Mathematics

4:09 PM

@DanielFischer Do you have any new thoughts about what we could possibly be overlooking about the parameter $a$? The $2 \pi$ restriction seems to apply to the entire family of infinite series.

ALannister

Hey guys - having trouble with this problem:

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Q: Prove that the ring $R$ has a left and right identity - missing a key detail?

Suppose that $R$ is a finite non-zero ring without zero divisors, and denote $R^{*}=R\setminus \{0 \}$. Consider the action of the multiplicative semigroup $R^{*}$ on itself by left translation; i.e., for $x \in R^{*}$, we have $x.y = xy$ by definition. I just showed that $R^{*}$ acts on $R^{*}$...

Daniel Fischer

@RandomVariable No, sorry. Haven't yet found the time to think about it more than cursorily.

Semiclassical

What was the numerical evidence for said restriction? @RandomVariable

ALannister

NM - figured it out.

Random Variable

4:26 PM

@Semiclassical Numerical approximations for the sums using Wolfram Alpha for values of $a$ greater than $2 \pi$. But the formulas themselves suggest that a restriction of some sort is needed since they don't make sense for large $a$.

Semiclassical

Hmm.

BAYMAX

Guys any idea about this situation - "I want to read the book online from google books but for me internet service will not be there tomorrow , any technique how i can access the pages while offline..."

Semiclassical

I'm not optimistic there's a workaround for that.

ALannister

Okay, didn't figure it out. Showed something similar, but not exactly. But, that question is now deleted. Need to post a new one...

DHMO

4:48 PM

@Vrouvrou j'ai deja prouve ici:

yesterday, by DHMO

Let me summarize:
Let $d \in D$ and $f = d(d,B) - d(d,A)$. $f > 0$ by definition.
consider $Y$ an open-ball centered at $d$ with radius $\frac f 2$
we need to prove that $Y \subseteq D$
let $y \in Y$
we need to prove that $d(y,B) > d(y,A)$
so $d(y,A) < d(d,A) + d(d,y) < d(d,A) + \frac f 2 = d(d,B) - \frac f 2 < d(d,B) - d(d,y) < d(y,B)$
QED

@Mahmoud congratulations!

> Every increasing sequence in the long line converges to a limit.

What the hell

I'm quite interest in knowing that $(n,\sin n)$ converges to

Semiclassical

Ugh, it's too early in the day for me to be feeling tired.

DHMO

Actually, I've figured it out.

It converges to $(\omega,0)$.

now it's getting interesting

ALannister

5:03 PM

New question:

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Q: Need to prove that a given nonzero ring $R$ with no zero divisors has both left and right identity

Let $R$ be a finite non-zero ring without zero divisors. Denote $R^{*}=R \setminus \{0\}$. Consider the action of the multiplicative semigroup $R^{*}$ on itseslf by left translation - i.e., for $x \in R^{*}$, we have $x.y = xy$ by definition. I already proved that $R^{*}$ acts on $R^{*}$ by bije...

abstract-algebra ring-theory group-actions

DHMO

@Vrouvrou au revoir

Semiclassical

Probably far too trivial a thought, but $1$ is always in $R^{ * }$ so it's always in the image of $R^{ * }$ acting on $R^{ * }$ from right or left.

Vrouvrou

@DHMO but i don't understand $d(d,A) + \frac f 2 = d(d,B) - \frac f 2 < d(d,B) - d(d,y) < d(y,B)$

Semiclassical

But if that map is a bijection, doesn't that imply that there's always a preimage of 1 w/r/t that mapping and therefore inverses? @ALannister

ALannister

$1$ would already have to be in $R^{*}$

I don't think there's any guarantee of that

Semiclassical

5:11 PM

Isn't $R^{ * }$ just everything in $R$ except $0$?

ALannister

It's not the real numbers, it's a ring.

Semiclassical

Sure, but every ring has a multiplicative identity. (that's what I mean by 1.)

ALannister

No it doesn't!!

Every ring is associative under multiplication and that's it.

Semiclassical

Per Wikipedia, one of the properties of a ring is that it's "a monoid under multiplication"

SteamyRoot

It depends on definition

Semiclassical

5:13 PM

Ugh, I was wondering if that's what was going on.

ALannister

sighs I suppose there is some argument among algebraists about that. But, as far as I'm concerned, to most human people, a ring is a semigroup under multiplication, not a monoid.

SteamyRoot

Though some people call a ring without multiplicative identity, a "rng"

whoa

ALannister

looks all innocent

SteamyRoot

Hah.

Semiclassical

I was on a different tab when that was up, so I'm in the dark.

Aaanyways.

SteamyRoot

5:14 PM

Good thing I only really work on groups and not rings :P

ALannister

Meh. Jessy was naughty.

Semiclassical

I guess the point is that it's straightforward for rings with identity.

ALannister

Yes, it would be.

Ugh. Only 4 people have even looked at my question!! I wanted to get this done this afternoon, but I guess that's not going to happen now.

I have to get a shower and go get ready for class.

Alessandro Codenotti

Hi chat

Sha Vuklia

hi

Alessandro Codenotti

5:33 PM

@Balarka is it true that every compact manifold is a CW-complex? We saw in class that this works for compact surfaces so I was wondering about the general case. I think you can cover it with a finite number of open $n$-balls and then do some fiddling with the borders where they overlap but I'm not sure how to formalize this last part (assuming it's correct)

Mike Miller

5:45 PM

@AlessandroCodenotti Homotopy equivalent to, yes. Homeomorphic to, open.

For smooth manifolds you're always triangulable so necessarily homeomorphic to a CW complex. But there are nasty topological 4-manifolds where things could go wrong.

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$Mathematics$ Mathematics