help-23

[tiny wraith]

Isn't 7 the first one

[spare forum]

Yeah hmm

[tiny wraith]

Maybe this is useful

Cunningham chain is probably what I was doing

nvm, it doesn't say that the last term has to be of the form 10n + 7

only that if there is one of the form 10n + 7 it has to be the last

[spare forum]

I have has work with modular arithmetic involving primes before so that might be useful too

fermats little theorem might prove handy though that is with exponentiaton

[tiny wraith]

Yeah the thing is that proving a certain prime will divide 2k + 1 after some iterations is impossible, since there infinitely many primes of the form qk - 1 for some prime q

and 2 * -1 + 1 = -1 mod q

unless you can do things with the divisors of 2k + 1

[spare forum]

q is any second or more member in one of those chains

For q>7

[tiny wraith]

Do you think this much theory is required?

[spare forum]

honestly, no, but i dont think that chasing those chains until you get to a prime bigger than your lowest answer is the way to go

And i dont know what else to do

i could write a program to brute force the chains for the first few like 50 primes maybe and work backwards from the answer

[tiny wraith]

Exactly what I was thinking

but I don't think it's the intended answer

[spare forum]

Yeah no shot but it is allowed

to help you find the correct way

[tiny wraith]

You only have to check primes such that p(2p + 1) <= 3 * 10^9

(maybe some tighter bound can be found, but this should be enough)

The smallest number n I've found is ~2.8 * 10^9

[spare forum]

whatt surely the chain starting from 3 is smaller

[tiny wraith]

$$3\cdot7\cdot5\cdot11\cdot23\cdot47\cdot19\cdot13\cdot9$$

[flat frigate]

Jelle

[spare forum]

Wait im stupid yeah

hm yeah im not really sure how to approach this

I guess ill start on the brute force algorithm if we wana do that

I do want to just tell them ten and grammar police my way to credit

but idk

[tiny wraith]

lol

even the algorithm isn't that simple

[spare forum]

yeah it would have to do some prime factorizing

Im just trying to bash stuff together and hope it works

Wait so these chains end if and only if 2p+1 is a composite number of the primes in the chain before

[tiny wraith]

Yes

[spare forum]

Btw if we can reduce the brute force method to a reasonable number of things we can do it

[buoyant shadow]

1 satisfies the conditions

[spare forum]

oh my god 1 doesnt have any prime factors

well then -1 too

[buoyant shadow]

well −1 is smaller idk if that counts

[tiny wraith]

What was the exact question

Also, I will find the smallest number > 1

[spare forum]

btw

Yeah

n has to be natural i forgot

[buoyant shadow]

there's nothing below 6,000,000

[spare forum]

well if n = 1 then no prime number p divides it (1 is not prime)

but the other solution is interesting enough for the problems sake

[tiny wraith]

My program says that 24809 will give a smaller chain, but there might be a bug

[spare forum]

Ill compute that if you want

[tiny wraith]

oh wait, it might not work

If my program works, it is the only possibility for a smaller n though

[spare forum]

24809 gives us (29)(59), and in that case wouldnt it just be faster to start at 29?

[tiny wraith]

Yeah, you are right, I didn't consider branching in my program though

[spare forum]

yeah youre gonna need a prime factorisation function and then compute all paths it might create

Whats fun is that youre gonna have to cut them off at some point because some might approach infinity

[tiny wraith]

I think I've done it

[spare forum]

lets see!

[tiny wraith]

It doesn't give any solutions, but there could be some bugs

uint64_t mainChain = 3ULL * 7ULL * 5ULL * 11ULL * 23ULL * 47ULL * 19ULL * (3ULL * 13ULL) * 9ULL; This is the smallest n we've found btw

[spare forum]

oh i dont know that langage ;-;

[tiny wraith]

It's C++

[spare forum]

still i think i can make sense of it just from standards

[tiny wraith]

Yeah, I haven't used any weird features

uint64_t is a 64 bit unsigned int btw

[spare forum]

thats what u meant

[tiny wraith]

hmm there were still some bugs actually, but I think I've got it working now and it just gives the number we've found already

[spare forum]

to what extent are you checking?

or to the max you said earlier

[tiny wraith]

Yeah, that max

this is the 'correct' code

these are the results it gives (n, starting prime)

If I remove the bounds all the numbers still seem to end in 5...

[spare forum]

hmm wait if thats the case can we prove that 5 is always a divisor and go from there?

assuming the code is correct ofc

[tiny wraith]

Given that it produces the number we've already found, I am relatively confident

It also means that every number is a multiple of 2775537765

[spare forum]

yeah so proving that 5 must be a factor solves the problem

Ok progress, can any of the theorems from earlier help

[tiny wraith]

Not sure, I can prove though that there were always be primes that are -1 mod 5

so 2 * -1 + 1 is still -1 mod 5

[spare forum]

Im looking into cunningham chains rn

this is what were doing but without branching paths

im pretty sure

[tiny wraith]

Yeah, but branching means that the cunningham chain has ended

[spare forum]

I can barely stay awake still idk what to do

[tiny wraith]

What was the original language btw?

Maybe send that question as well

[spare forum]

albanian

sure

[tiny wraith]

oh, yeah I don't speak albanian

don't bother lol

[spare forum]

yeah lmao

it makes less sense in albanian dont worry xd

[tiny wraith]

Every chain starting with a prime below 2 * 10^5 is divisible by the original solution

I might be able to optimize my program some more and give a larger bound, but this is probably enough

[spare forum]

yeah that should be enough to guarantee that lol

no need to check further the answer is the chain that starts with 3 then i guess

or 5 would be better for proof

doesnt really matter the n is the same

[tiny wraith]

!elliptic curve meme

rip

[safe radish]

🍎 = 154476802108746166441951315019919837485664325669565431700026634898253202035277999
🍌 = 36875131794129999827197811565225474825492979968971970996283137471637224634055579
🍍 = 4373612677928697257861252602371390152816537558161613618621437993378423467772036

[tiny wraith]

I saw this earlier today I think

so.....

