help-27

[ionic harness]

You're right idk why I wrote that

[zenith spoke]

np its ok, i just thought that it might be worth it to mention this since imo its easy to confuse injectivity with being well defined if one isnt careful

[swift dawn]

Why are you so helpful

[devout snow]

@ionic harness Has your question been resolved?

[ionic harness]

\begin{proof}[Proof of \textbf{80}]
  Suppose $A \subsetneq S$ has $\phi : A \to S$ a bijection and suppose $S$ is of finite cardinality.
  Then $|S| = |A| + |S \setminus A| = |S| + |S \setminus A|$ yields that $|S \setminus A| = 0$, i.e., $A \not\subsetneq S$ which is a contradiction. This doesn't feel rigorous but if true proves $S$ is infinite.

  Now suppose $S$ is infinite. Associate each element of $S$ with a number (justification as to why we can do this?). Then $f(x) = 2x$ is one-to-one and onto a subset of $S$ (since the odd-numbered elements don't get mapped to). Therefore we can construct a bijection between $S$ and a subset of $S$ if $S$ is infinite.
\end{proof}

[woven radish]

Coolempire2026

[ionic harness]

I think my brain is mush already

So maybe I shouldn't have added this last one

But here we are

This was the last one

If anybody can omment on my comments within the proof

Then I'll be done

[hot steeple]

As far as i know that works because you assume S is finite which means you can use that kind of argument with the sizes

it doesn't work with infinite cardinality

[zenith spoke]

for the second direction you have implicitly assumed that |S|=|\mathbb{N}| which isnt necessarily true if S is infinite, eg \mathbb{R} is infinite but |\mathbb{R}| neq |\mathbb{N}|

[ionic harness]

I knew that I was doing it as I did it and still did it 🤔

idk why I didn't think about why I couldn't do it

[zenith spoke]

yea i do things like that sometimes when i try to prove something but cant really prove the general case

i think its easier if you try the contrapositive

[ionic harness]

Maybe it's because they didn't formalize it with N_k in this book but I have no clue why I can use that kind of argument

[ionic harness]

Good point

That sounds much better

[zenith spoke]

if you can work on an equivalent statement but for finite set then why not

you have more tools there so it might be better

[heavy current]

if you want to do the second direction directly, one idea might be to ||choose a countable sequence of elements {a_n} from S||, ||and then delete a_1 from S to get A||; ||you can then define a bijection from S to A by mapping each a_i to a_i+1||, ||and anything not in {a_n} to itself||

I didn't check that this actually works but I think it will

[ionic harness]

Can anybody give me a hint as to what it says under the spiler

Since I can't read it without unspoiling th whole thing

[heavy current]

let me split the spoiler into several segments

[hot steeple]

if infinity is P and bijection is Q

then you've proved not P ⇒ not Q right

which is equivalent to proving Q ⇒ P right?

i always forget

always

i hate it

[zenith spoke]

so whats the contrapositive of "S is infinite if there exists a proper subset A of S such that there is a bijection between A and S"

[hot steeple]

If you can prove that assuming there is not bijection between a proper subset and the set

that the set is finite

[zenith spoke]

yea then you are done

[violet wind]

I've seen this used as a definition of infinite

[ionic harness]

Yes isn't this dedekind's definition?

[violet wind]

Is there a different formal definition of infinite they want you to use?

[ionic harness]

Or was his the real one and this was the fake one

[violet wind]

Idk the name

[ionic harness]

You know that's a good question let me read back

[zenith spoke]

so @ionic harness, prove that contrapositive which is what katharine said here (and in the message below it). As a hint, ||consider an injective function f:A->S (why does such a function exist to begin with) and prove that it cant be surjective||

[ionic harness]

Yeah their definition of finite and infinite are realy informal and lacking imo

I much prefer the book that we used way back when

[ionic harness]

This is about as far as I got but stopped trying to think of why I can grab an injective one

Other than saying it's trivial

[zenith spoke]

wdym by trivial?

[ebon coyote]

Thusly, they should have listed N as an infinite set, surely?

[zenith spoke]

it doesnt matter if there is more than one injective map

what matters is that there exists at least one so that you can continue the argument

[ionic harness]

Just start numbering the elements one by one, or the identity function for that matter, are injective, so I can construct an injective one

[ebon coyote]

It turns out this is the definition of a Dedekind-infinite set: https://en.wikipedia.org/wiki/Dedekind-infinite_set

[ionic harness]

Directly after

[ionic harness]

Yep that's what I thought

[ionic harness]

.

[zenith spoke]

yea, thats what is called the inclusion map. It basically embeds the subset A into S by mapping each element of A into itself but viewed as an element of S after being mapped

which is also the identity map of A if you restrict the original map f:A->S to g:A->A

[ebon coyote]

oh crap that's actually a good way to approach it

So it suffices to then show that you CAN get a subset of S that's also (regular-def.) infinite [this is starting to feel incorrect tbh tho]

[zenith spoke]

so now you know that there is at least one injective map A->S

now you want to show that this map cant be a bijection

in other words that it isnt surjective

[zenith spoke]

A set is said to be infinite if it is in-finite fr

[uncut crow]

a vector is any element of any vector space

[zenith spoke]

rereading this, it doesnt look right if i understood what you wrote correctly. numbering the elements of A is a map A->{1,2,..,n} where n=|A|, but thats not a map A->S. Tho you can always find an injective map A->S which is the map that i described to you in an earlier message

[ionic harness]

That was meant to then be mapped to elements of S

[zenith spoke]

I see, yea that works too. Ie, starting with a permutation of A and then injecting that into S in the obvious way (by mapping each element to itself)

tho thats unnecessary since any permutation of elements of A is just A since a set is unordered, so you might as well map A into S directly by mapping each element of A into itself

not trying to be nitpicky, but just clarifying things (i hope i am doing this job)

Anyways, now that you know that there exists an injection A->S, prove that it cant be surjective and you are done (why is it sufficient to do this?)

And sorry if i bothered you with anything, i didnt mean to do that

[ionic harness]

No I'm just very thinking

[zenith spoke]

yea take your time, and if you are confused about anything i said tell me

[ionic harness]

No just annoying at having to finish writing this

😂

[zenith spoke]

real

[ionic harness]

Oh

Am I dumb

Was the first half enough?

