elementary-number-theory

[worldly tree]

But not know why this helps

[winter bear]

anyhow each factor is also p_1^(k-1) (p_1-1)

right

when does this divide 16

[worldly tree]

Can be lots

[winter bear]

sure, which ones

[copper holly]

From

#❓how-to-get-help

Rule 4

If your question has not been answered for a minimum of 15 minutes, you may use the Helpers tag once. Please do not try to bump your question using this ping unnecessarily. Do not abuse this ping. Do not individually ping users with the Helpers tag without their express permission.

[winter bear]

start of with just which primes

[copper holly]

From

#❓how-to-get-help

Rule 2

Before asking a question, make sure that the channel you are using is not currently in use. An occupied channel would generally have an ongoing discussion or a trailing unanswered question. Likewise, when you are done using a channel, make this clear so that it's open for others to use.

[winter bear]

what

[sleek cove]

lol

[winter bear]

lmao bot broken??

[worldly tree]

(8,2)(16,1), (4,4)

(2^3,2)(2^4,1)(2^2, 2^2)

[winter bear]

thats only some of them, so as i said first focus on just what primes p_1 can be

[worldly tree]

p_1 can be 2

[winter bear]

right what else, because (p_1-1) can also divide

[worldly tree]

Hmm

3

3-1 = 2

5

5-1 = 4

[sleek cove]

any more or is that it

[worldly tree]

17

17 -1 = 16

[winter bear]

mhm those are the possible primes

[sterile pumice]

JohnDoeSmith:

[winter bear]

to satisfy that right

what must the powers be

[worldly tree]

For 3,5,17 needs be 1

[winter bear]

mhm, well either 1 or 0 right

[worldly tree]

Oh yes, zero then a can be 4

No

When 3,5,17 is zero

[winter bear]

basically figure out all the possible combos

[worldly tree]

(0,1,0,0), (0,0,1,0),(0,0,0,1)

This for the odd prime

[winter bear]

well are they alone enough to have phi=16?

[worldly tree]

No

17^d - 1 = 16 so d = 1

For 5 , c = 1

I can only have 4 from 2

2^a - 2^(a-1) = 4

2^a - 2^(a-1) = 2^2

a + (a-1) = 2

2a-1 = 2

a = 3/2 ?

When c = 1 a= 3/2 maybe? @winter bear

[winter bear]

umm a=3/2 doesnt make sense

again theres quite a bit of combinations, just figure out which one gives exactly 16

[worldly tree]

Can you maybe say what equation I am making equal? This i don't get

2^a * 3^b * 5^c * 17^d = 16 ?

This will not make sense

[winter bear]

apply phi to that

so basically you want to know when $\phi(2^a)\phi(3^b)\phi(5^c)\phi(17^d)= 2^4$

[sterile pumice]

JohnDoeSmith:

[winter bear]

and you know each of them yields a different power of 2

depending on their exponent

figure out how to combine these powers to get precisely 4 powers of 2

[worldly tree]

There is a algorithm way?

I can miss some

I not sure when I have all of them

[sleek cove]

start with p=2

[winter bear]

i mean tbh its kinda exhaustion

[worldly tree]

Okay I understanding now

One is 34

17 * 2

phi(17*2) = (17-1)(2-1) = 16

One is 2^5

31 - 16 = 16

[copper holly]

From

#❓how-to-get-help

Rule 4

If your question has not been answered for a minimum of 15 minutes, you may use the Helpers tag once. Please do not try to bump your question using this ping unnecessarily. Do not abuse this ping. Do not individually ping users with the Helpers tag without their express permission.

[worldly tree]

I can do this, thanking you @winter bear

[winter bear]

np

[worldly tree]

How many are there?

[winter bear]

oh umm i didnt count it myself lol

[worldly tree]

Is there algorithm way to know how many?

[sleek cove]

probably like atleast 5 idk

[winter bear]

i mean its just this but yeah its a bit tedious

[worldly tree]

So I cannot know when is time to stop

[winter bear]

i mean do something exhaustive

like start by listing out all the things for a certain value of a

for example

[sleek cove]

theres 6 for 16

[worldly tree]

maximum combination = 432*1

[sleek cove]

so you need 4 more

[worldly tree]

theres 6 for 16
@sleek cove How to know this?

[winter bear]

yeah i also get 6

[sleek cove]

uh i did them out idk if theres s way

you could cheat and look at a table if you want

[worldly tree]

How to know which combination to skip?

[winter bear]

i mean you kinda do it in your head, like lets use powers of 2

for 2^5, you cannot multiply with anything else right or else it will go over

for 2^4, you can only have one extra power of 2, so only 3

[sleek cove]

well if your phi(p^k) is greater than 16, you know it cant be right

[winter bear]

etc etc

[worldly tree]

Go over what?

[sleek cove]

like you have an upper bound on all of the exponents

[winter bear]

well phi would go over 2^4

[worldly tree]

Yes, what is upper bound?

[sleek cove]

like the bound on 17 is lower than the bound on 2's exponent

[winter bear]

i mean if you want 5 is the upper bound on power of 2 and 1 on the rest of them

[worldly tree]

You meaning that if I do 2^6 , then phi(2^6) = 2^6 - 2^5 ?

[winter bear]

and then you can go plug it into a python algorithim or smth to be fully exhaustive

[worldly tree]

But I am minusing

I can maybe go back down?

[winter bear]

you couldnt, write it as p^(k-1) (p-1)

to see it clearly

[worldly tree]

2^5(2-1)

[sleek cove]

yes, phi of 2^6 is that, but your phi n must equal 16, so the left side of the equation must not be greater than 16

aka p^(k-1) (p-1) must not be greater than 16

[worldly tree]

Oh

[sleek cove]

and specifically, it must divide it

[worldly tree]

So a < 6

[sleek cove]

yes

[worldly tree]

but why 3^b must be 1 ?

[sleek cove]

well try 2

[worldly tree]

Because it need 3-1 | 16 ?

3^2 - 3 = 9-3 = 6

[sleek cove]

and does 6 | 16

[worldly tree]

Okay now understanding

[sleek cove]

what if its 3^3 or higher?

[worldly tree]

It is going over 2^4

[sleek cove]

yup

same with 5, 17

so their powers must be 1

or 0

[worldly tree]

Okay now understanded

Thanking John and Sideurk

[hollow venture]

not sure if this is #advanced-number-theory or #elementary-number-theory but I'll try here:

Looking to make some progress on this question, focusing on part a. (I think I have an answer for part b, and part c is apparently extremely difficult)

[simple hearth]

Hint: rewrite the claim a^2 = 1 mod p without the mod notation.

