i noticed that invariance of domain required that the map be a homeomorphism too. i was a little surprised that there wasn't already a google-able question like this that i could find on stackexchange

Hi guys looking to get some sort of hint here, practicing for my FA comp. Let $H$ be a Hilbert space and let $M$ be a closed linear subspace of $H$. Let $f \in M^*$ be a linear functional defined. $f : M \rightarrow \mathbb{C}$. Show that there exists $g \in H^*$ such that $g(x) = f(x)$ for all $x \in M$ and $||f|| = ||g||$.

I am pretty sure the answer here is a one liner: "Apply the Hanh-Banach theorem, and we are done". But is there a way to do this without using Hanh Banach?

@PedroTamaroff Do you maybe have an idea what property should be satisfied so that a point $\in \mathbb{P}^2(\mathbb{C})$ ?

@masfenix Sure, we don't need Hahn-Banach if $H$ is a Hilbert space. Orthogonal projection.

Or Riesz representation theorem, since we're looking at maps to $\mathbb{C}$.

@DanielFischer How can we conclude that $$\mathbb{F}_{p^n}\cap \mathbb{F}_{p^m} \subset \mathbb{F}_{p^{\gcd (m,n)}}$$ ??

@DanielFischer well, so if its as easy as I think it is. You can define a $g: H \rightarrow \mathbb{C}$ by $g(x) = <x, y>$ for all $x \in H$. This gives us $||g|| = ||y||$

@masfenix And what is $y$ there?

$y$ is given to us. Let $f \in M^*$. This implies, that for all $x \in M$ there exists a unique $y \in M$ such that $f_y(x) = <x, y>$

@MaryStar Have you understood why we have $$\mathbb{F}_{p^n} \cap \mathbb{F}_{p^m} \subset \mathbb{F}_{p^{n-m}}$$ if $n > m$?

but my concern now is that, we've only $y$ that belongs to $M$.

@masfenix Well, $M\subset H$.

Yes, but $H*$ is isomorphic to $H$. I mean i dont know how to voice my concerns, but if we only have $y \in M$, dosnt that break the isomorphism?

@TedShifrin the "bounded distance away" is what I was looking for

@DanielFischer

@DanielFischer infact even the proof of Reisz Representation theorem only finds such a $y$ in a closed subspace. We know $T: H \rightarrow H^*$ is a isomorphism. But ... ahh im getting confused here.

I dont know how to voice my concern.

@masfenix I don't understand. By Riesz, since $M$ itself is a Hilbert space, we know that there is a unique $y\in M$ with $f(x) = \langle x, y\rangle$ for all $x\in M$. So define $g(x) = \langle x, y\rangle$ for all $x\in H$.

@DanielFischer but what if there arnt enough "$y \in M$" to construct such a $g$.

@masfenix You only need one.

@DanielFischer OH I see!. So basically (not related to his particular question), we have a space of linear functions $H^*$. Each linear functional is associated with a particular unique $y \in H$. But the keypoint is all the functionals could be associated with the same $y$.. you only need one. Its just that, if you pick a particular $f \in H^*$ then this $f$ is only associated with ONE $y \in H$.

I hope I make sense.

This clears up a LOT of confusion :)

@masfenix No, each $f\in M^\ast$ is associated to a different $y\in M$. The Riesz representation theorem gives you an antilinear bijection between $M$ and $M^\ast$.

But you have only one $f\in M^\ast$ to start with. That gives you one $y\in M$ with $f = \langle\,\cdot\,, y\rangle_M$.

And now, since $M\subset H$, you simply define $g = \langle\,\cdot\,, y\rangle_H$.

"Its just that, if you pick a particular $f\in H^\ast$ then this $f$ is only associated with ONE $y\in H$." <- That part is right.

I see. so this may be a stupid question, but does $M$ have an infinite number of elemnets?

Hello everyone. Anyone up for a couple of algebraic geometry questions?

@DanielFischer Sorry, i dont know why i cant seem to grasp it. But basically again, i dont understand how we can associate $y$ with $g$. We already said $y$ is associated with $f$. And since $g$ is a linear functional different then $f$, it can not be associated with $y$?

@masfenix Unless $M = \{0\}$. Every vector space over $\mathbb{C}$ or $\mathbb{R}$ except for the trivial vector space has infinitely many elements.

