Mathematics - 2024-05-09

Jakobian

12:55 AM

@SoumikMukherjee want another exercise?

EE18

A number theory question

Suppose ab = cd and a,c even, b,d odd. I suspect I can concldue that $a = c, b = d$?

Soumik Mukherjee

@Jakobian Yes. anytime.

I haven't solved the first one yet though, I am trying to focus on the ongoing PhD interviews more

Jakobian

Prove that if $\rho(x, y) = \max(|\frac{2x}{1+x^2}-\frac{2y}{1+y^2}|, |\frac{1-x^2}{1+x^2} - \frac{1-y^2}{1+y^2}|)$ then $\rho$ is a metric equivalent to the standard metric on $\mathbb{R}$

@SoumikMukherjee its okay I forgot what it was anyway since it was such a long time ago

Soumik Mukherjee

Something related to proper and perfect maps

Jakobian

@EE18 nope

EE18

1:01 AM

OK, back to the drawing board

Jakobian

@Jakobian and describe the completion of $\mathbb{R}$ with this metric

Soumik Mukherjee

Okay

EE18

Basically trying to show that for $\langle m,n\rangle \in \omega^2$, if $2^m(2n+1) - 1 = {2^{m'}}(2n'+1) - 1$, then $m = m', n= n'$

Soumik Mukherjee

@EE18 Are n,n'>0

EE18

$\geq 0$, all are naturals (in $\omega$)

I feel like I need to use arguments via even and odd

Maybe I'll use induction. Thought I could do something more direct

Soumik Mukherjee

1:11 AM

@EE18 divide by the least of 2^m,2^m' after you add 1 to both sides

EE18

And then I guess I can use even and odd arguments from there, I think I got you

Soumik Mukherjee

@EE18 This is different from the question that you asked though, 2*3=6*1 is a counter example in the previous question

EE18

For sure, I was just spitballing (clearly poorly) to try to solve this question

Thanks very much for the help Soumik :)

Soumik Mukherjee

Np:)

Jakobian

If ab = cd are a, b, c, d are like in your setting then a, c have the same amount of 2's but they can have different amount of other primes

That's why the original question wasn't true but this one is

EE18

1:24 AM

Hmm, interesting

How did you learn your number theory Jakobian?

Just picked it up?

EE18

1:43 AM

BTW Soumik, you might like this slick argument: math.stackexchange.com/questions/847425/…

Soumik Mukherjee

@EE18 The 2nd one?

EE18

Oh I meant the first :)

But ya I guess second works too, didn't even look

EE18

2:06 AM

I don't quite have any intuition for this (which I guess is called Cantor's pairing function?)

How are they coming up with this by tracing the point of the $\omega^2$ lattice?

These questions feel crazy. Should I be able to come up with these explicit bijections? These are the first exercises in Enderton I'm struggling with

Thorgott

I mean, they're telling you the bijection, so it's not like you have to come up with it

the idea behind the picture is pretty clear, I hope

EE18

Ya the idea behind the picture is, but not how it corresponds to that forula with the $\cdots$ in the second picture

Thorgott

whenever you have something like $1+\dotsc+n$, you're counting points in a triangular shape (1 in the 1st row, 2 in the 2nd, 3 in the 3rd, etc.) and the picture indicates why that might be relevant here

EE18

2:25 AM

I am still not totally sure I see it. Would it be possible to explain how the first term comes about and how the second comes about (the lone $m$)? I guess this is one of the cases of where it's hard for you to know what I can't see lol

EE18

3:08 AM

Got it, never mind

Jakobian

8:31 AM

@EE18 no idea

Most of it just comes naturally at some point, I think

At least the basics

@EE18 those are called triangular numbers - this is the reason

$T_n = 1+2+...+n$ that is

Daniel Donnelly

9:51 AM

Tird eye opening can't solve twin primes :)

Thorgott

10:15 AM

@EE18 nice!

Jakobian

12:11 PM

@Thorgott are you a coffee person

Daniel Donnelly

1:24 PM

@Shaun and room

I have an interesting conundrum

Consider $H= \{ x \in \Bbb{Z}/n: x^2 = 1 \pmod n\}$. Then what is the size of the pseudo coset $kH$ for any integer $k$ not jus units!