[hollow venture]

Sorry, I was doing another question. @simple hearth do you mind helping me through your logic with mod? It's my weakest area of this module, I just don't understand it very well

[simple hearth]

I suggest looking up the definition of a = b mod n.

[hollow venture]

ah I should say like, I understand working with it when we're talking about actual known numbers but my confidence gets a bit shaky when we're dealing with primes

oh hang on

ok I think I see what you mean

you can rewrite a = b mod n as a-b=kn i.e. a - the remainder when divided by n = some multiple of n

so here a^2-1=kp

[simple hearth]

yes

and a^2-1 should make something leap out to you

[hollow venture]

difference of 2 squares

[simple hearth]

perfect

[hollow venture]

(a+1)(a-1)=kp
a+1=kp therefore in "A-B=KN", A=a, -B=-1, N=p so a= -1 mod p as given

[simple hearth]

Yup. I'd have phrased it as: a^2 = 1 mod p means p divides a^2-1=(a+1)(a-1)

but the definition of prime is that if p divides a product, it divides one of the terms

so p divides a+1 or p divides a-1

but we are given that p does not divide a-1, so it divides a+1.

(I'm not disagreeing with your argument, I'm just showing you can avoid introducing k if you want)

[hollow venture]

yeah I see your logic, and my argument doesn't explicitly exclude the use of the (a-1) bracket which could be useful

[simple hearth]

yeah you jumped over that part a bit

[hollow venture]

for the next part, a composite number is just not prime right?

so 4, 6, 8, 9, 10, ...

[simple hearth]

a composite number is neither prime nor 1

correct

but one has to be slightly careful that 1 isn't composite or prime

[hollow venture]

I don't think I've ever looked but is there a special name for 1's status as neither?

or is it just that it isn't composite or prime

[simple hearth]

1 is called a unit

but usually you don't get into that unless you want to talk about whether -5 is prime, for instance

[hollow venture]

well, all negative numbers n have -1 and -n as factors so I don't think they're prime since that's clearly more than just 1 and n

[simple hearth]

ahh, but then isn't 5 not prime because 5 = -1 * -5 ?

[hollow venture]

👀

hhm

then I think exclusion of negative numbers would have to come from defining "prime" to consider just natural numbers

[simple hearth]

so, what's going on is that -1 is also a unit

so in order to say a number is composite it has to be a product of two non-units

and then you get a coherent answer

[hollow venture]

I suppose you could say -1 * -5 = -1 * -1 * 5 = 1 * 5

[simple hearth]

This is all completely gratuitous if you're only talking about Z, and indeed Conway even argued that in that context we should just declare that -1 is a prime, but if you want to generalize the idea past just the integers, you don't have a good analogue of the natural numbers

[hollow venture]

ok fine, primes are cool 😛

I'm finding number theory's quite a breath of fresh air vs all the probability and stats work I've been doing

pure maths best maths heh

in the next part of the question, I think I've seen something similar before where you use the fact that n has a prime factorisation to show it's composite

[hollow venture]

I've got uhhhhh... Something? But I'm pretty sure it's meaningless and doesn't tie a and n together @simple hearth

All that does is show that those numbers have remainders when you divide them by a prime

[simple hearth]

if the number was prime, would it be possible for this number to square to 1?

[hollow venture]

if e.g. a=2, 2^2=4, 3 is an odd prime so 4 = 1 mod 3

so yes

[simple hearth]

Look at what you proved in part a

[hollow venture]

right

I mean 3 does divide 2^2-1=(2+1)(2-1)=3

and 2 = 2 mod 3 not 1

oh

nah I think that's fine because -1 = 2 mod 3 so ye the proof would work

what're you getting at?

[simple hearth]

What did you provei n part a?

[hollow venture]

that if a^2 = 1 mod p then a = -1 mod p

[simple hearth]

so... here a^2=1

is a=-1?

[hollow venture]

if we're still talking about the a=2 example then no a=2, but 2=-1 so a=-1 with an extra step?

[simple hearth]

I mean the example in part b

[hollow venture]

well 2949^2=8696601=1 mod 67, so by the proof 2949 = -1 mod 67

unless I shouldn't be picking a p but that sounds weird

[simple hearth]

you should be looking mod the n you are given.

you're very close

[hollow venture]

oh

right

well that makes sense lol

buut 3953 isn't prime and it necessitates p being prime O.o

its factors are 59 and 67 which is why the pairing with 2949 is spicy but

[simple hearth]

Write down the contrapositive of part a

[hollow venture]

so if a =/= -1 mod p then a^2 =/= 1 mod p

[simple hearth]

and you can read that as "modulo a prime"

[hollow venture]

the first statement?

[sleek cove]

lmao

[hollow venture]

moving on from that question then as I really don't see how to finish it

Here, I know I need to use Hensel's lemma and I've tried that but I don't think I've actually used it correctly

This is what I've tried, but all that really is is showing that I can use the lemma then not really doing anything with that fact

[hollow venture]

err I think I've got the right answer but I don't think I've used the method correctly, I've just kinda gone straight to the answer

[hollow venture]

nvm, solved it

[hollow venture]

This is another question I'm having trouble with... I think i know the required definitions, just can't figure how to use them

[last grove]

can someone help explain why this is true? C = euler's constant here

and what does C(s) even mean

C * s?

[fervent crest]

It's probably a function of s

It's probably defined before that equation

[last grove]

that's weird

here's how they defined it

wait asec

maybe it doesn't mean anything at all

just another constant

[fervent crest]

Huh weird

What book is this

[last grove]

because they then show that C(s) = zeta(s)

Apostol's analytic number theory

[fervent crest]

Ah

[supple furnace]

This is standard notation

Euler Maclaurin I’m guessing?

It’s a function of s that does not depend on your x value by the way

That is why it is defined as such

[last grove]

I see...

also, what is its relationship to the Euler-Maclaurin formula

[supple furnace]

I’ve seen this C function come up as a result of a “weak” version of Euler Maclaurin (first order approximation of it).

[last grove]

thanks again

also... one last thing

how come it still equals zeta(s) even after they subtract x^(1-s)/(1-s)

OH

NEVERMIND

when 0 < s < 1, both zeta(s) and the limit diverge as x->infinity

sooo they are essentially the same when x->infinity for s:(0,1)

[sleek cove]

why does he use n^2/2 -t_n, instead of t_n - n^2/2

wait nvm my bad

because we want sigma - n

[sleek cove]

how is this analogous, its literally the same thing but infnty replaced with a

[simple hearth]

I think this is their way of admitting that both O notation and << notation are slightly uncomfortable, but at least you can always choose which you like better 😛

[sleek cove]

hmm I guess

and yeah they are like pretty weird ngl

like the whole concept of scaling functions and "Absorbing" smaller errors, or increasing the error to accomodate other functions, makes sense intuitively, but it just seems kind of arbitrary/odd

also this interlude is bleh, relearning calc with big O notation-hard pass lmao

[steady nest]

how would you go about doing this?

i did find an answer but i feel its perhaps too simple?

the question did say use the euclidean algorithm

[torn escarp]

the answer is simple

11! is just 11* 10! so it just simplifies very quickly

[steady nest]

?

