12:13 AM
Tomorrow I officially outlive Galois

12:37 AM
@AkivaWeinberger Either you're turning 21 and you've already lived longer than Galois did, and it's easy to check, or you went to the effort of working out how many years and days old he was, taking into account leap years :P

I am not yet 21

I just meant that he was around 144 days short of 21 (too lazy to work out the actual number)

@user69608 I really don't know, because I didn't solve it. I could solve it if I gave it enough time, but if that counts as knowing, then I know all mathematics. integral-calculator.com gives a perfectly OK solution in my opinion

12:57 AM
Hi @Present

Hey @BalarkaSen, what're your focused on these days (math-wise or other)?

Reading probability currently. I would say some mixture of probability and topology

1:12 AM
Do you have many people to talk to about the stuff you're reading?
I feel like my motivation goes up significantly if I have a few people to talk to about my stuff

Yeah, that's usually not a problem
Lots of both topology and probability people out there

Freaky

D:
Spooky

Not totally sure why tbh, but apparently it'll help me with what I'm working on (says my advisor)

1:15 AM
haha
Do you have a strong category theory background?

Oh you're probably the right person to ask, @Present. How do you intuit the shriek map for sheaves over locally compact spaces?
I was reading Verdier duality a few weeks ago and I didn't have anyone to ask much about
Stopped because high investment and low payoff

@Drathora Yeah. My masters thesis was titled 'models for an obstruction theoretic approach to the period-index problem', and I assign to any category which the category of algebraic stacks appropriately 'embeds into' an alternative index to the index of an Azumaya algebra over a scheme (which divides the true index), so lots of category theory is at play

Oh damn, yeah that's a little beyond me haha

@BalarkaSen I haven't thought too much about that stuff tbh, I'm not directly an AG man
@BalarkaSen I'll come back to you if/when I get into that though, perhaps you'll tell me the answer then though :)

OK
Thanks

1:20 AM
I will be reading plenty of AG though over the next two months I suppose, probably Ravi's notes

@BalarkaSen The tendency of the rate of intellectual profit to fall strikes again

I actually have a project in mind soon to extend this paper:
But I'd have to do some brushing up on my category theory first. It wasn't a large focus of mine in the past

@MikeMiller Haha, I suppose. Honestly I meant payoff in the sense of "what is the point, tho?" kind of sense

@Drathora What degree are you working on atm, or what level are you at generally?

I'm a PhD student in Computer Science

1:22 AM
Oh nice

Working in the area of Probabilistic Programming

Does that entail plenty of actual coding? Or are you mostly theoretical?

I am happy to read things I will not immediately use; it's a bit much to ask myself to read 50 pages of jargon before I get to even understand why is what I am reading is interesting to me.

I don't really do any programming. It's all just reasoning about probabilistic programs using measure theory stuff
I've recently acquired an undergraduate student minion actually

@BalarkaSen I'm just trolling

1:24 AM
He's going to implement my research results as an algorithm for his 3rd year assessed project

You should be careful not to take "what is the point" too seriously
here there be tigers

@Drathora Very nice, I wish I had a minion :')
I pickle the owls

well, sure, but I think it is fair to say the "point" of something is a human conceptualization of when that something is interesting to you

Haha, I did some fast talking to my supervisor to convince him that it was better this way than for me to do it myself

so when I say "man this stuff is pointless" I really mean I don't give a shit lol

1:26 AM
@BalarkaSen You'll easily offend many people by saying 'man this stuff is pointless' though

Of course I don't say that in public, I say that in my head
In public I will be like "Ah I see"

Sure
Hmm, has he said "Ah I see" to me before :P

hahaha
That's far too relatable
When someone grabs you and demands you listen to them talk about their poster

Actually the "Ah" won't be there

And there's no good way to escape even though you have no idea what the poster is even about

1:28 AM
It will be a more expressionless "I see"

Surely you've been on the other side of that though @Drathora. We all wish we could grab someone and force them to listen to our favourite math (or whatever interests)

Most likely haha

@BalarkaSen "Why does anyone care about the period-index problem?" - gulps

I don't care about why anyone would care. I would like to hear why it's interesting!

I like to think I'm a little more subtle about it though. My side of that usually consists of writing my problems that I can't solve on the shared office whiteboard

1:30 AM
But what is this mysterious problem anyway?
haha

And then bide my time until they snap and can't help but try to solve it
Working from home has thrown a spanner in that scheme

@BalarkaSen What sort of answer do you want? A long historical answer, or just a statement?

