discrete-math

[tough locust]

You have information about x

But you don't care about x, you care about x²

[weary tiger]

the information I have i that x have to be bigger than 3 ?

[tough locust]

I dunno, does it say that?

[weary tiger]

not excplicitly

[tough locust]

Yes, explicitly

[weary tiger]

but if not then left side is false, therefore the other statement can never be true

[tough locust]

It literally and exactly says x is greater than 3

[weary tiger]

so we need to find which x fullfills the left side condition / statement

[tough locust]

Gonna try my clue one last time

You have information about x

How do you get information about x²?

[weary tiger]

by squaring x

[tough locust]

?

What's that mean

[weary tiger]

i am confused

by solving the inequlity for y?

[tough locust]

$x > 3; x^2 > ?$

[vital dew]

gfauxpas:

[weary tiger]

x

9

or something like that I guess

[tough locust]

[weary tiger]

I am actually trying

[tough locust]

Something like that

I said that if 0<y<x

And 0<a<b

Then 0< ay < bx

Use this rule to prove that if x>3, x²>9

Or use any other argument

Why you're allowed to turn x>3 into x²>9

My way is a suggestion, there are other ways

You could also prove x²-9>0 and use a formula for difference of squares

[weary tiger]

did you remove the y?

[tough locust]

From the beginning of our conversation until now

I have been talking about x and x²

Do you think that x>3 implies x²>9?

[weary tiger]

yes

[tough locust]

Can you prove it?

[weary tiger]

with induction maybe ?

[tough locust]

Sure

[weary tiger]

let me see

[tough locust]

I gave an easier way tho

I said that if 0<y<x

And 0<a<b

Then 0< ay < bx

So if 0 < 3 < x

[weary tiger]

if y is between 0 and x

thats what you are saying right

[tough locust]

forget y

[weary tiger]

I said that if 0<y<x

I mean this part

[tough locust]

Oh

Yeah sure

So if 0 < 3 < x

And also

0 < 3 < x

Then by the rule i told you

You can multiply the inequalities together

[weary tiger]

is this because they have a common relation

[tough locust]

It's because

[weary tiger]

or do I just need a break

from life.. right now maybe

[tough locust]

If 0<y<x

[weary tiger]

I feel st

[tough locust]

And 0<a<b

Then 0< ay < bx

So if 0 < 3 < x

[weary tiger]

thats an equality rule

right ?

[tough locust]

And 0 < 3 < x

Then by the rule i just said

What can you conclude

[weary tiger]

then 0 < 3*3 < y

something like that you are looking for

[tough locust]

Y?

0 < 3×3 < x.x my friend

[weary tiger]

since you are multiplying, not sure about the y though, I just noticed the multiplication. Its easier with pen and paper

[tough locust]

Just pattern matching

[stray reef]

3.3

[tough locust]

There :P

I just multiplied 0<3<x together with 0<3<x

To get 0 < 9 < x²

[weary tiger]

as a student i feel like

if u just plugged in some numbers like

0 < 2 < 3

and 0 < 3 < 4

it would go down easier

not u im saying the guy with the question

yes, that helps for me as well

I tried first pluggin in some numbers like 1

which did not fullfill the x > 3

then I understood that the domain had to be bigger than 3, but wasnt sure if this was because of the x > 3, since the domain was just all integers

but since this is for some x

and I found 3 to be true,

then it should be true, since its not for all x, only for some x

so if the left side was true, with values bigger then 3, then that was good, but for the and clause, the right side also had to be true

so I just tried plugging in for values bigger than 3, but got thrown off by the y

[tough locust]

The trick is

[weary tiger]

since I didn't know which values I could plug in. Maybe I phrased my question wrong

[tough locust]

That since you have a statement about x² next to a statement about x

You want to get 2 statements about x² to compare the two predicates

[weary tiger]

why 2 statements

is it because if you can find one wrong

then the whole thing is wrong, isn't that for the all symbol and not some x

[tough locust]

x>3 is one statement, the other is x²+y²=5

[weary tiger]

yes

and I plug in 3

which didnt work, so i tried 4

[tough locust]

You can't plug in 3

x>3

[weary tiger]

and it wworked, but then the right side was false, since != 5, therefore the whole thing is false

for all values of y appareantly

and I am not talking about 4, since y could be whatever, this is what I was confused about

I thought x and y had to be equal if you choosed something from the domain

but I guess they are kind off from different domains

[tough locust]

It doesn't say x=y

[weary tiger]

I Understand that, but the notation threw me off, I had only seen for some x, not some x and y

thats where the confusion came from

[tough locust]

Okay

We still didn't finish thi

[weary tiger]

so if there where like 3 statements, like x > 3, x^2+ y = 5 and x+y+z^3

then z would been from another domain as well

[tough locust]

It does not say x²+y=5

[weary tiger]

haha

it does?

[tough locust]

No

[weary tiger]

It does not say x²+y=5

it doesnt, but its a statement?

[tough locust]

You can't just introduce new predicates like that

[weary tiger]

I get that, I just wanted to understand, what would happen with more variables

[tough locust]

You have no justification to say x²+y=5

[weary tiger]

I mean > 5

my bad

[tough locust]

You didnt finish the problem and you're asking for a harder one

[weary tiger]

😛

I kind of finished it

[tough locust]

x²+y>5 isn't helpful

[weary tiger]

But I am not sure if I would get it on the exam

[tough locust]

No, you really didn't

[weary tiger]

alright

[tough locust]

x²>9 so what

[weary tiger]

where is the y

[tough locust]

?

[weary tiger]

i scrolled up for that question and

[tough locust]

We didn't prove anything about y

[weary tiger]

cant he say x^2 <= 5

[tough locust]

No, because x²>9

[weary tiger]

i mean in the x^2+y^2=5

the equality was wrong, that was my bad

if x^2 > 9 and x^2 <= 5 then its not valid for integers

this is the original question

[tough locust]

How could x²<=5

We said x²>9

[weary tiger]

y^2 can either be 0 or a positive number greater than 0

can you show us this on paper @tough locust

[tough locust]

No

[weary tiger]

Why isn't i <= 12

or is this just shorthand

[pale epoch]

it's shorthand

both i and j are between 1 and 12

[thick sky]

having a little trouble understanding the "left side" part of this, where does that come from?

[sleek swallow]

Your first statement

They just substituted it in

[thick sky]

wouldn't that just produce what's in the "right side" statement?

