elementary-number-theory

[blissful venture]

python is dead slow anyway

[weak rapids]

is it just doing that algo for each congruence?

[blissful venture]

As in?

[wild zinc]

what does the zip function do?

[blissful venture]

Zip creates an array of tuples to iterate over, like

zip([1, 2, 3], [4, 5, 6]) = [(1, 4), (2, 5), (3, 6)]

But it returns it as a generator the value only when needed, so it uses less memory and is faster

[sacred junco]

@weak rapids the answer is 41?

No no thats wrong.

31 shud be the answer

[weak rapids]

@sacred junco from online calculator they are saying its 251. Really if someone can explain how to get it using extended euclidean algo, it would be much appreciated

[sacred junco]

Aaaahh.why am i wrong

Oh i got it

Ok listen.

Heres what i did.

No dont listen i cant explain on this platform..

Ok ive written it down for u

@weak rapids

[weak rapids]

@sacred junco thanks. Why is there a 1 in front of the mod on each line ?

Tryig to figure out patterns

[sacred junco]

Thats what the line previous to it asserts.

Ok listen

[blissful venture]

This is such a painful way of solving

[sacred junco]

U wont understand any of this rn.

@blissful venture theres always room for improvement.

Share ur solution..im really new to chinese remainder theorem.

[blissful venture]

You remember all those so's right? 👀

you know that x = 1 (mod 5)
=> x = 1 + 5k
=> 1 + 5i = 6 (mod 7)
=>i = 1
=> x = 6 (mod 35)
=> x = 6 + 35j
=> 6 + 35j = 3 (mod 8)
=> 35j = 5 (mod 8)
=> j = -1 = 7 (mod 8)

[sacred junco]

What will u do after that?

[blissful venture]

7*35 + 6 is your answer

gg easy

[sacred junco]

Whoa ok holy shit.....

But what for larger numbers?

[blissful venture]

I've had practice though, used to do this as high school trick for 4-5 equation in my head, so I have lot of practice simplifying fast

It's same for large numbers, it's gcd, so it becomes small fast

[sacred junco]

@weak rapids well as u can see i got owned.

So ill go wear my hat of shame

@blissful venture thanks for disclosing the method dude.

[blissful venture]

It's alright, I used to do it like you too

because the proof uses that

[sacred junco]

But then u evolved.

Falcon does it the same way

[wild zinc]

I don't think I've ever seen the proof so I had to come up with my own method :^)

[sacred junco]

Im trying to prove it myself.

[blissful venture]

Man, when I came across the proper proof in undergrad I was like woah

never needed that much rigor before for olympiads

[wild zinc]

initially it was just a slower version of that but then I realised I was being a dumbass

[sacred junco]

@blissful venture plz dm me the proper proof if u can.

[blissful venture]

^^

I dunno, you can google it, should be on wiki

oops

t!wiki CRT

[sacred junco]

Nvm ..ill do it myself.

[blissful venture]

[sacred junco]

It uses a certian proposition..

Which at some point will ne necessary to prove in a solid way

[blissful venture]

I wrote some code for multivariable CRT in undergrad

It was good stuff

It's really cool how you can extend the CRT, it comes in handy.

[wild zinc]

is there a better way to extend CRT to non-coprime moduli than just making the moduli coprime by dividing out some common factors?

[blissful venture]

But taking gcd of the moduli is fast, so finding common factors is also fast

You can do the usual stuff too, as long as you don't accidentally multiply/divide by a factor of the moduli while getting the inverse, but I found that I was more prone to mistakes there

[sturdy dirge]

$$
\left{
\begin{matrix}
x&\equiv1\pmod{5}\
x&\equiv6\pmod{7}\
x&\equiv3\pmod{8}
\end{matrix}
\right.
\iff
\left{
\begin{matrix}
x & \equiv 1\pmod{5}\
x & \equiv 27\pmod{56}
\end{matrix}
\right.
\iff
x\equiv 251\pmod{280}
$$
.

[sacred junco]

What the fuck do u guys smoke😂 😂

@sturdy dirge u using the crt in some other alien way?

[vapid tree]

How do u
x= 6 (mod 7)
x= 3 (mod 8)
And thus
x= 27 (mod 56)

For me I do
7a + 6 = 8b + 3

Then
7a + 3 = 8b

Through trial and error a = 3 b = 4
Thus 27 (mod 7 x 8)

But idk how to do it for big numbers

Is there a rule

[sacred junco]

I really dont know.

U can see my method is dumb and different

[vapid tree]

But did it work though?

[sacred junco]

I mean yeah

But its lengthy

[vapid tree]

I think i found a way that might just be pure luck

x = 6 (mod 7)
x = 3 (mod 8)

6/7 - 3/8

27/56

Idk

[vapid tree]

Yeah I think that actually makes sense

Since if we write
x = 7a + 6
x/7 = a + 6/7
x/8 = b + 3/8 -
x/7 - x/8 = a - b + 6/7 - 3/8
x/56 = a-b +27/56

And a b are integers

Then we can say
x = 27 (mod 56)

However I don’t think it works for all

x/56 = a + 27/56
x/5 = b + 1/5 -
51x/ 280 = a - b + 27/56 - 1/5
51x / 280 = integer +79/280

Uh.....

51x = 79 (mod 280)

51 = 51 (mod 280)

x = 79/51 (mod 280)??????

280(51)/51+ 79/51

[vapid tree]

(x/280)51 = Integer + 79/280
b = original numerator before x51 and simplify
a = amount of times numerator was subtracted by denominator
79 = 280a - 51b
b < 280
b has to end with 1

Oh wow that’s some big trial and error
b = 251, a = 46

x/280 = integer + 251/280

251 mod(280) = x

Finale

[sacred junco]

I'm having trouble showing that f is onto implies gcd(m, n) = 1

Can anyone help? <@&286206848099549185>

[vast vessel]

Please follow the min 15 min rule: wait a minimum of 15 minutes before pinging @helpers.
#❓how-to-get-help
Thanks.