@torn escarp srry i don't understand

$ 11! + 8 = (11) (10! + 2) -14 $

[sterile pumice]

suck2015:

[steady nest]

would you just do

$ 10!+2 = 14 $ ?

[sterile pumice]

suck2015:

[steady nest]

and then from this point, given that all n! , n being > 1 are all even ? so 2 must be the gcd?

because 10! is divisible by 7 and thus 10! +2 cannot be divisible by 7

[fervent crest]

Yes I think 2 is the answer

But what do you mean by 10!+2=14?

Remember what euclidean algorithm says:
$$\text{If }a_0=ba_1+a_2,\text{ then }\gcd(a_0,a_1)=\gcd(a_1,a_2).$$

[sterile pumice]

Whoever:

[steady nest]

@fervent crest oh i forgot to reply to this

yeah by the way of the euclidean algorithm, 10!+2 = 14(k) + remainder right?

but then at this point, we can just see that the only divisor possible is 2 because 7 won't work

[fervent crest]

Yeah

[last grove]

not a question but I made a meme
(source: Apostol's Analytic Number Theory)

[sacred junco]

i like it when women overpower me

[light flicker]

This would be better with the press B to be dominated meme

pretty good tho

[alpine jasper]

could've just put it as an exercise

[fervent crest]

the first thing that came into my mind is dominated convergence theorem

[sleek cove]

negative exponents are ugly

[stoic bear]

You're ugly

Jk

[sleek cove]

thats not what your dad said last night

[stoic bear]

I love you sideurk

<3

Wait what

[sleek cove]

[inner crest]

[granite wagon]

We have m belonging to N and α belonging to Z, then there exist unique numbers q, r belonging to Z such that α = mq + r with r being 0<= r < m. We know that α is congruent with r for mod m. We say that r is the least remainder of α for m and α belongs to the class of r for m. Thusly there exist m classes of mod m, the r (mod m), r = 0, 1,...., m - 1.

Does that mean that for unique α, r and q r is the least remainder of α congruent to r (mod m) iff α belongs to the class of r (mod m)? Is that a class or a specific case since these three symbols are unique. I am not proposing that they can't be arbitrary but for a specific case they aren't. So if for this α we have a class of r (mod m) does that mean that m is arbitrary and therefore that is the reason that there exist a class of α congruent with r (mod)?

[craggy solstice]

Basically an equivalence relation on a set partitions it into classes, modulo is an equivalence relation. Here our set is Z and we can divide it into classes d' such that 0<=d<m such that each class is a set of elements of the form mk+d {see that mk+d (mod) m is just d}
So every element which produces a remainder k modulo m is in the class k',like this we can partition Z into a set of m classes {0', 1' , 2',...(m-1)'} and all of these classes are non empty

f'=n' means that f=n mod m

i.e. they just make the same the equivalence class

[granite wagon]

I see

I don't know why this is not explicitly stated in my pdf..

That equivalence relations don't partition a set into classes.

Does that have something in common with dedekind cuts?

How to create the set of irrationals out of the rational numbers by creating small sets in between an irrational number or a unique rational through the process of creating two sets (well ordered with a least element) that are adjacent to said number and thus you partition the rationals through that process. I don't know if that makes sense at all.

[neon comet]

is there a trick im missing?

or do i have to do the 2^47 calculation?

prime powers -_-

guess i could break it up into 2 factors

lemme try that

[sleek cove]

how did you go from 27^47 = 2^47 mod 5

@neon comet

[neon comet]

27/5=5*5+2

[sleek cove]

ok

so you now want to reduce 2^47

[neon comet]

ye

but 47 is prime

so i have to make it 2 factors?

[sleek cove]

that would probably help

do you know any power of 2, st 2^k= 1 mod 5

[neon comet]

2^4

no

[sleek cove]

no?

[neon comet]

2^4=5*3+1

[sleek cove]

yes, so =1 mod 5

[neon comet]

ye

[sleek cove]

so now you can reduce 2^47

using properties of exponents

[neon comet]

2^4*2^43

[sleek cove]

but you are still left with the same situation

[neon comet]

ye its prime again

so i proceed?

repeat*

[sleek cove]

recognize that (2^4)^k = (1)^k = 1 mod 5

i mean you could, but thats inefficient having to repeatedly subtract 4 until you get a small power

[neon comet]

thanks for the help i think i can solve this now

[sleek cove]

why does O(x^2) get absorbed here

[light flicker]

because you're looking at x small

[sleek cove]

oh wait yeah

evaluating close to 0 my b

[sleek cove]

jfc this is annoying

why cant they just use a different letter/symbol for non infinite values

[carmine burrow]

hey, can someone explain why the gcd(a, b) = gcd(b, r) with euclidean algorithm, I get the rest of how it works, I'm just a bit confused with the proof for this statement (r is the remainder and a and b are both positive integerrs)

[swift shard]

Because
gcd(a, b) = gcd(a, b + ma)
For any integer m. This is exploitable, choose an m that makes the second argument as small as possible. This ends up being r. @carmine burrow

[sleek cove]

1/(O((z-a)^2) becomes O(1), because we are letting a approach z?

[soft kestrel]

Because the next term in Laurent series is a constant no ?

[sleek cove]

um, how do you know that?

[sacred junco]

https://i.imgur.com/bu7yvip.png I'm confused on how to read this; are these operations the same (both plus, or both minus) or what? Another problem I'm working on it doesn't seem to agree with it but I might be doing something wrong

the proof was too hard for me as well so that didn't clear anything up

it seems like one should be plus and one should be minus or something

[void widget]

it may be equal not sure

what about trying an example

[sacred junco]

it means the answers you get are both positive and negative

so for example, x^2 - Dy^2 = 4 and -4

[sacred junco]

umm not quite what I'm asking

I mean
are the solutions to x^2-Dy^2=-4 given by
fraction thing = - exponent thing
and the solutions to x^2-Dy^2=+4 given by
fraction thing = + exponent thing

[light flicker]

For the second plus, minus, it shouldn't matter what you take

Since, changing the sign of x,y doesn't matter

[sacred junco]

is that suggesting the solutions to x^2-Dy^2=4 and to x^2-Dy^2=-4 are the same?

[light flicker]

uh no?

[sacred junco]

then what doesn't matter?