I proved that $p \wedge q$ is a contradiction

[woven radish]

Coolempire2026

[ionic harness]

i.e. $\not (p \wedge q)$ is true

[woven radish]

Coolempire2026

[ionic harness]

So $\neg p \vee \neg q$ is true

[woven radish]

Coolempire2026

[ionic harness]

Meaning that $(\neg p \vee q) \wedge (\neg q \vee p)$ is true

[woven radish]

Coolempire2026

[ionic harness]

Assuming excluded middle

wth is going on

Finite and infinite obviously have excluded middle by definition

And a function is either bijection or not bijection

[ionic harness]

The reason I noticed is because in writing this I found I was just rehashing the first half

\begin{proof}[Revised proof of \textbf{80}]
  Suppose $A \subsetneq S$ has $\phi : A \to S$ a bijection and suppose $S$ is of finite cardinality.
  Since $S$ is finite, its subset must also be finite, and we can write $S \approx \NN_k$ and $A \approx \NN_\ell$, where $k < \ell$ since $A \subsetneq S$.
  Then $|S| = k = |A| + |S \setminus A| = l + (k-\ell) = k + (k-\ell)$ yields that $k = \ell$ which is a contradiction. 
  Therefore, $S$ cannot be finite when such a bijection exists, and vice versa.
\end{proof}

[woven radish]

Coolempire2026

[ionic harness]

Someone tell me I've gone crazy

[red grove]

It looks like you proved only one implication

[ionic harness]

It does look like that

[ionic harness]

But if I just switch the order of the first sentence, it looks like I proved the other

[red grove]

The problem is there

[ionic harness]

Assuming excluded middle

It should be the case

[red grove]

Well take p:true and q:false for exemple

[ionic harness]

O you're right 🤔

q is false so the above is true but the below isn't

[red grove]

Yep

So you have to go back to your previous proof

[ionic harness]

But then how does the proof even demonstrate one direction

Because if I rewrite like this

$\neg p \vee \neg q \equiv p \implies \neg q$

[woven radish]

Coolempire2026

[ionic harness]

I could also rewrite

[red grove]

You proved that if S is finite, then there is no strict subset in bijection with S

[ionic harness]

$\neg p \vee \neg q \equiv \neg q \vee \neg p \equiv q \implies \neg p$

[woven radish]

Coolempire2026

[ionic harness]

Ah those are contrapositives

Okay good

Thank you

[red grove]

Then now if you take S infinite, you just have to find a strict subset in bijection with S

[ionic harness]

Right, or show that if no bijections exist, then S is finite

Right?

[red grove]

Yeah, but that seems harder to prove it this way

[ionic harness]

We tried this one but it was difficult

Easy for N-sized sets but not quite for anything else

[red grove]

It’s doable (if you accept the axiom of dependent choice, which I think most sane people do)

[ionic harness]

What's that

[red grove]

In your case, it just means that every infinite set contains a N-sized subset

[ionic harness]

How does that help 👀

I'm learning here how much easier it is to work with naturals than reals 😂

Every idean I have only works with the idea of successors and predecessors

Or maybe it's just fields in general

As long as I can do f(x) = 2x everything starts working 😂

So annoying a set not endowed with multiplication

[woven radish]

Starlord

[ionic harness]

Mmm

They both contain subsets of the same size?

I can map the subsets to each other

I end up mapping every element to every element I assume

I have to think

Oh S is also a subset

[red grove]

That’s a good start, but you should choose these subsets in a specific way

[red grove]

No sorry that’s a typo

It’s just a set

Let’s say you name A a N-sized subset of $S \textbackslash {x}$.
\What N-sized subset of S could you choose so that you can then build a bijection between S and $S\textbackslash {x}$ ?

[woven radish]

Starlord

[ionic harness]

$A$ itself, if it doesn't contain x

[woven radish]

Coolempire2026

[zenith spoke]

Sorry I won't be back for long so I will write the solution I had in mind in spoilers.

||Let A be a proper subset of S and let f:A->S be an injective map, then f is not surjective. Indeed, if |S|=n and |A|=k then k<n and if f is surjective then for each y in S, there exists x in A such that f(x)=y.||
||choose any k distinct elements y_1,..,y_k of S, then there exists elements x_1,..,x_k in A such that f(x_i)=y_i for i=1,2,...,k.||
||Now there is an element s in S with s≠y_i for i=1,2,...,k since |S|=n>k and there exists an element a in A such that f(a)=s since f is surjective by assumption, but then f(a)=f(x_m) for some m in {1,2,...,k} since A has k elements, ie A={x_1,..,x_k} which contradicts the fact that f is injective. Hence an injective map A->S can not be surjective||

||You might want to prove that what I proved above is sufficient to conclude that there is no bijection A->S||

I split the solution into multiple spoilers so that you can get hints without directly getting the whole solution

[red grove]

This is the implication that he already proved though ?

Yep

[zenith spoke]

He proved that if there is a bijection A->S then S is infinite

[red grove]

Which is the exact same

[zenith spoke]

Wait lol

Did I prove the same thing

[red grove]

Yep

[zenith spoke]

There is something weird since I was following with him earlier

And I remember that he proved something else

That's why I typed that direction

lmao

[red grove]

What we’re trying to do is building a bijection between S and S \ {x}. So the subset A’ of S that would be adapted to A should allow us to prolong a bijection between A’ and A to one between S and S \ {x}. What should A’ be ?

[zenith spoke]

@red grove check this, I was proving the second direction

[ionic harness]

O

[zenith spoke]

Which I did

[red grove]

Only in the N-sized case

[zenith spoke]

That's what he proved

Thats why it was missing

But I did prove it for any infinite set

[ionic harness]

A' should be A\{x} can't be right

[red grove]

Almost there, wrong sense

[red grove]

But you’re still missing the other implication

[zenith spoke]

That's why he was trying to prove it after that, because he knew that his argument didn't work for all infinite set but just for countable ones

[zenith spoke]

Yeah because I only proved this direction which he needed

[ionic harness]

A'= A union {x}?

[red grove]

Exactly !

As both are countable, they are in bijection

[zenith spoke]

Anyways I gtg now

[ionic harness]

You're telling me a set with N elements and a set with N+1 elements are in bijection

[zenith spoke]

If you check this later and have any questions about it ping me and I will answer when I can @ionic harness

[red grove]

But S \ (A U {x}) = (S \ {x}) \A

[zenith spoke]

You may not need to check it if you already proved this direction

And of course you are forced to check it either way

[red grove]

N is not a finite number, countable means N-sized

[ionic harness]

Oh you mean N as in $\NN$

Okay

[woven radish]

Coolempire2026

[red grove]

Yeah

[ionic harness]

I was wondering why you chose a capital n for a number

[ionic harness]

That makes this a lot less trivial 😂

[red grove]

I didn’t know the latex shortcut for this one 😅

[ionic harness]

\bN for you

Okay I suppose this all makes sense so far

Glad to be doing combinatorics where I don't have to see infinite anything

[ionic harness]

Right and the rest are in bijection trivially

So we have a bijection from S \ {x} to S

[red grove]

Yep

[ionic harness]

This makes sense

Horrible, but sensible

[red grove]

And S \ {x} is obviously a strict subset of S

Which concludes your proof

[ionic harness]

Yeah

Man only doing the * and ** questions from a textbook really sends me running for my money

[red grove]

This is a reduced version of a more general proof, which might be why it seems weird

The actual property is that for any infinite set S, if D is an at most countable subset of S, and if S \ D is still infinite, then S and S \ D are in bijection

In this case, you apply it to D = {x}, as you only need one example of a strict subset in bijection with S

[ionic harness]

Yes this makes sense

Is there anything else I should write down before I close

[red grove]

Well you have a valid proof for both implications now, so I don’t think so

In any case, I must go to sleep, bye 👋

[ionic harness]

Thank you! 👋

.close

[devout snow]

Channel closed

Closed by @ionic harness

Use .reopen if this was a mistake.

