That's correct, right?

no.

b must be 1 for it to be a multiple of 3

is 9 a multiple of 3?

and how many times do 2,5 divide 450?

@Semiclassical Thank you so much for paying attention to my problem. In problem 26, Rudin says to use exercise 23 and 24, so I have to go this way: "Using 24, limit point compactness implies separable (A>C) and using 23, separable implies countable base (C>D)" And I am asked to show compact. So final implication path is "A>C>D>B". My question is: Can I deduce that because "A>C>D>B" holds, we get "C>D>B"?

Or is this deduction some sort of violation of the fact that implication is not associative?

@Semiclassical I think I fixed it now

@DarkRunner yeah, that's right

That should be it

So my method was just listing out the factors, and I found out pairing them up in a certain way gave a constant product of 1350, and then raised it to the number of pairs, which was 6.

neat.

My final answer was 1350^6, but the book solution was entirely different, and I'm dumbfounded

the way I'd do it is to say that if a=3b divides 450, then b divides 150

and the divisors of 150 are 1,2,3,5,6,10,15,25,30,50,75,150

How could you conclude tat if a=3b divides 450, b divides 150?

I don't understand that part

define divides

No remainder

150 divides 450, so 50 divides 150.

do you know division algorithm?

or, more simply: if a divides 450, then 450=na for some integer n.

but if a=3b, then we have 450=3nb means 150=nb. so b divides 150.

he be like that's too many symbols!!!

If 450=blah(3x) then 150=blah(x)

this is when i'd hand in pseudo code of some for loop as an answer

lol

(ab)/(ac)=b/c

So, since we can pair up 150 and 3 to get 450, all the factors of 150 that are multiples of 3 are all the factors of 450 that are also multiples of 3

Right?

no.

rip

the factors of 450 that are multiples of 3 are in one-to-one correspondence with the factors of 150

no restriction on the latter (other than positivity)

> no restriction

> positivity

hey you said no restriction

@anon now that's way more machinery

then is required

f(x)=3x ?

I mean, right, combinatorics boil down to bijection

but it usually isn't taught as bijections

knowing that every multiple of 3 which divides 450 is 3 times more than a multiple of 150...is a lot of machinery?

> 3 times more than

sigh

i also said multiple of 150, not factor of 150

derp

As an example, @DarkRunner, consider the factors of 12 that are multiples of 3. They are all 3 times some (any) factor of 12/3=4. You can see this in the graphic below.

@Semiclassical no, I mean phrasing it in terms of bijections is a lot of machinery

@LeakyNun okay, "one to one correspondence"

aka what I did in the first place...

I think the point Leaky's making, much as I disagree with it, is that intro combo is just taught as counting instead of finding bijections between sets. Not as much the terminology, I don't think

@Daminark I'm not necessarily complaining about the pedagogy

@anon I think I'm starting to understand

@anon

yes?

Do you know why this one-to-one correspondence happens?

because it's injective and surjective...

ew who likes math

@Fargle me

If x is a factor of 150, then 3x is a factor of 450. Conversely if 3x is a factor of 450, then x is a factor of 150. (The reason: 450=(blah)(3x) implies 150=(blah)(x).)

hi @Fargle. Happy almost new year.

Happy holidays to you, @anon.

@TedShifrin ich lese the sensual quadratic form

kennst du diesen Buch?

Heya @Ted! Happy New Year $-\varepsilon$ to you too.

Oi @Fargle, how've you been?

@DarkRunner just give your teacher:

factors = []
sum_this_list = []
for i in range(450):
    if 450 % (i+1) == 0:
        factors.append(i+1)
for number in factors:
    if number % 3 == 0:
        sum_this_list.append(number)
print (sum(sum_this_list))

@Salt because someone can't do range(1,451)?

you can't possibly fail

@Daminark Alive. Didn't quite graduate--seems the university had a vested interest in not telling me about one of my graduation requirements.

That's annoying. Which?