[spare forum]

Yesterday we were meeming about what 2+2 equals

“Proving” that it doesnt exist was fun

honestly im too tired to do this anymore xd

[tiny wraith]

Yeah, same ima go

We might be the same time-zone

it's 11:38 here

[spare forum]

well thanks for all the help

Same

[tiny wraith]

You're welcome, just submit 1 btw

[spare forum]

yeah lol thats funny

well see you!

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[last breach]

can someone confirm if I evaluated the problem correctly or not:

[delicate sphinx]

Looks good

[last breach]

thanks

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[lean otter]

I am confused

[pseudo raptor]

so are @lean otter

[lean otter]

I am no longer confused

Thank you

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[lean otter]

We're banned

[pseudo raptor]

<@&268886789983436800> are they really?

[white umbra]

What

[pseudo raptor]

^

[white umbra]

If they were banned, they couldn't talk on here

[lean otter]

pain

[safe radish]

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[fickle grove]

hi, why is the middle one positive?

[thin bridge]

range of the arccos function is [0,pi]

[fickle grove]

right

oooh wait

okay i get it

ty

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[lean otter]

@median vigil just type here idk what happened with that channel

[median vigil]

the answer would be in terms of square roots, yes

[lean otter]

but then i would have 2 answers?

as there are 2 roots

[median vigil]

one answer will be a negative value for time, which doesn't make physical sense given the problem

so you should discard it

[lean otter]

i got 3.697

which doesnt make sense

cuz if u look at the other one the max height is 18m and that took 1.8 seconds to reach its max height

[median vigil]

what about it doesn't make sense?

[lean otter]

how can it be 3.697 if the maximum is 1.8

[little barn]

That's when you get the max height, not when the ball hits the ground

The ball is in the air for longer than it takes to reach its highest point

[median vigil]

you would expect it to reach its maximum height about halfway through its time in the air (not exactly since it had some starting height)

[lean otter]

Oh I got it also wanna make sure the initial height is always the c when in standard form?

the c is always the starting point (height)

[little barn]

This depends on how the chosen coordinate system lines up with the physical situation

But c seems to be the initial height in this case

But a problem might say that the ball starts at t=10s, or t=-20min

etc.

[lean otter]

aigh thank you

appreciate it

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[bitter plover]

f(x)=x^2−4x+3. And i have to find the points of tangency of its tangents passing through (0,0)

[bitter plover]

I hope it makes sense because I translated it from Greek 😭

I don’t even understand what am I supposed to find

Where the tangents meet?

[lean otter]

you have a curve f(x), and a tangent line that is tangent at some point P

you know that the tangent line passes through the origin

you want to find what P is

so for that, take the derivative of f as a first step

[bitter plover]

But irs asking about many tangents

That pass through 0,0

[lean otter]

yeah that's fine you can figure that out as well by setting up your equation

[bitter plover]

I have two tangents from previous questions are these any useful?

Oh no I have only one

y=2x-6

[lean otter]

okay great that's your derivative

so refer to the point of tangencies by an arbitrary coordinate such as (h,k)

[bitter plover]

the derivative of f is 2x-4

[lean otter]

so the slope of those tangents is
[
\eval*{\dv[y]x}_{(h,k)} = 2h-4
]

[lean otter]

now, what is the equation of a line? @bitter plover

in point slope

[bitter plover]

Of a tangent ?

[lean otter]

yes

we are considering the point (h,k) for the point slope

[bitter plover]

y-f(a)=f’(a)(x-a)

[lean otter]

okay great

we said f(a) is k and a is h

[bitter plover]

Aren’t those zeroes

[lean otter]

so y - k = f'(h)(x-h)

[lean otter]

hm?

[bitter plover]

Because it’s supposed to pass through 0,0?

[lean otter]

okay, thats fine we will substitute (x,y) = (0,0)

[lean otter]

so you get
-k = f'(h)*-h

now, can you simplify this?

[lean otter]

you already know f'(h)

[bitter plover]

Ohhh

-k=2h-4)*-h

but then what

[lean otter]

not -k

on the right

[bitter plover]

Oopsies

[lean otter]

okay so the signs cancel and you get [
2h^2 - 4h = k
]

[lean otter]

clear?

[bitter plover]

Yes

[lean otter]

now we want to solve for h, but having the k there is preventing that

we want to find another expression for k in terms of h

[lean otter]

we go back to our original function of [
\m fx = x^2 -4x +3 ]

[lean otter]

(h,k) is a point of tangency, so it obviously exists on the function as well

Substitute f(x) = k, and x = h

what do you get?

[bitter plover]

h^2-4h=k-3?

Ohh so k =3

[lean otter]

cool but keep the k isolated

now you recovered an expression for k in terms of h

[lean otter]

plug in k = h^2 -4h +3 here and solve for h

I'll leave the rest to you since its just a quadratic

[bitter plover]

I still don’t understand what I’ll find

[lean otter]

the points of tangency

(h,k) describes the points of tangencies, which we restricted to include (0,0) as well

[bitter plover]

Ohhhhh

[lean otter]

which is what your question is asking for

[bitter plover]

But that’s one point?

No?

[lean otter]

its not

[bitter plover]

oh okay I got it

[lean otter]

they describe a set of points since they are arbitrary variables similar to x and y

[bitter plover]

okayy I see now

I think I can do the rest by myself

Thanks a lot!

[lean otter]

noice

[bitter plover]

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[tired snow]

I dont know how to solv this, A and B, i dont know the logic behind it. Any help appreciated

[dreamy elk]

1st one is about the sum of the angles inside a triangle, the second is 'randvinkelsatsen'

[tired snow]

Yeah i figured out the scond one which sats it is. but the first one im lost on

Im trying to find this in the math book i have, but i cannot

[dreamy elk]

All triangles fulfill that the sum of all inner angles is 180 degrees.

