help-28

[woeful pasture]

plz ping me

when help arrives

[grave elm]

any progress?

[woeful pasture]

so i know that

For ${u, a_n \in \mathbb{U}_n}$,
[ { [u][a_0], [u][a_1], \dots, [u][a_n]}]
is also an emueration of ${\mathbb{U}_n}$

[glossy valve]

k

[grave elm]

For what part is this?

[woeful pasture]

and every element has a multiplicative inverse

[woeful pasture]

a

[grave elm]

It already sounds overcomplicated, but continue

[woeful pasture]

so im thinking

if i say that ${[u][a_n]}$ is the inverse of ${[u][a_k]}$ for some ${n,k}$, then [ [u]^2 \equiv 1 \mod (n)]

[glossy valve]

k

[woeful pasture]

but

this isnt always true

so idk what to do next

[grave elm]

Have you tried using the hint?

How many elements are in that set?

You dont even have to find the exact number

upper bound is more than enough

[woeful pasture]

tbh i dont know how to proceed..

[grave elm]

Do you know how big this set is?

At least approximately

very approximately

compared to the size of N

[woeful pasture]

less thanor equal to U_n

[grave elm]

exactly

for our purposes, it actually suffices that it's finite

do you know the pigeonhole prinicple?

[woeful pasture]

not exactly.. but is it smth like if there are more birds than hole. there will be at least a hole with more than one bird

[grave elm]

Yeah, right

imagine that you are assigning positive integers (pigeons) to elements of Un (holes)

is it possible to do it in such a way that each hole has at most 1 positive integer?

[woeful pasture]

no

[grave elm]

Why?

[woeful pasture]

Un is less than N

[grave elm]

Exactly

We even know that there must be infinitely many integers falling to the same element of Un

the question asks only for 2 tho, so thats more than sufficient

there must exist 2 integers i, j with [u]^i = [u]^j

so that's part a)

[woeful pasture]

holy damn

[grave elm]

btw to notice such solutions, it often suffices to just try it on specific example. If you actually try assigning integers to elements of Un for some n, youll very soon notice that there isn't enough elements of Un to accomodate all of N

do you also need help with part b and c?

[woeful pasture]

gimme a sec to digest

so how do we know tho

that there doesnt exist a [u] such that it doesnt get mapped to any natural number

[grave elm]

It can exist

or at least we dont know that it doesnt exist yet

but what we know is that there exist some [u], to which at least 2 naturals get sent

and that's important

you have infinitely many objects and finitely many categories

obviously there must be a category which gets more than 1 object

we dont care about categories which get no object

they might exist, but that shouldnt bother us

[woeful pasture]

so this isnt true for any [u] in U_n?

[grave elm]

statement a) itself is true

for any [u]

Suppose [u] is in Un

then consider the set of all powers of [u]

this set must be finite - but was generated by infinitely many integer powers

therefore at least two different integer powers i, j must've resulted in the same element of U. [u]^i = [u]^j

like you have
u^1, u^2, u^3, u^4, u^5, ......

which are all equal (in some order) to the following elements of U
U = {u1, u2, u3, u4, u5, ..., un}

[woeful pasture]

ahhh

wait

i think i got it

so just for [u] alone

u^1, u^2, u^3, u^4, u^5, ... can map to whatever

but at least two will fall into the same box

hence that

[grave elm]

Yep, you just have to add three dots

[woeful pasture]

now

for part b

for some k

it must map to [1] as well

becuz [1] is in U_n

[grave elm]

Why?

How do you know that some boxes wont stay empty?

[woeful pasture]

hmm

fair

[grave elm]

Suppose [2]^2 = [2]^7, can you find k such that [2]^k = 1?

[woeful pasture]

this is mod?

[grave elm]

Yes

you dont even need to know n to do it, just use what you know about the properties of Un

[woeful pasture]

ill multiply both sides by inverse of [2]^7

[grave elm]

inverse of [2]^7 isnt 2^[7]

[woeful pasture]

inverse of it, srry

[grave elm]

this would work if [2]^2 * ([2]^7)^(-1) was a positive integer power of [2]

keep in mind that [2]^7 is just [2] * [2] * ... * [2] 7 times

and same for [2]^2

[woeful pasture]

k = 5?

[grave elm]

Can you prove that it works?

can you prove that [2]^5 = 1?

[woeful pasture]

i mean if it isnt one

then [2]^2 would be different from [2]^7, no?

[grave elm]

why?

[woeful pasture]

[2]^7 = [2]^2 * [2]^5

in general, if j > i, and [u]^i = [u]^j, then [u]^(j-i) = 1 satisfies the existence statement

[grave elm]

Yes, but im not fully convinced by your proof that [u]^(j-i) = 1 yet

[grave elm]

this is a nice equality and i can see that it's true, but how does it imply that [2]^5 is 1?

[woeful pasture]

if it is something else, then the equality wouldn't work, would it?

[grave elm]

why?

[woeful pasture]

each multiplication is unique

[grave elm]

Okay, now we are getting to a better argument

Yeah, sure, that works

or alternatively, multiply both sides by inverse of [2]^2

and it cancels (it also cancels [2]^7, since it's equal to [2]^2)

leaving 1 = [2]^5

[woeful pasture]

oh ye

[woeful pasture]

i proved this using that 😭

[grave elm]

now the last question

c)

[woeful pasture]

[u]^(k-1) [u] = [1]

so its the multplicative inverse of [u]?

[grave elm]

yep

[u]^(k-1) = [u]^(-1)

[woeful pasture]

thank u

[grave elm]

np

[woeful pasture]

i appreciate ur bearing with me 👍

.solved

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[wise gate]

Let $\mathbb T={z\in\mathbb C:|z|=1}$ and $\mathbb T^\infty=\prod_{n\in\mathbb N}\mathbb T$ with the product topology and the Haar measure

Is the space of continuous functions $\mathbb T^\infty\rightarrow \mathbb C$ dense in $L^p(\mathbb T)$, $1\leq p<+\infty$?