[light flicker]

the second plus or minus doesn't matter

because if (x,y) is a solution to your equation, then so is (-x, -y)

[sacred junco]

why is it there then?

[light flicker]

Because you're looking for all solutions?

[sacred junco]

oh I think I see what you mean now

so you would build the set of solutions the same way for either equation, +4, or -4, with both p/m for the second p/m, but the -4 or +4 determines the fundamental solution

I think I should rewrite that

[light flicker]

Right

That's the right idea, stated better than I put it

[sacred junco]

the set of solutions to x^2-Dy^2=4 and the set of solutions to x^2-Dy^2=-4 are constructed the exact same way, but have different fundamental solutions
and the construction uses both fraction thing = + exponent thing and fraction thing = - exponent thing

[light flicker]

Yep

[sacred junco]

ok thank you

[light flicker]

As a side note, this is strongly connected to something called the fundamental unit theorem in algebraic number theory

and they probably don't give a proof, but that theorem is at least one way you'd prove it

[sacred junco]

https://i.imgur.com/8eGT1eX.png this is the proof they give

[light flicker]

Oh yeah

so this does use the fundamental unit theorem

[sleek cove]

so for the RHS i treat it as a product of two functions u = zpi and v = inf product of w(z)

so for v' / v, this turns into the inf sum of w'(z) / w(z)?

[sleek cove]

nvm

[sacred junco]

number theory is officially sexy

if anyone already has a desmos/programmed plot of this lmk but I will try to do it myself using that or python matplotlib and the like

[slow yew]

We are learning about Euclid's division lemma.
And we have to prove a lot of stuff.
I am trying to make a collection of tips to help.
Show that the square of an odd integer is of the form 4q+1 for some integer q.
Show that the square of any positive integer is of the form 3m or 3m+1.
Show that the cube of any positive integer is of the form 9m, 9m+1, 9m+8.
Compare with a=bq+r.

1. For questions which ask to prove that the square or cube of something is of some form, follow the following tips and examples:
   a. The most important thing is look for the least base form that can prove the asked forms.
   b. Look at b of the form(s) to prove. It indicates b of base form. If b of form to prove is high such as 9, the base b is 3.
   c. The number of forms asked to prove, generally, indicates what our base form should be.
   d. There can be various cases:
   i. Asked to prove multiple forms: The constant (remainder) of the various forms would be different. The highest constant indicates the highest remainder of the base form.
   Un-square/Un-cube the constant if required. You will get the highest remainder of the base form. Add one to the highest remainder of the base form, you will get b.

ii. Asked to prove single form: Un-square/Un-cube the constant. You will get the highest remainder of the base form. Add one to the highest remainder of the base form, you will get b.
^ This is what I have so far, any improvement would be appreciated.

[sleek cove]

I think you're overcomplicating it

all you need to do for these problems is just consider the n cases when working mod n

like for n^2 = 0 or 1 mod 3, just do the three cases for n

also, why does it take 6 pages to prove that C is e^-γ smh

[sacred junco]

recs on best analytic number theory intro book?

[light flicker]

apostol

[simple hearth]

I'm a fan of Koninck and Luca if you want an easier intro and Overholt if you want a slugfest

I bounce back and forth between the two

[sleek cove]

i prefer my n=1 sample size and reccomend stopple

[sacred junco]

thanks yall

[slow yew]

@sleek cove did you reply to me?

[sleek cove]

yes

[slow yew]

I am in year 10, lol.

[sleek cove]

um ok?

[slow yew]

Meaning, I don't understand that.

[sleek cove]

all you need to do for these problems is just consider the n cases when working mod n

what dont you understand ab this

mod?0

[slow yew]

Yeah, mod, and the formula

Haven't been taught what mod is.

[sleek cove]

oh the picture is completely unrelated lol

[slow yew]

ok lol

[sleek cove]

uh so basically

lets stick w the n^2 being of the form 3k+1 or 2 examplep

so we know that every single integer is of the form 3k + 0,1, or 2

right?

[slow yew]

Yep

[sleek cove]

ok, so we know that every single integer squared, must be those three possibilities squared as well, right

[slow yew]

Yep

[sleek cove]

ok, so now all we have to do is, take each possibility and square them

3k+0 is trivial, as it squared is 3(3k^2)

thus in the form 3k+0

[slow yew]

Yeah, ik the process of proving of it.

The problem is choosing the correct b

in the base form, a=bq + r

[sleek cove]

erm im confused

why do you care about b

shouldnt all you care about is r?

[slow yew]

Umm, do we not need b so we can complete the base form and then square/cube it?

[sleek cove]

oh

well if the problem asks you to show/prove that something must be of the form nk+r

you set b = n

and only consider r

[slow yew]

What about 9m+1, 9m+8,9m?

Surely, I wouldn't set b=9.

Well, I could do that.

But it's tooo lengthy.

So, I'd set b=3.

And it'd work

See, I know generally when it is not squared or cubed, b=n

But these special cases.

[sleek cove]

well you're not wrong, however setting b=3 is probably a bit more work

you have to then worry about factoring

like, with each case when you set b=9, you probably have like 20 seconds of work

[slow yew]

Yeah, but we have an exam on Monday, and we are supposed to solve 10 of these questions in 20 mins :/

[sleek cove]

well

it would probably be worthwhile to learn some basic modular stuff

you already know the basics

[sacred junco]

10 of these?

you can reduce them all to multiplication and subtraction problems once you understand modular arithmetic lol

[slow yew]

Hmmm

[sleek cove]

we say a = b mod n, iff a - b = nk. and theres simplifications you can do with these as well, for example

what is 125 = ? mod 7

this is equivalent to 125 = 7k + x

[slow yew]

what is the constraint on k?

like positive integer only?

[sacred junco]

might want to clarify the = is an equivalence, not a literal "=" as to not confuse him

[sleek cove]

not necessarily

oh yeah, = isnt actually the normal equivalence relation

[slow yew]

oh

[sleek cove]

it means conguent mod n which means they have the same remainder when divided by n

so like 5 = 12 mod 7

because 5 - 12 = 7 (-1), which is equivalent to 5 = 7(0)+ 5

and that 12 = 7(1)+ 5

hence r is 5 here

so back to 125 = x mod 7

we notice that 125 = 5^3 = 25 x 5

but, we know that 25 = 4 mod 7, since 25 = 7(4) + 4

thus 125 is congruent to 4(5) mod 7

so now instead of solving
125 = x mod 7, we have
20 = x mod 7

and we repeat again getting 20 = 7(2) +6

thus 125=6 mod 7

this was just an example to show that using mod can quickly reduce large numbers into smaller, more manageable ones

so for the 9k problem, it is very fast to check all 8 possibilities

hmm that was a lot, and idk how fast you would be able to learn to use this efficiently

[slow yew]

I didn't get the

which is equivalent to 5 = 7(0)+ 5
part

[sleek cove]

well whats 7 times 0

its just showing that the remainder when 5 is divided by 7 is 5

remember, k can be any integer

[sacred junco]

hmm so many ways to explain modular congruence

[slow yew]

its just showing that the remainder when 5 is divided by 7 is 5
@sleek cove oh.