I believe it's got something to do with gcd

It clearly at a maximum of $|kH|\leq |H|$ since you can't magically spawn elements

But what is its exact value, I believe there is a simple formula

The formula for $|H|$ when $n$ is square-free is just $|H| = 2^{n - [2\mid n]}$ exactly, where $[2\mid n] = \begin{cases}1 \text{ if } 2 \text{ divides } n \\ 0 \text{ otherwise} \end{cases}$

Do you think this would be a good fit for a question? :/

Shaun

1:42 PM

@DanielDonnelly I suppose this is something you could use either semigroup theory or ring theory for. I can't decide right now whether you gain or lose information; probably gain. But still . . . one of the two.

@DanielDonnelly and hi :)

EE18

Question which may largely be for Leslie since it's specific to Enderton's text in some ways

This was a theorem I saw long ago (now revisiting as part of some other proof), and it struck me that, unless I'm misunderstanding, the phrasing of the final sentence is somewhat off.

We define for natural numbers $m,n$ that $m < n :\iff m \in n$. Thus isn't it more fair to say that rather than "strengthening this" we are in fact defining natural numbers as the set of all smaller natural numbers?

Incidentally, this leads me to the following...how are we guaranteeing that natural numbers only contain natural numbers? The definition above merely says nothing about non-natural numbers potentially being in $n$. Is it correct to say that we get around this because of $\omega$'s transitivity? If we had some $x \in n \in \omega$ then $x \in \omega$, so that $x$ is a natural (since $\omega$ consists precisely of the naturals)? All of this may just be a matter of which definitions my text works..

..with, in which case feel free to ignore

Jakobian

1:58 PM

@EE18 is a natural number defined to be a member of omega

EE18

No, but it was proven that $x$ a natural $\iff x \in \omega$

Jakobian

putting aside what the word strengthen means in this situation, since its irrelevant

then sure, you just proved all elements of a natural number must be natural numbers

EE18

Gotcha, but just wanted to confirm the second bit: the fact that every natural is the set of all smaller naturals is from $x \in n \iff (x \in \omega)\land (x < n)$ and that is because of how we chose to define things, right?

Jakobian

sure, this comes directly from definition

EE18

Got it, thanks Jakobian!

Jakobian

2:03 PM

I don't feel like I've explained anything, but confirming is good enough, I guess

EE18

:) indeed it is

Daniel Donnelly

@Shaun see my latest post

I asked the question

It's quite involved

But I think the final answer is something simple

0

Q: $|H| = 2^{n - [2\mid n]}$ for square-free $n$; what's size of $kH \subset \Bbb{Z}/n$ where $k$ is any integer! For subgroup $H \leqslant \Bbb{Z}/n^*$?

Consider the ring $\Bbb{Z}/n$ for square-free $n$. Then the multiplicative subgroup $H \leqslant \Bbb{Z}/n^*$ given by $H = \{ x \in \Bbb{Z}/n : x^2 = 1 \pmod n\}$ has size exactly $|H| = 2^{n - [2\mid n]}$ where $[2\mid n]$ is $1$ if $2$ divides $n$, else $0$. So since the size of $|H|$ is know...

Pizza

Consider the vector space $\mathbb{R}^3$ and the following system of vectors:
$$\mathbb{B} = \{(k+1,k,1-k),(-1,1,1),(0,k,k)\}, \ \text{where} \ k \in \mathbb{R}.$$Determine for which values of $k$, $B$ is a basis for $\mathbb{R}^3$.\begin{pmatrix}
k+1 & k & 1-k \\
-1 & 1 & 1 \\
0 & k & k \\
\end{pmatrix}

After I lay out the vectors, do I need to calculate the rank of this matrix to see if the vectors of B are linearly dependent?
If the answer is yes, I have to assign a value to k, but I'm not sure if it is correct as a procedure.

maybe I'm going the wrong way

John Zimmerman

2:24 PM

Are there problems in mathematics that are well posed but can never be given a proof?

Daniel Donnelly

Yes

Possibly Collatz is one

a proof that it is true

John Zimmerman

Doesn't that only take into account the current state of the system. The problem cannot be solved at this moment with these methods

however in 100 years new methods may solve it

ohh

Daniel Donnelly

It could be. They usually don't prove that something doesn't work; they just dead end but only sometimes prove that it's can't possibly be fruitful. Such as parity problem in sieve theory

John Zimmerman

there are such problems like the halting problem i think

Daniel Donnelly

Yes. You can state twin primes in terms of a very simple special case of the halting problem.

