help-36

[thin remnant]

$$
\text{To Prove: If } \gcd(M,N) = 1 \text{ then } f(MN) = f(M)f(N)
$$

[soft zealot]

Ren🍓

[thin remnant]

$$
\text{Given: } f(n) = \sum_{d \in \mathbb{Z}, d>0, d \equiv 1 \pmod 2, d \mid n} (-1)^{\frac{d-1}{2}}
$$

[soft zealot]

Ren🍓

[thin remnant]

$$
f(MN) = \sum_{\substack{d_3 | MN \ d_3 > 0 \ d_3 \equiv 1 \pmod 2}} (-1)^{\frac{d_3-1}{2}}
$$

[soft zealot]

Ren🍓

[thin remnant]

so $$
\text{Since } \gcd(M,N)=1, \text{ any } d_3 | MN \text{ can be written as } d_3 = d_1 d_2, \text{ where } d_1 | M \text{ and } d_2 | N.
$$

[soft zealot]

Ren🍓

[thin remnant]

considering parity, $$
d_3 \text{ is odd } \iff d_1 \text{ is odd and } d_2 \text{ is odd}.
$$

[soft zealot]

Ren🍓

[thin remnant]

and $$
f(MN) = \sum_{\substack{d_1 | M \ d_1 > 0 \ d_1 \equiv 1 \pmod 2}} \sum_{\substack{d_2 | N \ d_2 > 0 \ d_2 \equiv 1 \pmod 2}} (-1)^{\frac{d_1d_2-1}{2}}
$$

[soft zealot]

Ren🍓

[thin remnant]

$$
\text{Now Let } k_1 = \frac{d_1-1}{2} \text{ and } k_2 = \frac{d_2-1}{2}.
$$

[soft zealot]

Ren🍓

[thin remnant]

or $$
d_1 = 2k_1+1, \quad d_2 = 2k_2+1
$$

[soft zealot]

Ren🍓

[thin remnant]

subbing them back into d3, $$
\frac{d_1d_2-1}{2} = \frac{(2k_1+1)(2k_2+1)-1}{2} = \frac{4k_1k_2 + 2k_1 + 2k_2 + 1 - 1}{2} = 2k_1k_2 + k_1 + k_2
$$

[soft zealot]

Ren🍓

[thin remnant]

$$
(-1)^{\frac{d_1d_2-1}{2}} = (-1)^{2k_1k_2 + k_1 + k_2} = (-1)^{k_1 + k_2} \quad (\text{since } 2k_1k_2 \text{ is even})
$$

[soft zealot]

Ren🍓

[thin remnant]

$$
(-1)^{\frac{d_1-1}{2}} (-1)^{\frac{d_2-1}{2}} = (-1)^{k_1} (-1)^{k_2} = (-1)^{k_1 + k_2}
$$

[soft zealot]

Ren🍓

[thin remnant]

which implies that $$
(-1)^{\frac{d_1d_2-1}{2}} = (-1)^{\frac{d_1-1}{2}} (-1)^{\frac{d_2-1}{2}}
$$

[soft zealot]

Ren🍓

[thin remnant]

$$
f(MN) = \sum_{\substack{d_1 | M \ d_1 > 0 \ d_1 \equiv 1 \pmod 2}} \sum_{\substack{d_2 | N \ d_2 > 0 \ d_2 \equiv 1 \pmod 2}} \left( (-1)^{\frac{d_1-1}{2}} \right) \left( (-1)^{\frac{d_2-1}{2}} \right)
$$

[soft zealot]

Ren🍓

[thin remnant]

soooooo $$
f(MN) = \left( \sum_{\substack{d_1 | M \ d_1 > 0 \ d_1 \equiv 1 \pmod 2}} (-1)^{\frac{d_1-1}{2}} \right) \left( \sum_{\substack{d_2 | N \ d_2 > 0 \ d_2 \equiv 1 \pmod 2}} (-1)^{\frac{d_2-1}{2}} \right)
$$

[soft zealot]

Ren🍓

[thin remnant]

since that's just f(M) and f(N)

$$
f(MN) = f(M)f(N)
$$

[soft zealot]

Ren🍓

[lime crest]

i already told him thiss

[jagged flare]

ohh thats reallt clever

thanksA

.solved

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[tranquil pine]

im not sure how to go about thinking closure under addition.
I have f(x) = f(x+p)
g(x) = g(x+q)
add them:
f(x) + g(x) = f(x+p) + g(x+q)

f(x) + g(x) = f(x+p) + g(x+q)

so i think p = q
But im not sure

[tranquil pine]

oooh wait

if we put in x=x+p

f(x+p) + g(x+p) = f(x+p +p) + g(x+p+q)

f(x) + g(x+p) = f(x) + g(x+p+q)

g(x+p) = g(x+p+q)
g(x) = g(x+p)

so p = q right

[scarlet sequoia]

well, IF p = q then f + g is periodic

that's not a necessary condition

but you're starting to notice that if p and q aren't linked

then it's hard for f+q to always be periodic

[tranquil pine]

Yes thats whatbim trying

if p has to equal q then its not a subspace

but it has to for it to be true always

so its not a subspace right

[scarlet sequoia]

I don't get exactly what you're trying to say but

The set of p-periodic functions, i.e periodic functions with same period p, is a subspace

[ripe jewel]

not sure they have to be equal for it to be a subspace, what if p is an integer multiple of q, is that enough?