Yeah I don't know the statement

There's the original statement, and then there are many generalised statements, and slight analogous statements, so in some sense there are many 'period-index' problems. The classical problem is to answer the question:

What is the minimal $l\in\Bbb N$, dependent only on the underlying field $k$, such that $ind(A)|per(A)^{l}$ for all central simple $k$-algebras $A$?

I don't know what ind(A) and per(A) is

1:34 AM
Where the period of $A$ is its order in the Brauer group, and the index of $A$ is $\sqrt{\text{dim}_k(D)}$ where $D$ is the division algebra Brauer equivalent to $A$
In general one has that $per(A)|ind(A)$ and that these have the same prime factors

Trying to remember Brauer groups from past life, but failing
For a field it's $H^2(K; \overline{K}^\times)$, right?

It's the group of central simple $k$-algebras with respect to the tensor product, up to Brauer-equivalence: $A\sim B$ if there exists $m,n\in\Bbb Z_{\geq 1}$ such that $A\otimes_k M_m(k)\cong B\otimes_k M_n(k)$
@BalarkaSen Yep

I'm actually forgetting this, is what I wrote correct? $H^2(\text{Gal}(L/K), L^\times)$ should be the correct Brauer group of an extension, where the torus $L^\times$ is naturally a Gal(L/K)-module.
OK, seems good
@Alex Oh, huh

Well this is all generalised significantly

@BalarkaSen respect

1:38 AM
hahaha Mike

First to Azumaya $R$-algebras and the Brauer group of a commutative ring
Then to Azumaya algebras over a scheme and the Brauer group of a scheme

Oh, what's the Brauer group of a scheme?
Can you give me a self-contained definition

Where these are true generalisations, i.e. $Br(\text{Spec}(R))=Br(R)$ and if $R$ is a field, then this is the same as the one above
@BalarkaSen Sure, everything lifts really nicely
It'll take a min to write

Aren't simple algebras over $k$ just matrix rings on division algebras over $k$?
That's part and parcel of Artin-Wedderburn, no?

Central simple k-algebra: What it says on the tin.
Brauer equivalence: $A\sim B$ if there exists $m,n\in\Bbb Z_{\geq 1}$ such that $A\otimes_k M_m(k)\cong B\otimes_k M_n(k)$.

Azumaya $R$-algebra ($R$ commutative): An $R$-algebra with centre $R$, which is finitely generated and projective as an $R$-module, and such that $A\otimes_R A^{op}\cong End_R(A)$ via the homomorphism $a\otimes b^{op}\mapsto (c\mapsto acb)$.
Brauer equivalence: $A\sim B$ if there exists finitely generated projective faithful $R$-modules $E$ and $E'$ such that $A\otimes_R End_R(E)\cong B\otimes_R End_R(E')$
@BalarkaSen $A\cong M_n(D)$ for some division $k$-algebra $D$, and it's a twisted form of a matrix algebra over a field, in that $A\otimes_k k'\cong M_m(k')$ for some $k'/k$

1:48 AM
Isn't $A \otimes_R A^{op} \cong \text{End}_R(A)$ for f.g. projective $A$? It's true if $A$ is f.g. free, and projective is locally free -- am i missing something?
If these modules are locally isomorphic then they are globally isomorphic

@BalarkaSen $A$ is not assumed to be commutative

Oh duh

In every case above, the Brauer group is just the group of equivalence classes of central-simple/azumaya algebras over whatever, and in each case it's a true generalisation

@Alex Um um what happened there? Why is $A \otimes_k k' \cong M_m(k')$?

@BalarkaSen Theorem happened
@BalarkaSen In general it's true that an Azumaya algebra $\mathcal{A}$ over a scheme $X$ is a twisted form of the matrix algebra sheaf for some \'etale cover. In the field case, this means there is some finite separable extension $k'/k$ such that $A\otimes_k k'\cong M_m(k')$ for some $m$

1:54 AM
Oh there are no nontrivial finite dimensional division algebras over an algebraically closed fields, so suffices to take $k' = \overline{k}$
this is easy right?
No need to do etale crap

@BalarkaSen I was just giving the very general form (which is very useful for what I'm working on)

@BalarkaSen It's elementary, but I don't know how you define easy
I'd say easy is when it's both elementary, and takes very little work

Easy is something I understand

Well it's proved in an early section of a (great) textbook
'Central simple algebras and galois cohomology' - Gille, Szamuely

1:57 AM
Ah

The point of the thing I said above, is that then Azumaya algebras correspond (by non-elementary arguments) to PGL_n-torsors (for some n) over the scheme (which brings the pointed set $H^1(X,\PGL_n)$ into play, and as you mention above, then brings $H^2(X,\Bbb G_m)_{tors}$ into play)

Ah that makes sense, then the Galois cohomology comes up when you pass along $H^1(X; PGL_n) \to H^2(X; \Bbb{G}_m)$

sniped

OK I sort of see it
This is cool!

