help-39

[patent vessel]

With two roots in the denominator

Try subbing smth to remove the root

Wait no that’s wrong, if x^2 = 1/(1-t), how is 1-t in the denominator

[boreal slate]

i got 0.5x in the numerator

and then x=1/root(x-1)

[patent vessel]

Ha fine yes

[patent vessel]

.

[boreal slate]

yep, i did u=root(t-t^2)

worked out

thx

.close

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[maiden gulch]

hello i need help understanding this i am grade 6 so i am pretty dum lmao

[sterile python]

Terms are individual components that are seperated by + - signs, which is definition

[maiden gulch]

alr

[sterile python]

For question 1, you can see there are 3 components that seperated by + -

Which is 3 terms

For question 2, do you see any - + signs?

[maiden gulch]

ye

[sterile python]

Right, do you know if (a+b)/c will be a/c + b/c?

[maiden gulch]

not yet

[sterile python]

This form

[maiden gulch]

ah ok

[sterile python]

In which, you can see (2x+1)/ can be splitted out

[maiden gulch]

mm alr

[sterile python]

For question 3, im not sure if your teacher ask for fully expanded form or grouped form?

[maiden gulch]

sure

[sterile python]

Can you confirm with your teacher, but i will solve both of them

Idk if you learnt about algebraic identities so i assume it aim for grouped (no 1)

[maiden gulch]

alr

[sterile python]

Then base on the previous question what number of terms you think it would be for question 4?

[sterile python]

Yes so i was right, which is case 1 (grouped)

[maiden gulch]

mhm ok

[sterile python]

Can you try to solve the rest based on the previous solving?

[maiden gulch]

alr

[sterile python]

If you really understood you can type .close, else return here to ask, happy to help🙌

[maiden gulch]

Thank you

thx alot 🫡

[sterile python]

No worries!🙌

Type .close to close the chat if you finish💯

[maiden gulch]

.close

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[crystal dew]

<@&268886789983436800>

[pearl pond]

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[crystal dew]

holy speed

[shell brook]

ikr

makes me feel scared that they're watching my every move

waiting for me to slip up

but unfortunately for them, thats exactly why i made a group chat

[nocturne grail]

omeganato mod watches MY every move and waits until I misinform a helpee and then calls me out for it /s 😔

[shell brook]

deserved

[proper nova]

🥀

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[safe prairie]

for q6bi and 6c, i was copying what the lecturer wrote on the board most of the time, except for those in yellow which i just wrote. i dont understand how 5/4 came to be

[glass meadow]

,, \frac59 \times \left(\frac23\right)^{n-2} = \frac59 \times \left(\frac32\right)^2 \times \left(\frac23\right)^n = \frac59 \times \frac94 \times \left(\frac23\right)^n

[jolly parrot]

Nel

[safe prairie]

wow

alr thx

.close

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[pearl pond]

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[neon spoke]

can somebody explain to me how the teacher used this circle type of thing to solve the following function,{sinx>rooting3/2 and tgx<-1 and the answers were x e(-60,360k,240+360k),x e (90+180k,135+180k) and then U (360+360k,315+360k)

[wide pawn]

every point on the unit circle (the circle in the picture) is
(cosx,sinx)

so you can see that the places where the y>sqrt(3)/2 (where y coordinate sinx) occur in between 60 and 120

[neon spoke]

how does that work tho,i am confused

[wide pawn]

which part of it is confusing to you, the coordinate part or?

[proper nova]

$\sin x >\frac{ \sqrt{3}}{2}$

[jolly parrot]

1 divided by 0 equals Infinity

[neon spoke]

yeah like how do i know where it potentially lyes

like how did you get between 60 and 120?

[wide pawn]

you can see that they've marked angles right?

like on the line joining the coordinate and centre

[neon spoke]

yeah?

[wide pawn]

yes those are the angles corresponding to the coordinate shown

so that coordinate is (cos(60),sin(60)) = (1/2 , sqrt(3)/2)

if you increase the angle more, you can see that the value of sinx also increases

cause the points on the circle are going higher

[neon spoke]

alright

[wide pawn]

imagine a rod rotating

that rod is of unit length

and it's making an angle theta with the xaxis

so if you were to calculate the point of the end point of the rod (which is just how much u need to go in x direction and how much u need to go in y direction) u get (costheta,sintheta)

[neon spoke]

okayy

so sinx and cosx are really just end points?

[wide pawn]

yeah it corresponds to the end point of the rod

[neon spoke]

i see it kinda

[wide pawn]

u can see at 90 degree costheta becomes 0

and sintheta becomes 1

because the rod is vertical

and at 0 sintheta is 0 and costheta is 1

[neon spoke]

i can kinda understand it now

but how do i begin the thing?

like if i were to try and solve it on a fresh circle

[wide pawn]

as in?

[neon spoke]

like let's take this equation for example

sinx>rooting3/2

[wide pawn]

(it's called inequality btw!)

[neon spoke]

and i need to present it here

[wide pawn]

ah right

[neon spoke]

so what do i do first

then the thing you said

[wide pawn]

Now I'm not too sure how your school/institution requires you to do this so I would recommend checking with them/peers to be safe
but I would begin by plotting the points where sinx=sqrt(3/2)
and then draw their corresponding angles
after that I would shade the part in between (that's where sinx>sqrt(3)/2 in this case)
that's our particular solution, you can add the period of sin to get all solutions

[neon spoke]

okay uh basically we drew a circle

then put lines in different places starting from the middle

then shaded certain spaces out till we got to the only possible solutions and from there presented it

but i am genuinly so confused on how to do that

[wide pawn]

oh, when you were putting lines did you put all the lines like here?