[glossy valve]

d

[wise gate]

I know this is true if instead of $\mathbb T^\infty$ we choose a compact subset of $\mathbb C^n$

[glossy valve]

d

[wise gate]

And that it happens for more general topological spaces, but I don't really know this more general case in detail

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[young night]

\documentclass[12pt]{article}
\usepackage{amsmath, amssymb}
\usepackage{geometry}
\geometry{margin=1in}

\title{Geometric Properties of Linear Transformations}
\author{}
\date{}

\begin{document}

\maketitle

\section*{Task (d)}

\textbf{Problem:} Let ( n \in \mathbb{R}^{2 \times 2} ) be an invertible matrix satisfying the condition
[
|\det(n)| = 1.
]
Such a matrix represents a linear transformation that continuously bends slopes in ( \mathbb{R}^2 ) without changing the area of geometric figures.

\bigskip

\textbf{Question:} \textit{Explain geometrically how these matrices act on vectors in ( \mathbb{R}^2 ), and how their inverses correspond to reciprocal transformations. Provide a clear and rigorous description of the geometric effect of the transformation ( n ), and why its inverse ( n^{-1} ) precisely undoes this effect.}

\bigskip

\vspace{3cm}

\noindent\rule{\linewidth}{0.5pt}

\vspace{2cm}

\noindent\rule{\linewidth}{0.5pt}

\bigskip

\textit{Hint:} You may consider examples such as rotations, reflections, or area-preserving shears, and describe what happens to direction (slope), position, and area under both ( n ) and ( n^{-1} ).

\end{document}

[young night]

.close

[glossy valve]

ppq#7826
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[young night]

.reopen

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✅

[young night]

\documentclass[12pt]{article}
\usepackage{amsmath, amssymb}
\usepackage{geometry}
\geometry{margin=1in}

\title{Geometric Interpretation of Reciprocal Linear Transformations}
\author{}
\date{}

\begin{document}

\maketitle

\section*{Task (d) Proof}

Let ( n \in \mathbb{R}^{2 \times 2} ) be an invertible matrix such that ( |\det(n)| = 1 ). Then ( n ) defines a linear transformation ( T_n: \mathbb{R}^2 \to \mathbb{R}^2 ), where ( T_n(\vec{v}) = n\vec{v} ). Since ( |\det(n)| = 1 ), this transformation preserves the area of any region in the plane. It may rotate, reflect, or shear vectors, but it does not scale them uniformly. Geometrically, this means that ( n ) continuously "bends" or redirects the slopes of vectors without tearing, compressing, or expanding space.

Because ( n ) is invertible, the inverse transformation ( T_{n^{-1}}(\vec{w}) = n^{-1} \vec{w} ) reverses this action. That is,
[
T_{n^{-1}}(T_n(\vec{v})) = n^{-1}(n\vec{v}) = \vec{v},
]
for all ( \vec{v} \in \mathbb{R}^2 ). This implies that any geometric effect of ( T_n ) is undone exactly by ( T_{n^{-1}} ).

For example, if ( T_n ) rotates vectors counterclockwise by angle ( \theta ), then ( T_{n^{-1}} ) rotates them clockwise by ( \theta ). If ( T_n ) reflects across a line, ( T_{n^{-1}} ) applies the same reflection. Thus, the pair ( (n, n^{-1}) ) forms a reciprocal relationship in which the first transformation bends or redirects slope, and the second restores it exactly.

\textbf{Conclusion:} Matrices ( n ) with ( |\det(n)| = 1 ) define area-preserving, slope-altering transformations in ( \mathbb{R}^2 ), and their inverses precisely undo these effects, returning all vectors to their original position and direction.

\end{document}

[glossy valve]

ppq#7826
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[young night]

<@&286206848099549185> would this be good and fine

[gritty flax]

!15mn

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[cunning sapphire]

That from you or from our friend GPT?

Because there isn't much content despite this wall of text, and even less proof of the statements

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[terse spade]

Help

[terse spade]

How to do the last one?

[steel solar]

Transform this into a system of 3 linear equations by, say, subsituting v=1/x and u=1/y

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[terse spade]

No

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[candid flint]

Hi can someone pls help on this question

[noble mountain]

try to apply the arc angle theorem

[candid flint]

ok

Im actually rlly confused

[unreal wren]

Also, remember that isosceles triangles have 2 similar angles

[noble mountain]

^

[candid flint]

geometry is my weakest point

can you pls explain

[noble mountain]

Try to find the measures of the components of x

it seems like its made up of two angles

Can you determine the larger one?

[candid flint]

I know that the triange with the angle 59 is right angled

[noble mountain]

Actually yeah wait thats so much easier than what I said first

you know that angle is 59 so what is the other angle

besides the right angle

[candid flint]

31

[noble mountain]

good so thats one component angle of x

the other one you can determine

can you figure out what this angle is

[candid flint]

thats also 90

[noble mountain]

so that means the sum of the other two angles is also 90

[candid flint]

ohh

so 59 + ? =90?

[noble mountain]

If you notice, the legs of that right triangle are radii of the circle

which means they have the same length

what do we know about triangles with two equal sides

[candid flint]

yes

so basically the two triangles are the same

right?

[candid flint]

they have the same angles?

[noble mountain]

this is what you've found out so far

does any of this seem unfamiliar

you told me the angle opposite to the 59 was 31

[candid flint]

yes

[noble mountain]

lemme add letters

your goal is to solve for angle ABC basically

[candid flint]

yes so is DO the same as BO

[noble mountain]

Yes, but that isn't going to help us

[candid flint]

and AB is the same as AD

[candid flint]

oh ojk

[noble mountain]

We know that ABC = ABO + OBC

does that make sense

two parts of the whole

[candid flint]

yes

[noble mountain]

OBC = 31

Now we need ABO

We know that triangle AOB is a right triangle

Because angle AOB is a right angle

[candid flint]

and is ADC = ABC

[candid flint]

yep

[noble mountain]

not quite

[candid flint]

oh alr

[noble mountain]

we also know that AOB is isosceles

Why do we know that

Can you tell me why

What do we know about AO and BO

[candid flint]

they are thge same

[noble mountain]

so what does that mean for the triangle AOB

[candid flint]

oh so isoceles means that two sides are the same

[noble mountain]

yeah but more importantly since we know AOB is isosceles, we can also say something else to the two angles ABO and OBA

