help-26

[chrome glen]

They should change

[frail gyro]

how do we do that

[chrome glen]

Lemme try an example

[frail gyro]

alright

[chrome glen]

I get 24 dollars per foot

You get 3 dollars per inch

You make 3 dollars per inch * 12 inches per foot = 36 dollars per foot

[frail gyro]

yes

[chrome glen]

So you make 12 more dollars per foot than me

[frail gyro]

yea

oh okay so we need to know the conversion factor

of km to miles

or other way around

[chrome glen]

Depends which one youre converting

Lemme do a fraction example

No units

You have 3 numbers a b and c

$\frac{a}{c} \cdot \frac{?}{?} = \frac{a}{b}$

[thorny flame]

PapaBread

[frail gyro]

c/b

c's cancel out and b is n0w denominator

[chrome glen]

Yes exactly

So the same for units

[frail gyro]

ahhh

okok

[chrome glen]

If a is dollars

c is km

And b is miles

Thatd be exactly what you do to convert from $/km to $/mile

You multiply by km per mile

[frail gyro]

ahhhh okok

and then we need the conversion factor

[chrome glen]

Ye

[frail gyro]

so 1) convert to correct units

2. conversion factor

okok thats simple! Thank you bro!

You're the best

now i will ace my math test.

i cant believe in im grade 10 and i dont know this..

ok adios

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[chrome glen]

Adios

[topaz sinew]

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[neon iron]

how to be good in ln calculus

[past kayak]

Practice

[neon iron]

i tried to learn every trick

but there is alot of them

:/

[past kayak]

Calculus is a dense subject

But every "trick" has some justification behind why it works

[neon iron]

fax

thanks for pulling up bro btw 🙂

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[sweet bear]

Did i do somethting wrong

I feel like i cant just assume A=0

[loud raven]

Wow that’s advanced math

I don’t know man sorry

Could you help me please

I really need bro

Is this correct the length of.

X

[sweet bear]

Can you go to another channel please

Theres other channels available

[topaz sinew]

@sweet bear Has your question been resolved?

[sweet bear]

why

in the denominator it is squared

you mean it sho8uld be C(x-1)^3?

[sweet bear]

Why cant i do this

Also i think i need the A and the B term because of this

[neon iron]

yeah this is correct

[sweet bear]

so mine is right?

[neon iron]

yes

[sweet bear]

alright, i still feel like i did something wrong tho cuz when i put it in a calculator it gives this after doing partial fractions

[topaz sinew]

@sweet bear Has your question been resolved?

[grim jacinth]

That somewhat makes sense for the PFD

That is more easily integrable than the original problem

[sweet bear]

yea but i dont know hwo to get to that step

[grim jacinth]

Partial fraction decomposition

[sweet shard]

where's the original question

[sweet bear]

the top is what ive gotten it down to

ill send th e orgiinal

The top one

I did polynomial long division

oh did i do the partial fractions wrong

cuz it needs to be Cx+D right

because bottom is a quadratic

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[cyan lion]

can someone help me out with these? I'm not sure what to do

[wooden matrix]

1 is just show all the axioms hold

2 is just show it meets the definition

3 is just do the same algorithm like any other example

[topaz sinew]

@cyan lion Has your question been resolved?

[cyan lion]

I'm not sure what you mean in 1 and 2, I think I get 3 after looking it over

[vital relic]

review what the rules/axioms for a vector space is.

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[graceful elk]

I got the right equation but my answer is 102km/h

[graceful elk]

somehow its wrong though and its actually 108

[topaz sinew]

@graceful elk Has your question been resolved?

[grim jacinth]

340/x = 360/(x + 6)

@graceful elk Do you have work for your solving of this equation?

[graceful elk]

yes

this is how I did it

340=340x+2040

-340x

20x=2040

20x/20=2040/20

I got x=102

@grim jacinth

[grim jacinth]

You mean 360x on the LHS

[graceful elk]

360x

yes

mb

I got x=102

[grim jacinth]

That looks right to me

Although in the screenshot you sent above it doesn’t look like they asked for you to find speed

Just the equation

[graceful elk]

@grim jacinth

[grim jacinth]

Yes

x = 102 km/h

[graceful elk]

but its wrong

x=108

was right

idk if that was a mistake on the teachers end but ya

[grim jacinth]

They made a mistake

Oh nah

They’re right

They’re asking for the speed of the other person

Yanaisa is faster than Lantay (102)

@graceful elk

[graceful elk]

so

is my equation

wrong

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[jaunty vector]

I've simplified it to log4(1/3) + log4(7/8)
log4(7/24)
log(7/24)/log(4)

[wooden matrix]

you seem to have ignored the entire rest of the sum

and only looked at the terms explicitly written

[jaunty vector]

do you mean the ...

[wooden matrix]

yes

[jaunty vector]

how do i find the rest of the sums?

[wooden matrix]

wdym

the ... implies the rest of the terms follow the same pattern

[jaunty vector]

so i dont need to find the rest of the sums?

if so, is my previous work correct?

uhh. i multiply the logs together to get 1/3 and 7/8

[past kayak]

There are 11 whole other terms being added

[jaunty vector]

wdym

[past kayak]

The ... contains a bunch more terms

You can't just ignore them

[jaunty vector]

ok. uh so what should i do to the other terms?