Could you express this somehow?

[tired snow]

tbh, no, im really beginner in geomitry, i see what you say 180 degress in total

but i do not know the steps to calc it, what technique to use

[terse lichen]

all of the angles in the triangle add to 180

means first angle + second angle + third angle = 180

[dreamy elk]

Well, do it literally, here it would be $x+30^\circ + 2x + 90^\circ = 180$

[flat frigate]

Crystopher

[tired snow]

where did the 90 come from?

[dreamy elk]

The angle at the corner has a square thingy

that means it is 90 degrees

[tired snow]

oh

oh i see now

so the question wants me to calc the equation?

[dreamy elk]

Basically, yeah

[tired snow]

so basically all triangles, that need to solve X or whatever, as a equation, i add everything up to make the equation, then i solve it.

[dreamy elk]

Yes, in most cases it will be something like thatm unless it is some hard problem.

[tired snow]

Aye, lets see if i can do this on paper now hehe, and ill read up on the radvinkelsats abit more

so i would solve this one like this?

Also for the second, Randvinkelsatsen, would it be true that the middle one is twice as big as C? or is that rule false in this case?

[safe radish]

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[spark zinc]

no idea where to start

[humble helm]

replace

[potent seal]

Do you know the
Expansion for sin(a+b)?

[spark zinc]

yes

not gonna write it because its too long but yes

[lean otter]

not that long tbh

[spark zinc]

sinacosb±cosasinb

[lean otter]

okay so

find cos

using

$sin^{2}(x) + cos^{2}(x) = 1$

[flat frigate]

snow

[lean otter]

u have sin value

[spark zinc]

alright i'll try to use this and see if i can get the answer

ill be a couple of minutes

hang on

[lean otter]

thats not the only step

[terse lichen]

1. draw right triangles in their respective quadrants
2. find trig ratios for both of the angles.
3. use sum difference identities
4. solve

[spark zinc]

I don't want to become dependent on drawings, but i'll use it as a last resort. I know it can be done solely with math

[lean otter]

it can

[terse lichen]

sure thing, you do you my man, i just find it easier using drawings

[spark zinc]

cos(a)=12/13

i'm getting somewhere

[lean otter]

that's good

now use

[spark zinc]

wait dont tell me

i think i know

[lean otter]

ok

what

remember your final task is to find cos(B)

so you can use sin(A+B)

[spark zinc]

idk

im stuck

i'm not too familiar with trig identities so that may be why im struggling

everyone left lol

<@&286206848099549185>

:/

i think i might have found cos(B)

.close

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[hexed steeple]

a + b = 1
a² + b² = 2

a¹¹ + b¹¹ = ?

[desert juniper]

you would start by solving the system with the first two equations

[hexed steeple]

hm

[hexed steeple]

but how can I sove a¹¹ + b¹¹ w a+b square??

[lean otter]

is 11 even?

[hexed steeple]

yep

Eleven

Is a Russian olimpic question

[lean otter]

no i am asking if it is EVEN not ELEVEN

[hexed steeple]

?

[lean otter]

do you know what even and odd means

[hexed steeple]

no

im Brazilian lol

[desert juniper]

this is in the olympiads? looks pretty easy to be tbh

[hexed steeple]

not every olimpic question is hard

Some are for time

[hexed steeple]

OOOH

I got it

11 is an odd number

if u asking

[quasi sentinel]

get the value of ab first

[hexed steeple]

how

[quasi sentinel]

use (a+b)^2 = a^2 + 2ab + b^2

[hexed steeple]

oh

ab = -1/2

then

[quasi sentinel]

get the value of a^3 + b^3

[hexed steeple]

(a+b)³ = a³ + 3a(ab) + 3b(ab) + b³

a³ - 1,5a - 1,5b + b³

[lime dust]

a^3 + b^3 is not (a+b)^3

[hexed steeple]

I know

I'm getting into that

[quasi sentinel]

you know the value of a+b

[hexed steeple]

(I think(

yes

[quasi sentinel]

so what is a^3 + b^3

[hexed steeple]

(a+b)^3 - 3a²b - 3ab²

[quasi sentinel]

i meant the exact value

you need a number

[hexed steeple]

I was solving

I think

[quasi sentinel]

alright

take your time

[hexed steeple]

a³ - 1,5(a+b) + b³ = 1

a³ + b³ = 5/2?

[quasi sentinel]

now get the value of a^5 + b^5

using (a^2 + b^2)(a^3 + b^3)

[hexed steeple]

:cry

ok

oh

5?

cuz

a2+b2=2

a3+B3=5/2

5/2.2=5

[quasi sentinel]

they are not equal

[hexed steeple]

Oh

[quasi sentinel]

a^5 + b^5 is not (a^2+b^2)(a^3+b^3)

[hexed steeple]

Obv

lol

Wait

a⁵ + a³b² + a²b³ + b⁵
a⁵ + b⁵ + (ab)²a + (ab)²b
a⁵ + b⁵ - 0,25a - 0,25b
a⁵ + b⁵ - 0,25(a+b) = 1
a⁵ + b⁵ = 1,25

[quasi sentinel]

why did (ab)^2 become -1/4

[hexed steeple]

cuz (-1/2)²

oh

is 1/4

[quasi sentinel]

so a^5 + b^5 = 3/4

[hexed steeple]

isso aí

[quasi sentinel]

now get the value of a^6 + b^6

using (a^3+b^3)^2

[hexed steeple]

🤡

[quasi sentinel]

you're almost there

[hexed steeple]

a⁶ + 2.a³.b³ + b⁶

then

[quasi sentinel]

you know the value of a^3 + b^3 and ab

[hexed steeple]

a⁶ + b⁶ + 2(ab)³

[quasi sentinel]

yes

[hexed steeple]

a⁶ + b⁶ = 1 - 2(-1/8) = 1 + 1/4 = 5/4

then a6b6(a5b5) I guess

[quasi sentinel]

yes

[hexed steeple]

ok

Wait

[quasi sentinel]

make sure you made no mistakes

i mightve missed something

so check it before submitting

[hexed steeple]

989/32

I guesa

[safe radish]

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[limpid shell]

,rotate ...