[candid flint]

wait im confused

[noble mountain]

what are the properties of an isosceles triangle

[candid flint]

the angles are acute

[noble mountain]

what else

search up what an isosceles triangle is

[candid flint]

its a triagnel with two equalk side

sides

it has one unequal side which is the base

and two equal sides

[noble mountain]

thats one important property

whats the other one

[candid flint]

i realyl sorry im clueless

[noble mountain]

its the one we used to determine AOB is isosceles, because we know AO and BO are equal

[candid flint]

oh yes

and that is the two sides that make the triangle isocles

[noble mountain]

yes thats a defining characteristic of isos. triangles

We will use the other important property to finish the question

I don't wanna give like the answer to you so I want you to find it out

[candid flint]

is it that opposite angles equal to 90

[noble mountain]

Search it up

Not necessarily for isosceles triangles

[candid flint]

it has two equal angles

[noble mountain]

good yeah

so we are ready to put it together

[candid flint]

so 59 + ? = 31 +x

[noble mountain]

where did you get that from

[candid flint]

because of the two equal angles

so like ADC isnt equal to ABC?

[noble mountain]

No ADC is not equal to ABC

[unique perch]

pay attention to ∆AOB only

[candid flint]

oh alr

[noble mountain]

just so we don't have to scroll

[candid flint]

yep

thanks

[noble mountain]

what are the components of angle x

Like what angles make up angle x

I erased x from this picture but its in your original 1

[candid flint]

b - 31 = x

x + ? + 90 = 180

[noble mountain]

this is your original

[candid flint]

yes

[noble mountain]

see how theres an arc next to x

[candid flint]

yes

[noble mountain]

so in this picture angle x corresponds to angle ABC

[candid flint]

Yes

[noble mountain]

this one

not angle ABD

do you see the difference

its not the smaller angle its the larger angle

[candid flint]

yes

this is coz of the angles of the triangle below the isoceles

the one with 59 and 31

ADO would be a smaller angle

while ABO will be the larger angle

[noble mountain]

let me call you I think its easier to show while I screen share

[candid flint]

alr sure

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[unreal wren]

where did i mess up help 😭

[unreal wren]

how the heck was i able to get that f(x) was linear when the correct proof literally showed examples which arent linear??? Where did i go wrongggggg

oh wait, forgot to specify that b must be an integer by given and the way i used it in the first line of my proof

f(x+y) - f(x) - f(y) = k where k is an integer
plugging in x,y = 0 we get : k = -f(0)
to make things tidier set -k = b = f(0)
b must be an integer too, thats why c must be equal to a-b
(a-b) * x - b - cx = -b which is an integer, thus setting c = a-b always works cause f(x) must always be linear

[long walrus]

ok

your jump to “ f(x + y) = f(x) + f(y) ” (and hence the whole linear chain) sneaks in here

f(a+1) = f(a) + f(1) (you wrote b = f(1))

[unreal wren]

uhm okay

[long walrus]

but the problem only guarantees

f(a+1) – f(a) – f(1) is an integer

[unreal wren]

huh

[long walrus]

not that it equals 0. each time you add another 1 you could pick a different integer error term so the induction step

[unreal wren]

wait look at this

i wrote a "-b" in that statement that u didnt read 💀

[long walrus]

where am i looking

[unreal wren]

thats why there was a "b" theree

[unreal wren]

i have no idea

top line

[long walrus]

uh

ok

your very first line takes the given condition f(x+y) – f(x) – f(y) is an integer and rewrites it as the single equation f(x+y) = f(x) + f(y) – b (with one fixed integer b = f(0))

[unreal wren]

yes yes

correct

thank you for reading it

[long walrus]

that step is not valid

[unreal wren]

what how

[long walrus]

the integer difference can depend on the pair (x,y) so you cannot replace it by one constant b for every x and y. once you lock in that false “– b” the rest of your work forces f to behave additively and look linear but the original problem never said the same integer works for all pairs hence the linear conclusion is an artefact of that incorrect simplification

[unreal wren]

oh.

makes sense

[long walrus]

fiyuh

[unreal wren]

wellp thats stupid (of me)

[long walrus]

it happens

[unreal wren]

ty ash

[long walrus]

we all do it

np

[unreal wren]

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[last bay]

How to solve this?

Question:
Find all pairs of consecutive odd positive integers both of which are smaller than 10 such that their sun is more than 11.

[frail bone]

Do you want the explanation in other words?

[last bay]

What do you mean

Teacher solves this question in lecture but I don't understand

[frail bone]

I mean the question phrased differently/ more understandable

[last bay]

I myself is not able to understand

This question is from linear inequalities

[onyx glen]

the most difficult part of it is to decipher the language

and translate carefully from English to algebra

[last bay]

I think so

[frail bone]

Basically you are searching for consecutive odd integer pairs that are smaller than 10.
An example would be 3 and 5 and the sum of the pair needs to be over 11. (Obv 3 and 5 as a sum Isnt 11)

[onyx glen]

you need to:

• find all pairs of integers, such that
• said integers are positive, odd and consecutive,
• both integers in the pair individually are less than 10, and
• their sum is more than 11.

[last bay]

But the problem is not that I cannot solve. This question is oral but If this question came in my school test how will I do it properly

Because the format matters as well in school

I can write you the format. how my brother solve this question then you help me define that

[onyx glen]

well for that the best source is your teacher tbh

ask THEM what they expect for format

[last bay]

Because I am not able to understand the format by which it is solve

[frail bone]

Ig you can try out. Odd numbers under ten 10 are: 1 3 5 7 9 and then you try out revery combination then count all combinations that has a solution over 11

[last bay]

Ahm the teacher is not of my school. I study online

[onyx glen]

i could share how i personally would have written it
or you could share the worked solution that you are given but don't understand

which one would you prefer

[last bay]

Yeah let me send mine

And you try yours as well

Let me see how you will solve. If I like yours I will definitely try that

[onyx glen]

so which one first, yours or mine?

[onyx glen]

trial and error is no good btw

[last bay]

I am writing

[onyx glen]

ok then let's see yours first

[jagged cairn]

@odd cedar RAAAAA

[frail bone]

I mean to go sure you could do every equation with every combination to be exact sure

[onyx glen]

@last bay you there still?

[last bay]

I will tell you in a few hour

[onyx glen]

...