[silver solar]

Chinese remainder theorem?

although nah, Chinese remainder theorem is actually the converse...

maybe you can prove using contradiction

[sacred junco]

oh sorry, I didn't realize there were new rules

I used CRT for the converse

[silver solar]

like, if m and n are not coprime, there exists a pair that is not reached by f

[sacred junco]

hmmm ok I'll think about that

[slender berry]

@sacred junco

[sacred junco]

hi

[indigo iris]

Imagine the future if Math didn't exist

[elder bobcat]

does math exist

philosophy is a scary place

be careful or you might want to die

kek

w h y is math

[elder bobcat]

n u m b r e t h e o r y

m o i t e

[glad nacelle]

@ionic quarry please don't post in multiple channels

[ionic quarry]

apologies

[shell ore]

what's the subset thing that doesn't have a - in the middle?

[fathom sierra]

yeah $$\subset$$ literally means is a proper subset of @shell ore

[fathom sierra]

it's the equivalent of

$$x\in A$$ but when x is a whole set

[shell ore]

@fathom sierra hmm alright, any clue by the way how to get not less than sign?

[fathom sierra]

not less than, just put greater than or equal

$$\geq$$

[fathom sierra]

(\not<)

[shell ore]

can anyone help me with veninn diagrams?

I need to calculate this: (A△B)∩C

and also thanks @fathom sierra

[spare python]

Venn*

[unborn geyser]

I was goofing off in maths class looking at the multiplication table in my book thinking of the different ways of multiplying numbers that give the same answer and I came up with:

n*m = ((nx)/y) * ((my)/x)

[unkempt hedge]

How do I solve this generating function ∑f(x)x^n = x/(1-x)^3 + f(0) ? where f(0) is a constatn

[unborn geyser]

I finished my thinking:

I propose that n^2 = ((nx)/y) * ((ny)/x)

Where x, and y are factors of n, then ((nx)/y), and ((ny)/x) are factors of n^2

[ornate wigeon]

Yes, but x and y aren’t necessarily factors of n

Check n=3 x=2 y=2

[unborn geyser]

3^2 = 9 = ((32)/2) * ((32)/2) = 3 * 3 = 9

I was getting at is the factors of 20 are 1, 2, 4, 5, 10, 20

The factors of 20^2 (400) are 1, 2, 4, 5, 8, 10, 16, 20, 25, 40, 50, 80, 100, 200, 400.

I can make all of the factors of 400 only using the factors of 20 in the ((nx)/y) * ((ny)/x)

N is always 20, x and y are always a factor of 20.

(I can't seem to post all of the sums I want to in this chat :/) there seems to be a word limit, I only want a few more lines for my sums

[fair lynx]

For a, i put

Let s be the sup A, and t be the sup B.
then, t >= b, for any element b in B. Since
s < t,
and t is the upperbound of any b in B including b itlse,f then we an say that there is an element b where b = t.
Or,
s < b.
==> sup A < b.

but then what example would not work if sup A was <= sup B?

i'm assuming since it works with Sup A < sup B,
the problem occurs when sup A = sup B

oh wait I can think of an example, nvm

[spiral heath]

My brain is so tired

For the past few days I have been obsessively searching for hints towards an exact representation of a=sum(i=0,∞,1/floor(a^(2^i)))

The algorithm for approximating it is easy and seems to get one decimal digit correct for every three steps - just start with any random value for a, I use 2, then take the sum, average that with the previous value of a, repeat - N steps seems to produce N/3 accurate decimal digits

but attempting to find any other patterns is seemingly futile with my knowledge

[blissful venture]

I saw that before the edit

[spiral heath]

I've been trying to find patterns in the sequence of a^(2^i)

but I can't

not even a^i

there are some pseudo-patterns but they always start diverging eventually

ugh this place is always so dead except for the question channels

https://media.discordapp.net/attachments/488104216678760469/493161634072625164/latex.png

this! this is my number/formula/thingy

(latex is hard ugh)

I have no idea how to define this any other way or construct the sequence of floor(a^(2^n)) without knowing a already

[vast vessel]

\👀

I doubt there's a closed form

t!wiki Lacunary series

[unique vault]

📖 | ** https://en.wikipedia.org/wiki/Lacunary_function **

[vast vessel]

If a is an integer, this is just a Lacunary series

No closed form

oh wait

ur asking me to solve an equation

🤢

why is this #elementary-number-theory lol

well in any case I doubt there's a closed form for a unless it happens to be some nice trivial value like a rational numbre

[spiral heath]

how is it not number theory, I'm studying numbers...

I guess maybe it goes in pure math?

point is, no, I didn't ask you to do anything, my point is I feel like shit because I've been on a wild goose chase for several days like a typical amateur mathematician who knows basically nothing

[craggy plume]

"number theory", despite it's name, generally studies only integers.

(And stuff like divisibility and prime numbers.)

[spiral heath]

yeah.

[craggy plume]

This looks more like algebra.

[spiral heath]

when I think of algebra what comes to mind is abstract algebra

I hate the confusing, random terminology for what kinds of math are what

like how the only difference between geometry and algebra is whether you are looking at a picture or not, ultimately

[sacred junco]

@spiral heath

[spiral heath]

lovely, I didn't think to plot it in Desmos - I mean, I know what the number IS anyway, I have an algorithm to calculate it to arbitrary precision

I just wanted a faster way of expressing it, a series representation for the number not reliant on its own self

[sacred junco]

oh nice

[spiral heath]

also, it's clearly NOT 5/3 😄

[sacred junco]

Yeah

I know

I was just making a quick guess

ignore that

just scratch

[spiral heath]

I had a continued fraction representation for it up to 15 digits but I deleted it because what's the point 😄

[sacred junco]

But I just read it like a minute ago

so put it into desmos to get some start

[spiral heath]

yeah you can get it to arbitrary precision by starting with x=2, taking the sum of 1/floor(x^(2^n)), then making the average of that sum with x be new x, over and over again

three steps of that gives one decimal digit correctly, it seems

[sacred junco]

well yeah that is one way to do it

[spiral heath]

and every kind of mean seems to work about equally well so I just use the arithmetic mean