[neon spoke]

yeah kinda like that

just way less

[wide pawn]

ah alright

that probably means you put lines corresponding to only where sinx=sqrt(3)/2

[neon spoke]

i avoided sending this cs of my bad handwriting but this is the thingy

i payed 0 attention while she was explaining it

[wide pawn]

hmm yes that doesn't help much but I get the essence

[neon spoke]

so is it done the same way you explained it

or

[wide pawn]

I think it is but the solution you've copied down does not refer to sinx>sqrt(3)/2 infact the first one is for sinx>-sqrt(3)/2

that's why you've shaded most of the circle

[neon spoke]

maybe?

i'm not sure maybe i missed it

[wide pawn]

hold on I'll do sinx>sqrt(3)/2 for you

[neon spoke]

but can you just explain it for my next inequality

[wide pawn]

so as you can see, for points in between 60 and 120 the value sinx will be more than sqrt(3)/2

cause the y coordinate is sinx 's value

[neon spoke]

it's only up there cs the coordinates are only positive?

i get it now

[wide pawn]

you will present the solution as (60+360k, 120+360k)

[wide pawn]

yeah, it's only up there cause for values greater than sqrt(3)/2, that part of the circle is over that value (sticks up)

[neon spoke]

can i try to do it with tgx>-1?

[wide pawn]

tgx is tanx?

[neon spoke]

the thingy is only between 180 and 270 right?

[neon spoke]

yeah

[neon spoke]

or wait wasn't the circle different

[neon spoke]

my mistake

[wide pawn]

tan(180) isn't -1

first you'd have to solve where tanx is -1

you're right about 270

wait no

that's not right

yeah I'd recommend solving tanx=-1 first

refer to the unit circle for help

[neon spoke]

it's tan of 135

[wide pawn]

tanx=sinx/cosx
so where ever you put ycoordinate/xcoordinate and it comes to be -1, that's where tanx=-1

[neon spoke]

tgx<tan135

[wide pawn]

yes that's right

[neon spoke]

so between like 0 and 135?

[wide pawn]

No no, try doing it on the circle, also you haven't found all values where tanx is -1

[neon spoke]

so on this thing

[wide pawn]

yes

[neon spoke]

i think it's between 45 and 135?

am i looking right?

[wide pawn]

tan45 is 1

not -1

[neon spoke]

135 and 315?

[wide pawn]

yep

[neon spoke]

wait but isn't x smaller?

[wide pawn]

you can't apply the same logic as sinx and cosx here unfortunately

since tanx is a ratio of both

[neon spoke]

alr

[wide pawn]

you can notice that in between 135 and 315 there's places where tanx>-1, wherever that's happening that's your solution

so basically u just need to find that side

u can find this for a single value and then that whole thing in between is your solution

[neon spoke]

btw the solution is x=135+360k,315+360k?

[wide pawn]

uhh yes? but I'm not sure if you're required to put that in standard form

[neon spoke]

she put it like that so i think yes

but just one last thing

how do you get the U

[neon spoke]

like from here

[wide pawn]

ah that's because tanx is not defined at 90 degrees

so we skip that one

that's why we put a U to join the two intervals

essentially the values at the end of the intervals are not included

so if you have (1,5) then 1 and 5 are not included, but everything in between is

at tan(90) it becomes undefined due to it being 1/0

so we remove that from out solution

it also happens at 270

[neon spoke]

so how did we get the 360 degrees?

[wide pawn]

so keep that in mind

[wide pawn]

the 360 degrees is because for sinx, cosx, tanx, any combination of these
tan(x+360)=tanx sin(x+360)=sinx and so on
it's because if you think about it, if you turn 360 degrees in a circle, ur at the same spot, so you have the same value

so for any multiple of 360 u will be at the same spot again

hence we add 360k

to get all solutions

[neon spoke]

i get it now

thank you so much man <3

[wide pawn]

Np!

[neon spoke]

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[crystal summit]

Given is a trapezium, where the diagonals are perpendicular to each other. Prove, that the sum of the lengths of diagonals squared is equal to the squared sum of lengths of the bases.

[blissful cosmos]

I NEED HEE-EEEE-EEEEEEEEEEEEEEEEEEEEEAEEEEEEEEEEEEEEEEEEEEELP

AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA

[pearl pond]

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

[crystal summit]

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[heady kiln]

Hi

[daring nymph]

!da2a

[pearl pond]

Asking the actual question right away is more likely to get responses.

Asking "Can I ask...?" or "Does anyone know about...?" doesn't give people enough information to decide whether they can help, and answering can feel like a promise to help with the actual question, which they might find themselves unable to.

@heady kiln Has your question been resolved?

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[fair thunder]

I am stuck trying to figure out all the steps in doing this question. I’ve only gotten down to taking the derivative of the parabola.

[safe prairie]

Does super soaker spray fowards(right) or backwards(left)

[fair thunder]

forwards right

[safe prairie]

try solving for all these

[pearl pond]

@fair thunder Has your question been resolved?

[fair thunder]

I got down to here but I get the point being (17.17, 137.40)

[pearl pond]

@fair thunder Has your question been resolved?

[simple niche]

Let A be the point where Snoopy must fire to soak Red Baron.
Then A = (a, a²/2 - 10)

Since The stream of water is tangent to the Snoopy's Flight Path,
The equation for the water = d/dx [x²/2 - 10] = x.

The gradient of the water at point A = a.

The equation of the water at point A:
y - (a²/2 - 10) = a(x - a)

This equation passes through (8,0), so substitute x = 8 and y = 0 we get
-(a²/2 - 10) = a(8 - a)

Solve for a, then substitute that into the point (a, a²/2 - 10).

And there you have it.

[umbral tangle]

this was fun doing

[fair thunder]

I see, that makes sense😭

thank you!

[simple niche]

Beware: You will obtain two values of a.

So it's very important to substitute the points and choose the right one.

[simple niche]

Welcome.

[fair thunder]

.close

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[umbral tangle]

you can mess around with a as a slider here:

[fair thunder]

I was trying to do that but had no idea how to use desmos😭

[simple niche]

Kind of easy.

[umbral tangle]

oh lol. Hopefully you can understand what goes on in there.

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[trail falcon]

5/x=2

[trail falcon]

I don’t know how to solve

[tender wave]

multiply both sides by x

[trail falcon]

So 5x and 2x

[tender wave]

close

2x is correct

[trail falcon]

That’s what I’m saying how isn’t 5 like 5x

Ur multiplying it by x

[tender wave]

5/x * x = 5

[trail falcon]

Khhhhh

The two x cancel

[tender wave]

ye think of it like 5 divided by x, times x

[trail falcon]

And since 2 doesn’t have any x it becomes 2x

Now what do we do

We have 5 and 2x

[wooden flare]

how would you get x by itself?

