12:39 AM
@johnny09 The entries are individually unbounded on $\Bbb R^2$. The operator norm is probably yucky to compute.

@TedShifrin good point! out of curiosity, is it sufficient to simply look at the entries?

Well, can you show that all norms on $2\times 2$ matrices are equivalent?
I don't know how your actual question is arising.

i am studying nonlinear systems, and my question arises from a result that ensures that the state equation is globally lipschitz if and only if the jacobian is uniformly bounded

Ah, so the operator norm is most natural for that. But can you show that if the sup norm is unbounded, then the operator norm must be too?

in general show this or specifically to an example?

12:50 AM
Probably in general is easier :P
Then you don't have to compute norms explicitly.

interesting! i'll look in to that (can't give you an answer right now)
but thanks for the help!

That's fine. I wasn't asking for an answer. Just giving you guidance.

i know :) thanks again

Sometimes my guidance is faulty. :P
You're welcome.

@TedShifrin That's where you're wrong. Your guidance is NEVER faulty

12:52 AM
Nah, I have made my share of stooopid mistakes — including on main.
It's just that some other people make more than their share :P

Well, if nothing else, you are indeed an excellent source of guidance, Ted.

Thanks.

@TedShifrin im gonna dedicate my whole summer working on your book. But I also want to focus on diff geo and topology alongside your book, what resources would you recommend?

LOL, for diff geo, you can download my free book.

ok, i'll do that
for topology?

12:58 AM
For thorough point-set topology, I'm fond of Munkres. I also like Simmons's book on topology and analysis.
BTW, @topologicalmagician, working lots of exercises/writing proofs is more significant than just reading texts. That's why I am not a fan of "self-learning" — one needs feedback.

@TedShifrin, yeah that's why im on this site. To get as much feedback as I can. I also am trying to work on as many problems as I can in your book

There are a lot of proof problems as well as elementary and challenging computations.
I'm a big believer in being able to do challenging computations too :) Lots of folks on this site don't believe in computation.

hi chat

Oh no! It's Joe Shmo.

oh c'mon. i expect a more cheerful greeting than that

1:06 AM
Got to go
have a good day :)

Bye @topologicalmagician.

@TedShifrin take care

I was perfectly cheerful ...

suppose $\{f_j\}_{1 \le j \le n}$ are linearly independent functions in $[-1, 1]$ with the inner product defined by $(f, g) = \int_{-1}^1 fg dx$. Let $\{f_j\}_{1 \le j \le m}$ be even, $\{f_j\}_{m+1 \le j \le n}$ be odd
now comes the kicker

Then they're orthogonal.

1:09 AM
good evening

yes

hi @CaptainAMerica
Long time no see.

they are orthogonal

yeah, i'm on spring break

How many digits or limbs have you broken?
I cut off a finger ... what did you do in the meantime? :P

1:10 AM
wait...wat?

Yeah, almost a month ago I cut off the end of a finger while I was prepping and cooking. I do that every once in a while.

let the "matrix" $A = [f_1\ f_2\ \ldots\ f_n]$ be of dimension $[-1,1] \times n$. So $A: \mathbb{R}^n \rightarrow X$ is a linear operator

Despite being a semi-expert chef.

geez
this is why i don't cook

is this an ok matrix?

1:11 AM
I think you are a disaster without cooking. But you should learn to pay attention.
What are you doing @JoeShmo?

Eh...i'm working on it

No, matrices are discrete.

i didn't realize my attention span was actual problem until recently

i am copying over the question almost verbatim.
no, exactly verbatim
I know, I have no idea what he's talking about

Uh huh. What is $X$?
And how did we lose the easy question so fast?

1:13 AM
I have about a month and a half of school left
trying to get through finals
but i suck at studying

Oh, I'm shocked.

I have no idea.. this guy isn't a mathematician. he invents stuff as he goes and tries to convince us it's OK

The orthogonality question was trivial. Then you write this s**t.