[past kayak]

^

They all follow the same pattern

[jaunty vector]

wait do i try to find the pattern?

[wooden matrix]

The pattern is given

like for example the 3rd term is clearly ln(3/4)

[jaunty vector]

i meant evaluating the problem

i got log(7/24)/log(4)

[past kayak]

If I have 1 + 2 + ... + 9 + 10, is that sum equal to 22?

[jaunty vector]

yes

[past kayak]

...

The ... means there's more terms than just 1, 2, 9, and 10

[jaunty vector]

oh.. no then

[past kayak]

We just don't write them to save space

Similarly, log(1/2) + log(2/3) + ... + log(14/15) + log(15/16) isn't log(1/2) + log(2/3) + log(14/15) + log(15/16)

[jaunty vector]

ohh.. then what i should do instead of adding them up?

thats the part im stuck on (the ...)

[past kayak]

How would you solve 1 + 2 + ... + 9 + 10?

[jaunty vector]

figure out what terms go into the ...

[past kayak]

And how would you do that?

[jaunty vector]

look at 22 and add up 1+2+9+10

and see what terms is missing

[past kayak]

That's not what you'd do

There's a pattern

[jaunty vector]

+1

[past kayak]

Yes

Clearly, the whole sum is 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10

[jaunty vector]

do i have to find the whole sum?

[past kayak]

I don't know if there's any simplifications that can be made on your problem

I only wanted to address the ...

[jaunty vector]

it just said elevate and requires decimals or whole numbers

[past kayak]

You just have to calculate the sum

[jaunty vector]

since the whole sum is a pettern, i dont really have to find the whole sum?

[past kayak]

You do. How else would you calculate 1 + 2 + ... + 9 + 10 if you didn't have a formula like n(n+1)/2?

[jaunty vector]

can the pattern be multiplication?

[past kayak]

Considering nothing is being multiplied in your sum...

Look, try writing out the whole sum

[jaunty vector]

cool. i got it

thanks

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[vague sapphire]

Could someone explain part c

[grim jacinth]

It’s just algebraic manipulation (completing the square)

[vague sapphire]

oh i see it, i am just not familiar with non monic completing the sqaure so the B threw me of

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[frigid zenith]

I'm really not sure how to go about this one

[grim jacinth]

You can do change in position / change in time for the given intervals

You can find position at any given time given y = (stuff)

[frigid zenith]

How would that look like? Sorry, I'm still fresh with these kind of questions

[grim jacinth]

You can do:

position at t = 3.01 minus position at t = 3

Take that, and divide that by:

3.01 - 3

[frigid zenith]

so i'd essentially be diving 3 by .01 for that first answer?

3.01-0.01 =3. 3/0.01 = find this answer

woops sorry typo

but you get my point

and then for the other time intervals I would just need to substitute the .01 in my equation for the new time itnerval?

[grim jacinth]

$\frac{y(3.01) - y(3)}{3.01 - 3}$

[thorny flame]

Shen

[frigid zenith]

if im not mistaken that just gives y right?

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[near mist]

can someone please check for me if my calculation is correct or not?

[chilly kestrel]

that's correct

[near mist]

thank you very much

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[deft bloom]

I'm having trouble understanding this question

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[thorny bison]

hello i need some help with some homework

[thorny bison]

i have to use implicit diffrentation to find dy/dx of

ycos(xy)=y^2+6

[leaden tusk]

what have you tried

[thorny bison]

well if i remeber correctly

you want to do the d/dx to both sides

then split it into several d/dx's

then it turns into a big mess and i get lost

[leaden tusk]

if you're unsure of how to do implicit differentiation i suggest a youtube video

[thorny bison]

well we learned this in class today its just 1 problem took up an entire sheet of paper lmao

[thin vigil]

If you're getting lost in your work try doing each side separately first or writing down your steps neatly

[thin vigil]

Try the RHS first

[thorny bison]

but was i doing it right the way i described or is that not how i do this problem

[thin vigil]

Yes, you take the derivative wrt to x on both sides

Try d/dx [y^2 + 6] first

[thorny bison]

hmm

so i split that into two right

like d/dx y^2 + d/dx 6

[thin vigil]

Yes

[thorny bison]

then the 6 goes away and what happens to the y^2 ?

or do you just do the constant rule and put 2y

[thin vigil]

y is a function of x, so really it's [y(x)]^2

[thorny bison]

so d/dx [y(x)]^2

[thin vigil]

Yes keep going, you don't need to ask for validation every step, we'll tell you if something is incorrect

[thorny bison]

idk what to do with that lol

do you do the 2 d/dx[y(x)] ?

[thin vigil]

How would take the derivative of say sin^2(x)

[thorny bison]

no idea

sin(2x)?

im not sure

[thin vigil]

Have you learned about the chain rule yet?

[thorny bison]

yes

f'(g(x)) * g'(x)

[thin vigil]

Do you know when to use the chain rule?