[tender wave]

5=2x

divide by 2 on both sides

[trail falcon]

Wait how are they equal

[trail falcon]

U divide it

[wooden flare]

your finding a value for x that makes the equation true

[tender wave]

we had 5/x = 2 then multiplied both sides by x to get 5=2x

[trail falcon]

So we move the 2 to left so it becomes 5/2=x

Okay thank you

[wooden flare]

$5/x = 2$

$5 = 2x$

$x = 5/2$

[jolly parrot]

KB

[pearl pond]

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[safe prairie]

im working on q9b, can someone point out whats wrong from my methods?
i gave cartesian coord (0,0) to A
the red highlights are all shorthand for “Pr(reaching point(x,y) from (0,0))”

[safe prairie]

btw for 9a i got 6c4 / 7c4 = 3/7 but its probably not relevant for 9b

[summer imp]

I think you’re missing some parentheses in the first few lines where the red is and it’s making you lose some 1/2s

[safe prairie]

yeah its pr31 i think

ima work on it

yes it is
i think ill be cooked in an exam for conceptualising the solutions because even thought bottom left is a concise way to get a correct answer i took too long to come up with that

[pearl pond]

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[slender stream]

<@&268886789983436800>

[pearl pond]

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[median abyss]

I need help matey

I am cooked here gang

Helpers

<@&286206848099549185>

I know how to solve quadratic simultaneous equations but I don't get anything else in this question.

Pls help me

As I have an exam on this

@digital garnet it ain't that hard
Aren't u experienced u 18

Sorry

Man sleep

.close

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[sterile python]

@median abyss sorry for late reply but i used viète theorem hope ts helps😭

[sterile python]

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[crystal dew]

!nosols btw

[pearl pond]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[sterile python]

My bad i saw he slept so gotta help

[crystal dew]

you can help, but try to probe OP's understanding and work through it first

treat full solutions as a last resort

[sterile python]

Okie gotchu❤

Last resort?

[crystal dew]

if the OP is so badly stuck that nothing else works

[sterile python]

Okay will keep in mind i saw the OP slept and exam next so try to help but sure🙌

[pearl pond]

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[north talon]

Determine the smallest $a \in \mathbb{N}$ such that for every $x \in \mathbb{N}$, the function $f(x) = x^2 + a^4$ is not prime

[jolly parrot]

Copter

[tropic saddle]

doesnt hurt to first just try a couple small values a I suppose

[jolly parrot]

woody6978

[north talon]

ive noticed they fail at x = 1,4,9,...

is there something about that or just a coincidence

[north talon]

im not from the states, but yeah N doesnt contain 0 probably

[night frost]

oh ok

[jolly parrot]

woody6978

[north talon]

i mean those terms are always irr. so probably not

[pearl pond]

@north talon Has your question been resolved?

[pearl pond]

@north talon Has your question been resolved?

[north talon]

<@&286206848099549185>

[night frost]

try to find a way to get a factorization with the value of 'a' that you think works

[north talon]

i dont think any works?

unless sophie germain or something

[rustic gate]

did you happen to typo the question

x^2 + a^4 is so doomed

but x^4 + a^2 has an easy solution with sophie germain

[covert quiver]

lmao theyve been working on a probably open problem

[north talon]

nah💔

[north talon]

really😭

i may have been trolled

[dusty jungle]

you prolly cant factorise it cus discriminant is -4a^2 which is negative

idk

[night frost]

yes maybe but I wanted to factor it with the polynomial in a , like maybe a = Y^p something like that then x^2 +a^4 = x^2 +Y^(4p)=P(Y)

[pearl pond]

@north talon Has your question been resolved?

[pearl pond]

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[candid fjord]

I'm trying to get a better understanding of Cantor's Diagonalisation Proof. In the laymen's terms the proof was explained to me in, I can't for the life of me understand why the proof can't simply be mirrored to show exactly the opposite conclusion on the other side; that there are more natural numbers extending out to infinity then there are real numbers between 0 and 1.

My maths skills are basically hopeless, so I'm sure that my misunderstanding lies in some assumption I have. Not sure if it's possible to do this in a voice call or not.

[steep saddle]

what was the proof given to you

[humble lintel]

Basically any binary expansion of a natural number has a finite number of digits. If you try to mirror the argument, the diagonal number you generate is not an element of N

[candid fjord]

(Thanks for the help!) Well, it was explained verbally, but the jist was something like:

Two infinite lists, with the Natural numbers on one side, and the Real numbers on the other. And the idea is to go down through the list of real numbers changing a single digit of each number, but which digit changes by one place to the right for each number. This gives us a new number that's different than every other number in the list. Thus MORE numbers than the matched set of natural numbers.

Which, doesn't fully makes sense to me in the first place, but even if I just concede it, I don't understand why I couldn't do the same in reverse with the natural numbers.

It seems to me this only works because the infinite real numbers all have infinite trailing zeros. But if you assume the same is true for leading zeros with the natural numbers, then why can't I just do the same diagonal trick, but moving one digit to the left with each number in the set?

[candid fjord]

As I said, I'm sure I just have a mistaken assumption. So you're saying that infinite leading zeros on natural numbers are NOT the same as the infinite trailing zeros in real numbers?

And if that IS what you're saying, why not? What's the difference?

[glass meadow]

the infinite real numbers all have infinite trailing zeros
No, most real numbers do not end (in an infinite sequence of zeroes)

[candid fjord]

Well, not necessarily zeros no, but I mean that there are infinite... digit places?... to the right on real numbers

[glass meadow]

Yes, and those "digit places" contain non-zero digits

Unlike the leading zeroes

[candid fjord]

Well, most do, yes. Some don't. Some just have zeros going infinitely.

[glass meadow]

Yes

[candid fjord]

How does this differ from natural numbers?

[glass meadow]

Do you mean naturals?

[candid fjord]

I can just as easily imagine infinite zeros going left as I can going right.

What's the fundamental difference?