[sleek swallow]

As in:

$m \leq x \leq m+1$

$m + n \leq x+ n < (m+1)+n = (m+n)+1$

So then use the definition after that

[thick sky]

how does the floor get added to the n part though?

i get that this has to be true on a broader level, but not the mathematical proof of it

[vital dew]

Abhijeet Vats:

[sleek swallow]

Look at the definition closely. You have x+n, which is greater than m+n and less than (m+n)+1. Let x+n = y. Then:

m+n =< y < (m+n)+1 and that just means that [y] = m+ n = [x+n]

[thick sky]

is proving it this way rigorous enough to get full credit on a typical exam?

thank you for the explanation!

[sleek swallow]

I’m pretty sure it works. If your instructor wants you to include additional explanation, you’d probably have to do that.

[soft thorn]

Doesn't it?

I got 18 for both sides

LHS = 3 + 3 * 5 = 18

RHS = 3(25 - 1)/4 = 3 * 24/4 = 3 * 6 = 18

@neon thorn

[dry knoll]

in circle it should be n - l - 1 isn't it?

[sleek swallow]

Why?

[dry knoll]

@sleek swallow maybe im wrong, but can u explain about that index shifting?

[sleek swallow]

Try writing it out the terms explicitly

You should be able to see it.

[elder oak]

anyone knows any step by step tools that solves propositional equations , mega stuck on this
(p∨q) ∧ (r ∨ p) ∧ ( ¬q ∨ ¬r ∨ p)

goal is to simplify it.

[neat fog]

Can't you do this with boolean algebra?

[pale epoch]

what do you consider "simple"?

[elder oak]

i mean u can

[pale epoch]

this is already in CNF, so arguably pretty simple

[elder oak]

it becomes one huge pile of equations the way im doing it

so i miss something

hmm im gonna try again found some new absorption laws.

[sleek swallow]

If you wanna simplify, there is a p common to all of them so you can pull that out using distributivity

@elder oak

[elder oak]

will try.

thanks

[languid ledge]

Hello guys

Is it important to learn set theory?

[pale epoch]

important for what

to do mathematics? the basics are enough

[faint cargo]

hey i've got a quick question, so for proof by induction in the inductive step, i most commonly see people assuming that the statement is true for some n, and then prove that it is true for n+1, but is it also valid to assume that it's true for n-1 and then prove that its true for n? my teacher started using both and it's kinda confused me since the result seems to be the same

[stray reef]

yes

[faint cargo]

ok ty

[weary tiger]

Hello i do not understand this definition

If an induced subgraph has the vertex set S, isn't it just a spanning subgraph

And an edge induced subgraph is when a given an edge in the set of edges E = {e1, e2, ...} in a graph G, and e1 = uv , the edge induced subgraph H only requires that it has at least one edge in its edge set (e1 for example) E' = {e1, e'1,e'1...} and one vertex in u V' common with G?

[shut fjord]

Can someone help me understand part D of this problem

Can someone help me understand part D of this problem

[meager silo]

Hahaha 203? @shut fjord

its because d) is the contrapositive of the question

[stray reef]

what have you tried

[vital dew]

End:

[stray reef]

you can select m-k elements from A instead

or put another way, you can choose k elements of A that WON'T go in your subset

[wheat crypt]

Hello. Could someone explain to me why the following example is an equivalence relation? I'm not sure I understand how to put it into practice.

I'm not sure the thinking is correct, but the tuples are symmetric because m is related to n in m/n and p is related to q in p/q?

[pale epoch]

$(m,n) \equiv (p,q) \iff m \cdot q = p \cdot n \iff p \cdot n = m \cdot q \iff (p,q) \equiv (m,n)$

[vital dew]

Lochverstärker:

[pale epoch]

it follows from symmetry of "="

[wheat crypt]

👍 Ah okay. So it's then reflexive because m . q = m . q ?

[pale epoch]

yes

everything follows from = being a equivalence relation

[wheat crypt]

Ah okay, that makes sense.

[weary tiger]

Anyone know how he found i to be equal to 8?

for x^10

is it because of this step

[faint narwhal]

because taking i = 10 and i = 8 will multiply with the 1 and the 2x^2 outside

to give you x^10

[weary tiger]

oh okey, so you arent looking at the (2x)^i at all

[faint narwhal]

uh you are

[weary tiger]

because thats kind off referring to the [1+2x^2]

I think I understand what you mean now

Do you also happen to know why we are actually multiplying with -1 ^1

would we also do it if we had, 1 - 2x^2 ?

or do we do it since we have + between them

[faint narwhal]

doesn't matter

you only really care about the exponents here

[weary tiger]

ty

I got another question here

I am thinking about (-1)^i

does this come because of the 1 + 2x, and specifically the 1

so if it was (8 + 2x)^-6

then the -1 would be -8 instead?

and same procedure with (-8)^i

or have I misunderstood the -1, hope you understood my question. I am not wondering about the i anymore, so this is not related to the last question 🙂

I don't see whats happening from that step to the step with << x^10>>

Why is it 2^-6 * 2

And what happened to the x, we don't care about the x in the calculation?

[rich condor]

I’m not sure what the question is asking

O.O

[weary tiger]

Hi

could someone drop into an voice channel and explain something for me?

find coefficient of x10 in (1 + 2x2)(2 + 4x)−6.

can someone help me with this

Can wolpfram or symbolab do this one?

I think that becomes for all X

since you are negating the for some x

for all or for some

V this one turned upside down

what do they mean by discourse?

do i negate the p too?

only a name for the course?

i dont even know, the book does a shit job at explaining

No, I think you don't need to negate the p

I believe that ~[Ex whatever] becomes to All X whatever

shoot me if I am wrong

but I believe its like this and the other way

so ~ All X whatever becomes Ex whatever

but not E that way ofcourse

so its not the case that all student is taking a math course

I believe that is the case

since you are negating some x (some students )

and when you negate some x, then I believe it should become for all

oh ok bet, thx

wait a sec

nvm, I am empty battery. Was going to send you a picture

we might be wrong

i looked up an answer online

I agree with this one, but then I misunderstood the question

But yes, that makes sense.

Do you happen to know biniomial expansion?

ahh no i don

dont*^

[agile lake]

something going on in 35

33 and 35 are the same so not sure how you have different answers

[weary tiger]

do the parenthesis mean something?

if so 35 is wrong af

yea i noticed @agile lake

[agile lake]

35 is wrong

[weary tiger]

thanks for claryfying

i cant spell today dff

u know how i can get better at premises and conlcusions

i've watched tons of video and its not clicking

[agile lake]

Do you understand what all the quantifiers mean?

