help-27

[shell plover]

You want help with both 1 and 2?

[obtuse lantern]

Just check my answers

If there are errors and tell me

[shell plover]

alr give me a moment

[round raptor]

you can't have 0 even that would be a 4 digit number since set conatins 4 odd numbers

[obtuse lantern]

What letters are correct

And incorrect

[shell plover]

I think repetition is allowed here

[round raptor]

oh ok right then

[obtuse lantern]

.

[obtuse lantern]

.

[shell plover]

i dont think theres an error here, btw where are the questions to the answers that youve written below with pen

[obtuse lantern]

Additional Instructions:
For number 2
Letter
e. Find the probability of S in letter a.
f. Find the probability of P in letter b.
g. Find the probability of S union P in letter c.
h. Find the probability of S intersection P in letter d.

[shell plover]

i don't understand these parts? S and P are sets, not events how can we calculate there probability?

[obtuse lantern]

?

Idk

thats why im here

[shell plover]

i think these questions might ne wrong

what did u assume that the question meant?

And btw 1e is wrong

You included (0,0) in the favourable set

0×0 is not less than 0+0

[devout snow]

@obtuse lantern Has your question been resolved?

[devout snow]

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[normal bison]

i dont any property to help solve it

i tried using parametric form

but no use

[solemn hatch]

Well if P is any point, assume a point P on the circle that will make your calculations easy

[normal bison]

isint that already given that p is on the circle ?

[solemn hatch]

Yeah so P can be a lot of points

All P work I assume since the word "any" is used

And it's unlikely eccentricity and point both can be solved unless it's a limiting casr

My suggestion would be take the point P as (0,a) on x^2 + y^2 = a^2

[normal bison]

i let P to be in the y axis ,

[normal bison]

yea exactly

it didnt help

[solemn hatch]

See there will be a symmetry

On AP and A'P

AP/AQ will be 3/2

[normal bison]

what , how ??

[solemn hatch]

Draw the diagram and send

[normal bison]

i am prolly making it wrong ,

the pt q and q' , doesnt seem to be good

[solemn hatch]

Well let's just say you can prove triangles AOP and A'OP can be proved yo be congruent

[normal bison]

yea

[solemn hatch]

Generally PAP' is symmetric about y axis

And the ellipse is symmetric about y-axis too

[normal bison]

yess

[solemn hatch]

So both the intersections will be symmetric about y-axis too

So AP and A'P have the same length

So do AQ and A'Q'

So AP/AQ = A'P/A'Q'

[normal bison]

ohh , haa

[solemn hatch]

You can tell if you don't understand

AP is √2a

[normal bison]

yea ,

i am figuring out how to find the length of b , (semi minor )

[solemn hatch]

To find AQ you need to solve x/a + y/a = 1 and the ellipse equation

[solemn hatch]

You probably can't

Ig in terms of a you can

1 - y^2/b^2 = (1-y/a)^2

2y/a = y^2(1/a^2 + 1/b^2)

Too lengthy ig

Could be solved tho

[normal bison]

yea ,

thanks for helping!

[solemn hatch]

Welcome

[normal bison]

.close

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[vestal topaz]

I have a linear transformation T: C_4[x] -> C_4[x]
I want to turn it into matrix representation by plugging in the basis polynomials, but I am having a hard time figuring out what those would be
could I get some help on this?

[twilit field]

hi, for future helpers, I think it would be best if you posted your question straight up!

[vestal topaz]

Well I guess if I rewrite the question.
How do I figure out what the basis for C_4[x] is?

[runic prawn]

you mean helpees?

[pseudo basin]

no such thing as "the" basis

[uncut crow]

what are “those”

[twilit field]

I mean helpers.

[pseudo basin]

C_4[x] is a vector space like any other

[vestal topaz]

rather "a" basis

[pseudo basin]

can you cook up a basis for R_4[x] as a real vector space

[vestal topaz]

I don't really understand what would constitute a basis for polynomials to be honest, how do I isolate 4 polynomials such that each one would represent 1, x. x^2, x^3, x^4

[pseudo basin]

5, not 4

[vestal topaz]

yes

[pseudo basin]

{1, x, x^2, x^3, x^4} is in fact a perfectly serviceable basis for C_4[x] over C as well lmao

[vestal topaz]

how do I write it as a polynomial though? for example p(x) = a_0 * x^0 + a_2 * x^2 etc...

[pseudo basin]

write what as a polynomial

i... don't really understand what you're asking here

1 is a polynomial. x is a polynomial. x^2 is a polynomial...

so what's this "it" that you're talking about

[vestal topaz]

I mean what would the polynomial x^3 for example would look like as p(x)

what would the argument for p be

[pseudo basin]

x^3 is x^3 ...

if you want to be mega pedantic about it you could like... write it as ⁨0 + 0x + 0x^2 + 1x^3 + 0x^4⁩ but that's maximally dumb

[vestal topaz]

okay, now given this polynomial how do you write it as P(...) = ⁨0 + 0x + 0x^2 + 1x^3 + 0x^4⁩

what would ... be here

[mystic scarab]

x

(OF COURSE)

[pseudo basin]

it's all x all the way down lmao...

[mystic scarab]

If you're studying linear algebra, I really hope you've previously studied prealgebra, algebra and precalculus

[pseudo basin]

if you wanna use the name p for this polynomial then like

just write p(x) = x^3

but also understand that polynomials don't need to have letter names slapped on them

not inherently

i think you're overthinking something here

[vestal topaz]

okay I have totally embaressed myself, i understandd how dumb the question was initially

.close

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[prime whale]

I have a fair understanding of liner alegbra fundamentals, so this is where I would like to be met in the following question, if possible.

[prime whale]

Suppose i have a square matrix $A$. It is in an exponential $e^{A}$

[woven radish]

Kurrristian

[prime whale]

The matrix $A$ can be diagonalized: $e^{A} = e^{PDP^{-1}}$. Suppose now that I have $(e^{A})^{x}) = (e^{PDP^{-1}})^x$, where x is just some scalar

my immediate thought is, well, $PDP^{-1}$ should always commute with itself. So I could just distribute $e^{xPDP^{-1}}$

[woven radish]

Kurrristian

Kurrristian

[prime whale]

But there is the super basic definition of power, which is just multipying osmehting times itself over and over agian $(e^{PDP^{-1}})^{x} = (e^{PDP^{-1}}) (e^{PDP^{-1}}) (e^{PDP^{-1}}) (e^{PDP^{-1}})....$

[woven radish]

Kurrristian

[prime whale]

uh oh

i think im trolling

[young spade]

To be honest youre basically forced to go into the taylor series of e^x

[prime whale]

hold on i think i fucked something up bad

ill be back

.close

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[flat jungle]

I don’t even know how to start this question… how do I solve?