Whether or not the "next twin prime average" loops forever

that function

John Zimmerman

2:27 PM

@DanielDonnelly oh neat

Daniel Donnelly

You can code it with like $\gcd(x^2 - 1, p_n\#)$ where $p_n\# = p_n p_{n-1} \cdots p_1$.

and $p_i$ is the $i$th prime number

If the GCD is $1$ and $x$ is within a certain range, then bam, you automatically have a twin prime average

The range by sieve of Eratosthenes at least is $[p_n + 2, p_{n+1}^2 - 2]$. I calculated this all before which is why I can just draw on it

If the GCD is not one, then in that case you don't have a twin prime average. Increment $x$ or even increment it by $6$ and check again, in a while loop. That's it.

John Zimmerman

I came up with a problem that is beyond the reach of modern mathematical methods

Daniel Donnelly

The question is will this simple function always terminate. Who knows!

Nice!

Well it's beyond the reach of yours. But maybe Terrence Tao could solve it because he got tutored by Erdos a few times

Erdos genius must have rubbed off

That's not to discourage you. It's just that we're not at the apex of mathematical knowledge yet in our lives. That only comes maybe 20 years from now lol

John Zimmerman

I'm probably more suited to solve it than tao

only because my information path is much more developed

Daniel Donnelly

Yes, thats possible too

John Zimmerman

2:33 PM

@DanielDonnelly yeah :)

20 years !

Jakobian

@JohnZimmerman yes. Are all subsets of reals either countable or of size continuum

you can't give a proof of this in ZFC

Daniel Donnelly

That's definitely continue greater

$2^\textbf{continuum size}$

Jakobian

and you can't give a proof that its false either

Daniel Donnelly

@Jakobian I think you're joking Mr. Fine man

You can't prove it in 2nd order logic?

John Zimmerman

not joking

Daniel Donnelly

2:36 PM

SO what logic can you prove it in Homotopy Type Theory?

@JohnZimmerman you're very far my friend. Keep on going with your studies. One day you will crack the problem wide open

Reduce the problem to almost nothing is the first step I think

Maybe by deleting axioms

Jakobian

as far as I understand what you meant @JohnZimmerman I think I've answered you

Daniel Donnelly

@Jakobian can you answer my question above?

It's modular arithmeticy

Jakobian

and there are other things that you can prove can't be proven in ZFC, such as large cardinals

John Zimmerman

@Jakobian yes thanks that is a perfect answer

Jakobian

👍

John Zimmerman

2:45 PM

This is my very difficult problem

Problem: Consider a codimension one surface of revolution $S$ and an embedding $e:S \hookrightarrow X^n$ for $X^n=[0,1]^n$ with points $p,q$ elements of $\partial X^n$ where $\partial X^n=X^n-(0,1)^n$ for $\mathrm {sup}~ \mathrm{dist}_n(p,q)=\sqrt{n}$. Assume $S$ must remain a surface of revolution, have constant positive Gaussian curvature, and have a pair of antipodal corner points $p,q$ as cone points.

Find $\rho_n=\max \lbrace \mathrm{vol_n(S)}$ for $n=123456789.$

$n=3$ yields $\rho_3\approx 1/2$

Jakobian

@JohnZimmerman is this AI generated

John Zimmerman

No I generated it

Jakobian

so you're not a bot :P

John Zimmerman

No am not a bot lol

I'm real breathing person

Jakobian

thats what a bot would say

🤖

John Zimmerman

2:59 PM

fortunately A.D. Hwang gave a nice answer to the $n=3$ case

like many "hard" problems

I take advantage of high dimensions

Xander Henderson

Coming up with "hard problems" is not particularly hard. The goal should be to come up with interesting, well-motivated problems.

John Zimmerman

@XanderHenderson I find it interesting but I may be just biased

I might that bot bias

I find it well-motivated as well

Xander Henderson

@JohnZimmerman By "interesting", I mean "interesting to the mathematical community".

John Zimmerman

@XanderHenderson I was told the problem is "vexing" but I haven't received any "wow!" that problem knocks my socks off!" yet

however I think it's a few cents away from accruing interest from the mathematical community

Xander Henderson

3:14 PM

@JohnZimmerman I mean, part of your job is to convince other people that the problem is well-motivated and interesting. YOU need to explain to others how the problem fits into the corpus of mathematical knowledge, and why anyone should care about the solution.

Right now. it just reads like a bunch of gibberish to me (likely because I am not up-to-speed on whatever part of mathematics is relevant, but possibly because it is gibberish).