[sacred junco]

that is just a uhhh approximation algorithm

[spiral heath]

it's just... not what I want

yeah exactly

I want the exact number

naturally it will still require some form of approximation to actually use, of course, but at least I would know how it's related to other numbers

[sacred junco]

I am trying to think about a more elegant way to get it

well it would be nice if you could apress it as something simple and elegant

[spiral heath]

Well if there's a generalized way of solving this problem, I have a bunch of other double exponential constants of similar sorts we could work on as well and make a paper out of it 😛

[sacred junco]

Sounds interesting

I was going to give it a shot to kind of get a fresh prespective

[spiral heath]

For several days I was obsessively looking at the sequence of floor(a^(2^n)) - and even just a^n - to see if I could find patterns in the sequence

For several days I was obsessively looking at the sequence of floor(a^(2^n)) - and even just a^n - to see if I could find patterns in the sequence

if there was a pattern in the integer sequence, I would be able to get a out of it

[sacred junco]

I been sick the past week and stuck in bed working on a paper

[spiral heath]

but alas, there is nothing I can find that holds up

[sacred junco]

Want to work on something else

A general solution sounds hard tbh

[spiral heath]

of course, which is why I'm focussing on specifics for now 😃

I wonder... if we pretend a is a binomial, I wonder if there would be anything in Pascal's Triangle which could help...

[sacred junco]

thats not a bad idea

[spiral heath]

but I don't really know how to use it for arbitrary binomials anyway, much less numbers being symbolized by them

still, we would just have to look at rows in the Triangle which are powers of 2, and do... something? with the coefficients there

problem is that floor function, and the reciprocal

I don't have any idea how to put it all together.

[sacred junco]

I mean if you are dealing with a real number raised to a power inside a floor function

you are making things complicated

[spiral heath]

of course I am

it started with a sequence actually

[sacred junco]

That is oh

[spiral heath]

I was looking at Sylvester's sequence minus one (1,2,6,42,1806...)

and noticed this simple pleasant pattern in it (each number is x(x+1) where x is the previous number)

and discovered Vardi's constant squared in it, the "a" in floor(a^(2^x)) which makes the sequence (without knowing about it beforehand, or even that this was a sequence anyone else had studied - never heard of Sylvester till recently)

then I thought "well is there any such sequence where reciprocals of the numbers in it sum to the same constant 'a' that makes it?"

[sacred junco]

interesting

[spiral heath]

and indeed there IS such a sequence, but the problem is that there seems to be no easy pattern in the sequence itself by which you could construct it

so you have to start with the constant that makes it, instead of ending up with that

which is... kind of counterproductive

It goes 1,2,7,57,3347, and on upward - there are some rather complicated patterns in the lower values that completely fail in pretty much the next value - every time

I even tried it with powers that aren't themselves powers of two, just a^n generally and it looked like a sort of nudged fibonacci sequence (because a=1.6606... is close to phi), but again all the patterns I tried to find in it eventually failed

[autumn swift]

my class is actually graph theory, i think this channel is closest

need help on induction proofs

please ping me

[silver solar]

Graph theory comes closest to #combinatorics actually @autumn swift there

[autumn swift]

ok

[blissful venture]

How is Graph Theory combinatorics?

Wikipedia lists it under other

Though I think geometry might be closer

[sturdy dirge]

Taxonomy issues.

[silver solar]

Graph Theory is kind of all over the place tbh

[spiral heath]

I actually would place graph theory as a form of topology... :3

[craggy plume]

Graph theory is of course about algebra. You graph f(x)=y all the time. 😛

[glad nacelle]

Graph theory is probably combo tbh

Like I'd say it's a combinatorial structure, among others studied in the subject (hypergraphs, projective planes, etc)

[sacred junco]

Is product of (1-2/(p-1)) to p(n)
O(1/log(n))

Mind you not talking about all primes less than n

The nth prime

[spiral heath]

what?

@sacred junco what are you referring to there? (sorry to ping you)

whatever it is it sounds interesting

[sacred junco]

@spiral heath oh hello

So uh

Let me see here

I should take a snap of my white board

[sacred junco]

@spiral heath

[spiral heath]

that doesn't really help much

I can't figure out exactly what it is meant to be

oh the big pi is the product symbol

[spiral heath]

measurement of how long an algorithm takes to reach an answer

basically (or rather how quickly it converges)

I'm... really not sure what it's doing there though

or what the whole point of this equation is

also I didn't know you could just put the top number on the product symbol without a bottom one - I presume that means product from i=0 to n, or something, but it's confusingly written

[spiral heath]

maybe

but what she said still doesn't make a lot of sense and I don't understand the context of the equation

still at least now I have something to evaluate, see what it does

(then there is the question of what is the zeroth prime)

the thing is if this is prime numbers then it will provide a result of 0 if it includes any prime less than 5

and that makes no sense

@sacred junco your equation makes no sense and I have no idea what it's for, that's why I pinged you

[spiral heath]

and all those people tried to explain it to me and just made me even more confused. god fuck I feel stupid.

but yeah that equation multiplies to zero after like the second term so I seriously have no idea wtf it's all about

[sacred junco]

@spiral heath

hey

so the context

it is about PNT for arithmetic progressions

this product was showed up in my calculations

and I am trying to find a big O

[spiral heath]

but... the way you wrote it, it multiplies to zero

unless I just do not understand math suddenly

one minus two over the second prime (3) minus one is 1-2/2=1-1=0

so the product for all n of (1-2/(nth prime minus one)) equals zero

[sacred junco]

So anyway I have some notes

I will post them when I am done

Trying to be a little clearer

For review

[sacred junco]

So what I am asserting is this

Given a mapping P -> A
Where 0<a<p

For p>3

We can create a class of Arithmetic progressions

pn +- a

So results on Arithmetic progressions show that for example

pn + a will assymtotically account for 1/(p-1) of the distribution of primes

Thus +-

It would account for 2/(p-1)

And then the product of 1-2/(p-1)

would be account for the numbers not represented by such arithmetic progressions

all of this follows from the Poussin's results on arithmetic progressions and the PNT

featured here

https://en.wikipedia.org/wiki/Prime_number_theorem#Prime_number_theorem_for_arithmetic_progressions

So since this product must be O(1/log(n)^C) for some C

for the nth prime

then the number of primes that do not conform to any such arithmetic progression

would be n * O(1/log(n)^C)

which would be

O(n/log(n)^C)

which the limit of this would be infinity for any C

so given any mapping

P -> A

0<a<p

p>3

there are infinite quantity of prime numbers that can not be represented as pn +- a

any thoughts?

