help-39

[ivory basin]

I mean if you're picking uniformly from all elements in A*, then yes

[tardy reef]

P(A)=1/n
what does this mean tho?

[crystal dew]

that was what I'm confused about. A* doesn't look like the power set and we're talking probability, so I thought P(A) is some probability, assuming A here somehow means some element in A

[tardy reef]

yea, its a nightmare for the notation

A* is sometimes a notation for set complement, P(A) is most commonly used as powerset and if you want to say probability, it would be much preferred to explicitly state what you want

[wild fable]

baktep

[tardy reef]

still not watep 🥀

[green aurora]

I'm doing the probabilities of the set of parts

I don't think it can be done.

[tardy reef]

[ivory basin]

I'm just going to do this

!original please

[pearl pond]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[green aurora]

I solved it

A number greater than 0 cannot be rolled.

Because if for example I toss a coin it is impossible that nothing comes out

[crystal dew]

so that is the context

well, if you're done, then you may close the channel

[green aurora]

Thanks anyway

[crystal dew]

!done

[pearl pond]

If you are done with this channel, please mark your problem as solved by typing .close

[green aurora]

.close

[pearl pond]

Channel closed

Closed by @green aurora

Use .reopen if this was a mistake.

[wild fable]

math god saves the day

[pearl pond]

Available help channel!

Send your question here to claim the channel.

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[green aurora]

$P(A|B)=\frac{n_{A\cap B}}{n_B}$

[jolly parrot]

Goofy Joe

[green aurora]

Why is it done this way?

[proper nova]

what's $A \mid B$ here?

[jolly parrot]

1 divided by 0 equals Infinity

[proper nova]

describe that in words for me

[green aurora]

A knowing that it happened the event B

So the probability that A happens knowing that event B has happened

[unborn abyss]

yeah so if we restrict ourselves to the events where B happened, it's the probability that A also happened, right?

[green aurora]

Yes

[unborn abyss]

remembering that the probability of something is (number of successes) / (total events under consideration)

[green aurora]

Yes

[unborn abyss]

number of successes would be times when A happened and B happened

total events under consideration would be number of times when B happened

[green aurora]

Yes

[unborn abyss]

so.... n(A & B) and n(B) as in the equation

[green aurora]

$P(A|B)=\frac{|A|B|}{|B|}$

[jolly parrot]

Goofy Joe

[thorn agate]

its $P(A|B)=\frac{P(A\cap B)}{P(B)}$

[jolly parrot]

scoob

[green aurora]

Yes

But is the 2nd formula that is written correct?

[unborn abyss]

the one that you wrote? |A|B| isn't a symbol that makes sense as far as I'm aware

[green aurora]

I did the sub

$P(A|B)=\frac{P(A\cap B)}{P(B)} with A|B=C$

[jolly parrot]

Goofy Joe

[green aurora]

And so

$P(C)=\frac{|A|B|}{|B|}$

[jolly parrot]

Goofy Joe

[green aurora]

Because P(A) would be favorable cases over possible cases

[unborn abyss]

what does "A|B = C" even mean

[green aurora]

Substitution

[unborn abyss]

A|B isn't a thing on its own

it only makes sense in P( A | B ) "the probability of A given B"

[thorn agate]

to support these are venn :
this the chance B occurs given that A has occured, reducing your sample space to A

[green aurora]

P(A) = |A|/|OMEGA| , S = A|B => P(S) = |S|/|OMEGA| ?

[thorn agate]

i just flipped A and B

[unborn abyss]

"S = A|B" makes no sense, i do not know what you are trying to say

[green aurora]

I call S the set that has the elements A|B

[unborn abyss]

do you mean A intersect B??

[green aurora]

No

[thorn agate]

you mean event C such that P(C) = P(A|B)

[unborn abyss]

ok....
if A is "i wear a raincoat"
and B is "it's raining outside"

then what is A | B?

[green aurora]

It is the set that has the elements that respect it

[unborn abyss]

okay better idea - can you colour in A | B on this please

[green aurora]

Where are the elements?

[unborn abyss]

each pixel is an element

[green aurora]

???

The intersection

[unborn abyss]

like.... the circle that's labeled A? everything in that circle is an element of A

so this is what you're calling "A | B"?

[green aurora]

I'm wrong

This is the intersection

[unborn abyss]

that is indeed the intersection, written A n B

[green aurora]

Why isn't it A|B?

[unborn abyss]

-# well for one thing that's not a set like i've been telling you

[green aurora]

What's wrong?

[unborn abyss]

A|B is not a set

A|B is not a set of symbols that make sense on their own

A|B as a phrase only makes sense within P(A | B)

what i have highlighted is A ∩ B, the set of events where A is true and B is true

[green aurora]

So the formula with the substitution doesn't make sense?

[thorn agate]

-# its sometimes divisibility

[unborn abyss]

this one does not make sense, as it uses |A|B| which is not a symbol that makes sense

[green aurora]

So C cannot be equal to A|B, since C is a set and A|B is not?

[unborn abyss]

however, it is true that $P(A | B) = \frac{ |A \cap B| }{ |B| }$

[jolly parrot]

hayliänus austrǎlis

[unborn abyss]

that's correct

[green aurora]

So how do I know if a replacement is possible or not?

[unborn abyss]

where did C even come from? did you make it up?

[green aurora]

Yes

[unborn abyss]

you can relabel things, but $P(\rsq | \bsq)$ should really be thought of as something that takes two sets and computes a number

[jolly parrot]

hayliänus austrǎlis

[unborn abyss]

rather than soemthing that takes an object that looks like A|B

[green aurora]

👍

But $P(A\cap B)$ Is the chain rule?

[jolly parrot]

Goofy Joe

[green aurora]

Like $f(g(x))=f'(g(x))g'(x)$

[jolly parrot]

Goofy Joe

[unborn abyss]

no, those have nothing at all to do with each other

[green aurora]

$P(A\cap B)=P'(A\cap B)(A\cap B)'$

[jolly parrot]

Goofy Joe

[thorn agate]

where is bro cookin this from

[crystal dew]

...what?