[thorny bison]

nope

[thin vigil]

$$(f\circ g)'=f'(g(x))*g'(x)$$

[thorny flame]

Tronsi

[thorny bison]

would it be 2[y(x)] * y(1)

[thin vigil]

For the derivative wrt to x for y^2?

[thorny bison]

thats atleast how i think i do it

yes

since you said y^2 is the same as y(x)^2

[thin vigil]

Whats y(1)?

[thorny bison]

you do the outside as f' and the inside as g'

y(1) would be the derrivative of y(x) the inside ?

[thin vigil]

d/dx [y(x)] by definition is dy/dx. You can't just take the derivative of the x inside just like how d/dx sin(x) is not sin(1)

[thorny bison]

oh

[thin vigil]

So yes d/dx [y(x)]^2 would be 2y dy/dx

Now you need to do the LHS

Which is a bit more complicated but follows the same basic logic as the RHS

[thorny bison]

okay ill give it a try

d/dx y * d/dx cos(xy)

so y dy/dx * -sin(xy)

[thin vigil]

Well first you need to apply the product rule, remember unlike addition or subtraction, you can't just break multiplications or divisions apart for derivatives

As a reminder $$(f*g)' = f'g+fg'$$

[thorny flame]

Tronsi

[thorny bison]

ah

ill just have to stop by my teachers office hours so i can get some one on one learning thank you for your help my memory is just shit

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[neon iron]

heyy

[neon iron]

It’s a small exercice of limits if anyone could help me would be rlly nice

I’m trying to do the first one I have my theory next to me

[stuck lintel]

What's your particular issue at this time?

[neon iron]

right now

I did this

So I guess I already have a small part of the answer

but now I have infini - infini

And here Idk what I should do @stuck lintel

[stuck lintel]

let me see

when $x \to +\infty$, are you sure that $(3x-2)^2(5-x)^3 \to^? +\infty$

[thorny flame]

Remavas

[neon iron]

when x -> +infini its +infini-infini

and that doesnt exist so I need to go more deeper in my ex

[vale jacinth]

You understand that ax^n+bx^(n-1)+cx^(n-2)+…=x^n(a+b(1/x)+c(1/x)^2+…) right?

Eventually only the coefficient of the first term a really matters

[neon iron]

what?

[vale jacinth]

What “what” ?

And (a+b(1/x)+c(1/x)^2+…) goes to a when x approach infinity x^n will get bigger and bigger

[neon iron]

i don’t understand what ax^n+bx ect means

isn’t there anyway u can show it like this

[neon iron]

.

[vale jacinth]

So whether it diverges to positive infinity or negative infinity depends on the sign of a

[neon iron]

whats a here?

Im pretty sure it was a calcul that was 3-4lines

[vale jacinth]

3x^2-2x+4=x^2(3-2(1/x)+4(1/x)^2)

Now better?

3 is positive so it diverges to positive infinity

[neon iron]

oh nice

What no

I did it w someone else

we found -9x^5

[vale jacinth]

Not 3x^2+4 goes to positive infinity, but -2x goes to negative infinity so I don’t know which one it is, but just the sign of the coefficient of the first term

[neon iron]

and he told me we take the sign of the x w the highest exposant

[vale jacinth]

Then we said the same thing

Not exactly the same I mean the sign of the coefficient of x^n where n is the greatest one

[stuck lintel]

In this case

[vale jacinth]

Negative infinity

[stuck lintel]

mhm

And so when x goes to +infty, the lim is -infty - infty

Which gives you - infty

[vale jacinth]

No

I just considered the sign of the coefficient of x^n where n is the largest one

In this case -9

[stuck lintel]

and you can use the same though process for when goes to -infty

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[topaz sinew]

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[final sierra]

brushing up on logs, where do I go from here?

[final sierra]

I'm not sure how to simplify it further or where to find a list of log rules 😅

I think I can simplify the first term to just 2

actually I found a list of log properties and rules!

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[smoky light]

How do you convert a non-homogeneous second order ode to a first order system? Equation I'm trying to convert is this:

[smoky light]

Just confused on where the g(t) goes in the matrix

[topaz sinew]

@smoky light Has your question been resolved?

[smoky light]

<@&286206848099549185>

[sweet shard]

see example 4 here
https://tutorial.math.lamar.edu/classes/de/systemsde.aspx

[smoky light]

Thank you!

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[topaz sinew]

@hot sigil Has your question been resolved?

[hot sigil]

can anyone help with this?

[topaz sinew]

@hot sigil Has your question been resolved?

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[prisma prairie]

https://cdn.discordapp.com/attachments/496784996510466057/946241830176051260/unknown.png

[prisma prairie]

how do they get from c'(t) to |c'(t)|?

I tried doing it but the terms don't cancel as nicely?

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[burnt swift]

how would i go about solving this?

my first instinct is to cancel something out and use one of the formulae provided to me

but doesnt look like thats possible

[hearty narwhal]

try partial fraction decomposition

[burnt swift]

i did partial fraction decomp, now what?

i got 1/k^2 for one of them and the other is very similar

[hearty narwhal]

this is what you got, right?