[sacred junco]

<@&286206848099549185>

not so much of a goddess at math are you

😂

Huh

Maybe not

It is more about beauty

💋

@sacred junco

......

alright have that win then

lets focus on the math

Yeah so criticism

Of my argument

Or not

[vast dove]

I don't even understand it

[shell ore]

can anyone help me with this, what do we have to assume when proving that this is true

$$(A \setminus B) \cap C \subset A \setminus (B \cap C)$$

[shell ore]

is it that $$A \subset B$$ or something?

[shell ore]

<@&286206848099549185>

[somber rampart]

@shell ore let B = {}

then A intersection C is obviously not A \ C in all cases

so it can't be that B is just a subset of A or C

if A is a subset of B doesn't work because then we have {} on the left hand side and A on the right hand side letting C = {}

so we've narrowed down our options

of subset relations

[shell ore]

hmm alright

thank you!

[somber rampart]

with C = {} then A = {} as well

in fact I think the condition we're looking for is A = C

[shell ore]

so 1 of them just has to be {}

[somber rampart]

no it's just that {} is convenient for calculating

since it behaves nicely with intersection

and with set minus

so we can check for the degenerate case of {}

and see what happens

[shell ore]

what degenerate case?

[somber rampart]

of something being = to {}

[shell ore]

ah

[somber rampart]

A = C works for proving it

not sure if that's too strict a condition though

also didn't see the subset thought it was =

[shell ore]

how do you go about to figuring out that A has to be C etc? is it just by doing theese a lot

[somber rampart]

so let me go back a bit

yeah just thinking about the problem and testing out different ideas

[shell ore]

also quick sidenote, it says that it has to be for all A, B and C

[somber rampart]

there might not be any assumption then

I thought it was just = not subset

lets see

so let something be in A, not in B, and in C

then it is in A

and not in B, thus not both in B and C

and so we're done

yeah it works for all A, B, C

why were you looking for an assumption?

[shell ore]

because it says "what assumption do we have to make"

I tried running the question through google translate but that was no good, but it asked for an assumption from the start

I think it's that A not in B but in C

[somber rampart]

Ohhh, it's just the translation is a little weird

you start with the assumption that there's an element a in A, not in B, and in C

[shell ore]

yea, I'm always bad at translating theese questions

[somber rampart]

I thought you meant an assumption about the sets themselves

which would contradict your for all A, B, C

[shell ore]

is there a difference?

[somber rampart]

yes

[shell ore]

oh

[somber rampart]

here we're proving that if a is in (A \ B) intersect C then it's in A \ (B intersect C)

which begins by assuming a is in (A \ B) intersect C

and showing that given this a is in A \ (B intersect C)

make sense?

[shell ore]

uh, maybe..

okay no, I'm not actually following...

so because a has to be in both (A\B) intersect C and in A\ (B intersect C), A must be in C

[somber rampart]

we're not saying A must be in C

that's not what we're doing

you misunderstood the question a bit

we're not looking for a relationship between the sets and an assumption there

we're looking for an assumption to start with to do the proof

[shell ore]

it says prove that and with all and what we have to assume at the beginning

I don't understand what I missunderstood....

right

[somber rampart]

Ok lets do the proof first so you can see it works regardless of what you know about A, B , and C

[shell ore]

okay

[somber rampart]

We want to prove that (A \ B) intersect C is a subset of A \ (B intersect C)

it suffices to prove
if a is in (A \ B) intersect C then it's in A \ (B intersect C)

[shell ore]

yes

[somber rampart]

assume a is in (A \ B) intersect C (this is the assumption they're talking about)

then a is in (A \B) and in C

then a is in A, a is not in B, and a is in C

thus a is in A, a is not in (B intersect C)

thus a is in A \ (B intersect C)

[shell ore]

why are proofs so confusing...

[somber rampart]

you'll get used to them

look at each step in particular

do all of them make sense?

[shell ore]

sidenote, a is just some arbitary set right?

[somber rampart]

a is an arbitrary element in (A \ B) intersect C

yes

[shell ore]

okay

ooh now I think I'm understanding it better

because a has to be in A\B it can't be in B because then it wouldn't be in the A\B

[somber rampart]

mhm

[shell ore]

but it has to be in C because it has to intersect

riiiiiiight

so it would look like this pretty much written on paper?

[somber rampart]

mhm

with symbols if you want to use them

but yes

[stoic basin]

Rendering failed. Check your code. You can edit your existing message if needed.

[shell ore]

assume $$a \in (A \setminus B) \cap C then a \in A, a \not \in B, a \in C$$(edited)

[stoic basin]

Rendering failed. Check your code. You can edit your existing message if needed.

[shell ore]

okay I don't know how to fix that..

or can I just write that all that, would that be eligable as proof or does it have to be symbols?

@somber rampart so this would be correct? http://i.imgur.com/ggGxpTh.png

oleta että means assume

[somber rampart]

Yes ^^

[shell ore]

alright thank you! I'll see if I can do the other ones myself :p

[shell ore]

@somber rampart would you have time to help me with one more? or give a hint

I have to prove that if m is an odd number prove that m^2 is also odd

[fathom sierra]

one way is just letting m=2k+1; k in N

[shell ore]

yea I've got that, but idk what from there

:d

[fathom sierra]

And calculate m^2 from that

[shell ore]

hmm alright...

[fathom sierra]

m^2 = (2k+1)^2

[shell ore]

yea

I guess

but do I solve the (2k + 1)^2 then break out a 2 or something

and bam it's not even or something

[fathom sierra]

Expand that dope and show that it's equal to 2(something)+1

Yas

[shell ore]

right

how do I know if I'm done with the proof

I've gotten to m^2 = 2(2k^2 + 2k + 1/2)

[fathom sierra]

It's 2(2k^2+2k) + 1

Veru important distinction here

[shell ore]

oh

is it proven like that?

