discrete-math

[coral raven]

in line 4 we are working with =

[sleek swallow]

Are you seriously going to call people names now?

[coral raven]

not ==

[tawdry edge]

dude....

the variable m

what ever it is

is the modulus

you get it from the modulus

[coral raven]

no it's not, but ok

[tawdry edge]

yes it is...

you are wrong

[coral raven]

so your problem is with the idea that you can multiply m by c?

[tawdry edge]

you derive those formulas directly from the statement a == b (mod m)

[coral raven]

no

[tawdry edge]

how can you be so incredulous

yes you do

I'm done

[sleek swallow]

This person is trolling

[tawdry edge]

You people are helpess

[coral raven]

yeah ok

i give up

@mint bane you can come back now probably

[tawdry edge]

I'm not... you legit lack such a fundamental conception of language or math that you cannot explain yourselves.... at best all I see is regurgitation

[coral raven]

unoriginal

weak

[mint bane]

bruh

[coral raven]

right can you repost your question

it's buried

[errant bear]

what is a hommie

[tawdry edge]

Let $a = b (mod(m))$. And, $c = d (mod(m))$. Then, $ac = bd (mod(m))$.

[vital dew]

MaggyD

[coral raven]

maggy please stop

[tawdry edge]

How do prove this?

[coral raven]

oh it's a different question

i'll get to you later

[errant bear]

you multiply by c

[sleek swallow]

Sideurk, it's pointless

The guy is trolling

[coral raven]

i feel like the first two are right but the third is wrong?

[sleek swallow]

Doesn't actually want to learn shit. He says we're wrong without really saying why we're wrong lmao

[mint bane]

p l e a s e

[tawdry edge]

I don't know how to convey to you that I'm not trolling.... however, I will admit that I was sincerely flummoxed

[errant bear]

im trolling too

[coral raven]

nitezba

i think first two are right

third is wrong

[mint bane]

i thought so as well but why

[coral raven]

ok so you have that the non-t letters can be anything, 25^7

and that the t can only be one thing, 1^1

[tawdry edge]

I'm done debating, because my position was a matter of semantics on the previous question... I understand how you derived the proof, I just don't like it personally how that conclusion is made from the premise. That is my person issue to resolve.

[coral raven]

oh ok

but anyway

so you have the different things that it could be

all the letters

but how many positions can the t be in?

[mint bane]

yeah any position

[coral raven]

so how do you account for it being in any position

[mint bane]

^8

[coral raven]

no

[mint bane]

but then that doesnt guarantee you'll have a t

that's what im stuck at

[coral raven]

ok so imagine you had an 8-letter word with a t to begin with

and then 7 non-t letters

there would be 25^7 words, right?

[mint bane]

yeah

[coral raven]

or well strings

ok but the t could also be the second letter, not the first

and that would be another 25^7 letters

all of which are different

[mint bane]

but it's not 26^7 either is it

[coral raven]

to those where the t is first

no, it's not

so you have 25^7 strings

for each position of t

how many positions can t take

[mint bane]

:O

25^7 * 8

[coral raven]

yes

[mint bane]

thank you

how bout those other 2 tho

[coral raven]

have a go

[mint bane]

i think d is 7!

[coral raven]

not quite, how did you get there

[mint bane]

fix the first letter as t, then factorial cuz no repeats

wait no

[weak shuttle]

@weary tiger when you write a permutation in loops it's like a Concatenation of functions that feed from right to left

[mint bane]

P(26,7)

[coral raven]

that sounds more like it

wait

no

ah you gotta be careful

it's not 26 specifically

because what do you know about those 7 letters

[mint bane]

theyre not t

so 25

[coral raven]

yes

nice

[mint bane]

bet

ok now for the last one

[weary tiger]

I get that part. How do I read it though? '4 is the new number 1' or '1 is the new 4' or are both wrong?

[mint bane]

would the last one be 5 * 26C4

[coral raven]

no

how did you get that

[mint bane]

t can be anywhere in the set

[coral raven]

ok

[mint bane]

need to pick the remaining 4 from any 26 letters

[coral raven]

4??