3rd one please ...
and explain the process .

[limpid shell]

<@&286206848099549185>

<@&286206848099549185>

[safe radish]

@limpid shell Has your question been resolved?

[gentle hatch]

Question number?

[limpid shell]

3

ans is incorrect check it ... please ..and tell me the process

[gentle hatch]

im sorry im of no help here

[safe radish]

@limpid shell Has your question been resolved?

[safe radish]

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[warm patio]

F ( x ) = 2 - x^(2/3)
When i solve it for critical points i don't find
But actually in graph there's a local max point and it's (0,2)

[safe radish]

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

[warm patio]

F'(x) = 0 doesn't exist

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@warm patio Has your question been resolved?

[warm patio]

<@&286206848099549185>

[red delta]

f'(x) = 0

?

[warm patio]

Yeah to find the critical point

Any help ?

<@&286206848099549185>

. close

. close

.close

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[dense wadi]

Where did dy/dx come from

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[granite idol]

maybe they're using (d/dx) = (d/dy) * (dy/dx)

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[tepid spade]

hi! i have the dumbest question ever, right now i cant think. is {1} union with {0,1} {1,0,1} or {0,1} ?

[lime dust]

All unique elements from the union

[compact pelican]

Technically, {1,0,1} and {0,1} are setwise equivalent

[tepid spade]

is that {1,0,1} or {0,1} tho? 💀 english is not my first language sorry

[lime dust]

Unique = don’t repeat

[compact pelican]

It's better to say {0,1} because it doesn't repeat

[tepid spade]

ok, thanks

.close

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[supple elbow]

anyone that could

explain why h(t) has to be on top in the vector instead of the bottom

prof said i got it wrong bcs of that

[plucky elk]

Usually it has to do with the eigenvalues order

Try plugging your eigenvector in and see if it's satisfied for both t<0 and >=0

[supple elbow]

sorry im a bit confused still

[plucky elk]

Do you know how to verify if a function is a solution to a differential equation?

[junior smelt]

(Do you have the original question stated anywhere by any chance?)

[safe radish]

@supple elbow Has your question been resolved?

[supple elbow]

sorry ill have to translate it

[junior smelt]

,rccw

[supple elbow]

basically it says find the solution of (DE) while h(t)= .. and assume the system hs an inital velocity of 0

[junior smelt]

not sure, think I agree with the way you've done it tbh when running through it

(remember we kinda covered this earlier too, and that I think I moaned at their solution for looking a bit sus, did they explain that?)

Mind you, comparing the final solution you get does seem strange, so

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[dense wadi]

to solve the differential equation here

[dense wadi]

im nto too sure how to seperate the 2xy

so that one side is y the other is x

[pseudo scroll]

divide by 2x

[dense wadi]

uhm

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[barren root]

improper integral

[plucky elk]

What's the question

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[humble nexus]

can someone explain to me what this two symbols do idont get it😭

[humble nexus]

sry that it is in german

but is about combinning 0 and 1 idk

[marsh walrus]

its just union and intersect operator

[flat frigate]

jan Niku

jan Niku

[humble nexus]

i know it means union

but idk how these works when they combines together

maybe is should ask more specific: E means 1, and N means 0, so for the first one i circled Nk u(Ek n Nk+1) i think it means 0 union (1 followed by 0) but i dont get what the exact part i circled what it means

[marsh walrus]

it means, for every natural number, including 0

so k = 0, 1, 2, …

[humble nexus]

could you tell me how does the one above (the big interscet) works ?

[toxic shoal]

i think $E_k$ and $N_k$ are sets, you get one of these sets for every natural number $k$. i don't know what elements the sets have because i can't read german and i don't know if they're defined in the problem . the $\bigcap_{k \in \N_0}$ in front means what jan Niku said, so you get a new set whose elements are those elements that are in ALL of the sets $N_k \cup (E_k \cap N_{k+1})$

[flat frigate]

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

[toxic shoal]

fuck the emote broke it. oh well

[marsh walrus]

so the big intersect just says, the intersect of all sets N_l where l>k

for example, were k=5, then $\bigcap _{l>k} N_l = N_6 \cap N_7 \cap \dots$

[flat frigate]

jan Niku

[humble nexus]

let me think a bit need some time

[marsh walrus]

okay

feel free to ping

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@humble nexus Has your question been resolved?

[humble nexus]

i translated it

[humble nexus]

so can i see this equivalent to this

[hot thistle]

yes

[marsh walrus]

yea i think so

sorry took me a long minute to realize this is just a notation question still

[humble nexus]

😂 its fine

oh i get it

finallly

one more little question, just want to be sure that all sequence begins with 0-th (not 1-th), because the answer starts from 0?

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[lean otter]

having a blank moment, I get told that these 2 things are the inverse of eachother, how please?

[lean otter]

sqrt3/1 in what way becomes sqrt3/3 when described as the inverse please?

[drowsy karma]

$\left( \sqrt{3} \right)^{-1}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}$

[flat frigate]

Joanna Angel

[lean otter]

ahhhhhhh okay

thanks!

.close

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[lean otter]

It's obvious now you're mentioning it, I'm trying to speed run math basics and when I hit calculus I'll go back and then remember all this properly.

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[mild bridge]

part ii of this question

[safe radish]

@mild bridge Has your question been resolved?

[mild bridge]

<@&286206848099549185>

[mild bridge]

<@&286206848099549185>

[calm violet]

can you try the first steps of proving by negation/contradiction?