[native gyro]

,rccw

[flat jungle]

My handwriting is ass

Sorry

[pseudo basin]

and you don't have a printed version of this?

cause it's kinda unreadable

$\sqrt[3]{3} \cdot 93 xz^3$

[woven radish]

Ann

[pseudo basin]

is that what it says?

[untold ravine]

thats probably times 2 cubed

[pseudo basin]

let's let OP confirm

[devout snow]

@flat jungle Has your question been resolved?

[flat jungle]

2 cubed yeah

3sqrt(393) times 2 cubed

[winter patrol]

3sqrt
do you mean cube root, (which shortened would be cbrt)

3sqrt(393) means $3\cdot\sqrt{393}$

[woven radish]

ραμOmeganato5

[winter patrol]

regardless what is the question asking you to do with this expression

[devout snow]

@flat jungle Has your question been resolved?

[flat jungle]

read above

[flat jungle]

3qrt(393) multiplied by 2^3

i need help with that

[woven radish]

OverHeatVD

[ionic geyser]

what is the question asking you to do?

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[onyx elm]

I need help finding y so I can find the sides. I found x and then tried finding y, but I messed up in my calculations.

[torn vessel]

there is no side with length 2y-1

[onyx elm]

.close

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[severe prairie]

can someone help w b please

sorry for the mess

[hollow ice]

relate it with the real parts of other two of the solutions

and compare it with the sum of all solutions

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[pulsar tundra]

Question says,
"A rectangular container with two square sides and an open top is to have a volume of V cubic units. Find the dimensions of the container of minimum surface area. Record in terms of V."
The picture is what I have so far. Am I doing it right? (I haven't finished it yet. Wanna know if i'm on the right track)

[lyric flame]

you are on right track

[woven radish]

GODSWORD

[cerulean plover]

correct

[pulsar tundra]

I'm kimd of confused on how to isolate s after..

like this?

[cerulean plover]

you could have split the terms before differentiating like this. It'd be easier

[pulsar tundra]

ohhh is V a constant?

i think that makes a lot more sense

thanks a lot!!!

.close

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[quartz bloom]

I claim this expression:-

v(to)+∆v(to+∆t) +∆v(to+2∆t)+...+∆v(t1)

Is exactly equal to v(t1).

Why have they used ≈ in the text book.
Isn't this just vector sum ?
The approximation is not yet taken

[sand pumice]

it's not exact when \Delta t is arbitrary

as \Delta t becomes smaller and smaller (goes to zero), it becomes exact

[devout snow]

@quartz bloom Has your question been resolved?

[quartz bloom]

Can you elaborate please

Do you mean , we have to add smaller and smaller increment in v(to) , to get to v(t1) .
And unless we use that increment to be infinitesimal , we are approximating the increment

[sand pumice]

pretty much

[quartz bloom]

"Pretty much"...

[sand pumice]

well

[sand pumice]

yes

[quartz bloom]

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[tepid garden]

Prove
Lim(x->0) sin(kx)/kx=1

[tepid garden]

So our proffesor wants this

I showed him geometric proof, doesn't want it

Nor does he want l'hopital's rule

AND he wants it using squeeze theorem

The only "hint" he gave is
-1<=sin(kx)<=1

I'm lost

Omg u were typing all that im sorry

[hushed oyster]

$$\frac{1}{2} \sin(\theta) < \frac{1}{2} \theta < \frac{1}{2} \tan(\theta)$$Dividing all parts by $\frac{1}{2} \sin(\theta)$:$$1 < \frac{\theta}{\sin(\theta)} < \frac{1}{\cos(\theta)}$$Taking the reciprocal (which reverses the inequalities):$$\cos(\theta) < \frac{\sin(\theta)}{\theta} < 1$$As $\theta \to 0$, we know $\cos(0) = 1$. By the Squeeze Theorem, $\frac{\sin(\theta)}{\theta}$ must also approach $1$.

[woven radish]

Ajay

[hushed oyster]

$-1 \leq \sin(kx) \leq 1$ is too loose

[woven radish]

Ajay

[tepid garden]

That's the hint he gave

But how do we know that 1/2 sinx is less than x/2

[gaunt summit]

graph

[tepid garden]

Doesn't want it

Ik it to be true

Here's the proof I showed to him

[hushed oyster]

Consider a unit circle with an angle $\theta$ (in radians). For a small positive angle $\theta$, we can compare three areas:

1. Area of the Inner Triangle: $\frac{1}{2}(1)(1)\sin(\theta) = \frac{1}{2} \sin(\theta)$.

2. Area of the Circular Sector: $\frac{1}{2}r^2\theta = \frac{1}{2} \theta$

3. Area of the Outer Right Triangle: $\frac{1}{2}(1)\tan(\theta) = \frac{1}{2} \tan(\theta)$

[woven radish]

Ajay

[tepid garden]

No thoughts?

[tepid garden]

Yes but he wants another method to prove

For some reason

[hushed oyster]

What method?

[tepid garden]

Idk

He just doesn't want this one

[hushed oyster]

So not even a little geometry is allowed?

His hint doesn't work at all

Because taking Limit[Divide[1,kx],x->0] does not exist. From the left is gives -infinity and from the right infinity. So you cannot apply the squeeze theorem

[devout snow]

@tepid garden Has your question been resolved?

[tepid garden]

Exactly

I tried telling him that

And added that of it's x-> infinity then it's equal to 0 and it's easy to prove that using squeeze theorem

[devout snow]

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[mild magnet]

hi im currently using matlab for the first time and am rather stuck with something if anyone has experience using it would appreciate some help thank you

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[deft flower]

is this doable using complex analysis

[deft flower]

or any other method really

[cold haven]

$2∫_0^∞\f{√{1+x}}{\l(x^2+2x\r)^{3/4}}\di x$

[deft flower]

2x

[cold haven]

What’s in the place of ?

Thanks!

[woven radish]

Infinium³

[deft flower]

exactly

idk how to deal with the pole at 0

should the contour be like this ?