[weak shuttle]

@weary tiger the image of 1 is 4, the image of 4 is 8

[mint bane]

oh wow im stupid

5 * 26C7

[coral raven]

that looks a little better, uhhh

why 5

[mint bane]

t can be anywhere in the set

althought order doesnt matter

[coral raven]

ok

but how many letters in the string

5?

[mint bane]

wait im being dumb again

8 * 26C7

doing two problems at the same time lol

[coral raven]

almost there

one more thing, why C

[weary tiger]

Okay, ngl im not getting it. I'm gonna do some examples and come back

I appreciate it 🙂

[mint bane]

order doesnt matter

[coral raven]

doesn't it?

[mint bane]

nah, subsets of 8 characters that includes t

there can be repeats

[coral raven]

i mean since it's repeats

[mint bane]

oh wait

lol

[coral raven]

wait actually i think i screwed up 4 for you

misled

[mint bane]

8 *26^7...?

[coral raven]

yeah that looks good

and now i think 4

back to 4, whoops, i slipped up

d

there should not be P

[mint bane]

why not

[coral raven]

because repeats

i don't think you can P and C with repeats

it's just ^

[mint bane]

but d doesnt have repeats

hence its a permutation

[coral raven]

well it could

wait

oh

i'm a fool

[mint bane]

so it is a permutation

[coral raven]

aaaaa yes it's just 25P7

ok

[mint bane]

but then why cant e be a C

[coral raven]

because e can have repeats

right

[mint bane]

yeah but cant choose also work with repeats

[coral raven]

not really

[weak shuttle]

@weary tiger every loop in lin 3 is a permutation and function by itself. Now delta is the concatenation of these functions. so delta(4) is (27)4 which is like Identity(4) which is 4. then (14)4 is 1. then (18)1 is 8 but 8 doesn't appear in the next loops so they apply to 8 like the Identity so delta(4) is 8

[mint bane]

wynaut

[coral raven]

without repeats, how many ways to choose 2 sweet flavours from 2 sweet flavours?

2C2 = 2!/2!0! = 1

with repeats, how many ways to choose 2 sweet flavours from 2 sweet flavours?

2^2 = 4

[mint bane]

huh

neat

thank you very much then

[coral raven]

np

[tawdry edge]

1543x == 1 [mod 1847]

how?

it says tip: use the euclidean algo

but I am, I don't see how I can solve for x...

[naive gulch]

Are you familiar with Euclidian Algorithm and backwards substitution? Like where are you getting stuck with Euclids alg.

Also for the pigeonhole principle

I know we're using [N/ k] but I'm not too familiar with where

We have seven integers and we need to prove two pairs of them both = 11

But our goal isn't to show that [N/ k] = 11 right?

[tawdry edge]

Backwards substitution?

[weary tiger]

Ive been pulling my hair out. Im sure this isn't meant to be this confusing..

I'll come back here after some sleep

[burnt frost]

hi can anyone give me an example of two functions f and g that have these properties:

• f is not surjective
• g is not injective
• g o f is bijective
  and explain why

[steel jacinth]

not sure brodie

[coral raven]

when i read g o f is bijective i immediately just want to be lazy and make g o f = x

does that help? @burnt frost

[weary tiger]

I suggest f: {1} → {1,2} and g: {1,2} → {1} defined in whichever way you wish

whichever of the two ways you can define them, lol

Can someone help me with this problem?

[unreal stump]

Should be O(x^2),O(x!) and O(x log(x))

[weary tiger]

Why would it be those?

[unreal stump]

O(xlogx) is clear,right?

[weary tiger]

yea

[unreal stump]

now logx< x for x>=10(there's a lower x but this works)

so x log x < x^2 for x>=10

now that function < c x log x for some c

Which means function < c x^2

for all x>= x_0 for some x_0

[weary tiger]

ohh okay thank you

[unreal stump]

Similarly for O(x!)