[devout snow]

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[cinder yacht]

AGAIN…

Task 27. (0–1 point)
Complete the sentence. Choose the correct answer from the given options.

The number of all three-digit odd natural numbers whose decimal representation contains the digit 0 exactly once is:

A. 45  B. 50  C. 54  D. 81

[cinder yacht]

So it’s like 101,103,105,107,109

Five times in one 100

9*5 =45

And now this

And every number ending with 0 is divisible by 2…

So it can’t be 110,120,130 right?

Isn’t it just 45

I can’t think of nothing else…

,w 990/2

[woven radish]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by fuzzials

[cinder yacht]

Yea divisible

Isn’t it 45?

Guys

[crude niche]

yeah i got 45 too

[cinder yacht]

This is so easy..

[short hare]

ah?

[short hare]

ah...

well- I dont think your process was right

but you did hit the 25%. good for you

[crude niche]

it is correct

[short hare]

How is
"Isn't it just 45
I cant think of something else
990 is divisable by 2"

the process to solve this question?

[crude niche]

they have this
"So it’s like 101,103,105,107,109
Five times in one 100
9*5 =45"

[cinder yacht]

^^^

There’s no other way to put 0…

[crude niche]

yeah

[cinder yacht]

It can’t be at the beginning nor the end.

[short hare]

I see now.

great method!

apologies.

[cinder yacht]

Np.

[short hare]

hm, do yk where the 9 came from tho?

just making sure

[cinder yacht]

Yea cause there are 9 different hundreds that begin with other number

[short hare]

TRUE!

[cinder yacht]

100,200,300…. - 9 times

[short hare]

you solved it in a more sufficent way than I did

[cinder yacht]

How’d you do that ?

[short hare]

goodjob boo 🤩

[short hare]

ah- for the first digit,
there are 9 possibilities
for the second digit, there is 1 possibility (results from how a number cant start with 0 and also in this case the number cant end with 0 bc as you said it'll be even) so we need the 0 to be in the middle.

for the third digit, there are 5 possibilities (the 5 ODD numbers between 1 and 10)

i think combinations is a better word to phrase this situation

instead of possibilities

[cinder yacht]

Mhm

[short hare]

but yeah! your method was more creative I'd say.

alright fuz, any more questions ?

[cinder yacht]

No ty

[short hare]

you may close the channel, karen 🤨

[cinder yacht]

.close

[devout snow]

Channel closed

Closed by @cinder yacht

Use .reopen if this was a mistake.

[devout snow]

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[remote copper]

I slept early last night so here we go again

[remote copper]

speed of A is x distance Is d

[young spade]

!status

[devout snow]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[remote copper]

1

equation would be

D/(t-42/60)=x+14

x-2=D/(t+0.8/60)

Equate

[young spade]

You probably want to use distance/velocity here.

[remote copper]

(x+14)(t-42/60)=(x-2)(t+0.8/60)

,w (x+14)(t-42/60)=(x-2)(t+0.8/60)

[young spade]

and totally drop the /60.

You can find the answer in hours and then translate for ease of the algebraic expression

[topaz beacon]

thats why the 60 is there

[tropic sinew]

distance should be the invariant not speed

[remote copper]

(x+14)(t-0.07)=(x-2)(t+0.013)

[remote copper]

Is this wrong?

[tropic sinew]

careful 4.2 minutes is 0.07 hours not 0.7

[remote copper]

Ohh wait

[tropic sinew]

actually keeping the /60 is cleaner but this is fine too i guess

[remote copper]

What I do next?

[topaz beacon]

are we to assume D=xt?

i think thats the only way to get anywhere

wait it might not even do it

[devout snow]

@remote copper Has your question been resolved?

[remote copper]

we need something more?

[tropic sinew]

you have to use this

[devout snow]

@remote copper Has your question been resolved?

[devout snow]

Channel closed

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[devout snow]

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[heavy current]

Suppose $(M, g)$ is a connected Riemannian manifold, $S \subseteq M$ is a connected embedded submanifold, and $\widetilde{g}$ is the induced Riemannian metric on $S$. Prove that if $(M, g)$ is complete and $S$ is properly embedded, then $(S, \widetilde{g})$ is complete.

[woven radish]

higher!

[heavy current]

the fact that $S$ is properly embedded means that $S$ is closed in $M$, so if we think of $M$ as a metric space so that $(M, d_g)$ is complete, we have that $(S, d_g|_S)$ is complete

[woven radish]

higher!

[heavy current]

but this is where I'm confused: $d_g|S$ is the restriction of the distance function (read: metric, but I'm calling it a distance function to avoid conflating it with the Riemannian metric g) $d_g$ on $M$, while $d{\widetilde{g}}$ is the distance function induced by the induced metric on $S$

[woven radish]

higher!

[heavy current]

these aren't the same things, right?

to be clear, given points p and q in M, d_g(p, q) is given to be the infimum of all piecewise smooth curves from p to q

[heavy current]

let me elaborate on why this is concerning to me

[vagrant skiff]

one is the ambient distance, the other is the intrinsic distance

[heavy current]

the annulus, properly embedded in R^2

when I show that d_g|_S is complete, I'm saying that the ambient distance function on S, given in red, forms a complete metric space

but I need to show that the blue distance function on S is complete, correct?

if I'm misunderstanding something, then do let me know
but if I'm not, then I'm hopelessly lost

S being complete in the ambient metric doesn't seem to do anything for me when I'm concerned about the intrinsic metric

[heavy current]

Pure, I shall summon you to make all my problems go away

I'm about 93% sure I'm misunderstanding

[vagrant skiff]

higher! i'm outside at the moment.. not sure if i'll have time to make a proper reply

[heavy current]

ah, sorry

[vagrant skiff]

no worries

[heavy current]

ignore me then

[vagrant skiff]

if i can make time ill reply

[heavy current]

if it matters, I've proven that $d_{\widetilde{g}}(p, q) \geq d_g(p, q)$ for $p, q \in S$ already

[woven radish]

higher!

