Mathematics

12:30 AM

So, an "absorber" is defined an element $a$ of a magma $M$ satisfying that $am=ma=a$ for all $m\in M$. Obviously, you can only have one absorber per operation on the object, but is there a concept of a "second order absorber?" Ie an element $b$ satisfying that $bm=mb=b$ for all $m\neq a$?

user76284

You mean an absorber that absorbs all but another absorber?

Rithaniel

Yes, almost exactly. I'd say give each absorber a "weight" and the one with the lowest weight wins when they are combined.

user76284

I don't expect there to be a name for that

Are you looking for an example?

Rithaniel

I'm wondering if it's a topic that anyone has studied, or if there are any interesting facts related to them.

Secret

1:05 AM

If an nth order absorber is two sides, it is also unique

however there isn't a name for these things because the algebraic structure is basically a magma with absorbers extended with some extra elements that does not get absorbed

you still cannot define division or in any way invert such absorbers though because absorbers often means there is some noninjectvity going on

A special cases where you have a hierarchy of "nth order absorbers" is a semilattice, where ab=b if max(a,b)=b

Konformist Liberal

2:00 AM

Hi! At end end of a proof I saw this:
Exercise. See if you can refrain from translating this proof into
cohomology. - Does this mean, see if you can think of a proof which doesn't use any machinery from cohomology directly?

Akiva Weinberger

5:46 AM

@Rithaniel Kinda like how addition works with infinite cardinals?

(Actually - does $X+X=X$ rely on the axiom of choice?)

(I suppose it doesn't work for Dedekind finite sets...)

(And $\aleph_\alpha+\aleph_\alpha=\aleph_\alpha$ seems like a consequence of how ordinals work.)

Revision: Kinda like how addition works with aleph cardinals?

user76284

6:01 AM

en.m.wikipedia.org/wiki/Cardinal_number#Cardinal_addition says “Assuming the axiom of choice”

Prashin Jeevaganth

9:18 AM

Does this Math chat answer statistics questions?

Akiva Weinberger

9:59 AM

> "The essence of calculus, perhaps, is that an infinitely accurate approximation is no longer an approximation."

Leaky Nun

10:19 AM

@PrashinJeevaganth yes

@AkivaWeinberger inb4 Fabius function

Akiva Weinberger

@PrashinJeevaganth It depends on whether there's anyone around who knows statistics

which could just depend on who's online

Prashin Jeevaganth

11:52 AM

Hmm, http://itarget.com.br/newclients/sbgg.com.br/informativos/28-12-18/5%20Exercise%20type%20and%20activities%20of%20daily%20living%20disability.pdf

This is a research article that talks about finding an association between exercise and disability. I was thinking whether if u reject a null hypothesis: There is no association between exercise and disability for a given population, and perform a 1 tail test(idk if I'm right judging from the context of this article Figure 3), so your alternative hypothesis will be There is a negative association between exercise and disability. If we reject …

Silent

Determine all ring elements that are both nilpotent elements and
idempotents.

I think only such is zero element

because $a^n=a$ for all $n>0$. Am i right?

Leaky Nun

12:18 PM

yes

Silent

thank you! and on the same line of reasoning, we see that 1 is the only element that is both unit as well as idempotent in commutative ring with unity?

right?

Konformist Liberal

12:47 PM

Hi! At end end of a proof I saw this:
Exercise. See if you can refrain from translating this proof into
cohomology. - Does this mean, see if you can think of a proof which doesn't use any machinery from cohomology directly?

Leaky Nun

1:00 PM

@Silent yes

@KonformistLiberal yes

Leaky Nun

1:17 PM

reddit is down :c

Konformist Liberal

@LeakyNun But it says refain from translating, not try to translate it

Leaky Nun

well more context would help

or you could just follow your heart

user193319

1:38 PM

Elements in $\Bbb{Q}[\sqrt{2},\sqrt{3}]$ are of the form $\sum_{i,j=1}^k a_{ij} (\sqrt{2})^{n_i} (\sqrt{3})^{m_j}$ for $a_{ij} \in \Bbb{Q}$ and $n_i, m_j \in \Bbb{N}_0$, right?

I'm trying to show that $1,\sqrt{2},\sqrt{3}, \sqrt{6}$ span the space.

Leaky Nun

yes

user193319

2:11 PM

How do I argue that $\sum_{i,j=0}^k a_{ij} (\sqrt{2})^i (\sqrt{3})^j$ is spanned by $1, \sqrt{2}, \sqrt{3}, \sqrt{6}$? I can split the sum like $\sum_{i,j=0}^k a_{ij} (\sqrt{2})^i (\sqrt{3})^j = \sum_{i=0}^k a_{ii} \sqrt{6}^i + \sum_{i \neq j}^k a_{ij} (\sqrt{2})^i (\sqrt{3})^j$, but I don't think it makes things any clearer.

The h Bar

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$Mathematics$ Mathematics