@TedShifrin you can show there exists $\epsilon:\mathbb{Z} \rightarrow (0,\infty)$ such that for every smooth $f:S^n \rightarrow S^n$ of degree $k \neq 1,-1$ there exist points $x,y$ s.t. $f(x)=f(y)$ and $|x-y|>\epsilon (k)$?

@masfenix Different spaces, $f\in M^\ast$, and $g\in H^\ast$. If we let $R_E \colon E \to E^\ast$ denote the Riesz map of the Hilbert space $E$, then we have $$g = R_H (\iota (R_M^{-1}(f))),$$ where $\iota \colon M\to H$ is the inclusion.

@DanielFischer Thanks. I got it now. But if I can ask you to expand a little bit on the inclusion part or link to wikipedia for the relevant information?

@TedShifrin What is the property that should be satisfied so that a point $\in \mathbb{P}^2(\mathbb{C})$ ?

@DanielFischer We took $x\in \mathbb{F}_{p^n} \cap \mathbb{F}_{p^m}$ and concluded that $x\in \mathbb{F}_{p^{n-m}}$, right??

@DanielFischer So, is it right to say that on $M$ $M \subspace H$, we have that $g$ and $f$ are the same linear functional?

@masfenix Formally, $g\lvert_M = f$, the restriction of $g$ to $M$ is $f$.

@TedShifrin Also why $I_P(x^5+x^4+y^2,x^4)=I_P(y^2,x^4)$ ?

@MaryStar Right. Have you understood why that is the case?

@DanielFischer thankyou

Night everybody. Thanks for all of your help today! ^_^

@MaryStar If $u^{p^m} = 1$, then $u = 1$ since the Frobenius homomorphism $F\colon z\mapsto z^p$ is injective.

@robjohn ?

@MikeMiller thanks pal, good luck against the cowgirls ;)

@DanielFischer Or is there something wrong??

Anyone heard the idiom "Derivation is a tool, integration is an art"?

@skullpatrol Thanks, but we don't need it pal :) Last week was when we needed luck :P

@MaryStar Right. And since naturally $\mathbb{F}_{p^n} \cap \mathbb{F}_{p^m} \subset \mathbb{F}_{p^m}$, we have $$\mathbb{F}_{p^n}\cap \mathbb{F}_{p^m} \subset \mathbb{F}_{p^{n-m}} \cap \mathbb{F}_{p^m}.$$ Now the Euclidean algorithm then gets us down to $$\mathbb{F}_{p^n} \cap \mathbb{F}_{p^m} \subset \mathbb{F}_{p^{\gcd(n,m)}}.$$

@MikeMiller yep, it's coming down to crunch time...

...the real season.

Ho Ho Ho

hi

hello

Hii

ho ho ho

:-)

Hi @skullpatrol!!!!!

@HipsterMathematician how are you pal?

Not very good but fine @skull and how are you doing?

@HipsterMathematician Fine thanks.

@skullpatrol :D

@HipsterMathematician Are you going to join in on Winterbash?

Ah I don't think so @skullpatrol I don't come here often

@HipsterMathematician yes, I have noticed :(

@skullpatrol miss you pal

@HipsterMathematician yipyipyip me too

:DDDDDDDDD

};-)

@HipsterMathematician did you know, pedro is running in the election for mods coming up on dec 23

@FreeMind Yes?

I couldn't sleep without knowing that @skullpatrol

@HipsterMathematician lol

@HipsterMathematician anyway I gotta go out and get an oil change done, it was nice hearing from you pal; Merry Christmas and have a Happy New Year!

@skullpatrol it's always nice to see you again :D have a nice weekend!

Quick question: could someone confirm that my answer here is not wrong? I understand what the other answerers are saying about the value of the sum (generally speaking) depends on the order in which the sum is taken, but I thought that (because we're assuming absolute convergence) the order of summation didn't matter...

hi @robjohn ... You seem to be here at randomly distributed hours in the 24-hour day ... day or night :P

Well, @anorton, because of absolute convergence, in fact you can arrange your $(m,n)$ terms in any order you desire and you'll get the same sum. Take any bijection $\Bbb N\times\Bbb N\to\Bbb N$ and use it to order your terms.

It appears that's what user 116.... said, although he made the notation a bit more difficult than needed.

Ok, thanks. :)

math.stackexchange.com/questions/1064376/… Four answers in about a minute, hahah. Attack all the fruits

mmm...the accepted answer has no up votes :/

and how can anyone down-vote the question and up-vote an answer?