:d

[fathom sierra]

Ya (well there's a shitton of ways to prove it but that's one of the most basic ones)

[shell ore]

oh

okay...

should I use the equivalent sign like this

$$m^{2} = (2k + 1)^{2} \equiv 4k^{2} + 4k + 1 \equiv 2(2k^{2} + 2k) + 1$$

[fathom sierra]

tbh I never used it in my whole life

[shell ore]

oh..

[fathom sierra]

But I guess you can here yeah

[somber rampart]

Uh

IDK why you would

[fathom sierra]

i'll just shut my mouth

Well "never used it in my life", used it to denote congruence modulo w/e ofc

[somber rampart]

Yeah it's not for this, which is just regular equivalence

It's for equivalence relations which are not the regular notion, such as congruence modulo

[shell ore]

oh

okay I'll take it away then

I don't know should I use lim for this or what, but I need to prove that if a > b then (a + b)/2 < a, I've used words but I don't know if I'm supposed to use mathematical symbols and stuff for it instead

but I've written that if a would be equal to b then a + b would be 2 a which would be equal to a, but because a > b then (a + b) /2 can never come all the way to a which means it will always be less than a

[vast vessel]

a > b
a+a > a+b
2a > a+b
@shell ore

[mystic elm]

anyone able to jumpstart me on an epsilon delta proof?

[sullen rune]

sure I love torture

[mystic elm]

nice

i've been angsting over proving that the limit of the dirichlet function does not exist, i.e. when x is rational, f(x) =1, if x is irrational, f(x) = 0

i've been trying to prove the negation of the limit definition, but i'm hitting a brick wall when it comes to relating the resulting inequalities as epsilon isn't expressed in an inequality with x, as f(x) only takes on the values of {1, 0}

internet tells me to remember the density of rationals and irrationals, which would imply linking the two cases of rationals and irrationals, but i keep falling into proving them seperately, and thus having no common ground between them

[sullen rune]

is the limit just anywhere or

[mystic elm]

yeah

limit as x approaches a for every a

[sullen rune]

ok do you think you have the negation of the limit definition

[mystic elm]

ye
there exists an epsilon > 0 s.t. for all delta > 0, there exists an x s.t. 0<|x-a|<delta and |f(x)-L| >= epsilon

so in the case of a rational, |1-L| >= e, in case of an irrational, |L| >= e

[sullen rune]

gotcha, so the density of ir/rationals let's you say there's always an x that gives f(x)=0 or 1

[mystic elm]

ye

part of my issue is that i'm not exactly sure where i'm supposed to end up lol
yes, prove that there's an L that works for the negation, but how exactly do i get there

[sullen rune]

er I think you wanna show that no L works

[mystic elm]

then i want to be proving the positive rather than the negation and then reach a contradiction, correct

part of this might be that i'm completely flipped over heads over heels here

i honestly don't know

the past hours have not been kind

[sullen rune]

gotcha

so let's just pick an e

we're trying to get far away from 0 and 1, and with >= I think anything less than 1/2 will do

so like e=1/4

we can try to prove no limit for a general L

[mystic elm]

okay

[sacred junco]

Plumorant

[sullen rune]

nice

[mystic elm]

so trying to prove a limit exists

and reach a contradiction

[sullen rune]

nah

[mystic elm]

okay still going the other way

no limit

cool

[sullen rune]

yea I think so

[mystic elm]

okay
then we have,
x is rational
|1-L| >= e = 0.25
x is irrational
|L| >= e = 0.25

[sullen rune]

I don't really understand that

like you have x in {a-delta,a+delta}

so you can go for 2 cases for x: rational/irrational

which if we're using e=1/4 gives |1-L|>=1/4, and |L|>=1/4

and no L can satisfy both

er misspoke

L has to satisfy one of those

[mystic elm]

i feel kinda dumb
when proving two cases for x like that, can you assume that the limit would be the same for both?

ohokay

how do you go ahead and prove that L actually satisfies one of those conditions, or do you just say that an L exists such that one of the conditions are satisfied, and thus the proof is done?

[sullen rune]

yeah you gotta prove that all L satisfy at least one condition

[mystic elm]

next question is how

[sullen rune]

which is either so easy it doesn't need a written proof, or you can be like R\(3/4,5/4) satisfies the first, and R\(-1/4,1/4) satisfies the second

and the union of those is R so you're good

[mystic elm]

oh

[sullen rune]

oh and e=1/2 works woops

[mystic elm]

so just break up the inequality, and then give the range of L. we can choose any e > 0 in place of 1/4, and thus L will take on any value of [-e, e] or [-e+1, e+1]. therefore the negation is proved for all L, and thus the limit does not exist.

or would i be going back into the weeds by saying that

[sullen rune]

yeah looks like weeds

[mystic elm]

[sullen rune]

everything above says that the limit isn't L for all L

[mystic elm]

okay

apologies that i'm being a backwards human being, but if we walk back a step

why exactly would you take the union of the two cases of L

[sullen rune]

the line with R\?

[mystic elm]

ye

just recently covered most of this, so some of my basics are a bit shakey

as evident

apologies

[sullen rune]

uh, union is like the OR for sets

and we're looking at L that satisfy condition X or Y

so we're saying L is either in this set or that set

where being in the sets mean there's no limit

[mystic elm]

ahhh

that makes sense

that makes a lot of sense

dude you're amazing

thank you

[sullen rune]

np~

[mystic elm]

< 3

[sacred junco]

A grasshopper moves along a straight line. She knows how to make jumps of 11 cm and jumps of 17 cm, each in both directions (to the right or to the left). After a number of these jumps, she finds herself exactly 9 centimeters from her starting point. How much did she jump, at least? Somebody can tell me the solution

<@&286206848099549185>

[sullen rune]

you can simplify it a little bit

does it ever make sense to go say, 11 cm to the left and then a bunch of steps later go 11 cm to the right?