[glass meadow]

"most" is key here

The set of real numbers is written R

Among real numbers, there are some for which the decimal expansion (their "written" value) ends at some point

Note that I'm using "end" as in "the rest is trailing zeroes", that means the same

[candid fjord]

"Written" value. I mean, math doesn't actually deal with anything written, right? Like, the underlying math.

[glass meadow]

Like 0.35 and 0.350 are the exact same number

[candid fjord]

Well sort of

[glass meadow]

Decimal expansion is rigorous

[glass meadow]

Anyway, those numbers are the rationals

The set of rational numbers is written Q

[candid fjord]

0.350 has more information then 0.35 UNLESS we are just pretending that all the trailing zeros are there.

[steep saddle]

this isnt physics bro you can relax about sigfigs

[glass meadow]

No it does not

Not in this context

[candid fjord]

No, I can't relax about sigfigs because they're the whole point of this proof.

[steep saddle]

no they arent

[glass meadow]

I'm talking about exact values here

Significant digits only matter for approximate values, typically from physical measurements

Or if you're trying to approximate a number with a large or infinite number of digits

This is not what I'm doing

0.35 is an exact number

0.350 is also an exact number

They are the exact same number

[candid fjord]

Which MEANS that there are infinite zeros after that 0.35.

We're just not writing them for conveniance.

[glass meadow]

You can think of it like that if you want, yes

[candid fjord]

Which is the same for natural numbers

[glass meadow]

Well, let me finish

[candid fjord]

There are infinite zeros AHEAD of the number

We just don't write them

[glass meadow]

Yes that's right

[candid fjord]

Okay, so same on both sides, got it

So why can't we mirror?

[glass meadow]

So these are the rationals

The real numbers that are not rationals are called irrational

[candid fjord]

Oh boy

[glass meadow]

Yes?

[candid fjord]

I'm talking real vs natural, why are we into rational vs irrational?

[glass meadow]

As it turns out, there are the same "number" of rationals as naturals

However there are strictly more irrationals than rationals or naturals

[candid fjord]

Well, this is where part of my whole question stems from. This proof I'm not understanding is one that proves there are more Real then natural numbers. But I'm not convinced that it's true yet.

Now, we can go get into the proofs for the things you just said

But I'll probably end up having issues with them too lol

[glass meadow]

Since the rationals and irrationals make up all the reals, the fact that there are more irrationals than naturals also means there are more reals than naturals

[glass meadow]

The proof is basically the same

[candid fjord]

Right... but this one is so far, unconvincing to me.

So why would they be any more convincing?

[glass meadow]

Because your only objection was about trailing zeroes

Irrationals do not have trailing zeroes

[candid fjord]

My question, was why can't I go down the list of Natural numbers, changing one digit in each number as I move diagonally, thus creating a new number that DOESN'T appear in that list, thus proving the list is bigger than the matched list of reals?

[glass meadow]

Because you can only move diagonally a finite number of times

[steep saddle]

for every real between 0 and 1 there is a natural that is basically the same number without a decimal
so then we know that the number of naturals is at least equal to the number of reals, right

then we can add any number of trailing zeroes to those naturals
the difference is that adding trailing zeroes to a decimal doesn't create a new number

[candid fjord]

Why can I only move a finite number of times left on a natural, but I can move infinitely right on a real?

[steep saddle]

therefore there are more naturals than reals between 0 and 1

[candid fjord]

(Do you guys know of another server similar to this, but where people use voice? Because trying to type this all out is crazy.)

[steep saddle]

oh i got it flipped lol 💀

[candid fjord]

lol

Well, I mean, that's my entire issue.

I don't understand why we can't just flip this entire proof.

[glass meadow]

Ok I see what you mean now

[glass meadow]

This is not quite true

It's also not quite the point of the proof

[candid fjord]

I mean, either it's true in both directions, or it's not true in either. OR I'm missing something.

[glass meadow]

Do you at least agree that naturals are denumerable? As in you can enumerate them in an infinite list

[candid fjord]

Well, the way I see it, I can use this logic to prove EITHER set is larger.

Not exactly sure what that means.

[tropic saddle]

no explanation will work here if you cant agree with the fact that natural numbers have finitely many digits

[candid fjord]

((Again, sorry, my maths skills are TERRIBLE. I'm very much a laymen))

[glass meadow]

It means writing them down one by one, line by line, and every natural eventually appears in the list

[tropic saddle]

whatever number you would construct has infinitely many digits

[candid fjord]

Is there some reason or proof or explanation as to why I SHOULD agree that natural numbers have finitely many digits?

WTF is this? Is this spam?

[glass meadow]

<@&268886789983436800>

[tropic saddle]

just ping <@&268886789983436800>

[glass meadow]

If a natural number had infinitely many digits, it would be infinitely big

[candid fjord]

Well, not if all the leading digits are zeros?

[glass meadow]

If a number is infinitely big, well it's not really a number, it's just infinity

Oh again with the zeroes

047 is the exact same number as 47

[candid fjord]

Right

Like 0.3 is the same as 0.30000

[glass meadow]

If you remove leading zeroes, naturals have finitely many digits

[candid fjord]

The same is true on both sides

[glass meadow]

Do you agree with that?

[dim linden]

are you familiar with a construction of the natural numbers

[candid fjord]

Well, I don't know by what mechanism you feel you can just remove them?

[candid fjord]

Negative.

Sorry, definitely WAY out of my depth here.

[glass meadow]

The "mechanism" is equality

[candid fjord]

000000005 = 5

If they're equal, then I CAN use infinitely many zeros.

And thus, I CAN mirror the proof..

[glass meadow]

You keep jumping to conclusions

[candid fjord]

Yes

I'm sure I am

That's why I'm here

[glass meadow]

Yes, you sure can

[glass meadow]

No, you still can't

[candid fjord]

Why?

[glass meadow]

To begin explaining why I need you to agree with what I said just then:

If you remove leading zeroes, naturals have finitely many digits

If you don't, I'm just giving up at this point

[humble lintel]

The point is that any natural number has finitely many non-zero entries, the number of zeros don't really matter. If you try to do what you are proposing you will generate a number that isn't natural.

[candid fjord]

I mean... I can agree that IF removed, they have finitely many digits. But they don't ALWAYS have finitely many digits.

[dim linden]

something like 1111... is not a natural number. natural numbers are, by construction, finite.

