today i saw a question online, which was, if a number divided by a1 ,a2, a3 leaves the same remainder in each case, find the number. The answer is easy, but the author used a trick, he wrote the number is H.C.F (a1-a2 ,a2-a3, a3-a1), why is this true?
No, there are different kinds of symmetry. $y=x^3$ is symmetric about the origin. Oh, I see .... about the line $x=k$. That would require reflective symmetry, which an odd degree polynomial clearly cannot have.
Now, he knew immediately that I used inspect element, but someone else who was with me after I did my work didn't realize that and was like, Amin why?!
guys, I'm sorry to interrupt, but please help, do you know any other authors like Terence Tao that you can recommend? I'm searching for those who presents material (like proofs, key ideas, big picture, new definitions) in a clear and rigorous way
Also I created a response from the professor saying something like, thanks for reminding me about the pset but this is bad email etiquette so I'd recommend not making a habit out of this
@TedShifrin perhaps my puns are ahead of their time but I don't think I spend my time doing things which aren't appropriate
@Shobhit: Now that you stated the problem carefully and correctly, is it obvious to you that the mystery number $m$ must divide all of their pairwise differences? It should be.
I mean can we even classify this as a nerve problem?
(At least I tried)
Okay wait no much as some folk want to paint me as bottom of the barrel, I have standards and this was a quality lapse, why do you star my worst jokes?
Let $E$ be any real bundle over $X$. I have somehow produced a 'proof' that $2p_1(E)=0$:
By the splitting principle, we can write $E=E_1 \oplus ... \oplus E_n$ where $E_i$ are real line bundles. Then $p_1(E)=c_2(E \otimes \mathbb{C})=c_2(E_1 \otimes \mathbb{C} \oplus ... \oplus E_n \otimes \math...
Which can, if only explain a fraction of what's there, at least start an interesting conversation. Depends on how much time we're prepared to put on it
I mean i think ELI5 posts greatly overestimate the ability of 5 year olds
Usually the correct response to "ELI5 ____" is to tell the kid to play tag and that they'll learn it in 10-15 years, likely upon asking an ELI5 question
having a zero at $z_0$ means that the first coefficient in the taylor series vanishes. Because the derivative has also a zero at $z_0$, the second coefficient vanishes, too
here in Vizarain's textbook about approximation algorithm: Chapter 3, it is about proving the follwoing lemma: let V' be subset of V, such that |V'| is even, and let M be a minimum cost perfect matching on V', Then cost(M) <= 1/2 (OPT)
OPT = tour of TSP
TSP is Travling salseman problem
TSP in NP-complete, and we try to give an approximation algorithm for this problem in metric case.
@BalarkaSen I think it might be the meme of pullback vs. just evaluating at the new point. Like, the result I get is that composing Green's function w/ an orientation reversing isometry gives the negative of the Green's function, which is clearly absurd.
What's the difference between $\mathbb{R}^{2\times 2}$ and $ \mathbb{R}^{4}$? The context is that I have a 2x2 matrix manifold on this space. Are those spaces equivalent or is there some deeper meaning?
"matrix manifold" I mean a matrix lie group that can be described as a manifold on the above spaces.
@Skortya: The vector space of $2\times 2$ matrices is isomorphic to $\Bbb R^4$, of course, but matrices have some structure (e.g., matrix multiplication) you can't give on $\Bbb R^4$. So it's important to know what we (you) are talking about.
Quite well, I have a question for you since you've been a professor for a lot of time, are you familiar with Miles Reid undergraduate commutative algebra book?
I see, because he wrote he had to leave out some topics from the book to fit into a 30 hours undergrad course, but it seems to me one would have to work through the book at a crazy speed to cover it in a 30 hour course
It's interesting that some people have commented on my YouTube lectures about how slowly it all goes ... but these are advanced students who've forgotten that I was teaching it all to people learning it for the first time.
Ah, @Mathei do you happen to have an example of an integral domain in which all irreducibles are primes yet isn't an UFD? I was thinking about something like $\Bbb Z[\{2^{\frac{1}{2^n}}:n\in\Bbb N\}]$ but I'm not sure
Every element has a square root, because if $a^2=b$ and $b$ is an algebraic integer, then so is $a$ (you can write down a polynomial for $a$ explicitly)
So we have $b=a^2$, if $a$ is a unit, then $b$ is a unit, if $a$ is not a unit, $b$ is not irreducible
I think the ring of entire functions also qualifies. Iirc the only irreducible elements up to a unit are of the form $z+a$ for $a \in \Bbb C$ and these are also prime. Not sure how to show that, though
Okay, that follows from the Weierstraß factorization theorem
you can always construct entire functions with zeroes as you like (as long as they are discret) and $f \mid g$ in the ring of entire functions iff the order of zero for $f$ is $\leq$ than the order of zero for $g$ at every point
From there it follows that every irreducible element is associate to $z+a$
$\mathcal{O}(\Bbb C)/(z+a) \cong \Bbb C$ by the first homomorphism theorem, so $(z+a)$ is maximal, thus $z+a$ is prime
if you take any entire function with infinitely many zeroes, then you can't factor that as a finite product of irreducibles
The example of entire functions can probably be generalized to holomorphic functions on domains in $\Bbb C$, but I'm not sure if Weierstraß factorization works the same way for that (I think it does, but the proof is harder? not quite sure)
Right, we looked at some implications between "every (nonzero blablabla) element has a factorization into irreducibles", "if an element has a factorization then it is unique", "irreducibles are prime", "the ring satisfies ACC on principal ideals"
I spent some time exploring pretty obscure corners of ring theory and I still never heard of half of those properties
and as I said, if you've never done noncommutative algebra, then you won't encounter a lot of those properties even if you're an expert in commutative algebra
There are more important things than knowing a lot of properties that a ring can and can't have and sometimes one doesn't know the word although one has worked with the property itself