[frozen aurora]

why the IV quadrant?

you can just keep it in the first one

and i think it'll be easier if you do a full quartercircle instead of the smaller one you have

something like this

[deft flower]

this is absolutely impossible isn't it

[frozen aurora]

[deft flower]

oh I just saw your msg mb

[frozen aurora]

yeah that looks scary

[deft flower]

ohhh actually lemme try it

[deft flower]

😭😭😭😭

[frozen aurora]

what's the context of the question

[devout snow]

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

[deft flower]

my weird attempt of trying to solve the integral for the lemniscate constant

[frozen aurora]

lemniscate constant
Ah.

yea don't expect it to be pretty i think xd

[deft flower]

it's equal to half of B(1/4;1/2) but i'm nowhere near that right now

i don't even see how i'd get it using complex analysis but i'm trying cuz why not

[devout snow]

@deft flower Has your question been resolved?

[deft flower]

I don't see how I can solve it <@&286206848099549185>

[fickle magnet]

Diddy?

[deft flower]

yes 😔

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@deft flower Has your question been resolved?

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[idle dune]

help

[idle dune]

i don't understand how the answer for d is the answer for d

i have the answer btw just don't get how to get there

[twilit field]

you don't know how the answer for d is the answer for d?
I don't see the answer for d being shown here though. that box seems to be the answer for c.

[idle dune]

can you tell me the answer for d

sry it's like im testing u

but just so u know the answer and then explain it rather than basing an explanation off of the answer i give you

here's the whole picture with answers if needed

[twilit field]

I was just confused about the statement and hoped to establish more context for future helpers, sorry.

[idle dune]

no worries

.close

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[runic dew]

Following up from my previous post, I've written out an attempt at a proof, but something feels wrong. In particular, i feel like the last page uses faulty reasoning, and that i ought to use some sort of epsilon prime trick to do it properly.

[supple knot]

that's some handwriting

[polar chasm]

Not that bad tbh

the only issue is you're still using limit laws

When you handle the terms 1/n^2, 3/n, ... separately, you are secretely using laws about limit of sums, products...

You'll probably have to actually pick your N manually

[runic dew]

I see, how should we go about doing that?

[runic dew]

Im sorry 😭 ive never had good handwriting...

[polar chasm]

You'll need to bound that original absolute value thingy

So I'd start by simplifying it as much as possible

and manually subtracting the 1/2 (bring it to common denominator)

[runic dew]

This is what I initially tried, but the furthest ive gotten is simplifying it to (3n+9)/|4n^2 - 6n - 14|

Couldnt figure out how to simplify from there

[polar chasm]

Okay, this is quite good

[polar chasm]

The cool thing is that we only need it to be < epsilon, we dont care about what it actually equals

So if we somehow "round it up" and it's still < epsilon, we dont really care, it gets the job done

So you just need to "round it up" (find some upper bound), which is simple enough so that you can work with it

and it cant be a trivial upper bound, the upper bound itself must still -> 0

[runic dew]

I also tried doing this, but couldnt find anything that was actually an upper bound

[polar chasm]

I see, well, it's a fraction

so we will want to make the numerator slightly bigger and the denominator lower

Another good thing is that we dont care about how it starts, but we just need the upper bound to eventually overgrow it (so if we pick large enough N, the upper bound should overgrow it)

so e.g. we could bound 3n + 9 by 4n, because if we pick N >= 9, which we can, it will be an upper bound

and to ensure that we pick N >= 9, we will pick N = max(9, whatever we get later on)

Do you understand why we can do this?

[runic dew]

I do, and i also see (i think?) that if we pick N >= 9, then the denominator can be erased too, since |4n^2 - 6n - 14| > 1 for n >= 9, leaving us with the upper bound just being 4n

[polar chasm]

Yeah, sure, 4n is an upper bound

but that upper bound doesnt tend to 0 anymore

so we cant really pick N which makes 4n as close to 0 as we need

[runic dew]

Oh true

[polar chasm]

We need a better upper bound, we need the upper bound to still tend to 0

And are you trying to find an upper bound for the denominator or a lower bound?

Remember that we are trying to find an upper bound on the whole fraction, numerator / denominator.

[runic dew]

So we need to find a lower bound for the denominator?

[polar chasm]

Yeah, making denominator smaller makes fraction bigger (for positive fractions at least)

So if we want an upper bound on fraction, we need lower bound on denominator

[runic dew]

Oh i think i see it now

You take the 4n^2 term out of the abs function, since it's always => 0, and then the rest of the terms can cancel, since theyre always positive, we have an upper bound of 4n/4n^2, aka 1/n

[polar chasm]

it's not true that |a+b| = a + |b|, even if a is positive

so you cant really take it out

[runic dew]

Ah crap

[polar chasm]

in fact, for n >= 4, you can completely ignore the abs, since 4n^2 - 6n - 14 will be positive

[runic dew]

Oh right! And is it that since 4n^2 > 4n^2 - 6n - 14 for positive n, that is what makes those terms cancel?

Wait no that wouldnt make sense i dont think

[polar chasm]

It's the fact that 4n^2 - 6n - 14 will eventually be positive (its an upwards facing parabola, so in fact itll grow to +infty)

so the | | wont matter in the long term

this is the graph of 4n^2 - 6n - 14 and its absolute value

we can see that for n >= 4, there is no difference anymore

[runic dew]

Right right, although even with the abs gone, im not sure how we can upper-bound it further

[polar chasm]

You want to lower bound it

[runic dew]

My bad, thats what i meant

[polar chasm]

But try making it grow asymptotically slower

[polar chasm]

When I chose to bound 3n + 9 by 4n, my thought process was that I want to preserve the linear term (something with n), and I want to somehow delete that constant part. So I replaced 9 with n (it will be larger for n>=9) and got 3n + n = 4n

4n^2 - 6n - 14
Here, you'll want to get rid of -6n and -14 and you'll probably want to maintain that n^2 term, since that's what matters the most in the long term (you can change the coefficient though)

Try transforming that -6n - 14 into something with n^2, that'll be lower in the long term

[runic dew]

i see what you mean, would this have to be something like -6n^2 - 14n^2, which will be lower than -6n - 14 in the long term?

[polar chasm]

I guess that would work lmao

oh actually maybe not

anyway, something much simpler would be plain -n^2

[runic dew]

i was gonna say, that would result in 4n/-16n^2, which im not sure works as an upper bound

[polar chasm]

Yeah, the problematic part is that you made the denominator go negative

[polar chasm]

this should work though

[runic dew]

oh right, in the long term n^2 > 6n +14 because polynomials of higher degrees always grow larger than those of lower degrees in the long term

[polar chasm]

Can you say exactly the point from which it will be lower?