[weary tiger]

like i had to guess because he obviously got rich because he studied hard

yea we only covered the backwards E a nd the upside down A for now

[agile lake]

I assume this is a logic course so there's some pedantry around how you do the derivation but obviously iff means gthat r -> p, therefore p.

[weary tiger]

yea she's basing the course around logic and critical thinking

[agile lake]

Its been a few years so I dont remember all the names of the rules. but the usual idea is, you're given r is true, and you're given q iff r, which implies r iff q, then you conclude q.

[weary tiger]

q iiff r? u mean p if r

[bleak cypress]

I need help

Anybody with discrete

[patent juniper]

When you're given a string and told to do multiple things to it that affect the same places, do you do them in sequential order? For instance
"Swap the first and third characters in the string
Move the third character to the fourth position"
Do I move what was originally the third character or what is currently the third character?

[south olive]

what is the point of diagonal method?

[naive saffron]

I’m sorry, but what’s the diagonal method? @south olive

[south olive]

https://en.wikipedia.org/wiki/Cantor's_diagonal_argument @naive saffron , my teacher just try to teached but I didnt understand want she was doing.

[faint narwhal]

The point of the diagonal argument is to show that the real numbers are uncountable

[naive saffron]

Oh

That

Ok

I thought it was something different

[strong atlas]

why do you need (2,2) to be transitive? are all transitive relations reflexive??

[faint narwhal]

Well you see that we need to add in (2,1)

[strong atlas]

yes

[faint narwhal]

now you have (2,1) and (1,2)

[strong atlas]

ah i see

thanks

[strong atlas]

how would i do this?

[heady geyser]

https://gyazo.com/90ce3e0e2c3eac0ef55eae580a639f04

Can someone help me with part (b)

i dont have any idea how i would "guess" a formula

[pale epoch]

what did you get for (a)?

[heady geyser]

its at the bottom of the photo i sent

[pale epoch]

im retarded

my tip would be calculate a few more and try to see a pattern

that's the intention of the exercise at least

[heady geyser]

alright

[pale epoch]

could be a familiar series of numbers

||it's the triangular numbers||

[heady geyser]

thank you

so the formula is $\frac{n(n+1)}{2}$

[vital dew]

l spacebaR l:

[pale epoch]

yeah

tbh this is not too easy to see

[heady geyser]

yeah i wouldnt have noticed it's the triangular numbers

[pale epoch]

first time seeing induction proofs?

[heady geyser]

no

its just the guess that was messing me up

hindsight is 20/20 though so it makes perfect sense now

lol

[pale epoch]

the better way would've been to think about the recurrence instead of the actual numbers tbh

my tip was garbage

or just $a_n = a_{n-1} + (n-1) = a_{n-2} + (n-2) + (n-1) = \dots = a_1 + \dots + (n-1)$

[vital dew]

Lochverstärker:

[pale epoch]

$= \sum_{i=1}^{n-1}i$

[vital dew]

Lochverstärker:

[heady geyser]

just work it down till i get the base case in the expression?

is that the best way to approach this stuff

[pale epoch]

in general no

[heady geyser]

ive seen the eigenvalue method, but this isnt homogenous so that wouldnt work

[pale epoch]

i think that's a lot of work computationally

in general you can do recurrence relations with generating functions

[heady geyser]

generating functions?

[pale epoch]

yeah, i guess if you haven't covered them, then nevermind

[heady geyser]

hmm im on wikipedia and they seem interesting

[pale epoch]

but you can encode series in a (formal) power series

[heady geyser]

would that come up in a number theory course?

[pale epoch]

possibly

i covered it in a discrete class i think

[heady geyser]

im just taking an intro discrete class atm

[oak wigeon]

Hi! How would I approach this proof?

[gleaming zephyr]

induction

[oak wigeon]

i'm not too sure how to use induction in order to prove this

[sour arrow]

@oak wigeon
Induction always goes the same way.

Prove the base case, where you prove that P(2) is true. (By that, I mean you prove that 2! < 2²)

Then prove the inductive step, where you prove that IF P(n) is true, THEN P(n+1) is true

P(n) being "the statement is true for the number n"

[drowsy trout]

what exactly is the inductive step here though?
my first thought is to expand the factorial and separate the exponent
but where do you go from n!<(n+1)^n

is it even safe to assume you can divide by (n+1) given that n>2

oh wait
youve already assumed n!<n^n
so then n! is obviously <(n+1)^n
is that valid?

[sour arrow]

@drowsy trout
So you have assumed that n! < n^n

And you want to prove that (n+1)! < (n+1)^(n+1)

Divide both sides by n+1:
n! < (n + 1)^n

It's clearly the case that n^n < (n+1)^n, but n! < n^n, proving the inductive step

[drowsy trout]

so what i said?

[little pike]

is this just 7! * 6

[sleek swallow]

$(n+1)^{n+1} = (n+1)^n \cdot (n+1) > n^n \cdot (n+1) > n! \cdot (n+1) = (n+1)!$

[vital dew]

Abhijeet Vats:

[sleek swallow]

@drowsy trout

[soft thorn]

Yes

[balmy nacelle]

hey dudes is anyone here rn?

anyone wanna help me on this?

and like, pls don't just post the answer I kinda want to get there myself

if you can think of a way to guide me towards thinking about it right

[sleek swallow]

okay what do you need help with specifically?

Do you want to prove that?

[balmy nacelle]

yeah, I'm trying to think of examples

that would prove that statement

[sleek swallow]

That's the wrong thing to do

Specific examples may work but that doesn't mean the statement is true in all cases.

So, you need to be more general if you want to prove the truth of the statement

[balmy nacelle]

so does that work in all cases?

[sleek swallow]

Well, let's try and see if it does. We have to prove it if it does. We have to prove it if it doesn't. So, let's assume that it does and try to see if we hit a snag in the proof, yea?

What does it mean for two sets to be equal?

[balmy nacelle]

they have the same elements

[sleek swallow]

In other words, the elements of one set belong to the other and vice versa. So, they are subsets of each other.

Hence, we must prove that $A \cup (B -A)$ is a subset of $A \cup B$ and vice versa. We'll begin with the first statement, okay?

[vital dew]

Abhijeet Vats:

[balmy nacelle]

Sounds good

[sleek swallow]

Let $x \in A \cup (B-A)$. Then, $x \in A \lor x \in B-A$

[vital dew]

Abhijeet Vats:

[sleek swallow]

Understand?