[uncut matrix]

https://i.imgur.com/I8p21Bf.png
why would {a, i} ∈ C be considered false here?

are elements a and i not both in C?

[unreal stump]

{a,i} is very different from a and i

[uncut matrix]

does {a, i} not contain {a}, and {i}

it's talking about the whole {a, i}?

[unreal stump]

{a,i} refers to a set containing a and i

[uncut matrix]

so it would want C = { (a, i)} for it to be true?

[unreal stump]

Think of it as a different object

For instance if C={{a,i},e,u} we say C contains {a,i}

Because you can pick out an element and match it with {a,i} to conclude they are same

[uncut matrix]

hm

so A ∈ B would consider whether the full expression of A exists in B then

[unreal stump]

Yes

[uncut matrix]

meanwhile if you had something like A ⊆ B, you would interpret whether each element within A exists in B

[unreal stump]

Yes

[uncut matrix]

This was really the fundamental understanding that I lacked

thanks so much for clarifying it

gonna redo all these problems and give it another go

[lapis idol]

How can we find the total number of solutions to a + 18b + 24c + 48d + 120e + 240f = 240. I know how to solve these type of questions if the coefficients are 1. Any help is appreciated, thanks!

[stray reef]

are a through f positive integers?

or nonnegative integers?

[lapis idol]

nonnegative integers

[stray reef]

uh huh

you can do this through generating functions

or through some clever simplifications that may reduce the number of variables or make the coefficients smaller

[lapis idol]

I tried to use generating functions at first but I'm not sure what n would represent

I let F(x) = (1 + ...x^240)(1+...x^13)(1+...x^10)(1+...x^5)(1+...x^2)(1+x)/(1-x)^6

[stray reef]

you want the coefficient of $x^{240}$ in $$\frac{1}{(1-x)(1-x^{18})(1-x^{24})(1-x^{48})(1-x^{120})(1-x^{240})}$$ i think

[vital dew]

Ann

[lapis idol]

I'm a little lost. Can you please explain why it would be that?

Does it have to do with multiples of each coefficient?

[stray reef]

yes

[uncut matrix]

Sorry, does anybody know what B ∈ P(A) would be false for:
https://i.imgur.com/I8p21Bf.png

P(A) = {∅, {∅}, {∅, {∅}}, {∅, {∅, {∅}}}} if I'm not mistaken

wouldn't B = {∅, {∅}} be the third element in P(A)

[sleek swallow]

Nope

that third element should be the set containing the set containing \varnothing and {\varnothing}

To keep things simple, let $\varnothing = 0$ and let ${\varnothing,{\varnothing}} = 1$. Then, $A = {0,1}$. So:

$$\mathcal{P}(A) = {\varnothing,{0},{1},{0,1}}$$

[vital dew]

Abhijeet

[unreal stump]

It would just be 1

Not {1}

[sleek swallow]

Nope. 1 is an element of A. {1} is a subset of A

Sure and the power set of A is the set of subsets of A. 1 is an element of A, so {1} is a subset of A

[unreal stump]

mb

B=1

[sleek swallow]

@uncut matrix

As Buncho Drunk just said, B is just 1, in my notation. So, you can see that B cannot be an element of P(A)

[uncut matrix]

Ah, that's really helpful

Thank you!!

[sleek swallow]

You're welcome

[uncut matrix]

I should get into the habit of setting a larger set into a simple letter notation so I'm less likely to make errors & can just substitute the values into it after

good tip!

[sleek swallow]

Indeed, it's especially useful when you have complicated looking things and you're not sure if you're missing braces here or there

[lapis idol]

Thanks a lot! I got it now 🙂

[uncut matrix]

man I keep getting confused

this entire problem set is just a mess for m

https://i.imgur.com/I8p21Bf.png For this:
(∅, ∅) ∈ A x B = True
and ... {(∅, ∅)} ⊂ B X A = True

So, the reasoning for the first one is that:
A x B computes to {(∅, ∅), ...} so there certainly exists a (∅, ∅) in A x B

It is different from the other case for which B = ∅, {∅}, and P(A) = {∅, {∅}, {∅, {∅}, {∅, {∅}}}, where the third element of P(A) = {∅, {∅}} ... so essentially if we let B = 1, then the third element of P(A) = {1} and 1 is not equal to {1}

For the second one, since it's a proper subset, we test(?) individual elements of ∅ within a bracket, and since ∅ and ∅ can be found in at least one element in B X A = {(∅, ∅), ... },
it is true?