[last bay]

I gotta go

[onyx glen]

well then close this channel and open it again in a few hours lol

[last bay]

Ok

How can i

Close the channel

[frail bone]

With .close

[gritty flax]

".close"

[last bay]

.close

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[last bay]

This topic is of which grade?

[unreal wren]

That's the defining rule for bases

Wdym?

[severe bolt]

[unreal wren]

Wait 900_9 doesn't exist 💀

I just realized, I think you wrote something wrong?

Okk

That makes sense

The fact that 1000_9 = 729_10 is pretty much it's defining rule

nth digit of a base b -> the digit • b^(n-1)

Oowh

Ok, can you elaborate on what you mean by "why does it work"?

Ok

Owh, like a general formula for translating?

Lol just get used to it

That kind of question is pretty trivial for a help channel

Just practice translating those numbers and everything will be fine

Not really

I don't use other bases much so I never studied that seriously

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[cunning epoch]

What does this mean:

[thick hedge]

using 3 and 4

[cunning epoch]

So, 5 is true because of 3 and 4? Something like that?

[dense jungle]

3 and 4 both show a variable being equal to 8

substitution using steps 3 and 4

[cunning epoch]

Well, I kinda see what they are saying but it's still a bit confusing.

Maybe the English translation would be "See steps 3 and 4 for how we got the answer to 5"

[thick hedge]

more specifically

note KN=KM+MN=8

and KL=8

so what can I replace 8 with

[cunning epoch]

Good enough, I'll let sink in a bit. Thank You both.

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[solid ruin]

Can someone help me walk through the decomposition of
the permutation (231)(74)(2359)(16)(532)(167) in S_9
into a product of disjoint cycles

[solid ruin]

I got (16)(293)(47) but Im not confident

[vast fossil]

The given permutation sends 1 to 2 rather than to 6, so this is a different permutation. Could you show your work so we can spot the mistake?

[solid ruin]

I cant show work because i did it in my head but i see what ur saying let me retry

Can u please tell me the way u do it because i keep getting confused
heres what i do:
start with 1 and look at last cycle 1 sends to 6, check the cycle before theres no 6 check before it, 6 sends to 1, no 1 in the 2 cycles before it , go to first cycle 1 sends to 2, so overall 1 sends to 2
now start with last cycle theres no 2 go back theres 2 sends to 5 then third cycle 5 sends to 9 and then 9 to itself so 2 sends to 9,

[onyx glen]

lemme repro

[solid ruin]

Idk im explaining exaclty how i think its not clear

[fast peak]

you said yourself that overall 1 is sent to 2. so your decomposition should start with (12 ...)

not (16)

[solid ruin]

I know this is me retrying

[onyx glen]

btw

do we compose right to left like functions

or left to right

[solid ruin]

Im asking if my thought process is ok

Right to left

[onyx glen]

ok

well your thought process looks OK but it's also kind of a devil in the details thing

[vast fossil]

Usually it comes down to how you would usually find the decomposition of a permutation, but in this case you can do some simplifications before that

For example, note that the last two cycles commute and also rewrite (167) as (16)(67)

[solid ruin]

I dont think this simplification helps a lot no?

[fast peak]

well the two (16)s cancel

tho I also wouldnt really think like that. I would just do over and over again what you wrote above

[solid ruin]

I got (1 2 9 3 5
and now I got 5 sends to 2 so what is wrong here ?

[onyx glen]

3 doesn't get sent to 5

i wrote out each row of small digits right to left

[solid ruin]

Oh I had a typo in the permutation

[onyx glen]

ok what's the correct permutation then

[solid ruin]

Oops sorry
(235)(74)(2359)(16)(532)(167)
5 instead of 1 in first cycle

wait so ( 16) is correct rigt ?

[fast peak]

1->6->1

[solid ruin]

(16)(2935)(47
and now i have 7 sends to 6:/

[onyx glen]

no

1 gets sent back to 1, it's a fixed pt

[solid ruin]

Oh thats how it works? ok

OK then (2935)(476)

[onyx glen]

yup correct

[solid ruin]

Great

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[torn jolt]

guys

i'm having some issues understanding generated subgroup

[torn jolt]

i don't quite get the definition

[vast fossil]

Which part of it is confusing exactly? Do you understand what's being written? The generated subgroup is the set of all products you can get with the elements of A and their inverses

Alternatively, products that you can get with powers of elements of A and their inverses

[torn jolt]

i don't understand what it is exactly

like what's A

is it just a normal element

[vast fossil]

It's a set of elements of the group

[torn jolt]

okkk

[torn jolt]

but like

what does that mean

really

i can see that it's the product of powers and inversers

but what significance does it have?

[vast fossil]

If you are interested in another characterization, the generated subgroup is also the smallest subgroup that contains A

[torn jolt]

is like the smallest subgroup that meets the qualifications of a group?

[torn jolt]

hmmm

[vast fossil]

No, that would be the trivial subgroup

[queen gull]

smallest subgroup containing A

[vast fossil]

But if you consider ones that contain A, the smallest one is the generated subgroup

And it makes sense, because if I tell you a subgroup contains elements of A, then it should also contain their powers, inverses and their products

And there shouldn't be any other type of elements in there if it's the smallest such subgroup

[torn jolt]

got it!!

that was easy i was a bit confused

thanks guys!

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[dull flint]

In a 𝜟ABC if a/1 = b/√3 = c/2 then-
A) A+B-C = 90°
B) the triangle in acuted angled
C) A, B, C are in A.P
D) the triangle is obtuse angled

[dull flint]

I guess I need to somehow show that A, B, C are in AP

Idk I'm not sure how to show that

How to use that condition to show it

<@&286206848099549185>

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[hushed vigil]

@dull flint

[dull flint]

.reopen

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✅ Original question: #help-28 message

[slate glacier]

Damn, was thinking but it's fine, sorry for the delay

[candid hazel]

use the sine law

[dull flint]

Doing

[slate glacier]

It's pretty straightforward after that

[slate glacier]

Let us know when you get it mate

[dull flint]

I got sinA√3 = 2sinB = sinC√3

[slate glacier]

I'm not sure if you were supposed to do that

[dull flint]

Are you saying to use the sine law which have R

[slate glacier]

No what I think you're supposed to do is find the angles

So like a/sinC = b/sinA = c/sinB

[hushed vigil]

wait

[dull flint]

A+B+C = 180° also if I let given ratio = k

[hushed vigil]

isnt it a/sinA = b/sinB = c/sinC?