[green aurora]

I don't understand what it means that it's the chain rule

[crystal dew]

it's... not though

[unborn abyss]

did someone tell you to use the chain rule in the context of probability? or did you make that up?

[crystal dew]

why is there a chain rule in probability anyway

[thorn agate]

is it 2 am thought

[green aurora]

On Google it comes up

[crystal dew]

what did you search?

[green aurora]

Chain rule probability

[crystal dew]

why did you search for that anyway?

I don't see anywhere in this convo where a chain rule is mentioned

[green aurora]

It connects to the formula above

[crystal dew]

that formula is called conditional probability

not the chain rule

[green berry]

its this

[green aurora]

Exactly!

[crystal dew]

huh, ok.

but this is just a rearranging of the conditional probability formula!

[unborn abyss]

.... and what you wrote down was not even close to that???

[crystal dew]

yea this is the calculus chain rule and is definitely not applicable here

[green aurora]

I thought it could be done

[crystal dew]

let's not mix things up

[green aurora]

But why is it called the chain rule?

[crystal dew]

I have no idea honestly. this formula, to me, has always been the conditional probability formula, rearranged or not

[tropic saddle]

you are "chaining" the conditions

[unborn abyss]

it makes more sense with multiple events

[crystal dew]

I stand very much corrected. TIL and thanks

[drifting vale]

Consider this: 50% of people you know go to your school. 60% of people you know also like burritos. What is the probability that a given person you know likes burritos if they already go to your school?

"Going to your school" becomes the new universe

We only consider these cases

And the intersection (both things happening) is some percentage of that universe

You want to find which percentage of B the intersection is

[green aurora]

But

[drifting vale]

So you divide the intersection by B

[green aurora]

$P(A\cap B)=P(A)P(B)$

[jolly parrot]

Goofy Joe

[cursive wraith]

Only when A and B are independent

meaning no matter whether A is realized or not, it doesn't change how likely B is to be realized

$P(B|A) = P(B)$ in that case where $A$ and $B$ are independent

[jolly parrot]

Rafilouyear2026

[cursive wraith]

so, you can hopefully see the link between this formula

[cursive wraith]

and the formula for the specific case when A and B are independent

[green aurora]

But if A | B, isn't A automatically dependent on B?

[cursive wraith]

what does "A | B" mean

[green aurora]

A knowing B therefore A depends on B

Or not?

[cursive wraith]

I don't really know how to interpret your question
I'm saying that, when A and B are independent, asking whether "A is realized knowing B is realized" is kinda the same as asking when A is realized, probability-wise

The most basic example is 2 coin flips

A = "tails on first flip"
B = "tails on second flip"

Does knowing whether you flipped tails or not on the first flip tell you anything about the second flip's outcome?

[green aurora]

No

[cursive wraith]

exactly, that's what independence means

So the probability of B being realized doesn't matter on knowing A is being realized

P(B|A) = P(B)

In the opposite case where P(B|A) and P(B) are different, they are dependent on each other

Still on 2 coin flips
A = "tails on first flip"
B = "heads on one of the two flips"

Can A being realized or not affect B's outcome?

yes

how so?

[green aurora]

Yes

[cursive wraith]

ok

If A is realized, then B only has 1/2 chance of happening

[green aurora]

I don't know if heads came out on the first roll that's why it's possible

[cursive wraith]

if A isn't realized, then B is also happening

[cursive wraith]

yes but knowing whether A or not(A) happens influences how likely B is

[green aurora]

Yes

[cursive wraith]

so B does depend on A somehow

P(B) = 3/4, whereas P(B|A) = 1/2 and P(B|not(A)) = 1

[cursive wraith]

So this is not true

[cursive wraith]

And this is the only true formula concerning P(A and B)

[green aurora]

Thanks

.close

[pearl pond]

Channel closed

Closed by @green aurora

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

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• 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.

[still nymph]

yo can i get help

[drifting vale]

Probability is the ratio of the number of favorable outcomes to the total number of possible outcomes

We have 4 favorable outcomes and 14 outcomes overall

[still nymph]

yea

[drifting vale]

Just find the ratio

[still nymph]

1/4?

[drifting vale]

4/14

Simplify and get the answer

[north coral]

Teach don’t tell answer

[still nymph]

so u divide right

how did u get 4/14 sir

[drifting vale]

Again, probability is favorable outcomes divided by the number of all outcomes

[north coral]

Imagine you have 3 chocolates out of which 2 are white chocolate what’s the chance of getting a white chocolate

[still nymph]

1%

.

[drifting vale]

Okay. This has to be rage bait

[north coral]

Yeah

Ur profile says undergrad

[thorn agate]

Nd bro is playing arena breakout

[still nymph]

yes yes but im doing my assignments

[bronze ore]

what

[thorn agate]

Uh

[robust lynx]

How did you acheive this answer?

!show

[pearl pond]

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

[thorn agate]

No help in live test

[robust lynx]

live test!?

[drifting vale]

Did you just doxx yourself?

[thorn agate]

He posted a ss of live assignment

Ig

[robust lynx]

<@&268886789983436800> racism/troll whatever

[still nymph]

so in total

no

if u mix

red paint

with white paint

what happens?

the red takes over MORE

[bronze ore]

?

[thorn agate]

:facepalm

[drifting vale]

Bro is a modern day philosopher

Let him cook

[robust lynx]

why are we talking about paint now?

[still nymph]

so please support me further through my question please (:

[still nymph]

so as why are we talking abt choccolate?

[thorn agate]

What is the probability trump invades greenland

[robust lynx]

because it’s an example that will help you solve your question.

[still nymph]

ok

i used paint

as an example of u saying im racist

but im not.

[radiant terrace]

Explain this more clearly to me?

What does "the black take over the white" mean to you?

[drifting vale]

If you mix 2 liters of white paint with 1 liter of black paint, the mixture will be 1/3 black and 2/3 white. It'll be closer to white than to black

[still nymph]

what does it mean'

it means

[robust lynx]

this was the example though

[still nymph]

the black is a darker shade and it takes over the lighter shade white

[radiant terrace]

Why?