[sleek swallow]

Try just using substitutions. Let $\varnothing = 0$, ${\varnothing} = 1$ and ${\varnothing, {\varnothing}} = 2$. Then:

$$A = {0,2}$$

$$B = {0,1}$$

$$A \times B = {(0,0),(0,1),(2,0),(2,1)}$$

$$B \times A = {(0,0),(0,2),(1,0),(2,0)}$$

Now, you want to know if $(0,0) \in A \times B$. That's certainly true. Moreover, it is certainly true that $(0,0) \in B \times A$ so ${(0,0)} \subseteq B \times A$.

[vital dew]

Abhijeet

[sleek swallow]

@uncut matrix

[uncut matrix]

thanks again

I'm trying to wrap my head across this but I'm always getting mixed up somehow

[sleek swallow]

Perhaps a break is in order

[uncut matrix]

for instance, there's another question which says (∅, ∅) ⊆ A x B is false

[sleek swallow]

Well, there is a difference between (0,0) and {(0,0)}

These are not the same objects

[uncut matrix]

wait, scratch that, I think I understand this one - it's false because even if you take ∅, there are no elements of ∅ in A x B, only objects of some bracket combination ()

but yeah I'm more or less just confused by when the brackets appear and when they don't

[sleek swallow]

You'll figure it out soon enough

[uncut matrix]

for reference here were all the questions lol
https://i.imgur.com/mqxH9rm.png

man ...

for {∅} is an element of A
For ∅ = 0, {∅, {∅}} = 1, A = {0, 1}
Then {∅} = {0} and since A only contains 0, and 1, it is false?

thanks for helping me again and again by the way

despite how hopeless I am at this lol

not gonna sleep until I give all the remaining ones a shot

will try to substitute as much as possible

[sleek swallow]

You should rest

There is no sense in banging your head against the wall when you're bummed out by the work you're doing

You can try again when you have rested

[uncut matrix]

alright, I guess I'll go to sleep

this is really just too confusing

thank you for taking your time to help me today

[sleek swallow]

You're welcome

[exotic shadow]

Can anybodey help with these?

[weary tiger]

For 1) What does it mean for a graph to have no cut edges?

Also if it asks to prove this for every integer k, the first thing to try is induction I guess

For 2) observe that an odd cycle cannot be bipartite

So if a a bipartite graph has a subgraph which is not bipartite this contradicts pretty obviously the fact that the graph is bipartite

[exotic shadow]

Like this. As far as I understand

[stray reef]

say G is a 2k-regular graph, so every vertex has degree 2k

imagine you have an edge which, when cut out, makes the graph fall apart into two components

(connected components to be more precise)

within each component, one vertex has degree 2k-1 (the one incident to the cut edge) and the rest have degree 2k

so we have a graph with a single odd-degree vertex

which is impossible

[exotic shadow]

Ohh Okay

I got it now. Thank you!!!

[dense sapphire]

Did anyone struggle with recurrence relations and generating functions and then get it? What was key? Rosen is not helpful, and MIT's 6.042 is losing me as well. are there any videos or anything you watched that helped

[unreal stump]

Try generatingfunctionolgy

It's a book by Herbert wilf

[weary tiger]

can implications, i.e. -> or <-> be used in an algebraic style proof regarding sets?

[unreal stump]

wdym

[weary tiger]

So for instance the statement (C ⊆ (B − A)) → (A ∩ C = ∅)

in an algebraic style proof can I say
C subset (B-A)) -> (B-A) intersect C = C

not for the proof but like is it valid

is it "algebraic"

then to culminate different implications to show that they then imply the result

[proud birch]

would there be a difference among these 2 sentences?

