help-49

[twilit saddle]

So that together it would show that every element of N is accounted for by a function input

[runic hamlet]

well ok x wouldnt be hit here

you need to "start" from the codomain to show surjective

and then decide which case of the piecewise hits the specific element you chose

[twilit saddle]

I think I wrote it wrong, gimme a sec

f(n) = {n, n < x
{n+1, n >= x

So if n<x, the function uses n and if n>= x, the function uses n + 1

[runic hamlet]

yes I got that

[twilit saddle]

So I'd go:

Case 1: Let y be (the thing I need help with)
Let p = y
Then f(p) = y
Case 2: Let y be (the thing I need help with)
Let p = y - 1
Then f(p) = (y - 1) + 1 = y
Therefore, f is surjective

[runic hamlet]

the function doesnt hit x

[twilit saddle]

In the particular problem I'm taking this from that's the whole point. it's' actually f: N -> N - {x}. I should have been more specific

Sorry

[runic hamlet]

let y be in N-{x}. if y>x, then y-1>=x and therefore f(y-1)=(y-1)+1=y
if y<x then f(y)=y

thats how could write it down

[twilit saddle]

That's the thing I'm confused about. Why am I using y > x and y < x when that might not be consistent with the part of the function I'm using under it

for instance, if I had a piecewise function where one part is {n-1, n<=3 for example. I cant use y<=n then for that case because of y = 3. When y = 3, n = 4 which no longer works

[runic hamlet]

well it will just depend heavily on the function

I cant really give you a more specific answer than that

[twilit saddle]

Thanks for the effort, this is my first year dealing with rigor so I'm still learning the ropes of how to communicate it.

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[drifting root]

How do I put this in exact form?

[drifting root]

this is the base question, it wants the answer as an exact form

[carmine sigil]

I don't know of an easy identity here, but it might be possible to use triangles here.

[drifting root]

hmmm

the maths catered bot I use says this

do you think this is the easiest way to get exact value

[carmine sigil]

You could use this as well

But you need to be a little careful with how you apply it!

[drifting root]

yep! its just cause these are both arctan values its easy to do right?

[carmine sigil]

You'll need to do arctan(tan(arctan 4 - arctan 3/5)))

[drifting root]

ahh

[carmine sigil]

Then apply the formula to the inner tan

And then you have arctan((4-3/5)/(1+4*3/5))

[drifting root]

aight i think i got it from here

thanks for help :)

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[dense arch]

hello, im currently reducing this matrix to REF to find the rank. I noticed that these 2 bottom lines are the same but *(-1) so i removed it. is that correct? and would the rank be 3 then?

[runic hamlet]

well it would be better to just leave it as a row of zeros

but the rank is 3, yes

[dense arch]

one of them becomes a 0 row?

[runic hamlet]

you are adding the third row to the fourth

or vice versa

so the fourth row becomes a zero row

[dense arch]

ah yeah

can this system be solved?

it would have infinite solutions no?

[midnight plank]

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[dense arch]

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[primal nexus]

Hello, looking for help for someone to teach me this in a simple and effecitve method cause my math teacher is making me dizzy D:! i have unit test soon so it'll be nice thanks!!!

[midnight plank]

@primal nexus Has your question been resolved?

[primal nexus]

someone 😭

[wide isle]

,rotate

[wide isle]

a) have you calculated length of AB? AC? have you tried drawing the points, maybe connecting them into a triangle?

!status

[midnight plank]

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

@primal nexus Has your question been resolved?

[primal nexus]

Hello

am on step 1

Sorry it took a while to replay

reply

nobody was replying me

also what do u mean the length of AB and AC?

[wide isle]

you have the coordinates of A and B; you should be able to calculate length of AB

line segment?

[primal nexus]

the length?

[wide isle]

distance from A to B?

[primal nexus]

Oh

Do i have to use distance formula or smth @wide isle

[wide isle]

absolutely 🙂. You should have a formula, maybe with something like a sqrt, some x's, and y's.

[primal nexus]

Oh i didnt know i had to use distance formula

That's kind of weird

[wide isle]

well, exercise a) asks you for distance from A to B, and to calculate some fraction 🙂

were you given such formula?

[primal nexus]

hmm

well yeah

there are distance formula

d = root of(x2 - x1)^2 + (y2 - y1)^2

but its a bit strange hold on

lemme get smt

my friend told me dis for that problem

Idk what he meant tho he went offline and never told me about it

@wide isle

can u decode his ancient encryptions

😭

[wide isle]

I think your friend is pretty advanced 😄

and his advice might not help us

[primal nexus]

Oh yeah i think

[wide isle]

his method is probably shorter, but without understanding it, we'll get stuck on the next exercise

[primal nexus]

because u need to understand the first one

(i think)

[wide isle]

Let's use the old method, of calculating both segment lengths and then dividing

[primal nexus]

which was tihs one

how to rotate dis

i thikn hes basically saying that instead of us finding that fratcion distance of the line segment, exercise 2 tells us to find the fraction OF the line segment

wait do i make sense

uhmm

Like the fraction is unknown but dis time they gave us the other value which exercise 1 didnt

[wide isle]

,rotate

[wide isle]

yes, this is exercise works differently.

[primal nexus]

I understand hwo to do that one

but is there a fast way to do number 2?

cause for number 1 my friend taught me pretty well

I thought there'd be a nice way to do number 2 efficiently

What do u think?

[wide isle]

Well, I dont know how familiar you are with vectors. The problem 2) can be quickly solved with some affine combinations.

[grand pond]

Wild123

[wide isle]

And then the segment AC will be 1/2 of AB.

[primal nexus]

Huh

Hm

So is there a general formula for part 2?