[still nymph]

so there for the dark is taking 1%

[drifting vale]

But it doesn't mean it's more black than white

[still nymph]

bc what if the choccolate melted?

or if its being melted?

or

what if we chop it up and melt it..

[bronze ore]

Are we using laundry color theory in a probability problem about beads

[north coral]

The ADHD goes crazy

[thorn agate]

What if one is colorblind

[still nymph]

so there for no racism is involved.

[radiant terrace]

So your point here is just that these two colors visually mix a certain way?

[north coral]

R you doing math or philosophy

[still nymph]

si

[drifting vale]

Got you there! It's black and white anyway

[still nymph]

which helps answer 1%

[thorn agate]

Oh 👍

[north coral]

Still no

[still nymph]

oh

[robust lynx]

I’m still very confused on how you got the answer 1%. Please do show your steps.

[still nymph]

ok

so 3 white

2 black

[thorn agate]

What if they become grey

[north coral]

@still nymph Good luck with life if this is seriously a doubt 🫡 🫡 🫡

[still nymph]

they are greay

grey

[robust lynx]

not relevant. Please stay on track.

[still nymph]

they arent grey

[thorn agate]

What is left relevant here

[still nymph]

he clerified its black and white.

this is my question

[robust lynx]

The black and white chocolates.

[still nymph]

what does this mean.

[north coral]

Chance of getting 4

[radiant terrace]

If I have 4 cats and one is a tabby cat you'd say the odds of me picking up a tabby cat if I pick up one of my cats randomly is 1 in 4 right?

[drifting vale]

It means the probability of pulling 1 specific card out of 4 cards, in this case a card with a number 4

[still nymph]

so

does this mean i choose 4 1/3-1/4 times?

[drifting vale]

Do you know fractions?

[radiant terrace]

I'm just talking about the bead problem here for what it is worth

Ah I see the card one too.

[still nymph]

im not on beads now

im on this

[radiant terrace]

The same style of thinking I mentioned earlier applies to this too though

You have 4 cards, you pick one card randomly once here

You wanna know what the probability is that the card you drew is 4.

[still nymph]

1/4

[radiant terrace]

Yep

If you have 27 friends at a party

And 4 of them like pizza

What is the probability that if you walk up to one randomly and chat with them that you will be talking to a pizza enjoyer?

[still nymph]

wait

what kind of party is it?

[radiant terrace]

Doesn't matter

[still nymph]

yes it does

[radiant terrace]

Nah

[still nymph]

yes

[radiant terrace]

Pretend it's your birthday party then

[still nymph]

nope

i have no friends

[drifting vale]

Maybe the real probability is the 27 friends we made along the way

[still nymph]

so it cant be

[radiant terrace]

Pretend they are my friends and it's my birthday then

[still nymph]

well i dont know u so ur not invited.

but you helped me

so you are invited

but i didint invite your friends

so there for its only me

and you

and my mom and dad

[radiant terrace]

Nah

[still nymph]

why?

explain

[radiant terrace]

The question is just 27 people at a party, 4 like pizza, and you talk to one randomly.

[still nymph]

1/27 odds

[proper nova]

try to make $\Omega$

[jolly parrot]

1 divided by 0 equals Infinity

[radiant terrace]

Close

Why 1 on top?

[still nymph]

bc

the 1 kid likes pizza

so do i

[radiant terrace]

But there are 4 pizza enjoyers at the party

[still nymph]

oh so 4/27?

[proper nova]

so like

[radiant terrace]

Yep

[proper nova]

try to imagine what $\Omega$ is here

[jolly parrot]

1 divided by 0 equals Infinity

[proper nova]

$\Omega$ is just a set of all possible outcomes

[jolly parrot]

1 divided by 0 equals Infinity

[still nymph]

i dont need your eqation bro

leave

[proper nova]

and see how many elements/outcomes is in $\Omega$

[jolly parrot]

1 divided by 0 equals Infinity

[still nymph]

is this racist?

[proper nova]

-# rude ahhhh

[radiant terrace]

Yeah don't be rude

[still nymph]

im being honest

i dont need his help good sir

[radiant terrace]

They may not know this terminology/notation though tbf

[still nymph]

im only in middle school

[radiant terrace]

If you're not going to interact here in good faith what is the point?

[still nymph]

u guys are not good support\

[proper nova]

then leave

[thorn agate]

Are you over 13

[still nymph]

scoob

[drifting vale]

Well the question then is, is he black or white paint?

[still nymph]

we went over this before

im in 9th grade

middle

im 15

remember

[proper nova]

okay

[thorn agate]

I don't rlly but ok

[still nymph]

bro

[radiant terrace]

But you are kinda just being rude and wasting peoples time when they are trying to help you.

[still nymph]

not true

[radiant terrace]

I'm not asking you.

[proper nova]

didn't you just been rude with me?

[still nymph]

somone came in here saying some $omega/ solve

or smth

and im respectfuly saying i dont need what hes tryna do

so please leave.

[radiant terrace]

There's a more formal way to describe these kinds of problems that they think might help you to use

[still nymph]

are you calling me autistic?

[proper nova]

-# respectfully?

[radiant terrace]

Nope

[proper nova]

-# since when lmao

[still nymph]

bc i am

[radiant terrace]

Getting fairly tired of the trolling though.

[thorn agate]

Respectfully

[still nymph]

i am not trolling sire

[proper nova]

if you think this is not a good source of help

[radiant terrace]

I disagree

[proper nova]

then you can leave

[proper nova]

we don't need rude people like you

[radiant terrace]

It's the same trick as the last couple questions.

[still nymph]

si but

how.

do i do it again

i forgot bc we got off topic

[radiant terrace]

What would you get if you tried the same trick?

[still nymph]

what trick?

huh im confused

[proper nova]

$\text{probability} = \frac{\text{correct outcomes}}{\text{total outcomes}}$

[jolly parrot]

1 divided by 0 equals Infinity

$\text{probability} = \frac{\text{correct outcomes}}{\text{total outcomes}}$

[radiant terrace]

Yep

[proper nova]

-# that is only when all outcomes are equally likely to happen btw

[drifting vale]

That's what I told him in the very beginning

[still nymph]

@carmine schooner

can you help me

[thorn agate]

18 horse , 5 bay
Chance bay = favourable/total outcomes =
No. Of bay/total horses

[pearl pond]

Please do not ping individual helpers unprompted.