[slate glacier]

yeah I just imagined a triangle

It's the vertex opposite to the side yeah?

[hushed vigil]

yup mb

[dull flint]

Yes

[slate glacier]

okay

so a/1 = b/sqrt(3) = c/2

now multiply all of them by 2, what do you get?

[dull flint]

2a=2b/√3 = c

[slate glacier]

okay

now can we write 2a as a/1/2?

[dull flint]

Yes

[slate glacier]

and 2b/sqrt(3) as b/sqrt(3)/2?

[dull flint]

Yes

[slate glacier]

okay

so a/1/2 = b/√3/2 = c/1

And the sine rule says

[dull flint]

Okay I see angless on comparing

[slate glacier]

AB(a)/sinC = BC(b)/sinA = CA(c)/sinB

Yes

Get those angles, I believe that's all you'd need

Okay, so check (a)

Is it correct?

[dull flint]

No

Neither B

[hushed vigil]

wait what

[slate glacier]

Eh, why not B?

[dull flint]

Acute angle is what we should we two angles sum up to 90°

Yes then it's true

Mb

[slate glacier]

wait wait

[dull flint]

mb I am getting confused

[slate glacier]

I might be wrong, but it's a right-angled triangle right, so it shouldn't be obtuse

I guess you were right earlier

[dull flint]

Then option c is left

And it fives 2A = C+B

But it means A, C, B are in AP not A, B, c

C*

[slate glacier]

no no

2B = A + C

wait

well B is whatever the middle angle is

You've got 30, 60 and 90 yeah?

[dull flint]

C, A,B ^

Yes

[slate glacier]

wait A was 30

B was 60

and C was 90

so B is the middle one

2B = 120, and A + C = 120

[dull flint]

I was having doubt in this

That's why I got order wrong

[slate glacier]

Ahhhhh that's fine man

[dull flint]

Okay so option c

Thanks

[slate glacier]

And (d) is .....?

[dull flint]

Incorrect

[slate glacier]

Yes

[dull flint]

Right triangle can't be obtuse

[slate glacier]

Yes

So the correct answer/answers is/are?

[dull flint]

Spent 40 minutes on this question 😭😭🤫😭

[dull flint]

Correct is C only

[hushed vigil]

more time spent = more knowledge gained

[slate glacier]

Correct!

[dull flint]

true

[maiden vapor]

Hi Blackido, have you solved this?

If not, I could help…

[dull flint]

Yes just now

[slate glacier]

Thanks guys for introducing me to the Sine law, was my first time knowing about it and using it

[dull flint]

oh because of me your learnt that

[hushed vigil]

damn dude

[slate glacier]

Yes

[hushed vigil]

good job

[dull flint]

I mean it helped 2-3 people here to learn

[slate glacier]

I saw the formula on google and worked with what I understood

[dull flint]

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[gaunt lake]

How do u work this out? I've been trying to revise for a math test and I don't understand it

[maiden vapor]

Which one?

[gaunt lake]

Honestly all but mainly the fourth

[maiden vapor]

Do you know all of the exponent rules?

These rules

[gaunt lake]

Not rlly, I didn't understand it

I know most of them

But not all

[maiden vapor]

Well, you only need some of them for these questions. For example, the first question can be solved immediately by applying the 'Power Rule' from the table.

[gaunt lake]

So it'd be g-12?

[maiden vapor]

That's correct

[gaunt lake]

For the second one, would it be 51x3y4? I got that before but I was unsure from the y

[stiff pier]

Might wanna use carrot symbols or superscripts to avoid ambiguation. Do you mean 50𝑥³𝑦⁴?

[gaunt lake]

Yeah

Sorry I don't know how to write it down here

[stiff pier]

There are a couple mistakes there.

[gaunt lake]

Was 4 meant to be a negative?

[stiff pier]

$17x^3y^{-2}\times3xy^{-2}$

[glossy valve]

𝙸𝚗𝚏𝚒𝚗𝚒𝚞𝚖³

[stiff pier]

𝑦⁻² = 1/𝑦²

so you can rewrite

$\frac{17x^3}{y^2}\times\frac{3x}{y^2}$

[glossy valve]

𝙸𝚗𝚏𝚒𝚗𝚒𝚞𝚖³

[stiff pier]

@gaunt lake are you understanding everything im doing so far

[gaunt lake]

Yeah

[stiff pier]

i will continue

$\frac{17x^3\times 3x}{y^4}$

oops

[glossy valve]

𝙸𝚗𝚏𝚒𝚗𝚒𝚞𝚖³

[stiff pier]

Are you able to figure out to do with the numerator?

[gaunt lake]

Is it 51x3?

I didn't know how to put it up

[stiff pier]

It’s fine i understand what you mean.

not quite, you’re close, and you did the multiplication with the coeffecient correctly, but you need to distribue the 𝑥 part to the 𝑥³

$\frac{3\times17x^3\times x}{y^4}$

[glossy valve]

𝙸𝚗𝚏𝚒𝚗𝚒𝚞𝚖³

[stiff pier]

so multiply 3 by 17

and then 𝑥³ • 𝑥

what would you get?

[gaunt lake]

Would it be 51x4? Or different?

[stiff pier]

bingo

that is correct.

so we have

$\frac{51x^4}{y^4}$

[glossy valve]

𝙸𝚗𝚏𝚒𝚗𝚒𝚞𝚖³

[stiff pier]

and then we can rearrange the bottom 1/𝑦⁴ into 𝑦⁻⁴

so you can either write

$\frac{51x^4}{y^4}$ or $51x^4y^{-4}$

[glossy valve]

𝙸𝚗𝚏𝚒𝚗𝚒𝚞𝚖³

[gaunt lake]

Why would y be negative though?

[stiff pier]

Are you asking why the exponent law works or are you curious how to apply it?

[gaunt lake]

No because I thought that when u multiply both negatives it would make a positive

[stiff pier]

That’s for constants/numbers, not exponents

[gaunt lake]

Ohh

[stiff pier]

UNLESS

the base is the same

$x^{-1}x^1=x^0=1$

[glossy valve]

𝙸𝚗𝚏𝚒𝚗𝚒𝚞𝚖³

[gaunt lake]

Alright, thx!