[scarlet sequoia]

p/q = rational number

anyways

[ripe jewel]

oh even that weak? cool

[scarlet sequoia]

the fact that you're allowing all different kinds of periods

should tell you that if you mix two functions

[tranquil pine]

Woops

[scarlet sequoia]

with drastically different periods

[ripe jewel]

anyway you just need one counterexample

[tranquil pine]

lemme think

cant find one

[ripe jewel]

hint: one category to consider is the set of all period functions added to ||a constant function||

[tranquil pine]

omg

[ripe jewel]

wait

does my hint even work

[tranquil pine]

wait no

[ripe jewel]

lol no

because you need

p/q not rational

[tranquil pine]

i didnt know about that condition

[ripe jewel]

we said it above

briefly

[tranquil pine]

i knoww but i wanted to find a different way cause i didnt know it lmao

but okay now i know it

[ripe jewel]

hm, a different way

[tranquil pine]

can i take a sin wave with p = 1 and one with q = pi

[ripe jewel]

well how about try adding cos(px)+cos(qx) and seeing what p and q you can choose after a trig identity

but from the onset dont choose p and q yet

[tranquil pine]

hmm

lemme google the identity lmao

Okay

2cos((px+qx)/2)* cos((px-qx)/2)

[ripe jewel]

so one has period 2pi/(p+q) and the other has period 2pi/(p-q)

multiplied together they should have ... no, this is too complicated

[tranquil pine]

fr

[ripe jewel]

let's use a known property of trig

ANY period of cos(x) has to be a multiple of 2pi

lets make cos(x) one of them for simplicity

and the other cos(qx)

so for cos(x)+cos(qx)

the period has to be both:

• an integer multiple of 2pi
• an integer multiple of 2pi/q

so let cos x have period m2pi

[tranquil pine]

n2pi /= m2pi/q

[ripe jewel]

and let cos(qx) have period n2pi/q

you have to find a q that contradicts

[tranquil pine]

n /= m/q

q = pi

[ripe jewel]

unless m is divisible by q

[tranquil pine]

q = e

[ripe jewel]

so pick m a multiple of pi

so pi is rational

[tranquil pine]

huh

[ripe jewel]

or e is whatever

that's the contradiction

[tranquil pine]

m a multiple of pi?

[ripe jewel]

like 6pi

[tranquil pine]

how

[ripe jewel]

I meant

q=1/pi

anyway you get a contradiction

[tranquil pine]

n/m /= 1/q
m/n /= q

[ripe jewel]

if you try to fix it

so we just found that q has to be irrational

[tranquil pine]

yes

[ripe jewel]

nice

so we proved the theorem in this specific case

so can use a counterexample of cos(x) and cos(ex)

or whatever

[tranquil pine]

but does this axler guy really make it so hard? 😭

there must me an easier example

[ripe jewel]

never seen the book

[tranquil pine]

but cant think of one

[ripe jewel]

well perhaps because all counterexamples are going to be of the form

irrational = ratio of integers

not sure it will get easier than that

[tranquil pine]

real

[ripe jewel]

it really do be that way sometimes

[tranquil pine]

hate trig 😔

[ripe jewel]

🫂

trig gets more fun once you do complex analysis

[tranquil pine]

i did a bit of it

i like fourier guy

but basic trig stuff annoying

[ripe jewel]

I didnt appreciate or understad trig until I started drawing circles and reconstructing the trig function curves based on the circles I drew

I did it until it became intuitive

[tranquil pine]

lmaoo

[ripe jewel]

I low-key recommend this to anyone having trouble with trig

[tranquil pine]

i just dont know the identities by heart

[ripe jewel]

neither do I

well

[tranquil pine]

most of them you can derive with complex things

[ripe jewel]

i know most of the common ones

[tranquil pine]

Yeaa

pythagoras

[ripe jewel]

but ive been doing math for decades

[tranquil pine]

and stuff

[ripe jewel]

im 38

[tranquil pine]

oi

[ripe jewel]

an old man

👴

[tranquil pine]

Did you know the dinosaurs

[ripe jewel]

almost

[tranquil pine]

real

did u study math?

[ripe jewel]

yeah

[tranquil pine]

Was it a good choice? Job wise
Im thinking about studying it too

imma make a new channel, got a new wuestion lmao

.close

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[harsh solstice]

🙏

No trig

[bold turtle]

I mean this reads like an Olympiad problem; I'd be surprised if trig were necessary

Have you tried drawing a diagram?

[harsh solstice]

Ok

[harsh solstice]

I did try

[harsh solstice]

Trigonometry is not necessary, of course, but it can be used (by a person who isn't aware of my boundries) and I prohibit its use

angle MDB = angle MAD = angle DAN = angle NDC - some things I concluded

Let F = the point of intersection of the circle with the extension of DE

angle FMD = angle FND = 90 deg
angle MDF = angle MAD = angle DAN = angle DFN.
By the RHS theorem triangle FMD is congruent to triangle FND.
Hence MD = DN

[harsh solstice]

Note: I need not do this to prove MD = DN.

MN || BC

[bold turtle]

Yh I'm drawing blanks beyond that, it's certainly a combination of cyclic quadrilaterals and similar triangles

[final saddle]

@harsh solstice Has your question been resolved?

[harsh solstice]

<@&286206848099549185>

[final saddle]

@harsh solstice Has your question been resolved?

[harsh solstice]

<@&286206848099549185>

[harsh solstice]

BD^2 = BM * BA

(power of a point).

Q.E.D.

Thanks.