[balmy nacelle]

x is an element of A, and x is an element of B-A?

[sleek swallow]

In this case, x is just an object that is an element of A U (B-A)

Nope, or

It doesn't have to be in both sets

[balmy nacelle]

oops

[sleek swallow]

Are you clear so far?

[balmy nacelle]

yes

[sleek swallow]

Okay, suppose that $x \in A$. Then, $x \in A \cup B$. So, $A \cup (B-A) \subset A \cup B$. If $x \in B-A$, then $x \in B \land x \notin A$. So, $x \in A \cup B$ and the subset relation between both sets still holds.

[vital dew]

Abhijeet Vats:

[sleek swallow]

This proves that $A \cup (B-A) \subset A \cup B$.

[vital dew]

Abhijeet Vats:

[sleek swallow]

Do you follow so far?

[balmy nacelle]

one sec trying to wrap my brain around the first part, you can start typing the next one out though

[sleek swallow]

Well, I wanted you to do the next part, actually.

The next part would be to show that $A \cup B \subset A \cup (B-A)$.

[vital dew]

Abhijeet Vats:

[sleek swallow]

If that holds, then you've shown equality between the two sets.

But yes, take your time in absorbing what i've said above.

[balmy nacelle]

when you do this are you just thinking through this logically and getting it or are you following certain steps?

[sleek swallow]

A more terse proof would've been the following:

$x \in A \cup (B-A) \implies x \in A \lor x \in B-A \implies x \in A \lor (x \in B \land x \notin A) \implies [(x \in A \lor x \in B) \land (x \in A \lor x \notin A)] \implies [x \in A \lor x \in B] \implies x \in A \cup B$

[vital dew]

Abhijeet Vats:

[sleek swallow]

Well, I'm using the idea of mutual containment. That's a standard proof technique for showing that two sets are equal to each other.

^Don't read the terse proof above. Just fully understand the more detailed proof i've written above for you.

A large portion, if not all, of set theory is based off of logic. So, in principle, you could use logic as a way of deducing set theoretic statements. At least, that's the way I learnt to do it.

[balmy nacelle]

So the third statement you make, that starts with "So, A union (B-A).." How did you make the conclusion that that is a proper subset of A union B?

[sleek swallow]

Let A & B be two sets. Then:

$A \subset B \iff [\forall x: x \in A \implies x \in B]$

[vital dew]

Abhijeet Vats:
Compile Error! Click the reaction for details. (You may edit your message)

[sleek swallow]

So, really, what i'm showing is that if i take any element in A, then that element also has to be in B in order for A to be a subset of B

In a similar way, I've taken an arbitrary element, x, from A U (B-A) and i'm showing that that element also belongs to A U B

[balmy nacelle]

so, is it because x is an element of A in the A union (B-A) that you can make the conclusion that that also belongs in A union B?

[sleek swallow]

Well, that's the general idea.

[balmy nacelle]

ok cool, i understand

as for the next step you wanted me to do

didn't you already prove that?

[sleek swallow]

Nope. We haven't proved that $A \cup B \subset A \cup (B-A)$

[vital dew]

Abhijeet Vats:

[balmy nacelle]

but we proved it the other way around tho right?

Abhijeet Vats:

ok that didnt work

https://cdn.discordapp.com/attachments/496785905474994186/679911938121269278/340191992636243968.png

that being the other way around

[sleek swallow]

Yes

We need to prove it both ways

[balmy nacelle]

ok so, i'm not sure how to work the bot that you're using so bare with me as I try to type this out lol

[sleek swallow]

Sure

[balmy nacelle]

so x is an element of A or (x is an element of B and not an element of A), but because it's a union we know that x is an element of A union (B - A), and since we know that x is an element of A union B, then A union B has elements in A union (B-A)

does that look okay?

[sleek swallow]

Nope. You need to begin with the statement that x is in the union of A and B.

So, let $x \in A \cup B$. Then, $x \in A \lor x \in B$. If $x \in A$, then $x \in A \cup (B-A)$. If $x \in B$, then $x \in B-A$ (since $x \notin A$). Hence, $x \in A \cup (B-A)$.

[vital dew]

Abhijeet Vats:

[sleek swallow]

That proves that $A \cup B \subset A \cup (B-A)$

[vital dew]

Abhijeet Vats:

[balmy nacelle]

Well, was I almost right? It looks like the only difference is that I didn't start with x being an element of A union B

[sleek swallow]

Nope. Your proof was not very good because it wasn't very clearly formulated.

It was nearly incomprehensible. It also started with the wrong statement. You had to begin with x in A or x in B.

But don't give up. This stuff takes time to learn. Keep going, you're doing better than most.

[pale epoch]

no

[stray reef]

uh

[pale epoch]

i'll let ann take this now

[stray reef]

ok you typo'd

should be A(k) and k^3 < 3**^**k

anyway

anyway

there are a number of ways you could lay this out

uh.

what

what??

this is what i did last time someone asked me to do this exact problem

er

there's a typo in there hold on

this is proving the inductive step

also typo fixed

no

you don't HAVE TO do anything

and i didn't do any expansion either

i multiplied and divided by k^3 lol

is it

i thought this was relatively straightforward

[loud basalt]

Correct me if I’m wrong ann

That n^a<a^n for n>a

For a>e

[stray reef]

i think so? it looks correct to me at a glance.

the step itself has nothing to do w induction

if you want i can attempt to explain to you the informal reasoning behind doing what i did

i am working with (k+1)^3 which i don't know much about

if you squint, it looks sort of like k^3, which i do know something about

so this step basically... the introduction of k^3

[pale epoch]

i mean you start with (k+1)^3 and want to arrive at 3^(k+1) by only using < and = signs

you only have information about k^3

[stray reef]

so the goal is to introduce the variable on the other side of the equation
no...

the goal, if any, is to relate A(k+1) to A(k)

[sonic herald]

So I'm having a bit of trouble on a problem, it goes as follows: Prove that between every rational and every irrational number that there exists another irrational number. I'm not familiar with my proof techniques so this one's been a struggle. Any help would be appreciated. Thanks.

[sleek swallow]

You could try to explicitly construct such a number. Let a be a rational number and b be an irrational number. Then, consider (a+b)/2.

@sonic herald

[little nacelle]

guys

whats the neutral element

in min plys alebra

[analog sonnet]

,rccw

[soft bobcat]

so

I'm finishing calc ii and one of my options for next year is discrete math

How is it compared to like lin. alg and calc iii and that stuff?