"D
:D
Well it turns out that in many cases the Brauer functor is $\Bbb A^1$-invariant, i.e. for every $X$ the projection $X\times \Bbb A^1\to X$ gives an isomorphism $Br(X)\to Br(X\times \Bbb A^1)$ - so originally I was hoping that $H^1(-,\PGL_n)$ was also $\Bbb A^1$-invariant, but I've created a counterexample to this, so that dream died

2:02 AM
oh no enter Voevodsky
I'm leaving

:')

What are simple examples of G-torsors over X x A^1 which are not equivalent to some torsor over X x A^1 which is trivial in the second factor? I don't know an example for any G
Shows my incompetence with algebraic geometry

I'll have to be careful in answering that
Let me think for a moment to say something precise

Thanks, seems like something I should keep an example of in my head, is all

Yeah so I can give a concrete example for $\PGL_2$ over $\Bbb A^2$ but I'd have to think for awhile to give it explicitly as a torsor, rather than by equivalenjce

2:07 AM

When you say equivalent, you mean isomorphic right? not Brauer equivalent?

Yeah isomorphic as torsors
I am a simple man
I googled something and Morel-Voevodksy has a result that if $G$ has simplicial dimension zero, the map $X \times \Delta^1 \to X$ induces a bijection on the level of $G$-torsors over $X$ for any simplicial sheaf $X$. Isn't the whole point of A^1-homotopy theory that with appropriate Grothendieck topology, Sch/S is a simplicial site where $\Bbb A^1$ is $\Delta^1$?
@BalarkaSen Does that mean this is true if $G$ is a finite group?
I am guessing what words mean at this point

2:24 AM
For nisnevich $G$-torsors for G finite etale, $H^1_{nis}(-,G)$ is A^1-invariant yeah, but I'm interested in \'etale-locally trivial $\PGL_n$-torsors (since these are the ones corresponding to Azumaya algebras)

Gotcha. So actually your kind of examples must be the simplest ones
This is very cool, let me know if you can explicitize your example

Well my example is explicit, just not presented as you might like

haha I like everything I just have to be able to understand it

Let $H$ be the quaternions, and let $P$ be defined by:
$$0\to P\to H[x,y]^2\stackrel{g}\to H[x,y]\to 0$$
where $g(1,0)=x+i$ and $g(0,1)=y+j$
One can show that $P$ is projective of rank $1$ and not free, and upon restricting scalars $End_{H[x,y]}(P)$ is an Azumaya $\Bbb{R}[x,y]$-algebra of rank $4$, so corresponds to a $\PGL_2$-torsor

What's so hard about the 2-category of G-equivariant A^1-invariant Nisnevich sheaves over Azumaya gerbes, where G is an algebraic groupoid scheme?
2

2:27 AM
@Alex Wow, that's super explicit. I like this, let me read and try to understand

@MikeMiller There are actually generalisations of what I'm working on, that are more memish than that sentence

I upped the ante

The Brauer group of a 1-motive M is the Brauer group of its associated Picard S-stack M:
The Picard S-2-stack of Gm,M-gerbes on M is the Picard S-2-stack of Gm,M-gerbes
on M:
GerbeS(Gm,M) := GerbeS(Gm,M).
We've strayed too far from god

indeed
mike was right to upgrade to gerbes

Actually upgrading to gerbes is reasonable, and Grothendieck tried that (for solving the period-index problem)

2:30 AM
lol sure my dude
the pickup artist himself
of course he tried everything

Well, he also actually defined the Brauer group of a scheme, whose definition I recalled above (In his three paper series le groupe de brauer)

lmao
why do we take this wannabe incel seriously
im sure the reason he went nuts at the end is because he failed at a pickup stunt