[stiff pier]

np!

do you need any more help? Otherwise you can type .close to close the channel

[gaunt lake]

If u don't mind, how do u do this last question?

[stiff pier]

$\left(\frac{a}{b}\right)^{-c}=\left(\frac{b}{a}\right)^{c}$

[glossy valve]

𝙸𝚗𝚏𝚒𝚗𝚒𝚞𝚖³

[stiff pier]

so first do that.

[gaunt lake]

So it'd be (27/64) 2/3?

[stiff pier]

*if you would like me to explain why that identity holds i can do that aswell.

[stiff pier]

Yes.

$\left(\frac{27}{64}\right)^{2/3}$

[glossy valve]

𝙸𝚗𝚏𝚒𝚗𝚒𝚞𝚖³

[stiff pier]

then distribute the power onto the numerator and denominator

[gaunt lake]

Like 27^2/64^3?

[stiff pier]

Not quite.

you distribute the entire thing onto the numerator and denominator

[gaunt lake]

So, both 27 and 64 would have 2/3?

[stiff pier]

a power of ²⁄₃ yes

[gaunt lake]

27^2/3 and 64^2/3? That's what I understood

[stiff pier]

Do you mean $\frac{27^{2/3}}{64^{2/3}}$

[glossy valve]

𝙸𝚗𝚏𝚒𝚗𝚒𝚞𝚖³

[stiff pier]

if so then yes

and then use the property to convert the fractional power into a radical

$a^{b/c}=\sqrt[c]{a^b}$

[glossy valve]

𝙸𝚗𝚏𝚒𝚗𝚒𝚞𝚖³

[gaunt lake]

So 3√27^2 and 3√64^2?

[stiff pier]

yup

$\frac{\sqrt[3]{27^2}}{\sqrt[3]{64^2}}$

[glossy valve]

𝙸𝚗𝚏𝚒𝚗𝚒𝚞𝚖³

[stiff pier]

you can also rewrite it this way:

$\sqrt[3]{\frac{27^2}{64^2}}$

[glossy valve]

𝙸𝚗𝚏𝚒𝚗𝚒𝚞𝚖³

[gaunt lake]

[stiff pier]

if you need help with anything else feel free to ask

[gaunt lake]

No that's all, I'll try the third myself this time, thanks! (⁠つ⁠≧⁠▽⁠≦⁠)⁠つ

[stiff pier]

❤️

[gaunt lake]

How do u close this?

(⁠─v─⁠|⁠|⁠）

[stiff pier]

.close

[gaunt lake]

.close

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[silver walrus]

Guys, how do I solve this Quadratic Inequality using Test Point here's the problem btw: x² - 5x + 6 > 0

[azure onyx]

have you found the roots of this quadratic expression?

[silver walrus]

Yea, It's ( x + 2) (x + 3) then x= -2 and x = -3

[azure onyx]

wrong signs for the roots.

this will give you x^2 + 5x + 6.

[silver walrus]

Wait wha

[azure onyx]

you found (x + 2)(x + 3) as your factors. but expanding these factors gives you x^2 + 5x + 6, not x^2 - 5x + 6 (note the sign of the 5x).

[silver walrus]

Oop alr

So like when factoring it's like -2 and -3 and you plus it to have -5?

[azure onyx]

if it's what I think you mean, then yes.

[silver walrus]

While for 6 it's multiplication -2 x -3 it becomes 6 becusse of positive

[azure onyx]

but to avoid any confusion, please show us your new factors.

[silver walrus]

Is this correct so far?

[azure onyx]

those are the correct factors/roots, yes.

[silver walrus]

So to make a test point is it like this

[azure onyx]

that's the correct setup, yes. you now need to pick a point within each interval and test it against the quadratic.

[silver walrus]

So like in x < 2 i need to pick a number greater than 2 which is 3. And for the middle automatic 0. Lastly for the x > 3 it's 4?

[azure onyx]

x < 2 means x less than 2 though.

[silver walrus]

Oh so like 1?

[azure onyx]

and what do you mean by "for the middle automatic 0"?

[azure onyx]

yes, that would work.

[silver walrus]

That's what my teacher told me because in the middle you automatically put 0

[azure onyx]

I... don't exactly get why that is, but if your teacher told you to do it, I suppose they must have a reason?

[silver walrus]

It's like in this example

[azure onyx]

0 is a valid test point in that example because the middle interval is (-1, 3), and 0 fits in there.

but 0 does not fit in between 2 and 3!

[silver walrus]

Oh

So like what number?

[azure onyx]

whatever number between 2 and 3 is a valid test point here.

[silver walrus]

Is it 1?

[azure onyx]

is 1 between 2 and 3?

[silver walrus]

No, i guess a decimal or none?

[azure onyx]

well, there aren't any integers between 2 and 3, so whatever test point you come up with must be a decimal number.

[silver walrus]

Oh

So like you can't do a Test Point only a critical point one?

[azure onyx]

I'm sorry?

[silver walrus]

Like this on my notes

[azure onyx]

well I guess the methodology is similar here, but why can't you do a test point on the interval 2 < x < 3 just because no integers fit in between 2 and 3?

you have more than just the integers to work with here, I presume?

[silver walrus]

It's because like if I integers in the equation it's many like 2.1, 2.2 like trial and error

For me

[azure onyx]

you only need one such test point.

if you want to go with 2.1, sure, do 2.1 as a test point.
likewise, if you want to use 2.5 as a test point, there's nothing wrong with that.

there's no trial and error necessary here.

[silver walrus]

Oop alr

[azure onyx]

so long as your test point is strictly between 2 and 3, any number will work. of course, pick one that doesn't make your head spin 10000 times.

[silver walrus]

Alr

So the next thing to do is substituting?

[azure onyx]

yes.

[azure onyx]

what's ☑️ or ❌?

[silver walrus]

True or False

[azure onyx]

true or false, according to what?

oh whether it satisfies the inequality?

[silver walrus]

Yea, like if there is a true that the answer for that Quadratic Inequality

[azure onyx]

this is the first time I've seen this being used, but sure, if it works.