[still nymph]

bru

she is the most help

):

[proper nova]

-# unless you have permissions from her

[still nymph]

shes my besti

ofc i do?

[carmine schooner]

this. ingrain the reasoning into your head

[still nymph]

HAIIII

[carmine schooner]

hello !

[still nymph]

can you help me

[radiant terrace]

Do they actually have permission to ping you?

[still nymph]

they are not making sense ):

[thorn agate]

-# Bsharp enters the battlefield

[carmine schooner]

not technically, but i won't hold it against them

[radiant terrace]

I will

[still nymph]

atleast in my head

[carmine schooner]

but yeah everyone else has said what i would have anyway

[thorn agate]

-# Chill

[proper nova]

-# ||onto someone who wants me out because there is an omega set which represents the set of all outcomes possible||

[radiant terrace]

Great, so this time I did mute you rather than somebody else. Moving forward don't blatantly lie, troll, waste peoples time and etc when those people are giving you free help.

[drifting vale]

Yes, trying to teach you some sort of tricks? When you're not a magician or anything
Crazy

[radiant terrace]

.close

[pearl pond]

Channel closed

Closed by @radiant terrace

Use .reopen if this was a mistake.

[proper nova]

great job mods

-# yk after he kicked me out i didn't want to help him anymore

-# im generous enough to still help in this channel

[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.

[ember ledge]

Could anyone explain 2 (b) for me? Struggling with Depth first search

[ember ledge]

My idea was a-b-c-d-i-j-k-e-f idk?

[bitter herald]

do you know about like how preorder/inorder/postorder traversal work

[ember ledge]

i thought i did

we have this from class slides

i understnad this

but im kinda still struggling with the task above

[bitter herald]

u are almost right

u r just like

missing the last two

[ember ledge]

a-b-c-d-i-j-k-e-f-g-h ?

[bitter herald]

yeah

good job

[ember ledge]

welp that was easz

[bitter herald]

lmao

this is just preorder traversal

[ember ledge]

so basically follow the path in alphabetic order and backtrack if not possible

chatgpt confused me lol it said smth different but i guess i was right

[bitter herald]

yay

[ember ledge]

thank you for the confirmation

[bitter herald]

anytime

good luck

[ember ledge]

thanks!

[pearl pond]

@ember ledge Has your question been resolved?

Channel closed

Closed by @ember ledge

Use .reopen if this was a mistake.

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.

[verbal tree]

How can we show that if f:R-> R smooth is such that f^(n) (0)=0 for n odd, then there exist g smooth with f(x)= g(x^2) ?

[toxic lichen]

p sure this may not hold without extra assumptions

because of some kind of nasty but smooth bump function that turns out not to be even

[verbal tree]

Hmmm

Some book kinda says it does tho

Im gonna say the précise result it says

[autumn fossil]

If we defined f(x) = e^(-1/x^2) for x > 0 and f(x) = 0 for x <= 0, then this would be smooth, all derivatives at 0 would be 0 and it's clearly not even

[toxic lichen]

yeah that's the exact kind of bump function i was talking about

[toxic lichen]

yeah do that

[verbal tree]

Oh

[toxic lichen]

can you translate the handwritten text

[verbal tree]

I changed 3 to 2 to make the problem simpler

[toxic lichen]

mmmmmm sus

[verbal tree]

Rtilde has the structure given by ...

[verbal tree]

Ill rewrite it

[toxic lichen]

what writing system is this

[autumn fossil]

is it structure?

[verbal tree]

How can we show that if f:R-> R smooth is such that f^(n) (0)=0 for n not divisible by 3, then there exist g smooth with f(x)= g(x^3) ?

[toxic lichen]

how tf is it structure

[verbal tree]

Yes

[autumn fossil]

looks roughly czech

[toxic lichen]

ʌtɯclūu

[autumn fossil]

Anyway, same counterexample works but with f(x) = e^(-1/x^2) (the og one works as well actually)

where did u get the g(x^3) from? There seems to be no g in the question

[verbal tree]

g is supposed to be f(x^(1/3))

It is the composite of f by the inverse local chart

[toxic lichen]

x^2 isn't one to one btw while x^3 is

wouldn't that fuck things up

[verbal tree]

Ye i figured

[autumn fossil]

<@&268886789983436800> scam

[toxic lichen]

<@&268886789983436800> scam

[pearl pond]

@verbal tree Has your question been resolved?

[pearl pond]

@verbal tree Has your question been resolved?

[pearl pond]

@verbal tree Has your question been resolved?

[verbal tree]

.close

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[silver jackal]

ii)a)

[dense axle]

PT:TR=4:3

if the triangle is similar, what would it be

[silver jackal]

?

didn't understand

[dense axle]

the tringle is similar so the ratio between each side should be same

[silver jackal]

so 4:3?

I'm dumb excuse me

idk tbh the answer

ohhhh

I understand ty @dense axle

I didnt the question well mb

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[vestal thistle]

Hello, could someone check if this proof looks good please?

\begin{Theorem}
Suppose $A$ and $B$ are sets. Then $\powerset{A \cap B} = \powerset{A} \cap \powerset{B}$.
\end{Theorem}

\begin{proof}
First, we prove that $\powerset{A \cap B} \subseteq \powerset{A} \cap \powerset{B}$.
Let $X \in \powerset{A \cap B}$.
Then, by the definition of the powerset, $X \subseteq A \cap B$.
Let $x \in X$.
Because $X \subseteq A \cap B$, we have that $x \in A \cap B$.
Then, by the definition of the intersection, $x \in A$ and $x \in B$.
Since $x \in X$ implies $x \in A$ and $x \in B$, we have that $X \subseteq A$ and $X \subseteq B$.
Then, $X \in \powerset{A}$ and $X \in \powerset{B}$.
Thus, $X \in \powerset{A} \cap \powerset{B}$.
We have that $X \in \powerset{A \cap B}$ implies $X \in \powerset{A} \cap \powerset{B}$,
hence $\powerset{A \cap B} \subseteq \powerset{A} \cap \powerset{B}$.\\