I don't want to interrupt any active discussion, but I have a notation question. If I want to describe a function $\text{set}(f, x \to y)$ where it takes the function $f$ and produces a new function $g$ whose value at $g(x) = y$ instead of $f(x)$, but is equal to $f$ for all other elements in the domain of $f$. Is there an already accepted way to write that? It's important to a minor part of a question I'm going to ask.

there may be an "accepted" way of writing it, but why not just explain your notation in the question...

...nobody ever got down-voted for clearly explaining something :-)

before using it

Well, is there a way to use latex in code blocks or code strings?

Hello friends

Hello.

@MikeMiller do you believe my conjecture?

@JorgeFernández Oh, I thought you said it was true. I was not altogether surprised when you said it, so I suppose that means I believe it could be true.

it is true

Overkill, but you could probably kill it with prime gaps.

you can kill it with prime number theorem

Any text recommendation for someone who is trying to relearn trig/geometry. I took the courses in high school but I didn't really care back then so I got only the basics.

@JorgeFernández Is it a conjecture if you know it's true?

@JorgeFernández Interesting.

But at the same time I don't want to redo the course with the same textbook as a I used when I was a freshman in high school, so something of mediocre difficulty maybe? I'm taking calc now for reference

why not just use youtube?

khan academy

a refresheenr in trigonometry

that book is awesome, it goes to the point and proves all the shit you neeed to know for real math

Trigonometry Refresher (Dover Books on Mathematics), A. Albert Klaf, Mathematics

that^ too :-)

@Axoren have you seen this?

@skullpatrol Never liked that.

I could use youtube, but khan academy isn't very in depth? At least for the few things I used it for. I understand the stuff khan academy would teach me, I mean go more in depth than that, but nothing too crazy.

I have a couple dover books on other subjects, I might get one for geometry since they're cheap

@MikeMiller never mind, I don't know if it is true anymore, the solution that I found was wrong, allthough it is true no polynomial of degree $2$ satisfies it.

@JorgeFernández It should be true. One can prove that for a polynomial of degree $>1$, any $c$, and large enough $|n|$, $|f(n+1)-f(n)|>c$. So pick $c>70,000,000$; there are infinitely many primes within this of each other, so pick a pair greater than $f(n)$ (or less than, if the leading coefficient is negative). It'll miss one.

oh wait, nevermind, it wasn't wrong, I'm just a retard and read it wrong

@skullpatrol I've read everything but the comments, there's nothing about writing LaTeX inside of a code-block.

@MikeMiller LEL.

@Pedro It's easy to prove things when you've got big theorems.

For example: System.out.println($\varempty$);

@MikeMiller What's the original question?

So I can sleep.

@PedroTamaroff If a polynomial $f \in \Bbb Z[x]$ has all primes in its image, it's of the form $x+c$ or $-x+c$.

@MikeMiller Ugh. That's going to take a while.

@PedroTamaroff But it's true.

Is mathjax supposed to automatically render in this chat?

Nope, read the 14-starred chat entry on the far right.

Nope. Take a look at the $\LaTeX$ in chat link on the right.

Oh woops, thanks!

Did anyone ever figure out how Cleo finds solutions to those integrals so quickly?

yeah, it's true, the point is the number of values a polynomial of degree $k$ can take that are between $-n$ and $n$ is $O(n^{1/k})$ while by the prime number theorem the number of primes between $-n$ and $n$ is $O(\frac{x}{\log(x)})$

does this satisfy you

this is because the polynomial grows too fast

which is basically what you said

Yeah, cool argument.

Since we were talking about book recommendations earlier. Does anyone know a good text on Group Theory and Abstract Algebra?

I can strongly recommend Dummit and Foote's book calle Abstract Algebra, also Paolo Aluffi's book algebra chapter zero, and for finite group theory Isaac's books called finite group theory.

Would you recommend these in any order? Or all of the above as parallel reading?

in that order, although aluffi and dummit and foote overlap, but I think Dummit and Foote is the best to start with.

although you can read Isaac's book after reading the first 200 books of Dummit and Foote

Thank you very much, Jorge.

no, you have to read their first 200 books. Isaac's is really advanced shit.

sry

no one has been able to read the book, that is mainly why Dummit and Foote write books, so one day people are able to read Isaacs.

although they only have 1 published book right now, so it's going to take a couple hundred years.

fortunately their sons can follow their work

it is said the book contains all the mysteries of math. but no one has read enough Dummit and Foote books to understand Isaacs.

yeah, it should say pages instead of books.