It seems fun definitely but i need other opinions lol

[pale epoch]

did you have proof-based classes yet?

[soft bobcat]

Not yet

[pale epoch]

sometimes discrete is an intro to proof

[soft bobcat]

I'm in HS still

[pale epoch]

and well, discrete differs a lot from place to place

[sleek swallow]

Maybe check the syllabus for the course?

[pale epoch]

could be combinatorics, could be intro to proofs, could be graph theory

[soft bobcat]

I'll take a look at the syllabus (if I can find it. I don't go to the college, but I would be going for this one class)

[pale epoch]

from my experience most students like it

especially graph theory stuff

[soft bobcat]

"Introduction to Discrete Structures"

[pale epoch]

that's more like a computer science class

[sleek swallow]

Oof

[sleek swallow]

May still have proofs

Eyyyyy nice

[pale epoch]

yeah, its usually intro to proofs for CS students

[sleek swallow]

That's very fun

Make sure you enjoy yourself

[soft bobcat]

Hmm

[pale epoch]

yeah, this is intro to proofs

[soft bobcat]

I'll look into it

Should I learn some Linear Algebra before it?

[pale epoch]

i.e. absolute basics if you want to do anything with mathematics

not required

[sleek swallow]

No lol

You learn linear algebra afterwards

[soft bobcat]

Ohh

[sleek swallow]

Linear algebra uses quite a bit of the material in that course above

Well, most of math does

[soft bobcat]

My school says if you finish AP Calc BC, you can choose between Lin. Alg, Calc III (If you get a 5 on the ap exam), and discrete math so I wasn't sure of the like "hierarchy". Poor word choice, Ik, sorry lol

[pale epoch]

if you are in highschool, then this would be a first taste of "what math really is"

[soft bobcat]

Hmm

That sounds like something I would find extremely interesting

[sleek swallow]

It's not a good idea to do lin alg and calc iii

Do discrete structures first

[pale epoch]

i would assume that the lin.alg. class is then computational in nature

i.e. garbage

[soft bobcat]

I can find the syllabus and put it in the proper room

[pale epoch]

unless maybe you want to be an engineer

[sleek swallow]

Yea sure just put it here

I mean, to be fair, computational linear algebra is pretty important

As much as i hate to admit it

[pale epoch]

we taught computers how to do everything we need, so we dont have to

[soft bobcat]

Can I put it here or should i put it in the lin alg room

[sleek swallow]

Put it here

[sleek swallow]

Okay yea looks quite computational

[soft bobcat]

Ik we can only use Scientific Calculators so

[sleek swallow]

A theoretical class would start off with a detailed study of vector spaces

[pale epoch]

this is a lot of stuff

[sleek swallow]

If you want to go ahead with it, you can. Personally, though, I found it really hard to understand LA when starting with matrices.

[soft bobcat]

Yeah, Never done matrices

I have no clue how to do anything with them. But I'm leaning towards discrete math

I do really want to learn proofs. I think after the AP Exam, my calc teacher is gonna start teaching us how to do proofs

[sleek swallow]

Hey, don't let me deter you. Perhaps you might understand it better by starting with matrices. But it didn't work for me.

[pale epoch]

i mean, this covers at least real vector spaces later

[sleek swallow]

Go for discrete structures then. Though, in a way, I'm rather skeptical about the quality of the content that will be presented.

[pale epoch]

but ye, kind of a weird order

and a lot of stuff

[sleek swallow]

It covers both real and complex vector spaces

[pale epoch]

even QR-decomp and least squares

[soft bobcat]

Hmm. Another option is to take Discrete math first semester and Lin. Alg second semester. I can take calc iii freshman year in college if that's a better option

[sleek swallow]

What does calc iii consist of?

[soft bobcat]

I can look into that

[sleek swallow]

If it's multivariable, then you should take linear algebra to reduce the number of things you have to memorize

[pale epoch]

honestly if you do anything with math you will need woth lin.alg and calc 3

and discrete as well

although that is the easiest to pick up without any class in it

at least the syllabus you posted

[sleek swallow]

But idk, if i were you, i'd just take discrete

[sleek swallow]

Like, take discrete in the first semester, then go into each topic in depth on my own in the second semester

[soft bobcat]

That sounds like a good idea

[sleek swallow]

Like, for example, I'd learn the content from my teacher for, say, Set Theory. Then, I'd get a set of notes on Set Theory online or maybe a small book (Naive Set Theory by Halmos maybe?) and just work through that in the second semester

[soft bobcat]

Okay. It's a semester long class I'm pretty sure

I'm still getting used to that lol

But would it be a good idea to do Discrete math 1st semester, Lin. Alg second semester and calc III freshman year of college?

[sleek swallow]

I personally wouldn't.

Discrete Math 1st semester, I'd pick up a linear algebra book that starts with vector spaces for the 2nd semester & work through that, then calc iii in my freshman year of college

^But that's my personal take

[pale epoch]

is your goal even to study math?

[soft bobcat]

Yeah. I want to major in it when I get to college

Should I try to slow down a bit?

[pale epoch]

just try to get a taste of proofs

all serious mathematics is proof-based

[soft bobcat]

Okay.

I've learned

[pale epoch]

if you like it, good

[soft bobcat]

My math teacher does proofs for us in class.

[sleek swallow]

Discrete math 1st semester, then in-depth exploration into your discrete math topics in semester 2

[pale epoch]

avoid classes that are not proof-based

[soft bobcat]

Okay.

[sleek swallow]

Cos the thing is, you have to remember that your classmates will be taking the class as well

And the teacher would, ideally, want them to pass the class. So, the content may not be presented in the most honest way

[pale epoch]

implying teachers care

[sleek swallow]

Oof rekt

Loch handing out rekts left-right-center

[soft bobcat]

I think my teacher does lol

[pale epoch]

honestly the main goal of the teacher - at least at university - should be keeping integrity

[soft bobcat]

He proves everything he can

[pale epoch]

ofc he wants students to pass

[soft bobcat]

He also teaches things past calc at times

[pale epoch]

but if the students don't put in the work, it's not his problem

[sleek swallow]

Alright

[soft bobcat]

He's gone over some more advanced diff eq stuff, taught a bit of number theory, just basic things of later classes because he thought it'd be fun

[sleek swallow]

So if he's honest about the way he's presenting the content, then you're gonna have fun in discrete

But like i said, use the 2nd semester to delve more deeply into each topic you've discussed in class

[soft bobcat]

Okay, thank you. That helps a lot.