John Zimmerman

@XanderHenderson if it was gibberish how would Andrew D. Hwang have been able to answer it?

Xander Henderson

In particular, the choice of the magic constant $n=123456789$ seems downright bizarre me.

@JohnZimmerman I have no idea. But arguments from authority are unconvincing to me.

I mean, who the heck is "Andrew D. Hwang", and in what context was this question "answered"?

Soumik Mukherjee

@XanderHenderson The book I am reading says me to convince myself, shall I file a case against the author?

Xander Henderson

@SoumikMukherjee I can't tell if you are trying to be funny or not, but I suspect that the context is wildly different.

Soumik Mukherjee

Trying to be funny ofcourse:)

John Zimmerman

3:23 PM

genealogy.math.ndsu.nodak.edu/id.php?id=32043 @XanderHenderson

Xander Henderson

@JohnZimmerman The more salient question is "In what context was this question 'answered'?" Knowing who the answerer is might matter, if that person is some widely acknowledged expert in the particular field in which a problem is asked, but the context and content of the answer is much more important.

EE18

Enderton makes me the following promise (so nice of him! Apparently it has to do with needing to invoke regularity which hasn't been done yet, but that's not the crux of my question). My question is, does the following summarize what he will do (the English language can be informal and i want to be sure I understand): he will construct a function card with domain set of all sets (but this can't be right, there is no set of all sets!) which is such that it obeys the properties (a) and (b)

Xander Henderson

That was kind of the entire point made by the Bourbaki collective: individual people don't matter, only their ideas.

EE18

John Zimmerman

He answered the following question @XanderHenderson. What is the volume of the largest surface of revolution with constant Gaussian curvature that can be placed inside the unit cube?

Xander Henderson

3:27 PM

@EE18 Enderton is saying that the tools to properly present a definition don't exist yet, but that they will be introduced later in Chapter 7. Until then, you are to take the definition given (called a "promise") as a given.

@JohnZimmerman That looks like a completely different question to me, and you still haven't addressed the context and content of the answer. But I am also not really interested in this problem that you are so very excited about, so I am going to stop replying, now.

@EE18 Basically, Enderton is saying "Trust me. There does, in fact, exist a function $\operatorname{card}$ such that $\operatorname{card}A = \operatorname{card}B$ when $A\approx B$, and if $A$ is finite and $A\approx n$, then $\operatorname{card}A=n$."

(I presume that $A \approx B$ means that there is a bijection between the two, and that $n$ is a von Neumann ordinal, or something similar).

EE18

OK, that makes sense. Only thing is that I guess no such function can actually exist, right? Else its domain is the set of all sets, a contradiction...

Xander Henderson

@EE18 Why is the domain automatically the "set of all sets"?

That seems like a bit of a leap of logic...

In any event, the whole point of that passage is that you are not supposed to think too deeply about it right now, and that it will all be made rigorous in Chapter 7.

I have no idea what the content of Chapter 7 is, but you could, I suppose, read ahead...

Thorgott

@Jakobian nope, never drank

EE18

Oh for sure, I won’t dwell too much on it that’s well taken

Thorgott

@EE18 it doesn't have to be a function

EE18

3:37 PM

The reason that’s the domain is that the claim in Enderton’s promise is that card is defined for all sets A, right?

I guess it’s an existence statement Thorgott? For all sets A there exists a set A card A such that…

Jakobian

@Thorgott surprises me there are people who have never drank coffee in their life

Thorgott

not an existential statement, it's a definition

Xander Henderson

@EE18 That passage is very informal, and I think that you are overthinking it.

Thorgott

or, it will be a definition in chapter 7

Xander Henderson

My language was also pretty informal, and you are overthinking it.

Thorgott

3:38 PM

I don't think EE18 is overthinking. This is an important distinction to understand.

EE18

The reason I want to be somewhat formal is Enderton then goes on to a bunch of theorems involving this promise

So understanding exactly what the promise is saying feels important

Jakobian

@XanderHenderson well said. Coming up with a problem that can't be answered without possible whole lot of mind stretching is not really what interests most people

Thorgott

You can define for any set $A$ a thing depending on $A$, even though this definition won't yield a "function on the set of all sets" (cause there is no such set)

Xander Henderson

@Thorgott Yes, the distinction is important to understand, but the whole point of that passage is that the correct tools to provide understanding are currently out of reach. Don't think about it now, it will be explained later.