[mild bridge]

thanks for helping

So i was thinking about it

and intuitively, when a function is continuous, then it's image cannot be countable due to the fact that when we are taking values from the set of real numbers, which is an uncountable set, we can't get a countable image of the function

unless it's a constant function

[sick rivet]

can you help me

[mild bridge]

does that make sense?

[calm violet]

please use another channel

ok let's start off from the definition of constant function, what's a constant function?

[mild bridge]

for example f : R -> R, f(x) = 2

right?

it always outputs the same value

[calm violet]

In more formal terms?

[mild bridge]

for all inputs

[mild bridge]

um

i don 't know more formal

i thought this was it

[calm violet]

Ok; here's a standard formal definition; A function f: U -> V is constant if there exists a c in V such that for all x in U, f(x)=c.

[mild bridge]

ah ok

[calm violet]

Now, formally, how would you negate the definition; so what can we say about f: U -> V if f is not constant?

[mild bridge]

we change the for all part to there exists?

[calm violet]

ok

can you write that out?

[mild bridge]

or just say that c doesn't exist

there doesn't exist a c in V such that f(x) = c for all x in U

[calm violet]

Now, if V has more than one value what does that day about how f maps values from U to V?

[mild bridge]

if the function were to be constant?

then it wouldn't be surjective

nor injective

[calm violet]

Alternatively, wlog, consider f: R -> R such that f(0) = 0. If f is not constant, what does that say about how f maps the other values?

[mild bridge]

that f is the identitiy function?

and every element maps to its equal

[calm violet]

But f=x^2 is not the identity function, nor is it constant (f(0)=0, f(2)=4)

[mild bridge]

oh yeah

[mild bridge]

then i dont know

[calm violet]

Ok the only thing you can conclude for general f, is to look at the formal definition; first assume that for all values x in U, we also have f(x) = f(0)=0; does that satisfy the definition of constant function?

[mild bridge]

yes

[calm violet]

sorry, I require here actually that U has more than one value

[mild bridge]

there exist a value c in this case zero such that all x's map to it

[calm violet]

Right, but this would contradict our assumption that f is constant

So what is the logical negation of this statement?

[mild bridge]

what contradicts it?

[calm violet]

We had assumed that f such that f(0)=0 is not constant, but if for all values x in U, we also have f(x) = f(0)=0, then f is constant, which is a contradiction.

so the negation of "for all values x in U, we also have f(x) = f(0)=0" must hold

what is the negation?

sorry, edited the quoted part

[mild bridge]

for some values in U?

[calm violet]

For some values in U, what?

[mild bridge]

f(x) = f(0) = 0

[calm violet]

(by the way, f(x)=f(0)=0 is actually 2 logical expressions)

f(x)=f(0) and f(0)=0

[mild bridge]

sorry I'm really confused

can you explain to me the general idea of how we are trying to prove that statement?

I'm a bit lost rn ngl

[calm violet]

Ok consider the statement: "for some x in U, we have f(x) = f(0)=0" is this true of f that is not constant?

[mild bridge]

no it's not

[calm violet]

We are proving that assuming f is not constant for continuous f leads to a contradiction

hence f is constant

to do so we need to have a clear idea of what it means for a function to be not constant

[mild bridge]

wait

i thought it was the other way around

[calm violet]

sorry, yeah , typo, edited

[mild bridge]

the only way for f to be continuous and the codomain countable is for f to be constant as well

which makes sense intuitively actually

[calm violet]

yes, but intuitive doesn't cut it

[mild bridge]

and we are trying to prove this by showing that when f is not constant, then the codomain is not countable?

[calm violet]

yes

[mild bridge]

ok

[calm violet]

Alright let's continue

[mild bridge]

ok

[calm violet]

If f:R->R such that f(0)=0 is not constant, there exists x in R .... (continue the sentence)

it needs to be something involving f(x) and f(0)

[mild bridge]

such that f(x) = \ = f(0) = 0

[calm violet]

alright

So you know that the range has 2 distinct points f(x) and f(0)

have you learned about the intermediate value theorem?

[mild bridge]

yes i beleive

[calm violet]

you have to use it; how would you use it?

[mild bridge]

it say that when a function is continuous and we have two values f(x) and f(0), then there also exists a value y eg such that f(g) is between f(x) and f(0)

right?

[calm violet]

yes

[mild bridge]

so our assumption is that the function is continuous, and it is not constant

which would mean we have two values at least, namely f(x) and f(0)

and according to the theroem, we have f(0)>f(y)>f(x)

or otherway around

[calm violet]

well we can assume that wlog

[mild bridge]

what does wlog mean?

[calm violet]

without loss of generality

in your course maybe you shouldn't say that

[mild bridge]

my stuff isn't in english so i'm unfamiliar with some terms

[calm violet]

but it's like when I said wlog assume f(0)=0; we expect that the reader can easily extrapolate from the f(0)=0 case to the f(x_0)=c case

in this case we expect that the reader can easily supply the proof for f(0)<f(y)<f(x) once we supply the proof for f(x)<f(y)<f(0)

[mild bridge]

yes

we don't have to apply the proof if it's proven in class

[calm violet]

alright, so can you continue from "according to the theroem, we have f(0)>f(y)>f(x)"

[mild bridge]

we can just refer to it

[calm violet]

what next

[mild bridge]

hmm

[calm violet]

what is y in the first place?

[mild bridge]

so we are trying to show that the codomain is uncountable

[mild bridge]

an element from the domain

[calm violet]

It's more specific than that

[mild bridge]

between x and 0?

[calm violet]

yeah I should have caught that earlier

hmm actually yeah though come to think of it that's not impt

alright continue from "so we are trying to show that the codomain is uncountable"

[mild bridge]

yept

um

what'S the formal definition of a countable set actually?

[calm violet]

it has a bijection with the set of whole numbers

that's the formal definition

[mild bridge]

so that means every element from the whole numbers set maps to an element from our set

and we need to use this definition right?