Or any point for which you're sure that it'll always be lower from then on

[runic dew]

can you can work it out by plugging n^2 - 6n - 14 into the quadratic formula, and setting n as being greater than whatever the highest solution is?

or alternatively doing -n^2 + 6n + 14?

[polar chasm]

Sure, that would work

but it might be more convenient to just try some n

10^2 > 6*10 + 14 = 74

[runic dew]

oh right

[polar chasm]

and its pretty clear that it'll continue being bigger from then on

so you can just pick N at least as large as 10

[runic dew]

and then when we collect the terms of the denominator, we get 4n/3n^2?

[polar chasm]

yeah, so now we know that
|a_n - 1/2| < 4n / 3n^2

[runic dew]

which is then equal to 4/3n, and that can be used to find a value for N?

[polar chasm]

Yep

if you make 4/3n < epsilon, |a_n - 1/2| will also be < epsilon

[runic dew]

oh my god youve saved me 😭 thank you so much 🙏

[polar chasm]

np

[runic dew]

you and others here are life savers 🥹

.close

[devout snow]

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[devout snow]

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[restive river]

I have calculated that the no of sylow 3 subgroup has to be either 1 or 7. Now how do I say that it is indeed 7 and not 1?

[restive river]

Should I consider calculating the no of sylow 7 subgroup?

[fickle magnet]

Only one value for that

[tame quest]

What's the topic guys

[restive river]

Sylow theorem for group theory

[fickle magnet]

Sylov group

[fickle magnet]

Is there a w in the name!?

[restive river]

Yes

Are u perhaps from Norway?

Anyway that's not that important

[restive river]

Yup but it could be 1 or 7

[fickle magnet]

Yea calculate late n7

[fickle magnet]

Thats for n3

[restive river]

Yup

We get n7 must be 1

[fickle magnet]

Yes

They said they are non cyclic

[restive river]

What's the link ?

With it being non cyclic

[fickle magnet]

When both subgroups are normal the group is cyclic

[restive river]

And it would be normal iff it is unique

Ryt!

?

[fickle magnet]

Ryt

[restive river]

So we get k is unique which sylow 7

Now we can't take n3=1

Cause if we do we will get cyclic

[fickle magnet]

Yes

[restive river]

Therefore it has to be

7

[fickle magnet]

N3=7

[restive river]

And since order of H I.e sylow 3

Each subgroup contain 2 elements of

Order 3

Therefore 7*2= 14 in total

[fickle magnet]

So no of elements is 2*7

[restive river]

✨

[fickle magnet]

Yaay

[restive river]

Thank you 🩷

[fickle magnet]

Btw where are you studying group theory from

[restive river]

Yt lecture mostly

I have a book

[fickle magnet]

Actually i just know what i studied myself from bits of info on internet

[restive river]

By a local writer

[fickle magnet]

Which

[restive river]

S.K mapa

[fickle magnet]

Ok ill check it out

[restive river]

Okay

[fickle magnet]

R u studying for iit jam?

[restive river]

Tho i found better resources

[fickle magnet]

Oh

[restive river]

I am preparing for sem exam

Later

Thanks for the help

[fickle magnet]

All is well

[restive river]

.close

[devout snow]

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[fickle magnet]

@versed juniper what happened?

[restive river]

@fickle magnet https://judsonbooks.org/aata-files/aata-20250801.pdf

I recently found this book it's pretty good

[fickle magnet]

Okay

[devout snow]

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[rain sluice]

I could use help with this problem:
Find the limit of the sequence:

[woven radish]

Lauren

Lauren

[rain sluice]

$a_n = tan^{-1} (n^2)$

[woven radish]

Lauren

[woven vale]

okay, what have you tried so far?

[rain sluice]

nvm, I thought I typed it wrong and the bot didn't recognize my code as correct... now it posted it 3 times 😄

Ok so, I figured out that the answer is:

$\frac{\pi}{2}$

[woven radish]

Lauren

[rain sluice]

But I wonder how should I do it if it were an exam work or something similar. Do I need to provide a proof or step by step explanation how I do it?

Because I have no idea how to do that specific part correctly

Yes

I thought it had to be continuous at/around some point though, not infinity

[fickle magnet]

Do you know the graph of tan inv x

[rain sluice]

I didn't know it. I looked it up when I came across this problem. It's actually from a book.

I think I can get the feel on what the correct answer is. At least in this particular case. But I don't know if simply stating the answer would be enough in a test. So that's my question: do I need to prove my answer in this case or do it step by step somehow? And if so, then how?

[thin fern]

actually you're right mb 🥶

but the basic idea still applies

[thin fern]

You definitely need some explanation at least

(assuming this is a calculus class)

[rain sluice]

How the explanation would look like?

[thin fern]

You could say $\lim_{n \to \infty} n^2 = \infty$

[woven radish]

Ari

[thin fern]

So $\lim_{n \to \infty} a_n = \lim_{x \to \infty} \arctan(x)$

[woven radish]

Ari

[thin fern]

and that's pi/2

[rain sluice]

Thank you very much. Anything else or shall I close?

[woven vale]

i will say: usually it's a theorem proven in class that arctan goes to pi/2 as n goes to infinity, but in the rare case it's not, you have to argue that

• arctan is increasing
• the range is at most pi/2,
  then since arctan is increasing and bounded above, the limit is the lowest upper bound (pi/2)

but 99% chance your teacher proved this and lets you use it without these steps

[rain sluice]

I see. Maybe I should ask my teachers just to be sure.

.close

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[narrow wigeon]

Maybe this isn't exactly a math problem, but a theory question.

" On the other hand, the set of all decimal expansions is not countable.
How come?
Well, if we had a list of all decimal expansions, we could easily construct a number that cannot be on it. Just make its entry j places beyond the decimal point differ by 2 from the entry in that place of the number that is jth on your list . Then the number we create differs from every number on your list and therefore cannot be on it. All of this means that we cannot list all the real numbers or decimal expansions!
Neither you nor I could actually do the infinite number of acts necessary to construct such a number, but we can imagine it done.
To visualize this imagine the first three numbers on your list were
.101010∗
.314720∗
.71324∗
A number not on the list by the construction given would start by .335 since this sequence differs from the first number by 2 in the first place, differs from the second number by 2 in the second place and from the third member by 2 in the third place. The number we ultimately get will definitely differ from the three numbers above. With the rest of its digits similarly determined in reference to the next numbers in sequence on the list, we can deduce that this number cannot be on the list anywhere, so long as the number of its digits is the same as the length of the list.