[sleek cove]

yeah i probably didnt explain it very good lol

[sacred junco]

one common first explanation to counting (not quite arithmetic, but counting) in a modulo is a clock

if it is 5 o'clock

and 12 hours pass, it's 5 o'clock again

adding 12 was negligible

5 and 17 are congruent

[slow yew]

o.0

[sacred junco]

same with any multiple of 12

[slow yew]

I see.

[sacred junco]

5 hours and 17 hours are the "same"

[slow yew]

"same", yes.

[sacred junco]

just as, in modulo 12,
5 and 17 are congruent

they both leave a remainder of 5 when divided by 12

[sleek cove]

congruence is different from traditional equality, but it allows tou to substitute much larger numbers for smaller ones, because they are "congruent" in the world mod n

[slow yew]

Oh, so modulo is all concerned with remainders.

[sleek cove]

yes

[slow yew]

I see.

[sleek cove]

hence why it is a very good tool for these kinds of problems

[slow yew]

it's kinda like why 6/3 = 12/6. They both have quotient 2. And so we can replace 12/6 by 6/3.

[sleek cove]

sort of, like the bigger number that we replace, will be of the form
xk_1 + r, and when we substitute a smaller number, yk_2 + r, the k_2 must be smaller than k_1, and it gets "absorbed"

also, what would you do if asked to show that say, every "something" must be of the form 6k + r_1, 6k + r_2, etc

you wont be able to use the trick that you did by replacing 9 with 3, as it relied on 9 being a square

or what about larger numbers, like 13k + r

you would have to use modular arithmetic, which you already were doing but you just didnt kno, it was in disguise

[slow yew]

oh.

[sleek cove]

i mean, doing them all out staying in the form of a=bq+r is fine 100%, and will work

its just that taking advantage of modular congruence reallllllyyyyyy saves a ton of time

so I do think its probably worth learning/getting used to it by monday, atleast the basics

[slow yew]

Yeah, it would help in MCQs but for the written section, I would have to use a=bq+r. Anyways, I will still learn mod arithmetic.

[sleek cove]

i mean technically modular arithmetic is using a=bq+r

but yeah, it would probably help with your understanding of the problems and like what to consider etc

[slow yew]

embrassed I guess.

[slow yew]

@sleek cove I read up on mod arithmetic.

Could you show me how it applies to the questions I posted above?

[sleek cove]

yes

[slow yew]

I understood your examples with numbers

[sleek cove]

which one

[slow yew]

any

[sleek cove]

k 1 sec

alright lets do the top one

so we are trying to show that
n^2 = 4q+1, for n odd

which is equivalent to n^2 - 1 =4q

so this is a hint that we should probably be working modulo 4, right?

[slow yew]

Yeah, it's like the form difference of two number = modulus * k

[sleek cove]

yup

so, we know that n is odd right

[slow yew]

Yep

[sleek cove]

ok, so can n = 0 mod 4?

[slow yew]

Hmmm

= is modulo congruence?

[sleek cove]

yes sorry

[slow yew]

yeah, you should put in braces

[sleek cove]

we will be working congruent from here on out

[slow yew]

Oh ok.

Well, I think no.

Odd numbers aren't divisible by 4

I guess

[sleek cove]

"i guess"? be confident

[slow yew]

Ok

They ARE not divisible by 4

because if they were, it would imply they are divisble by 2

[sleek cove]

alright good lol, so what are the possibilities for an odd number to be congruent to, mod 4

[slow yew]

we get a remainder

1,3

[sleek cove]

alright good

[slow yew]

I am not purely doing mod arithmetic here, I am kinda using a=bq+r

[sleek cove]

so we have 2 cases to consider n = 1, 3 mod(n)

[slow yew]

yeah

[sleek cove]

and yeah they are equivalent, just easier to work with mod usually

now, all we have to do is square n, right?

[slow yew]

Wait

if the remainder was 2.

[sleek cove]

mmhm

[slow yew]

why wouldn't it work out in mod?

I mean, according to mod, why would it not be an odd integer?

[sleek cove]

well it would imply n = 4k + 2

[slow yew]

I can see in a=bq+r (let's just call this euclid's lemma or lemma)

Yeah

But is there a pure mod reason?

[sleek cove]

which is n = 2(k+1)

[slow yew]

Yeah

[sleek cove]

but n was odd, remember

[slow yew]

Yeah

So are lemma form and mod form equal?

Like, we use one to supplement the other?

[sleek cove]

yeah, by definition, a = b mod(n) iff a = nk +b

so like, i guess we are using the lemma to check for the remainders here, but we havent actually needed to use it yet

[slow yew]

so like, i guess we are using the lemma to check for the remainders here, but we havent actually needed to use it yet
@sleek cove ? those two sentences seem contradicting.

[sleek cove]

like

this is how i would arrive at where we are now in the problem

so we know we are working mod(4), so 0 cant be a possibility because that would mean n is even, and 2 cant work for the same reason

like, in my head, im using the lemma

but you shouldnt have to like write it out

[slow yew]

Yeah

Yep.

[sleek cove]

ok so we know n is 1 or 3 mod(n)

right

[slow yew]

Yeah

[leaden compass]

[sleek cove]

and we want to check n^2 so whats the obvious thing to do

[slow yew]

Umm, square

[sleek cove]

right, so square both possibilities

[slow yew]

Yep.

[sleek cove]

then we get n^2 is either 1 (again) or 9 mod 4

but, what is 9 mod 4?

[slow yew]

1

[sleek cove]

mmhm

and we're done

[slow yew]

wait a second

that's like half page shorter than my proofs

[sleek cove]

you can shorten it if you like, cut out a bunch of words, and like some "trivial" stuff too

[slow yew]

I'll do this proof on a paper, and approach you if i have any problems

[fervent crest]

Just to format everything nicely:
Suppose $n$ is an odd integer, then $n$ is equivalent to either $0,1,2,3$ modulo $4$. If $n$ is equivalent to $0$ or $2$ then $n$ is in the form of $4k$ or $4k+2$, both of which are even. Therefore $n$ is equivalent to $1$ or $3$. But $1^1\equiv1\equiv3^2\pmod{4}$, so in any case $n^2\equiv1\pmod{4}$.

