Mathematics

8:00 PM

I'm just saying: why aren't we looking at some fancy function $g(d(f(x))$ to get our norm

where $d$ is our metric

It might help to say this formally. So assume there were a norm $\|\cdot \|$ which induces the metric $d$. Then $d(x,y) = \|x-y\|$. So $d(x,0) = \|x-0\| = \|x\|$

MatheinBoulomenos

if $\|.\|$ is a norm, then $d(x,y)=\frac{\|x-y\|}{1+\|x-y\|}$ is a bounded metric

Daminark

So $d(x,0)$ must succeed

MatheinBoulomenos

but that is not usually considered the metric induced by the norm

Sha Vuklia

@MatheinBoulomenos I meant it the other way around: if we have a bounded metric, then we can't get a norm out of that

Daminark

8:03 PM

I guessed that your question was less philosophical/convention-based and more formal. If you were just asking why we define the metric induced by a norm the way we do, the answer is that it's kinda the natural thing to do in the contexts where everything is clear so we rolled with it

Sha Vuklia

I understand why we define the metric induced by the norm the way we do, because it works

but the point is that the other way around it fails

and I don't see why we have to look at $\Vert x\Vert = d(x,0)$ to say that it fails

Daminark

Ah yeah in that case check what I wrote. If d is induced by a norm, then my computation shows that d(x,0) is the norm

Sha Vuklia

btw, in all honesty: my head is spinning because of "induced by" and "induces" :p

I'm constantly messing up the direction

@Daminark lemme check

user193319

If you aren't in a vector space, what does $d(x,0)$ even mean?

Not all metric spaces are vector spaces.

Daminark

@user193319 the question was, if you're in a metric space that's also a vector space, why does it suffice to look at d(x,0) in order to determine that this metric is/isn't induced by a norm?

Rick

8:07 PM

Is there a tutorial or book I can use a reference.

user193319

Ah, okay. Sorry.

topologicalmagician

Guys

How do I show that if the limits as $x-x_o$ approaches 0 then $(x-x_0)$ approaches dx implies that $(dx)^2$=0?

Rick

for Mathematical notation and it's proper use

topologicalmagician

by dx I mean differential distance

it seems obvious

Daminark

Wut

2

Are you working when hyperreal stuff?

*with

If not please do not use dx for this

topologicalmagician

8:12 PM

We were taylor expanding vector quantities, and the professor described dx as the infinitesimal distance, and that we should neglect the higher order terms (dx)^n because they are zero. I understand that, but I wanted to prove it rigorously

Sha Vuklia

@Daminark but here you are assuming that we found a norm that induces the metric $d$ in a specific way

topologicalmagician

she said that that $\delta x$ approaching 0 is dx

Daminark

There's no rigor with infinitesimals unless you go way into a dark rabbit hole filled with drugs

Sha Vuklia

maybe it wasn't possible to find a norm that induced $d$ in that way, but maybe there was a norm that induced $d$ in a different way

Daminark

@ShaVuklia remember that our definition of "induces" is to induce in that particular way

Sha Vuklia

8:14 PM

oh lol, right, that was my question in the first place:p

Daminark

So if you can write a metric as some crazy function of a norm, that doesn't qualify

Sha Vuklia

whether we meant in specifically like that or more generally as a composition

alright, well then it's clear i guess

topologicalmagician

@Daminark but I thought math was all about rigour

Daminark

@topologicalmagician the rigorous way to do this is to not talk about infinitesimals, or to start thinking about hyperreal numbers

But I feel like unless you have a particularly vested interest in learning about the hyperreals, they're not especially useful

So for the generic person my answer is to just stop using infinitesimals

topologicalmagician

but its not me thats focusing on infinitesimals, anyways if I wanted to prove it through the hyperreals, how would I prove it?

Daminark

8:18 PM

I dunno, hyperreals are kinda weird. It's not you focusing on infinitesimals, it's your professor who's doing Taylor's theorem in a shitty way

topologicalmagician

LOL

MatheinBoulomenos

@topologicalmagician you should see such a derivation just as a heuristic (cause that's what it is), if you want to prove the particular statement rigorously, it's probably easier to think of a proof using the limit definition of derivatives than delving into hyperreals

Daminark

My advice is that in class just roll with what your professor is doing, if you're interested look up hyperreals and see how they're done, and if you're not (most analysts/mathematicians probably aren't) then you can learn stuff like delta-epsilon

Yeah as Mathein said

Curio

I have no idea about how to find k if I know that k+1, 2k+1, 3k+1 and 4k+1 are prime numbers

@Daminark I found the number using a java programme, but I should get it via only math

Should I use congruences?

Modular arithmetic I mean

topologicalmagician

@Curio in modulo what would you be working with?

Mike Miller

8:31 PM

it is quick to check that if those are all prime numbers, k must be 0 mod 2, 3, and 5. furthermore, k must be 0, 1, or 4 mod 7

a list of the smallest numbers for which that is true are 60, 120, 210, 270, 330

the first four do not work but the fifth does

the smallest such k is 330. i have no reason to believe there are not more such k.

Jossie Calderon

How can I generally find $b, k : b^2 - 1 = 12k$?

topologicalmagician

over the integers,youre essentially asking when is b its inverse modulo 12

@JossieCalderon

Jossie Calderon

8:47 PM

Yes

topologicalmagician

that is the case when gcd(b,12) =1

Jossie Calderon

I've concluded it would have to be a representative of the reduced residue class of 12

Thanks for the input!

topologicalmagician

In infinitesimal analysis, how would I prove that differential distance is an nilpotent infinitesimal?

I meant

In infinitesimal, nonstandard analysis, or synthetic differential geometry, how would I prove that differential distance is an nilpotent infinitesimal?

@JossieCalderon no problem

Lozansky

9:29 PM

If $H = \langle xyx^{-1}y^{-1} : x,y \in G \rangle$ for some group $G$, how can I prove it's normal?

My approach right now is to pick an element $g \in G$ and show that $g xyx^{-1}y^{-1}g^{-1} \in H$

Astyx

Hi chat

It seems I barely missed @ShaVuklia :/

@Lozansky That works

Curio

@MikeMiller why that? I mean, how did you get it?

Lozansky

@Astyx Could you give a hint how to rewrite the product?

Curio

Thanks

Astyx

$x\mapsto gxg^{-1}$ is a group morphism

Your trying to find the image of $xyx^{-1}y^{-1}$ by this morphism

Lozansky

9:37 PM

@Astyx Are you saying, I just have to show it is a group morphism and I am done?

Astyx

Well, am I ?

If that makes sense to you sure

But does it ?

Lozansky

Yeah I think it does

Astyx

Explain it to me ?

Mike Miller

@Curio You should be able to verify everything I said yourself. For the congruences all I used is that each of your terms is prime. For the examples I checked by hand.

Lozansky

Okay so let $\phi : x \mapsto gxg^{-1}$, then $\phi(xy) = gxyg^{-1} = gxg^{-1}gyg^{-1} = \phi(x) \phi(y)$ so it's a group homomorphism

Mike Miller

9:42 PM

I will leave the details to you.

Lozansky

And if $\phi(x) = \phi(y)$ then $gxg^{-1} = gyg^{-1} \Rightarrow x = y$ so it's injective

Astyx

You don't really need to show it's an isomorphism, although it is

Lozansky

Ah okay

Astyx