Mathematics

skullpatrol

12:01 AM

force of habit, i guess

old habits are the hardest to break :-)

Ted Shifrin

You're about the only regular here whose math knowledge I know basically zero about.

ÍgjøgnumMeg

12:15 AM

Is it common for people in grad school to attend undergraduate courses to flesh out their background?

N. Maneesh

12:37 AM

@SubhasisBiswas Thanks

manooooh

1:01 AM

Hello!

How can we deduce that if $\gcd(n,7)=1$ then $n^{6}\equiv1\pmod{7}$?

Is it because of Fermat Little's Theorem?

Thanks!

Ted Shifrin

2:09 AM

That's precisely Fermat's little theorem, @manooooh.

manooooh

3:53 AM

@TedShifrin good to know. Thank you Ted!

user76284

5:32 AM

plato.stanford.edu/entries/quine-nf/#4 says "extensionality cannot be defined in a language without equality", but doesn't this contradict en.wikipedia.org/wiki/…?

RaphX

7:03 AM

Hi guys

I have a doubt regarding matrices

r′=r−s^(r.e^3)/s3

How will I write that as a vector?

By the way its not s to the power of or e to the power of , they are s cap and e cap respectively

e cap is a basis vector

christmas_light

7:33 AM

hey chat, if someone of you knows a bit homogeneous space and want to earn a bounty, please give a look at this question, it would help me alot
https://math.stackexchange.com/questions/3234925/understanding-projection-of-vector-field-in-homgeneous-spaces

Secret

10:24 AM

May 11 at 16:52, by Tobias Kildetoft

If you lecture long into the void, the void starts lecturing back.

The mystery of the lecturing void continues

May 11 at 16:55, by Ted Shifrin

@anakhro: Generally lack of responsiveness, lack of interaction. With too small a class, people are quieter, in general.

Hmm...

I want to create a mathematical object that behave somewhat like this

However, it is not clear what context needs to be abstractised to produce said object

A "void lecture" ,assuming Ted does capture most of the gist, is one where you transmit information into it, and getting only total silence as a response

Thus one way to abstract this is the zero map, which is trivial

What if, you can produce "a zero with intricate structure", abstracted from how in most practical case, what seemed silent is full of implied information in nonverbal form

and said information may be recovered given the correct tools

One of the easiest nontrivial way to imbue a structure on an element that otherwise behave like zero is to have a semigroup where there is a subsemilattice. Then you have elements that behave like an identity except for a few elements.

Likewise, nth order absorbers are also straightforward to implement

Apr 4 at 0:30, by Rithaniel

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 $m\neq a$?

In addition, suppose we have a strange zero function $\mathscr{o}$ such that:

$$\int_a^b \mathscr{o} dx > 0$$

but $\mathscr{o} + f = f$ for some functions $f$, then we will often end up with divergence issues e.g.:

$$\int_a^b f dx = \int_a^b \mathscr{o} + \mathscr{o} + \cdots + f dx$$

Luke

1:55 PM

Hey folks, how do you remember that $f(A\cap B) \subseteq f(A)\cap f(B)$ is the only operation where in general we don't have equality? I always know there's one thing prohibiting the image from being a lattice homomorphism, but I keep forgetting which one.

Karl Kronenfeld

2:59 PM

@Luke: one way to think of the equality is as a generalization of the definition of injective functions. Indeed, the definition of injective functions is literally the same as $f(\{a\}\cap\{b\})=f(\{a\})\cap f(\{b\})$.

Balarka Sen

think of two horizontal disjoint disks at different heights in 3d getting floopsquished into a venn diagram in 2d by projection

taking that as f makes those set theoretic equalities/inequalities obvious for me at least

Luke

3:33 PM

@KarlKronenfeld ah, nice, this is even more minimal than the example I came up with. Thank you!

user131753

@Luke Why not simply use the definition of $f(A)\cap f(B)$?

user131753

@Luke In any case, another way is $\emptyset =f(A\cap A^c)\subseteq f(A)\cap f(A^c)$, so the reverse inclusion does not hold in general.

Luke

@user170039 The problem is not „understanding it on a deeper level“, I just needed a minimal example or mnemonic so I get the correct fact with minimal effort / distraction when proving other things.

Leaky Nun

4:20 PM

euclidea.xyz

fun game

Transcript for

$Mathematics$ Mathematics