Grothendieck had a bunch of children
he's a massive player
He even abandoned them like a true chad

yeah im just saying he probably went nuts because he couldn't get some person to sleep with him or something
grothendieck was the rooshV of his generation

lmao, I'm sure you're joking, but I don't know the reference

2:34 AM
its an intelligent guess nothing more
why else would he go nuts

Well Serre and someone else gave their reasons they're confident about

deligne probably became a better pickup artist than him

He hated those young new found contraptions in ANT
Bloody kids these days with their valuation rings

@Alex This is great! Someday you should give me more details on this

@BalarkaSen I can send them to you now if you want :D

2:37 AM
Yeah for sure, assuming of course I understand

It's all elementary, but long

OK, cool

Looks like I've never emailed you before
got it
See page 27 bottom to page 30

Thanks!!

I better get to reading this Levine paper, I'll be back later :D

2:44 AM
@Alex Ah, Raghunathan has some things in this direction

What hasn't Raghunathan done?? Nuts

Am I correct?

@Alex i remember trying to read this a long time ago

sorry, I got the mistake.

1 hour later…
4:04 AM
RREF doesn't guarantee that a system is maximally simplified, since different assignments between variables and columns yields different RREFs, correct?

1 hour later…
5:10 AM
@user10478 Incorrect. Renumbering the variables and changing the columns of the matrix is a different matrix.

2 hours later…
7:13 AM
@Drathora @CalvinKhor Hello!

hey ^^

Can we do some inequality? (I don’t mean discrimination, I meant mathematical inequality lol)

lmao
Post it and I'll see if I can help

You surely can :-)
We want to prove that $$\frac{(2k+2) (2k+1)}{k+1} \gt 4$$
Where $k$ is a natural number

oh
Just simplify the left side and you'll get your answer pretty quickly

7:18 AM
Now, can I do this $$(2k+2) (2k+1) \gt 4(k+1) \\ (k+1) (2k+1) \gt 2k+2 \\ 2k^2 +3k +1 \gt 2k+2$$
$$2k^2 +3k \gt 2k+1$$
Which we can prove by induction I think?

Why not just simplify the left side straight away by noting that $k+1$ is a factor of the numerator
$(2k+2)(2k+1) = 2(k+1)(2k+1)$

Oh yeah!
So, we get $$4k+2 \gt 4$$ ??

yup
And now whether the inequality holds or not depends on which side of the "is $0$ a natural number" fence you land on :P

Lol
Didn’t get you
I think we can factor out 2, and we have $$2k+1 \gt 2$$ which we can argue by $$2k \geq 2 \\ \therefore 2k+1 \gt 2$$
$k$ is a natural number

Yeah, you can essentially just say that the inequality holds when $k$ is greater than one half
Which holds for all natural numbers $k$,if we say that $0$ isn't a natural number

7:31 AM
Thank you so much

@Knight ey m8, u good?

@CalvinKhor What is m8?

mate

Aha! You’re a fashionable man, eh?
F9, you?

lol
whats F9

7:32 AM
(I’m fashionable too)

Fine...? hahaha

Yes

good job, brain

You used 8, I used 9
Lol

lol
where are you getting these problems from, are you prepping for an exam?

7:35 AM
@CalvinKhor yes
I want a Univeristy
To study

sounds good 👍

by the way actual problem was $$\frac{(2n)!}{(n!)^2} \gt \frac{4^n}{n+1}$$
^^^ we have to prove that

Should I show my work so that you can validate it?

cuz its not true for large $n$. You can do that yes :)

7:38 AM
@CalvinKhor What?

Is that $(2n)!$ or $2(n!)$, I assume the former?

Yes former

oh, that might change things

Sorry for unclearity

np :) haha

7:39 AM
Okay, here we go: $$\frac{2n (2n-1) \cdots n!}{n! \times n!} = \frac{2n(2n-1) \cdots (n+1)}{n!}$$
We have $$\frac{2n(2n-1)\cdots (n+1) }{n!} \gt \frac{4^n}{n+1}$$ the statement is true for $n=2$ (for n=1 they become equal)
Let the statement be true for some $n=k$, so we have $$\frac{2k(2k-1)\cdots (k+1)}{k!}\gt\frac{4^k}{k+1}$$ we must notice that k is a natural number greater than 1, therefore $\frac{4^k}{k+1} \gt \frac{4^k}{k+2}$ and hence $$\frac{2k(2k-1) \cdots (k+1)}{k!} \gt \frac{4^k}{k+2}$$