Number of upper division hours.

but range(450) is prettier

print(sum(i for i in range(1,451) if i%3==0 and 450%i==0))

@Salt there you go, in one line

On the bright side, I only have one part time semester to go.

hi

that works

@MeowMix hi

thats some ugly code ya got there

well, you only need to run it once

unless youre code golfing, then i guess its pretty :)

i meant leaky’s

@MeowMix :P

I do come from PPCG

yours is quite tidy

i can appreciate a one-liner

@MeowMix I can do better

im sure you can

im not much of a python boy thougg

so i cant say id understand it at first glance

python is for people who can't understand stuff at first glance, you can't go wrong

it's why i use it

print(sum(i*(i%3==450%i<1)for i in range(1,451)))

i do think c++ is a valid substitute for most languages

49 bytes for you

though, keyword most

#include <stdio.h>
int main() {
	int sum = 0;
	for(int i=1;i<=450;i++){
		if(i%3==0 && 450%i==0){
			sum += i;
		}
	}
	printf("%d\n",sum);
}

there you go, in C

I don't know C++

man, python really has the superior loops

most valid C is valid in C++

python loops are ugly imo

well, we all know that lisp loops are the prettiest

asm loops are the prettiest :D

the isometry group of the quadratic form $3x^2+6xy-5y^2$ is generated by $\begin{bmatrix}1&2\\0&-1\end{bmatrix}$ and $\begin{bmatrix}2&-1\\3&-2\end{bmatrix}$

cmp ecx, loop_max

inc ecx

jne .loop

the hell

then youd have your looped code before those 3 lines and after a .loop label

.poop

it could be called .poop too, yes

now i'm interested

@Salt Much appreciated, but this is not for school

fair

So basically, just to clarify, we take the number, N. To find the product of all the multiples of k that are factors of N, we divide N/k. We take the product of all the factors of N/k, because all the factors of N/k will simply be 1/kth of the factors of N that are multiples of k (one-2-one correspondence). Then, to compensate for the fact that we have found the product of all the factors of N/k, we multiply this product by k raised to the power of the number of factors of k.

Damn

> spends some time to write a long paragrph

writes “2” for “to”

^this gave me a chuckle

@DarkRunner k raised to the number of factors of N/k

am i right?

Huh: "Every natural number has a unique representation in terms of divisor products."

divisor products?

Product of the divisors of a given integer

if p is a natural number coprime to the given integer N, then how do we write p as a product of divisors of N?

I mean in the sense of: If $\pi(n)$ is the product of the divisors of $n$, then $\pi$ is a bijection from the natural numbers to themselves.

hmm

It's mentioned on A007955

W. Lang gives a proof, but it looks pretty involved

All terms of this sequence occur only once. See the second T. D. Noe link for a proof. - T. D. Noe, Jul 07 2008

Yeah. I mean, that goes in hand with the representation being unique

an explicit formula would be $\prod p^{e(p)}\mapsto \prod p^{(1+\cdots+e(p))\prod_{q\ne p}(1+e(q))}$ I think

Though the proof of each term only showing up once seems much simpler

sspectra.com/math/DivisorProduct.pdf a one-pager

not as nice as a one-liner, but it's the next best thing

right

it's the other direction, showing that each natural number has some representation, that is the hard part

anyways. neat theorem.

so, what's the inverse function?

No clue.

"Sometimes called the "divisorial" of n."

come to think of it, I'm a little surprised the inverse function isn't tabulated on OEIS

@GFauxPas Illustrating how to do a matrix multiplication

hmm, $\pi(n)$ feels a bit like cyclotomics

Actually, no. My reading of it is wrong

the key point is that it's divisor products, plural

so you can express 4 as $\pi(2)^2$

yes, the $q$-analog of $\pi(n)$ (divisor products) is in fact $\Phi_n(q)/(q-1)^{\omega(n)}$

but not $\pi(n)=4$ for some $n$.

So every $\pi(n)$ is distinct, but you may need to multiply a few $\pi(n)$ together to get a given natural number. So, no bijection :/

I see

Still weird, though

And interesting that you still have a sort of 'unique factorization' theorem

I wonder what percentage of sequences in OEIS with the property that each term in the sequence occurs only once "factorize" natural numbers

my q-idea was all wrong, don't mind me

Hi

anyone there

not i

Does a vector space always admit a basis ?

yes

Thanks

hi

my question is : is that minimum necessarily positive?

nvm, it is.

The meme squad has assembled

Clouds are gathering

Morning

mournings

morning

morgen

Is $A\implies B \implies C$ same as $A\implies C$?