[calm violet]

the set of whole numbers is defined as the smallest set N such that
0 in N
if x in N then S(x) in N, where S is injective

[calm violet]

edit: you don't really need to use this definition not really, what are some things you know that are uncountable?

[mild bridge]

real numbers

um

irrational numbers

i think

[calm violet]

Alright consider the tangent function on the interval (-pi, pi)

what is its range?

more specifically; is (-pi, pi) countable?

[mild bridge]

no

cuz the interval that we have is uncountable already

[calm violet]

given my definition of countable, how would you prove that (-pi, pi) is not countable?

[mild bridge]

and if all of them elements map to some elements in the codomain

which makes the codomain uncountable as well

[calm violet]

You can use that R has a bijection with the power set of N

[mild bridge]

is that valid

[calm violet]

yes, that reasoning is valid

[mild bridge]

what is the power set of N?

[calm violet]

the power set of a set S is the set containing every subset of S (including S)

Cantor's diagonalization proof tells you that there exists no surjection from any set to its power set. (hence no bijection)

[mild bridge]

oh

ok so elements in N are like 0,1 ,2 ,3 ...

and in it'S power set they are (0,0) , (0,1) , ....

[calm violet]

no...

[mild bridge]

isn't the power set N x N

[calm violet]

the power set is {{}, {0}, {1}, {2}, ..., {0, 1}, ... , N}

what you gave me was the cartesian product

[mild bridge]

is the power set the cartesian product plus every (N, ∅)

[calm violet]

in the first place sets (and subsets) aren't ordered tuples thereof

and, e.g. {2, 4, 6, 8, 10, ...} is in the power set but has more set elements than the cartesian product has tuple elements

[mild bridge]

ok

got it

so every combination of x elements from the original set where x is the number of elements in the set

[calm violet]

well, x can be infinite and any of the distinct infinites in set theory. How do you talk about a combination of that?

[mild bridge]

then you have infinite elements in the power set

[calm violet]

It's more subtle than you think, but that's for another time; can you just use the facts that I've given you to show that (-pi, pi) is uncountable?

[mild bridge]

hmm

[calm violet]

the facts regarding which sets of N, powerset of N, R, and (-pi, pi) have bijections

and the formal definition of countable

[mild bridge]

there exists no surjective function z , such that z: N -> (-pi, pi)

[calm violet]

why?

spell out the reasoning

[mild bridge]

and if all of them elements map to some elements in the codomain
which makes the codomain uncountable as well

this is what I wrote bnefore

but idk how to show it mathematically

[calm violet]

Alright, the proof is this: we know that there exists a bijection between P(N) (P(S) is the powerset of S) and R, and R and (-pi, pi); hence there exist surjective h: (-pi, pi)-> R and g: R-> P(R). Recall that the function composition of two surjective functions is itself surjective. Hence if surjective z: N->(-pi, pi) existed, then g.h.z (where . is function composition) is a surjective function from N to P(N) which contradicts the fact that no set surjects to its power set

While that was the case for the tangeht function, we can, for example, rescale it to show that every interval is uncountable

Can you use the fact that any interval is uncountable to complete the proof?

[mild bridge]

question

[calm violet]

yea?

[mild bridge]

I honestly don't understand how the power sets having bijections says anything abour countability

[calm violet]

What's the definition of a set being countable?

[mild bridge]

a set being countable means that there exists a function such that this fucntion can map every element of this set to the set of natural numbers

correct?

[calm violet]

Well, the identity function maps the set {1} into N (it maps 1 in {1} to 1 in N) but {1} is finite, not countable.

[mild bridge]

aren't finite sets countable?

[calm violet]

Well, in my texts we draw a distinction between countable sets, which are the smallest infinite sets, and "at most countable" sets, which are either finite or countable.

But got it, we can use that definition

[mild bridge]

ok

[calm violet]

As for the power set, they are the standard way by which we prove that the real numbers are uncountable

I think in your course you just assume that outright

[mild bridge]

we have proven that R is uncountable in class

I was sick for just a single week

and I missed most of this stuff

and we don't have proofs that are done in class anywhere else

[calm violet]

I think if you're a math major when you have time you should try to supplement your gaps with a course or book that discusses mathematical logic and builds up the number system from sets up through N and Z to Q; (and also builds relations, functions, equivalence classes)
As for uncountability of reals, see: https://en.wikipedia.org/wiki/Cantor's_diagonal_argument

[mild bridge]

do I need to know the proof of how the real numbers are uncountable for this question?

[calm violet]

Anyway, assuming that any interval where the endpoints are distinct is uncountable, can you complete the proof to your qn?

[calm violet]

you don't but you needed to be familiar with how countability is related to bijection, which I was testing

if you couldn't assume that intervals were uncountable

[mild bridge]

ok

so

assuming that f is not constant

since f is a function that takes elements from an uncountable set, namely R

we would have an interval in R, say (a, b) where f(a)> or < than f(b)

since (a, b) is an interval of R, it is uncountable

[calm violet]

what happened to the ivt?

[mild bridge]

oh

so ivt state that since this function is continuous

then there are also other elements between f(a) and f(b) in the codomain

and since the codomain is an image of an uncountable interval, it itself is also uncountable

assuming f is not constant

[calm violet]

how did you arrive at this line?

[mild bridge]

that make sense?

[calm violet]

no

what facts did you use?

[mild bridge]

because f is not constant?

so the outputs have to be different values?

our assumption

[calm violet]

The function f that is zero everywhere except for f(0)=1 is not constant, but has no such interval

[mild bridge]

but then it wouldn't be continuos

[calm violet]

How do you know that it isn't continuous?

[mild bridge]

No I'm saying f is in our case continuous

[calm violet]

Right, but you are saying that my function isn't continuous.

My counterexample

How do you prove that?

[mild bridge]

but f(x) = x^2 is continuous but two different values from set R could give the same output

[mild bridge]

um

it just looks like it

and there's method

epsilon delta

[calm violet]

Will your professor accept this for an answer?