Next, we prove that $\powerset{A} \cap \powerset{B} \subseteq \powerset{A \cap B}$.
Let $X \in \powerset{A} \cap \powerset{B}$.
So $X \in \powerset{A}$ and $X \in \powerset{B}$.
Then, by the definition of the powerset, $X \subseteq A$ and $X \subseteq B$.
Let $x \in X$. 
Since $X \subseteq A$ and $X \subseteq B$, we have that $x \in A$ and $x \in B$.
So $x \in A \cap B$.
Since $x \in X$ implies $x \in A \cap B$, this means that $X \subseteq A \cap B$.
Thus, by the definition of the powerset, $X \in \powerset{A \cap B}$.
We have that $X \in \powerset{A} \cap \powerset{B}$ implies $X \in \powerset{A \cap B}$,
hence $\powerset{A} \cap \powerset{B} \subseteq \powerset{A \cap B}$.\\

We showed that $\powerset{A \cap B} \subseteq \powerset{A} \cap \powerset{B}$ 
and $\powerset{A} \cap \powerset{B} \subseteq \powerset{A \cap B}$.
Therefore, $\powerset{A \cap B} = \powerset{A} \cap \powerset{B}$.
\end{proof}

[jolly parrot]

Mor Bras

[rancid depot]

looks good to me

maybe one critique is to say this is true for all x in X

[vestal thistle]

Where?

[rancid depot]

"Since x in X implies x in A and x in B [for all x in X]..." 3 lines up from the bottom of the first proof paragraph

i dont really think its all that necessary

[pearl pond]

@vestal thistle Has your question been resolved?

[rancid depot]

<@&268886789983436800> scam

holy

<@&268886789983436800> some kind of raid it seems

[pearl pond]

@vestal thistle Has your question been resolved?

[shut elm]

Looks good to me

[pearl pond]

@vestal thistle Has your question been resolved?

[worldly jewel]

You can get a much more concise proof simply by expanding and equating the definitions

X ∈ ℘(A ∩ B) ⟺ X ⊂ A ∩ B ⟺ X ⊂ A and X ⊂ B ⟺ X ∈ ℘A and X ∈ ℘B ⟺ X ∈ ℘A ∩ ℘B.

[vestal thistle]

Thank you for your comments!

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[amber pagoda]

i was trying to solve it like this

[rigid mist]

two angles subtended by the same chord at the circumference of a circle on the same side are equal

consider joining the two bottom points with a chord to see it more clearly

[amber pagoda]

like this

[rigid mist]

yes

use the theorem

[amber pagoda]

btw what does subtended mean i lowk forgot

[rigid mist]

"GEOMETRY
(of a line, arc, or figure) form (an angle) at a particular point when straight lines from its extremities are joined at that point."

in simpler terms, when an arc or a chord makes an angle by its endpoints connected to that point with lines

for example, here, chord BC subtends x angle at A

and chord AD subtends 60 deg at C

[amber pagoda]

ah

[amber pagoda]

so there is nothing to do with triangles

[rigid mist]

there is

hmm actually not

why not use the theorem in triangle ABC and BCD

[amber pagoda]

what if we do x+80+60=180

interior angles

x+140=180

x=180-140

x=40

idk how i did that

[rigid mist]

correct

[rigid mist]

what do you mean

[amber pagoda]

i dont understand it like

last time when i did smthing similar lik this i just made a isocles traingle and tries to solve like that

[rigid mist]

what could this blud be possibly talking about

[amber pagoda]

lol

https://cdn.discordapp.com/attachments/908078029778083872/1462591441107816489/image.png?ex=697162ed&is=6970116d&hm=57c5126e3d5f5faca498d4e7439a5b06f3ad90632fe0e9398ffe8e9550b1ad30&

[rigid mist]

the conditions must have been different in that question

[amber pagoda]

#help-39 message

[rigid mist]

you could do that because they're all equal (radius)

which is not the case here

[amber pagoda]

like the vertices

[rigid mist]

no the lines

[amber pagoda]

ah

[amber pagoda]

so what am i basically doing here

[rigid mist]

since chord AD subtends two angles on the circumference of the circle, they'll be equal

[pearl pond]

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[amber pagoda]

.reopen

[amber pagoda]

alright thanks mate

.close

[pearl pond]

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[subtle crystal]

I am dumb, why is distance of b from line ap less than d

[tropic saddle]

its less than the height of afp which is less than one of its sides which is d

or apply a bunch of pythagoras I guess

[pearl pond]

@subtle crystal Has your question been resolved?

[subtle crystal]

Yea that's just restating the problem
How is it less than height of afp aka d

[tropic saddle]

distance of a to b/distance of b to ap=distance of f to a/height from f to ap

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[quartz folio]

If I have y= sin of the picture below, how would I interpret it and graph it? Does it mean that the period is 2x faster (meaning the x values are multiplied by half), and then I shift everything to the right by pi/3?

[inland pecan]

SIn(A+B) = sinAcosB + cosAsinB, you could start with that as well... plotting the addition is quite easier

[jolly parrot]

scoob

[still hamlet]

in general
f(ax+b)=f(a(x+b/a))
so you translate -b/a units, and stretch by a factor of 1/a

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[fading ledge]

How do I solve it?

Absolutely convergence no because nth term tends to 1

Not 0

[cinder flower]

what is the problem?

[toxic lichen]

correct conclusion but wrong reasoning

this series fails to converge absolutely but its general term DOES tend to 0

although...

now that i am looking at the first few terms written out here

the very picture is wrong

[tropic saddle]

something is certainly wrong here

[toxic lichen]

very

so much inconsistency between op and his image and in the image itself

[toxic lichen]

and also no question actually asked

[fading ledge]

What is wrong?

[fading ledge]

What?