[steep saddle]

you can mirror everything except for the repeating decimals

thanks discord wrong reply

[glass meadow]

Let's use a slightly different language then

Every number has an infinite number of equal representations

Simply due to the fact that you can add arbitrarily many leading zeroes to it

[candid fjord]

Same as real numbers with trailing zeros, yes?

[glass meadow]

Same as rationals with trailing zeroes

But also rationals (and irrationals) can have leading zeroes anyway

[candid fjord]

Yeah sure

[glass meadow]

From now on, I will only consider the main representation of a number

That's the representation that has no leading or trailing zero

[candid fjord]

Okay but what mechanism lets you do that?

[glass meadow]

Again, equality

[candid fjord]

But if the outcome CHANGES based on getting rid of those leading zeros, are they even equal?

[dim linden]

the outcome doesnt change

[glass meadow]

I mean if you disagree that 034 = 34 I can't help you

[humble lintel]

I think the best way to explain this is to embed the naturals into the padics, do the diagonal argument going the mirrored direction like they wanted and show where it fails. But I'm not going to open that can of worms.

[candid fjord]

And it seems like you're saying that the proof can't be reverse if we use the representations you want to use. But if they're equal, then why can't we just use the representations that I want to use, and show me where it breaks down using them?

If they're equal, then DON'T use the main representations, use the ones with infinite leading zeros, and with those, show me how it's wrong.

[tropic saddle]

the number you would construct doesnt start with leading zeros. it would start with leading 6s (or whatever other number the proof you read used)

zeros and other digits are not the same

a number starting with infinitely many nonzero digits is simply not a natural number

every natural number is finite

thats their whole point

[candid fjord]

Okay, this is interesting. Why not?

Hmmm

[glass meadow]

Right, if you use the representation with infinitely many leading zeroes, the natural you would construct by doing the "diagonalization" thing would need infinitely many non-zero leading digits

[candid fjord]

In the same way you need to do on the other side with the reals, right?

Like, you need infinite trailing... digit places..

[glass meadow]

Except with trailing digits

[candid fjord]

So same on both side, no?

[glass meadow]

The point is that a number with infinitely many leading non-zero digits would be infinitely big so is not actually a number

[candid fjord]

But how don't we have the same issue with the infinite amount of decimal places in the real numbers?

Why is infinite places okay on one side, but not the other?

[glass meadow]

Because the number is still bounded?

[steep saddle]

its just how theyre defined

[glass meadow]

A real number is somewhere on the number line

A number with infinitely many leading non-zero digits could not be on the number line

[candid fjord]

So we simply defined these two sets in a way where one is bigger than the other. But this isn't actuall related to fundamental mathematics? It's just our symantic use of words that create arbitrary bounds for these sets?

But that number line goes out to infinity doesn't it?

Which means there are an INFINITE number of possible digit places

[dim linden]

that doesnt mean you can have every digit place filled at once

[candid fjord]

What? They are filled by default

Like with trailing zeros?

[dim linden]

yes but leading zeros are different

[glass meadow]

I feel like you're just going to keep asking more questions no matter what answer we give you, but your new questions will require deeper mathematical knowledge than you have

[candid fjord]

This is likely true.

[candid fjord]

So I keep being told, but don't understand why.

[glass meadow]

I would invite you to learn undergraduate maths the long way until you get to the construction of the various number sets

[candid fjord]

Yeah, almost certainly not going to happen, lol. But I sincerely appreciate the attempts at help!

[glass meadow]

Well I'm afraid you won't get a satisfying answer then

[candid fjord]

Yeah, I couldn't find one anywhere else either, hence why I came here.

Curiosity be damned! Guess this proof is just going to remain above my head.

[glass meadow]

I mean you can probably still understand the proof if you're willing to concede a few things

[candid fjord]

Well, I mean, I already DO understand it if I concede things that I don't think I should be conceding, lol.

And no one has yet been able to explain why I SHOULD concede them.

[glass meadow]

I'm not sure you quite do, considering you didn't appear to know what irrationals are

[candid fjord]

I mean, I know what they're claimed to be. But the issue is that every proof just uses other assumptions that I have to concede before moving forward with them. Too often, assumptions involving infinity, or imaginary numbers, where the math all works IF you assume that these things are real....

But anyway, this is all much more a meta problem I have with math explanations, not really the scope of this question.

[glass meadow]

Fair enough

[candid fjord]

Alright, guess that's that. Thanks again, sorry I'm useless 😛

.close

[pearl pond]

Channel closed

Closed by @candid fjord

Use .reopen if this was a mistake.

[spiral pivot]

@candid fjord not useless

Anyway, each natural number has an infinite number of zeroes to the left. But the construction in cantor's won't result in an infinite padding of left zeroes, so you don't get a natural number from the process

That's the gist

[pearl pond]

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[daring nymph]

How to do no 12

[pearl pond]

@daring nymph Has your question been resolved?

[daring nymph]

<@&286206848099549185>

,help

[jolly parrot]

A brief description and guide on how to use me was sent to your DMs!
Please use ,list to see a list of all my commands, and ,help cmd to get detailed help on a command!

[daring nymph]

yea idk 😭

[balmy monolith]

think whats the minimum conditions u will need to apply on c for it to be midpoint...

i didnt put c as a mid point intentionally in picture

hek not even on line at that matter xD

[daring nymph]

for this do i use d= (x2-x1......................... so on or the midpoint forumla

cause i used the midpoint one and i got -5 , 6

[balmy monolith]

thats wrong

but welp the question is saying the length formula so lets just use length formula ig

[daring nymph]

but then i will just get smth like d = rounded .... or d= squareroot of somthing over smth

[balmy monolith]

uhh say u have 3 points how do u make sure they are in a straight line and not making a triangle instead???

[daring nymph]

like uhh u cant? or u just have to get c in the middle of a and b

[balmy monolith]

if the distance AC + BC = AB they will be in a straight line

and to make sure its in middle AC = BC

that 2 condition makes C to be the midpoint

[daring nymph]

wait so do i do

ac= cb on the distnace forumla

so first find the answer of ac and then cb

[balmy monolith]

yes calculate AC using the formula then calculate BC

they will be equal write that ... after that calculate AB

and check AC + CB = AB

if both condition satisfys

u have verified

and welp

done the question

:D

[daring nymph]

got it

thx

.close

[pearl pond]

Channel closed

Closed by @daring nymph

Use .reopen if this was a mistake.