I don't plan on reading 200 books, lol. I'll gladly buy and read 200 pages, though.

I knew what you meant.

sry

My biggest issue is that most of my Abstract Algebra is pieced together out of need rather than being taught, so I have a really shaky background. As a computer scientist, I have to go out of my way to learn higher-level mathematics because it doesn't fit with my course work. Except for things like group theory.

I love computer science.

hello everyone

hi pal

hello

anyone up for a commutative algebra question?

askaway

I have a statement that I need to disprove with a counterexample:

"If $I,J$ are ideals, then $\sqrt{I}\sqrt{J}=\sqrt{IJ}$"

I guess I'm struggling to understand what $\sqrt{IJ}$ is

i.e., what elements inside of that look like

...any hints?

anyone? lol

Consider two singleton ideals to make your job easy. And for simplicity, consider the ring of integers.

Singleton being $|I| = |J| = 1$.

@Mike @Pedro That looks like an immediate consequence of Cohn's irreduciblity test.

@Axoren ok, wouldn't $\sqrt{I} \sqrt{J}$ just be $\sqrt{1}$?

how would that be different from $\sqrt{IJ}=\sqrt{1}$?

No, I meant their cardinality would be 1 each.

A set containing one element of the Ring.

ah, gotcha

sorry- it's been a long day

Well, I worded that wrong.

suppose $I=\sqrt{x^2}$ and $J=\sqrt{x^3}$. What would ${IJ}$ look like? Just $<x^2, x^3>$?

Let me rephrase.

I guess the product of the two ideals is what I find confusing/frustrating

It's the set of all finite sums of pairs of elements of I and J.

@TedShifrin still here?

When I said the ideals contain only one element, I meant that they were spanned by an element. Like the set $\{nz\}$ for integers.

Pick a single element $n$ and you end up with an ideal which is "multiples of $n$".

Multiplying two such ideals is like saying "the set of all finite sums of the products of multiples of $n$ and $m$".

If you pick those two ideals such that $n = m$, you end up with "the set of all finite sums of multiples of $n^2$" which is just "the set of all multiples of $n^2$".

This should be a good enough hint.

... unless I misspoke again :S

ok, so $IJ=h_1(x)x^2+h_2(x)x^3$ where $h(x) \in k[x]$, right?

just so I can understand what I'm working with

2 and 3 are coprime (and thus, prime) so they'd be small enough but distinct enough so that I understand what i'm doing

It's only going to be harder to disprove if your ideals are different.

Perhaps you want $I=\langle x^2\rangle$ and $J=\langle x^3\rangle$, instead.

And I already gave you a much better option.

@daOnlyBG How nitpicky do you want me to be with this statement?

be as nitpicky as you like

I'm new to a lot of this- sorry if my questions seem trivial

I could try using similar ideals

if $I=<x>, J=<x>$, then $\sqrt{I}=<1, x>$, right?

same with J?

*same with $\sqrt{J}$

?

And then what is the product of $\langle 1, x\rangle\langle 1, x\rangle$?

Try that out @daOnlyBG. First thing is to check that $\langle 1,x\rangle$ is contained in the radical of $I$.

So is a power of every element of $\langle 1,x\rangle$ an element of $\langle x\rangle$?

(That question is itself rather complex... What is an element of $\langle 1,x\rangle$, after all?)

an element of <1,x> is any polynomial $1\cdot h_1(x)+x\cdot h_2(x)$

setting $h_1(x)=0, h_2(x)=1$ would give you $<x>$

@Axoren In standard ring theory notation, the square root symbol means "radical" ideal, which differs from a solution $I_0$ of the "ideal equation" $I_0^2=I$.

@daOnlyBG Ok, what if $h_1\ne 0$?

But it's equivalent to taking all the possible roots of elements of that ideal.

Isn't it?

if $h_1≠0$, then no, I don't believe <1,x> is contained in <x>

so no, not every element of the former is contained in the latter

just one

er- just under one condition

Here, I will do some nitpicking @daOnlyBG

ok

@Axoren That's true, but a priori raising an ideal to a power does not relate to this.