[wide isle]

Yes. For a segment AB, if you want a point C to be at distance $\frac{x}{y}$ of AB from A, you do $x_C=(1-\frac{x}{y}) \cdot x_A + \frac{x}{y} \cdot x_B$

[grand pond]

Wild123

[wide isle]

But this goes into affine combinations; I don't think it's pre-uni levels of math, or whether your teacher will accept this.

[primal nexus]

😭

[primal nexus]

what should i do then?

there's gonna be multiple steps are there D:

[wide isle]

frankly, I don't know if your teacher gave you the formula I just did.

Or, maybe you've already been taught some basic basic vector calculus?

like, finding $\vec{AB}$

[grand pond]

Wild123

[primal nexus]

what is vector calculus

i only knowing finding lengths uses distance formula

[wide isle]

ah

[primal nexus]

on the bloom taxonomy, i am the type to only remember and execute, to find the pattern and exploit it..

understanding geometory is too hard for me..

so yeah thats that

[wide isle]

familiar with finding the equation of a line, then: something like AB: fraction = fraction ?

[primal nexus]

finding the equation?

what does finding the equation mean

[wide isle]

I don't think I am well equipped to help. I'll let another better prepared helper ... to help. Sorry.

I know posts already replied to by other volunteers usually don't get much visibility; I'll tag <@&286206848099549185> to let them know I am backing off, since I cannot help.

[primal nexus]

D:!

wild123 noooo!!!!

😦

I am still waiting for help..

anyone ?

https://tenor.com/view/eating-ramen-noodles-chopsticks-bowl-dinner-time-gif-14875828643102595359

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[primal nexus]

.reopen

[midnight plank]

✅

[primal nexus]

i just realized i still need hepl

Anyone please 😢

[last arch]

Alright so uh

[primal nexus]

oh wait

[last arch]

You've heard of the mid point formula right

This kinda works like that

[primal nexus]

thats the wrong screenshot my bad, i already know how to do that right when u ping me and i rechcekced

sorry

[last arch]

Ok so

Actual qn now fr

[primal nexus]

BRO

there we go

its sendign wait

this one

if u know how to

😭

[last arch]

The point C is on the line

So the fraction of the distance is just the x or y coordinate of point C - the corresponding coordinate of point A divided by the difference of the corresponding coordinate of A and B

[primal nexus]

Hmmm

processing..

[last arch]

It's like the reverse of this

[primal nexus]

Uh

reverse!?

How do i do that then

[last arch]

This

[primal nexus]

its asking us to find the coordinate instead of the fraction cause they already asked us to find the fraction right

[last arch]

So -2-(-3) = 1
And the total difference between the x coordinates of A and B is
5-(-3)=8
So point C is located 1/8th of the distance from A to B

Should get the same if you do it with the y coordinate

[primal nexus]

huh

WUT

[primal nexus]

i used a - f(b-a) formula for part 1here

so idk what ur speaking off

[last arch]

Oh

Well it's the same thing

So which part do you need help with

[primal nexus]

Part 2

I juts dnot know how to do it

like the general formula or smtt

I alr know part 1

[last arch]

b.)?

[last arch]

Aint that like the same thing

But for B

[primal nexus]

Noo uhm

[primal nexus]

Dis one

[last arch]

But you said you did part a.) already right

[primal nexus]

OH NO

i mean]

I already did the entirety of

[primal nexus]

Dis

[primal nexus]

But idk how 2 do dis

[last arch]

Yea so

This is the answer for part 2 sub question a.)

It's 1/8th of the distance

I wasn't answering part 1😭

Using the lerp formula which you used in part 1 you can also use this approach
-2 = -3+f(5-(-3))

Where you solve for f

Which also gives you 1/8

@primal nexus

Or as a ratio 1:8

[primal nexus]

Oh

How do u do that !

@last arch

[last arch]

You know basic algebra like
solving for x in x-3=0 right

[primal nexus]

Yes

x-3=0
0-3=x
3=x!

[last arch]

So just solve for f

In this equation

[last arch]

-2 is the x coordinate of C

-3 is the x coordinate of A

And 5 is the x coordinate of B

[last arch]

I'm basically using the formula to find f

[primal nexus]

oh

ur doing a - f(b-a)

but finding F?

[last arch]

Yea

[primal nexus]

is it really that simple

[last arch]

Perchance

[primal nexus]

okay what about the other questions

[last arch]

You've found the ratio AC:AB and by doing qn b.) you find the ratio of CB:AB

So you can find the ratio of AC:AB using these two ratios

Yipee

[primal nexus]

juts by doing

XX:XX ratio

no calculcations?

[last arch]

Gimme a sec

If we had ratios A:B=1:2
and B:C=4:2

We can find A:C

Firstly we make both B values in the ratios the same

Let's make the B value 2 in B:C

We divide both values by 2 in B:C to get 2:1

A:B= 1:2 B:C=2:1

So A:C=1:1

@primal nexus

[primal nexus]

theres a ratio formukla?

[last arch]

Wouldn't call this a formula

But don't you learn this in primary school

[primal nexus]

Uhm

please dont bully me

😭

[last arch]

My bad😭

Just saying please don't cancel me🙏

[last arch]

But do you recall now

[primal nexus]

UHm not really

[last arch]

Rip

It's like saying a car is two times faster than a bike and a plane is 4 times faster than a car so the plane is 8 times faster than the bike

[primal nexus]

So we divide

AB and BC for AC?

its ratio?

A:B B:C for A:C?

[last arch]

We make the B value common for the two ratios

So we can find A:B:C

Which basically yields us also A:C

Oh actually for b.) you could just do 1-1/8 to get the answer

Mb

[last arch]

For part d.) use the fact that the sums of the two segments is equal to the whole segment

[primal nexus]

Huh

[last arch]

Because 1 represents the whole length of the line

And 1/8 is the length of the segment we found

And the rest of the line is the length of CB

[primal nexus]

Do i do that for the rest of the questions?