[lament quiver]

you can draw this as a picture

*try to

that's terrible advice but it will show you something

[proud birch]

ohh okay, I just thought they were the same sentence just different ordering lol

[lament quiver]

yeah they're not

[weary tiger]

E = there exists
upside down A looking thing means for every
the C with the line in the middle means "x IN N", and N is the set of all natural numbers

[weary tiger]

so it means

1. For every x in N, there exists a y in N such that x < y.
2. There exists a y in N, that for every x in N, such that x<y.

If you need more help it would probably be good to look up the difference between existential and universal statements

[proud birch]

thank you king

[weary tiger]

@proud birch but yeah the universal quantifier (upside down A) and existential quantifer (backwards E) mean very different things and can result in substantially different claims as well as the difficulty of proving said claims. I.e., every British person is stupid, or there exists a british person who is stupid, the latter can be easily proven, the former cannot

[weary tiger]

Likewise when disproving a statement, you have to basically use the other quantifier, i.e. to disprove the statement
Every British person is dumb
You simply have to show there exists a single British person who is not dumb

However for the statement
There exists a British person who is dumb
To disprove this
You have to either

1. Exhaust every British person showing they are not dumb
   or 2. Show that it is not possible for a single British person to be dumb by traits of a generic particular where the particular is a British citizen

[proud birch]

i see, that makes the notation very much more clearer

[weary tiger]

I forgot to add that the latter part of my message was about DISPROVING a statement

[balmy hornet]

How many bitstrings of length less than n are there? in my notes i have "there are 2^𝑛 bitstrings of length n." so would but be 2^n-1?

[proven garnet]

If a(n) denotes the number of bit strings of length n. You want a(1) + a(2) + ... a(n-1)
which is $$2 +2^2 + \ldots 2^{n-1} $$ and that's a standard sum of a geometric series

[vital dew]

andreO

[sleek swallow]

You forgot about the empty string

[proven garnet]

depends on convention if we wanna include it or not

[balmy hornet]

wait im confused. i dont know what you mean by the inclusion of empty string. so them 2^n-1 is correct in the seme that it is the typcial formula?

[proven garnet]

The sum comes out to $2^{n}-2$

[vital dew]

andreO

[proven garnet]

If we wanna consider the empty string (which i don't think we should in this case), you just gotta add 1 to it to get $2^n - 1$

[vital dew]

andreO

[balmy hornet]

ah ok that makes sense then

thank you!

[uncut matrix]

https://i.imgur.com/kH5F80J.png
Can anybody explain these three please?
(∅, ∅) ∈ A x B
{(∅, ∅)} ⊂ B x A
(∅, ∅) ⊆ A x B

I'm just terribly confused and have been for a few days now

[errant bear]

is this purposely annoying

whoever gave u this exercise should commit sudoku

[uncut matrix]

basically, yes

I'm starting cs soon and discrete'll come in second semester

don't want it to kick my ass so I'm doing questions for it in advance

haven't honestly struggled in it that much (even with basic proofs) until this one collection of questions idk

[errant bear]

i mean i wouldnt worry about this problem. literally ur never gonna need to deal with stuff as convoluted as this

[uncut matrix]

I mean it's going to be really shitty if it comes up on a final and I'm not able to solve it

[errant bear]

i mean, are you able to deal with AxB, where lets say A={1,2,3}, and B={4,5,5}?

you dont need to worry about nested empty sets

[uncut matrix]

yeah ofc

[errant bear]

yea i mean, i doubt your prof is gonna be a dick and give you problems like these

[uncut matrix]

it's when they nest proofs that it becomes some clusterfuck

[errant bear]

i wouldnt worry ab these imo

[uncut matrix]

I mean this is in the problem set though

[errant bear]

i mean, if you want, the first one you asked is true

[uncut matrix]

unless they'll change it for next year*\

hm

[errant bear]

wdym problem set. is this from the class? which you're taking next year/sem?