[burnt swift]

yes

[hearty narwhal]

notice how, when you write the sum, the terms cancel out

[burnt swift]

ohhhh

its like a telescopic series

[hearty narwhal]

yes

exactly

[burnt swift]

did you get

1-1/(n+1)^2

[hearty narwhal]

yep

that should be it

[burnt swift]

:3

thank you!

.close

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[icy pilot]

why can't there be a negative number in the radical? (for example in the equation below)

[icy pilot]

$$f(x)=a\sqrt{x-h}+k$$

[thorny flame]

geoxcaliber

[icy pilot]

Second question: If there is no h or k visible, does that mean their both 0? My teacher said that's true of the h, not sure about k

[tulip prism]

Well, for example, what do you think the square root of -4 would be?

[topaz sinew]

@icy pilot Has your question been resolved?

[icy pilot]

so for square root it can’t be anything

but for cube

if you had say -64 that’d work bc -4 will get you that?

[tulip prism]

Yes, with the cube root it works

In general, if you take the nth root, your radical must be positive if n is even, and can be anything if n is odd

for n positive natural number obviously

[icy pilot]

Thanks

do you have any tips for finding the 'a'? in this case, -2

it went +1 then -2

if you did 1 / -2 that'd be -.5. Would you do -2/1? if so, why? does the top down movement go first

oh, rise over run

[icy pilot]

Can somebody explain how

[worn inlet]

u want to find the a right?

generally speaking, if you know the vertex of a parabola, it's much easier to find it using the vertex form

[icy pilot]

Yes

[worn inlet]

so it would be a(x-1)^2 +8

now sub one of the pairs in, I am using (3,0)

a(3-1)^2+8=0

4a=-8

a=-2

[icy pilot]

I see, thanks

$$\sqrt{x+2}=10$$

[thorny flame]

geoxcaliber

[icy pilot]

can you explain how this becomes

$$x + 2 = 100$$

[thorny flame]

geoxcaliber

[icy pilot]

what takes place

for the radical to be removed in the left side

so

does the square root half the 2

then the 2 exponent bring it back? or, better question, what happens first, exponent or square root?

$$\sqrt{x+2}^2=10$$ I meant

[thorny flame]

geoxcaliber

[worn inlet]

ummm you sqare both side

sorry for replying late

10^2 will give you 100

[topaz sinew]

@icy pilot Has your question been resolved?

[icy pilot]

it's fine

Did you read my question? that isn't what I asked

[worn inlet]

ummmm

square root

and power of 2

are invertible

like you can reverse them

by using the other one

[icy pilot]

oh okay, thank you

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[lime verge]

Hi ! Would you happen to know how to determinate if an angle between two 2D vectors is positive or negative ? Thank you in advance.

[sweet shard]

See
https://youtu.be/4hIh8ujylWE

[lime verge]

Thank you but this is not what I asked.

[lime verge]

For example, on this picture, the angle between V1 and V2 is positive. The angle between V2 and V1 is negative. I would like to know if there is a formula which would allow me to know which way it goes.

In a 2D plane, of course.

[topaz sinew]

@lime verge Has your question been resolved?

[lime verge]

Welp

[sweet shard]

Why would one be negative and the other be positive?

Are you assigning it by a rule you made up?

[past kayak]

If anything, I would think that it'd be the other way around, since positive angles go CCW.

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[lilac grail]

Hey can I get some help?

[lilac grail]

Marginal cost/economic topic related questions

<@&286206848099549185>

[leaden tusk]

read

[lilac grail]

I'm not really sure how to go about calculating it with out prices

<@&286206848099549185>

[leaden tusk]

you already pinged helpers, and read

• After 15 minutes, feel free to ping @Helpers.

[lilac grail]

sorry...

<@&286206848099549185>

[leaden tusk]

bro

stop pinging helpers

[vernal vale]

👀

[lilac grail]

I'm sorry it's just that it's been >15 minutes,

[vernal vale]

ye they waited

[leaden tusk]

they pinged 3 times

[lilac grail]

It's fine, I'll just leave the server...

[vernal vale]

wait

i wanna answer your question

im just trying to remember basic econ

yea ngl i dont remember

wed write the linear combination right?

so like jak can catch 8 fish or find 12 pineapples

then each fish he forgoes catching gives him opportunity to find another 1.5 pineapple

so $\frac 1 8 F + \frac{1}{12} P = 1$

[thorny flame]

jan Niku

[vernal vale]

assuming he uses his whole day

[lilac grail]

So for the second it would be 2?

[vernal vale]

second?

[lilac grail]

Daxter

[vernal vale]

well theyd each have 2 costs, i believe

so 4 total answers

i think

no one is helping so i wanna try

you have some marginal cost for fish

its in terms of pineapples

and visa versa

since pineapples are the thing you give up to get fish

make sense?