I know.. I don't even know what he's asking
then he goes on to claim that $A$ has an SVD
because its a regular linear operator (HA HA!)

1:15 AM
I'm trying to plan out my math for the summer. My guidance counselor said she thinks I should do independent studies for the rest of high school because of the progress I've shown on my own time.

I don't even know what the codomain is

She's hoping I can cover a lot of college math now.

@CaptainAmerica: Save real learning for college, but, yes, I've given you plenty of guidance on stuff to do. Working on Spivak for a year would be great for you.

Yeah, I want to finish spivak
this school year was stupid

Finishing Spivak is a serious long-term commitment.

1:17 AM
i know. I'm taking a lighter course-load next year and my math will be independent. so spivak will literally be part of my school day
depending on how much i do over the summer

OK, have the school system pay me for my time helping you :P

lol, your more fun than my actual math teacher
*you're
she's always taking points off for stupid stuff

your math teacher likely doesn't know math

I wonder if JoeShmo is suggesting that I likely know some math :P

no

1:20 AM
OK, good.

i dunno, she has a Dr. in front of her name. maybe that counts for something

Probably a Dr. in Education.

JK. Orders of magnitude more math than I know, FWIW

ah, ok

nasty and cynical

1:20 AM
:)

i'm so bored
stuck at home over break

But you still don't pay attention even when it isn't boring

Dr. in nonense

everything is boring, tbh

I resemble that remark.

1:21 AM
wdym
are u saying you're boring?

No, I'm suggesting that I'm not.
resemble = resent

i still suck at english...

quick silly question: but how do we show that $||x|-|y||\leq|x-y|$ for all $x,y\in\mathbb R$?

No, you suck at humor.

yeah, that too

1:22 AM
@johnny09 you got the inequality backwards

It's true in normed vector spaces, @johnny09. Write $x=y+h$ and use the usual triangle inequality.

I always interpret "I resemble that remark" as "That sounds like me."

or no

It's right.

you confused me
you conned me

1:23 AM
I think i'm going to make a low B for english. My teacher says my papers "aren't detailed" enough and I think i've given up at this point

That's called the reverse triangle inequality.

same @Rithaniel

@TedShifrin and @JoeShmo Thanks a lot!

So how has life been for you Ted? Besides you cutting off the tip of your finger>

Let's try this a different way, for my own sanity. Suppose $X_1, X_2 \subset \mathbb{R}^n$, with $X_1 \perp X_2, AX_1 \subset X_1, AX_2 \subset X_2$. Suppose $A:\mathbb{R}^n \rightarrow \mathbb{R}^n$ is bijective. Is there a basis under which $A \sim \begin{pmatrix} A \restriction_{X_1} & 0 \\ 0 & A \restriction_{X_2} \end{pmatrix}$?
Answer: I think so. orthogonolize the bases for $X_1, X_2$ and there you go

1:26 AM
You don't need orthogonal bases for the subspaces, @JoeShmo.
Any old bases will do.

OK. Good.

Pretty good, @CaptainAmerica. Life is more interesting these days. And my AoPS teaching is almost done.

because $X_1 \perp X_2$ are orthogonal under the current basis correct? w.r.t the standard inner product

Nice. have anything planned for summer?

Half the time I see questions being asked in here and I don't even know what the words or symbols on the screen necessarily mean. I need to get past that.

1:28 AM
You don't even need orthogonality. All you need is complementary subspaces. If they are invariant, any basis of $X_1$ and any basis of $X_2$ give you such a matrix.
@Rithaniel: Often I don't understand, either, if that makes you feel better.

Fair enough.

hey me either, if that helps

sorry, invariant is not what i wanted. suppose $X_1 \oplus X_2 = \mathbb{R}^n$

Sometimes there's stuff I just know nothing about. Sometimes people write garbage.

instead

1:29 AM
Perhaps I should try my hand at answering questions on the proper website.