.close

[final saddle]

Channel closed

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

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[warped grail]

How to figure this out when there's a coefficient

[worthy wren]

which one

[warped grail]

the first

i assume first will help with the second

[terse crypt]

hi dj

[fast smelt]

"has x intercepts" means to solve for when "y = 0"

[worthy wren]

well just use the factored form

[warped grail]

someone helped me on one without a coefficient

[terse crypt]

long time no see

[warped grail]

yes

[worthy wren]

are u aware if xyz = 0 then either x, y, or z are 0

[warped grail]

yes

[fast smelt]

then what does it mean for x(x-1)(x-2)^2 = 0?

[warped grail]

something is 0

but i did one before and it was all math error because the entire function was above the x axis

[terse crypt]

this one aint

[terse crypt]

use this

[warped grail]

how

[fast smelt]

which somethings?

are u aware if xyz = 0 then either x, y, or z are 0

[warped grail]

that's what i don't know how to figure out

[terse crypt]

from this

can u see that the product is 0 if either ${(x)}$, ${(x-1)}$, or ${(x+2)^2}$ are zero

[fast smelt]

if xyz = 0, then either x = 0 or y = 0 or z = 0
if x(x-1)(x-2) = 0, then either ...

you can fill in the rest

[soft zealot]

k

[terse crypt]

by comparing the expression given in the question to this

[warped grail]

what

https://tenor.com/view/yawn-boring-pelican-gif-23394211

[terse crypt]

https://youtu.be/J7MjMM_mi4k?si=2sS5U6vsAhnghHzG

[warped grail]

so are there 3 x intercepts

so what i should have

x = 0
x - 1 = 0
(x + 2)^2 = 0

how do i deal with the last one

[warped grail]

isn't the answer just x

wtf

so is it 3

0, 1, and - 2

[terse crypt]

Yup

U actually watched the video and learned the thing

Damn

[warped grail]

well

no i guessed

[terse crypt]

Nice job :D

[warped grail]

the video explains things that i already know lol

[terse crypt]

At least it activated ur memory T_T

[warped grail]

well

you guys kept telling me something is 0

which is obvious

but never told me how to figure out WHICH something is 0

so

i just

i don't know

guessed basically how to do it and i think it ended up being correct

video still didn't help lol

[warped grail]

with this

[terse crypt]

Wdym last one

[warped grail]

0 = (x + 2)^2 i don't know how to do that

but i guessed

and got it

so it's fine

[twilit cradle]

you figure it out by setting each thing to 0 and solving for x

[warped grail]

You told me this in the other channel

[twilit cradle]

ye

[warped grail]

I got that part loud and clear from all parties 😭

the issue is not setting each thing to zero it is how to solve when there's a thingy

whatever

i got it and it's fine

man

i don't know how to do 7

[signal crater]

what happened

[warped grail]

I don't know how to rearrange or whatever when there's an exponent

[signal crater]

could you give an exampple

[warped grail]

hold on

the question is different and i'm trying to figure that out first

i don't know what sort of answer it wants from m

me

[warped grail]

but like if i were to be looking for the x intercepts

and the question were y = 2x^9(x^2 - 1)

i KNOW i have to set the shit to 0 and i have known this since before i opened this channel

but i don't know how to solve it when there's an exponent.

these assignments are urgent and i wasted half an hour getting told something i already knew before i opened the channel It's so over for me I'm so cooked

I just need to know how to solve when there's an exponent

i don't know what to do to the other side

what is the opposite of an exponent

of anything over 2

[signal crater]

i dont get whats wrong though

theres 3 intercepts

you just solve them

[warped grail]

did you read my messages

[signal crater]

which part are you stuck at

[signal crater]

so do you have to see the x intercepts?

where it becomes 0

[signal crater]

which part of the equation? the 2nd one

[warped grail]

what do i do with the ^9

[signal crater]

do nothing

its just 0

like x^9 = 0

so its alawys 0

[warped grail]

if i were doing like

0 = 2x^9

wait

[signal crater]

its just 0 same thing

[warped grail]

yeah

[signal crater]

divide by 2

[warped grail]

so is x always 0 when it's outside the brackets

unless one of the brackets is 0

also can you at all tell what question 7 is asking

[signal crater]

yea

[warped grail]

like what adjective or whatever am i supposed to put there

[warped grail]

tysm that's what i needed to know basically

originally

[warped grail]

what sort of asjective i mean

it's obviously a polynomial function but i don't think that's what it's asking

because oit says it already

in the question

[signal crater]

so the a(n) part?

[warped grail]

yeah

[signal crater]

im not sure, but if i had to guess, its maybe the highest power

if you expand it

[warped grail]

so like

i think it's 11

[signal crater]

once you multiply the terms, whatever the highest power of x is

i would say that yea

[warped grail]

so is there a word for that

i only know up to quintic lmao

[signal crater]

uhh im not sure

just write like 11th degree polynomial

[warped grail]

okay thanks very much

[signal crater]

11 is kinda crazy even for me but it is what it is

no problem!

[warped grail]

yeah i don't know what it's looking for because the lesson doesn't have anything about like "types" or whatever

unless it means symmetrical or asymmetrical ??

[signal crater]

yea teachers do that

[signal crater]

it is what it is

[warped grail]

i hate online school

[signal crater]

lmao

[warped grail]

never been more frustrated in my life

probably not true but ykwim

[signal crater]

no worries, i understand you

[warped grail]

is this symmetrical

i'm wondering if it's looking for that as an answer

[signal crater]

i think it is

[warped grail]

haven't gotten any questions about that yet

it was mentioned in the last lesson not the one for this assignment

hdljhgfkdfjhgkhg

[signal crater]

lol

well good thing is if you get stuck you can always ask here

or ask chatgpt to explain

[warped grail]

chatgpt is evil

can't do math

I think it's looking for "odd" or "even"

[signal crater]

nah this actually checks out

[vital crag]

!nogpt

[final saddle]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[warped grail]

what

yeah this is from the lesson

was jsut showing

does "changes sign __ times" mean to write how many turning points

[final saddle]

@warped grail Has your question been resolved?