[sturdy dirge]

9=17*(-4)+7*11 thus 4 + 7 = 11.

[sacred junco]

So the sollution is @sturdy dirge 9

Oh thank

@sturdy dirge

[sturdy dirge]

You are welcome.

[sacred junco]

And for this exercice somebody can tell me the solution (like this i see if my solution is true)

A frog (negligible size) moves by jumping in a grid pattern (infinity) 1 cm × 1 cm. It jumps from point to point, where a point designates an intersection of two lines of the grid. She can do 4 types of jumps:
6 cm jumps to the right,
Jumps 8 cm upwards,
Jumps 9 cm to the left,
10 cm jumps down.
Among all grid points, a square of 10 × 10 points is colored red. How many red dots, among the 100, can the frog reach as it starts at one corner of that square?

<@&286206848099549185>

[sturdy dirge]

Maybe 15.

[sacred junco]

No 😢 i'ts not working i have wrong answer)

@sturdy dirge Did you have some idea

[sturdy dirge]

Horizontal jumps are multiple of 3 and vertical jumps are multiple of 2.

[swift shard]

Doing right, left gives 3cm left, so a 3cm left jump is possible. Then, do right, gives a 3cm right jump.

Up, down gives a 2cm down jump. Do that 3 times, then up, to get a 2cm up jump.

I believe that's the best you can do

[sacred junco]

So the result is

?

@swift shard

[sturdy dirge]

$$
\left(1 + \left\lfloor\frac{10}{2}\right\rfloor\right)\left(1+\left\lfloor\frac{10}{3}\right\rfloor\right)=4\cdot6=24
$$.

[swift shard]

Yap, that looks pretty good

[sacred junco]

No

i'ts not working

[swift shard]

It didn't take 24?

[sacred junco]

No i didn't take 24

[sturdy dirge]

More data is required.

[sacred junco]

The 24 is y

But x

I don't know

[sacred junco]

Hello

[spiral heath]

Hello also 😃

[woeful ravine]

100!/10^24=(Is there a way to do it without a calculator)?

I tried looking into factorial congruence... but I didn't understand any of it.

[silver solar]

The result is 134 digits long...

[sacred junco]

Yes

Euler theorem

[last nova]

is there a way to prove:
(3^(2^69))-1
is the sum of two perfect squares?

$$
(3^(2^69))-1
$$

[last nova]

oh gawd

[sacred junco]

Lol

[last nova]

be able to help me out?

[sacred junco]

Hmm it's easy

[last nova]

this is my first time using lol

[sacred junco]

Wait

What is 'perfect square'?

[last nova]

i guess prove it is equal to:
a^2 + b^2
where a and b are integers

[sacred junco]

Easy.

[last nova]

nice

i've been stuck on it for the last week

what u got?

[sacred junco]

Just has to prove that it's possible with any number with the form 2^n(4k+1)

[last nova]

what theorem is that?

[sacred junco]

Wait a minute

[last nova]

be careful btw

3 to the 2 to the 69

so its 2 to the power of 69

and 3 to the power of that

altogether minus 1

[sacred junco]

https://en.m.wikipedia.org/wiki/Fermat's_theorem_on_sums_of_two_squares

That and

For a, b of which is sum of two square

a×b is also like that

[last nova]

hmm sorry not sure what u mean on that part

[sacred junco]

Last part?

Well it's some identity

https://en.m.wikipedia.org/wiki/Brahmagupta–Fibonacci_identity

[last nova]

so how do i apply that to the question?

[sacred junco]

Maybe factorize

Hmm, and also

[last nova]

how do u factorise a second power?

[sacred junco]

I don't like spoonfeeding

Do you know what the power there means?

[last nova]

sorry im desperate now lol

let's assume i dont

[sacred junco]

So you don't know what is 3^(2^69)?

[last nova]

its 3 to the 2 to the 69?

[sacred junco]

2 power 69 first

And power 3 with that

[last nova]

yes

[sacred junco]

And you know that

3^(2^69)=( 3^(2^68) )^2

Right?

..?

[last nova]

wait

how did u get that

though im starting to see it now

[sacred junco]

Oh right you need to learn working with it..

It's 3^(2^68×2)

And distribute it

[last nova]

dont u get 2^68 + 2^68?

oh hang on

OH i got it alright thanks

68+1 nice

alright

[sacred junco]

Good

Now to factorize stuffs

It's in the form of

x^2-1

Right?

[last nova]

yep

OH

x^2 - 1^2

so we have (x+1)(x-1)

[sacred junco]

Yupp

[last nova]

the braghm-fibo theorem?

[sacred junco]

Yep!

x+1 is already sum of two squares

[last nova]

sick!!!

[sacred junco]

Right?

[last nova]

yep

wait hang on

[sacred junco]

So x-1 left.

[last nova]

hang on why would x be a square

the 2^68 is still there

[sacred junco]

x = 3 ^ (2 ^ 68)

[last nova]

ye

[sacred junco]

And do what is done again

[last nova]

oh hang on

can we repeat it over with both x+1 and x-1?

[sacred junco]

x+1 is already sum of two squares

So don't need to for x+1

[last nova]

ah right

[sacred junco]

Think :

[last nova]

ok i get it

[sacred junco]

Got it? Good!

[last nova]

so we need to prove the right side is equal to 1?

ahshit what am i saying

forget i said that

why would it

hm how would we bring the braghm theorem here now though

[sacred junco]

Well did you get that

x+1 is sum of two squares

[last nova]

yea

can be reduced to 2^67

[sacred junco]

So it is enough to show that

x-1 is also sum of two squares

What is x-1?

[last nova]

doesnt that break down into (x+1)(x-1) again?

where the power is lower by 1?

[sacred junco]

Yup

[last nova]

alright

now we have 2 sums of

WAIT

[sacred junco]

So.. what happens?

[last nova]

we can drop that braghm theorem in there

[sacred junco]

How?

[last nova]

sum after splitting x-1 and the orignal sum of squares

[sacred junco]

I mean you need to show that x^2-1 is sum of two squares

[last nova]

oh

hang

alright u lost me again

i feel so dumb right now

[sacred junco]

..?