[daring nymph]

.reopen

[pearl pond]

✅ Original question: #help-39 message

[daring nymph]

Find the area of the right triangle with vertices r(4,4), S(-2,-2) and T (10, -2)

[tribal peak]

uh

so this takes a while

so id prolly take this step by step

do u know the distance formula?

between points

[daring nymph]

yea ik the distance and the midpoint formula

Like my teacher be giving stuff which she never gave an example of yk

[tribal peak]

k

so since this is a right triangle, which side is your base?

in other words, which of the two sides that isnt the hypotenus is longer, and whats it length?

and if u can, which sides do you think are the longest to shortest(i.e, basically create a list of RT,RS, etc)

[pearl pond]

@daring nymph Has your question been resolved?

[daring nymph]

Yea I got no idea for that

That was the question there was no other info

[daring nymph]

.close

[pearl pond]

Channel closed

Closed by @daring nymph

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

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• 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.
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[turbid lark]

what did i do wrong for the last part

[shell brook]

Good handwriting

[west sapphire]

your answer for a_n looks right

[turbid lark]

is this sarcasm

😔

[west sapphire]

i think it's good handwriting

[turbid lark]

ik, i cant find the sum tho

[west sapphire]

what did you try?

oh wait, was this your attempt?

that's finding the limit of $s_n - s_{n-1}$, which is the limit of $a_n$ (the sequence, not the series)

[jolly parrot]

Bungo

[graceful timber]

it’s asking for the sum of a_n. not what a_n approaches. take the limit as n goes to infinity of s_n. that should give the infinite sum of a_n.

[turbid lark]

thank you

i got it now that makes sense

.close

[pearl pond]

Channel closed

Closed by @turbid lark

Use .reopen if this was a mistake.

[vernal jasper]

My prof is always so picky about L'hoptials for series!

[pearl pond]

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[sharp smelt]

nvm

made a mistake

😔

.close

[pearl pond]

Channel closed

Closed by @sharp smelt

Use .reopen if this was a mistake.

[shell brook]

Okay wai

[sharp smelt]

actually, I might be cooking with this

.close

.reopen

[pearl pond]

✅ Original question: #help-39 message

[shell brook]

Ohhhh

This betta be worth it

[sharp smelt]

Define $P_n= {0,1/4,1/9, \dots , \frac{1}{n^2}, 1}$.
Then $U(f,P_n)≤ \sum_{i=1}^{n} \frac{1}{i^2} - \frac{1}{(i+1)^2}$
The limit of the upper bound is $0$. Moreover, $L(f)≤U(f,P)$.
So $0≤U(f)≤0$. Thus $U(f)=0$

[jolly parrot]

wai

[sharp smelt]

wait, this goes to π^2/6 does it not 😔

[autumn fossil]

Your points in partitions arent even in increasing order btw, and the total sum would be 1 (if u wrote it correctly) by telescope

[sharp smelt]

Well, another idea. so for fixed n, 1/ni^2 should work, right

as then it will be less than 1/n

[autumn fossil]

Idts, use ur answer to b

[sharp smelt]

will do then

thanks

.close

[pearl pond]

Channel closed

Closed by @sharp smelt

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[sharp smelt]

worked it out a bit, the idea seems to work :)

[pearl pond]

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[tranquil raft]

new to intermediate value theorem so im tryna figure out where to begin on this question

cos sqrt(x) = -2+e^x on the interval (0,1)

[feral sedge]

Maybe the answer is 42

[tranquil raft]

what

[thorn shale]

rewrite the equation then test f(x) with varying x values

[frank violet]

it's a joke, anyway what's even the task here, what are we trying to find. Does it say solve for x in that interval

[tranquil raft]

not solving for x

just trying to show theres a solution

[tranquil raft]

rewrite the whole thing as f(x)?

[feral sedge]

Depends on what you have in mind I guess

[thorn shale]

yeah I would shift everything to one side

[tranquil raft]

alr lemme cook

somthin went wrong

theres gonna be something obvious i screwed up on

[thorn shale]

you used 1 degree

[patent vessel]

Did bro really write degrees

[tranquil raft]

god bless

[patent vessel]

Also pro tip: you don’t need a calc for this

cos(1 radian) =cos(57 deg) approx cos(60 deg)=0.5

[tranquil raft]

so is it just for the second part

the top wouldnt be effected

or is that part off

[thorn shale]

your f(0) is right

[tranquil raft]

i should prob show what i just wrote

so thats a valid solution right?

[thorn shale]

yeah but you probably need a concluding statement applying the ivt

[tranquil raft]

wdym?

[thorn shale]

you have to explain what your findings mean so i would write a conclusion:

the function is continuous, so you can say f(0)<0<f(1) or f(0)f(1)<0 (many ways to describe ig) to explain that f(x)=0 exists, so there exists c∈(0,1) with f(c)=0

[tranquil raft]

ok that makes sense. but is all the math correct? i just need to explain it?

[thorn shale]

yes the math is right

[tranquil raft]

yay

[thorn shale]

maybe I would write a <0 or >0 after the end calculations

but that's just presentation sake

math is all good

[tranquil raft]

thx

.close

[pearl pond]

Channel closed

Closed by @tranquil raft

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[pearl pond]

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[empty rover]

can someone explain the steps here?

[patent vessel]

First step is cuz sin^2 x + cos^2 x = 1

Second is solving the quadratic in sinx

[empty rover]

but how they get sinx = -0.5 and sinx = 2?

[patent vessel]

Use the quadratic formula

[shell brook]

quadratic formula

[patent vessel]

Or factor the equation as (sinx-2)(2sinx+1)=0

[torn flicker]

yeah and you factorise it

and you have to reject one of the solutions from the quadratic

[empty rover]

how I factorise into this?

is it I expand 2(1-sin^2x) first?