[warped grail]

i guess so...

[final saddle]

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[exotic crane]

how does this prove all squaring functions are continuous

i dont get it intuitively

[tropic tapir]

It doesn't prove it's true, it just states it's true

[exotic crane]

how

[tropic tapir]

You want the proof of the statement?

[exotic crane]

how does it state that

not proof

i dont think i can understand it

[tropic tapir]

Ah it says "This true statement says that..."

It's stating the statement is true

Not proving it

[exotic crane]

i am actually not into math but this book is the unified approach of linear aljebra vector calculus and differential forms so it sort of is motivating me to get used to proof reading

[whole halo]

are you having trouble with this specific statement or for statements in general

algebra

[exotic crane]

yeah this

[whole halo]

this is strange to look at at first,s o you have to take each variable one at a time

first, you can see theres epsilon and delta

does this remind you of anything related to continuity?

if you can match this up with something youre familiar with, that would give us an intuition we already have

[exotic crane]

i forgot the epsilon delta definition

lol

[whole halo]

thats a shame

then lets try taking each variable one at a time

[exotic crane]

is it really important to understand stuff like this ?

[whole halo]

yea if you ever use it

[exotic crane]

then should i not be studying real analysis instead

lol

that would take months

[whole halo]

??? thats nonsense

thats like saying

[exotic crane]

i am more into applied math

[rocky tusk]

no this is elementary calculus

[whole halo]

"I dont need to read, if I need to understand this sentence shouldnt I be studying english instead"

[tropic tapir]

It's not too hard, you'll understand it in a couple minutes here. Don't worry

[whole halo]

epsilon-delta as an idea is important for more than real analysis

[whole halo]

this is a lie

you just dont know it

I didnt take months to learn this

[brittle sequoia]

help help me i need help plssssss gimme the answer

[whole halo]

we gotta get through this part first

this I think is the hardest part, so once we're over this we'll be a long way there

[exotic crane]

okay

i am with you

[whole halo]

usually most proofs only have like up to two of these for alls or there exists

this has three so its hard to consider

what you can think of is that:

x and epsilon can be anything you want

then delta is found depending on the x and epsilon you choose

[exotic crane]

oh

[whole halo]

for example, if you have
for all A, for all B, there exists C

you can choose anything for A
then you can choose anything for B
then there should be a C you can get

[exotic crane]

that makes sense

[whole halo]

since you needed A and B first, the C you have must depend on that earlier A and B

[whole halo]

same thing here, we begin with a real number x and a positive number epsilon
based on these two numbers, we have a number delta

[exotic crane]

right

[exotic crane]

okay

[whole halo]

thats already a bit of the hard stuff out of the way

keep in mind the statement is saying you can always find some delta

"there exists" a delta

no matter what x and epsilon you choose

does that make sense

[exotic crane]

yeah it does

[whole halo]

now we can move to what the statement is claiming

[exotic crane]

like a delta is always defined

[whole halo]

yep

[whole halo]

sometimes you can prove a statement wrong by choosing a specific x and epsilon, then showing you cant define a delta that works

[exotic crane]

yeah this is the thing i didn't get actually

[versed crater]

I really hate this definition of limits as an intro, I prefer the sequential definitions more, it feels a lot more intuitive and natural to come aboit

[whole halo]

thats what I internally think of, but that theyre equivalent is a bit difficult to internalize

only recently proved them on my own then Ive forgotten it again

[exotic crane]

oh

yeah

[versed crater]

I think understanding the sequential version first makes understanding this version easier

It’s just really weird that both 𝛆 and δ are both very close to 0

[sage meteor]

it just looks like buzzwords to someone that is not familiar

[whole halo]

so for this, the important part is the "if-then" statement

are you familiar with "implies"?

[exotic crane]

yeah i don't get the correlation between delta and epsilon here

[exotic crane]

yeah

[whole halo]

so we know that our statement should apply to points where |y - x| < delta

the statement says that, then, we also have a guarantee that |y^2 - x^2| < epsilon

are you good on the "implies" part of this statement

[exotic crane]

implies = means

thats how i interpret it

[whole halo]

sure, not exactly clear but I do that

also, its implies that the y can be any real number here

[exotic crane]

yes

[whole halo]

however, if |y - x| >= delta, the statement does not consider them at all

it is only considering the x and y for which |y - x| < delta

does that make sense

[exotic crane]

yes

[whole halo]

thats good

we can now do something convenient to help us along:

[exotic crane]

with a function intuition we can think of it as a specific range of delta

or is that a bad way to think of it ?

[whole halo]

a specific delta-sized interval

you can see the statement uses a for-all for x and for y

[exotic crane]

yeah

[whole halo]

if you think about it, the position of "for all y" in the statement does not matter

it will still allow you to pick any real number y regardless of whether you defined or didnt define delta

as long as its before the |x - y| < delta

does that make sense

[exotic crane]

yes

choice of y does not depend on delta

[whole halo]

so because of that, we can move "for all y" to be next to the "for all x"

makes it easier to see that x and y here are almost indistinguishable

[exotic crane]

yes

[whole halo]

so then |y - x| < delta must mean "any x and y that are less than delta apart," right

[exotic crane]

yes

[whole halo]

then the statement claims that, then, |y^2 - x^2| < epsilon

or that |y^2 - x^2| must then be less than epsilon, right

[exotic crane]

yes

[whole halo]

now heres where the importance of continuity comes in

first, we need to move the y back to its original spot

[whole halo]

theres actually a subtle mistake here otherwise which I almost fell into

[exotic crane]

what mistake

[whole halo]

imagine you can pick delta already knowing x and y

[exotic crane]

hmm

[whole halo]

we'll see later this goes against continuity in general

[exotic crane]

then ?