@JoeShmo: That matrix form precisely requires invariance.
That's how you learn to improve, @Rithaniel.

To be fair, it's easy to write garbage when you haven't put enough thought into the question to begin with. (But again, that leads to learning to improve)

If you want to write "finished rough drafts" and ask for us to take a look, that's fine if we have time, @Rithaniel.
In general, it's best not to post answers that are wrong, although people do do that frequently.

Ah, yeah, that's what has kept me from posting for the most part. The fear that I'll say something demonstrably incorrect, and that the person I'm answering would take it as fact.

Well, as I said, you can run the occasional one by us.

1:34 AM
without using the definition, How do I prove that $\frac{1}{2nx+1}$ is not uniformly continuous on $(0,1)$. $\frac{1}{2nx+1}$ converges to a constant function, If that limit is a discontinuous function, I could have manipulated that.

@N.Maneesh: Once again (I think it was you) you're confusing uniform convergence with uniform continuity. It's about a sequence of functions. Each individual function is not uniformly continuous, but the rest of your words suggest you're talking about uniform convergence. Sigh.

@TedShifrin sorry! I will correct my notations.
without using the definition, How do I prove that $\{\frac{1}{2nx+1}\}_{\mathbb N}$ is not uniformly continuous on $(0,1)$. $\{\frac{1}{2nx+1}\}$ converges to a constant function, If that limit is a discontinuous function, I could have manipulated that.

No, you haven't corrected anything.
You're talking about convergence of a sequence of functions $f_n$. And you're talking about uniform continuity of an individual function $f_n$. Which one is it?

uniform convergence of sequence of function.

So rewrite ... it has NOTHING to do with uniform continuity, as I said.
DogAteMy: IT's the middle of the night.

1:42 AM
without using the definition, How do I prove that sequence of function $\frac{1}{2nx+1}$ is not uniformly converging on $(0,1)$. $\{\frac{1}{2nx+1}\}$ converges to a constant function, If that limit is a discontinuous function, I could have manipulated that.

Ted, is it ever useful that two subspaces are $\perp$ for the representation of the matrix?

Correct.
So does each graph stay within $\epsilon$ of the graph of the limit function if you make $n$ large enough?
@JoeShmo: Only if you care about orthogonal (unitary) linear maps.

@TedShifrin Should I draw graph?

Yes, @N.Maneesh.
Seeing what's going on in a picture is usually very illuminating.

Hello.
Giving the sequence of natural numbers: 6,7,8... I am trying to find what number is on the position 1997, without counting palindromic numbers. Any hint?
i.e the sequence is like: 6,7,8,9,10,12...
Notice 11 is missing since it is a palindromic number

1:48 AM
You obviously need a way to count all the palindromic numbers in that range.

Yeah...

Question from curiosity: Why do you start at 6?

Random :)

@Rithaniel: You have to start somewhere.

True, but position 1997 changes depending on where you start.

1:50 AM
Yes, yes.

@TedShifrin I'm back in New York briefly!
Until next Monday

Ohhh, DogAteMy. Happy Pesach.

^_^ Thanks

Happy Pesach to all!

Chag saméach

1:51 AM
חג שמח!

LOL, it was funnier with "chat."

happy chatting

Anyhow, @JoeShmo: Have I clarified the linear algebra situation?

@Odestheory12 Isn't 6 a palindrome?

I dunno what to do with 1-digit numbers.

1:52 AM
:)

not this time, unfortunately. i have no clue what he's asking

Presumably palindromes need more than 1 digit.
@JoeShmo: As a rule, it's best to ask the instructor.

he doesn't know himself

ROFL.
Not my problem.

i know.. thanks for the help, as always

1:53 AM

I guess I have to suppose 1-9 aren't palindromic.