[last arch]

If you have to find both of the segments yea

Find the length of one

And use this method for the other

So AC:AB is 1:8 and CB:AB=7:8

So AC:AB:CB=1:8:7

So what is AC:CB

[primal nexus]

Hmmm

what about the next question?

[last arch]

Gtg

[primal nexus]

alr tysm

<@&286206848099549185> can someone help?

[edgy schooner]

what is the question

[primal nexus]

ned help on this work

for b c d

@edgy schooner

GUYSS 😭

[edgy schooner]

do you know the distance formula ?

[primal nexus]

Yes

d = root ov (x1 - x2)^2 + (y21 - y2)^2

[edgy schooner]

yeah

do you know the section formula?

[primal nexus]

No 😭

[edgy schooner]

Hmm

then I guess just find AC, BC and AB by distance formula

[primal nexus]

What do i do after?

Also which part are u doing

[edgy schooner]

then go according to the question

[edgy schooner]

All parts

[primal nexus]

HUh

all 4 in one formulas?

[edgy schooner]

yes

for (a) AC/AB
for (b) BC/AB
for (c) AC/BC
for (d) AC/(BC+AC) = AC/AB

[primal nexus]

what aboue

a + f(b - a) and i find F

[edgy schooner]

what does that even mean?

[primal nexus]

@edgy schooner

does it work

[edgy schooner]

well I don't know that particular notation

[primal nexus]

Hmmmm

[primal nexus]

So i can juts use the same formula and answer it for all parts

A B C D
is what ur saying

[edgy schooner]

yeah

[primal nexus]

they seem to be asking for other stuff

Can u read real quick

[edgy schooner]

All of them are asking for a ratio or fraction

except (d)

Its asking us for a conjecture

[primal nexus]

how do i do that

[edgy schooner]

its pretty simple actually, so lets assume there are 2 parts of a whole, i.e. part A and part B

So, whole = part A + part B

[primal nexus]

mhm

[edgy schooner]

right?

[primal nexus]

Yus

[edgy schooner]

now we have to find partA:whole ratio

which will be same as partA:(partA + partB)

now partA:partB is given

[primal nexus]

oh

[edgy schooner]

so we know part A and part B

[primal nexus]

since we did part a and b

and we juts plus them

and its A: (the answer?)

[edgy schooner]

no, I'm not talking about parts of the question

Its a conjecture

[primal nexus]

oh

wut do id o then

[edgy schooner]

Given partA:partB, partA:whole = partA:(partA + partB), that's it

[primal nexus]

o..

[edgy schooner]

otherwise suppose partA:partB = m:n, partA:whole = partA:(partA + partB) = m:(m+n)

[primal nexus]

part b and part a are from the other parts right

[edgy schooner]

yeah

[primal nexus]

and thats

the conjecture

aka ansewr

for part D?

@edgy schooner

[edgy schooner]

yeah

[primal nexus]

alr

tysm

i love u gang

ill give u a kiss if i cuold

TYSMMM

LOVE U @edgy schooner

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[shell summit]

Studying for my final, I have a question about this. I'm not sure why this is wrong or what they're asking for the second part, because the derivative of the function is just -1/2

[shell summit]

that's the function i made from the question, which if you plug in x=7, you get 5, which is correct for the first part of the question

[slender walrus]

where's your first equation coming fron

where did you get the -1/2 in
y = -1/2 (x-7)+5

[shell summit]

i got -1/2 from dividing 5-7/4-0, y2-y1/x2-x1

@slender walrus

[slender walrus]

you're not applying slope formula correctly

[shell summit]

y-y1 = m(x-x1) this formula?

[slender walrus]

no

albeit missing (),
m = (y_2-y_1)/(x_2 - x_1)
but you didn't use the correct values

[shell summit]

y2-y1 / x2-x1

7,5 | x1 = 7, y1 = 5
0,4 | x2 = 0, y2 = 4

4-5/0-7

oh i see

ok i got it right now, thank you

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[twin ridge]

I'm doing this, and there's the obvious method of solving it.

[twin ridge]

But this chapter is about the Pigeonhole Principle. How does that apply here?

[midnight plank]

@twin ridge Has your question been resolved?

[twin ridge]

<@&286206848099549185>

The Pigeonhole Principle is that if there are more items than containers, then one of the containers has to have multiple items

Is this talking about prime numbers?

Or relatively prime I guess.

[spring wave]

first u assume that one is not any of those nums

also what obvious method?

[twin ridge]

Any of which numbers?

[spring wave]

8 numbers

[twin ridge]

I misread the question at first.

[spring wave]

because if 1 is any of those nums we are done

[twin ridge]

I thought it said 8 consecutive for some reason

[spring wave]

oh

[twin ridge]

So it can't be 1.

[spring wave]

now divide the remaing numbers in 7 sets

such that

[twin ridge]

And if 2 is in there, then any other even number is also done

[edgy schooner]

at least 2 numbers will be even

[twin ridge]

Oh that makes a lot of sense.

But 8 isn't divisible by 6

[spring wave]

in a set, any two numbers have one divide the other

like

[edgy schooner]

there are 7 odds in 1 to 14, if we discount 1 there are 6

[spring wave]

(2,4,8)

U r trying to do PHP method right?

[twin ridge]

Right.

[spring wave]

do this

[edgy schooner]

so, there should be atleast 2 evens

[twin ridge]

But not all evens are divisible by each other

[spring wave]

yeh

[spring wave]

create 7 such sets

[twin ridge]

I am not sure I follow what you're asking

[spring wave]

like in
(2,4,8) if u choose any 2 nums, one divides the other

[edgy schooner]

(4,6,7,9,10,11,13,14)

I don't think the question means to be completely divisible

Am I missing something?