[fading ledge]

general term tends to 0? How?

yes it tends to 0

[autumn fossil]

it depends on whether u look at the left side or the right side of the equality

interestingly, they are different

[tropic saddle]

well on the right it would tend to 1/2

so still not 1

[toxic lichen]

and yet we are supposed to brain wash ourselves into thinking they are the same

[toxic lichen]

uncrop and send full page please

[fading ledge]

Ok

[autumn fossil]

just out of interest, what's the hint?

the question is wrong though, the LHS and RHS arent the same, so I'd just skip it or solve it for the LHS and RHS individually

[fading ledge]

I guess LHS should be (-1)^n-1 into n

@autumn fossil @toxic lichen

What should we do now?

For absolutely

n/2n+1 and limit will be 1

As I stated at the start of the question and ann denied

[autumn fossil]

it'd be n * (-1)^(n-1)

[fading ledge]

Or maybe another helper saying 1/2

[autumn fossil]

limit wont be 1

[fading ledge]

Ohh true

1/2

My bad

Denominator

So it is divergent

[autumn fossil]

yeah

[fading ledge]

For checking conditional convergent

I will use limit comparioson test?

1/n

[autumn fossil]

try it

at worst it wont work

there is a much easier way than all those tests though

Terms of all convergent series (that is both absolutely and conditionally) tend to 0

[autumn fossil]

in this case, the terms will oscillate roughly between 1/2 and -1/2, they certainly wont tend to anything, so not even 0

and so it cannot be convergent (not even conditionally)

this is either known as divergence test, or in this specific case, it's part of alternating series test

[pearl pond]

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[crimson nebula]

Me proof for "if (zn) is a divergent sequence and yn is a convergent sequence, such that yn≠0, then a sequence define by xn:=zn/yn is divergent ".

Kindle check

[pearl pond]

@crimson nebula Has your question been resolved?

[crimson nebula]

Please helpppp

[inner granite]

The proof is alright.

[crimson nebula]

Thank u

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[vestal pelican]

I'm so lost at this limit

[sudden arch]

Are you allowed to use L'hopital

[vestal pelican]

Yes

[sudden arch]

Write this as e^{x ln (tan 2x)} and try to use L'hopital

(Not sure if will work)

[vestal pelican]

Ok may i please know where did u get that

[sudden arch]

It's a commom trick

to turn powers into something you can apply L'hopital

[vestal pelican]

Ah nvm i got it now thanks

I was so lost

[sudden arch]

How did you do it

[vestal pelican]

I did it without l hopital

[sudden arch]

Nice

[vestal pelican]

With ln

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[digital summit]

.

Can someone explain converses, inverses, and contrapositives better to me? My problem is that I cannot determine if the statements are true or false. For example, one problem I did that asks "Rewrite the postulate in if-then form then write the converse, inverse, and contrapositive and state which ones are true," I got the statements right:
Of there is a plane, then theres atleast 3 noncollinear points.
Converse: If theres atleast 3 noncollinear points, there there is one plane.
Inverse: If there is not one plane, then there is not atleast 3 noncollinear points.
Contrapositive: If there is not atleast 3 noncollinear points, then there is not one plane.

I said that all the statements are true.
But, the converse and inverse are false? Could someone explain.
My thinking for the converse is that if there are atleast 3 noncollinear points, then one plane exists.. and for the inverse if there is not a plane, then therefore 3 noncollinear points dont exist because then there would be a plane if they did.
(i'm new to geometry) :)

[digital summit]

Also, the question Do two lines that intersect have to be in the same plane? also confuses me. I said that they did not have to be in the same plane, but every other source says they do. What if two planes intersect, with lines on each of them, that intersect at the line created by both planes?

[glass meadow]

Who said the converse and inverse are false here?

[glass meadow]

Two intersecting lines do indeed share a plane

[digital summit]

some website i looked up for answers to Big Ideas Math: Geometry

[digital summit]

could you explain how? Why can't they not share a plane

[worldly jewel]

If you are given a statement of the form P ⟹ Q...

The converse is Q ⟹ P,
The negation is P and not Q,
The contrapositive is not Q ⟹ not P, which is logically equivalent to the original statement.

[glass meadow]

Share a link?

[digital summit]

I lost it 😂 do you know any websites that do have a reliable answer key though? (for every question would be nice)

[glass meadow]

Well I don't really get what you're doing exactly, geometry or logic?

[digital summit]

its kind of a mix

its reasoning and proofs

[glass meadow]

For the same reason that 3 points share a plane. A line can be defined by any two points on it, so two intersecting lines can be defined by their intersection point and another point on each of them. Any three points share a plane, so the three that define the two intersecting lines share a plane, and every other point on either line are also in that plane.

[glass meadow]

It feels like you're not very comfortable with geometry so if you're going to study logic I would recommend either doing so without the side of geometry, or study basic geometry first

[digital summit]

ohhhhhh

[digital summit]

i get it now my bad i was thinking a different dimension for some reason from the previous question

[digital summit]

I am trying to learn Geometry more so right now but also learn the logic parts of it. I am trying to get ahead on my school classes to get into a higher class next year, but its kind of tough learning Geometry on my own.

[digital summit]

what do you think I should do then?

[glass meadow]

I don't know, search for various resources online or try asking in #book-recommendations

I probably wouldn't study logic and geometry at the same time using the same resource, I would study one and then the other

Otherwise your lack of understanding of one topic could make the other topic harder to understand

https://discord.com/channels/268882317391429632/488120190538743810

[digital summit]

truew

i do already have 2 textbooks currently

Big Ideas Math: Geometry and Geometry: Seeing, Doing, Understanding

I am trying to just do geometry honestly but the textbooks involve some logic i think

[glass meadow]

In that case, understanding converse, inverse, and contrapositive is probably all you need

Like, just remember that if you have some statement in the form of an "if ... then ...", its contrapositive is the same statement, and so they are either both true or both false

[digital summit]

yeah

[glass meadow]

On the other hand, the inverse and the converse are not the same statement, so they could be true or false independently of the original

(and the inverse is the contrapositive of the converse so they are the same statement)

[digital summit]

i never thought of it like that, but that makes much more sense

so the inverse and converse are either both true or false?