Should be an inequality, not equality in your third line

Sorry
Inductive step: if $n=k+1$ then we have $$\frac{(2k+2) (2k+1) (2k) (2k-1) \cdots (k+1)}{(k+1) k!}$$ and we want to prove that it is greater than $\frac{4^{k+1}}{k+2}$

at the top, last term should be k+2
i think?

oh!
I’m S N I P E D

yeah cuz u divided out $(k+1)!$

7:50 AM
Yes
So, multiplying both numerator and denominator of LHS (after correction that the last term is k+2) we have
$$\frac{(2k+2) (2k+1) 2k \cdots (k+2)(k+1)}{(k+1)^2 k!}$$
Now the extra terms on LHS are $$\frac{(2k+2)(2k+1)}{(k+1)^2}$$ and on RHS the extra term is $4$
(Extra terms mean what are some extra things that are appearing after $n=k$ formulation has been done)
Now, we want to prove $$\frac{(2k+2)(2k+1)}{(k+1)^2} \gt 4 \\ \frac{2k+1}{k+1} \gt 2$$

and this is the part above

I'd advise writing this as LHS > RHS as a result of the inductive hypothesis

@Knight \begin{align} \left.\frac{\frac{(2n+2)!}{(n+1)!^2}}{\frac{4^{n+1}}{n+2}}\middle/\frac{\frac{(2n)!}{n!^2}}{\frac{4^n}{n+1}}\right. &=\frac{\frac{4n+2}{n+1}}{4\frac{n+1}{n+2}}\\ &=\frac{(2n+1)(n+2)}{2(n+1)^2}\\ &=\frac{2n^2+5n+2}{2n^2+4n+2}\\ &\ge1 \end{align}

Rather than hopping like this

I am back with yesterday question again on geometry. But not the same.

7:57 AM
\middle/ is something I will use in the future, thanks lol

I couldn’t understand what does that slanted line mean

it means everything on the left divided by the stuff on the right

May...be... later looks busy

@Knight Do you know what $1/3$ is?

Knight I believe you've made a small error
Your "extra term" on the right should be $4(n+1)/(n+2)$

7:58 AM
@robjohn lol, yes
@Drathora Hmm.. can you please explain it?

Okay so go back to the line before you use your inductive hypothesis

15 mins ago, by Knight
Let the statement be true for some $n=k$, so we have $$\frac{2k(2k-1)\cdots (k+1)}{k!}\gt\frac{4^k}{k+1}$$ we must notice that k is a natural number greater than 1, therefore $\frac{4^k}{k+1} \gt \frac{4^k}{k+2}$ and hence $$\frac{2k(2k-1) \cdots (k+1)}{k!} \gt \frac{4^k}{k+2}$$

Now do your $k+1$ case
And when you apply the inductive hypothesis, write it as "by inductive hypothesis, LHS > RHS"

For RHS?

Wait I'll just type it properly

8:01 AM

So in the k+1 case we have:

I know this is fibonacci addition law
I don't know it before I google it
the problem is the exercise
1.21
and I really don't want to use induction I want intutive visual proof
I have tried this one but failed to understand the multiplications in the additional law

@robjohn Thanks

It is applied in monotone path so I don't have idea
How do I see the pattern this time?
Yesterday I draw lots of diagram and finally found a pattern about it which was another question and this time is to find identity

@Knight That says that $\frac{\frac{(2n)!}{n!^2}}{\frac{4^n}{n+1}}$ is increasing. Since it is $1$ for $n=0$, it is $\ge1$ for $n\ge0$

8:04 AM
This time is different

$\frac{(2k+2)(2k+1)(2k)...(k+2)}{(k+1)!}$
One moment NoseBleed, I'll take a look at that in a second if nobody beats me to it

$f_{n-1}=f_nf_{m-1}$
I think

So this is where we're at with our $k+1$ case right Knight?

@Drathora ok you can take as much as you want

@Drathora yes

8:07 AM
Okay, so it's worth noting that you didn't actually need to do this simplification this time, you could have just simplified to $\frac{(2k+2)(2k+1)(2k)!}{(k+1)^2 (k!)^2}$ and used the inductive hypothesis from here

tag me when you see the pattern

@Drathora yes

Call that thing I just wrote LHS to save me typing time
So we have:
$LHS > \frac{(2k+2)(2k+1)}{(k+1)^2} \times \frac{4^k}{k+1}$ by inductive hypothesis
As everything here is positive
Are you happy with that?