[vagrant skiff]

If I understand correctly, you're not misunderstanding anything. Near the limit point x, the red and blue distances are comparable (bi-Lipschitz). So once the sequence is sufficiently close to x, cutting across the hole doesn’t matter anymore as all points lie in a small neighbourhood where intrinsic and ambient distances differ by at most a constant factor

[heavy current]

Pure, I don't know what that means
bi-Lipschitz?

this?

is it obvious that red and blue are such though?

oh, Pure #2 is here

[arctic field]

i think all you need to show is that the map $\id : (S, d_{\widetilde g}) \to (S, d_M\rvert_S)$ is uniformly continuous

[arctic field]

then cauchy sequences are preserved

and you can find the limit under d_M|S

[vagrant skiff]

It is as Coriol Anus said via Hopf–Rinow

in simple terms, locally your submanifold S looks like a flat Euclidean plane sitting inside M, and the metric tensors g and g~ vary continuously

the hole doesn't matter locally

[arctic field]

at some point all the points in the cauchy sequence will lie in the same chart

[heavy current]

hmm, can I just argue that the identity is Lipschitz because of this inequality?

with constant 1, that is

so it's unif cont?

[arctic field]

yes

[vagrant skiff]

yes

[heavy current]

I see

[arctic field]

blud what is VIVII

[heavy current]

you two are literally isomorphic

[arctic field]

oh

forget i asked that don't answer

[vagrant skiff]

6 7

[arctic field]

blud answered me anyway

[vagrant skiff]

too late

[heavy current]

[uncut crow]

[heavy current]

okie dokie, this problem is solved

thank you @arctic field and @vagrant skiff for the help

and thank you @uncut crow for the moral support

.solved

[devout snow]

Channel closed

Closed by @heavy current

Use .reopen if this was a mistake.

[uncut crow]

[heavy current]

[vagrant skiff]

[arctic field]

[devout snow]

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[eager stone]

i just dont get these type of questions at all even tho i know what all of these words mean in theory, can someone help

[heavy current]

hmm, what do you find particularly confusing?

[eager stone]

in the actual exam i wrote everything basically inverse except the 5th question..

[heavy current]

I’d need to hear your reasoning for these problems then

I’m not sure where you’re lost

if you know all the definitions and a few basic properties, it shouldn’t be too bad

[heavy current]

e.g for question i), what incentivized you to respond the way you did?

if you’re just guessing, then that would be a problem

[eager stone]

sorry i dont really remember my reasoning at the time but is the 1st one about how col space changes with matrix manipulation and row space doesnt?

[heavy current]

yeah, that’s the question :p

do you remember what elementary matrices are?

[eager stone]

damn ok.. 😭 i definitly knew that

[heavy current]

^^

[eager stone]

nott really

[heavy current]

an elementary matrix is any matrix you get by performing one row operation to the identity matrix

left multiplying a matrix by an elementary matrix corresponds to doing a row operation

and row operations don’t change the row space of a matrix :p

on the other hand, they have no obligation to preserve the column space

[eager stone]

i just remmeber it as col(R)=/ col(A) so i guess i got confused.. dumb mistake

[heavy current]

I don’t know what R is there

[eager stone]

its the reduced row echolen form of the A matrix

i dont think it is an offical like term it is just what we use while solving questions

[heavy current]

ah, then sure

but then you got it backwards

col space can change, row space cannot

in any case, the rest of the problems are quickly solvable using some basic tools you’ve seen in linear algebra

[eager stone]

can i get help on the 3rd one also? why is it non invertible

[heavy current]

do you remember how the determinant changes under row operations?

specifically, under the operation of swapping two rows?

[eager stone]

it is multiplied with -1 right

[heavy current]

yup

what does this tell you about det A?

remember: A consists of all 7s, so swapping two rows still leaves you with A

[eager stone]

oh det(-A)=det(A)?

[heavy current]

mm, not det(-A)

[eager stone]

-det(A)..?

[heavy current]

-A is a different matrix, with entries all -7

[heavy current]

yes!

[eager stone]

okkk that makes a lot of sense

[heavy current]

-det(A) = det(A)

so it’s 0

[eager stone]

smart

[heavy current]

I have to go to bed now @eager stone

it’s… rather late for me

[eager stone]

thanks a lot for the help!!

[heavy current]

but my suggestions for ii) and v) are the same: use the rank nullity theorem; for iv), note that a nontrivial null space means the matrix is non-invertible (hm, you can use the rank-nullity thm here too :p)

okay, good night, and good luck!!

[eager stone]

tyty goodnight

[vast zenith]

have you seen invertible matrix theorem?

[eager stone]

the one with det orr?

oh this im not sure

i think it was after this exam

[vast zenith]

it’s a list of equivalent statements for when a matrix is invertible

but it’s really useful for situations lik these

for example with question 3 you could see that the matrix is not row equivalent to the n x n identity matrix (which is listed as b in the theorem) and deduce that it’s not invertible

since it’s not invertible the determinant is 0

[eager stone]

ohhh that is really useful

thankss i will make sure to memorise it

[vast zenith]

it might be overwhelming but if you learn it you won’t have to memorize much

I think David lays book on linear algebra does a great explanation

[eager stone]

i know some of the properties but i get confused on this type of questions easily

i will try to learn it fully..

.close

[devout snow]

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[normal bolt]

I am not getting approach for both the questions

We can start by 24 first

[lavish sigil]

Do you know how displacement on chained pullies work?

[normal bolt]

using constraints.

[lavish sigil]

Uh

Well

Okay

Have you started balancing the forces around

[normal bolt]

T = mg
2T= Kx_o (x_o is initial displacement due to m block)

=> 2mg = Kx_o

and constraint equation will be x_o = 3x

[lavish sigil]

What is x

[normal bolt]

further displaced by me

for small oscillations

[lavish sigil]

I don't generally consider gravity in the mix, it'll cancel out anyways

If the middle pulley moves by $x_2$ upon $m$ being disturbed, with string tension $T$, we have that $kx_2 = 2T$. Similarly, if the left pulley moves by $x_1$, you have that $2T = kx_1 + kx_1$. And the total displacemenet $x$ of $m$ would be $2x_1 + 2x_2$

[woven radish]

jewels!

[lavish sigil]

You get T = (k/6)x which would be the magnitude of the restoring force on m

[normal bolt]

where are you

i am waiting

[devout snow]

@normal bolt Has your question been resolved?