[mild bridge]

fuck no

I'm just saying why i thought it was not continuous

but if I were to prove it

I would try to use the method

i think it'S called epsilon delta in english

[calm violet]

alright, so try to prove that in your free time

[mild bridge]

is it not possible?

[calm violet]

it's possible; but you do realize that to claim this statement, that your proof needs to tell me why functions like what I just gave you are not continuous.

So you can't just tell me oh, it's obvious

considering that it's so difficult for you

[mild bridge]

this here is also a counter example to what I said

[calm violet]

no, that's a different thing

[mild bridge]

ok nvm

so

we have a uncountable codomain R

[calm violet]

Ok let me start

[mild bridge]

we have a continuous non constant function f

ok go ahead

[calm violet]

you don't have to use uncountability of the domain

[mild bridge]

ok

[calm violet]

Assume that continuous f is not constant. Then there exist a =/= b (wlog assume a<b) such that f(a)<f(b). By IVT, since f is continuous, 1) ___, then f(a)<=f(y)<=f(b) (<= means less then or equals here).

1. fill in 1)
2. Does an interval exist in the range? What interval is it that you know surely exists in the range?
3. What does an interval existing in the range of f tell you about the countability of the range? What about the countability of the domain?

[mild bridge]

there exists some value in the interval (a, b) such that ....

[calm violet]

(btw i use range here to mean the image of the domain under f)

Alright 1 is correct so far

[mild bridge]

2

for

2

since the function is continuous

we know for sure that ( f(a) , f(b) ) is an interval in the range

i mean range

[calm violet]

1 is not complete; you haven't even mentioned y

[mild bridge]

um

[calm violet]

actually wait I think it's wrong

[mild bridge]

D :

i thought i was cooking for once

what is wrong

[calm violet]

there is an "there exists" involved, and a "for all" involved

Wait I'm sorry

I fucked up

[mild bridge]

for all values y in the interval (a, b) we have f(y) between f(a) and f(b)?

[calm violet]

Assume that continuous f is not constant. Then there exist a =/= b (wlog assume a<b) such that f(a)<f(b). By IVT, since f is continuous, for all y such that f(a)<= y<= f(b), then 1) ___.

1. fill in 1)
2. Does an interval exist in the range? What interval is it that you know surely exists in the range?
3. What does an interval existing in the range of f tell you about the countability of the range? What about the countability of the domain?

[mild bridge]

1. there exists a value c in interval (a, b) such that f(c) = y

[calm violet]

yep

[mild bridge]

for all y

ok dope

IM UNDERSTANDIN

then for 2

[mild bridge]

i think what i said before is right

[calm violet]

yeah

[mild bridge]

since for all y the interval f(a), f(b) there are values, c in our uncountable domain such that f(c) = y

then the range must also be an uncountable set

[calm violet]

don't say uncountable domain

yet

[mild bridge]

why not

[calm violet]

actually yeah, you didn't need to directly use that the domain is uncountable

[mild bridge]

isn't the domain R

[calm violet]

Yes, but you didn't directly have to use that R is uncountable

actually my bad, I shouldn't have asked you that last part of q3

[mild bridge]

why not

[calm violet]

You just say that (f(a), f(b)) is uncountable, and then that (f(a), f(b)) is a subset of f(R), hence f(R) is uncountable.

Assuming that you don't have to prove that any open interval is uncountable

Then say that this contradicts the assumption that f(R) is countable, hence f must be constant.

[mild bridge]

ok

it all makes sense

except for the part with powerset

i have no clue what that is

I'll have to learn that

i'm gonna start organizing all this and writing it down

thank you so much for the help bro

really appreciate it

[calm violet]

Like the proof can be quite short

[mild bridge]

yeah it actually seemed very logical in the end

and simple

[calm violet]

Assume that continuous f is not constant. Then there exist a =/= b (wlog assume a<b) such that f(a)<f(b). By IVT, since f is continuous, for all y such that f(a)<= y<= f(b), then there exists a c in the domain such that f(c)=y. In particular, y lies in the range of f. Since y was arbitrary, [f(a), f(b)] lies in the range of f. But [f(a), f(b)] is uncountable, so the range of f (which is f(R)) is uncountable. We have proved the contrapositive of the statement.

[mild bridge]

[ f(a), f(b)] lies in the range of f

or am i trippin

[calm violet]

right oops

[mild bridge]

instead of [a, b]

ok got it

it is actually very short

now I feel even more of a dumbasas

[calm violet]

Umm your foundations have big gaps

Fix them and you should do fine

[mild bridge]

yeah

the thing is

I was planning on giving up on the course a whle ago

but then decided to get back to it and at give it a good try

[calm violet]

Are you a math major?

[mild bridge]

I'm studying business mathematics

but i kinda got scammed into it

I thought the hardest thing we gonna is like interest calculations

and then here I am

it's basically just maths with some business classes on the side

[calm violet]

Ah no wonder you don't have the foundations for this course

It's an analysis course right?

[mild bridge]

I even checked all the classes before taking it

[mild bridge]

yes

but the names are the same as the ones from non maths related classes

the maths that the cs majors learn in our uni has the same class name as what we learn in this course

they are both called analysis I, II, III

and I asked my homie to show his hw and stuff

and I was like yeah this shit ez

boom

scammed

tricked

bamboozled

[calm violet]

cs majors should have sufficient foundations for this; they typically cover this in a "discrete structures" class

lmao

[mild bridge]

but I'm gonna give it a good try

I also picked up a book for analysis to go through

[calm violet]

mm

[mild bridge]

maybe itll help to build these foundations

[calm violet]

the thing about math proofs is that it's deceiving in that it's hard in two ways

1. When mathematicians write proofs they have in the back of their mind a waaaayyy lengthier formal model in the back of their mind that's actually logical and systematic in a mechanical and precise way.
2. When they then write the proof down, they write down in prose the most important aspects of how they reason through the formal model, and there is some formal shorthand involved too. So for example in my proof above I said "since y was arbitrary", which is shorthand for going down to the formal definition of interval [f(a),f(b)]={c: f(a)<= c<= f(b)}, and saying that our constraints on y was exactly what they were if y were a member of the interval (anc not an endpoint).