[twin ridge]

Does "between" mean noninclusive?

So has to be 2-13?

It says there's two integers k and l in the subset where k divides l.

[spring wave]

consider
(1,2,4,8)
(3,6)
(5,10)
(7,14)

[twin ridge]

Which means that l/k gives you an integer.

[edgy schooner]

7 divides 14 lmao

[twin ridge]

I know. But I have to prove this is true for every single subset of 8 integers.

[spring wave]

you dont even have to discount 1

[twin ridge]

You also don't need to think about it though. If it's in the set, it is given that the statement is true.

[spring wave]

@twin ridge did you get this?

[twin ridge]

I do. There are 4 subsets that have at least 3 numbers each that are divisible.

If you add 1 to all of them

I'm not really sure how that fits in the PHP though

Oh wait

Is it because the bottom half of the set can be doubled to get the top half?

[spring wave]

ok

is it fine if i give soln?

[twin ridge]

I think I got it in my head now, but I'm curious what yours is.

[spring wave]

consider sets
(1,2,4,8)
(3,6,12)
(5,10)
(7,14)
(9)
(11)
(13)

we have 7 sets and 8 pigeons

so atleast 1 sets has 2 pigeons

[twin ridge]

I was thinking something similar, I had it written out.

[spring wave]

oh, btw

r u doing math circles?

[twin ridge]

I am not sure what that means.

This is a Combinatorics class.

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[tardy spruce]

.reopen

[glad brook]

how would you write the correct notation for the final answer?
(-1,2)

[gleaming latch]

(f^-1)(2) = ?

[clever turtle]

do you guys know the aswer to B+B+C

[glad brook]

so it would be f(^1)(2)=(-1,2)?

[midnight plank]

@glad brook Has your question been resolved?

[astral steeple]

no, f^(-1)(2)=-1

[glad brook]

oh thank you

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[dense holly]

Hello

[dense holly]

Can anyone help me with a physics question

it has math in it

How big does a balloon with helium in it have to be for it to lyft up a person with the mass of 60kg

@gray widget

[midnight plank]

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[grand pond]

why is math so hard anyways?

[midnight plank]

@tidal turret Has your question been resolved?

[midnight plank]

@tidal turret Has your question been resolved?

[tidal turret]

i) w1 = v1 + v2 + v3

ii) w2 = v1- v4

iii) w3 = v1 + 3v2 + kv3 + 2v4

[grand pond]

why is math so hard anyways?

why is math so hard anyways?

[tidal turret]

,w rref {{1,1,1,0},{1,0,0,-1},{1,3,k,2}}^T

k needs to be different than 3 so that we got three linearly independent vectors

If k is not 3, then {w1.w2.w3} forms a basis with three linearly independent vectors

[grand pond]

why is math so hard anyways?

[tidal turret]

I guess the values of k such that dim(T_k) = 3 are R - {3}

pretty much anything but k = 3, because then we get only two pivots when finding the range

[tidal turret]

.solved

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[gaunt ibex]

i dont know how to approach this problem

[gaunt ibex]

I saw Markov chain in class but im so lost

[midnight plank]

@gaunt ibex Has your question been resolved?

[merry pewter]

yeah you should find your transition matrix and then find the steady state of that matrix

it will be the left eigenvector corresponding to the 1 eigenvalue

[merry pewter]

you should probably have more states than this btw

[gaunt ibex]

yea ur right

[merry pewter]

alr now do the thing i said above

[next rover]

do they mean "fired"

ah it's like it's not in a steady state at the beginning

but then everyone will be employed

wtf

[merry pewter]

no, they always have at least a 40% chance to get a job

so the markov chain is irreducible

[next rover]

either there is an unknown chance of being fired

or it's the 5% that they are talking about in the beginning

[merry pewter]

i assume by "worker" they mean someone who is currently employed

[next rover]

we can't solve it if it's not what they mean right?

[merry pewter]

err by the other interpretation, i guess they would mean that it would be 5% chance that they would be in state U1 or U2 in the steady state

not sure if there's enough information

in that case

[next rover]

well i don't get it

i can't solve this with 6th grade algebra

i mean there's no solution

and "5% according to the task" can't be right either

[merry pewter]

is this in the other interpretation?

[next rover]

no it's what you said i think

5% is that you get fired

given that you can be fired

[merry pewter]

alr, then what is the issue

[next rover]

don;t you see this interpretation makes the answer 5%?

[merry pewter]

with the interpretation that someone has a probability of being fired being 5%? no?

[next rover]

no the other one

[merry pewter]

oh i see, right

[next rover]

a third intepretation could be like, there's 5% chance you become U1 from any state

nonsensically

oh also 5% that you go W→U1 without "given W"

that worked

[midnight plank]

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[gaunt ibex]

wait thats the solution

[gaunt ibex]

i got 13.05% will be unemployed

[next rover]

that's not the solution then

wolfram means =

[midnight plank]

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[twilit field]

I'm trying to prove $\sup(\bigcup_{i=1}^{n} A_i) = max(\sup(A_i))$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

Would induction work here?

so something like this :
\
It's trivially true that $\sup(A_1) = \max {\sup(A_1)}$. Now let $\sup(\bigcup_{i=1}^{n} A_i) = max(\sup(A_i))$. This is our inductive hypothesis. Let $\bigcup_{i=1}^{n} A_i = G$. Let $\sup(G)=g$. We then have $\sup(G \cup A_{n+1}) = max(\sup(G), \sup(A_{n+1})$. But $\sup(G) = max(\sup(A_i))$. We thus have $\sup(\bigcup_{i=1}^{n+1}) = max(\sup(A_{n+1)}, \max(\sup(A_i))$