This means that the set of all decimal expansions cannot possibly be listed. The decimal expansions are uncountable."

I'm not sure if I should ask here but I am genuinely confused. I was reading a section about countable sets.
Can someone help explain why a set of decimals is not countable? I never studied math in english so this is hard to grasp.
Specifically why this author is trying to justify it by saying "the number we create differs from every number on your list and therefore cannot be on it"...
Maybe any concrete examples?

[acoustic leaf]

by 'countable' we mean a set of objects where we can make a function that assigns each of those objects to exactly one natural number (so one of them is assigned to 1, then to 2, another to 3, and so on). people often think of this as laying out those objects in an "infinite list" where their position is determined by the natural number associated with them

what this is claiming is, given any function which associates a natural number with a decimal expansion ("lists out" decimal expansions), it must not be able to reach all decimal expansions, since we can always find a decimal expansion that isn't on the list

so there can never be a function which lists out every decimal expansion, which is what we mean by "they are uncountable"

[devout snow]

@narrow wigeon Has your question been resolved?

[narrow wigeon]

So like indexes of the elements of a list in programming?

How I prove it in my head is that even if you have a list of lets say,
[0.1, 0.2]
indexes 0 and 1
you can still find 0.00000000001 for example in those... but it wasn't included in the list and apparently that makes it so its uncountable

Okay well so far i've only seen examples of decimals between 0 and 1, but what about numbers like floats, such as a list of
[5.1, 7.324234534, 39999.30] would these also be uncountable?

Also the author proceeds to say
"Actually every real number between 0 and 1 has a decimal expansion, but some, namely those rational numbers that end with all 00’s, appear twice as decimal expansions. The reason is that 0.99999... is really the same as 1.00000. Since these are a subset of the rational numbers, this difference is of no particular importance."
which just kills me.

[acoustic leaf]

"uncountable" is a description applied to a set of objects

[narrow wigeon]

Sry I am just used to viewing the logic through the lens of a programmer
But maybe its completely different

[acoustic leaf]

a set (think of this like a data type) is countable if we can make a function which takes in an index (natural number = integer at least 0) and every member of that type is given as an output at least once.
for example the integers are countable by the function
f(0) = 0
f(1) = 1
f(2) = -1
f(3) = 2
f(4) = -2
f(5
and so on. simularly the rational numbers (fractions of two integers) are countable because we can find a function which takes in indices as input and returns every single rational numbers once

[devout snow]

Channel closed

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[acoustic leaf]

however, there is no function which can take in only indices as input and returns every possible decimal expansion. it can return some of them, but not every single one of them. at least some (in fact, nearly all) decimal expansions will be "left out" and never outputted by that function

[devout snow]

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[restive yacht]

can someone explain to me what sin cos tan are and how to use them

[silver veldt]

whats the difference between all these help channels?

[next saddle]

these are occupied! if you have a question, post it in an available channel such as #help-21｜아리스킨충1, or you can help any of these helpees here in the occupied section :)

ope dont post it there now but you get my point

[next saddle]

what course (general) are you taking, something more geometric or algebraic?

[devout snow]

@restive yacht Has your question been resolved?

[restive yacht]

Geometric

[hollow pulsar]

if u wanna learn trigonometry its better to watch youtube

there are a lot of good videos that explain it well

[restive yacht]

thanks

.close

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[sand fable]

I dont understand why this is wrong? Its not like the answer is a^2 and b^3.45 etc... Dude I am struggling so much with these questions, I got the two other ones right but only after like 15 minutes... is there a place where I can quickly learn the rules regarding this nonsense.

[supple knot]

You didn't simplify completely

,tex .exp rules

[woven radish]

riemann

[sand fable]

hell yeah thanks for those thats huge. I just realized I could do it another way anyways... is this valid?

I dont see why it shouldnt work but it seems a bit too convenient somehow

[pseudo basin]

it is valid but your radical symbol is too short there

[devout snow]

@sand fable Has your question been resolved?

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[devout snow]

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[hazy halo]

can you walk me step by step throught he proof of the Guassian correlation inequality

[pseudo basin]

aren't you the same person who just posted a basic question about exponents

[hazy halo]

im hid friend

[pseudo basin]

...

[hazy halo]

his*

[pseudo basin]

<@&268886789983436800> some kinda trolling going on here idk

[hazy halo]

. . .

[copper harbor]

Please don't troll in the help channels, it makes it harder for the volunteer helpers to give help to people who need it

[hazy halo]

i swear that was my friend. He saw me on this discord server and wanted to try something. I took back my ipad.

[copper harbor]

okay well Im going to close this

.close

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[restive river]

Need help with this proof . How to go about it

[wicked turtle]

let S be a p-sylow subgroup of H, what can you say about S as a subgroup of G?

[devout snow]

@restive river Has your question been resolved?

[restive river]

Ik that the no of sylow p subgroup will be of the form 1+kp and it must divide the order of group

But idk what you are asking for

[wicked turtle]

focus on what the sylow theorems say about conjugacy

i don't think you need counting here

[restive river]

H= gkg-1

For two sylow subgroup

In group g

[wicked turtle]

yes

suppose S is a sylow p-subgroup of H
argue why it's also a sylow p-subgroup of G

[restive river]

Cause H IS A SUBGROUP OF G

And for s to be a subgroup of H

It has to divide the order of H

Which further divided order of G

Well to be more clear

[dire iris]

dont you have this backwards?

[restive river]

We can use the sylow s first theorem

[wicked turtle]

well the "backwards" part is what we need to prove

but we can't just assert that it's true

[dire iris]

you assume S is a sylow p supgroup of G, then show it is a sylow p subgroup of H

[restive river]

I see

[wicked turtle]

sure, and i'm saying we can prove it by starting with the reverse statement and then using the sylow conjugacy theorem

[dire iris]

oh i see

i think you can do it this way as well though fyi

[wicked turtle]

if S is sylow in G then every sylow is conjugate to S, so...

[restive river]

Yup

[wicked turtle]

you'll need to use that |G/H| is coprime to p and you'll also need the normality of H

[restive river]

I don't remember normality

[restive river]

Ohh

[wicked turtle]

it's one of the hypotheses

the statement isn't true if H is not normal

[restive river]

gHg-1=H

[wicked turtle]

yep

[restive river]

How do I use the first one

The co prime

We know p is prime

Therefore [G;H] CAN BE ANYTHING

[wicked turtle]

you want to use that to show that a p-sylow subgroup of H is in fact p-sylow in G as well

[wicked turtle]

anything that is not divisible by p

recall the definition of a p sylow subgroup

[restive river]

If it has order of the form p^n

[wicked turtle]

probably you recall it involves something about the HIGHEST power of p dividing H

[restive river]

??