[devout snow]

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[spring salmon]

How do I start

[harsh sierra]

since you know the answer is with e, try to get something that looks like the definition of e as a limit

[spring salmon]

If ican get 1^(infinity) form then getting e is possible

Btw how is that term outside the function written ? Is it (1/x^x)²

[lavish sigil]

I'd start by taking that logarithm of the whole thing

[frozen cedar]

I think it would be x^(1/x²)

[lavish sigil]

That would turn it into a sum which is easier to manage

[earnest valve]

No?

[lavish sigil]

It is 1/x^(x^2)

[earnest valve]

Its (1/x)^(x^2)

[frozen cedar]

yea

mb

;/

I thought it had the negative next to the 2

[spring salmon]

Are these two the same

[lavish sigil]

No

[earnest valve]

Let that whole expression be equal to some L, take log on both sides and see if u can simplify it

[earnest valve]

No

[lavish sigil]

The right expression is 1/x^(2x)

[spring salmon]

Ouu

Then these two?

[steep mulch]

Yes

[earnest valve]

Yea they are the same

[steep mulch]

Notice the other expression is also raised to x

[spring salmon]

Yessyes

Well I got 1^(infinity) type

Don't know how to simplify afyer that

[lavish sigil]

Insert an e^ln in there

$\lim f = \lim e^{\ln f} = \exp \lim \ln f$

[woven radish]

jewels!

[spring salmon]

How do I simplify this 😔💔

[lavish sigil]

Uh

What have you done

[spring salmon]

I applied the first formula my teacher gave me

[pastel pasture]

Okay but what happened to the (x + 1)^x

[spring salmon]

I brought x^x inside and divided it with each term inside the {brackets}

[pastel pasture]

So that would leave you with ((x+1)/x^x)^x

[spring salmon]

Well the question was this so I gave each term in the {brackets one x each bcz there were x terms total in there

[pastel pasture]

Oh

I misunderstood what you were saying

[storm raptor]

I still think you might be better off doing this

[ \exp \lim_{x \to \infty} -x + \prod_{k=1}^{x-1} \left ( x + \frac 1{2^{k}}\right ) ]

[woven radish]

gingerbread house

[lavish sigil]

It's fine, I think she's expected to use the squeeze theorem now

[storm raptor]

yeah that probably works better

I expected some type of Riemann sum but that's probably not happening

[spring salmon]

Ouuu i have to leave now but If i don't get it till tomorrow I'll ask again

.close

[devout snow]

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[devout snow]

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[restive river]

help me here. the correct answer is B. I have no idea how.

[lavish sigil]

What is the defining property of an isotope?

[unkempt junco]

why are we doing chemistry? 😭

[hardy narwhal]

look

its negatively charged

an isotope means same number of protons

so look

B has the same number of protons

[pure kelp]

the isotope of an element has same number of protons but different number of neutrons. Thus, they have same atomic number (Z) but different mass number (A). in your figure it looks like the white balls in the nucleus are representing protons, and the darker ones are representing neutrons.
there are three protons, this implies the atom must have 3 electrons to be neutral. however it has 4 electrons which means it is negatively charged with relative charge of -1, to get the ion, thats a different isotope and the opposite charge, we need to have the same number of protons but different number of neutrons and 2 electrons less (+1 charge)

[hardy narwhal]

and the total charge on it is +1

the diagram shows a -1 charge

hence its oppositely charged and also an isotope

[hardy narwhal]

how

how do u say this much for ts

😭

[devout snow]

@restive river Has your question been resolved?

[devout snow]

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[deft flower]

I know this is not a maths problem but a physics one but it's just 12th grade level so it shouldn't be that hard. the problem is question 5 where they ask for both the value of r and L but we don't have the value of E and I can't find a way to get their value without needing E. we can use the graph shown in the picture and we have R = 90 ohm

[tender kite]

Someone translate that

[tender kite]

Which question is it

[deft flower]

5)a

[deft flower]

some expressions are messed up but you can refer to the original pic

[tender kite]

Ok

[devout snow]

@deft flower Has your question been resolved?

[stoic lotus]

You cannot work out E without L and vice versa

$\begin{cases}
R+r = 200L \
E = 12L
\end{cases}$

[deft flower]

so we're missing infos in the exercise?

[modern lance]

-# You have to enclose it by the dollar sign or [ ... ] now for some reasons

[stoic lotus]

I don't speak French, but from what I can gather, it is implied that E is given as E

r is given as r

Just some arbitrary values

It's pretty common in physics problems

[woven radish]

a handsome russian dude

[stoic lotus]

Thanks

[stoic lotus]

In fact I haven't seen numbers in my physics problems for quite some time now, it's all just arbitrary values and physical constants 🥀

Better get used to it if you're planning to attend uni

[deft flower]

yeah ofc but here they're asking for the value

ig it's just missing info since there's too few data and too many unknowns

well thanks tho !

.close

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[blazing bolt]

Prove that (2n)!/(n!) is divisible by 2^n

[rain summit]

oof

[vagrant skiff]

what have you tried?

[blazing bolt]

Tbh idk if I'm right or idk where to begin

[rain summit]

hm

[blazing bolt]

But I have some sort of idea

[rain summit]

2n! is just 1.2.3...2n

n! is just 1.2.3...n

better to cancel first

[blazing bolt]

Id think if it will real work

[stoic lotus]

Did you mean (2n)!
?

[blazing bolt]

Yes

[rain summit]

i think cancelling should give you some space

[blazing bolt]

The question was given cancelled only I shortened it to this

[rain summit]

(n+1)(n+2)(n+3)...(2n)

so they gave you as this

[topaz beacon]

dude stop

[blazing bolt]

Yea exactly like this

[rain summit]

if you count how many even numbers are in this list

[blazing bolt]

n/2

[topaz beacon]

match the terms like this

--------------
  1   2   3...```

[tender wharf]

You have n even terms from 1 to 2n

[rain summit]

then you should be able to prove that this series is divisible by $2^{\frac{n}{2}}$

[blazing bolt]

[n/4] are divisible by 4

[woven radish]

1 divided by 0 equals Infinity

[spring salmon]

Is this a part of binomial theorem?

[topaz beacon]

oh my god

[spring salmon]

The chapter

[blazing bolt]

I don't think so

[spring salmon]

Hmm

[vagrant skiff]

bruh

[blazing bolt]

It's more like PnC ig

[spring salmon]

Looks really familiar

[rain summit]

idk where's the other $2^{\frac{n}{2}}$ go

[woven radish]

1 divided by 0 equals Infinity

[glossy dew]

i believe theres a funny proof using some sort of grouping

[rain summit]

really?

[topaz beacon]

this will get you the answer quite quick

[glossy dew]

i just cant prove it

[tender wharf]

Sire let's backout

[blazing bolt]

Wait guys calm down

[pseudo basin]

the question says to prove $\binom{2n}{n}$ is divisible by $2^n$?