Didn’t get what you did

Since we know that $\frac{(2k)!}{(k!)^2} > \frac{4^k}{k+1}$ by the inductive hypothesis

8:11 AM
You have multiplied both sides by the extra terms?

I'm just doing induction on the problem you asked
If you prefer we can do the induction you set up

I’m good with your set up

Okay, so yeah obviously we can prove the inequality in the problem holds for $n=2$
Then we assume it holds for $n=k$
And now we're up to the inequality I've written that begins with "LHS"
So now it remains to show that $\frac{(2k+2)(2k+1)}{(k+1)^2} > \frac{4(k+1)}{(k+2)}$

I think it is
We have $$\frac{(2k+1)}{(k+1)} \gt 2\frac{(k+1)}{k+2}$$
(I mean we have to prove ^^^)

yup
So just move each denominator across, noting that they're greater than 0
And then multiply out these quadratics
Then the answer should become clear

8:21 AM
We have $$2k^2 +5k + 2 \gt 2k^2 +4K + 2$$
Oh yeah!

And by answer I mean the fact that this inequality is true of course

5k is greater than 4K

Or alternatively, $k > 0$
You can do it perfectly fine the way you tried to set it up also, just be careful that you're actually trying to prove the right thing

@Drathora How you got this?

As in how did I think to show it this way?
I just took my LHS when $n=k+1$, and immediately tried to put it into a form where I could see the LHS from when $n=k$ within it
So that I could use the inequality of my inductive hypothesis
Then after using that, you see what else you need to do to get to the inequality you're looking for

8:28 AM
When $n=k$ we had our LHS as $$\frac{(2k)!}{(k!)^2}$$ and when $n=k+1$ we have $$\frac{(2k+2)!}{(k+1)!^2} = \frac {(2k+2)(2k+1) (2k!) }{(k+1)^2 (k!)^2}$$
What you did after this? (I want to learn from you, I know whatever you did was right)

I applied the inductive hypothesis
We assume that when $n=k$ the inequality holds
And when $n=k$ the LHS is equal to $\frac{(2k)!}{(k!)^2}$
So we know that $\frac{(2k)!}{(k!)^2} > \frac{4^k}{(k+1)}$

Yes

So we know that $\frac{(2k+2)(2k+1)}{(k+1)^2} \times \frac{(2k)!}{(k!)^2} > \frac{(2k+2)(2k+1)}{(k+1)^2} \times \frac{4^k}{(k+1)}$

Yes

Since the fractions we're dealing with are greater than 0
So now we want to show that the RHS of the above is greater than $\frac{4^{(k+1)}}{(k+2)}$

8:35 AM
Okay!

And we aim to do that by showing that $\frac{(2k+2)(2k+1)}{(k+1)^2} > \frac{4(k+1)}{(k+2)}$
If you wonder why, just look at what happens if you replace the LHS of the above inequality by the RHS, in the RHS of the inequality on the like "so we know that..."

Yes got it
Thank you so much Drathora

No problem ^^

:-)

@Drathora thanks for fixing knight's proof though I was asked :)
meanwhile just got told in a comment "you can't also ignore my query like an escapist"(??)

8:45 AM
haha
@NoseBleed you still here?
Any path from $1$ to $n+m$ passes through $n$ or $n+1$

There are $f_{n+1}$ paths from $1$ to $n+1$

Number of paths from $n+1$ to $n+m$ is $f_{(n+m) - (n+1) + 1} = f_{m}$

So there are $f_{n+1} f_{m}$ paths from $1$ to $n+m$ that pass through $n+1$.

There are $f_{n}$ paths from $1$ to $n$.

Number of paths from $n$ to $n+m$ is $f_{(n+m) - n + 1} = f_{m+1}$.

However, $f_{m}$ of these paths go through $n+1$, and hence are already accounted for in the earlier part.

So there are $f_{m+1} - f_{m} = f_{m-1}$ paths from $n$ to $n+m$ that do not pass through $n+1$.
Think of $n$ and $n+1$ here as "checkpoints". You take how many ways you can get to them, and multiply it by the number of ways you can get from them to your goal
Since each combination of those two things makes a unique path
And then the bit you have to watch out for is that some of your paths that go through $n$ also go through $n+1$, and hence were already included in your count.