[silver walrus]

Oop alr

Got this equation

[azure onyx]

your first interval looks wrong.

are you saying you got a negative value for the first interval?

[silver walrus]

Yea

[azure onyx]

then why is it true?

[silver walrus]

Oh wait

[azure onyx]

wait, no. it shouldn't even be negative in the first place.

[silver walrus]

It's greater than right?

[azure onyx]

show me your calculations for the first interval only.

[silver walrus]

Wait

[azure onyx]

wait, nvm. it's this one right?

[silver walrus]

Yea but I know how now

It's because -2 > 0 is False🥹

[azure onyx]

well first of all, that's all sorts of wrong.

in fact, why are you testing for > when your original inequality is asking for <?

[silver walrus]

Wait wha

[azure onyx]

secondly, you got (-1)(-2) for the first interval. you have two negative numbers multiplied together; their product should not be negative in the first place.

[silver walrus]

Ohh

[azure onyx]

your original inequality asks for the quadratic being less than 0.

so I find it weird that you're testing for numbers greater than 0.

unless this is some fixed algorithm/method that you're taught, in which case I yield, but I will say if it is, this specfic step makes zero sense.

[silver walrus]

Oh, it's because my teacher told me it's fixed on the less than

[azure onyx]

...what?

I suppose then, I'll yield to your teacher.

[mighty rose]

I don't think so

[azure onyx]

my point still stands that the method is all sorts of weird, but if it works for you and is what your teacher wants, I'll stand down on this.

[silver walrus]

Alr

[azure onyx]

wait nvm. I just realized I was looking at his other example.

[azure onyx]

okay, I get that. but if that's the case, now both the first and second intervals are called into question.

[azure onyx]

first of all, there's this problem.

second of all, the second interval sees you having -1/4, and you said that it's true that -1/4 > 0.

might want to reconsider that?

[azure onyx]

this looks much better, and is correct.

[silver walrus]

Yea, I got em. I was wrong on the 1st Interval so it's True

So after this it's the solution set right?

[azure onyx]

yes.

I would have thought people used this method to make sign charts, but I guess different teachers teach differently.

[silver walrus]

Yea, it's also in the book lol

Alr so final answer?

[azure onyx]

uh? why are you taking the false interval?

[silver walrus]

Oh wait

[azure onyx]

that looks better.

[silver walrus]

Alr nice

Thank you

[azure onyx]

glad to help!

[silver walrus]

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[snow berry]

hi

[lime trellis]

hello

[astral sinew]

!nosols

[full forum]

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.

@snow berry Has your question been resolved?

[full forum]

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[glad marten]

I don’t know what I’m doing wrong I have been trying to this problem for like two hours . I know how to do it but I can get pass the first step because my long division won’t work

[onyx glen]

why are you dividing by x+3 though

[glad marten]

I found the factors of the bottom equation in the fraction so I can just divide the top equation twice by those two factors

[onyx glen]

bottom expression

[glad marten]

Yea

[slate violet]

yes that's correct so far

[onyx glen]

but why bother factorizing the bottom

you can and should just do long division once

[glad marten]

I’m supposed to get a remainder of zero

[slate violet]

oh yeah you mixed the technique up with synthetic division

[glad marten]

I dont UNDESTAND how to do that and I know I would mess it up on a test

[onyx glen]

but the principle is the exact same as if the divisor is linear

[glad marten]

Wdym

[onyx glen]

at each step you kill the highest-power term

[slate violet]

https://www.youtube.com/watch?v=-N2GkEU7xA4

[onyx glen]

it doesnt matter that your divisor is linear, quadratic, or 92nd degree

[glad marten]

I have only done it with two numbers before

[slate violet]

time to practice then

[glad marten]

It’s already 12 am 😭 since I spent so long on this study guide I’m going to be up to LILE four now

[onyx glen]

why are you doing math at midnight?

[glad marten]

I started at 9:30

[slate violet]

https://www.purplemath.com/modules/polydiv3.htm
fair enough, come back to this once you've slept then

[onyx glen]

do you have something catastrophic like 17 tests each 4 hours long coming up tomorrow?

yes

[glad marten]

I have a freshman not doing any work in my finance presentation so I also needed to finish that up

And I completed every other problem fine but this one and then I got to do more pratice and then this study guide again

[onyx glen]

proper sleep is like 1,000,000 times more important than any studies

[glad marten]

I wouldent be so worried if I was able to ask my teacher a question about this but my quiz is third block so I can’t

Is there a way to do this witu the method I’m doing

I just know I’m not going to be able to learn another way of division even if it is similar

I suck at division

I can’t do the normal long division I doubt i will be able to learn another way with 3 numbers instead of 2 in one night

And even if I did I only one practice problem for it

It doesn’t make sense to me why this is not working because when we where given just the two factors for the bottom expression and given the option to find an expanded version with 3 numbers I was able to use this method just fine

So why would it not work now

I know there is a way to do it with the box method which I used to know last year but the answer key doesn’t use that version so I do know where to even start

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[solid trellis]

hi

[onyx glen]

image uploading?

[solid trellis]

no

i just need some help

why bother

[onyx glen]

then you should post your question

[solid trellis]

going to like college studying or doing anything really

if were going to all die one day

[onyx glen]

oh you're going existential are you

[solid trellis]

you wont have a legacy you arent important

people will eventually forget about you

[onyx glen]

im afraid this server isnt the right place for you mate

you should maybe speak to a therapist about this

[solid trellis]

you know i did the math like lets assume someone lives up to 79 years old and sleeps 8 hours everyday the day they turn 18

they have like 14,800 days remaining

time is short man

[onyx glen]

cause like existential dread and all that but like

it aint gonna do you much good talking about it on here

you should talk to a close friend or ideally a therapist about this sort of feeling

[solid trellis]

do u agree or disagree tbh

ok

thx man

[onyx glen]

don't call me "man" please.

i am not a man.

[solid trellis]

ok sorry

i use man as gender neutral

m'am

[onyx glen]

a habit worth breaking

[solid trellis]

ok sorry

[dense lantern]

Type shit bro

[full forum]

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[drifting summit]

hey im cofnused

in order for the cauchy integral formula to work, doesnt the top function have to be analytic everywhere in C?

so isnt for example f(z) = e^{2z} + 5z^{3} / (z+i)^2 not analytic everywhere in |z| < 2?