[blazing bolt]

There comes Ann

[woven radish]

Ann

[zinc flame]

friendly reminder if you cant solve the question on your own first then you shouldnt be the one helping lolol

[blazing bolt]

Say her after long time

[pseudo basin]

any "use this method" / "use anything BUT this method" kind of prescriptions?

if not then i have something that can cook

[rain summit]

go on

[glossy dew]

i dont think thats the same as (2n)!/n!

[topaz beacon]

this isnt true, see 4C2

[rain summit]

GET THE PAN

[pseudo basin]

my bad yes

[rain summit]

basically

[pseudo basin]

thought one thing, wrote another; (2n)!/n! in fact is what i was looking for.

i can think of two different methods to do it.

[rain summit]

$\prod_{i=n+1}^{2n} (i)$ is divisible by $2^n$

[topaz beacon]

oh my god please just look at this

[woven radish]

1 divided by 0 equals Infinity

[tender wharf]

Again

[rain summit]

i think that only get you to $\left \lfloor \frac{n}{2} \right \rfloor$ even terms

[woven radish]

1 divided by 0 equals Infinity

[blazing bolt]

GIF of n/2 no. are divisible by 2
GIF of n/4 no. are divisible by 4
..............
....... And so on

[sand dove]

you have 2n terms in total, not n terms

[turbid bane]

Hi

[glossy dew]

agree

[sand dove]

so n even terms, each divided by their half

[rain summit]

cancelling only gives you n terms

the question asks $\frac{2n!}{n!}$

NOT $\frac{(2n)!}$

[pseudo basin]

and thats enough innit. n factors, each of them 2, times some other uncancelled shit.

[sand dove]

Look at garlic's way of partitioning please

[topaz beacon]

i am going to ask you to kindly back off, youre being more harm than help @rain summit

[glossy dew]

point point

[pseudo basin]

aight so where is OP actually

[sand dove]

Are you still there @blazing bolt? 😂

[pseudo basin]

oh sniped

[sand dove]

Sorry for the disturbance

You can choose whoever to walk you through the solution among the helpers still here, I think it's better if only one helper took care of this from now on

[blazing bolt]

Yes

Oh
Well, I'll like to with Ann

[sand dove]

Well if Ann is fine with that, I'll leave you to it

[pseudo basin]

okay right so

there are two ways:

1. the algebraic, fraction-cancellation way (rehash of what garlic did)
2. the more abstract way involving permutations and combinations

which one do you want

[pseudo basin]

ok

let me think about how to formulate this most cleanly

aight hold on do i actually cook with this

cause unfortunately i might not

[blazing bolt]

Take ur time

[pseudo basin]

or less so than i thought i would

[turbid bane]

Hard question

[pseudo basin]

i mean i can do #1 in the meantime

[blazing bolt]

Well I have also begun to learn abstract maths out of JEE

[pseudo basin]

or give a demonstration of it or something

[blazing bolt]

It's an olympiad question
Ig from Spain and 19_9year

[pseudo basin]

ok so this is gonna be kinda weird and way less elegant than i thought

lets say we have a set of 2n balls, n red and n blue, labelled 1 through n within each color\

so every number from 1 to n is represented on one red ball and one blue ball

[blazing bolt]

Wow i see u thought out of the box

[pseudo basin]

now what i want to say is that (2n)!/n! is the number of ways to place n balls from that entire collection out in a row

but then i have to somehow argue that we can partition these arrangements into subsets of size 2^n (which means some kind of n systematic binary choices that i was hoping to encode in ball colors)

and you wanna say that you can basically just look at the ball numbers and leave the color of each ball up to a binary choice

but if you got both balls of the same number in the row already it doesnt work

and im trying to figure out how to resolve this issue on the fly

[turbid bane]

Hey

I got an idea, you still need help?

[blazing bolt]

Yes

[turbid bane]

Ok so

[tender wharf]

Number one colour even and one odd

[pseudo basin]

dont quite see how that helps us yet

[turbid bane]

Typing it out 1 min

[tender wharf]

Pull 2 as a factor from each even

[pseudo basin]

oh what no

ok you're doing the fraction thing

that is a workable approach but does nothing to help my combinatorial conundrum

[tender wharf]

Hmm..

[sand dove]

@tender wharf see this message, we're interested on the second point rn

[blazing bolt]

Then we'll have [n/4] even integers left

[tender wharf]

Ic

[pseudo basin]

lets forget the fraction thing since you wanted the other approach

[blazing bolt]

Sure

But......I'll be only having 13mins i would have to sleep at 1am

[pseudo basin]

maybe i can re-explain the algebraic way

so you at least walk away w something

cause rn you're getting confused on that

on the numerator, sort the even factors separately from the odds. so you have (1 * 3 * 5 * ... * (2n-1)) * (2 * 4 * 6 * ... * (2n))

and the odd stuff. just dont touch that anymore

and then with 2 * 4 * 6 * ... * (2n), cancel that out factor-by-factor with the n! a.k.a. 1 * 2 * 3 * ... * n that's on the bottom

[turbid bane]

Had the same idea and stuck there

[pseudo basin]

get left with 2 * 2 * 2 * ... * 2

with n factors of 2 there

and that's your 2^n laid bare

[blazing bolt]

Oh actually

Well ty

I'll surely come back tomorrow for abstract method

.close

[devout snow]

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[turbid bane]

So it's solved?

[sand dove]

Well the helpee was content with method 1 of showing divisibility by 2^n

[devout snow]

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[drowsy parcel]

can anyone give me the proof why it is subspace...
I don't want a short answer like ' If v is a span then it is also a subspace'
i am asking you to prove this

1. closed under scalar multiplication
   and
2. closed under addition

[lost laurel]

Is this a test?

[solid ruin]

does anyone know wheres 9lana

[drowsy parcel]

no

[solid ruin]

he was teaching me maths

[lost laurel]

okay, cool

[lost laurel]

Let's start with closure under addition

[iron sun]

Span is by definition the set of linear combinations of those vectors

So it’s closed

Under both addition and scalars

[drowsy parcel]

i know this..
i just want you to prove this to me

[iron sun]

U should try

[iron sun]

This is also a proof btw

[drowsy parcel]

i meant like this

[covert valve]

https://tenor.com/view/lebron-elite-elite-ball-elite-ball-knowledge-ball-knowledge-gif-9337371460101398527

[iron sun]

Let v and w be two linear combinations

Is v+w a linear combination?

Try and show this urself

Then we can also introduce a scalar a

[covert valve]

I am currently studying, but to satisfy this orally, a subspace of a vector space must satisfy three properties, it contains a "zero" vector meaning 0's inside the bracket notation, it is closed under vector addition and it is closed closed under scalar multiplication.