[lilac grail]

sort of

[vernal vale]

like just assume they spend a full day doing stuff

they arent sitting around doing nothing

they can either spend time fishing, or spend time getting pineapples

only so many hours in the day

if you decide that youd like to have 3 pineapples, thats fine

but you have to spend time finding them

during that time, you could have caught 2 fish

so thats your marginal cost for pineapple

to increase your production of pineapple by 1, you give up 2/3 fish

you usually just think about it like

well i do

as creating this function

this gives you how much it costs to produce either a fish or a pineapple, since they each take time and you only have one day

then you think how many of one do i have to give up to get one of the other

thats marginal cost

idk do you have specific confusion? @lilac grail

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[daring oak]

could someone help me with this one pls

[static void]

u subsitution right

[daring oak]

what would I sub u for?

[static void]

uh

e^x + pi/2?

[daring oak]

and then where would i go from there?

[static void]

evaluate the definite integral

i could be wrong idk

[daring oak]

by finding the derivative of u?

[static void]

yes

remember to change your limits

[daring oak]

du would just be e^x, right?

[static void]

yes

du = e^x dx

@daring oak i think im wrong

so we'll have to see if someone else knows

[daring oak]

okay, i was a little stuck, thank you

anyways

[mortal thunder]

._.

can't you just differentiate the options?

1. for example, is outright incorrect !

[daring oak]

like im supposed to do everything in the most efficient way possible

(i mean the question itself is not graded, its just practice)

please ping me if anyone knows

[neon iron]

Not sure but the first step is propably to take that adding pi/2 rotating by 90 degress we know that tan(x+π/2) = -cot(x) so ur function f(x) can be written as -cot(e^x) that's all I can think of until now

And e^x is a value between 0 and π/2

Including 0 and π/2

[wheat kindle]

Am I misundertanding? Why isn't it just d by the fundamental theorem of calculus? No need to work anything out.

[neon iron]

Fun fact I just read option d now and was like

[neon iron]

Something about what she means here

Probably

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[latent tusk]

Hello, my friend and I struggle with this question ://

im still stuck on questions a & b

[sweet shard]

show your current work

[latent tusk]

hold on

so far, im done on question a

but idk about question b

[cerulean shale]

Nice

well you see...

I’m not good at math, good luck friend.

[latent tusk]

huh 💀

well, thanks lol

<@&286206848099549185>

[cerulean shale]

Are you bad or good at math Wally?

[latent tusk]

I'm pretty neutral i guess?

[cerulean shale]

oh okay cool

[crisp anvil]

a is correct

For b you just have to set up linear equation, A = B

[latent tusk]

ohh

well, they want me to solve 3x+2 and -x/4 - 21/4 simultaneously and idk how lol

[crisp anvil]

So that means you now have two equations $y=3x+2$ and $y=-\frac{x}{4}-\frac{21}{4}$

[thorny flame]

Mango

[crisp anvil]

I would multiply both sides of C by something so that you get nice integers instead of annoying fractions

[latent tusk]

ah alright

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[latent tusk]

Yep

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[neon iron]

Hello there! So I got super sick and missed school and apparently we just started a new unit and I don’t even know where to start.

[neon iron]

Actually, it’s 1:53 am and I gotta go to bed. 😂

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[toxic dust]

In a meeting, there are $12k$ people, each one of them know $3k+6$ people, and for any 2 people, there are $n$ people know them, how many people in this meeting?

[thorny flame]

謝墨離

謝墨離

[toxic dust]

And I need to prove it

My teacher tell me to use a chart

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@toxic dust Has your question been resolved?

[vale jacinth]

Just a little thought. I denoted those people vertices, I connected them with blue edges if they know each other, red edges otherwise. I tried calculating the number of triangles who have two blue edges and one red edge in equivalent ways. But I got nowhere… that number could be 12k(9k-7)n/2 right? But I didn’t find another equivalent way to calculate that number so that I could get an equation…. Maybe discrete-math channel or combinatorial-structure channel could help you…

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[true crow]

In how many ways can you choose the first 20 positive integers such that no two are consecutive integers

[topaz sinew]

Please don't occupy multiple help channels.

[true crow]

。

Anyone know

Bro

In how many ways can you choose the first 20 positive integers such that no two are consecutive integers

[sweet shard]

Did you learn pigeon hole principle yet

[true crow]

wrote question wrong

this is question

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[mellow venture]

how can u find if the series sin(npi+(1/n) is convergent or divergent

[mellow venture]

yes

[mortal thunder]

First, notice this is simply: $$\sum_{n=1}^{\infty} (-1)^{n} \sin \qty(\frac{1}{n})$$

[mortal thunder]

now, use ratio test

[mellow venture]

what if it was sin(2npi+1/n)

[mortal thunder]

same thing

[mellow venture]

but with factor contant 2?

Constant

[mortal thunder]

not quite

$$\sum_{n=1}^{\infty} \sin \qty(2n\pi + \frac{1}{n}) = \sum_{n=1}^{\infty}\sin \qty(\frac{1}{n})$$

[mellow venture]

because 2npi is always 0?

[mortal thunder]

because 2pi is the period

[mellow venture]

for cos(npi+1/n) it would be cos(1/n)? which would just be divergent

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[static crest]

I am stuck at I and II
I know the definition of union
But I don’t know what assumption we still have to make before applying the definition of union

[topaz sinew]

@static crest Has your question been resolved?