[broken berry]

Maybe someone here will know the answer. Because someone said that here is the best place for my question.

I rephrase my question:
How many all possible combinations are there if we choose 2 blank squares from 30 squares, where each blank square can have an image from the set of 12 images, and it can be in 4 diferent positions (the image is rotated for 90 degrees each time)? (We can overlay each square with one image) My solution is C(30,2)124. So C(30,2) combinations times 12 different images times 4 positions of image. Is this correct?

[exotic sun]

Quick question, whats the term for stars and bars but without caring about the ordering?

[tacit willow]

Hey

 

Harry ­ Alisa, Ann, Don  

Karl ­ Alisa, Monti, George  

Alisa ­ Harry, Karl, Ann  

Monti ­ Karl, Monica, Don  

Monica ­ Monti, Don, George  

Ann ­ Harry, Alisa, George  

Don ­ Harry, Monti, Monica  

George ­ Karl, Monica, Ann 

 

Draw the graph representing these eight guests and their friends. Find a seating arrangement where each person sits between two friends (i.e., no-one sits next to one of their adversaries). ```

Could anyone help me with this one?

[lyric pumice]

@broken berry Hello. The correct result appears to be the following.

[tacit willow]

anyone?

[stray reef]

@tacit willow are they sitting in a row or in a circle

[tacit willow]

i figured it out, thanks 🙂

[broken berry]

Thank you @lyric pumice

[minor zephyr]

I am stuck on part b

.... I am especially confused with what the hint is trying to get me to do? I tried counting the operations within those constraints

https://cdn.discordapp.com/attachments/488104216678760469/680918206135009314/wolf.png

Although I don't see how this is any better than just computing total number of operations to begin with since they would both be cubic....?

[little nacelle]

uh hello guys

poo[

[weary tiger]

How would i write:
-15, 18, -21, 24, -27,...
as a closed-form expression?

I know how to write it as a summation but idk about closed form expression

[soft thorn]

It's a sequence though?

[weary tiger]

huh

its infinite

i forgot to add the ...

lol

[soft thorn]

Still a sequence though

[weary tiger]

wat does dat mean

the directions says to write a closed form expression first index is 0

[soft thorn]

How would you write it as a summation for a start

[weary tiger]

n = 0 to infinity (15+3n)(-1)^(n+1)

[soft thorn]

So the nth term is given by?

[weary tiger]

huh

[little nacelle]

hey guys

can somoen explain

how this is 1

in min plus algbra

(0 * a + 1 * 1) + b = ?

how is that 1

[soft thorn]

a_n = (15 + 3n)(-1)^(n + 1) no?

[little nacelle]

😦

[weary tiger]

yes

thats the closed form expression?

[little nacelle]

can someone help

[soft thorn]

Yes

[weary tiger]

wtf

it was that simple?

[soft thorn]

Yes

[weary tiger]

how do i rotate

?rotate

-rotate

anyways for summations, its only complicated like that when i have to adjust bounds?

[soft thorn]

, rccw

[soft thorn]

Perhaps

[little nacelle]

@soft thorn

i have a question

[weary tiger]

Am I doing this induction right

im not sure what to put after the plus sign

[weary tiger]

pls

[sleek swallow]

What is the proposition you’re attempting to prove? @weary tiger

[weary tiger]

I figured it out

i was just having trouble with the algebra tbh

lmao

[sleek swallow]

Alright nice

[weary tiger]

But i have another problem:

Write this in closed for expression:
-15, 18, -21, 24, -27, ...

is that simply:

a_n = (15+3n)(-1)^(n+1)

[sleek swallow]

If it is, prove it. If it isn’t, prove it.

[weary tiger]

so just plug a value in and if i get the expected answer its correct?

[sleek swallow]

I would say that you should attempt to find a recurrence relation for your sequence and attempt to show that your closed form satisfies it.

But tbh, if it does just work for the given terms and the pattern continues in the way it does, i guess you could just write a few lines for why your closed form works

[weary tiger]

Give a recursive definition:
The set S of ordered pairs (m,n) where m and n are positive integers and m is divisible by 3 and n is divisible by 7

Im assuming the base case is: (3,7) in S

am i wrong

[soft thorn]

You can start at 0 can't you?

[sleek swallow]

Positive integers though

[soft thorn]

True

Can't read

[weary tiger]

How would i do the recursive step

if x in S, then

[sleek swallow]

Well, m is divisible by 3 and n is divisible by 7

So what do those statements mean?

[weary tiger]

m = 3k and n = 7k?

[sleek swallow]

What's k?

[weary tiger]

just a placeholder variable, cuz doesnt m is divisible by 3 mean that there exists an integer k, where m = 3k

[weary atlas]

https://i.imgur.com/66InInm.png

Idk where to start for f

[stray reef]

i can't read this

"for every positive irrational number x, there is a positive irrational number y such that y < 1/2 and y < x"

is this it

@weary atlas

[weary atlas]

Yes sorry @stray reef

[stray reef]

try proving the result with the additional assumption that x < 1

[weary tiger]

can someone explain why the set of divisors of the integer a and the set of divisors of the integer −a are the same set.

[vital dew]

Crudichien:

Crudichien:

Crudichien:

[weary tiger]

thank you so much!

[blissful zenith]

can someone explain what this means in words? ∩n∈N[a_n,b_n] = ∅

[weary tiger]

https://twitter.com/SiriusVisuals/status/1232069956621217792?s=20

Sirius (@SiriusVisuals)

Ops? Trying to re-brand to SiriusVisuals and incorporate a S and V into my logo. Disregard colors. Purely looking at the design.

For the left one I tried matching the left angle of the S to the right angle of the V. But, the logo does look thick and short in length.

Twitter

[past dew]

Sorry I misread it it aucatally only have one base case then its proving it

[mortal light]

would this be the right channel to ask about context-free grammars?

[sleek swallow]

That really sounds like a logic thing. Go got #foundations & talk to the qualified individuals there.

[weary tiger]

I need some guidance

@sleek swallow Master

[sleek swallow]

Uh didn't we go through this before?