Why

[last nova]

i thought you can only factorise x^2-1

but not transform it entirely

[sacred junco]

Hmm

So are you getting it?

[last nova]

OH

WAIT

that

x^2 - 1

can we prove its odd

and then apply fermat

ah shit no we cant

[sacred junco]

Brahmagupta-Fibonacci is necessity

U can do it in another way tbh

(3^69)^2-1=a^2+b^2

[last nova]

3^(2^69)

second power

[sacred junco]

Latex it please.

[sacred junco]

U know that( a)^b^c=(a)^c^b

[last nova]

wait

[sacred junco]

Because thats what im doing.

[last nova]

really?

ohmy gawd

is this what lies beyond high school math

i feel so dumb here

[sacred junco]

Sarcasm?

[last nova]

not at all

maybe im just not taking the good math

[sacred junco]

$$
(3^(2^(69)))-1
$$

[sacred junco]

😐

Well what i was saying was.

Factorise the eqn.

And just show that such integers exist.

Yup I was suggesting that

And factorization part is the first hurdle for him I guess

Then its sorted i guess.

Ahh..then explore and enjoy my friend.

Best of luck

[last nova]

how would u get 69 squared after swapping them around

3^69 whole squared is 3^(69+69) no?

[sacred junco]

Wdym?

[last nova]

is his point to make it 3^(69^2) instead?

[sacred junco]

No

[last nova]

ah

then?

[sacred junco]

Factorize.

[last nova]

factorize what?

[sacred junco]

Factorize 3^(2^69)-1

[last nova]

thats what u were saying with 3^2^68 right?

[sacred junco]

Do you mean x?

Then ye

[last nova]

yes so we reduce it to a square

so what was atmosphere trying to say?

[sacred junco]

Just keep factorizing.

[last nova]

so now u mean factorise 3^(2^68) - 1 right?

[sacred junco]

Yep

[last nova]

and make it into a sum of squares right?

[sacred junco]

I mean you can keep factorise

Right?

[last nova]

yes

keep reducing them down

but now we have a bunch of sums of squares

all multiplying each other

[sacred junco]

Yup

And then you can involve

https://en.m.wikipedia.org/wiki/Brahmagupta–Fibonacci_identity

[last nova]

and we have to do it 30 times?

or just stop at the first two sum of squares created?

[sacred junco]

Wdym by 30 times

You just need one pair of (a, b) right

[last nova]

hang on

So we get here?

[sacred junco]

Yup

That's correct

You can keep going down

[last nova]

so now we apply brahm?

[sacred junco]

Well, now or after whole factorisation

[last nova]

so that's what i mean by 30 times

we keep breaking down the 3rd term of minus one by factorization?

[sacred junco]

Yup

But that's monotonous

Right?

[last nova]

then we apply brahm to 30 pairs of square sums?

[sacred junco]

All the process is nearly same

Yup

[last nova]

ok!

lemme try

[last nova]

sorry back

we have a bunch of sums of squares multiplying each other now

[sacred junco]

Id suggest googling a similar type of question.tho

U can find even better methods.

[stuck lintel]

Prove that $$\forall m,n \in \mathbb{Z}, S(m,n) = \frac{1}{m} + \frac{1}{m+1} + \frac{1}{m+2} + \frac{1}{m+3} + \dots \frac{1}{m+n} \notin \mathbb{Z}$$

[stuck lintel]

huh mudkip left

[sturdy dirge]

No.

[vast vessel]

Yes @wild zinc

[stuck lintel]

@sturdy dirge can you help :p

there's a solution for this in a book but it's too advanced for me to understand

and mudkips gone so welp

[sturdy dirge]

@wild zinc ?

[stuck lintel]

wtf

???????

@wild zinc

[vast vessel]

what're u guys doing

[stuck lintel]

he wasnt here like 10 seconds ago

??

[vast vessel]

ur obviously just going crazy

[stuck lintel]

probably

want to help though

[vast vessel]

well

I recommend going to the local insane asylum for help

[noble jay]

@vast vessel wow hold on there m8

[vast vessel]

wut @noble jay

[noble jay]

lemme think about it first

[vast vessel]

k

[stuck lintel]

oh hey it's dusty :^)

hopefully after im done with this book i can finally solve your problems

[sturdy dirge]

Interesting https://projecteuclid.org/euclid.ndjfl/1224257540 .

[stuck lintel]

huh that's one long proof

but thanks i guess

[sturdy dirge]

You are welcome.

[spiral heath]

What is a "morphic number"? And what is an "automatic number"? I just saw those two terms in a paper and Google doesn't really tell me much.

[sturdy dirge]

Where is the paper ?

[north orbit]

How many integers $$ N$$ less than 1000 can be written as the sum of $$ j$$ consecutive positive odd integers from exactly 5 values of $$ j\ge 1$$?

[north orbit]

I'm having trouble understanding the solutions to this question.

I've gotten to $$N = j(2k+j)$$

[north orbit]

So if N is odd, all js must be odd

Not sure what to do from there

[spiral heath]

oh sorry I forgot I asked that question

oh well

[open rivet]

Is there a way to quickly tell what the order of a congruence class is other than plugging in powers one at a time and seeing where it gets you?

like for this, naturally, 10 would work because 11 is prime, and 11-1 is 10

but the true answer is 5, which divides 10

if the subscript (11 in this case) is not prime, is there a fast way about this

like this one

obviously its 2, but i only knew that by checking it

[swift shard]

No formula for doing these, I'm afraid

[open rivet]

alright, thanks

[stuck lintel]

How would you find the last three digits of a_n in the sequence a_1 = 2, a_2 = a_1 + 2^2, a_3 = a_2 + 2^3, etc

[flint canyon]

for all positive numbers a,b. If a|b and b|a then a = b

I have that b = an, a = bm,, a = b(k)

can I do that?

nvm

I see that an = bm where (b)=(a)

[cerulean pulsar]

Does i^3 = 1 ??