[static crest]

<@&286206848099549185>

[topaz sinew]

@static crest Has your question been resolved?

[static crest]

<@&286206848099549185>

Can anyone please help

[topaz sinew]

@static crest Has your question been resolved?

[static crest]

<@&286206848099549185>

Please help me I have been stucking here for a day already

Oh

So the question is about showing the union of the power set of C and the power set of D is a subset of the power set of the union of C and D

And the question is like fill in the blanks

Each of the roman number represents some words, expressions or even calculations

I guess (I) is the assumption part of the question but I am not sure what exactly it could be

[static crest]

<@&286206848099549185> but what assumption I still have to make in (I)? It doesn’t really sounds like I have to pick an element inside S

[sweet shard]

S is an element of the left hand side

$S\in \mathcal{P}(B) \cup \mathcal{P}(C)$ should do it

[thorny flame]

riemann

[static crest]

Oic

Suppose C,D are sets.
Pick any object S.
Let $S\in \mathcal{P}(C) \cup \mathcal{P}(D)$
Then $S\in \mathcal{P}(C)$ or $S\in \mathcal{P}(D)$, by the definition of union.

[thorny flame]

Trenton

[static crest]

Do u mean by this?

[static crest]

Suppose $S\in \mathcal{P}(C)$. Then S is a subset of C by the definition of power set.

[thorny flame]

Trenton

[static crest]

We verify that $S$ is a subset of $C\cup D$ (according to the definition of subset relation and union):

[thorny flame]

Trenton

[static crest]

Pick any object $x$. Suppose $x\in S$.

[thorny flame]

Trenton

[static crest]

Since $S$ is a subset of $C\cup D$ and $x \in S$. Then by the definition of subset relation, $x \in C\cup D$.

[thorny flame]

Trenton

[static crest]

Therefore by the definition of union, we have $x\in C$ or $x\in D$.

[thorny flame]

Trenton

[static crest]

@sweet shard But now I stuck at VIII. I am not sure how to turn the things into power set, or is there actually any theorem to apply.

[static crest]

<@&286206848099549185> can anyone help me about what should I do next ? What should I make in (VIII)?

[topaz sinew]

@static crest Has your question been resolved?

[elfin sparrow]

@static crest

You've just shown that $(x\in S)\implies(x\in C\cup D)$. This is the definition of a subset.

[thorny flame]

tatpoj

[topaz sinew]

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[static crest]

.reopen

[topaz sinew]

✅

[static crest]

Yes I shown that but I am going to figure out what VIII is. So we need to use the power set notation.

Yet I just don’t know what I can do next to make the prove logically make sense at VIII

Can anyone help me?

[static crest]

<@&286206848099549185> can anyone help me please

[static crest]

<@&286206848099549185> please help me

[static crest]

<@&286206848099549185> sorry for pinning again and again but I really don’t know what should I do next. So I know x in S and x in C U D. But why we can conclude that S is in the power set of (the union of C U D)?

I mean the part (VIII) in the question

[sweet shard]

What's your full proof so far? I don't know what's remaining to show

[static crest]

Suppose C,D are sets.
Pick any object S.
Let $S\in \mathcal{P}(C) \cup \mathcal{P}(D)$
Then $S\in \mathcal{P}(C)$ or $S\in \mathcal{P}(D)$, by the definition of union.

[thorny flame]

Trenton

[static crest]

Suppose $S\in \mathcal{P}(C)$. Then S is a subset of C by the definition of power set. We verify that $S$ is a subset of $C\cup D$ (according to the definition of subset relation and union):

[thorny flame]

Trenton

[static crest]

Pick any object $x$. Suppose $x\in S$. Since $S$ is a subset of $C\cup D$ and $x \in S$. Then by the definition of subset relation, $x \in C\cup D$. Therefore by the definition of union, we have $x\in C$ or $x\in D$.

[thorny flame]

Trenton

[static crest]

This is my question

So now I finished (I) to (VII) already

And I stuck at (VIII)

I guess (IX) is “$S\in \mathcal{P}(C\cup D)$”

[thorny flame]

Trenton

[static crest]

So I just don’t know how to relate the aforementioned to (IX)

[sweet shard]

Your VII is probably the same as VIII

Or x in S and S is a subset of the union

[static crest]

Ok thank you, get it

[static crest]

.close

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[static crest]

.reopen

[topaz sinew]

✅

[static crest]

So I figured out some statements successfully:
(I): if $(x\in A \cap B)$ then $(x\in A)$

[thorny flame]

Trenton

[static crest]

(II): Suppose $x\in A\cap B$

[thorny flame]

Trenton

[static crest]

(III): $x\in A$ and $x\in B$

[thorny flame]

Trenton

[static crest]

(IV): $x\in A$

[thorny flame]

Trenton

[static crest]

(V): For any object $x$, if $(x \in A)$ then $(x\in A\cap B)$

[thorny flame]

Trenton

[static crest]

(VI): Pick any object $x$. Suppose $x\in A$.