[weary tiger]

did we?

oof

well we kinda did then u stopped talking 😂

[sleek swallow]

Oh lmao sorry sorry

I got busy

[weary tiger]

no problem lol

[sleek swallow]

Uh okay:

$S = {(m,n) \in \bZ^+ \cross \bZ^+ | (m =3k) \land (n = 7r) \land (r,k \in \bN) }$

[vital dew]

Abhijeet Vats:

[weary tiger]

yes

Base Case would be (3, 7) in S

[sleek swallow]

Well yea

I mean, the above works as a representation for S?

[weary tiger]

yea

i need help with the recursive step

my prof gives us examples like this, then gives us hard questions 😦

[sleek swallow]

Yeap sure, so if $m \in S \implies m+3 \in S$ and $n \in S \implies n+7 \in S$

[vital dew]

Abhijeet Vats:

[weary tiger]

ahhh

it was that simple 😭

[sleek swallow]

Yeap. Obviously, you have to specify that the base case (3,7) belongs to S as well

[weary tiger]

would the recursive step apply for like (6, 21)

[sleek swallow]

What do you think?

[weary tiger]

im not too sure cuz 21 is 7 + 7 + 7, but 6 is 3 + 3

idk if it applies to those cases

[sleek swallow]

Does (6,21) belong to S?

[weary tiger]

yea

[sleek swallow]

Okay is 6 divisible by 3?

[weary tiger]

yes

[sleek swallow]

So all we're saying is that (9,21) also belongs to S

Actually, my phrasing above is kind of wrong

[weary tiger]

(6, 21)

[sleek swallow]

??

[weary tiger]

u said (9, 21) belongs to S but im asking if (6, 21) belongs to S 😂

[sleek swallow]

You answered your own question

You just told me that it does

[weary tiger]

oof

ok ok i get it

[sleek swallow]

Anyways, you need to fix the phrasing above. My phrasing is not correct but the idea just extends itself

[weary tiger]

phrasing for which?

https://media.discordapp.net/attachments/496785905474994186/681738434238218308/340191992636243968.png?width=1443&height=82

this one?

[sleek swallow]

Yes

Because m does not belong to S

S is a set of ordered pairs

[weary tiger]

can i use any variable for that?

like x and y

[sleek swallow]

Sure but the set that it belongs to needs to be changed and the recursion just needs to be modified

[weary tiger]

this is the format were supposed to write it in

[sleek swallow]

Basically:

$(m,n) \in S \iff (3 | m \implies 3 | (m+3)) \land (7|n \implies 7|(n+7))$

[vital dew]

Abhijeet Vats:

[sleek swallow]

Where 3|m just means that m is divisible by 3

Saying that $x \in S$ and $y \in S$ is wrong because they're the components of each ordered pair and they don't belong to S. That's the mistake that I made above and I've corrected that.

[vital dew]

Abhijeet Vats:

[weary tiger]

ahh so (x,y) in S

[sleek swallow]

Yes

Then use the most recent recursion that i gave to you

And obviously, include the base case

Cos the base case is the thing you'll use to construct other elements

[weary tiger]

If (x,y) in S, then (x+3, y+7) in S

that can work right?

[sleek swallow]

Urm okay that's a bit weird

[weary tiger]

i dunno, our prof wants it in these mongoloid terms

lmao

i gotta use english not logic

[sleek swallow]

Because if we begin with (3,7), then you're saying that (6,14) is in S while (6,7) is not in S.

Oh

You can just replace $\land$ with and lol

[vital dew]

Abhijeet Vats:

[sleek swallow]

Like, in my recursion above, as long as the first component of the ordered pair is divisible by 3 and the second component is divisible by 7, it's in S

Idk i'd honestly just go with the very initial statement I made about S:

$S = {(m,n) \in \bZ^+ \cross \bZ^+ | (m =3k) \land (n = 7r) \land (r,k \in \bN) }$

[vital dew]

Abhijeet Vats:

[weary tiger]

alright

Thank you

[sleek swallow]

You're welcome.

Your professor certainly has a weird way of doing this sort of thing.

[weary tiger]

yeah..

heres the second example he gave us

[sleek swallow]

Ya that works because you're not working with ordered pairs

With ordered pairs, you need two definitions.

[weary tiger]

The next question is about strings

this shit

so like a, abc, aaa, abbac

[sleek swallow]

Hmm base case is just a

[weary tiger]

yep i got that already haha

recursive step seems like a big headache

[sleek swallow]

Then the recursion is just going to involve the length of the string

So if the length is divisible by two, you take the length and add 1 to it and that produces a string that does appear to be an element of S.

[weary tiger]

so:
x in S and |x| + 1 is divisible by 2

[sleek swallow]

Yea that works. Then, you just need to include your base case in there and you should be done

[weary tiger]

hmm thats the if statement

whats the then statement 😂

[sleek swallow]

If x in S, then |x|+1 is divisible by 2

[weary tiger]

oh lmfao

[sleek swallow]

Then you also need to say something additional about the first letter in the string

[weary tiger]

yeah was just about to say that

[sleek swallow]

So it's more like, If (x_1 = a) and (|x|+1 is divisible by 2), then x in S

[weary tiger]

kk

[faint narwhal]

what is 12 * 17 mod (29)?

reduce that down, what do you get

uh, how'd you get 209, 17 * 12 is 204

what is 204 (mod 29)

riht

so what does 17x * 12 ≅ 3 * 12(mod 29) become

yes

but how'd we know that the inverse of 17 was 12

[sly pewter]

It's p much established that a = b (mod m) then ac = bc (mod m)

[faint narwhal]

The idea is that (mod 29) doesn't care about multiplies of 29

so when you write 1 = 17(12) + 29(-7)

And take this equation (mod 29)

the right hand side becomes 17(12) + 29(-7) = 17(12) (mod 29)

[sly pewter]

29*c = 0 (mod 29) ya

Nearly like definition of mod

[minor zephyr]

I just realized that this said "is this a more efficient way" and not "is this the most efficient way"

With that said, how could you prove that x^{2^k} can't be computed from x in fewer than k multiplications?

[sly pewter]

I think there could be a way

Actually induction proved me wrong duh

I thought there would be a legit way to do it in shorter step, but this integer world says no

[weary tiger]

How do I prove that my closed form expression is equal to a_n

[minor zephyr]

it looks like you started the induction. Basically, since $$a_k = \frac{k(k+1)(2k+1)}{2} $$ for some $k\geq 1$, you want to show that $$a_{k+1} = a_k + 3(k+1)^2 = \frac{k(k+1)(2k+1)}{2} + 3(k+1)^2 $$ $$ = \frac{(k+1)((k+1)+1)(2(k+1)+1)}{2} $$

[vital dew]

kxrider:

[weary tiger]

so im on the right track

or no

[minor zephyr]

yeah, you have all of the steps written out. The proof is just a bunch of algebraic manipulations

[weary tiger]

oof

i forgot all my algebra

😂

[minor zephyr]

Well, like "prove P(k+1)" is not a proof of P(k+1). The proof involves showing the equality I wrote above.