[glass meadow]

Yes

[digital summit]

also, in a high school geometry honors class though, do you typically have to learn logic?

to prepare for other classes like algebra 2 or calc

[glass meadow]

Uh I don't know, I'm not a teacher, but I would imagine you don't learn actual logic, just the very basic stuff (like contrapositive)

[digital summit]

yeah

[glass meadow]

I think the people I've seen doing logic are always in uni or above

[digital summit]

okay then ty

is it okay if I dm you or ping you for help sometimes?

[glass meadow]

No, just open channels here

It's better for both of us

[digital summit]

alright no worries. thanks for the help though!

yeah

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[naive dirge]

Let m and n be integers. If the four numbers m+2, 2m+1, n, 8m-101 form a geometric sequence in this order, find the value of n.

[pearl pond]

Please don't occupy multiple help channels.

[naive dirge]

It looks a bit sketchy

But I went trying to find m first

[vast prairie]

why did u square it??

[naive dirge]

Tho I cannot find any

[naive dirge]

It is like a3/a1=r^2

That’s why

[vast prairie]

ohh why don't you just divide by the term before it

then no squaring needed right?

[naive dirge]

It has n in it

[vast prairie]

oh wait my bad i thought it was a typo haha

you're right sorry

so what seems to be the problem?

[naive dirge]

I cannot figure out a possible m. I even expanded it but it was too difficult to factor it.

[naive dirge]

I think I have to find the possible m first. But I’m unable to do so

[vast prairie]

oooh im not that good but let me see if there's a way

[naive dirge]

I got thing like this when I expanded it

It is a bit nasty so I give up factoring it

[vast prairie]

isnt this divisible by 3?

and im not gonna lie

its nasty ;-;

if u use the quadratic formula you'll get some long decimal

are u sure this is the question? cause if so i give up 😂

we're gonna need a smart helper aha

[naive dirge]

Yes. It is from a high school competitive math institution

[naive dirge]

Sure

[vast prairie]

i mean its possible to just go for it but we're gonna be using weird numbers 😭

[naive dirge]

The answer is accordingly -25

[naive dirge]

Yes. I think you are going to need some talent like you have to be able to tell some possible m from what you have got

[plush basin]

-0.268 and -4.69

[dense jasper]

,w (2m+1)^3 - (8m-101)(m+2)^2 expand

[jolly parrot]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by civil_service_pigeon

[dense jasper]

idt your algebra worked out

you should've ended up with $81m^2+378m+405=0$

[jolly parrot]

Civil Service Pigeon

[dense jasper]

your $(2m+1)^3=(8m-101)(m+2)^2$ is fine though

[jolly parrot]

Civil Service Pigeon

[pearl pond]

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[naive dirge]

.close

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[inland laurel]

I am facing problem to count the cases in the problem

A fair die is thrown 3 times. The chance that sum of three num bers appearing on the die is less than 11

[west sapphire]

what's the expected value of the sum?

[inland laurel]

less than 11

[west sapphire]

no i mean, the mean

[toxic lichen]

bungo is asking for expectation not favorable

E[3d6]

[inland laurel]

36

[feral sedge]

if you roll a die 3 times, you would expect the sum to be 36?

[west sapphire]

how did you obtain that?

[toxic lichen]

do you know meaning of expected value yes or no

[inland laurel]

no

[toxic lichen]

then say so, don't make guesses

average value of roll of one dice is?

[inland laurel]

6/2= 3

[west sapphire]

do you know the meaning of average yes or no

[inland laurel]

yes

[west sapphire]

show how you computed it

[toxic lichen]

he did, as 6/2 (which is quite wrong)

[inland laurel]

21/2?

[west sapphire]

i was hoping that was just an unrelated trivia fact

[toxic lichen]

why 21, why /2?

[feral sedge]

again, if you roll a die, you expect the average outcome to be 10.5?

[inland laurel]

no

[inland laurel]

we can expect that total sum of values for 3 throws can be
(1+2+3+4+5+6)*3

21*3 = 63

[west sapphire]

we can't expect that

indeed the largest possible result would be 18

[inland laurel]

there are total 6 outcomes for one throw so there would be 18 outcomes for 3 throws

[west sapphire]

this is also not true

[mystic copper]

i dont think so 🙁

[inland laurel]

uh

[plush bramble]

How do you get a sum of 1 with three die

[inland laurel]

its not possible

[plush bramble]

So definitely less than 18

[inland laurel]

can we do P(sum < 11) = 1 - P(sum >= 11)

[west sapphire]

perhaps, can you justify why this is useful?

[inland laurel]

we may get correct answer because total sum - (not favourable) should give favourable

[west sapphire]

ok, so what is your answer?

do you know P(sum >= 11)?

[inland laurel]

but there are many cases for this

[west sapphire]

is it easier to calculate than P(sum < 11)?

[inland laurel]

ig

[west sapphire]

ok, go for it

[inland laurel]

can't count

😭

this is not a diffiuclt question at all god why i am not able to do

[west sapphire]

the idea behind finding the expected value (mean) was to use a shortcut that will make it very easy

i believe that is what was intended with this question

(the fact that they specified 11 and not some other number makes that pretty clear)

[pearl pond]

@inland laurel Has your question been resolved?

[pearl pond]

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[vast prairie]

i need help (specifically explanation) with cie a levels further mechanics, circular motion. i have no idea how they even begin to solve this question

[vast prairie]

thanks

i was searching how to rotate haha

so i dont know why they did any of the things they did in the working, basically i just need concept explanations and tips to solve questions and if there are any steps to solving these

[pearl pond]

@vast prairie Has your question been resolved?

[vast prairie]

i wish it was ;-;

[thorn agate]

hm..

what do you not understand

@vast prairie

[vast prairie]

hiii mb

hmmm

first of all

when we get a question like that, what's the first thing we do

why did they find Flim and what even is that

[thorn agate]

the terms are not similar but i might give an idea

[vast prairie]

okayy

[thorn agate]

you have a particle on disc which is rotating, for the particle to remain on its position the centripetal force should be what is required for rotation at distance r from the hole(center)

[vast prairie]

wait the particle is also rotating with the disc right?