[whole halo]

anyways wait I gotta draw this

[exotic crane]

okay

@whole halo you there ?

[whole halo]

just wait a bit

[exotic crane]

okay

[whole halo]

@exotic crane ok this is taking a bit

first, I gotta say the definition here (I think) should just show that y = x^2 is continuous

not uniformly continuous

not sure about that but we'll figure out the intuition and have that tell us whats happening

[exotic crane]

okay

[whole halo]

alr so this is the statement we began with, right

the y in R is still there in its original position

now to "use" this statement, what you do is replace something in for each variable as you go along

that'll work for the x and epsilon

so lets say we pick x = 2

then we pick epsilon = 4

are you with me so far on this

[exotic crane]

yeah

[whole halo]

this reduces how much we need to consider this statement

now its just:

oh whoops wrote down the 4 over the delta by accident

there we go

[exotic crane]

yes

[exotic crane]

U could've just wrote this directly

[whole halo]

I dont know what you mean by that, but I think Ill say yes to that

[exotic crane]

anyways carry on

[whole halo]

to picture whats happening here, we can consider a graph of y = x^2

[exotic crane]

okay

[whole halo]

heres the area we're considering

[exotic crane]

yea

[whole halo]

now we're not putting a number into delta yet

so we'll draw |2^2 - y^2| < 4 first

you can see its |f(2) - f(y)| < 4, with f(x) = x^2, right

[exotic crane]

yea

[whole halo]

so that means f(y) can only be up to 4 away from f(2), right?

[exotic crane]

yes

[whole halo]

so based on f(2) = 4,

f(y) can be anywhere from 0 to 8, right?

[exotic crane]

yea nice

[exotic crane]

YES

[whole halo]

now to the "there exists"

to do this, we look at the whole statement then choose a delta that works

again this is within the hard part, so we'll need to focus

[exotic crane]

okay

[whole halo]

but not that much, I picked a really easy epsilon to start us off

so looking at this,

[exotic crane]

okay

[whole halo]

we are beginning the statement with |2 - y| < delta

this indicates some y which is within delta of 2, right?

[exotic crane]

yes

[whole halo]

something like this, right?

[exotic crane]

yeah

[whole halo]

now here, I picked a delta of 0.5 (ish)

lets see if this guess works

[exotic crane]

okay

[whole halo]

with this, we now have a statement with just one quantifier

(a quantifier is a "for all" or a "there exists")

(not sure on that though, Ive never needed to use that word)

[exotic crane]

what word ?

[whole halo]

the word quantifier

[exotic crane]

it's just the name of those proposals i think

[whole halo]

it likely is, but maybe some definition is gonna pop out and go "actually no thats something else"

idk

go take a look at this

are you able to understand this statement on its own?

if not, we can refer to the graph

[exotic crane]

yes I can but graph just makes it 100 times easier

[whole halo]

so you understand both already

[exotic crane]

yeah kind of

[whole halo]

you can see here that for |2 - y| < 1/2,

[exotic crane]

Does delta have to be small ?

[whole halo]

delta just has to be small enough

thats the key, that there is a delta is enough for us to do this

maybe the delta is always extremely small

maybe just delta = epsilon works

if it exists, we can use it

if it doesnt, the function is "not continuous at x"

[whole halo]

anyways back to this

you can see we've picked a small enough delta so that |2^2 - y^2| < 4

right?

[exotic crane]

yeah that's why we defined epsilon first and on that we defined delta

[whole halo]

thats the key

for most functions you can imagine,

you can sort of see that if we choose an x,

then an epsilon (around that x),

we should be able to always pick a delta,

so that for any y that is within delta of x,

f(y) is within this epsilon of f(x)

epsilon is locked in place, chosen before delta

[exotic crane]

yea

[whole halo]

think I typed too fast there

did you read up to here yet

[exotic crane]

yeah

i did

[whole halo]

alr alr

epsilon is chosen first, so delta has to be small enough

[exotic crane]

yeah

i get it

it restricts delta

sort of

[whole halo]

thats good

heres a question

we know that delta = 1/2 works

do you think delta = 1 works?

[exotic crane]

it does

[whole halo]

are you sure?

[exotic crane]

we are still within our choice of epsilon

[exotic crane]

oh

lol

i didnt care to actually compute it

[whole halo]

3^2 = 9 > 8

[exotic crane]

yeah

[whole halo]

so any number close to 3 wouldnt work like that

[exotic crane]

right

[whole halo]

what number do you think is the "biggest" (supremum) we can square

since 3 doesnt work

[exotic crane]

2root2

?