[warped violet]

Yes

And then introduce a new variable

Say $u \coloneq \sin(x)$

[jolly parrot]

USS-Enterprise

[warped violet]

And you get a quadratic in u

Then you solve the quadratic for u, then substitute u back in for sin(x)

[empty rover]

eh, is there an easier way

cus like that, I gotta do extra steps

[patent vessel]

That’s a very short way

If you practice you’ll get very quick with it

[warped violet]

Sure, guessing is very simple if luck is on your side. Probably won't be though

[empty rover]

is it (-2sinx-1)(sinx-2)?

[warped violet]

What is

[empty rover]

?

[empty rover]

anyone?

is this correct?

[warped violet]

What is (-2sinx-1)(sinx-2)

I don't know what you mean by that

[outer canyon]

Yes its correct

[outer canyon]

He was asking if he did the factors correctly of the qn he sent before

[outer canyon]

This.

[empty rover]

so my sinx = -1/2 or sinx=2

but why we reject sinx=2?

[outer canyon]

range of sinx is [-1,1] only

sinx = 2 is not possible for any real x

[warped violet]

Oh I see. I didn't really read everything. Thanks

[outer canyon]

no prob

[empty rover]

oh then what about rang e of cosx and tanx?

[outer canyon]

cosx also has same range [-1,1] and tanx range is all real ( - infinity, infinity)

[empty rover]

just for info, may I know why?

like is it related to the sin and cos and tan graph?

[warped violet]

Not really "related"

How did you learn about the trigonometric functions?

[empty rover]

in class

[warped violet]

...

With the unit circle?

[empty rover]

no with the unit trigonometry

like they taught us

about quadrants

reference angles

compound angle formula

the pythagorean identities

how to do qnB4c?

anyone?

[desert hedge]

u need to find the 3rd quartile value and minus the first quartile value

Its like median but instead of (n+1)/2 its for first quartile ((n+1)/4)th term

[empty rover]

but how would I find the first and third quartile?

[desert hedge]

1st step write the values in assending order

like how we do median

[empty rover]

done

[desert hedge]

the 3rd quartile's formula to find the position is (n+1)*3/4

[empty rover]

what's n?

[desert hedge]

if it comes as a decimal take the whole number values which the decimal is inbetween and divide by 2

[desert hedge]

number of values

[empty rover]

so first quartile is n+1/4?

[desert hedge]

not 2 but 4

yh

thats the term

[empty rover]

why 4 tho?

[desert hedge]

quart means 4

also if the position comes in decimals u have to get the values in the position of the whole numbers the decimal is inbetween

add them up and divide by 2

[empty rover]

wait I dont get what you mean by this

[desert hedge]

example

i have 5 values

if i do the (5+1)/4 it gives me 1.5

there is no position as 1.5 so i take the value of the first position

and add it to the value of the second position

then divide that value by 2

since 1<1.5<2

i take the first and second value

[empty rover]

does this apply for Q3 as well?

[desert hedge]

yes

[empty rover]

but it wouldnt be (n+1)/4 right?

[desert hedge]

yh it would be (n+1)*3/4

[empty rover]

wut

[desert hedge]

but the decimal position concept stays the sam

[empty rover]

why's it multiply now?

[desert hedge]

3rd quarter

what is 3 quarters

3/4

[empty rover]

3/4

[desert hedge]

yh

[empty rover]

but why only this case we multiply when the other two its just divide by 2 or 4

[desert hedge]

what happens really is (n+1)*1/4

[desert hedge]

because (n+1)*1=n+1

and 2/4=1/2

[empty rover]

actually

in a way

first quartile can be (n+1) x 25% right?

[desert hedge]

yh

[empty rover]

but how come we must n+1?

when there's only 9 terms?

[desert hedge]

its the formula

[desert hedge]

formula

(9+1)/4

[empty rover]

ok so lets just say the position is at 2.75, do I take 2nd or 3rd position?

[pearl pond]

@empty rover Has your question been resolved?

[pearl pond]

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[empty rover]

how do I know if I need to use nPr or nCr?

[torn flicker]

hmmm

if there are restrictions use nPr

like part (c)

[empty rover]

wdym by restrictions?

[iron basin]

"selected" calls for nCr

you would rarely use nPr after a while

[empty rover]

so technically this qn is all nCr?

[iron basin]

if you knew how to use it yeah
although for c) and d) you might have to use nCr more than once along with some extra calculation

[empty rover]

ok, by the way n stands for the total and r is the targeted group right?

[iron basin]

yes

[simple niche]

Whenever you see Choose, C.

Whenever you see Arrange, P.

[pearl pond]

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[pearl pond]

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[tacit gorge]

!show

[pearl pond]

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

[tacit gorge]

!status

[pearl pond]

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

[sharp smelt]

4

I'm wondering if ussing tagged partions is all I need to do here

[eager jewel]

Not helpful but I think our teacher taught us this proof from some rectangles under the graph and some inequalities

[sharp smelt]

yeah, no 😭

I've proven a few more similar results, wish to use them

[eager jewel]

Yeah I don't recall the proof..just mugged up the trick which is 1/n = dx and k/n =x

[tropic saddle]

||squeeze and done||

[sharp smelt]

I thought I had an idea, but the only thing that seems to work is tagged partitions

[sharp smelt]

That was my original idea

\min{x_n}≤f(k/n)≤max{x_n}

and then use equipartitions

which does work I suppose

[random ermine]

let P_n be the uniform partition on [0,1] with subintervals of length 1/n

[sharp smelt]

.

[random ermine]

L(f, Pn) < sum inside limit < U(f, Pn)

[sharp smelt]

so this works

[random ermine]

show that the lower and upper bounds converge to L(f) and U(f) resepctively

then you are done

let eps > 0

[sharp smelt]

yea, got it

[sharp smelt]

that's given so I'm done

[random ermine]

there is a partition Q st U(f) < U(f, Q) < U(f) + eps

[random ermine]

u can use this result?

[sharp smelt]

yea

I forgort

[random ermine]

for large enough n you can fit P_n inside Q

hmm or not quite

anyways

[sharp smelt]

got it

thanks

.close

[pearl pond]

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[pearl pond]

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[dim parrot]

I didn't understand the part

[proper nova]

,rccw

[pearl pond]

@dim parrot Has your question been resolved?