[iron sun]

And ask, is av a linear combination

Then you’re done

[covert valve]

Someone here can represent my reasoning mathematically

and prove it.

Ok bye bye off I go!

[iron sun]

An alternative proof is just to observe that the span of some collection of vectors W in a vector space V

Is the smallest subspace of V that contains W

[drowsy parcel]

while proving closed under scalar multiplication..
do i multiply first vector with n and 2nd one with n and add it?

[iron sun]

What is n?

[drowsy parcel]

any real number

[iron sun]

Well write it out here

What is an arbitrary linear combination for starters

Of ur given vectors

[drowsy parcel]

[1,-4] and [0,2]

[iron sun]

You want to introduce an arbitrary linear combination of those vectors

[covert valve]

The "n" in ℝⁿ just tells you how many dimensions the space has. When he wrote ℝ² - that means each vector has 2 components indicating that its a flat plane. (x,y)

[iron sun]

??

Are u using ai

[covert valve]

No?

[iron sun]

We never said anything about R^n

[drowsy parcel]

? asking or saying?

[iron sun]

Saying

[covert valve]

He used that in his proofs. You asked what "n" meant and I assumed you asked that because you saw that next to ℝ.

[iron sun]

No

[covert valve]

To substitute the 2

[ebon coyote]

If you meant in this, though, he'd clarified this to mean "any real number" - so "R^n" is nonsensical

[drowsy parcel]

ok
I just want to know what are the steps to prove that the v is a subspace closed under scalar multiplication..using the actual values

[iron sun]

Okay

[drowsy parcel]

if i am clear,please understand this...

[iron sun]

You’re clear

But I think I’m not being clear

Okay so to show closed under scalar multiplication

Do u know what that means, in general?

[drowsy parcel]

multiplying with any scalar to both and add the results?

[iron sun]

No

This is not clear enough

[drowsy parcel]

yes..i forgot

[iron sun]

Okay

Let’s revise

Let V be a subset of some vector space

To show it’s closed under scalar multiplication

We show that for any vector v and scalar k in V, that kv is also in V

Makes sense?

[drowsy parcel]

to prove only one vector out of those two that are in the question?

[iron sun]

Forgot the question for a moment

[iron sun]

I want to be on the same page on u regarding this first

If it doesn’t make sense then so
won’t the question

[drowsy parcel]

e.g. please

[iron sun]

Let V = {0} contain only the zero vector of say R^2

To show its closed under scalar multiplication

I let k be an arbitrary scalar

And v an arbitrary vector in V

I want to show that kv is also in V

Okay but v must be of the form v = 0

So kv = 0, for any k

Hence kv is in V

This shows what we wanted

[drowsy parcel]

?

[iron sun]

v is an arbitrary vector in V

But what was V?

[drowsy parcel]

Set of vectors

[iron sun]

Which in my example ?

[drowsy parcel]

0

vector

[iron sun]

Mhm

So what does an arbitrary vector in V look like then?

[drowsy parcel]

1,2,3 ....n

[iron sun]

Hm, what makes u say that?

[drowsy parcel]

you should have taken R^2 vectors

[iron sun]

?

[drowsy parcel]

idk

[iron sun]

Okay let’s do this line by line

[iron sun]

Does this make sense?

[drowsy parcel]

no

[iron sun]

Why not? Something in particular?

[drowsy parcel]

why only 0

[iron sun]

It’s an example

You asked for one

[drowsy parcel]

isn't R^2 has two components?

[iron sun]

Ah I see the confusion

I specified that 0 was the zero vector in R^2

0 is shorthand notation for (0,0)

[drowsy parcel]

ohk

[iron sun]

Okay awesome

[iron sun]

Does this fist like make sense now?

[drowsy parcel]

yes

[iron sun]

Read onwards from here

[drowsy parcel]

ok got it

[iron sun]

Great

[drowsy parcel]

what if there are two vectors like in the que

[iron sun]

Does the general definition of closed under scalar multiplication make sense before we continue?

[drowsy parcel]

yes

[iron sun]

Okay so for the question

The key addition here is that V is the Span of two concrete vectors in R^2

So a vector in V in the question

Is a linear combination of (1,-4) and (0,2)

Do u know what that means?

[drowsy parcel]

multiplying scaler with both and adding them

[iron sun]

Mhm

Could you “mathematically/symbolically” write out an arbitrary vector in V

We’ll use it

[drowsy parcel]

,?

[iron sun]

I want to introduce an arbitrary v in V

what does v look like?

Symbolically

[drowsy parcel]

two vectors?

[iron sun]

This is in words

[iron sun]

No?

[iron sun]

You’ve said urself that you want to prove this concretely with the values

So write out v concretely

I’m not the one that wanted this

[drowsy parcel]

let the questions vectors

[1,-4] and [0,2]

[iron sun]

Mhm

Now what does v look like given those?

(If v is in V)

[drowsy parcel]

is it a linear combinations of thise two?

[iron sun]

Yes

[drowsy parcel]

ohh

so that is the definition of closed under scalar multiplication?

[iron sun]

Part of it

Compare it to my example

[iron sun]

We’re kinda on this step

[drowsy parcel]

it also has to be in the span?

[iron sun]

V is instead Span of two vectors

[iron sun]

Yes

[drowsy parcel]

ok

got it

i think

[iron sun]

we want to show kv is a linear combination of (1,-4) and (0,2)

[drowsy parcel]

ok

[iron sun]

That’s why it’s good to write out v symbolically

Since we know v is already a linear combination

[drowsy parcel]

ok

thank you aslan

closing this now

.close

[devout snow]

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[iron sun]

Closed under wizard hehe

[devout snow]

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[jade relic]

There’s more solutions (answer) to this question how do I find em

[jade relic]

<@&286206848099549185>

[fast garden]

tan repeats every pi

think about that

[jade relic]

OH YEAH

I’m so stupid

[fossil locust]

tan theta = -2 and the period of tan is... ?

Yeah exactly

[jade relic]

Uh

Idk

[fossil locust]

It's this

[tender wharf]

Write in general solutions form

[devout snow]

@jade relic Has your question been resolved?

[devout snow]

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[fervent helm]

what does linearly dependent mean? And how to solve this?

[pseudo basin]

concise but unhelpful answer:

a set of vectors is linearly dependent if there exists a linear combination of them with not all weights 0 which nonetheless adds up to the 0 vector

more targeted counter-question: have you done any linear algebra before yes/no

[covert shale]

if two vectors are linearly dependent, one can be expressed as a multiple of the other
to be a little handwavey. you should know the definitions of linearly independent/dependent already

[fervent helm]

yes

[pseudo basin]

right so then

yeah linear (in)dep is like the first concept you should learn in linalg

unless all you did was matrixfucking

[fervent helm]

well I have no idea

Can somebody explain?