[whole halo]

yep

so then 2 sqrt 2 - 2 is the biggest the delta can get

[exotic crane]

yeah

right

[whole halo]

alr thats good

now heres another important example

we need to see a discontinuous function so we can verify that this definition breaks when it should

[exotic crane]

uh

signum function lol

[whole halo]

youve seen it, youve heard of it

now youll have to prove it

[exotic crane]

hmm

[whole halo]

so here, "discontinuous" means we have to go against this definition:

,,\forall x\in\mathbb R,\forall\epsilon>0,\exists\delta>0,\forall y\in\mathbb R,|x-y|<\delta,|\operatorname{sign}(x)-\operatorname{sign}(y)|<\epsilon

[soft zealot]

mtt

[whole halo]

Ive swapped out f(x) = x^2 for f(x) = sign(x)

[exotic crane]

yeah so we can let x = 0

[whole halo]

good

lets put that in

[exotic crane]

yeah

[whole halo]

then what epsilon?

[exotic crane]

1

[whole halo]

dangerous

[exotic crane]

why ?

[whole halo]

thats the biggest epsilon you can get away with

but yes 1 works

[exotic crane]

lets say 1/2 then ?

[whole halo]

sure 1/2

this is our statement afterwards

x = 0, epsilon = 1/2

oh right thats important

there should be a 0 < in this statement which tells us continuity

but for this specific function it doesnt actually matter

[exotic crane]

a what

[whole halo]

here

to ensure we're only looking at y which is different than x

since if x = y, then f(x) = f(y) so |f(x) - f(y)| < any epsilon we picked

[exotic crane]

yes

[whole halo]

this doesnt matter though since the rest of the statement would still be wrong anyway for y different than x

if you think about it, adding this 0 < would make the statement less restrictive

because its considering one less possible y than before

so this 0 < (which is usually there) can be left out without consequence

I think they do that for metric spaces in the book I was using

oddly enough for lipschitz continuity too

[exotic crane]

what book did u use for learning all this stuff ?

[whole halo]

metric spaces, I had a PDF I can show you

sometimes it can appear terse because the guy already expects you to have experience in proofs

[exotic crane]

i see are you a math undergrad ?

[whole halo]

yea

Ill be doing a math major

[exotic crane]

i see , did you read any books about proofs before that

[whole halo]

I dont remember

I got familiar with proofs through experience, but from where I dont know

[exotic crane]

nice i am about to go to university in a month aswell

[exotic crane]

i think thats what i am about to do aswell

[exotic crane]

lol this proof becomes obvious after that since there will not be a delta around 0 for which epsilon is 1/2

if iam correct

[whole halo]

its intuitively obvious since no delta you can pick can limit y to be 0 only

as thats the only value where |sign(y)| < 1/2

[exotic crane]

yea

[whole halo]

if we were to prove this contradiction, this is what we'd do

already we picked an x and a epsilon

next, we assume delta can be anything
thats so we can show no possible delta works
does that make sense

[exotic crane]

yeah

[whole halo]

since we have to prove this for any positive delta, not just a single number,

the only thing we know for delta is that its positive

now do you want me to put in an example number for delta or nah

[exotic crane]

why do those proofs seem like we already know the answers intuitively now we are trying to prove them using axiomatic logic

[whole halo]

yes thats partly what proofs are like

intuition lets you lead the rough direction of where the proof is

experience then shows you prior ways youve been able to write down the intuition through a proof

in our case, Im going to rely on dividing by 2

[exotic crane]

nah i get it

[whole halo]

the proving part?

[exotic crane]

you could show me how we would write it

[whole halo]

sure

let y = delta / 2

note that |-y| = |-delta/2| < delta

then |-sign(y)| < 1/2

so |sign(y)| < 1/2 and so y = 0

hence y = delta / 2 = 0 and so delta = 0

however delta > 0

contradiction

[exotic crane]

damn

thats cool

lol

[whole halo]

do you see some of the weirdness here:

we sort of went around and found y and then found delta

[exotic crane]

yeah

[whole halo]

doing this is a good sign that we're zooming in on something specific

less things to consider and the proof gets easier

at the end, we can say delta = 0 which is enough to confirm a contradiction

intuitively, its what we agreed on

since the only sign(x) that can be within 1/2 of 0 is x=0,

[exotic crane]

yeah

[whole halo]

that means only y=0 works

if we pick any other y then it wouldnt have worked

[whole halo]

tying y to delta this way places y to be strictly between 0 and delta

[exotic crane]

yeah

[exotic crane]

yeah that's a good motivation

[whole halo]

its very commonly /2 as well

sometimes you can be a bit weird and go for /3

[exotic crane]

would it matter

[whole halo]

no

[exotic crane]

lol

good

[whole halo]

oh also another important thing to notice here

for this,

you know that "nonzero y doesnt work"

now usually some people want to go for a proof by contradiction and do that

they create a nonzero y then show it doesnt work

in general its a bad habit to go for proofs by contradictions, because youre spending a lot of time around wrong statements

[exotic crane]

lol

[whole halo]

in general, wrong statements let you conclude anything, because with them you cannot prove apart wrong from right

when you accept a wrong statement, thats a lot of time having to do math where youre not sure how much of it can be salvaged - given one of the statements is not correct

[exotic crane]

my proof was root 2 is irrational and it was a proof by contradiction

i did not like it very much

very boring

[whole halo]

proofs by contradiction are still good

they can be very efficient because then you get to use the statement directly to show that it is wrong

[whole halo]

like this one, its a popular classic

pythagoras (didnt actually) got drowned for that

[exotic crane]

lol

[whole halo]

so here,

some people may go:

split into cases

assume y is nonzero

then |sign(y)| = 1 < 1/2

thats no good

then use that y = 0

instead I prefer that:

|sign(y)| < 1/2

so y = 0

[exotic crane]

this is best

literally

lol

[whole halo]

you can see they both accomplish the same thing, but one of them is assuming a wronger version

then going with it

the other one doesnt use a flaw

if the statement youre using is going to narrow you down, state it directly

the more contradictions you layer, the harder it is to track whats happening

[exotic crane]

is that why proof by contradiction is not considered the best

[whole halo]

so really when you do a proof by contradiction,
youre using something specific in the wrong version of the statement that can specifically be helpful

theyre considered the best if doing a direct proof is impossible or impractical

in general you'd prefer doing a direct proof above a contradiction

but if the contradiction is short and easy, it takes precedence

[exotic crane]

i see

[whole halo]

theres another thing that you can hear out if you want

[exotic crane]

how did you learn analysis ?? and what did you learn before that ??

