Okay so, is it possible for a number to be a perfect square AND a perfect cube?

@Dragneel Yeah, $2^6 = 64 = 8^2 = 4^3$

Sure. Try thinking in powers of two.

Ohhhh so you'd have to have two different factors

More generally: Every sixth power of a given integer is both a perfect cube and a perfect square.

That makes sense. Now I'm asked to prove it. Should be pretty straightforward.

So a million in particular is also an example.

Because $10^6$?

Right.

What exactly are you supposed to prove?

Suppose $m$ is a perfect square and a perfect cube. Prove that it is also a perfect $6$th power.

Ahh.

Yea

We have that $m = a^2$ for some integer $a$ and $m = b^3$ for another integer $b$

Another way to phrase that: if a given integer is both a square and a cube, then the cubed root is itself a perfect square.

Can't we just write $m=sqrt(a)^3$

let's look at each prime factor

Nevermind

ill just do it myself for an exercise

If $a^2=b^3$ then $b$ divides $a^2.$

I'm just going to prove it with an example. I don't know how else to prove it haha

Eh, it should have a simple enough proof.

suppose $a$ has a prime factor $p$

then the amount of that prime factor in $m$ must be even

and thus, $b$ must also have that prime factor, otherwise the cube won't be equal

however, the amount must be a multiple of 3

thus, the amount of that prime in $m$ is a multiple of 6

apply this for each prime and you find that it's a power of 6

Hey y'all -- I've got a question about k-algebras in alg. geometry

go ahead and ask, then. (Not sure there's be any experts on that right now)

I'm supposed to show some results about the Zariski topology on Spec($A$), where $A$ is a finitely generated, reduced, k-algebra.

i can't fucking believe this.

My professor said that $A$ is isomorphic to $k[x_1, ..., x_n]/I$, where $I$ is a radical ideal, but I'm not sure if this is any arbitrary radical ideal or not. I'd like to talk about the elements of $A$ as they related to points in the open sets of the Zariski topolgy, but I'm not really sure what they look like

Well done @MeowMix

@Dragneel what do you mean?

oh, my solution?

I found another well-written proof for this here at #5 math.lsu.edu/~adkins/m4023/4023s08ps2a.pdf

@MeowMix Yea you solution :)

@MeowMix ?

all of my code and work was deleted

for a very hard project

O_O

written in MIPS assembly

and included notes i had taken over the span of a few months

No possibility for data recovery?

nope

the clusters were overwritten

Oh wow.

Hey guys

hi @Dami :[

Hahaha :P

@Daminark I saw your comment on devices and devisors

Remind me?

@Daminark oh woops. Wrong person. I was thinking of Danu

Oh kek

Is it true that if we have the disjoint union of two things and we take the cartesian product of that

then it breaks into the cartesian product of two things ?

My immediate inclination would be to say yes @Adeek

Hmmm how am I supposed to prove this: "Let $n > 0$ with prime factorization $n={p_1}^{n_1} {p_2}^{n_2} {p_3}^{n_3} ... {p_k}^{n_k}$. How many positive divisors of $n$ are there? Prove your answer."

I understand that the formula for finding the number of divisors is $({n_1}+1)({n_2}+1)...({n_k}+1)$

But how would I begin proving it, is my question.

Proof by contradiction maybe

@Dragneel Look at which primes we can choose

for $p_1$, we can choose from $1$ of them up to $n_1$

so that's $n_1$ possible amounts of primnes

but we can also have none at all

which is an extra case

then repeat this for every prime in the factorizaton

and thus we have $\prod_{i=1}^k (n_i + 1)$

make-o sense-o?

i guess you could call this one a "combinatorial argument" because you're multiplying the cases together

It sure does @MeowMix

That's true

I actually used that in a math competition. I didn't actually know the formula, just used that reasoning :^)

hah!

That's awesome

what text are you reading?

Book of Proof

ah, interesting

It's a free book online

finished with induction? :P

Nope. We just started haha

oh boy

I find induction easier than all the prime-related problems and the gcd garbage

well, once you get the hang of it, it's like the first tthing you'll think of whenever you have to do a proof for all natural (or integer for that matter) cases

Takes so much practice though

@AkivaWeinberger Want to discuss the QQ questions?

Sure, though I haven't looked at them more than like a minute

umm, shall i make a room for it?

Why not

@MeowMix By the way, let $d(x)$ be the number of divisors in $x$

and suppose $m$ and $n$ are coprime

Do we have $d(mn)=d(m)d(n)$?

