusk]

unc does not play around 🗿

[lone fiber]

op just send the video already

[stark wedge]

i think an error of this magnitude, if it is repeated and can't be attributed to a momentary brainfart (however large), should put the professional fitness of the teacher into great question.

[lone fiber]

wait wdym somewhat

[stark wedge]

i am not interested in elaborating on any relationship between this mistake, the bullshitness of the rule used, and u-substitution.

[lone fiber]

mb

[stark wedge]

ok now i don't understand you.

[lone fiber]

wait

even if you use a u sub it works?

[vocal tusk]

i was joking

[stark wedge]

u := what

[lone fiber]

well not a 3 at the bottom

[stark wedge]

what are you trying to do

[vocal tusk]

i think the best way is to just expand the square and integrate each term

[lone fiber]

inside the brackets

[stark wedge]

are you trying to justify why the wrong thing is right

you get like

$\int \frac{u^2}{10.8t+3.6} \dd{u}$ and then you cannot proceed

[rocky lotus]

Ann

[lone fiber]

oh yea my bad

i mean you can complete the square

[stark wedge]

easiest is to just expand the quartic and rawdog the decimals.

like ok

[lone fiber]

oh yeah ofc

[stark wedge]

since OP has departed,

if nobody else has any closing remarks, i'm .closing this now

[jade ivy]

I don't feel that strongly about this either way, but I don't think such off-the-rails discussion should be humored without creating another channel.

[stark wedge]

ok in that case

[lone fiber]

Uh yeah we should close this channel

[stark wedge]

.close

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[waxen bronze]

could somone explain completing the square better to me?

[wraith heart]

,tex .cts

[rocky lotus]

riemann

[waxen bronze]

.close

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[modest tundra]

I need help with a geometry question:
Given a circle (O;R) and a point A outside the circle. From A, draw two tangents AB and AC to circle (O) (with B and C being the points of contact). Draw diameter BD of circle (O). Let H be the intersection of two lines AO and BC. Draw line CK perpendicular to BD at K.

Let I be the intersection of AD and CK. Prove that HI is perpendicular to AB.

[stark wedge]

got a diagram?

[modest tundra]

i got one

but the picture is really blurry so you cant make out the points and stuff

[stark wedge]

you're right

[modest tundra]

can you draw it yourself? sorry for having such a blurry picture lol

<@&286206848099549185>

[vale dock]

@modest tundra Has your question been resolved?

[waxen berry]

Big hint: Try proving CKB similar to ABD

[modest tundra]

okay

[waxen berry]

One sec i think i made a mistake

Not ABD

[modest tundra]

the angle BCK is not equal to DAB

[waxen berry]

BCD

Yeh

I meant BCD not BAD

[modest tundra]

okay

i still dont see anything lol

[waxen berry]

Ok, ill tell you second last step
KI = IC

Try getting to there

[modest tundra]

thats what i figured out the first time i look at the question lol

i need something that leads to it

[waxen berry]

Use DO = OB

Prove that O corresponds to I in BCD similar to KCD

[modest tundra]

but KDC is not 90 degrees though?

[waxen berry]

DKC is 90

I havent ordered the letters in the triangle

[modest tundra]

im still confused sorry

[modest tundra]

how can you prove this statement

[waxen berry]

Show ODC sim KDI and IDC sim OCB

Then you can show KI to IC is BO to OC

[modest tundra]

IKD is 90, the triangle ODC does not have any 90s

so it cant be similar

[waxen berry]

Replace the O by H

Sry

[modest tundra]

HDC and KDI

i need another angle to complete it

similar to idc and hcb

HCB case... thats not a triangle

[modest tundra]

anyone can reopen the channel if interested... im leaving this problem for now

.close

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[simple pebble]

I have a logic question, I'm not sure if this Discord is the correct one for it, but I'm trying my luck (It's more of a Logic question, which I guess is related to math)

I'm trying to understand and be able to demonstrate the following questioning about simultaneous resolution.

Let’s say we have a common pool that currently holds no token.
Two actions happen simultaneously:
Player A attempts to take a token from the common pool.
Player B attempts to add a token to the common pool.
What would the end state be?

Thank you very much for your input! 🙂

[mystic harness]

Clearly B gets frustrated with A and this ends in a brawl

All joking aside, this feels like its missing some rigid rules

[glass kelp]

One token gets replaced? If player b is not taking the token from play A

[simple pebble]

I'm wondering because, if B can take the token from the pool, it would mean that A has resolved its action already, which would mean it wasn't resolved simultaneously.

I'm afraid I don't have much more rigid rules to add to this. Just wondering if there is a field about handling logic events like this and a way to explain them.

[vale dock]

@simple pebble Has your question been resolved?

[vale dock]

@simple pebble Has your question been resolved?

[vale dock]

@simple pebble Has your question been resolved?

[ebon glade]

https://en.wikipedia.org/wiki/Dining_philosophers_problem

there is this problem which is somewhat related

but its not really a logic problem from the start

its just an ill-posed problem

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[wanton sinew]

is it possible to use synthetic division with something like this (x^3 +2x -5)/(x^2 +1) or does the numerator x have to be to a degree of 1?

[tidal terrace]

1.yes
2. you will do -2 to the powers in the result instead of 1

[wanton sinew]

ok that's what I thought but chatgpt seemed to think that wasnt allowed. Thank you!

.close

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[tidal terrace]

but for non linear divisors i prefer long division

[storm perch]

Can you DM me an example?

I'm curious, I was ever only taught that it works on linear factors

[wanton sinew]

i don't think it would work with my example if ur confused because I did x^2 +1 so it's roots aren't real

[storm perch]

Yeah

I mean, if you wanted to I suppose you could do it on an expression that has integer roots

[wanton sinew]

yea

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[small crystal]

why is it C? I don't get it

[storm perch]

Did you do the sum or nah

[rocky lotus]

The following error occured while calculating:
Error: Undefined function ln

[storm perch]

,w 80/ln(9)

[rocky lotus]

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Requested by exiled_hype

[small crystal]

nah i did integral and was trying do estimate

[storm perch]

Yeah so uh

💀

[small crystal]

40 looks about right for overestimate but 60 is lil too much

[storm perch]

Try it out then come back

[small crystal]

😭

yo but how? im not given height should i plug in like everything in f(x) to do it like 0.5 * f(0.5)

[storm perch]

You are given the height

Remember how riemann sums work?

[small crystal]

yea i js plug into f(x)

lemme try

[storm perch]

Ok

[small crystal]

$$0.5\left(f\left(0.5\right)\ +\ f\left(1\right)\ +\ f\left(1.5\right)\ +\ f\left(2\right)\right)\ =\ 60$$

[rocky lotus]

ruby

[small crystal]

i did this since its equal interval

i got 60

[storm perch]

yep

[small crystal]

tried to use dumb trick and failed 😭

thanks for helping 🙂

.close

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[gritty carbon]

I have tried to demonstrate that given two integers a,b with a not equal to zero, then for (ax + b) mod n doesn't exist two equal results for all integers x in interval [0, n).

[gritty carbon]

To demonstrate that i tried this:

But if x1 and x2 are different, in the interval [0, n) there aren't value that are equal for the module of n

But there is something wrong in the prove, in fact if we take for example

I can't understand what is wrong in the prove, because for me i only applied some properties of modular arithmetic
Sorry if the question is stupid

[cloud coral]

wait a minute, this doesn't even seem true