@daOnlyBG It is true that $\langle 1,x\rangle$ is not contained in $\langle x\rangle$, and you gave a valid proof. But what I was asking was, for every polynomial $p(x)=1\cdot h_1(x)+x\cdot h_2(x)$ in $\langle 1,x\rangle$, does there exist $n$ such that $p^n(x)\in\langle x\rangle$.

(Note that this is equivalent to asking whether $p(x)$ belongs to the radical of $I=\langle x\rangle$ whenever $p(x)$ is in $\langle 1,x\rangle$)

I see the equivalency

and no, I don't believe an n exists

you'll always have an x^0 term

or, you'll always have a "1" term

in $p^n(x)$

That's true. Now, to be precise enough about that, I recommend picking a counterexample.

Think.. what is the easiest choice of h_1 and h_2 that will work?

1 and 1?

Even simpler.. h_1=1 and h_2=0. :)

OK, let me recap what we have so far

$I=<x>$ and $J=<x>$

@KarlKronenfeld I think you're making your way to the multiples of $n$ example I was talking about earlier, by going a roundabout way to it.

thus, for the radical ideals, we have:

$\sqrt{I}=<1,x>$ and $\sqrt{J}=<1,x>$

@Axoren k[x] is basically Z (intrinsically), so of course the examples will look similar.

@daOnlyBG Well, that's your claim. :P

I'm just wondering why we didn't start with Z which seems like a much simpler ring to work with.

OK, so what would the radical ideals be for <x>?

@Axoren Because @daOnlyBG chose polynomial rings, no big deal. The simplicity of the ring is in the eye of the beholder anyway. They are basically the same when we're just playing with ideals.

@KarlKronenfeld Fair enough.

@daOnlyBG We have to look for a condition on $p(x)$ under which $p(x)^n$ is a multiple of $x$ for some $n$.

ahhh

That is the same as defining the radical of $\langle x\rangle$.

I see my mistake

simply let $p(x)=x$

What then?

well, for any $n$, $p^n(x)=x^n$

so I guess $\sqrt{I}=<{x}>$

Well, you came upon the right result. But I am not very satisfied with your reasoning. :P

I'm not sure what result I came across to be honest

The result: $\sqrt I=\langle x\rangle$.

I have a cool way of proving it, that will be less dry than a typical proof. It should still be easy to grasp.

well, $\sqrt{I}=<x>$ because by definition, $\sqrt{I}=f$ such that $f^n \in x$

use \{ \} for set delimiters

mmk

\{f : f^n\in\langle x\rangle\}

and \langle and \rangle for the < and >

by the way, I'm open to hearing your proof

I want to hear you out first.

Your proof, as it stands, is incomplete.

oh I know

one second

so $\sqrt{I} \sqrt{J}$ should equal $\langle x^2 \rangle$?

with angles around that x^2, I agree.

Here, I'll show you my proof that the radical of $I=\langle x\rangle$ is $I$.

A property of $I$ is that $I$ is the set of all polynomials $p(x)$ such that $p(0)=0$.

right

If $q(x)$ is in the radical of $I$, then $q^n(0)=0$ for some $n$.

So $q(0)=0$.

so $q(x)\in I$ in the first place.

Therefore, $\sqrt I\subseteq I$.

The other inclusion $I\subseteq\sqrt I$ holds for all ideals $I$.

very cool

That's a really nice proof, I'm starring that too.

OK

But, @daOnlyBG . You've investigated a single choice of $I$ and $J$, but they didn't disprove what you were trying to prove, right?

no they did not, unfortunately

...right? lol

I mean,

it's my understanding that $\sqrt{IJ}=\langle x^2 \rangle$ as well

What if you chose a different power to be "squaring"?

@daOnlyBG That is actually not true.

alright- I'm really unsure of what $\sqrt{IJ}$ is, then

You can probably guess what IJ is.

IJ is just the product of all the elements in I with all the elements in J?

It's a little bit more complicated than that in reality. But since $I$ and $J$ are principal (of the form $\langle p(x)\rangle$), you can actually say that.

A finite sum of those products.

In your previous example of $x^2$ and $x^3$, I'm pretty sure it would have been something like $\langle x^2, x^3, x^5 \rangle$

@KarlKronenfeld correct me if I'm wrong? I don't usually work with the ring of polynomials.