[sullen rune]

-i

[dry acorn]

@stuck lintel a_n=[a_(n-1)+2]^2

[stuck lintel]

?

[dry acorn]

what?

[stuck lintel]

what

what’s that

[dry acorn]

you had a problem above

[stuck lintel]

proving that the sum of reciprocals of consecutive integers can’t be an integer?

[dry acorn]

no, "How would you find the last three digits of a_n in the sequence a_1 = 2, a_2 = a_1 + 2^2, a_3 = a_2 + 2^3, etc"

[stuck lintel]

oh right

so how would you do it for a_1000 or smth

[dry acorn]

hold on, i think i typed the formula wrong

theres suppoesed to be underscores

alright, there it is

where n is the given integer

[dry acorn]

@stuck lintel hey sorry to ping you, but I typed the formula wrong, this is definitely it.

[steady matrix]

Does anyone recognize this notation? I'm curious what it means in "clear words"

[silver solar]

Do you know interval notation?

[steady matrix]

Yes, it's including 0 to but not including a

right?

[silver solar]

It's a map from a Cartesian product of intervals into another interval

Yeah

And [0, infinity) is just any non-negative number

[steady matrix]

Allright, thanks 😃 Is there a way to visualize the product of intervals?

I know what an interval is, but the product of several is new to me

[flint canyon]

I am trying to figure out the format of the following:
For all integers a,b,c,d,n (n > 0), if ab(modn) and cd(modn) then (a+c)(b+d)(modn)

Is this saying ab(modn) congr cd(modn) then (a+c)(b+d)(modn) ?

[silver solar]

@steady matrix Rectangles, my friend

And in this case, a rectangle with two opposite sides of length a, and two opposite sides of length infinity

You could also call it a strip

so... an infinite strip of width a

[steady matrix]

Wow, that's so much easier than what I envisioned, haha

Thanks a lot man 😃

[flint canyon]

<@&286206848099549185>

[sacred junco]

Wait 15 mins before that ping my friend.

[flint canyon]

It was 15 mins

54-38 = 16

Unless Im delusional

[sacred junco]

Quick maths

Lol

[flint canyon]

Whats the rule for the 2nd ping (: /s

[sacred junco]

U cant.

Read the rules oi mate

[elder bobcat]

p0ng

ew

[flint canyon]

hi

Need help with induction on the following: 3^n - 1 is even

[sullen rune]

3^(n+1) - 1 = 2*3^n + 3^n - 1

[flint canyon]

Kind of needed the steps, but thanks lol, pretty quick

Like do I do
3^(k)-1 )+ 3^(k+1 )-1?

then try to proof

how'd you get to 2*3^n + 3^n-1 , specifcally the 3^n-1 part

[flint canyon]

I guess all I get is an answer and no explanation 😦

[noble jay]

@flint canyon its also known that the power of an odd number (positive integer power) is also an odd number, so odd number -1 = even number

better than induction

[sacred junco]

@flint canyon still need the steps?coz i can help

[wide shuttle]

So... I recently found a way to simplify the sums of integer powers

[flint canyon]

@sacred junco I got it for now, I might ask another induction problem I’m having trouble with later though, thanks!

[wide shuttle]

V(x,y) = \sum{n=1}{x}{n^y}

im still new to latex

V(x,y) = x^(y+1) / (y+1) - \sum{n=2}{y+1}{V(x,y+1-n) * ( y+1)Cn }

its derived from the idea of adding the differences of consecutive numbers to the y'th power.

ex: (1^2 - 0^2) + (2^2 - 1^2) + (3^2 - 2^2) + (4^2 - 3^2)= (1) + (3) + (5) + (7) =4^2,

sum(2n-1) = 2sum(n) - 1sum(1) = xx

so... moving terms around a bit

sum(n) = (xx - sum(1)) / 2 = (xx+x)/2

sum(n^y - (n-1)^1) from n = 1 to n = x.

[vast vessel]

use MathBot

[elder bobcat]

latex ez

[wide shuttle]

I love the support, thanks guys!

[elder bobcat]

lol

[wide shuttle]

❤

[crimson ibex]

$$\sum_{n=0}^{10} n^{2} $$

[crimson ibex]

@wide shuttle Discord supports Latex just add $$ at the beginning and end of the math

[silent lantern]

Does the last nonzero digit of a factorial eventually become periodic?

If so, find an explicit formula

[sacred junco]

7th without bertrand's postulate.

@solar gale

[silent lantern]

thats separate to a question right?

[sacred junco]

What?

Also the end says x union N

[solar gale]

u mean intersection

[sacred junco]

Oh lol yes.

[solar gale]

was there any theory before the questions?

[sacred junco]

Basically only the floor functio

Fact thag n consecutive no.s are divisible by n!

Tau function

And most things related to gcd and lcm

Thats all

Tho im 100% sure this involves use of the floor function

Just cant figure how to use it in this case.

[silent lantern]

Find a way to bound it

[sacred junco]

Wdym?

In what resepct?

To bound

@solar gale u solving it rn mate?

[solar gale]

er

[sacred junco]

Yes?

[solar gale]

idk ask a8

[sacred junco]

@silent lantern help lol

[silent lantern]

I’m doing french

[sacred junco]

Ok ill go die then

Bonjour

[silent lantern]

👍

[swift shard]

1 ∈ X, so X ∩ N contains 1 at least

I'm fairly sure that's the only thing it contains

[sacred junco]

Id like a more rigorous formal proof please.

[silent lantern]

@sacred junco but the answer is empty set

[sacred junco]

Hahahaha

[silent lantern]

Except for 1

[sacred junco]

Answer is {1} btw.

[swift shard]

Of course, but one has to play with the idea first before getting anywhere

[sacred junco]

Ah ok.

Yeah ur right for that part.

[swift shard]

Basically, X doesn't contain any other integers. Why?

Idunno atm but we can find it I'm sure

[sacred junco]

Uh because 1/x+1/y cant ever be an intger.

For x,y>1

No no sorry

[silent lantern]

Use an inductive proof

[sacred junco]

x,y>2

[silent lantern]

Show that when n is a prime it doesn’t work