[exotic crane]

sure

[whole halo]

learned it in college, pretty much
before that I remember actually getting some experience when my high school AP calc teacher had us do epsilon-delta proofs

a whole worksheet of them, a bit tedious

wait no it wasnt actually those, it was limits

no epsilon-delta used but still tedious

when I got to college, I dont think they give you any stepping stones
they show you the proofs then they either make the proofs easy or make the proofs normal (i.e. hard if you have no experience)

this is what shocks people who get comfortable with the high-school math
so you have to already be used to a bit of the college math to really get comfortable with it

[exotic crane]

i am doing a pure sciences degree will eventually major in math or physics i don't really have any heavy mathematics in first semester lol

i just graduated highschool few months ago

[whole halo]

oh then theres a good chance youll never need to see analysis

[exotic crane]

i will but not soon

3rd semester probs

[whole halo]

in a way you still need to learn proofs, so its important

need to be able to read formal logic statements

[exotic crane]

anyways i plan to learn math on my own like i have been

[exotic crane]

yeah

[whole halo]

yea, and as you always will

[exotic crane]

i need this

[whole halo]

specific but very important'

[exotic crane]

it seems to me its really important to be able to read really great texts and books

[whole halo]

great?

[exotic crane]

uhhhh yea

there are alot of cool proofs about stuff out there righ

r

[whole halo]

there certainly are cool proofs, but you still gotta learn basic proofs

you mean the flashy elegant proofs?

[exotic crane]

uh

yeah

lmao

idk

i am more into applied math though

[whole halo]

go do the basics first lol

also if youre getting into applied math, you dont have to go all in on these proofs

just the basics should work out

depends on how math-heavy it is

[exotic crane]

i am interested in proofs because alot of good authors use proof based writing to couple up intuition and rigour

idk if you know this book hubbard and hubbard

its a unified approach of linear aljebra vector calculus and differential forms

its the best book out there for vector calculus i think personally

but again it demands some basic familiarity with proof reading

[whole halo]

that makes sense

oh the one thing I needed to mention earlier

[whole halo]

still want to hear it?

[exotic crane]

yes

[whole halo]

you can still get the goods for contradiction without having to have a wrong statement

what you can do instead is write an opposite version of the statement, with the "not" operator

logical not

are you familiar with logical not?

[exotic crane]

no

[whole halo]

when you have true/false statements,

you can combine and invert their trueness or falseness

this is done with the OR operator (logical or)

the AND operator (logical and)

and the NOT operator (logical not)

written here from right to left

the not operator flips a statement between true/false

this is usually said as "inverting"

the result is the "inverse"

the NOT operator can also be called an "inverter" in specific cases

so if you had "for all x, P happens"

the logical negation of that would be "there exists an x where P doesnt happen"

[exotic crane]

lol why operator names analogous with logic gate names

[whole halo]

because theyre the same

[exotic crane]

then i get it

lol

[whole halo]

the important part here is how to negate for all and there exists

[exotic crane]

yeah that makes sense

[whole halo]

when you negate this, you flip between "for all" and "there exists" then move to the next one

you can see that this happens here

[exotic crane]

i see

[whole halo]

same here

if "there exists an x where P happens"

then the opposite is "there does not exist an x where P happens"

which is really "P does not happen for any x"

sometimes youll see "does not exist" as this

[exotic crane]

yeah

right

lol

[whole halo]

so if we want to avoid proving by contradiction,

instead of assuming this,

we can instead assume the NOT of this and then directly prove that its not continuous

[exotic crane]

yeah

[whole halo]

so there exists a real x and a positive epsilon,
so that for all delta,
there exists a real y,
where the statement is false

thats what we did, wasnt it

we chose x, epsilon, y

and we let delta be general

[exotic crane]

yes

[whole halo]

thats good

the logical not of this => operator is interesting

the opposite of "A => B" is specifically "A is true, but B is false"

in other words, => is only good for allowing a true statement to 'spread'

its not a very strong operator, it needs A to be true for its trueness/falseness to hold weight

[whole halo]

now by itself, does this make sense to you

we ignore when A is false here, we are only considering that
A is true, but B is false

[exotic crane]

yeah

[whole halo]

thats good, because that also lines up with what we did

[exotic crane]

can we read it like this

first we said if A is true then B is true

the logical not would be

if A is true then B is false

right ?