[grand pond]

ƒ(Why am. I here)=I don't Know

[hollow oyster]

personally I would have done with contradiction for these type of questions may be wrong here

[twilit field]

How would you have done this?

so something like this :
\
It's trivially true that $\sup(A_1) = \max {\sup(A_1)}$. Now let $\sup(\bigcup_{i=1}^{n} A_i) = max(\sup(A_i))$. This is our inductive hypothesis. Let $\bigcup_{i=1}^{n} A_i = G$. Let $\sup(G)=g$. We then have $\sup(G \cup A_{n+1}) = max(\sup(G), \sup(A_{n+1})$. But $\sup(G) = max(\sup(A_i))$. We thus have $\sup(\bigcup_{i=1}^{n+1}) = max(\sup(A_{n+1)}, \max(\sup(A_i))$

[grand pond]

ƒ(Why am. I here)=I don't Know

[hollow oyster]

i would have assume that sup of (cup A_i) \neq max(sup (A-i))

[rose cobalt]

I think you should prove $\sup(G\cup A_{n+1})=max(\sup G,\sup A_{n+1})$ first?

[grand pond]

afi

[twilit field]

I have proven this for two sets

[rose cobalt]

Oh ok then this seems right

[twilit field]

Thanks!

.close

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[twilit field]

.reopen

[midnight plank]

✅

[twilit field]

I now have to determine if this theorm is also true for $\sup(\bigcup_{i=1}^{\infty} A_i)$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

Not too sure o how to proceed here

I think it may be possible to construct a counter examople

[rose cobalt]

You should try that

[twilit field]

Maybe (0,n)?

It doesn't have a supremum

but each $A_i$ is bounded from above

[grand pond]

ƒ(Why am. I here)=I don't Know

[rose cobalt]

But then the maximum of the supremums doesn’t exists as well

[twilit field]

yes, so counterexample

Does that work as a conter example

[rose cobalt]

Not sure

[twilit field]

Anyway, got to go now

sorry

Thanks for the help!

[rose cobalt]

But what about the case $\sup(\bigcup_{i=1}^{\infty} A_i)$ exists?

[grand pond]

afi

[twilit field]

It doesn't always exsit, I just showed that

That's what was asked of me

[rose cobalt]

Hmm ok

[twilit field]

Thanks

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[rose cobalt]

The problem is that if we suppose the supremum of the union exists, the theorem could be true (its actually false) but anyway

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[pale barn]

can someone review my working for d, including the z-table values

[unique stream]

what does it say on the right

[pale barn]

= 0.25

srry for messy handwriting

[unique stream]

a new hieroglyph for 5 :D

[pale barn]

XD

[unique stream]

I'm primarily confused about P(X<=x) = 0.6 ===> P(Z<=z) = 0.25

what conversion did you do there

[pale barn]

oh I was looking at the z value for it

probably should’ve just been same as x

but I’m unsure if I did the correct z value or not

[unique stream]

did you mean this conversion?

where Z is normalized, with mean = 0 and std = 1

if so then yes the probability remains the same

[pale barn]

section d kinda confuses me as I don’t really know how to do it

can you walk me through it

[unique stream]

sure, your end value is right depending on the accuracy of your table, only the way written is a bit odd

so you generally always want to convert the given random variable

[pale barn]

what do you suggest proper notation will be

[unique stream]

into a normalized new variable with mean 0 and std deviation 1

because then you can utilize the z-score table

as it specifically only works for a normal dist. random variable that is normalized

[unique stream]

I'd explicitly write that you introduce a new variable Z which is defined as Z = (X-mu)/sigma

Here I wrote out the conversion

note that if you have any probability of the form P(A<B), the probability remains the same if you add some value or multiply with a factor
P(A<B) = P(A+C<B+C)
P(A<B) = P(A*C<B*C)

[unique stream]

is the conversion and introduction of Z mostly clear?

you have to do the conversion to this new normalized variable Z since the z-table gives you values for the equation P(Z<=...) = ... where Z is normalized

[pale barn]

would you set z to 0.25 and solve for x

[unique stream]

exactly!

well 0.2533

but yeah

maybe your table only has 2 digits

[unique stream]

hm is that the only table you're given

since that's the inverse table of what you need

[pale barn]

oh oops

they give us so many tables lol i got confused

[unique stream]

yeah :D

[unique stream]

np

kind of stupid with the tables ik

0.2533 :)

is it clear why or unclear

[pale barn]

is it because 0.6 is part of tens

[unique stream]

it's because P(Z<=z) = 0.5 + P(0<=Z<=z)

[pale barn]

so will the new value be 21.27

as final

for x

[unique stream]

you can visualize it with the area under the normal dist

if the area for P(Z<=z) shall be 60%

then this area P(0<=Z<=z) must be 10%

because each side of a normal distribution contains 50%

[unique stream]

yop!

[pale barn]

awesome :D

ty!!!

[unique stream]

np :]

[pale barn]

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[last slate]

How can i calculate matrixs

[last slate]

And how can i do derivates for buisness math

[midnight plank]

@last slate Has your question been resolved?

[last slate]

Nope

[main owl]

this server to to help with a specific problem

not to teach you material

[last slate]

Oh mb then tab can someone help me solve a practice sheet

The teacher didn't give us any solution for it he's really Terrible

This is it basically

[midnight plank]

@last slate Has your question been resolved?

[last slate]

Can anyone help?

I just need the solution no explanation

[last slate]

Dawg

Is anyone here?

[midnight plank]

@last slate Has your question been resolved?