Thorgott

For example, if I give you a fixed set $B$, then for any set $A$, the set $A\times B$ is defined

this statement makes perfect sense even though there is no "set of all sets"

Jakobian

3:40 PM

and if a problem is hard, you probably won't obtain solution

and problems become interesting mostly because they can be solved in some way in my experience

or they are important for some field of math

EE18

For sure, I follow now what you mean thorgott re there needn’t be a function specifically

Xander Henderson

@Jakobian Or because a solution will have applications to other areas.

EE18

I’m hoping to still clarify the other point, on existence versus definition. What do you mean in saying there is no existence claim being made?

Xander Henderson

Oh, you got there faster than I did.

Heh.

Thorgott

well, you can interpret it as an existence claim if you want to

but these properties don't specify anything unique

EE18

3:42 PM

They seem to apply to different things. A definition is just a name for something right? This seems to be saying that there is a thing card A for each A?

Thorgott

so that's an odd thing to do

Xander Henderson

@EE18 In principle, you can write down any definition you like, but you do not know, a priori, that any actual object of any kind will actually satisfy that definition. I think that Enderton is actually making a kind of existence claim, but he is not proving or justifying that claim (yet).

Thorgott

Enderton is saying that there is a way of associating to each set $A$ a set $\mathrm{card}(A)$ and this will satisfy certain properties, but this will be done by specifically defining $\mathrm{card}(A)$ at a later point

Xander Henderson

@Thorgott This.

Thorgott

@EE18 also, rather than worrying about whether it's an existence claim or not, the quantifiers here are wrong

EE18

3:45 PM

Am on mobile so I can’t see what you’re replying to, will check when on computer

Thorgott

$\mathrm{card}(A)$ is not meaningful individually for each $A$, but rather as an association rule $A\leadsto\mathrm{card}(A)$

EE18

But your message before that makes sense. We will show that there is such an association (probably many possible) and then we will define card A as a particular choice

Jakobian

are we talking about the notion of cardinality in ZF

EE18

In ZFC if that matters, but ya

Jakobian

well, others probably already explained this but here I go

Xander Henderson

3:51 PM

@Jakobian It is less about understanding cardinality, and more about understand Enderton's presentation.

25 mins ago, by EE18

Jakobian

there is a sort of equivalence relation of sets where $|A| = |B|$ means there is a bijection $f:A\to B$. Cardinal numbers are the "canonical representants" of this equivalence class in the sense that for each set $A$ there exists a unique cardinal number with the same cardinality as $A$

EE18

@Thorgott Am on computer now, but fail to see how my order of quantifiers is wrong. Isn't Enderton basically leading us to "for all sets $A$, there exists a set card $A$ with the following properties, and that we can speak of a specific card $A$ will follow from later arguments when we get more specific about card $A$ properties"

Jakobian

I don't know, simple is relative

I find the definition of least ordinal in bijection with $A$ to be a pretty simple definition of cardinal numbers

so I disagree with Enderton

Xander Henderson

@Jakobian I find it difficult to either agree or disagree with Enderton without knowing how the theory has been developed up to that point, or how it will be developed after that point.

And, according to the pdf I just found, Enderton doesn't formally define ordinals or orderings until chapter 7. So that may very well be exactly the definition he gives.

Thorgott

@EE18 he doesn't list properties of the set $\mathrm{card}(A)$, he lists properties of the association rule $A\leadsto\mathrm{card}(A)$

I guess the way of phrasing this is harmlessly interchangeable if we have uniqueness, but in the absence of uniqueness, this way sounds wrong

also, for what it's worth, if you set up in a context in which it is reasonable to talk about the "collection of all sets" (and this is done regularly), then card will indeed be a function from this collection to itself. it's a hard pill to swallow (so I probably shouldn't philosophize about it at this point), but "size issues" are tyically only a problem of relative nature, not of an absolute one.

Pizza

4:17 PM

Consider the vector space $\mathbb{R}^3$ and the following system of vectors:
$$\mathbb{B} = {(k+1,k,1-k),(-1,1,1),(0,k,k)}, \ \text{where} \ k \in \mathbb{R}.$$Determine for which values of $k$, $B$ is a basis for $\mathbb{R}^3$.
\begin{pmatrix}
k+1 & -1 & 0 \\
k & 1 & k \\
1-k & 1 & k
\end{pmatrix}\begin{align*}
\text{det}(\mathbb{B}) &= (k+1)\begin{vmatrix} 1 & k \\ 1 & k \end{vmatrix} - (-1)\begin{vmatrix} k & k \\ 1-k & k \end{vmatrix} \\
&= (k+1)(0) - (-1)(k) \\
&= k
\end{align*}$\text{basis means}$: spans entire space & linearly indepdnent.