[dim parrot]

Can anyone explain?

[worthy idol]

What part do you not understand? Is it the step to get to the final line?

[dim parrot]

Of course

Highlighted part

[worthy idol]

If you look at the function ii)

and the line just above the bit you circled

You can see you've got the same function you started with, but negative & squared

[dim parrot]

Yes so?

[stray canyon]

so thats the value ? idk

[dim parrot]

What?@stray canyon

I'm not getting what value you are saying

Say properly in words plz

[stray canyon]

do you understand the lines before ?

[dim parrot]

Of course so i highlighted

[stray canyon]

oh thats the conclusion that you did not understand

i thought it was the value of d/dx f(x)

[dim parrot]

Of course yes

It is the value of f'(z)

f'(z)=-f(z)^2

.close

[pearl pond]

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[pearl pond]

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[spiral pivot]

. @hard furnace please use this channel now

[hard furnace]

ok

[hard furnace]

so

i think

[spiral pivot]

as I explained before, a ray is the sort of line piece that extends from a point infinitely but only in one direction.

[spiral pivot]

so of these options, which fits that description?

[hard furnace]

B

[spiral pivot]

that's correct.

what about the other 3?

do you know what those are describing?

[hard furnace]

all of them mention having 2 sides and since a ray is just one side they would be incorrect

[spiral pivot]

The first option is a line with a hole in it.

The third option is a line segment (without endpoints, so it's an "open" segment)

The fourth option is a line segment (with endpoints, so it's a "closed" segment)

[hard furnace]

can you help me with this question?

[lilac quartz]

@hard furnace have you found out the corresponding distances?

[hard furnace]

thats what i need help with

can you help me?

please?

[smoky gull]

do you know about the distance formula? @hard furnace

[pearl pond]

@hard furnace Has your question been resolved?

[hard furnace]

yes

[smoky gull]

okay then just find the distances AC and BC and then ping me

[hard furnace]

so once i get the answer for A and C i just add them to get AC right?

[smoky gull]

okay so what i meant was find the distance between point A and point C

[hard furnace]

is C (0,2)?

[smoky gull]

yes but thats given in the Question isnt it}

[hard furnace]

so then its all distance formula from there right?

@smoky gull is it 1.5/27?

[smoky gull]

nope

should i just give you the working @hard furnace ?

or need hints?

[hard furnace]

can you give me some hints?

[smoky gull]

sure

[hard furnace]

i did the math and it said it was wrong when i did the check answer

[smoky gull]

distance between two points is given by

[hard furnace]

the distance formula?

[smoky gull]

so here for the distance AC, x1 = -1.5, y1 = 0

can you tell me what x2 and y2 will be?

[hard furnace]

x2= 0 and y2 = 2

[smoky gull]

good job!

now you can find d (here it is AC)

similarily can you do the same for CB?

[hard furnace]

yes

x1 = 0 and x2 = 4.5

y1 = 2 and y2 =8

and then do distance formula right?

[smoky gull]

yes!!

fantastic!

now after this your AC should come out to be 2.5

and BC should be?

BC comes out to be 7.5

[hard furnace]

so its 2.5/7.5?

[smoky gull]

yupp

so the answer after reducing the fraction should be?

[hard furnace]

1/3

[smoky gull]

good job

thats correct

[hard furnace]

can you help me with this one as well?

[smoky gull]

perimeter of a triangle is simply the sum of lenghts of all its sides

so find all the sides of the triangle - AC, AB and BC

[smoky gull]

use this formula too

[hard furnace]

which vertices belong to which letter?

[smoky gull]

In the same order

A(-4,0) B(-1,6) C(3,-1)

[hard furnace]

k

7.07106781187 is what i got for AC

[smoky gull]

seems right

[hard furnace]

7.1 if rounded

[smoky gull]

sure but dont round off just yet

[hard furnace]

right i still have AB and BC

[smoky gull]

take AC as 7.07

[smoky gull]

yup find those

and then add all three and then round off

[hard furnace]

21.83

[smoky gull]

thats AB?

or is that the sum

if thats the sum then its correct🎉

in the Q it says round off to nearest tenths, so your final answer should be?

[hard furnace]

it was indeed the sum

thanks for the help

.close

[pearl pond]

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[pearl pond]

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[plush bramble]

what are you trying to do

why do you have a slider

[cerulean cloak]

-# is this some sort of test 😭

[unborn abyss]

did you close and resubmit just to try to get your name near the top

dont do that

[cerulean cloak]

judging from the time i don't think it is

[unborn abyss]

different channel

[cerulean cloak]

oh shit

the time still seems pretty far apart (though yes it is the same question)

i'd personally assume good faith

[pastel umbra]

If it's the same question, and it seems to have been answered...?

[cerulean cloak]

judging from this i don't think it's answered

[pearl pond]

@hard furnace Has your question been resolved?

[hard furnace]

sorry

i was doing other things waiting for a response

actually ill do it tomorrow

.close

[pearl pond]

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[pearl pond]

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[empty junco]

what is this quesiton asking

[sharp vigil]

find s(t) with the given info (different function for parts a and b)

[empty junco]

the object as an velocity function and an acceleration function, but the acceleration funciton isnt the derivative of the velocity function

in this case

is it asking to "antidifferentiate" (without integration, since im calc 1, just analytically)

[sharp vigil]

treat a and b as completely different questions, find the position function for them separately

[empty junco]

since v'(t) != a(t) in this case

ok

to verify, is position the antideriv of velocity?

thats what im assuming here

[sharp vigil]

well there are many antiderivatives of velocity, but the position is the one satisfying the correct initial condition

[empty junco]

ok

[pearl pond]

@empty junco Has your question been resolved?

[empty junco]

ok so

i understand that 4a) is cos t + e^t

right?

but for 4b) idk what to do

im assuming that its giving us the position at s(0) or v(0) to imply that there is a constant in the antiderivative

help

[cloud zephyr]

they are indeed providing the initial conditions for v (when you integrated one time) and s(when you integrated two times)

[cloud zephyr]

would this be a good enough hint? @empty junco

[odd coral]

have you double checked 4a

[empty junco]

no

sorry for afk

im dumb