[uncut matrix]

nah it's for a class in jan

im starting in the fall

[errant bear]

how do you have the exact questions

[uncut matrix]

I found the class site by googling online

it's from like 2019 iirc and all the problem sets were on it

[errant bear]

aight

i mean, okay do u see why the first is true

[uncut matrix]

yeah, some guy earlier taught me about when you examine whether something belongs in a set, you look at it as a whole
so since (∅, ∅) exists as the first element in A x B, (∅, ∅) ∈ A x B is true

the other ones really confuse me though

with subsets/strict subsets you are apparently suppose to look at whether each element in the set exists or not

[errant bear]

there is no "first element" of AxB

sets arent ordered

[uncut matrix]

true

[errant bear]

but okay

next one

what do u think

[uncut matrix]

next one looks at B x A which is {(∅, ∅), (∅, {∅, {∅}}), ({∅}, ∅) and ({∅}, {∅, {∅}})}

so seeing if {(∅, ∅)} is a strict subset of that

for subsets/strict subsets im suppose to see if all the elements in the set exist in another (and then that one other condition for strict subsets)

wait

[errant bear]

[uncut matrix]

yeah so the issue is with {(∅, ∅)}

I don't know how to break it down to elements and compare it to elements of B x A

would it just be ∅?

or would it be treated as (∅, ∅) because of the { }

my intuition is telling me the latter but

intuition

[errant bear]

what is/are the element(s) of {(∅, ∅)}

[uncut matrix]

oh, it's just (∅, ∅)

[errant bear]

yes that is the only element

[uncut matrix]

it's not ∅ because there's no non-nested ∅ in the set

so since B x A contains one element that's (∅, ∅) and not all elements of B x A exist in {(∅, ∅)} it's true that {(∅, ∅)} ⊂ B x A?

[errant bear]

yes

[coral raven]

what is (∅, ∅)

exactly?

i'm confused

[uncut matrix]

idk, it's just written in the problem set

[errant bear]

wait

[uncut matrix]

I'm assuming its {∅, ∅} essentially

[coral raven]

that's not a set

no repeated elements

[uncut matrix]

which is just {∅} yeah

idk maybe the prof decided to write it as (∅, ∅) to fuck with people

[errant bear]

wait lol

ok i mean either way

its the same. no?

[coral raven]

OH

looking at the original thing

[uncut matrix]

[coral raven]

it's an element of the cartesian product

it's like coordinates

[errant bear]

stop messing me up u nerd

[coral raven]

look it's not as if it were on purpose

[errant bear]

lol okay ignore the past 10 messages carry on beyond

[uncut matrix]

so could I consider (∅, ∅) as just {∅}?

[errant bear]

no

i mean

no

[uncut matrix]

so just treat it as one element I'd find from the cartesian product?

I guess that makes sense since (∅, ∅) belongs in A x B but

next one is (∅, ∅) ⊆ A x B

A x B = {(∅, ∅), (∅, {∅}), ({∅, {∅}}, ∅), ({∅, {∅}}, {∅})}

since I'm trying to test whether (∅, ∅) is a subset of A x B, I need to examine whether each element existing in (∅, ∅) also exists in A x B

if I were to consider (∅, ∅) as a set(???), then I would look for whether ∅ exists in A x B

A x B only has nested elements, so ∅ does not exist

therefore false?

[errant bear]

hmm

[coral raven]

none of that is true

[coral raven]

starting here, all of the below is false

[errant bear]

phi is a set, you you cant take the interval im p sure

[coral raven]

it's a cartesian product element

like (1, 1) in R^2

to test whether (∅, ∅) is a subset of A x B, you test whether ∅ is in A and whether ∅ is in B

if both are true, it's in AxB

[uncut matrix]

so since A = {∅, {∅, {∅}}) and B = {∅, {∅}} it's true?

since ∅ does exist in A and B

[coral raven]

yes

[uncut matrix]

answers say otherwise though
https://i.imgur.com/WLTIiuN.png

[coral raven]