[wicked turtle]

ok if that doesn't sound familiar then tell me your definition of "S is a p-sylow subgroup of H"

without a definition there ain't no way you're going to prove this

[restive river]

Let G be a group of order p^n.m , where p is prime number and gcd (p,m)= 1 then any subgroup of g of order p^n is called a sylow p subgroup of G

[wicked turtle]

so far so good

now suppose S is a p-sylow subgroup of H
argue why it's also a p-sylow subgroup of G

[restive river]

Hmm let me think

Okay so from the above
defination we get
Let H=p^n.m
o(s) of the form p^n where p is prime and gcd of (p,m) is 1 .
Now H is a subgroup of G therefore o(H)/o(G)
Therefore order of G is of the form p^n . m.k let's say.

Now we already have s being of the order p^n we have to show that gcd (m.k , p) = 1

Okay i am stuck here

Okay i think it's straightforward

S is a subgroup of H and H is a subgroup of G

[wicked turtle]

use [G:S] = [G:H] [H:S]

[restive river]

So it must be a subgroup of G

[wicked turtle]

can p divide [G:S]?

[restive river]

Idk

[wicked turtle]

well if it did, it would have to divide either [G;H] or [H:S], right?

[restive river]

Yes

[wicked turtle]

and does it?

[restive river]

Idk I am confused

H is normal

Ryt

So [G:H]=2?

[wicked turtle]

no no

[G:H] = 2 implies H is normal, but not vice versa

[restive river]

I see

[wicked turtle]

recall you are given

[restive river]

Ohh p does not divide [G:H]

Hence p can't divide [ G:H] AND [H:S]

[wicked turtle]

well you want to show that it divides neither

and conclude therefore that it does not divide [G:S]

so why does p not divide [H:S]

[restive river]

Idk tell me why

[wicked turtle]

we assumed that S is a p-sylow subgroup of H, right?

[restive river]

Yes

[wicked turtle]

so if H = p^n m (where p does not divide m) then |S| = p^n

and therefore [H:S] = m

so...

[restive river]

Yup

And we already had in defination

gcd(p,m)=1

[wicked turtle]

ok so far so good

p does not divide [G:H] and it does not divide [H:S]

therefore it does not divide [G:S] = [G:H] [H:S]

therefore S is a p-sylow subgroup of G

agree?

[restive river]

Yup

[wicked turtle]

ok good

[restive river]

Understood so far

[wicked turtle]

now use the conjugacy theorem for sylow subgroups

all p-sylow subgroups of G are conjugate to S

therefore...?

[restive river]

S=gHg^-1

[wicked turtle]

no..

S isn't conjugate to H unless S is all of H

in general S is properly contained in H

[restive river]

I don't understand this statement

[wicked turtle]

how can S be conjugate to a group that is larger than S?

conjugation is an isomorphism

[restive river]

Ohh yes wait

We can do contradiction

Here

Okay nvm i couldn't make it make sense

I got S=H

Since H is normal

[restive river]

What if S is improper

Subgroup

[wicked turtle]

it will be, in general

[restive river]

If S is improper we get the theorem proved

Anyway

[restive river]

It can't

[restive river]

.close

[devout snow]

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[wise violet]

I did this problem wrong

So I’m doing it again

My diagram I just drew is there in pen writing

Here are the identities we learned related to this problem

[hollow ice]

so how is the second attempt going? did you use the identities here? whatd you get

[wise violet]

This is my second attempt

I labeled the triangles based on what I was given and then I just did secant of theta

So 3.42/-1

[hollow ice]

well, you also need one identity to use there

[wise violet]

Oh 😭

[hollow ice]

coz what you calculated was sec(theta)

not sec(-theta)

[wise violet]

Yea

💀😭

[hollow ice]

well, luckily for you tho, since cos(x) = cos(-x), you also get sec(x) = sec(-x)

but you gotta mention that explicitly

[wise violet]

Oh

If i were not lucky, how would the negative theta change the problem

[hollow ice]

tho personally, if you need to solve it using the identities, dont use a diagram to do this

[wise violet]

Oh

[hollow ice]

well, if sec and csc were flipped in the problem, youd have to use csc(x) = -csc(-x), meaning csc(-x) would be +3.42 instead

[wise violet]

Huh 😭😭

Sorry I don’t really understand :(

[hollow ice]

well, if the problem said If sec(pi/2 - theta) = -3.42, then find csc(-theta) then youd state csc(theta) = sec(pi/2 - theta) = -3.42 from identity 18(or 19)

and then use the other identity csc(-theta) = -csc(theta) to say csc(-theta) = -(-3.42) = 3.42

[wise violet]

Ohhhh

I see I see

Okok I think this all makes sense

Thank you!!

[devout snow]

@wise violet Has your question been resolved?

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[cobalt frost]

I received this question on a midterm review. I was trynig to solve it, but it doesnt make sense. by the alternating series test, the terms of f must go to 0, however, a simple contradiction f = x+2 just keeps oscillating. Is this question incorrect?

[versed juniper]

Wouldn't it be 1/n + 2 then which would basically be an alternating harmonic series with a known sum

[cobalt frost]

hmm. but it still doesnt go to 0 right, which is one of the requirements of the alternating series test?

and wouldn't it constantly just have -2 + 2 -2 + 2 which diverges

[versed juniper]

math is hard

[void fox]

yes this is false as stated

[cobalt frost]

also since yall are here, there are 2 more things i need yall to confirm. there's no way this could be 5x-1 in the exponent right?

[cobalt frost]

thanks

[cobalt frost]

its 5n-1 right? cause i dont see any feasible way to find an anti derivative of x^5x-1

[frozen aurora]

yeah it has to be 5n-1

[cobalt frost]

this problem set has way too many typos

[frozen aurora]

yeah sorry to hear that man

[void fox]

[frozen aurora]

wild ahh

[cobalt frost]

and in this one, the x is supposed to be inside the brackets of the sin right?

cause if not i dont see a point in asking about the existence of second order derivatives when its essentially a linear function

[versed juniper]

has to be...