[whole halo]

not really

this isnt very easy to word

theres a specific reason that doesnt work

if we always know that A is true, then yes the opposite of "if A is true then B is true" is "if A is true then B is false"
because then its just that the opposite of "B is true" is "B is false"

now suppose A is false

the => operator like I said earlier is not strong

it only chains trues together

as a result, it wont tell you youre wrong if you begin with a wrong statement:

false => false is true
false => true is true
true => false is false
true => true is true

the idea here is that the only reason => can be wrong is if youre about to lead a true statement into a false statement

it does not care if the false statement is wrong in the first place

[exotic crane]

i see

hmm

[whole halo]

this also means the opposite of => is specifically

A is true and B is false

as thats the only scenario that => is wrong

if you think about it, this lets you negate the statement and get a strong one in return

"if A is true, then B is false" doesnt do this

it still allows A to be false and will still be true

[whole halo]

if a series of true statements breaks, we'd want to focus on a scenario like this specifically

since negating => lets us do that, that I think is why => acts like that

ok I think I can show you an example of us doing this

@exotic crane in the proof we did earlier,

[exotic crane]

yes

[whole halo]

did we care about this kind of y?

[exotic crane]

what did we do

[exotic crane]

no

[whole halo]

thats interesting

so really when we were proving this,

we still had that |x - y| < delta

thats all we focused on proving

this made sense, right?

[exotic crane]

yeah

[whole halo]

that mirrors what happens if you invert =>

its just "first is true, second is false"

[exotic crane]

hmm

makes sense

can i add u ?

[whole halo]

sure

dont expect a timely response lol

[exotic crane]

i am not but can i just drop any random thing i don't get ?

lol

[whole halo]

sure why not

[exotic crane]

i admire your patience

[whole halo]

thanks

[exotic crane]

.close

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[tranquil pine]

Hey can u help with this question

i wonder whats the answer

like how to find all integers (x, y) such that:

x³ − y³ = 7

Can u do it?

Anyone?

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[craggy plank]

Hiii, don’t delete your post next time. Claim a new channel

[tranquil pine]

okay sorry

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[iron ivy]

i can provide the proof being referenced here if needed. I am trying to apply the theorem and with my solution i am not sure where to proceed from here..

[tired walrus]

well you want to know whether 211 is a prime itself or a product of some smaller primes not on your list

right

[iron ivy]

Well I mean i know that 211 is a prime, and that it can't be a product of smaller primes. But what i am not understanding is that how i would just do it with 2,3,5,7

[tired walrus]

can't be a product of smaller primes

no, it could be a product of some other primes that weren't on your list

[iron ivy]

i said that it can't be a product of primes cause i know that 211 is a prime

what i was about to point out was that, although from this list i can see that 211 is not a product of 2,3,5,7. How do i make sure while staying within the given range that it's not a product of primes above 7 either?

cause for $m = p_1 p_2 p_3 ... p_n +1$

[soft zealot]

Sammy

[severe verge]

make it small enough to be checkable ig?

like you can take 2 3 and 5

which makes 31

and √31 < 6 so it's checkable

[tired walrus]

you check it lol

[iron ivy]

how do i check that?

[tired walrus]

like the same way you would check a number for being prime

to check whether n is prime you divide it by all primes up to sqrt(n) and see if you get a remainder of 0 at any point

if you never do then it's prime

here you do the same thing you just start at 11 since you already know 2, 3, 5, 7 won't fit

[iron ivy]

yea but sqrt(211) would be around 14.5, and the primes below 14.5 would then also include 11 and 13. But those weren't in the original list

[tired walrus]

yeah so just check those

,calc 211 mod 11

[soft zealot]

Result:

2

[tired walrus]

,calc 211 mod 13

[soft zealot]

Result:

3

[tired walrus]

no 0's here that's it

that's all the checking you need

[iron ivy]

but what about as the numbers grow bigger? with 211 it's possible to check for primes above 7 and below 14.5

[tired walrus]

you're not expected to give a general and efficient procedure here

you're expected to just solve the problems in front of you that's it lol

[severe verge]

there are primordials+1 that are not prime

so you have to manually check

you don't have to take every number from that given list to make a new prime

[iron ivy]

I don't wanna come across as difficult but the theorem 3 they are referencing does not include the info that you can check against primes < sqrt(m). So i can just bring other info i know?

[severe verge]

what is theorem 3?

[tired walrus]

you kinda are making it more difficult for yourself

[iron ivy]

let me show you

[tired walrus]

thm 3 is most likely euclids proof of the infinitude of primes

[severe verge]

yea

[tired walrus]

in which the key supposition is that we start with a list of what's meant to be all the primes in existence

and run ourselves into a contradiction wherein their product + 1 can't be divisible by any of them yet also has to be

[tired walrus]

now that we know it's true, we still can (and do, and are asked to) use the "product + 1" trick to construct a new prime

but what follows is bookkeeping

[iron ivy]

yea

[tired walrus]

like the practical side of it

thing is that now we're explicitly starting with only a list of some primes

so running thru the logic of the theorem no longer results in a contradiction as such

but only that prod+1 is not divisible by any of the primes on our specific list

which leaves two possibilities:

1. it is prime itself (now not contradictory, as our prime list lays no claim to completeness)
2. it is a product of primes not on our list

[iron ivy]

yea this was the thing i was trying to get across but couldn't word it

[iron ivy]

so i suppose that at this point you have no option but to append your approach and deviate from the theorem

?

[tired walrus]

i would not consider it a deviation from the theorem

do not treat mathematical theorems as some kind of religious gospel or dogma

you're using the theorem as a springboard, not a prison

[iron ivy]

hmm alright

well thanks for the support guys

i will close this now

.close

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[robust imp]

HELP

[robust imp]

nvm sorry i was dumb

i realise now

do u wanna see it

[tender grail]

yeah

[robust imp]

alr wai

[tender grail]

ty

[robust imp]

.close

[final saddle]

Channel closed

Closed by @robust imp

Use .reopen if this was a mistake.

[final saddle]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[timber plume]

What is this range saying

[dusty quarry]

only take angles up to one full turn