There you go, @N.Maneesh. Now you're done.

for every n function doesnot completely lie inside any $\epsilon-$ neighbourhood of $f(x)=0$. So, Not uniformly continuous

Right, @N.Maneesh
@N.Maneesh: Well, that's a correct statement when $0<\epsilon<1$.

Can I able to do the same problem in short time, without graphing? It was appeared in a competitive exam.
@TedShifrin Thank you very much.

1:57 AM
You have to visualize the graphs in your head or on paper. You don't need a computer for this.

okay

if we know that a function is continuously differentiable, then can we draw any conclusion about Lipschitz continuity?

@johnny09: On a closed interval, sure. Otherwise, ???

so we only have to check cont. differ. on a closed interval
if it's on an open set

No, on an open set it may fail.

2:06 AM
okay it makes sense
just to be clear, the conclusion will be implication, right?
cont. diff. on a closed interval implies lipschitz cont?

huh?
yes.

okay! thanks again

or bounded derivative will work on any interval

3:03 AM
can we take the norm inside an integral?
for example, $||\int_{x_0}^{x}f(y)dy||=\int_{x_0}^{x}||f(y)||dy$

3:34 AM
need not be@johnny09

i am more interested to know if it's possible
eg if $f$ is continuous then would that help?

1 hour later…
4:45 AM
Yoooo

2 hours later…
6:23 AM
[Random]
Mar 11 at 9:48, by Secret
(will need to figure out how to define distance between ordinals, because of the fact that they do not commute under addition)
Mar 11 at 9:50, by Alessandro Codenotti
What you're thinking of is that $\omega\setminus n$ and $\omega$ are isomorphic
Mar 11 at 12:19, by Akiva Weinberger
@Secret You could think of them as surreals, where addition and multiplication are commutative and subtraction and nonzero division are defined

7:06 AM
Ok so this diagram I posted a few years ago is wrong. Here's why:
Recall that the natural sum $\alpha \# \beta$ and natural product $\alpha ⨳ \beta$ are defined to be the maximal order type of the disjoint union and direct product of well partial orders $\alpha, \beta$ respectively. In ordinals this is defined to be the componentwise sum of their cantor normal forms
Therefore, the natural sum give some kind of measure of the upper bound of the "distance" between two given ordinals

7:27 AM
Now:

1 hour later…
8:39 AM
0

Question: Let $f: \mathbb{R} \to \mathbb{R}$ be an increasing function. Suppose there are sequences $(x_n)$ and $(y_n)$ such that $x_n<0<y_n$ for all $n\geq 1$ and $f(y_n)-f(x_n) \to 0$ as $n \to \infty$. Prove that $f$ is continuous at $0$. Attempt: We consider two arbitrary sequences $(a_n)$...

Is there anyone to help me out? :(
I just want some verification.

9:36 AM
waddup y'all

10:13 AM
@MatheinBoulomenos Can we say $\frac23$ minimum and $\frac95$ maximum there?