[cobalt frost]

yea ty guys

.close

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[robust bobcat]

Can someone explain why desmos is telling me that cosx<1 for all x and why since<1 for all x. They both equal one at times so this cannot be true

[wheat pawn]

post the exact thing?

[wheat pawn]

exact values are too small to be represented

put the equality

[robust bobcat]

Wdym? You want ≤?

[wheat pawn]

no, =

[robust bobcat]

Why is that? Interval is too big

[robust bobcat]

Equality doesn't give us anything

[wheat pawn]

should give you any value of the form kpi for sin or kpi+pi/2 for cos

[robust bobcat]

Where? In desmos?

[devout snow]

@robust bobcat Has your question been resolved?

[void fox]

it's just too small for desmos to draw

In general you cannot trust graphing calculators on things like this

[devout snow]

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[vast basin]

How do I solve this I tried to integrate but can't seem to prove it

[fair storm]

You're done with the verification part, right?

[vast basin]

No

[fair storm]

Oh, just take the derivative of y = sin(1/x) for that part.

Have you learnt how to differentiate sin(x)?

[vast basin]

$\frac{-cos(\frac{1}{x})}{x^2}$

[woven radish]

codo0160

[vast basin]

What do I do next?

[fair storm]

$\dfrac{-cos(\frac{1}{x})}{x^2} = \dfrac{x\cos{\left(\frac{1}{x}\right)} + 2\sin{\left(\frac{1}{x}\right)}}{x^2\cos{\left(\frac{1}{x}\right)} - 2x^2\sin{\left(\frac{1}{x}\right)}}$

[woven radish]

Erebus

[vast basin]

Oh ok

I'm there

[fair storm]

Try showing this is true

If it doesn't match up, there might be something wrong with the question.

[vast basin]

It doesn't

Then it means it's not a solution

[fair storm]

Yes.

[vast basin]

But what do I do to the condition?

[fair storm]

Well, a way would be to try finding y first and using the condition to find the required function.

But I think the question might be wrong.

Because the first part failed.

[vast basin]

Yes

[fair storm]

Have you try checking the original source?

[vast basin]

I think maybe it was a trick question

Because that's the same assignment I got

So I'll conclude that it's not a solution

[fair storm]

Well, it'll be even trickier if you found a solution for y. 😆

Let's try to integrate dy/dx.

[vast basin]

I don't know how

What of this

I got 1 = -x + c for question 1

[fair storm]

I think it should be y = -x + c

From how I think of it, since it's asking for other possible answers for y, I think we would just need to find x and find y (not 1).

$y = 1$ is a \text{solution} $\implies (x - 1) \cdot 0 + x - 1 = 0 \implies x = 1$\
\
Solve $(1 - y)\dfrac{dy}{dx} + (1 - y) = 0$

[woven radish]

Erebus

[fair storm]

There might be other possible solution apart form y = 1

[vast basin]

But what happens to c?

Can't it be $y=-x+1$

Since the coefficient of x in y = 1 is zero so we can assume c = 1?

[woven radish]

codo0160

[fair storm]

We're not talking about c yet.

c can be any constant

It's just that we're looking for alternative solutions for y

apart from the given solution which is y = 1

[devout snow]

@vast basin Has your question been resolved?

[vast basin]

No

[vast basin]

.close

[devout snow]

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[sleek steppe]

shouldnt there be a f^(2n+2) term in the addition of F''(x)+F(x)?

[void fox]

What are F and f here

[sand dove]

f^(2n+2)(x) = 0

f is a polynomial of order 2n

[sleek steppe]

oh right

thanks

.close

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[gray lance]

i need to prove this:

[gray lance]

its more cs (big O) but yeah

dont really know where to start

[sand dove]

Theta(...) means both are a big O of the other

[sand dove]

change of log base rule

$\log_b a = \frac{\log_c a}{?}$

[woven radish]

Rafilouyear2026

[devout snow]

@gray lance Has your question been resolved?

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[dusk tide]

how would you solve questions like this?

[void fox]

write the number as 10n+m

n and m are your tens and units digits respectively

[dusk tide]

ok so

[void fox]

and then you can set up equations according to what they said

[dusk tide]

n is my ten tens
m is my unit

10(tens) + (units)

[void fox]

mhm

[dusk tide]

if it says the two digits are reversed is it

10(units) + (tens)?

[void fox]

yup

so 10m+n

[dusk tide]

ok what do i do with these two equations now?

10m+n = 3n + 18??

[twilit field]

solve them as a system of equations.

[dusk tide]

{ units digit + tens digit = 9
{ 10m+n = 3n + 18

is that correct

[void fox]

that's one of the equations you can write down with the given information

[twilit field]

you would want to write the first equation in m and n as well.

[void fox]

mhm but for the first equation you may simply write n + m = 9

[dusk tide]

oh so

n + m = 9
10m + n = 3n + 18

so

[twilit field]

then just use your favourite method to solve the system.

[dusk tide]

something is not right

m = 9 - n

10 ( 9 - n ) = 3n + 18

90 - 10n = 3n + 18

72 = 13n ???

[twilit field]

I see you rearranged for m in the first equation and substituted it into the second equation.

but during the substitution, the + n originally on the left side in the second equation is suddenly missing.

your original second equation was 10m + n = 3n + 18, but after substituting it became 10(9 - n) = 3n + 18.

fixing that should give you the right solution.

[dusk tide]

n + m = 9
10m + n = 3n + 18

m = 9 - n

10 ( 9 - n ) + n = 3n + 18

90 - 9n = 3n + 18

72 = 12n

n = 6???

and then

[twilit field]

well, yes. not sure why you need to put three ?s after the final answer. hope you don't do that on tests.

[dusk tide]

i dont only on my copybooks

[twilit field]

sure.

now m.

[dusk tide]

so commanding

m = 3

[twilit field]

correct.

[candid maple]

to cross-check, you should check that m and n form the two-digit number that was asked for in the question, including by swapping the two digits and checking the condition.

[dusk tide]

so
n = 6
m = 3

10 (6) + (3) = 63

[candid maple]

so 63 is your number. its digits does add to 9 as required.
now, swap the two digits and make sure that the second condition is also respected.

[dusk tide]

so

30 + 6 = 36

[candid maple]

18 more than 3 times the tens digit of the original number (6).

[dusk tide]

very cool concept i like

is it okay if i ask another question here

[twilit field]

yes.

[dusk tide]

would it be possible to do like

{ a/b = 3/5
{ 3a = 20 + b

and then

[void fox]

yes

[candid maple]

that sounds like the right idea.

you may want to write this on paper or a drawing program.