(∅, ∅) exists in AxB

{(∅, ∅)} does not

[uncut matrix]

gotta love it when they just write this crap without any explanation

[errant bear]

what. but its not a set

[coral raven]

they are different things

[errant bear]

kaish ur trolling

[coral raven]

in {1}, 1 exists

{1} does not

you don't understand how set inclusion works

[errant bear]

(phi,phi) is not a set

how is it a subset

[coral raven]

??

ok let's start over

what answer are you confused about

specifically

[errant bear]

(∅, ∅) ⊆ A x B is the statement we r on

[coral raven]

ok

so it's false

[errant bear]

yes

[coral raven]

because

it's not a set

so it's not a subset of anything

[errant bear]

...yes

[coral raven]

are you good with that

if so, what's the next thing you're confused about

[errant bear]

what. thats what i said

okay nvm

[coral raven]

oh bugger i confused you for the other person

i'm such an idiot

[errant bear]

i am everyone

[uncut matrix]

so it just doesn't exist because you can't access its elements because its not a set?

o

[coral raven]

it's not a set, so it's not a subset, so that top right thing is false

[uncut matrix]

okay

uhmm I might have a few more questions but these 3 were def the big ones

but yeah, thanks a bunch for clarifying everything

[coral raven]

ok cool

[uncut matrix]

one sec

next one I was wondering about concerns:
{{∅}, {∅, {∅}}} ∈ P(A)

with A = {∅, {∅, {∅}} so

im getting confused so I'm gonna use substitution to calculate the power set

let 0 = ∅ and 1 = {∅, {∅}}

[errant bear]

okay

[uncut matrix]

then P(A) = {0, {0}, {1}, {0, 1}}
P(A) = {∅, {∅}, {{∅, {∅}}}, {∅, {∅, {∅}}}}

[errant bear]

ok

[uncut matrix]

P(A) has the following elements:
∅, {∅}, {{∅, {∅}}}, {∅, {∅, {∅}}}

but does not contain {{∅}, {∅, {∅}}}, so it's false?

[errant bear]

ye

[uncut matrix]

damn, think I'm finally starting to get it

🙏

alright last one ... seems pretty simple but
∅ ⊆ A x B
for A x B = {(∅, ∅), (∅, {∅}), ({∅, {∅}}, ∅), ({∅, {∅}}, {∅})} as calculated earlier

∅ isn't a set, so it cannot be a subset

therefore false

wait im stupid

∅ is technically {} is it not?

then it could be a subset potentially

I still think it's false because A x B only contains nested elements and is not empty

[errant bear]

phi = {}

the empty set is a subset of every set

[uncut matrix]

wait

so ∅ is a subset to every set, but ∅ may not be an element in every set?

[coral raven]

yes

[carmine grove]

Someone just randomly sent this and went away.. what the hell does this mean and which channel do I even put this thing to ask this question

I'm just curious

[uncut matrix]

man, thanks so much again for everything you two

[violet kayak]

@carmine grove maybe it's me lacking mathematical knowledge, but I honestly think it was a shitpost instead of something serious

nothing seems to be coherent in it

[still wagon]

i agree

[jolly saffron]

What book would you recommend for getting started in graph theory(from mathematical perspective)?

And what to read next?

[carmine grove]

Oh LOL okay

[urban saddle]

I have this ecuation $a + b + c +d = 12$ and I want to see how many no negative integers are solutions

[vital dew]

Jackieto

[weary tiger]

HINT: Use a binomial coefficient related to an unordered selection with repetition

Assuming that a,b,c and d are all non-negative integers

[spare matrix]

I don't know where to ask this, but: Is it typical to denote the rational numbers by $\mathbb{R}_0$?

[vital dew]

veryhappyperson

[spare matrix]

I always thought you would call them $\mathbb{Q}$, but this might be some local thing.