[midnight plank]

@last slate Has your question been resolved?

[last slate]

No

[pearl idol]

we don't do that here

Which problem specifically do you want help on?

And what have you tried already?

[last slate]

Damn alright then

[oblique meteor]

nobody is gonna put effort for you if you don't put in effort for yourself

[last slate]

The funny thing is I'm trying dawg the professor's Lectures r useless since they give us no actual explanation and solves the problems WRONG i tried to look up vids but I didn't find any my finals r coming up soon if u can recommend me vids or some type of material i need it desperately

Today just today i learned the matrixs and a bit about how to calculate the critical points it took me 9 hrs

I feel exhausted sorry

[oblique meteor]

What's the specific topic you need lectures for? I might be able to help.

[last slate]

But ig that's what happens when I'm too broke to afford a good education from a good uni

[last slate]

Business math basically derivatives and two unknown in one function with matrixs and how to calculate critical points and maxs and mins

Basically

I'd greatly appreciate any help

[oblique meteor]

undergrad?

[last slate]

Ye I'm a business major

Accounting

Could anyone tell me Pythagoras thereom

@last slate

Or anyone else

[last slate]

The teacher gave us these questions without actual answers and I'm just here trying to solve them but idk if it's right or wrong at this point

[last slate]

Sorry dawg I'm not ur guy ahahaha last time i heard abt this i was in middle school

[last slate]

No problem,mate

[oblique meteor]

for matrices refer to this

https://youtube.com/playlist?list=PL49CF3715CB9EF31D&si=-X7zrrYx3X5Tur6o

[last slate]

I will

Thanks

[oblique meteor]

I sent the wrong link. check again

[last slate]

It worked

Oh

Ok

Thank u

[oblique meteor]

For derivatives
https://youtube.com/playlist?list=PL4C4C8A7D06566F38&si=u1DnLWu7mxd2QX8W

[oblique meteor]

Hypotenuse squared=base squared+perpendicular squared

[last slate]

Thanksate

[last slate]

Alright bet bet

Thank u

It's too indepth tho but it's ok I'll take what i need

Thank u so much

[midnight plank]

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[midnight plank]

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[native elm]

can someone help me'
i have to find the volume of the solid of revoution generated by revolving the region bounded by x=y-y^2 and the y axisaround the line y=-1
i used the washer method here
the correct answer is π/2
but i did not get that

[surreal moon]

!show

[midnight plank]

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

[native elm]

ok 1 second

i set up the integral as you can see on the right hand side

that integral however according to desmo does not equal π/2 which is what the answer key say

@surreal moon

i dont see where i went wrong here exactly

[surreal moon]

,w plot x=y-y^2, x=0, y=-1

[grand pond]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by spelwerdzrite

[native elm]

?

[surreal moon]

I was trying to plot this

So i could see better

[native elm]

oh

WAIT

IM kind of

dumb

my drawing is wrong

I WAS DOING X=-1

LOL

TGATS WHY I WAS CONFSUED

ok

@surreal moon
so would the corect setup be 2π * the integral from 1 to 0 of (y+1)(y-y^2)dy
using shell method

?

[surreal moon]

I was figuring something like this would be this issue. It's why i try to graph it first

[surreal moon]

Why are your integral bounds 1 to 0?

[native elm]

because isnt the max value of y 1 and the minimum 0?

[cerulean brook]

that looks right

[native elm]

ok let me desmos it

YES

its right

gg

ok thanks guys i appreciat e the help

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[surreal moon]

Oh. Yes it is. I was looking at wrong region

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[grim monolith]

how would a graph like this look like

[grim monolith]

its x = sqrt(y)

[slow thorn]

try to square both sides

[grim monolith]

sqrt(x) = sqrt(sqrt(y))?

wym try to square both sides

[radiant roost]

i would start by graphing x=sqrt(y) and x=0

are you stuck on graphing x=sqrt(y)?

if you swap the roles of x and y, it becomes y=sqrt(x)
so you can graph y=sqrt(x) and flip it along the line y=x

or you can square both sides, like blurple_galaxy said

they mean raise each side to the second power

(x)^2 = (sqrt(y))^2

[midnight plank]

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[twilit field]

The Cut property of reals is : If A and B are non-empty disjoint sets , with $A \cup B = \R$, and $a<b$ for all $a \in A, b\in B$, then there exists $c\in \R$, such that $x \leq c$ whenever $x \in A$ and $x \geq c$ whenever $x \in B$
\
Prove this using the completeness axiom.
\
We define $A = {x : x \geq c }$ and $B$ to be the set of all it's lower bounds. $c$ is the supremum of the set of lower bounds, and the infimum of $A$. Thus $x \leq c$ whenever $x \in A$ and $ \geq c$ whenever $x \in B$.

[grand pond]

ƒ(Why am. I here)=I don't Know

[fervent ember]

if A and B are given, we do you define A? you should prove that such a c exists, so you cannot not start with an existing c so what should your defined A be if you dont even know that such c exists?

[twilit field]

hmm

so I have no control over what the sets are

[fervent ember]

no. you only know that the union is R and that "one is smaller than the other" (in the given sense).

[twilit field]

got it

let me think about this for a while

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[twilit field]

Thanks

[fervent ember]

look at the supremum of A and/or the infimum of B (you need to show that they exist).

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[limber quiver]

i need the formual

[keen herald]

You have the formula, assuming you mean the quadratic one

[lavish venture]

look at the lower half of the screenshot you sent

[limber quiver]

uh

my friend send this

she said to solve it

[fallow scarab]

What problem

[lavish venture]

you mean derive the quadratic formula?

