@Semiclassical I didn't do this on purpose. I thought that since $f(x,y,z)$ and $u(x,y,z)$ both depend on the same variables then one can define $g(x,y,z)=u+f$

if this is wrong then let me know, and i'll edit the question to avoid any issues

It's about the concept of Subspace from linear algebra

So, I've already done exercises, tests and so on...

But I'm still do not grasp the very elementary notion about this structure, which means that I'm not grasp a few things:

yes

?

automatically**

which is the neutral element?

why not inherited? how can they be different?

if $W \subset V$ is a linear subspace,

@M.N.Raia I) Commutativity and associativity are automatically inherited because, well, it's the same operation. If a + b = b + a for all elements of V, and W is in V, then a + b = b + a for all elements of W too.

then elements in $W$ are elements in $V$

inherited** (sorry)

II) Your subset might be closed under addition and scalar multiplication, and not have 0 or negatives in there, because it might be empty!

ops

$\{\}$ is not a vector space

a+o=a

@JoeShmo That's my point.

I see

You require further, on top of closure under addition and scalar mult., that the set be nonempty---and you might as well stipulate that 0 be in the subspace specifically, because the closure under scalar mult. will automatically mean 0 had better be in there anyway.

so you want to define a vector space as the algebraic structure that respects these rules

If you've checked closure of scalar mult, then you don't need to check whether (-a) is in W, since -a = -1a.

these rules axiomatize vectors that you might encounter in (say) physics

But you do need to check whether the set is nonempty, and 0 is the easiest element to check. And also must always lie in a subspace if it is indeed a subspace.

Ok

If you have established the closure properties, you don't need to look for 0 specifically: if something else (say, a) is in W, then by closure of scalar mult, 0 = 0a is also in W. But as I said, 0 is the easiest to check in pretty much all cases, so most people just say to look for that.

Conversely, if your set is nonempty, but doesn't contain 0, you can immediately eliminate it as a subspace, because it can't be closed under scalar mult!

@Fargle that's my point, I don't understand this theorem properly.

@M.N.Raia I'm not sure why they point that bit out (the "In particular," sentence).

I'll read again your commetaries

Except, maybe they're pointing it out just because it is kind of special that closure under scalar mult buys you a lot less checking.

If you were just dealing with abelian groups, where there is no scalar multiplication, you would indeed have to check whether or not -a is in the subset for any a in the subset.

Your comment is interesting. My whole doubt concerning the 0 element of subspaces is just like you comment. You said:

"you would indeed have to check whether or not -a is in the subset for any a in the subset"

My point is: if -a is already in the large structure V why isn't automatically in the subset?

More generally, why a subset of V isn't already a subspace of V?

Because a subspace must be a vector space in its own right, with the same operation.

"in its own right" what suppose to mean this, technically? Have his own 0 element?

Well, it has to satisfy all of the vector space axioms. One of those, in particular, is that it must have a 0, yes.

It turns out that no matter what, it'll be the same 0 as in the big space.

The classic example is, say, a line through the (0,0) in the plane. It contains the zero vector; if you add any two vectors on the line, it stays on the line; if you scale a vector on the line, it stays on the line.

But, if we took a pair of different lines through (0,0): zero vector is still there, scaling still works, but addition doesn't---if you add a vector from one line to a vector from the other line, you'll land somewhere in between. The operation from the big space doesn't make the pair of lines into a vector space, because in any vector space, if you add two elements it needs to give you back an element of the vector space itself.

More briefly, if that pair of lines is W, then I just found a and b in W such that a + b is not in W.

But one of the axioms for vector spaces says, "if a and b are in V, then a + b is an element of V".

(Albeit, this is usually hidden behind "there is a binary operation + on V"...)

Hmmm

that's a good comment.

and explanation.

This is what I meant by "closure under addition".

Similarly, you can work out what I meant by "closure under scalar multiplication".

Once you have those two things, most of the other properties are things you get for free.

Commutativity, associativity, distributivity.

Even opposites come for free with those two closure properties.

The only remaining axiom is the one that says "There exists an element 0 such that yadda yadda".