[thorn agate]

yes

[vast prairie]

what is the Flim for

[thorn agate]

if omega is small enough particle will tend to move towards the centre because of tension from B, so friction will act outwards

[vast prairie]

wait wouldnt it move towards the centre no matter what

[thorn agate]

it's to determine the bounds of omega where the particle slips

[vast prairie]

which brings us to my second confusion how would the particle move AWAY from the centre 😭

[thorn agate]

nope

[vast prairie]

so when friction is at Flim it starts slipping? either out or inwards?

[thorn agate]

i dont want to go in pseudo force and shit

but think it as if you have attached a string to a stone then you move it in a circular path

the more you increase the speed the more force you feel the string is applying at your hand

[vast prairie]

so basically its possible for the particle to slip outwards?

[thorn agate]

yes as the angular speed increases

[vast prairie]

so high angular speed ->outwards, lower angular speed -> inwards

hm

[thorn agate]

Fc = mromega^2 the more you increase the angular speed more centrepetal force is required to keep the same radius

[vast prairie]

ohhh

so at some point it cant keep it and it just goes out

btw

now how did they get that calculations for the inward and outward?

why is inward 60=F+mromega^2 while outward is 60+F

[thorn agate]

so let's consider omega is small such that particle tends to move inwards
so friction is applied outwards
as T is greater than Fc required
you have T - friction = Fc

for other case the object tends to move outward so friction is applied inwards the Fc reuqired is greater than T so you have Fc = T+ friction

this would be easier to see with centrifugal force

where you just balace Tension and other forces
T = Fcentrifugal + friction(both act outward)
same for other

[vast prairie]

but i dont need centrifugal in this case right?

what is centrifugal? is it the opposite of centripetal? like the force away from the center

[thorn agate]

its the same as comparing with centripetal but yh

[vast prairie]

okayy

[thorn agate]

yh

[vast prairie]

thanks man i think i have a rough understanding now

ill try to do some practice questions and come back if i get stuck again

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[stone imp]

Idk how to start on this

It’s supposed to be ax^2 btw

[still hamlet]

do u know the general form of a quadratic or vieta's formulas

[cinder flower]

you are so pretty

[stone imp]

I think the general form of a quadratic cause I don’t recognise the other

[stone imp]

Thank you ❤️

[still hamlet]

okay so can you tell me what the general form is

[stone imp]

I don’t remember it was a couple months ago I was working with this

[still hamlet]

wrong image mb

you have k[x²-(sum of zeroes/roots)+(product of roots)]

[stone imp]

Where’s the imagine

[warped violet]

Which forms do you know

[stone imp]

Wdym by forms I don’t do maths in wnglis

[warped violet]

Which language

[stone imp]

Swedish

[warped violet]

Also you said general form here

I am asking which forms do you know

General form is one

$f(x) = ax^2 + bx + c$

[jolly parrot]

USS-Enterprise

[stone imp]

Oh yeah I know that one

[warped violet]

Any others

@stone imp

[pearl pond]

@stone imp Has your question been resolved?

[pearl pond]

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[unborn charm]

how do I show that every normal subgroup of a group is the kernel of some homomorphism ?

[cursive wraith]

Have you seen quotient maps yet

[unborn charm]

I have

[cursive wraith]

In the quotient maps, what is the kernel

[unborn charm]

the normal subgroup

oh lol makes sense

[cursive wraith]

So

If you have a normal subgroup

Which homomorphism should you look at

[unborn charm]

the quotient map cause that itself is a homomorphism

[cursive wraith]

The quotient map given by the quotient G/N yes

[unborn charm]

hmm

this map isn't the only one the kernel of which is the normal subgroup is it

[cursive wraith]

No it isn't

But you weren't tasked to prove that every normal subgroup is the kernel of a unique homomorphism

You were tasked to prove it's the kernel of some homomorphism

[unborn charm]

yes I got that part

[cursive wraith]

So show there's at least one that works

If N is my normal subgroup

And f:G->H is a homomorphism with ker(f) = N

[austere maple]

Look at the point of intersections with the x axis, 0 and -4, then apply the form of roots quadratic funtions in terms of its roots x²-(v+b)x+vb where v and b are the roots

[unborn charm]

excuse me? we are in the middle of something

[austere maple]

Mb

[stone imp]

Ohhhh thank you 🩷

[austere maple]

I'm new here

[cursive wraith]

This channel is no longer attributed to <3, as you can tell by the name of the channel

[austere maple]

Mb vro 💔

[cursive wraith]

Keep in mind for next time

[cursive wraith]

Anyways I can create $F:\begin{cases}G\to H\times H\ x\mapsto (f(x),f(x))\end{cases}$

[jolly parrot]

Rafilouyear2026

[cursive wraith]

That's another homomorphism

With kernel Ker(F) = N

Tons of examples I can make up

[cursive wraith]

I edited to write ker(f) = N

[unborn charm]

hmm

makes sense

I am taking a first course in group theory and sometimes the algebra gets too abstract for me to know what is supposed to be done, do you have any suggestions😭

but yes I should take this elsewhere, thank you

.close

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[dusky willow]

For quesrion ii is the mark scheme wrong

,rotate

[dusky willow]

like why isit doing 55.2 sin 25

55.2 should be the force

[west sapphire]

how does this relate to your problem in the screenshot? which part are you asking about? i don't see a 55.2 there

not sure if this is yo

"25 degrees to the vertical", but you've drawn it as 25 degrees to the horizontal

not sure if this is the source of your issues

[dusky willow]

Oh ye mb just realised that lol

Thank you

Have a good day or night

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[neat lily]

Confused whether a negative sign would come. From where I am studying it seems it should but I can’t understand why

It should just be pq cos(alpha) right

With the vector going into the plane

[pearl pond]

@neat lily Has your question been resolved?

[pearl pond]

@neat lily Has your question been resolved?

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[bright apex]

further

[bright apex]

how to break it firrher

further

ping me if u reply

[pearl pond]

@bright apex Has your question been resolved?

[mortal flower]

your 2nd line is incorrect

[bright apex]

how

[mortal flower]

you can't just split the fraction like that

[bright apex]

i can split it

[mortal flower]

do you know how to factor the numerator?

[bright apex]

1/(n+1)^2 is just n^2/n^2(n+1)^2