[sleek cove]

i mean most people here should know how to solve these problems, but sure

[slow yew]

🙂

[sterile pumice]

Whoever:

[sleek cove]

4k=odd

whoever crank confirmed

[fervent crest]

Bold of you to assume 4k is even an integer

[leaden compass]

i think you're an integer

replace number with ass snack though

[fervent crest]

I think you are a dense, totally disconnected subset

[leaden compass]

i think you're dense in C

[sleek cove]

btw using modular arith is probably allowed, since it really is just a=bq+r, you should ask your prof/teacher before if you can use it @slow yew I dont see why you wouldnt be allowed

[leaden compass]

i think you're dense in PP

[stoic bear]

Archsys try math pickup lines irl

[leaden compass]

would you like to see my P^2

[winter bear]

smh interrupting help

[sleek cove]

we finished lol

[leaden compass]

john

we are helping

[sleek cove]

shush loser john

[winter bear]

smh

[leaden compass]

yeah nerd

smh cranks these days

[sleek cove]

go back to whatever losers do

[leaden compass]

yeahhh

[winter bear]

ok back to complex analysis

like the loser i am

[leaden compass]

smh only losers do CA

[sleek cove]

know your place

you dont have permission to thonk me

[winter bear]

wtf

[fervent crest]

wtf

[sleek cove]

i didnt mean bruh

sh

[winter bear]

mfw ghost ping

ull get killed

[sleek cove]

sh

sh

[rotund zinc]

how i get ping

[sleek cove]

ask @fervent crest

[fervent crest]

I have a question: How do I prove $$|\ln\text{lcm}(1,2,\dots,n)-n|<\sqrt{n}\ln(n)^2$$

[sterile pumice]

Whoever:

[fervent crest]

For n≥3

[winter bear]

ah a classic

sideurk you should be able to do this

[sleek cove]

is this an actual problem

ok

[winter bear]

since you are doing ANT

[sleek cove]

uh well by me doing ant

[winter bear]

dont actually btw its a meme

its RH equivalent kek

[sleek cove]

its really me taking 2 hours reading a 6 page proof of showing the exact value of C

[winter bear]

2hrs for 6 page

is fast

[sleek cove]

no

well maybe

i mean I could get the general ideas/concepts in like a 5 minute readthru

but like, actually following allong myself for each step and like confirming the steps is what takes so long

[stoic bear]

Why iis ANT useful again

Analytic

[leaden compass]

john smh you did rudin in 3 days

how many pages of rudin did you do per hour

[winter bear]

smh it wasnt 3 days it was like a wekk

[leaden compass]

i think you're a week

weak*

[sleek cove]

i think hes actually a weak loser

[leaden compass]

no you

[sleek cove]

no im a weeb, loser *

😎

[leaden compass]

no you're a weak weeb loser

[sleek cove]

unironically im actually kinda strong

[light flicker]

r u bigger than me

[slow yew]

guys, isn't using modulus congruence a short hand for 'remainder = dividend mod divisor'?

[sleek cove]

probably

[slow yew]

oh.

[sleek cove]

i was talking to zoph

uh yeah sorta

[slow yew]

Ok

[fervent crest]

I think it is inspired by that idea, but in general you can have 5=3 mod 2

But like, 3 is not the remainder of 5 when divided by 2

[sleek cove]

like my body is probably bigger than zoph chans, but pp is another story 😔

[leaden compass]

you mean your pp is not as big as you?

[sleek cove]

its relative

[slow yew]

I think it is inspired by that idea, but in general you can have 5=3 mod 2
@fervent crest I thought it like: 5 mod 2 = 1. You see, I simplified the mod expression on one side

[light flicker]

smh im pretty big

[sleek cove]

ok lets doxx our bodies

[leaden compass]

@sleek cove pp's and relatives should not be in the same sentence

[light flicker]

I'm 170 lbs

[leaden compass]

mfw real alabama hours

[stoic bear]

usually written the other way @slow yew

[winter bear]

im 220

get fucked

[stoic bear]

Go to #chill smh, lets help this person

[sleek cove]

im 187 this morning 6'

[leaden compass]

what bmi is that

[light flicker]

yea but I'm only 5'7" get beat

[leaden compass]

,w 187lbs to kg

[winter bear]

alright lets move to chill

[sleek cove]

oh deadas

[winter bear]

lmfao

[sterile pumice]

Display results online and refine query

Requested by Archsys#2878

[leaden compass]

,calc 84.82/1.83/1.83

[sterile pumice]

Result:

25.327719549703

[leaden compass]

mfw overweight

technically

[winter bear]

move to chill

[leaden compass]

[stoic bear]

You say a = b mod c, if a can be written as a cn + b for some integer n @slow yew

[sleek cove]

yeah, most of it is muscle

[slow yew]

@stoic bear yeah

I thought it like: 5 mod 2 = 1. You see, I simplified the mod expression on one side

= is not equivalence here

it is normal equal

[fervent crest]

mod can be used like that as an operator

[slow yew]

Yep

[fervent crest]

But we usually talk about modulo as in a=b (mod n) means n divides b-a

[stoic bear]

[slow yew]

@fervent crest Yep.

[leaden compass]

whoever i think you are an operator

on R^2

[fervent crest]

I'm a multilinear alternating operator on R^n x R^n x ... x R^n

Don't ask me what that means Idek myself

[slow yew]

wot

[winter bear]

you are the zero operator whoever

[slow yew]

@sleek cove Also, we can't use modulus in our exam :/ We have to use the long process of using the lemma :/

Guys, is there a zero sum property?

[fervent crest]

No

[slow yew]

Like a zero product property?

No? Ok.

[leaden compass]

whoever

you are the determinant

[slow yew]

Can I not do this: ax + b = 0x + 0?

and equate a=0

b=0

[fervent crest]

?

Depends on how you interpret the question

If by ax+b=0 you mean x is some number satisfying this equation, then x=-b/a if a≠0 and x can be anything if a=0 and b=0

But if you mean ax+b as a polynomial is 0 (which means ax+b is 0 for all x), then it is true that a=b=0

[slow yew]

Yeah, it's from a polynomial question.

Let me send the question

If the polynomial x^4 – 6x^3 + 16x^2 -25x +10 is divided by another polynomial x^2 -2x +k, the remainder comes out to be x+a.

I would subtract the remainder

and then divide

and equate remainder = 0

[sleek cove]

nvm thought you meant solution

[slow yew]

@sleek cove , Show that the cube of any positive integer is of the form 9m, 9m+1, 9m+8, we are not permitted calculators, so taking n = [0,8] mod(9) is a bit hard

[sleek cove]

well, the same ideas will work

you just have to like write them out in lemma form if you were to do it

[slow yew]

yeah

[sleek cove]

and go through all 9 8 7 possibilities

0 and 1 are generally trivial

[slow yew]

Yeah.