If you already know your set is nonempty, then by the closure of scalar mult, 0 must be in it, so you have no checking to do.

But usually you don't already know your set is nonempty, and so you check the zero vector, because that's probably easier than checking to see if (1,2,35,7) is in there.

Personally I like to check 0 first, because if it's not in there I don't even have to muck about with the closure stuff---I already know at that point that I have a non-subspace.

Hmmm

That's right, thank you a lot.

Happy to be of help. Also I like the smell of my own...well.

Hello !

Can someone please clarify me with the above doubt

@AkivaWeinberger Thank you verymuch

heyo

if $f(x) =g(x) $ for all $x$ in the common domain of f and g, where $f,g$ are continuous and differentiable then is it true that $f'(x) = g'(x)$ ?

yes

you might also say that $f=g$

Can someone please help !!

So, projections onto convex sets are unique. Somebody said that projections onto non-convex sets are not unique. This seems wrong, because a projection onto a non-convex set seems like projecting onto the convex hull of the non-convex set.

Can we verify if a binary number is prime in base 10 without converting it to base 10?

Being prime is independent of the base

@Mikhail How would you project the center of a circle on its circumference?

@AlessandroCodenotti how would you project (0,1) to the lower half plane?

I don't see your point

@AlessandroCodenotti you didn't understand what i meant

Here's an example: We know that if we convert binary number 111 to decimal we would get 3 which is a prime number. My question is that is it possible to check 111 is a prime number in base 10 without converting it to its decimal form and then checking for primality?

Being prime has nothing to do with decimal representation

If we used base 3, we’d still get the same set of prime numbers. The labels would look different, eg 12 =3*4 in base 10 would become 110=11*10 in base 3

111 is 7

oops yes

i meant to write 11

But beyond that, nothing changes. Being a prime is a property of the integer itself, not of how we represent that number

Though yeah, multiplication and division work the same, you're just expressing the numbers with different symbols.

Now, a number being a palimdrome is different.

To the extent that doing multiplication in a different base seems strange, it’s because we’ve committed to memory the multiplication table for decimal representation

Also, concatenation of digit works differently. 1:1=11. In binary this number is 3. In trinary it's 4. In quartary it's 5, ect.

im sorry if i am not able to express myself properly

If the idea you’re trying to express rests on the notion that being prime has something to do with decimal representation

Well, you’re being perfectly clear about what you mean. But that statement is simply not true

At least, we believe you're being perfectly clear. :P

The number after the number after the number one is prime

And that’s true whether you label that as 11 (base 2) or 10 (base 3) or 3 (base 10).

what i mean is if you think of 111 as a base 10 number it is not prime but if you think of it as a base 2 number then it is prime as 111 in base 2 is equal to 7 in base 10

Ah, so 111 as in "the number one hundred and ten places after the number one."

Ah. So two different numbers which have identical representations in different bases

think of it more as a string of numbers consisting of 0's and 1's than a binary number

Hi all!

Is that correct: $$\nabla\times(\vec{d}(\vec{r})\rho(\vec{r})) =\vec{d}\times\nabla\rho - \rho\nabla\vec{d}$$

so what i want to know is that without converting this number 111 to 7, can we know if it is a prime number?

does that make sense?

Doubtful

A big thing about prime numbers is that checking a number to be prime is not exactly easy.

$\rho$ is a scalar function and $\vec{d}$ a vectir function of $\vec{r}$ (all over $\Bbb R^3$.

"-" is strange

I know.

this sounds like a problem out of additive combinatorics which is hard

Like, can you tell me, at a glance, if the number $7^k-1$ where $k=\prod_{i=0}^7 (i+7)$ is prime?

Arithmetic combinatorics, woops

An interesting question: suppose you take a prime number and write its binary representation

Now treat said binary form and interpret it as a decimal form. How likely is it for the resulting number to also be prime?