[limber quiver]

ye

[lavish venture]

do you know how to complete the square

[limber quiver]

uh no

[lavish venture]

riemann do you have a thing for that

[fallow scarab]

,tex .cts

[grand pond]

riemann

[limber quiver]

uh

[lavish venture]

at the last line just set it equal to zero

[limber quiver]

oh

[lavish venture]

and solve for x

[limber quiver]

ok

[marble stirrup]

Um but delta is better solution !?

[lavish venture]

$\left(x+\frac{b}{2}\right)^2 + c -\left(\frac{b}{2}\right)^2 = 0$

[grand pond]

knief

[limber quiver]

so

it equal to 0?

[lavish venture]

isolate x

note that the two terms on the right here are just constants so you should move them to the other side first

and i hope you know what to do from there

[limber quiver]

.close

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[gusty birch]

Given two sets $A$ and $B$, prove the following:
If $|A| \leq |B|$ and $|B| \leq |A|$ then $|A| = |B|$.\

I'm having some trouble, the first assumption ($|A| \leq |B|$) implies that there exists a set function $f : A \to B$ that is injective. The second assumption ($|B| \leq |A| \Rightarrow |A| \geq |B|$) implies there exists a surjective function $g : A \to B$. If i can somehow show that $f = g$ then that means that $f = g$ is both surjective and injective, i.e. bijective, and by definition if there exists a bijection between $A$ and $B$ then they have the same cardinality.

[grand pond]

woomy

[surreal moon]

It is not necessary that f=g initially

But you can construct a bijective function from your f and g

[gusty birch]

Or would a proof like the following suffice:
$|A| \leq |B| \leq |A|$ by squeeze theorem $|B| = |A|$
My only issue is that we are dealing with infinities and infinitesimals so this might not hold.

[grand pond]

woomy

[surreal moon]

'fraid not

[gusty birch]

hmm im trying to think of a way how

[surreal moon]

cardinalities are not exactly the same thing as numbers, so not a real squeeze theorem to use here.

If anything, you are proving squeeze theorem right now

[gusty birch]

oh lol yea

[surreal moon]

Btw this proof is known as Schröder–Bernstein theorem.

It's named after two people, so you know it's a banger

[gusty birch]

yep! its an exercise in week 42 of challenge week in the olympiad server

(exercise 4)

[surreal moon]

Ah olympiad. Then yeah I hope you have fun

1 and 2 are simple if you have formal set theory knowledge

You can get 2 for free if you do 3

and really, you get all for free if you do 4.

I suppose the problems are in increasing abstraction

[gusty birch]

i think i found a way to construct the bijection

[surreal moon]

https://tenor.com/view/come-on-give-it-to-me-gif-9237972

[gusty birch]

can i create a function $h$ that takes values from $g$ and then replaces all points where it wouldnt be injective with values from $f$

[grand pond]

woomy

[gusty birch]

not sure it even makes sense though

ill think about it for some more time

[midnight plank]

@gusty birch Has your question been resolved?

[gusty birch]

thanks for the hint! ill try to solve the rest by myself :) if i have any more difficulties ill ask! @surreal moon

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[glossy compass]

im not exactly sure what part B means

or like how to solve

[tribal temple]

What did you get for part A?

(intermediate value theorem )

[midnight plank]

@glossy compass Has your question been resolved?

[midnight plank]

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[main owl]

linas integrating is fine

its probably actually deriving what your suppose to integrate

im saying their integration was fine so the mistake must be the original formula

they integrated

yea

i honestly dont know so im not gonna answer that but that has to be whats wrong

[wet pollen]

One moment.

It's how Lina calculated the area of the triangle.

So the area of triangle is base times height divided by two.

But Lina tried to simplify the problem by only using one-half of the parabola in Q1.

However, that simplification resulted in the wrong height

Yes.

Looks good.

Hmm. That results in a negative value though.

Nevermind, I was mathing wrong in my head.

[midnight plank]

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[midnight plank]

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[twilit field]

Assume $\R$ posses the cut property, let $E$ be a non-empty bounded above set, prove $\sup(E)$ exists .

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

I'm kind of confused, doesn't bounded above mean by defn that supremum exists

[surreal moon]

What is the cute property?

[twilit field]

If A and B are non-empty disjoint sets , with $A \cup B = \R$, and $a<b$ for all $a \in A, b\in B$, then there exists $c\in \R$, such that $x \leq c$ whenever $x \in A$ and $x \geq c$ whenever $x \in B$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

And bound above means there exist a number that's greater than all the elements of the set

[surreal moon]

Okay so A and B partition R (not topologically though)

Oh and that extra requirement with a and b

[twilit field]

I guess I could consider cases, ( half open and half closed ) intervals

like if the interval is (a,b]

[surreal moon]

Perhaps, but have proven that A and B must be intervals?

[twilit field]

no

[surreal moon]

Oh you mean across A and B

[twilit field]

No, no

I mean A=(- \infty ,b]

B=(b,\infty)

then b is an upper bound and supremum of A

But that isn't rigorus enough

[surreal moon]

By b here, did you mean c?

[twilit field]

yea, my bad

[surreal moon]

Cool cool just wanted to be sure

Damn i proved this before

[visual tiger]

yeah, you could prove A is an interval by showing A is convex

but in what's asked for you to prove

it's a bit useless

[twilit field]

Convex sets are introduced in the next chapter I think

[visual tiger]

it's easier showing c = sup A or c = inf B work

(btw unnecessary but can you prove sup A = inf B)

[twilit field]

I think I've done that before

[runic hamlet]

no. in Q the set {x: x^2<2} is upper bounded (for example 100 is an upper bound) but the set has no supremum in Q

[twilit field]

ah yes,messed up my defns

[visual tiger]

that is why completeness of R is important

oh we're proving sup exists for bounded above sets

[runic hamlet]

your first thought should be how to get from some random upper bounded nonempty subset to two such sets A and B

so that you can use the cut property

[twilit field]

This is a differen question

A subpart to the same question that I haven't done before

There I assumed the compleness axiom to prove the cut property

here I assume the cut property to prove the completeness axiom

[surreal moon]

Ohh

[twilit field]

That's what the book asks me to do

[surreal moon]

So first, why would this not work for Q?