@Silent depends on context

usually yes

@LeakyNun Which context? Actually I had this question because of this

@Silent the context of the sentence, obviously

in that context, yes

@LeakyNun, when does that not hold? Sorry, but it is not obvious to me.

@Silent do you notice that $A \implies B \implies C$ is not a well-defined notion?

it's an abuse of notation

and one must know its context to know what it means

@LeakyNun Yes, so I think you mean that $(A\implies B)\implies C$ is same as $A\implies C$, but $A\implies (B\implies C)$ is not same as $A\implies C$?

nonononono

what they mean by $A \implies B \implies C$ is $(A \implies B) \land (B \implies C)$

oh!

you get very different meanings if you do $(A \implies B) \implies C$

you can

remember this

@LeakyNun But then that means here that every separable metric space is compact! But $\Bbb R$ is separable, but not compact!

@Silent let me read it

ok

@Silent do you actually not know how to prove it, or are you just confused in the implications?

@Narcissus I take back what I said about the new Elder album. I retroactively like it now.

I have solutions to those problems. i am confused because they are very similar. @LeakyNun

Yes, confused about implications.

A⟹C⟹B⟹D does not hold

because, as you said, C⟹B does not hold

@LeakyNun i am sorry, but that is according to Daniel Fisher's notation. You have use my notation, in which, following Rudin's hint, A⟹C⟹D⟹B holds, and from previous doscussion with you we can deduce C⟹D⟹B that is C⟹B .

There was a confusion of notation between me an Daniel Fisher, I am really sorry about that.

you can't deduce that

because the final step D⟹B uses A

why don't you actually post your solution

@LeakyNun So, there A⟹C⟹D⟹B means something other than (A⟹C)∧(C⟹D)∧(D⟹B) ?

well

there's a point after which abuse of notation doesn't really hold much water

@jublikon by differentiating it

@LeakyNun I am using this solution

yes :D I agree haha

@jublikon seriously

$\displaystyle \lim_{h\to0} \frac {\ln(x+h+k) - \ln (x+k)} h = \lim_{h\to0} \frac {\ln(1+h/(x+k))} h = \lim_{h\to0} \ln \left( 1+\frac h {x+k} \right)^{1/h} = \frac1{x+k}$

for $x_0 \in \mathbb{R}$

@jublikon because I don't want to type $|\alpha-1|$ 1000 times

@jublikon if you have a shifting lemma, I would gladly use it and drop the $k$

shifting lemma, i.e. $f(x)$ is differentiable iff $f(x+k)$ is differentiable

@LeakyNun unfortunately we do not have the shifting lemma in the scriptum :(

note that the name is just due to me, lol

i.e. I just made up the name

yeah, I see. But cannot find anything that looks like that in there..

composition?

it is not a VERY detailed script I think

do you have chain rule?

yes

then you have the shifting lemma :P

ah okay, I see !

thanks

lol

@LeakyNun Will you please let me know what does actual logical path look like as suggested in hint of exercise 26 instead of A⟹C⟹D⟹B? I will be obliged.

@Silent alright, reading

(A⟹D)∧((A∧D)⟹B)

brb

back

$\displaystyle \lim_{h\to0} \ln \left(1+\frac hx\right)^{1/h} = \lim_{h\to0} \ln \left(\left(1+\frac hx\right)^{x/h}\right)^{1/x} = \frac1x \lim_{h\to0} \ln \left(1+\frac hx\right)^{x/h} = \dfrac1x \ln e = \dfrac1x$

@jublikon

Why use $\arg z=v$ when you could just write $v$=something instead?

I am looking at $\frac{1}{x} \lim_{x \to 0} \ln (1 + \frac{h}{x})^{\frac{x}{h}}$

HI everybody

What doe this means in elementary terms:

Let K be a subfield of C and f,g $\in$ K[x]. Let z be a complex root of f and assume f is irreducible in K[x] and g(x)-z is irreducible in K[z][x]. Then f(g(x)) is irreducible in K[x]

@jublikon $x$ is fixed

let $t=h/x$

@AlexKChen do you know what a field is?

Nope very sorry (I know what finite field is though, courtsey you)

where does that question come from?

statement

Problems from the book, by Gabriel Dospeniscu and Titu Andreescu

is it from a course?

This is from chapter 21 of that book, called Cohn or something irreducibilty criteria

I'm uploading a picture, you'll see

how much time do you have?