2 hours later…
12:08 PM
1

Find the the splitting field of $X^5-2$ over $\mathbb{Q}$ and find it's degree. My approach: The roots of $X^5-2$ are $\{\sqrt[5]{2},\sqrt[5]{2}\omega,\sqrt[5]{2}\omega^2, \sqrt[5]{2}\omega^3, \sqrt[5]{2}\omega^4\}$ where $\omega=e^{2\pi i/5}$. It's quite easy to show that splitting field of $X... With$\omega$the primitive 5th root of unity, why is the degree of$\mathbb{Q}(\sqrt[5]{2},\omega)$over$\Bbb{Q}$clearly at most$20$? 12:27 PM is 11 the only prime number who's square is a palindrome? 12:42 PM A palindromic number or numeral palindrome is a number that remains the same when its digits are reversed. Like 16461, for example, it is "symmetrical". The term palindromic is derived from palindrome, which refers to a word (such as rotor or racecar) whose spelling is unchanged when its letters are reversed. The first 30 palindromic numbers (in decimal) are: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 22, 33, 44, 55, 66, 77, 88, 99, 101, 111, 121, 131, 141, 151, 161, 171, 181, 191, 202, … (sequence A002113 in the OEIS).Palindromic numbers receive most attention in the realm of recreational mathematics... the 4th palindrome square number with prime root is after 10000$101^2=10201$Having said that, I am curious however, whether there is a formula that generate all such numbers next question are there a finite number of prime numbers whose square is a palindrome? 3 No idea, can it be proved? 2 that's what i want to know 12:49 PM That sounds like a much, much harder question and here's another question that came into my mind is there any prime number whose digit sum$\ne$2 and whose square is a palindrome? there is no number whose square is prime 2 sorry typo @s.harp$\sqrt{2}$Do you count one-digit palindromes? If yes,$2^2=4,3^2=9$12:53 PM haven't had much sleep :P 2 gotchaaaaa lol @Thorgott excluding them$307^2=94249$hmm Can anybody see what I’m doing wrong? 12:58 PM i wish i knew how to code in PARI/GP now That I should almost certainly not be ther Correct answer should be all 4: a,b,c,d right? My sol manual says only a Yes, the empty set works for all four I checked$\{\frac1n\}\cup(2,3)\cup(3,4]$Oh wait, I read distinct and disjunct 1:03 PM What does disjunct mean? same as disjoint lets add some more restrictions is there any prime number that is a palindrome, whose digit sum$\ne$2 and whose square is a palindrome? i think 3 is the only answer if we count single digit palindromes$\{0\}\cup(1,2)\cup(2,3)$also seems to work as example where all five sets are pairwise distinct. @Thorgott do you think that 3 is the only answer? @Thorgott Thanks for this 1:20 PM I can't find an easy answer, but I see no reason to suspect such a number doesn't exist. @Thorgott could this be a new conjecture? Let$p : \forall q \in \Bbb{N}, p|q \implies p=q$Anything can be a conjecture 2 Conjecture is a conjecture Conjecture direct product with conjecture is a conjecture More generally: 🌀(conjecture) is a conjecture Let$p^2 = \sum_{\{1...n\}} a_kb^k$such that$a_k=a_{n-k}$Let$p = \sum_{\{1...m\}} c_kd^k$where$m < n$Then$p^2 = \sum_{\{1...m\}}\sum_{\{1...m\}}c_{k+\ell}d^{k+\ell} = \sum_{\{1...n\}} a_kb^k$For simplicity let$d=b$. Then we have:$p^2 = \sum_{\{1...m\}}\sum_{\{1...m\}}c_{k+\ell}b^{k+\ell} = \sum_{\{1...n\}} a_kb^k$Then$\max_{k,\ell} (k+\ell) = n$Thus for all$k,\ell$:$c_{n-k-\ell} = c_{k+\ell}$sorry typo$c_kc_{\ell}=a_q = a_{n-q}$for all$k+\ell=q$sorry typo$\sum_{k,\ell < n} c_kc_{\ell} = a_q = a_{n-q}$In particular, p does not divide the above sum 2 hours later… 3:30 PM If there is some spaces that are topological isomorphism to itself under Fourier transform other than Schwartz space,$L^2$, and Distribution space? references are appreciated, I googled but none was found. it seems that the space$W={f\in L^1: Fourier_transform(f)\in L^1}$is one, anyone see about this? 3:45 PM Does any number exist except palindromes such that when we reverse the order of the digits, it is divisible by the original number? any help? impossible by definition a number whoose digit is in ascending order, when reversed, becomes smaller/larger and proof? The polynomials$\sum_{k < n}a_kb^k$and$\sum_{k < n} a_{n-k}b^k$do not factorise into each other except when$a_k = a_{n-k}$And given any function passing through n points, its mirror image about the y axis is not identical to itself unless it has y axis symmetry 🌀🌀🌀 Also, given any number$a$, its digit reversal$a'$is not equal to$a$unless palidrome, let alone being divisible Perhaps, it will be a more valid question to ask what is$\gcd (a,a')$which I do not have an answer except when$a,a'$are coprime ah, i think 11|(a-a').. or a\equiv a' mod11 no, I'm wrong, 11 should be 9 1 hour later… 5:10 PM In this answer (math.stackexchange.com/a/3195148/109355) how does it make sense to plug the constant$C_2$in as the dependent variables$u$, since the arbitrary constant(s) and the dependent variable in a general solution are different entities. 2 You're right$2|5$is a false statement. Here are some true statements using |, see if you can see why they are true: $$a|b\wedge b|c\Longrightarrow a|c\\\{n|ab\Longrightarrow n|a \vee n|b\}\Longleftrightarrow \text{n is prime}$$ but n|32 => n|8 or n|4 => n=4,2? thus we have a counterexample I think$\{a|b \implies a=b\} \iff a \in \Bbb{P}$seemed to be more accurate 5:28 PM What could$\chi[-1,1](t)$signify in the context of averaging kernels? I just need one more verification https://math.stackexchange.com/questions/3197311/if-f0-1-to-mathbbc-be-continuous-with-f0-0-and-f1-2-then-ft this one 5 hours ago, by Mathphile are there a finite number of prime numbers whose square is a palindrome? If$a$is prime, then$a|b \implies a=b$In addition:$a^2|a^2,a^2|a$That means, if$a^2$is a palidrome, and$a \neq 11$, then$a^2$have an odd number of digits really nothing much can be said since criteria for square palidromy don't have a formula 5:55 PM Let$A$be a compact subset of$\Bbb R - \{0\}$and$B$be a closed subset of$\Bbb R^n$. Prove that the set$A=\{a · b | a ∈ A,b ∈ B\}$is closed in$\Bbb R^n$. what does divisibility of a real number even mean o wait I go like$x_n$be convergent sequence of$A$then$x_n=a_n\cdot b_n$, but this approach seems to be not working Well B is closed, thus all sequence of B converges to some limit point in B. Meanwhile A does not have an infinite cover, meaning any sequence has to be eventually constant or something Thus the product ab has to converge to some point, so the target set should have all its limit points 1 Find the the splitting field of$X^5-2$over$\mathbb{Q}$and find it's degree. My approach: The roots of$X^5-2$are$\{\sqrt[5]{2},\sqrt[5]{2}\omega,\sqrt[5]{2}\omega^2, \sqrt[5]{2}\omega^3, \sqrt[5]{2}\omega^4\}$where$\omega=e^{2\pi i/5}$. It's quite easy to show that splitting field of$X...