[brittle coral]

which part of this explanatiob did you struggle to understand?

[warped fulcrum]

For example
(3 4 5) and (6 8 10) are linearly independent because 2×(3 4 5)=(6 8 10)

[pseudo basin]

here's another example:

[fervent helm]

how do I use it in vectors?

[covert shale]

for some visual intuition:
a and b are linearly independent
b and c are linearly dependent

[fervent helm]

I mean how do I use it in my original problem

[pseudo basin]

let u = (1,1); v = (-1,1) and w = (0,1). these vectors satisfy u + v - 2w = 0, therefore they are dependent.

[pseudo basin]

in your original problem it is most convenient to use the specifics of when two vectors are linearly (in)dependent

[warped fulcrum]

If the 2 vectors are linearly dependent, then for some k,
x²=-2k(x-1) and b=3akx²

[pseudo basin]

namely the criterion FOR TWO VECTORS ONLY can be stated as follows:

• if either of your vectors is the zero vector, then the pair is dependent automatically.
• otherwise, if one of your vectors is a scalar multiple of the other, then the pair is dependent.
• otherwise, the pair is independent.

[fervent helm]

okay and similarly a and d are dependent too right?

[pseudo basin]

so step 0 is to check if there exist any x's that make V_1 = 0 or V_2 = 0

[fervent helm]

there aren't

[pseudo basin]

ok so you move on then

in fact V1 and V2 are never 0 for any x are they

so make $\vec{V}_1 = k\vec{V}_2$ and figure out what values of $x$ make $k$ exist

[woven radish]

Ann

[fervent helm]

ok

3ax^2 =kb(x-1)

and then the D of this quadratic should be greater than 0 right?

[warped fulcrum]

Thats not certain because we dont know if there are real values of x

[devout snow]

@fervent helm Has your question been resolved?

[devout snow]

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[normal bolt]

how do i take approach to it

[pseudo basin]

combinatorics

but more specifically a route is a sequence of instructions for which way to go

and it will consist of m-1 ↓ and n-1 →

[normal bolt]

m-1 means the person has crossed 1 lines out of total m lines?

[pseudo basin]

no, m-1 means m-1

[normal bolt]

formula?

[pseudo basin]

maybe a concrete example would help?

draw a layout with 5 NS streets and 3 EW streets.

see how many steps in each direction take you from corner to corner

[normal bolt]

3 and 5 respectively

[pseudo basin]

incorrect

show diagram

[pseudo basin]

wrong number of streets

you have 6 NS streets

and you have 4 EW streets

[normal bolt]

mark them please

[covert shale]

as a concept consider the picture below. notice that under these constraints, the red, yellow, and blue routes are all shortest paths
consider being in the position of the traveling man. at any step in your journey you can travel North or East if you want to take the shortest path

this is the wrong orientation but the idea is here

[normal bolt]

but it depends in which direction person goes

depends on which corner the person is standing

if i fix one corner to solve question

then according to this picture i go east or north

[covert shale]

yeah, and according to your question you go east or south instead

[normal bolt]

yes

[covert shale]

imagine writing out your route as a series of moves of either east or south

you'll always have to make some number of east moves and some number of south moves

but you can do them in any order since the grid means the destination will be reached regardless

so how many ways can you write such a route (as a string of easts and souths)

[normal bolt]

person has to move at least one time east and one time south. like this he covers total of m streets and n streets but when the path is created by moving so m and n repeats so is it : (mC2 nC2)/m!n!

[covert shale]

we can actually say exactly how many times the man must move east or south

we must cover (m - 1) North-to-South streets to reach the destination, there are m streets and we already occupy the first

similar logic for the other direction, we must make (n - 1) south moves

[normal bolt]

got it

(m-1+n-1)!/(m-1)!(n-1)!

(m+n-2)!/(m-1)!(n-1)!

[covert shale]

[normal bolt]

awesome explanation

[devout snow]

@normal bolt Has your question been resolved?

[devout snow]

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[drifting tartan]

when doing gaussian elimination do you have to start by making the first row begin with a pivot

[drifting tartan]

or is it not necessary

[stoic lotus]

It's not necessary

But it's usually done. Just like you wouldn't write -a + b
You'd probably want to write b - a, even though it's the same

When performing gaussian elimination, you really want there to be a pivot in the first row for convenience. But you don't technically need it

[devout snow]

@drifting tartan Has your question been resolved?

[drifting tartan]

oh i see thanks

[drifting tartan]

im doing a linalg course at uni and the first step is to do this

so do you reckon id lose marks if i didnt since its not necessary

[devout snow]

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[willow vector]

When i wanna a 5 cm rising on 100 cm it's easy because when i rise by 5 cm on 100 cm distance the rise gonna be 5 cm but what if i wanna get a 5 cm rise on just 60 cm instead of 100 ? how to calculate that

[covert shale]

what exactly do you mean

[proud perch]

https://tenor.com/view/vegeta-dragonballsuper-dragonball-meme-language-of-gods-gif-14707991

[willow vector]

let's say we have a wooden board that's 100 cm in length -> when we rise the other end by 5 cm we get a 5 cm rise because it's 100 cm but how to figure out how much cm i need rise to the the same 5 cm rise ( angle ) but just on a 60 cm long board ?

[willow vector]

yes that's me mostly 😂

[covert shale]

so, this angle? just use trig

[thin basin]

do you want percentage increase?

[deep abyss]

Superfast diagram wow

[thin basin]

(final - intial) / intial * 100%

[willow vector]

yes but i wanna have it in cm rise at the end of the board not in angle measurement if that's possible

[covert shale]

you want the angle in centimeters?

[pseudo basin]

i mean uhhh

the elevation at the end is... 5 centimeters...

[covert shale]

^??

[willow vector]

yes the elevation

[pseudo basin]

but you yourself said you want the elevation to be 5cm

so there's nothing to calculate

[willow vector]

i want to know the elevation in cm of the 60 cm board to be the same like 100 cm board and 5 cm elevation

[pseudo basin]

do you mean that you want the 60cm and 100cm board to be the same ANGLE off the ground?

[willow vector]

i want to get the same elevation on the 60 cm board so the rise how to be different then 5cm not ? like higher number

[covert shale]

this?

[willow vector]

basis is 5cm elevation now how to use the same elevation and calculate how much i need to rise at the end of the board of different board lengths without any angle just elevation in cm

[willow vector]

yes without angle just same elevation

[pseudo basin]

i apologize for asking this, but is English your native language?

[willow vector]

not native but i use it for like 15 years with friends over internet

[pseudo basin]

the thing is

[covert shale]

this you can do easily by scaling the side lengths

[pseudo basin]

you're making approximately zero sense right now

[pseudo basin]

but op says no angles...