[thorny flame]

Trenton

[static crest]

(VII) $x\in A$

[thorny flame]

Trenton

[static crest]

(VIII): $x\in A$ and $x\in B$

[thorny flame]

Trenton

[static crest]

(IX): $x\in A\cap B$

[thorny flame]

Trenton

[static crest]

(X): $A\cap B\subset A$

[thorny flame]

Trenton

[static crest]

(XI): $A\cap B=A$

[thorny flame]

Trenton

[static crest]

(XII): For any object $x$, if $(x \in A)$ then $(x \in B)$.

[thorny flame]

Trenton

[static crest]

(XIII): $x \in A\cap B$

[thorny flame]

Trenton

[static crest]

I doubt (XIII) is correct or not

And also I cant figure out what (XIV) is

Can anyone help me?

[static crest]

<@&286206848099549185> can anyone help me?

[elfin sparrow]

I probably can, let me read through what you've got already, one sec

[static crest]

Thank you

[elfin sparrow]

VII is not correct

In part VI, we assumed $x\in A$

[thorny flame]

tatpoj

[elfin sparrow]

That, by itself, does not prove that x is also in B

The wording suggests we should be using the subset relation

I think this would make more sense:

Since $A \subset B$, we have $x\in B$ by definition of the subset

[thorny flame]

tatpoj

[elfin sparrow]

Wait, hang on, I'm sorry

Oh wait, yeah, what I said is correct

[static crest]

Ok

Let me change it

[elfin sparrow]

Does that make sense to you?

[static crest]

Yes. I just forget to mention A is a subset of B

[elfin sparrow]

IX and X are good. Give me a sec to read through the last part

[static crest]

Ok thank you

[elfin sparrow]

XIII is not correct.

[static crest]

Yes I think that is kinda strange

[elfin sparrow]

In this part, we are trying to prove that $A \subset B$, so making the assumption $x\in A \cap B$ doesn't get us anywhere

[thorny flame]

tatpoj

[static crest]

It is repeated and appearing in both the assumption and conclusion

[elfin sparrow]

Since we're trying to prove that A is a subset of B, we should assume x is in A, and then prove that x is in B.

Does that give you a clue as to what we want in XIII?

[static crest]

Ok let me try

Thank you

[elfin sparrow]

Sure thing

Sorry you had to wait so long lol

[static crest]

No, it is my words too lengthy

[elfin sparrow]

For XIII?

[static crest]

What if I write $x\in A$ for XIII

[thorny flame]

Trenton

[elfin sparrow]

Exactly

[static crest]

Ok

[elfin sparrow]

This is how we prove that A is a subset of B. Assume x is in A, then prove x is in B

[static crest]

So for the XIV part I just apply the assumption of $A\cap B=A$ and $x\in A$ ?

[thorny flame]

Trenton

[elfin sparrow]

Yes. I think it's sufficient to just put $A \cap B = A$ for XIV

[thorny flame]

tatpoj

[static crest]

Ok thx

Then for the XV, I apply the definition of intersection, such that $x\in A$ and $x\in B$

[thorny flame]

Trenton

[elfin sparrow]

Yes

[static crest]

Umm but then I am not sure what should I put in the XVI

Which is the “In particular”

[elfin sparrow]

Remember what our goal is.

We want to prove that A is a subset of B. We assumed x is in A, now we need to show x is in B.

[static crest]

Umm so I specify “for any object $x$, if $x\in A$ then $x\in B$”

[thorny flame]

Trenton

[elfin sparrow]

Not quite

[static crest]

Umm

I just change it into conditional

[elfin sparrow]

I think, since we just stated $x \in A$ and $x \in B$, now it makes sense to say, "In particular, $x \in B$."

[thorny flame]

tatpoj

[elfin sparrow]

Since what we wanted to ultimately show was that x is in B

[static crest]

Oh i see

Thank you so much

[elfin sparrow]

No problem

[static crest]

.close

[topaz sinew]

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[static crest]

(I): Suppose $x\in A$

[thorny flame]

Trenton

[static crest]

(II): $subset$ relation

[thorny flame]

Trenton

[static crest]

(III): $x\in B$

[thorny flame]

Trenton

[static crest]

(IV): $x\in A$

[thorny flame]

Trenton

[static crest]

(V): $since$

[thorny flame]

Trenton

[static crest]

(VI): $x\in B$

[thorny flame]

Trenton

[static crest]

(VII): $and$

[thorny flame]

Trenton

[static crest]

Sorry can you help me for this

I got confused in this question

The question include $x\not\in B$

[thorny flame]

Trenton

[static crest]

But I cant find where can I deduce that

[topaz sinew]

@static crest Has your question been resolved?