If you didn't know about this and hadn't taken a mathematical logic introductory course, it's unlikely that you know the rules of the game of said formal, logical model

[mild bridge]

I definitely don't I can tell you that for sure

[calm violet]

mm

[mild bridge]

like sometimes things make sense logically

but when it actually comes down to showing it mathematically

im completely lost

like in this case, it made sense to me that a continuous function with an uncountable domain will have and uncountable range

but I couldn't even start writing it down mathematically

[calm violet]

yeah it makes sense intuitively

[mild bridge]

btw do you feel like another question?

[calm violet]

but there are a lot of counterintuitive things that hold in maths that you need to go down formally to show

yeah np

[mild bridge]

ok dope

lemme translate it

[calm violet]

uncountability itself is kinda unintuitive

[mild bridge]

true

[calm violet]

yep unintuitive stuff like this haha

[mild bridge]

lol

like wtf

ok

[calm violet]

ok do you know about the density of the rationals in the reals?

[mild bridge]

does it help if we look at the definition of continuity

[calm violet]

you need that

but can you review your course materials for the density of the rationals in the reals?

You should be able to assume that or have proved it in class

[mild bridge]

gimme a sec

[calm violet]

It's the statement that between any two distinct real numbers there exists a rational

[mild bridge]

I can't seem to find it

is there like a general name for it?

[calm violet]

Yeah that's the standard name

density of the rationals in the reals

Does your course materials talk about the archimedean property of real numbers?

They typically prove the density from the archimedean property

[mild bridge]

ye

the archimedean axiom

for every two real numbers a, b with a > 0 there is a natural number n such that a*n >b

is that it?

[calm violet]

Yeah

Ok so here's how to prove density from archimedean

actually I'll need to think about this a bit

https://www.math.ucdavis.edu/~hunter/m127a_19/rat_dense.pdf
the last theorem here

Ok I wanna go now, so I'll give you questions that outline the approach you should take

[mild bridge]

ok

[white umbra]

I can step in if you want

[calm violet]

1. If the function were to be continuous at any point, where intuitively is it likely to be continuous? [Hint: Consider the graphs of x and 1-x]
2. The density of the rationals implies that when |x-y| is not 0, then there is a rational number between x and y. Since an interval is uncountable and the rationals are countable, this also means that there is an irrational number between x and y. So whenever you are considering every point at most some distance delta away from a particular point, as in the epsilon-delta definition of continuity, you need to show that the image of rational numbers and irrational numbers under f both satisfy what you need them to satisfy.
3. Recall the function I talked about earlier g that is zero everywhere except for g(0)=1. You said that it is not continuous; I want you to answer your exercise question for g: where is it continuous? where is it not continuous? (your course notes might have a worked example that's like my function) Use the epsilon-delta definition and its negation to show that it is continuous where you think it is, and not where you think it is not.
4. Use the technique you used to show that g was not continuous at the point(s) that it wasn't continuous to show that f is not continuous at the point(s) that you think it isn't continuous. If you can't identify all the points where you think it isn't continuous, you should already be able to identify some by analogy to g, then this should give you confidence in identifying all the points where f is not continuous.
5. At this point you should have a clear idea of what point(s) are continuous. Now prove it.

[mild bridge]

ok man

ill give it a try

@calm violet thanks a lot

[calm violet]

np

[mild bridge]

hey bro, you there?

[white umbra]

Yeah. Do you have a specific question about what's been said so far?

[mild bridge]

yes

[mild bridge]

in the fist paragraph here

he says think of the graph

would that graph basically have gaps everywhere?

cuz intuitively, say we have a rational number

[white umbra]

yeah it would have a ton of jumps

[mild bridge]

we can go to the right or left of this number

by a really tiny irational number

[white umbra]

it would be jumping from x to 1-x and back to x all the time

[mild bridge]

and suddenly it jumps 1 away from it

[white umbra]

But

There's one point where x and 1-x are actually the same

What point is it?

[mild bridge]

so you said that it jumps back and forth

i guess the middle point

cuz in the middle

it can't jump back and forth

just jumps back to the same location

[white umbra]

Yeah exactly, what point is that?

Like what is the value of x at that point

[mild bridge]

idk where the middle point would be in the range

oh wait

0.5

[white umbra]

yup!

So at 0.5, x and 1-x are the same thing, but everywhere else, they're different.

Using this knowledge, where would you guess that the function is continuous?

[mild bridge]

wdym mean by x and x-1 are the same thing

actually

i don't get it

if we have x = 0.5

[white umbra]

Careful, it doesn't say "x-1"

[mild bridge]

oh shit

yeah nvm

it makes sense

[white umbra]

awesome!

[mild bridge]

the function is at 0.5 continuous

which means when we approach 0.5 from the left and right side

we get the same value

that right?

[white umbra]

yeah, the same value as f(0.5)

what about everywhere else besides x=0.5?

[mild bridge]

it wouldn't be continuous

[white umbra]

Yeah

So that's the answer, but we need to prove it

[mild bridge]

from a rational value we could move an infinitesimal irrational number to either left or right

and then 1 - x makes it so that there is a gap there

[mild bridge]

yup

[calm violet]

there's the case for 0.5 too, it starts jumping as soon as you move left or right

different magnitude, but it does

[mild bridge]

oh

[calm violet]

rn i suggest you go to step 2

[mild bridge]

then it's not continous at 0.5?

[calm violet]

to develop better intuition

[white umbra]

Well let's try proving that it is continuous at 0.5.

[mild bridge]

if it also jumps away from it when slight to its sides?