I tried to do this myself, can anyone confirm if this is correct?

EE18

@Thorgott Maybe my phrasing is poor, but I think I am talking about a sort of implicit uniqueness right. I am saying that eventually Enderton will specify a unique association rule between $A$ and the sets which could have served as card $A$ ?

Thorgott

Enderton will define something unique, of course, my point is the listed properties don't characterize this thing uniquely

EE18

Oh OK, that makes sense. on that I definitely agree

you're just saying that it doesn't even make sense to write down card $A$ (since that suggests/requires some unique thing) until we specify more properties

Pizza

I can also add these notes above. In $n$ dimensions:
- less than $n$ vectors are never enough to cover the entire space (therefore never a basis).
- more than $n$ vectors are always linearly dependent (therefore never a basis).
- if there are exactly $n$ vectors, then not even one
all vectors are linearly dependent AND DO NOT cover the entire space
OR
they are linearly independent and span space.

right?

leslie townes

4:45 PM

EE18 yeah, enderton is using this as a kind of informal guide. at this moment in the book, i would think of "Card" as an "an object to be defined later with these properties," not "an object that enderton has just defined with these properties," let alone "an object just defined that is characterized by those properties"

and in mild defense of enderton, i would think of it that way because that is what he expressly tells you to do

Sine of the Time

@Jakobian actually, I don't listen to music

@SoumikMukherjee long time no see indeed. Are you happy Guki won the candidates?

Pizza

5:04 PM

Hi @SineoftheTime

@Pizza sup

How is it going?

fine, how about you?

Pretty good

what courses are you following?

Pizza

5:10 PM

If you mean all the courses that are there now:Algebra and Geometry
Analysis 2
Physics 2
Electronic Computers

Sine of the Time

so you're studying multivariable calculus?

Pizza

Yes

Sine of the Time

nice

Jam

Suppose i have 3 balls of colour red 2 balls of blue kai 1 white ball. How many combinations of 3 balls exist order doesnt matter.

Pizza

@SineoftheTime did you do anything about algebra and geometry?

Sine of the Time

5:12 PM

yes

by algebra I think you mean linear algebra

Pizza

you can take a look at what I sent above (if you want)

@SineoftheTime yes

Sine of the Time

what message are you referring to?

Pizza

Where I said hi to you, there is another my message above, above again

The one where I wrote "consider the Vector space" etc

Sine of the Time

ok let me take a look

Pizza

Thank you!

Sine of the Time

5:24 PM

the determinant should be $2k^2-k$

Pizza

@SineoftheTime oh yes I checked now

Sine of the Time

so if $k^2-k=0$, the three vectors don't form a basis

Pizza

Ok but the next steps shouldn't change right?

Sine of the Time

I don't understand what you're trying to do if $\det \neq 0$

Pizza

Wait so I also have to consider when k is different from 1/2

Sine of the Time

5:29 PM

right

Pizza

@SineoftheTime I mean that if k is different from 0 or 1/2 then the determinant can never be 0

Sine of the Time

yes

so you have a basis

Pizza

and therefore I can do what is written next, right?

Sine of the Time

you don't need to check if the span $\Bbb R^3$

Pizza

@SineoftheTime $\text{basis means}$: spans entire space & linearly indepdnent?

its wrong?

Sine of the Time

5:32 PM

since you have 3 linearly independent vectors, you already know thay span $\Bbb R^3$

no it's correct

Pizza

aaaaaaa

using this

if there are exactly $n$ vectors, then not even one
all vectors are linearly dependent AND DO NOT cover the entire space
OR
they are linearly independent and span space.

Sine of the Time

yes

Pizza

but as you said "since you have 3 linearly independent vectors, you already know thay span $\Bbb R^3$"

so i can say this : they are linearly independent and span space.

Sine of the Time

yep

Pizza

5:46 PM

@SineoftheTime I was doing something, that's why I didn't reply to you, thank you very much anyway

so the solution is $k \in \mathbb{R} \setminus \{0,\frac{1}{2}\}?$

Sine of the Time