It's nice to find equivalences to axioms

[twilit field]

I'm being asked to effectively prove two alternate statements are equivalent

[twilit field]

Consider the set $x^2 \leq 2$, the supremum is $\sqrt{2}$ which isn't in $\Q$

[grand pond]

ƒ(Why am. I here)=I don't Know

[midnight plank]

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.

[runic hamlet]

the original question makes sense to ask. it really achieves nothing to discuss that here

just focus on the actual question

[surreal moon]

Okay, so we can "cut" $\bQ$ into $A=(-\infty,\sqrt{2})$ and $B=(\sqrt{2},+\infty)$, but the cut property fails because we have no desired $c=\sqrt2$ in $\bQ$.

[grand pond]

SWR

[twilit field]

yes

[surreal moon]

so, what's the takeaway here? If we had root 2, what can we know?

Hint: let $B={x\in\bR:\forall(y\in E)(x>y)}$

[twilit field]

then there would be no "gaps", so to speak

[grand pond]

SWR

[twilit field]

What si this B for?

oh, it's a cut?

[surreal moon]

Btw this proof is super useful since it lets you construct the reals from Dedekind cuts of Q

[surreal moon]

Kinda. We can make a cut, because we are saying R has the cut property

[runic hamlet]

you first have to show that it is indeed a cut

[twilit field]

So I have to show that there is a number $c$ such that all numbers in say $E$ are less than or equal to it and all numbers say in set $F$ are more than or equal to it?

[grand pond]

ƒ(Why am. I here)=I don't Know

[surreal moon]

No. You are assuming R has the cut property, so (after you find A), you assert that such a c exists

[runic hamlet]

proof sketch:
E is given
construct A,B from E so that they form a cut
obtain the number c
show that c is the sup of E

thats what you want to do

[twilit field]

Let $A=E$ , let $B= {x : x > a;\forall a \in E}$

[grand pond]

ƒ(Why am. I here)=I don't Know

[surreal moon]

Do A and B make R?

[runic hamlet]

make a sketch

how could E look like

how should A and B look like

[twilit field]

A should be one half of the line, B the other half

[surreal moon]

In a way, sure. How you chose A, is that true tho?

[twilit field]

I don't follow

[surreal moon]

Ponder on this question for a moment and tell me what you think

[twilit field]

Depends on what A and B are'

[surreal moon]

They are how you defined them here

[twilit field]

If E is an open interval, it won't be the entirety of $\R$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

If I'm allowed multiple cuts , $A=E$., $B = R\setminus E$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

If this is wrong, I'll close the channel and think about this a bit more deeply

I anyway have a class soon, so will have to close this

. close

Sorry

And thanks

.close

[midnight plank]

Channel closed

Closed by @twilit field

Use .reopen if this was a mistake.

[surreal moon]

@twilit field these are certainly disjoint, and Union to R, but do they meet every requirement for cut?

(But we can contemplate tomorrow)

[midnight plank]

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[tidal turret]

\textbf{Exercise 24.} Let $B = { \mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3, \mathbf{v}_4 }$ be a basis of a vector space $V$ and let
[
S = \langle \mathbf{v}_1 + 2\mathbf{v}_2, \mathbf{v}_2 + \mathbf{v}_3 \rangle \quad \text{and} \quad T = \langle \mathbf{v}_3 + \mathbf{v}_4, \mathbf{v}_1 + \mathbf{v}_2 + \mathbf{v}_4 \rangle.
]
\begin{enumerate}
\item[(a)] Find a basis and the dimension of $S \cap T$ and $S + T$.
\item[(b)] Find a vector $\mathbf{v} \in S + T$ such that $\mathbf{v} \notin S$ and $\mathbf{v} \notin T$.
\end{enumerate}

[grand pond]

why is math so hard?

[untold rune]

linear algebra?

since 4 vectors are bases of V, they are linearly independent.

$S \cap T = \langle \mathbf{v}_1 + \mathbf{v}_2, \mathbf{v}_3 \rangle$

[grand pond]

Silly Cyn

[untold rune]

$S + T = \langle \mathbf{v}_1 + 2 \mathbf{v}_2+\mathbf{v}_3+\mathbf{v}_4, \mathbf{v}_1 + 2 \mathbf{v}_2+\mathbf{v}_3+\mathbf{v}_4 \rangle$

[grand pond]

Silly Cyn

[untold rune]

hm..

[midnight plank]

@tidal turret Has your question been resolved?

[untold rune]

did you get it?

[midnight plank]

@tidal turret Has your question been resolved?

[tidal turret]

.cloe

.close

[midnight plank]

Channel closed

Closed by @tidal turret

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[midnight plank]

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[obtuse totem]

and ans

I remembered the angle being itheta +2pik, where k=0,1,2,3,...n-1

n being the power of the exponent

so the last two solutions dont make sense to me

[radiant roost]

why do i feel like i is missing from the exponents?

[obtuse totem]

it is!

which is not right

[radiant roost]

even if we put i in the exponent, look at the first answer
2^(1/8)e^(3pi*i/4) is not a fourth root of -1+i

so i don't know what's going on...

[obtuse totem]

this is my ans

[radiant roost]

those 4 roots are correct

but i don't know the def of principal branch in this case so i'm not sure which one is principal

[obtuse totem]

also did i plot this right?