[viral jasper]

Residue theorem

[onyx glen]

no, the numerator doesnt have to be analytic on the entirety of C

[onyx glen]

also missing brackets

[viral jasper]

Oh. You meant the numerator specifically

[onyx glen]

also the idea's to break up the contour |z|=2 into two smaller contours of which one encircles the pole at i and the other at -i

[drifting summit]

ahh

so basically we use smaller closed contours contained entirely within |z| = 2

that way, the function is analytic everywhere in this smaller contour

uhh i remember there was a theorem that stated that the integral over a smaller contour is the same as the contour that encircles it but i forgot the exact theorem name and what it started

[young bridge]

uh 1+1

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[full forum]

Please don't occupy multiple help channels.

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[delicate fjord]

I need to prove this with induction

[delicate fjord]

This is what I have so far

.help

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[delicate fjord]

.help rotate

[full forum]

No command called "rotate" found.

[stiff pier]

,rotate

[delicate fjord]

Im not really sure how to prove the A(n+1)>=B(n+1) equation

Or if the ones I wrote are even correct

[delicate fjord]

Ty

[queen gull]

you need to prove A(n+1) >= B(n+1)

[delicate fjord]

Oh right mb

[queen gull]

to do that
$A(n+1)=A(n) \cdot \left(1 - \frac{1}{4(n+1)^2} \right) \geq \frac{n+1}{3n} \cdot \left(1 - \frac{1}{4(n+1)^2} \right)$

[glossy valve]

ExpertEsquieESQUIE

[delicate fjord]

Wait I put the (n+1) part from a into b?

[queen gull]

you know $A(n) \geq B(n)$ by the induction hypothesis

[glossy valve]

ExpertEsquieESQUIE

[delicate fjord]

Im still confused

If I say sth as trivial as
A(n) * x >= B(n) * x
x will just cancel itself out
And you’ll end up with
A(n) >= B(n)
Again

[queen gull]

sure but why would you cancel x

[queen gull]

and here if you cancel x you end up on the left with A(n+1)/x

but you want to relate A(n+1) to B(n+1)

[delicate fjord]

Because I performed a trivial operation on both sides didn’t I
How does it add anything to the proof?

[queen gull]

why do you think its trivial

[delicate fjord]

Which would just be A(n), no?

[queen gull]

yes, which leaves you with nothing

so you don't cancel x!

the whole point is that you can use the induction hypothesis to determine a lower bound for A(n+1) = A(n) * x

[delicate fjord]

If I apply the same operation on both sides it will not change their equivalence or whatever relation like >= they have (unless it’s multiplication with something negative)

[queen gull]

that does not mean its trivial

[delicate fjord]

What do you mean?

[queen gull]

this is what we want to do, since A(n) is a part of the product the defines A(n+1)

[queen gull]

can you show the right side by here is greater then B(n+1)=(n+2)/(3n+3)

[delicate fjord]

So
(n+2)/(3n+3)=< (n+1)/(3n) * ( 1- 1/(4(n+1)^2)
?

[queen gull]

that is not true

[delicate fjord]

I was trying to figure out whether that’s what you asked for

[queen gull]

the right side I mentioned is $\frac{n+1}{3n} \cdot \left(1 - \frac{1}{4(n+1)^2} \right)$

[glossy valve]

ExpertEsquieESQUIE

[delicate fjord]

Like that?

[queen gull]

yeah

[delicate fjord]

Now that I’m thinking about it

I’m realizing a different problem

This isn’t a normal equation
It uses =< rather than =
Can I still prove it like a normal equation?

[queen gull]

yeah its an inquality

but as long as you do equalities its fine

but it gives you freedom to take bounds

[delicate fjord]

Bounds?

[queen gull]

yeah, smaller or bigger things. depends if you are working with <= or >=

[delicate fjord]

Sorry all of this is a lot

We’re expected to catch up on a ridiculous amount of content while the professor is not really interested in actually teaching

I feel like I’m asking for help with very basic things here

[queen gull]

Yeah, take things slowly

its good to go over the details

I don't know your professor but I assume he doesn't have the time to get into full detail on a lot of things

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@delicate fjord Has your question been resolved?

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[digital void]

Quick question that i prolly shouldve already known. Can a sequence by itself have an infimum/supremum, or is it only for the set it represents?

[grave elm]

supremum in what sense?

The least upper bound of the set of all a_n?

[digital void]

yes, but can we just avoid using the term 'set'?

[viral jasper]

The image of the sequence is a set where inf/sup apply

[umbral dome]

well if you think of a sequence as being a function N -> R then it doesn't really make sense to talk of the supremum of that, but a lot of people would just use the term "supremum of a sequence" to mean the supremum of the image of that function

[digital void]

i think im more confused now lol. lemme rephrase maybe

are we allowed to write something like sup(a_n) ? and does this always mean the supremum of the set thats represented by a_n?

[viral jasper]

I, personally, would not be confused if you wrote $\sup(a_n)$

[digital void]

okokok i think i understand. its just a convention/shortcut

thanks

.solved

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[glossy valve]

SWR

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[brisk minnow]

Hi I need help on this again 😭

[brisk minnow]

Idk how to fully structure it

[queen gull]

where did you get to

[queen gull]

this is not the definition of the generated subgroup by a single element

[brisk minnow]

Is that not what a cyclic subgroup is?

[queen gull]

it would be $<g> = \left{ g^n \mid n \in \bZ \right}$

[glossy valve]

ExpertEsquieESQUIE

[brisk minnow]

why is it not equal to 1?

[queen gull]

there is not meaning to what you wrote, because g^n=1 is an equation

in terms of set builder notation this is meaningless

[brisk minnow]

Ohhh wait

Yea

A cyclic group is g^n

subgroup

So what ever I’ve said is irrelevant 😭

I’m so confused now cause how do I even start this

Surely I thought cyclic subgroups was the right idea

[queen gull]

do you remember what I said last time?

[brisk minnow]

U said smth to do w proving by contradiction

[queen gull]

you have the subgroup of G, $<g>$, and it has subgroups of the form $<g^a>$ for $a \in \bN$