[dusk tide]

yeah

[void fox]

b = 3a - 20

[dusk tide]

huh

if i put b on the other side it will be negative

hold on wa

[candid maple]

the b is positive on the right side, if I'm seeing things correctly. so all you want now is to move the 20 to the other side by subtracting both sides by 20.

[dusk tide]

i tot

a = 15

[candid maple]

you may wish to show your working for checking.

[void fox]

yes it is correct

[dusk tide]

its sending

sorry if its a bit messy cuz i had a big brain moment during solving (i ususally get exciting solving complex questions cuz its fun and interesting lowk but each to his own)

[candid maple]

I'm sorry to say that I can only read the first line (45 = 20 + b), but from what I can make out from the second line (correct me if I am wrong, I apologize for my vision), I see
+b = 20 + 45 (?)
which does not follow from the first line.

[dusk tide]

oh yes

+b = -20 + 45 (i added strokes when i was reversing the signs, my fault)

after that

b = 45 - 20 ( i rewrote it so it looks more manageable for my eye)

b = 25

then

a/b = 15/25 = 3/5
which is true

[candid maple]

if that is the case, then you are entirely correct. please forgive my poor vision.

[dusk tide]

its okay

its my fault for not being clear enough lol
i was rushing because i unlocked a whole new dimension of reality while solving this

[candid maple]

I am glad that happened (?). it always feels nice to solve a problem you had been on for a while.
but also it's not your fault.

is there anything else that you would like to ask though?

[dusk tide]

yes

so

[dusk tide]

looks familiar

[candid maple]

please allow me some time to read the image.

[dusk tide]

looks like y = mx + b

[candid maple]

it is reminiscent of the slope-intercept form of a linear equation, yes.
but in this case this does not matter. you can again create two equations with the two pairs of values, and solve the resulting system as you did for the past two questions.

[dusk tide]

4 - n = (1)m

[candid maple]

now the other one.

[dusk tide]

uh

-2 - n = -2m?

so

{ 4 - n = m
{ -2 - n = -2m

[candid maple]

looks to be correct. now just solve the system as before.

[dusk tide]

i got -8 = 0 :)))

can you also do something like

(4 - n) + (-2 - n) = m - 2m

[candid maple]

can you show your steps, perhaps? maybe something went wrong rather horribly.
I did get m = 2 from a preliminary attempt earlier.

[dusk tide]

maybe the way i did it was too complex

[candid maple]

this does not help, as we still have two variables in a single equation.

[dusk tide]

oh yeah

oh i did it wrong

its corect

m = 2

had a little oopsie daisies

anyways

4 - n = 2

n is also 2

[candid maple]

that is correct.

so would there be anything else you need help with?

[dusk tide]

huh

all i can do is
1085 / 5 = 217 km per hour

[candid maple]

if you let the speed of the wind be w and the speed of the plane be p, can you form a system of equations with w and p?
keep in mind the relationship between speed, distance and time when doing so.

[dusk tide]

w is TBD
p is 217

uhh

i actually dont know the equation for

speed distance over time right

[candid maple]

p is not going to be 217. form the equations with w and p as they are.

[candid maple]

speed is indeed distance per time, yes.

[devout snow]

@dusk tide Has your question been resolved?

Channel closed

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[dusk tide]

.reopen

[devout snow]

✅ Original question: #help-27 message

[dusk tide]

@candid maple would it be like

.

[candid maple]

double-checking that what I am seeing is
P + W = 217
P - W = 155

I haven't checked the numbers, so if the numbers are correct, then this is the right method.

[dusk tide]

its correc

ti got

i got 31

[candid maple]

well done! is there anything else you would like to ask?

[dusk tide]

NOPE!

hopefully...

[candid maple]

alright, then I believe you may close the channel. all the best for your next questions!

[dusk tide]

thank yoU!

also yeah thats it

frfr

thanks

how do i close

.close

[devout snow]

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[simple hatch]

how do i solve for x?

[simple hatch]

ik i need to like, cube the entire thing and solve that but i'm doing something wrong when i cube it

cause my calculator's giving me imaginary numbers and the answer should be approximately 0.003684

[versed juniper]

what are you putting in the calculator

how would you even put this into a calculator

[simple hatch]

it has a polynomial solve function

[versed juniper]

oh wow

[sacred sedge]

a CAS type

[simple hatch]

so i put in x^3 + 0x^2 -(number)x + 0

cus if i cube it'll be -ve still

but... it seems to give me imaginary numbers

so something's not right

[versed juniper]

i don't think you cubed it right...

[simple hatch]

probably, that's basically my question

[pseudo basin]

are we required to use a calculator here

[versed juniper]

you should have x^4/3 terms and such

[simple hatch]

how do i cube it cus clearly i'm doing it wrong lol

[versed juniper]

well do you know that (a + b)^2 isnt a^2 + b^2

[simple hatch]

not really, i'm looking for like

do i cube the number before the x or whatever

[versed juniper]

why not just move the decimal term to the right side

then cube

[simple hatch]

yeah ofc

[versed juniper]

oh ok

[simple hatch]

oh wait so

take out x

outside the brackets

then continue solving?

[pseudo basin]

here is something to relieve headache

[simple hatch]

i've not done this in ages so i'm really rusty

[pseudo basin]

replace the big scary decimal number with something like A

for compactness

$x - Ax^{1/3} = 0$

[woven radish]

Ann

[simple hatch]

tried and true method

[pseudo basin]

add $Ax^{1/3}$ to both sides and get $x = Ax^{1/3}$

[woven radish]

Ann

[pseudo basin]

do you see how this can simplify further

[sacred sedge]

since you have Cas, just define A = that decimal

[simple hatch]

honestly no, my brain is fried rn

i've been doing this for a few hours

[pseudo basin]

why are you doing math with a fried brain

girl go take a break

[simple hatch]

but it should be so simple 😭

like i'm just trying to find when 2 depths = the same depth

so when i minus 1 from the other and get 0

anyway this is a good jumping off point at least

ig i should take a break

i may return in... an hour potentially. thanks for the help lol

.close

[devout snow]

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[hasty hedge]

Can someone e help me out with without this problem

[main atlas]

Yes

Basicly you need to see if the operations made to pass from one expression to another are “legal”

Or maybe another way, is to give some values to x , and see that both expressions give the same result

But the first method is better, as its important to know what operations you can and cant do