[sleek cove]

i mean, all youd be doing is showing your work

[slow yew]

That's the problem lol

[sleek cove]

fair lol

but atleast you can check your answers/intuition

[slow yew]

Yeah

I mean use b=3 would work perfectly for this

but that's thing, you don't know whether you can use, it might not work

while b=9 is a safe option

[sleek cove]

if your number isnt a square/cube/etc, you probably wont have any shortcuts ye

[hearty wadi]

How?

[potent dust]

telescoping

maybe it's easier to see as $$\sum_{k=1}^n (k+1)^3 - k^3$$

[sterile pumice]

Tubular Cat:

[potent dust]

write out a couple a terms of that sum and you should be able to see how the middle terms cancel each other

Show that the cube of any positive integer is of the form 9m, 9m+1, 9m+8, we are not permitted calculators, so taking n = [0,8] mod(9) is a bit hard
n={-4, -3, -2, -1, 0, 1, 2, 3, 4}

@slow yew

[potent dust]

<@&268886789983436800> title and date looks like an exam

[celest plinth]

got em

[hearty wadi]

@potent dust thanks. Authors of that book really had to write it like you did, and probably explain it too (as they usually do)

[sleek cove]

[dawn warren]

Yeah

[sleek cove]

where does the 2 in C_2 represent? does it come from like the twin primes for N, N + 2?

like what does C_k represent

[dawn warren]

It presumably represents primes of form like n,n+k

[sleek cove]

ya i see it ty

[atomic sable]

I dont understand what "decimal equivalents" mean in this graph

[sacred junco]

that the number in normal form

@atomic sable

[light flicker]

Base 10

[simple hearth]

The phrase "normal form" scares me, even though its benign here

[atomic sable]

Oh right

Its the numbers that we normally use

i feel so stupid right now lol

[timid olive]

that the number in normal form
you mean hex?

[last grove]

hey, I am working through Apostol's ANT right now and I came across a tricky problem:

I am able to show that it is equal to x*ζ,(2) + O(log x), but not x**/**ζ,(2) + O(log x)

here is my work

does anyone know a trick for solving this?

here are two relevant identities that I used

[torn escarp]

you probably need to look at how those theorems are proved and make something on your own with abel summation

[last grove]

oh my cow... thanks man

this will make life a whole lot easier

[torn escarp]

👍

[last grove]

hey

does anyone know how this is derived?

[simple hearth]

Factor the direchlet series of the multiplicative function mu(n)

using the EUler product

[last grove]

ok thanks man

I don't know what dirichlet series are right now, but I'll look into them later

https://proofwiki.org/wiki/Reciprocal_of_Riemann_Zeta_Function

this page really helped though

[sleek cove]

which book is that

wait nvm lmO

[last grove]

lmao

someone made an entire answer key for the boo

[simple hearth]

which book?

[last grove]

Apostol's Analytic Number Theory

https://greghurst.files.wordpress.com/2014/02/apostol_intro_to_ant.pdf

I'm still going to try all the problems first before looking at the answers though

[simple hearth]

I strangely have not spent much time with Apostol

[last grove]

how did you learn number theory?? 😃

[simple hearth]

I learned a lot of analytic number theory in bits and pieces along the way, but I think Overholt is my preferred text

though Konic and Luca has also helped

[silent cairn]

Is apostol’s analytic number theory low level enough to be considered elementary?

I mean I don’t really care if ppl are talking about it here, but I thought that book was decently advanced

[simple hearth]

elementary in the context of number theory literally means "not using analysis"

it doesn't mean easy haha

but I try not to die on that hill

[silent cairn]

Sure, but by that frame aspostol isn’t elementary right?

I just make that comment because it’s in the #elementary-number-theory channel haha

[simple hearth]

I mean, I think it is at the easier end of analytic number theory books

[bleak musk]

@simple hearth At what point in number theory do you start using analysis? I'm asking you because you seem knowledgable and friendly 😅

[simple hearth]

Number theory doesn't exactly have a standard order. So you could do it quite early, or only after a long time.

[bleak musk]

then let me rephrase

what sorts of problems lend themselves well to analysis techniques, and what background do you need to start learning analytic number theory

[simple hearth]

Generally questions about what happens "on average" lend themselves to analysis

so things like the prime number theorem, that the number of primes less than x is approximately x/log(x)

I have a video intro to analytic number theory if you like

[bleak musk]

omg really? Yes I would love to see that!

[simple hearth]

https://www.youtube.com/watch?v=oVaSA_b938U&t=1280s

I'm obviously biased, but I think it's a lot of fun

[sleek cove]

shameless plug

[simple hearth]

haha shame is overrated 😛

[last grove]

loaosodasdasd

TIS FINALLY OUT

[sleek cove]

this vid has been out

[simple hearth]

part two soon

[silver zealot]

1/a + 1/b = 1/c

solve for natural numbers a,b,c

nvm found out what I wanted

https://en.wikipedia.org/wiki/Optic_equation

[sacred junco]

how can i prove the the final sum of digits of any product of 9 will be 9?

if we add the digits of any product of 9 until the result is a single digit number, it becomes 9.

anyway to prove it?

[bleak musk]

@sacred junco you can write any number $n$ in base 10 as

$n = a + 10b + 100c + ...$

Where $a, b, c, ...$ are integers. You're concerned with the sum $a + b + c + ...$. As a hint, what happens if you let $n$ be divisible by 9, and subtract $a + b + c + ...$ from both sides?

[sterile pumice]

Nicholas:

[humble umbra]

@sacred junco https://en.wikipedia.org/wiki/Digital_root

[opaque dome]

hello i have a bit of a confusion

part b confuses me

[light flicker]

What have you thought about?

[wispy creek]

silverman

[sacred junco]

@bleak musk @humble umbra i fell asleep. thank you for those... i just saw those and will come back for any questions

[opaque dome]

ah sorry internet went out

In the answer key it says that it is true

but there's a counterexample

$15^2 + 8^2 = 17^2$

[sterile pumice]

funnynumberatyahoo.com:

[opaque dome]

since b = 8

this implies that

there exists n such that the equation $\frac{n(n+1)}{2} = 8/4$ is satisfied

[sterile pumice]

funnynumberatyahoo.com:

[opaque dome]

however the n is not of any significance since the answers to that equation aren't part of the natural numbers

or even the integers

so how that's supposed to be true confuses me

[sleek cove]

is b = 8 = 4(T_n)?