[static crest]

<@&286206848099549185> can anyone help me about it

[static crest]

<@&286206848099549185> can anyone help? I am not sure about the assumption part (I) is correct or not

[loud dawn]

Just trying to match these up

god this cookie cutter fill-in-the-blank thing is so lazy

Ok so it looks like

man @static crest are you sure you can't just write your own proof

genuinely, seriously, it might be easier to just write a proof yourself and then try to fill in these stupid blanks

[neon iron]

Yeah fill in the blank math is just 😮‍💨

[static crest]

I also want to… but such thing is my homework

[loud dawn]

ok so just write your own proof, and loosely follow whatever framework they're using

if they say proof by contradition, do a proof by contradiction

and then it'll be easy to fill in the blanks

[static crest]

Ok

[static crest]

Ok I tried but I don’t get what kind of contradiction the question want me to do

So there is actually three assumptions

$A\subset B$ $A\subset C$ $B\subset C$

[thorny flame]

Trenton

[static crest]

But neither of them such that the complements the sets are C\B or C\A

[static crest]

<@&286206848099549185> can anyone help

I really cant think how to make contradiction

[static crest]

<@&286206848099549185>

[static crest]

<@&286206848099549185> So how can I use the method of contradiction to show that C\B is a subset of C\A?

Can anyone give me some hints

[static crest]

<@&286206848099549185> sorry for pinning again and again but I really want to get this problem solved. I am going to have exam very soon but I still not familiar with this stuff

[mortal thunder]

@static crest

Do you know what goes into (I) ?

[static crest]

Umm

How about “Suppose $x\in C$ without B”

[thorny flame]

Trenton

[mortal thunder]

Is that fine?

[static crest]

I am not sure

[mortal thunder]

$x \in C - B$

[thorny flame]

Ansh

[static crest]

Ok

[mortal thunder]

or C \ B

according to your notations

what do you think would make most sense for (II)?

[static crest]

Oh so (II) should be “complement”?

[mortal thunder]

okay

how does that make sense?

[static crest]

Umm so we have $x\in C$ and $x\not\inB$

[thorny flame]

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

[mortal thunder]

I mean,

if you put "complement" in "II", what is your answer for "III"?

[static crest]

Umm can it just be this?

I mean “x in C and x is not in B”

[mortal thunder]

By the definition of "compliment", we have "x in C and x is not in B"

?

[static crest]

Umm idk if it make sense

[mortal thunder]

Do you not read english and know a little set theory?

[static crest]

Umm you mean I use the set language?

[thorny flame]

Ansh

Ansh

[mortal thunder]

so combine that with your "by definition of compliment" and it does not make any sense

[static crest]

Umm

How about apply subset relation definition

All but then I don’t know how to continue, seems I am in the wrong track

[mortal thunder]

Anyways, if you cannot guess, can you read the next line for me please?

[static crest]

Pick any object x, suppose $x\in C-B$. By the definition of subset relation, we have $x\in C-A$.

[thorny flame]

Trenton

[static crest]

Alright I guess it is quite problematic, as I am using the assumption part

[static crest]

Perhaps the set I choose in (I) already not correct

[mortal thunder]

^^"

I think you probably gave too much time to this question?

[static crest]

Yes…..

[mortal thunder]

Let's talk after an hr if you need to chill

[static crest]

Umm

Chill?

#chill

Do u mean this

[mortal thunder]

assumption part was that A subset of B. For your proof, you're picking an x from C - B, and by definition of set difference, x is not in B. Then the context hints you that you'll be verifying that x is not in A either, by proof of contradiction in the next paraphrase

[mortal thunder]

no, I mean, give your brain some time to chill if you've been working on this for a while.. It's common to stress out and skip trivial arguments when you're overheated

[static crest]

I see haha

[static crest]

Ok

[thorny flame]

Trenton

Trenton

Trenton

[mortal thunder]

no breaks? 👀

excuse me?

[static crest]

I think I am ready now lol. Just eat some candies

XD

[mortal thunder]

No you're clearly not

[static crest]

Oh I make a mistake

[mortal thunder]

$$x \in C - B \implies x \in A \land x \notin B$$?

[thorny flame]

Ansh

[mortal thunder]

what?

[thorny flame]

Trenton

Trenton

[static crest]

So the first line should be “Suppose $x\in A$”?

[thorny flame]

Trenton

[mortal thunder]

x has already been defined

as an element in C - B

[static crest]

Oh so the next line we can apply the definition of the subset relation

[topaz sinew]

@static crest Has your question been resolved?

[thorny flame]

Trenton

[static crest]

@mortal thunder is this what you mean?

[mortal thunder]

Till here, yes

I'd have to read after that to reassure you but I'm busy rn

[static crest]

Ok thx

[mortal thunder]

btw yes, that makes sense

that'd verify x is not in A

which was the intended approach

VIII should be x \notin A

[static crest]

Ok

Therefore $x\notin A$

[thorny flame]

Trenton

[static crest]

Hence by the definition of complement, we have $x\in C-A$.

[thorny flame]

Trenton

[static crest]

It follows that $A\subset B$.

[thorny flame]

Trenton

[thorny flame]

Trenton

Trenton

Trenton

Trenton

Trenton

Trenton

Trenton

Trenton

Trenton

[static crest]

When you are free, feel free to take a look. Thanks

[topaz sinew]

@static crest Has your question been resolved?

[topaz sinew]

@static crest Has your question been resolved?