With $\omega$ the primitive 5th root of unity, why is the degree of $\mathbb{Q}(\sqrt[5]{2},\omega)$ over $\Bbb{Q}$ clearly at most $20$?

6:30 PM
hmm, so carryovers can be produced in 3 steps by multiplication:
1. The product of a pair of digits
2. The diagonal sum (summing all terms of the same powers of base b)
3. The sum of the carryovers result from 1 and 2
So that means, we need to know given a pair of digits $a,b$, under what condition will the product $ab$ produce $n$ carryovers
and that boils down to given a fixed $a$, the minimum $c$ such that $ac=b$ where b is the base
i.e. when $a\gcd (a,b) \geq b$
That took care of the 1st type of carryovers
As for the second type: we are basically solving:
$\sum_{S\subset \Bbb{N}} s = b$
That is basically a variation of the subset sum problem which is NP complete
Thus the fundemental reason why number theory is hard, is because carryover dynamics are governed by a NP complete problem
Ok, what about sum of two numbers:
1. The sum of two digits
2. The sum of the carryover result from 1 (can produce a chain reaction depending on the digits)
Well here, the dynamics is a bit more tame. Given any two digits $b_1,b_2$ in base $b$, $b_1+b_2 < 2b$ thus we will only get one carryover for each pair
Thus in order to start a chain reaction, we need $b_3+b_4 \geq 2b-1$
If a number is normal, how likely will that be
Will continue to investigate tomorrow. I think I am starting to get the hang of carryovers

7:45 PM
Where, $n \in \Bbb{+Z}$ , the digit root of $n!$ is equal to the digit root of $n$ where, $n \ne 9k, k \in \Bbb{+Z}$ for a finite number of values.
How can we prove/disprove this?