As many as you can give me :)

so we are familiar (let's say we are) with the real numbers and complex numbers

we know many of their properties

Yes

but one might wonder, what properties are retained with other types of numbers

Like ?

so we might want to define other types of numbers

this is an abstract setting: we want it to work for all kinds of numbers

firstly we might want to define what a number is

we don't actually define what a number is

but a set of numbers, when they satisfy certain properties, we call them a ring

What properties ?

if they satisfy other properties, we call them a field

so let's say we have a set of numbers

we aren't just concerned with the numbers though; they need to interact

and we want them to interact using the two basic operations: addition and multiplication

Can I interrupt a bit ?

yes

I know a bit about integers modulo prime and very little about finite fields.

So no need to introduce them again.

Please tell what happens after that.

are you sure?

OK, so you can give analogies with integers modulo p and inverses modulo p. I know them well.

if you like it then you shoulda put a ring on it

@AlexKChen I'm doing this in the abstract setting

OK, sure, then go on.

it works with every possible kind of numbers

so we have a set of things we call numbers, as well as two binary functions we call addition and multiplication

ok

what's binary function btw ?

let our set be called R

then a binary function is a function of the type R^2 -> R

must be commutative ?

no

ok

now we want those numbers and the two functions to satisfy some properties

we want the two functions to be associative and commutative

ok

we want multiplication to distribute over addition

what does that mean ?

"distribute over addition" ?

that means, (x+y)*z = x*z+y*z

ok

we want additive identity and multiplicative identity to exist, i.e. a number we call 0 such that x+0=x for all x, and a number we call 1 such that x*1=x for all x

ok

we want negatives to exist, i.e. for every x there is y such that x+y=0

ok

we call those numbers together with the addition and multiplication a ring

"the [set] forms a ring under [addition] and [multiplication]"

@AlessandroCodenotti buongiorno

?

Hi

@AlexKChen do you get it?

what does "buongiorno" mean ?

it means good morning in italian

oh ok

So that's the defination of ring

now if multiplicative inverses exist for every non-zero element, we call them a field

so integers modulo composites are ring, and modulo primes are field ?

that you can spell buongiorno is more impressive...assuming that's how you spell it

@AlexKChen yes

and every field is a ring

of course

now let's say we have a field K

OK

we can define polynomials over K

how ?

a polynomial is defined to be something of the form a_n x^n + ... + a_1 x^1 + a_0 x^0

they are not functions. they are just formal objects

where n is a natural number

"they are just formal objects" I don't understand

Gimme an example

I thought polynomials were functions

@AlexKChen but here we are not considering them as functions

think of em as combinatorial objects....at least, i usually see them as physical objects with lots of little pieces

@LeakyNun we're treating them (like say) circles ?

nah, like legos

if we treat them as functions, we start running into problems like "why is 0 and x^2+x different polynomials over Z/2Z if they agree at every point in the domain?"

because a function is uniquely determined by its domain, codomain, and the image of each point in the domain

OK, so we can treat them as tuple of numbers: (Coeff of x^n, Coeff of x^n-1, ..., constant coeff) ?

precisely

now we can add polynomials together

we can also multiply polynomials together

and you can verify that the set of the polynomials over K forms a ring under polynomial addition and polynomial multiplication

ok

yeah of course

we call that ring K[X]

X is numbers from ?

no

K is a field

we denote by K[X] the ring of monovariate polynomials over K under addition and multiplication

"polynomials in X"

(polynomials with one variable)

only blue legos

Give me an example except common things (polynomials in R, Q, integers modulo P)

of what

a nontrivial example of a polynomial in K[x]

you need K to exist for K[X] to exist

x

It's just an ordered n-tuple of elements of K

if K is Z/2Z, K[X] contains x^3 + x + 1

the coefficients are only 0 and 1

I wrote "except common things (polynomials in R, Q, integers modulo P)"

but you don't seem to understand that K[X] depends on K

No I seem to understand.

let's say K is Q[sqrt(2)], i.e. {a+bsqrt(2) | a,b in Q}

I was asking for a nontrivial example of K

yeah that

Z/2Z is trivial? you take that back

So this is a good polynomial:

Do you know complex analysis? The meromorphic functions on a domain form a field

it heard you

(2+sqrt{2})x^3 + sqrt{2} ?

no

2+sqrt3 isn't in K