Might as well ping @Dragneel also, why not

that proof seemed to suggest so

because the factors are independent

I believe a function like that (defined on the positive integers and satisfies the above equation for coprime $m$ and $n$) is called multiplicative. @MeowMix

They're important in number theory

It's a homomorphism /s

actually, wait, is it?

I think $\phi(n)$, the number of numbers between $1$ and $n-1$ coprime to $n$, is also multiplicative

@MeowMix No, since it doesn't work for all $m$ and $n$

Plus, a ring homomorphism would need to be additive (and if it were a group homomorphism it would need to be additive since integers are only a group under that)

^

Also I love how the prereqs in math classes are all listed as additional notes

?

So the system doesn't actually know that they're prereqs, and thus doesn't enforce them

I could totally add in right now to a bunch of classes that I'm entirely unqualified for

I won't, but it's hilarious

Sorry I just noticed chat. Also, yes. We're asked to find the number of positive divisors of $n$ INCLUDING 1 and $n$

@Dragneel Yup.

Which makes things easier I would assume

no, $\Bbb N$ isn't a group

fuck that message sent late cuz my internet got kerfuffled

There's a name for it

I think a rig is a ring without negation?

There's a name for it that nobody uses except to chuckle at the pub

(Get it? Ri-n-g without n-egation is ri--g. Algebraists are hilarious. /s)

snaps Eyy @Akiva

Also, a magma is a think that exists, but I don't remember what it is

but I think it's like $(\Bbb N,+)$?

i thought a magma was like a less strict group

or $(\Bbb N,\times)$, for that matter

But that might be a monoid

oh,

it just has to be closed under the binary operation

that's pretty god damn loose of a definition

@MikeMiller can you help me on some question ?

I have some ideas but I don't know how to tie in together

Oh I was thinking of $\mathbb{Z}$

@MeowMix Ah. So I think the above two are monoids

IIRC, they have associativity also

@MikeMiller

So I was thinking of working seperately on both spaces

@Adeek What a wonderful drawing. Want me to put it on the fridge?

/me secretly thinks it's gibberish

on $I \times \{0\}$ ?

Oh, no, semigroups are magmas that are associative, and monoids are semigroups that have an identity

and on the other one $\{0,1\} \times I$ ?

so monoids are groups that need not have inverses?

Yeah then $(\Bbb N, +/\times)$ are both monoids

And rings without identity are rngs ("rungs"); without negation, they're rigs, and without both, they're rgs ("rugs")

And a fish with no eyes is a fsh

Although, fun fact: IIRC, in any rig, if $x=1+x^2$, then $x=x^7$.

This is easily seen to be true in a field like, say, $\Bbb C$, but it's not obvious why this would be true in a rig

you can apply binomial thm in rings, right? cuz distributive works

wait, no

because

ugh, it's weird

For commutative rings, I think you can

(Or if $ab=ba$, if you're trying to expand $(a+b)^n$)

@Adeek A fibration is defined in terms of the homotopy lifting property, yeah? So try that.

@MikeMiller but see the issue is

@MikeMiller that we need the top thing to be of the form M and bottom is MxI so is IxM homeomorphic to

I mean is $I \times X$ homeomorphic to $(I \times \{0\} \cup \{0,1\}\times I)\times X$ ?

?

?

I mean the Homotopy lifting property doesn't get it right away @MikeMiller since we don't have the required form

Hi everyone

I'm trying to read this formula

Do you think the || supposed to be logical or?

source: ncbi.nlm.nih.gov/pmc/articles/PMC4916476

@Adeek Is X an arbitrary space?

Anyway yeah your point is the thing we need.

yeah @MikeMiller X is arbitrarily

Is it true that $I times \{0,1\} \cup \{0\} \times I \cong I \times \{0\}$ ?

"Suppose $a,b,s,t,u,v$ are all integers such that $sa+tb=21$ and $va+vb=10$. Show that the $gcd(a,b)=1$." So, I started by using Contradiction but I have no idea where to go from there. Here's what I got so far: $a$ and $b$ have a common divisor $E>1$. Then $a=Ex$ and $b=Ey$ where $x$ and $y$ are random integers. Then, $21=s(Ex)+t(Ey)$. Becomes $21=E(sx+ty)$. This implies $E|21$.

On second thought, I think Contradiction was the wrong approach

@Adeek the first one looks like 3 sides of a square in $\Bbb R^2$?

yeah exactly

is the three sides of square homeomorphic to 1 side ?

@AlessandroCodenotti ?

Should be

okay awesome

now I am good with what I have in mind

I guess the homeomorphism we construct should send one side of the square to

half of one of the sides

and other half to similarly