8:04 PM
0

Here is an example from the book Groups, Graphs, and Trees (Ignore example 1.15; I accidently included it in my snippet of the pdf): From my understanding, the symmetry group of a graph $\Gamma$ consists of all bijections $f : V(\Gamma) \to V(\Gamma)$ such that $v,w$ are adjacent vertices if a...

8:42 PM
Are there a finite combination of positive integers x, y, z, where x and y are palindromes which do not have the digit 1 and z is any natural number such that $\log_{x}z=y$ ?

@user193319 the definition you describe is not the "right" notion of a homomorphism for multigraphs

guys?

@MatheinBoulomenos Ah, thank God you are here! Another question: why is the degree of $\mathbb{Q}(\sqrt[5]{2},\omega)$ over $\Bbb{Q}$ clearly at most $20$, where $\omega$ is the primitive 5th root of unity?
Here's more context:
1

Find the the splitting field of $X^5-2$ over $\mathbb{Q}$ and find it's degree. My approach: The roots of $X^5-2$ are $\{\sqrt[5]{2},\sqrt[5]{2}\omega,\sqrt[5]{2}\omega^2, \sqrt[5]{2}\omega^3, \sqrt[5]{2}\omega^4\}$ where $\omega=e^{2\pi i/5}$. It's quite easy to show that splitting field of $X... 8:58 PM @user193319$\Bbb Q(\sqrt[5]{2})$has degree$5$over$\Bbb Q$.$[\Bbb Q(\sqrt[5]{2},\omega):\Bbb Q(\sqrt[5]{2})] \leq 4$, since the minimal polynomial of$\omega$over$\Bbb Q$has degree$4$Are there a finite combination of positive integers x, y, z, where x and y are palindromes which do not have the digit 1 and z is any natural number such that$ \log_{x}z=y$? can anyone help me with this question? @MatheinBoulomenos I don't understand why$[\Bbb Q(\sqrt[5]{2},\omega):\Bbb Q(\sqrt[5]{2})] \leq 4$@MatheinBoulomenos I don't understand why$[\Bbb Q(\sqrt[5]{2},\omega):\Bbb Q(\sqrt[5]{2})] \leq 4$the minimal polynomial$f$of$\omega$over$\Bbb Q$can also be considered as a polynomial with coefficients in$\Bbb Q(\sqrt[5]{2})$, since$\Bbb Q \subset \Bbb Q(\sqrt[5]{2})$. Since$f(\omega)=0$, the minimal polynomial of$\omega$over$\Bbb Q(\sqrt[5]{2})$must be a divisor of$f$Heya @Mathein! Hi @Ted! 9:06 PM Ah, of course! Thanks @MatheinBoulomenos why isn't anyone replying to me here? @Mathphile take any palindromes x and y that do not have the digit 1 and put$z=x^y$I spy with my little eye Some nerds What's going on everybody? hi @Daminark WHat's up? 9:18 PM "up" is a kind of quark, I guess 9:31 PM Lmao, that was a good one 9:51 PM why is the the digit root of prime numbers always 1 or 8? 23 silly me The only restrictions on digital roots of primes I can think of is that they can't be multiples of 3 for every digit 1,2,4,5,7,8, there are infinitely many primes with that digital root (as an application of Dirichlet's prime number theorem) are there a finite number of pairs of prime numbers$a, b$such that$ab\$ is a palindrome?
a, b should be different
anyone know hoe to prove/disprove this?