[weary tiger]

o ok

[minor zephyr]

which involves adding the formula for a_k with 3(k+1)^2 and simplifying and stuff.

[weary tiger]

my brain isnt functioning right after doing this assignment for 5 hours

[minor zephyr]

relatable

[weary tiger]

@minor zephyr can u help me with one more thing

For the sequence -15, 18, -21, 24, -27
Can u confirm if the closed form expression is:
a_n = (15+3n)(-1)^n+1

i just wanna make sure

[minor zephyr]

assuming the sequence starts from 0 (i.e. a_0 = -15) looks fine to me.

[weary tiger]

yup

and for 0, 2, 6, 14, 30, 62, 126, ...
it would be:
a_n = 2^(n+1) - 2
?

[minor zephyr]

yea

[weary tiger]

Thanks man 😄

[minor zephyr]

npnp

[weary tiger]

my head be pounding

[hybrid plume]

hey what is the difference between partitioning and the urn modell ?

exactly the stirling numbers and the formulas with binomial coefficient

[wheat crypt]

Hello. I am currently studying finite state automata, and I came across these questions:

[sleek swallow]

What’s A o B supposed to mean?

[wheat crypt]

Concatenation of A and B

[sleek swallow]

So concatenation of two elements, one from A and the other from B?

[wheat crypt]

Yeah.

Couldn't it be true for all of them except for (e) assuming that one of the sets were concatenating with is only an empty string?

Actually scratch that.

I read it wrong.

[sleek swallow]

Hmm d) also looks problematic

[pale epoch]

so does c)

[sleek swallow]

Ye

[wheat crypt]

yeah. I thought of splitting the string after a 001 substring and count for odd '1's.

[pale epoch]

well

you are basically looking for '001' as a substring followed by at least one '1'

[sleek swallow]

That’s not entirely the best way to do it

This is basically asking you to find two strings that, when concatenated together, produce the string in each part. So if you can find them, you’re gucci

So for example, for a), you can form it by taking a string 10011 from A and a string 1 from B. Concatenating them gives you the string in a)

[wheat crypt]

Hmm, ah makes sense. So it's safe to say that d and e aren't members.

[sleek swallow]

Like loch said, c) also seems to have issues

[wheat crypt]

Hmm, yeah. If i were to separate them, i could get different results. 0010000 and 1 or 1001 and 0000. I feel like i'm missing something and somewhat wrong in thinking that.

[pale epoch]

concatenation is not commutative

[wheat crypt]

Ah got it. so the cut would be at 1001 for A.

[sleek swallow]

No

Then 11 would belong to B, from what you’re saying

Last time i checked, 2 isn’t odd

[wheat crypt]

for c)

[sleek swallow]

Oh okay

My bad i thought you were talking about a)

[pale epoch]

i mean, you can also cut at any later time

[sleek swallow]

The point isn’t where you cut

The point is if you can cut at all

[wheat crypt]

Ah, right. That makes sense.

Thanks for the help guys.

[sleek swallow]

You’re welcome.

[weary tiger]

can someone please help me with this

im not sure how to solve a)

or b) for that matter

[patent juniper]

How many bit strings of length 8 contain exactly 3 1s?

[balmy sandal]

What have you tried?

[patent juniper]

Nothing, I'm not sure how to solve this. Wouldn't it be like, 2^5?

[balmy sandal]

Why?

(Also, pick a room and stick with it!)

[weary tiger]

Can someone help me with the algebra lmao

idk how to make that equate to a_n+1

[balmy sandal]

What part do you need help on?

Well, have you tried actually doing the addition?

[weary tiger]

nah man my algebra is long forgotten

where do i start

😂

[balmy sandal]

Well, first, how do you normally add a number to a fraction?

If I asked you to add 2 to 3/7, how would you do it?

[weary tiger]

14/7 + 3/7

so make it 6(n+1)^2/2?

[balmy sandal]

Good so far. What does that give you now?

[balmy sandal]

And now you add together the stuff on the numerator. Do you remember how to multiply out brackets?

Because I should hope you do.

[weary tiger]

uh

u mean like FOIL

?

[balmy sandal]

What does that get you? Show all your steps.

[weary tiger]

uhh

oh wait made a mistake

should be 13n instead of 7n

[balmy sandal]

Good. Now you just have to factorise the top.

[weary tiger]

it doesnt seem very factorable

[balmy sandal]

Indeed, that's the slight problem, isn't it.

Here; how about this: How would you calculate 5(x+5) + 3(x+5)?

[weary tiger]

15(x+5)

[balmy sandal]

Why?

[weary tiger]

cuz they share the same variable

[balmy sandal]

And how did you get to your answer? Go step by step.

[weary tiger]

wait

its 8(x+5)

not 15

lmao

[balmy sandal]

And how did you get to your answer? Go step by step.

[weary tiger]

i just substituted x for (x+5)

since theyre the same variable

then added 5x + 3x

that sounds hard

why not just factor like terms or factors

its not factorable

sure it is

each term in that expression has x+5 as a common factor

no i mena

the one im doing rn for homework

thats an example

5(x+5) + 3(x+5) = (x+5)(3+5)

https://cdn.discordapp.com/attachments/496785905474994186/681996178526830609/image0.jpg

[balmy sandal]

@weary tiger I'd appreciate if you didn't butt in while I'm trying to teach a student.

[weary tiger]

this one

id appreciate not using discrete math channel to teach algebra

It has to do with induction

😂

so go to proofs

[balmy sandal]

Or maybe to #help-8.

[weary tiger]

can u continue pls 😂

im srry

[flat echo]

Can anybody help me?

[weary tiger]

@balmy sandal u still on 😂

?

[sleek swallow]

@flat echo there is a combinatorial way to interpret that sum that leads into a very nice answer

[flat echo]

How? I'm studying the binomial theorem, but I can't with this exercise

[weary tiger]

i went over that very vaguely in Calculus 3

u learn that in discrete?

[flat echo]

I haven't started with university yet haha

[weary tiger]

oh damn

u self teaching urself?

[flat echo]

I have just finished with highschool

Yeah