$\langle x^2\rangle \cdot \langle x^3\rangle$ is simply $\langle x^5\rangle$.

OK, so my counter example doesn't work

$\sum_{i=1}^nx^2p_i(x)\cdot x^3q_i(x)=\sum_{i=1}^nx^5p_i(x)q_i(x)=x^5\sum\text{blah}$

@daOnlyBG Don't lament. You only have established $IJ= \langle x^2\rangle$, which we will call $K$.

We still have to compute $\sqrt K$.

I see where my mistake was. I was somehow allowing $0 \in I, J$ when I've been looking at this.

Maybe overusing it when looking at it intuitively, because $0$ does belong to any ideal.

But root extraction and divisibility are pretty weird with $0$.

Not 0, 1.

Ah.

$\sum x^2 \times 1$

well, isn't $\sqrt{K} = K$?

I see @Axoren

@daOnlyBG I'd disagree.

How about $x$?

Oh, my bad- that theorem you gave earlier is true only for one ideal at a time, as opposed to the product of two ideals

however, the product of two ideals is a new ideal in and of itself, no?

so why doesn't it hold for K?

No, the proof applied to the specific ideal $I$ we have here and not necessarily $K$.

$I$ was $\langle x \rangle$ at the time.

@daOnlyBG Because $K$ is not the set of all $p(x)$ such that $p(0)=0$.

oh okay- I see now

alright, so let's compute $\sqrt{K}$

From that, you should know at least what $\sqrt K$ contains.

$\sqrt{K}$ should be any $f$ such that $f^n$ (for some n) is contained in $\langle x^2 \rangle$, right?

Yeah, as long as $f$ is a polynomial.

right

$\sqrt K$, rather than $K$

right

corrected

thus, $x \in \sqrt{K}$

yay

but $x \not\in \sqrt{I} \sqrt{J}$

So, then this is a counter example.

lol thanks for the formatting correction

and thank you very much for illuminating this topic

fun stuff

If he had picked $I = J = \langle x^2 \rangle$ it would have failed, am I correct in that?

No.

I mean in that it would have failed to disprove the question.

Not failed to be true.

Yep.

$\sqrt I=\sqrt J=\langle x\rangle$.

I was expecting that same thing to happen with $x$

$\sqrt {IJ}=\sqrt{\langle x^4\rangle}=\langle x\rangle$.

Are you sure that last one is right?

Yes.

Why doesn't $\sqrt{\langle x^4\rangle}=\langle x^2\rangle$?

$x^n\in\langle x^4\rangle$ with $n=4$ implies $x\in\sqrt{\langle x^4\rangle}$.

Well, what about $(x^2)^2$?

so $x^2\in\sqrt{\langle x^4\rangle}$ too

gentlemen, I'm going to trouble you with one more question, but only when you're ready. I don't believe it's as complicated

ok

Go ahead, I'm going to have to mull over radicals of ideals. When they're of things other than integers, I'm apparently careless with them.

OK, so here is the prompt I've been working with

Let me type it out..

"Let $f,g \in \mathbb{C}[x,y].$ Prove that $V(f,g)$ is finite if and only if $f,g$ have a non-constant common factor in $\mathbb{C}[x,y]$ by the following steps:

(a) prove that $V(f)$ is finite when $f$ is non-constant

(b) using (a), prove that if $f,g$ have a nonconstant common factor $h \in \mathbb{C}[x,y]$, then $V(f,g)$ is finite

Wait, $V(f)$ is the set of all $(x,y)$ such that $f(x,y)=0$, right?

yes

I also suspect the prof made a typo, as in part (c), I'd like to that we're supposing f and g do have a non-constant common factor

But $V(x)=\{(0,y):y\in\mathbb C\}$.

So part $a$ would be wrong. I suspect the prof made numerous typos.

Maybe infinite instead of finite.

for part (a) it could be that we're to assume that $f\in \mathbb{C}[x]$ only?

no way

Again, my counterexample came from $\mathbb C[x]$.

It really would be true if it said infinite

sorry- it did say infinite

yes, it's infinite

ok, where do you need help?

originally, I just needed help with (c), but now that I made a serious mistake, I might have to redo things on my own

I could try working on it on my own for a bit

if you guys are still here in a bit I will let you know how it's going

ok. I have stuff to do for like an hour, so feel free to @ ping me.

thanks Karl!

@robjohn What happened? No progress on that question? I still have no result :|