I wouldn’t be surprised if that’s an open problem

I don't see how is this different than just checking for primality in base $2$, $7$ is prime in base $10$ iff $111$ is prime in base $2$, run any primality testing algorithm on $111$ doing all arithmetic in binary and you'll find out whether it is prime (hence whether $7$ is prime in base $10$) without converting it

The point is that 111 (base 10) is a different number than 111 (base 2)

Sure, but the question is whether 111 (base 2) is prime

And this can be checked without converting it to any other base

we should invent a new notation

Usually the base is written as a subscript when needed, $111_2=7_{10}$

If the point is that there’s nothing more intelligent to do than “check for primality in whatever base you’re in”

Then I agree

out of the first 10 primes, 3 are primes when converted to binary and interpreted as decimal, etc

first 100 primes, first 200 primes, first 300 primes, etc

I mean, there’s a little bit of info there. For instance, 100_b is composite regardless of b

But I doubt there’s anything systematic

(After taking a shower, coming back, and looking at the example I gave, I immediately noticed that it's obviously divisible by 2)

Though, after plugging it into wolframalpha, $(\prod_{i=0}^7 (i+7))-1$ is prime.

@Rithaniel and 3 for that matter

I don't see it divisible by 3 immediately. What knowledge are you using there?

that 6 is divisible by 3?

So, you know that , $6\mid 7^k-1$ for all $k\in\mathbb{N}$?

right

I feel like this is using some number theory fact that I don't remember.

$7^k-1=(7-1)(7^{n-1}+7^{n-2}+...+1)$

Ah, polynomials

The thing that I never think to look for.

I only see the TeX script instead of the format that the text is meant to be in, is something wrong with my browser?

@NikilKumar No, you have to enable mathjax. See the “Latex in chat” in the room description

@Rithaniel alternatively: 7=1 mod 6, so 7^k=1^k=1 mod 6 as well

Why isn’t Mathjax enabled by default?

(X,M,u) be a measurable space. Suppose $A \in M$, then is it always true that $A$ has a measure (finite or infinite)

@henceproved Isn’t that a measure space, not a measurable space?

@henceproved Write down the definition of measure space and measure

@NikilKumar The room description should be on the top-right corner of your screen

hi chat

Ah, yeah, that makes it actually pretty obvious, Semiclassical.

I say this with the benefit of hindsight, mind. Back in high school the only way I knew how to see stuff like that was via the binomial expansion

then i learned modular arithmetic and was like "oh duh"

Hi , Is $\mathbb{P}(X^2\leq9|X>2)$ = $\mathbb{P}(2<X\leq3)$ ? X is a continous random variable , P is probabilty.

Looks fine to me

@Elsa you need to divide by $\Bbb P(X>2)$

Oh, right. P(A|B) = P(A & B)/P(B)

Yeah , that is what I was looking for. Thanks

Hi. Can anyone direct to me to a detailed explanation of the urn problem in combinatorics? I want the general case with $p$ ball colors. Then we could count the number of ways to draw with/without replacement...

today's smbc is a bit too real for me: smbc-comics.com/comic/box

verse-and-dimensions.fandom.com/wiki/The_Box

Everything Lies Within The Box

(PS, I found this weird subculture from some weird youtube recommendation which people came up all sorts of fictional cosmology beyond the observable universe)

Hello!

How can we prove that if $B\subset A$, then $A\subset A\cup B$?

I took $x\in A$ and tried to conclude $x\in A\cup B$ but I couldn't :(

@manooooh $A \subset A \cup B$ iff for every $x \in A$ we have $x \in A \cup B$ iff for every $x \in A$ we have $x \in A$ or $x \in B$ which is true, here it seems strange because we didn't need that B is a subset of A

@KonformistLiberal uhm, so you say that $A\subset A\cup B$ is a trivial proof, isn't it?

@manooooh I don't know if something goes wrong if we don't have law of excluded middle - axiom of choice or something - but besides that, it seems like a trivial statement, yes

@KonformistLiberal thanks! <3

@manooooh <3

I have a quick question too: I am trying to visualize (S^1 smash product I) and it looks like S^2 but it must be homeomorphic to reduced suspension of S^1 which looks like S^1

Sorry reduced suspension of the interval*

(okay got it)

Hi @loch @Semiclassic

Hi

Sure is quiet 'round these parts these days.

You know how a p-series diverges for N less than or equal to 1

if you don't sum over the positive integers consecutively is it much less straightforward to find the sum