[opaque dome]

yes

and then T_n = 8/4 = 2

and since it's a triangular number there must exist sum n s.t. n(n+1)/2 gives an answer, right

[sleek cove]

is 2 a triangular number

[opaque dome]

no i dont think so

[sleek cove]

so how is b=8 a counterexample if it doesnt satisfy the requirements

[opaque dome]

I had wolframalpha solve the equation (n(n+1))/2 = 2

Aren't they saying that every b = 4T_n?

or am i reading this wrong

[sleek cove]

yes lol

that is correct

[opaque dome]

OHHH

wAit

i read it wrong

lmaooo

[sleek cove]

yes

[opaque dome]

it was the other way around

i was so confused

[sleek cove]

uhhuh

[opaque dome]

thankss

im so tired i need to get to bed

[sleek cove]

not for every b in N, but for every n in N for T_n

[opaque dome]

yess

[sleek cove]

yea lol

[opaque dome]

eee ill just go to bed now then

gn

[sacred junco]

hello

[sacred junco]

hey guys, can help with some pointers maybe on how to start this question?

[sacred junco]

gladly

[nimble cedar]

Well, 5^12 = 17 mod 64, and -5^5 = 11 mod 64

[sacred junco]

yup, got that from the table, so -5^5 * x^5 = 5^12 mod 64

still strugglin

[nimble cedar]

So what is (-5^5)^-1 mod 64?

Or perhaps more to the point as I look up some stuff...

What is the multiplicative inverse of 11 mod 64

[sacred junco]

35 if my math is right

so we can multiply both sides by 35?

[nimble cedar]

That would get you to x^5= 17*35 mod 64, which is at least a start

So x^5 = 19 mod 64

I think

It is, because that is also -5^7 mod 64 like I thought I could do before

But then talked myself out of

[sacred junco]

given that, looking at the table -5^7 = 19mod 64

what did you talk yourself out of

[nimble cedar]

I talked myself out of saying 5^12 mod 64 divided by -5^5 mod 64 would be -5^7 mod 64

But that would have been fine

[sacred junco]

at this point, would it be fair to do something like this?

and we have x = 36 mod 64

[nimble cedar]

Hmm? I think 36^5=0mod64

[sacred junco]

x = 35*mod64

[nimble cedar]

Yes

Aside from making it faster to get to x^5=19mod64 I don’t know how the table helped

[sacred junco]

the R table or the table inc with the question?

[nimble cedar]

And I feel like it should have

The one with the question

[sacred junco]

an answer is an answer so i'm happy with it

[nimble cedar]

Maybe that’s all it was there for...but I’d like there to be a theorem we didn’t use that would have made it easier

The trick is being able to get answers later

[sacred junco]

the course offering i'm taking doesn't even have prerecorded lectures, just notes and theorems :/

otherwise i'd be trying to apply theorems, just dk where to start

[nimble cedar]

Well, this is related, but probably not what you would be expected to deal with in an elementary number theory class... https://math.stackexchange.com/questions/934798/modular-nth-roots-e-g-x5-equiv-6-pmod31

[sacred junco]

I'll ask straight up this time: can i get some help on this?

[torn escarp]

What's a universal exponent and what's U_n?

[last grove]

geb1 ok

[sleek cove]

charmichaels function/theorem i think

and i think U_n is
{ k in Z/nZ such that (k,n) = 1 }

[torn escarp]

seems like a reasonable guess

[sleek cove]

yea its just a matter of breaking down 18900 if thats the case

[sacred junco]

b^-3/b^-6 = 1/b^3
How? Isn't it supposed to be b^3?
-3 - (-6) = 3

yah

where did u get that?

[sacred junco]

where did u get that?
@sacred junco

@sacred junco wait a minute

what u said is the reciprocal of whats on the screenshot

lmao

Still I don't get it.. why is it inverted?> lmao
@sacred junco

wdym

what you asked and whats in the screen shot are inverses of one another

b^-3 / b^-6 = b^6 / b^3 =** b^3 **

Oh no, now that I notice it, I'm confused af
My bad, I'm sorry

no worries lmao

[digital swan]

n = p * q

Get start base
b = floor(n / 2)

Calculate v for our hyperbola
v = b^2 mod n

Find solution x, y?
y^2 - x^2 - 3x - 2 = v

Result:
p = 4x + 2(y - x) + 3

Does anyone know how i can get the x, y integers without n prime factors?

[sacred junco]

Is it correct?

[opaque dome]

hey so i was going through this book

and this is kind of confusing

i don't get what this notation means

i get the | notation, it means that b is a divisor of a

but the .

and the ->

i get the feeling that --> means the same thing as =>

[last grove]

-> means "imply" you're right

[opaque dome]

but i dont get why they'd use .

:/

[last grove]

waitwait

im going crazy

[opaque dome]

damn why didn't they use > that is up

like it's twisted up

is there a symbol for it in latex?

[last grove]

hmm

[opaque dome]

let's expand this then

earlier they said this

I was thinking you could "expand" out the b|a as a = bc but nvm

wont do much good

[bleak musk]

@opaque dome I think they're using it to mean "and"

if b|a and c|b, then c|a

for the imply, you can use $\Rightarrow$

[sterile pumice]

Nicholas:

[bleak musk]

capital r gives the double line

lower case r gives $\rightarrow$

[sterile pumice]

Nicholas:

[opaque dome]

oh okay

ty

[sleek cove]

what is this supposed to be lol @sacred junco

also this isnt number theory

[bronze veldt]

@sacred junco No, it is not correct. $12^3$ divided by $4^{-3}$ is not $3^6$. By the way, these kinds of questions don't belong in this channel. Please use #prealg-and-algebra instead, or other people may think you are trolling and will ignore your question completely

[sterile pumice]

MDragon:
Compile Error! Click the reaction for details. (You may edit your message)

[sleek cove]

thats a 4 and a 12, huh

[bronze veldt]

I think so :p

[sleek cove]

hmm yeah i see it now

[last grove]

hey all, is this identity correct? Mu is the Mobius function

[sacred junco]

ye

Im pretty sure

[torn escarp]

mu and n are multiplicative functions, so is their convolution, so looking at the identity on just a power of a prime is enough to confirm it

look at n=p let's just say, 1+1/p = mu(1)/1 + mu(p)/p = 1 -1/p

not quite right

@last grove

$(\mu \star N)(n) = \sum_{d|n} \frac{n}{d} \mu(d) = n \sum_{d|n} \frac{\mu(d)}{d}$

[sterile pumice]

Merosity:

[torn escarp]

now looking just at a power of a prime,

$\frac{(\mu \star N)(p^k)}{p^k} = \frac{p^k-p^{k-1}}{p^k} = 1-\frac{1}{p}$

[sterile pumice]

Merosity:

[torn escarp]

so that's the actual result