[vital dew]

veryhappyperson

[weary tiger]

No its not typical

In fact I dont think anyone uses it

Maybe they justify it somehow but seems weird

[proven garnet]

You want non-negative integer solutions to $$a + b + c + d = 12$$

This is the same as permuting  

12  $*$'s  and 3 $|$'s  ( $|$ are separators indicating which variable they belong to)

Example possible configuration: $**|***|*|******$ 
corresponds to $a =2, b=3, c=1, d=6$ 


In general, if you wanna find the non-negative integers solutions to 
$$a_1 + \ldots + a_r = n $$
it's the same as permuting $n$ $*$'s and $(r-1)$ $|$'s. That's the number of permutations of $n$ things of one kind and $r-1$ things of a second kind and is given by $${n + r -1 \choose r-1}$$

In your case, we have $n =12$ and $r = 4$
$$----------------$$

The number of non-negative integer solutions of 
$$a_1 + \ldots + a_r = n$$ is also the same as the coefficient of $t^n$ in $$(1 + t +t^2 + \ldots) (1 + t +t^2 + \ldots)(1 + t +t^2 + \ldots)\ldots (1 + t +t^2 + \ldots) \text{ ($r$ times) }$$
$$= (1-t)^{-r} $$

[vital dew]

andreO

[weary tiger]

If S and T are both finite and non-disjoint, i don't think 1) is well defined. Shouldn't it be disjoint union instead?

[weary tiger]

This is correct

See Inclusion-Exclusion Principle

[naive gulch]

Could someone give some pointers here? We have N and k, is our goal [N, k] = 11?

Err, are the N objects the integers, and the k boxes the "possible" pairs out of seven integers?

[weary tiger]

Not sure what that notation is but it seems simple to me to prove this by grouping pairs of integers that sum to 11 together

There will be 5 pairs, so if you choose 7 numbers total, you’ll have to have chosen both numbers in some two pairs

[naive gulch]

Gotcha let me try that

[weary tiger]

Looks like pigeon-hole principle @naive gulch

[weary tiger]

When taking mathematical notes by the book, do you include examples?

[stray reef]

if you find it helpful, go for it

[weary tiger]

But is it practical thing to do?

[lilac delta]

can someone explain how the onto part works?

[weary tiger]

Onto means that the codomain = image of the function

[lilac delta]

how did they get -1 from +1

[weary tiger]

For any y in the codomain Z, there exists an x in the domain such that

y = x+1

equivalently

x = y - 1

What is not clear about it?

[lilac delta]

ok

i've got, f(n) = n^3

how can i check if it's onto?

y = n^3?

[near prawn]

ur missing info

[lilac delta]

well then help me

thats why im here

[weary tiger]

y = n^3

Express n in terms of y

[near prawn]

i meant the question is incomplete

whats ur domain and co domain

[lilac delta]

its the set of all integers

for both

[weary tiger]

We're trying to help you. But are you sure you know what's the definition of an onto map?

[lilac delta]

yep

[sleek swallow]

What is the domain/codomain of your function?

@lilac delta This is going to affect whether it is onto or not.

[lilac delta]

the set of all integers

[sleek swallow]

For both, yes? Okay, so you have:

$$f: \bZ \to \bZ$$

$$n \mapsto f(n) = n^3$$

Now, let $y \in \bZ$. If you agree that $y^{\frac{1}{3}}$ is defined for any integer $y$, then you must ask yourself if $y^{\frac{1}{3}}$ is an integer for every $y$ or not.

[vital dew]

Abhijeet

[lilac delta]

is this 1 to 1?

no, right?

[sleek swallow]

?

[lilac delta]

this graph

is it one - to - one?

[sleek swallow]

You're jumping around

[weary tiger]

A graph is one to one (injective) if every horizontal line in R^2 cuts the graph at most one

[eternal roost]

Could anyone help me with this one, thinking its A but not sure e

[neon frost]

This is bipartite right? I'm not really sure how the K and M points affect it

if at all

[weary tiger]

They don't affect the bipartition of the graph since the definition is related to edges

This graph isn't bipartite

[neon frost]

Are NO and PJL not the two groups I can make to show it's bipartite though?

[vernal linden]

Hey!
I have this question: "There is a function f: N -> N, prove that the function is surjective if and only if every two infinite and different sets A, B subsets of N exists f-1[A] /= f-1[B]. ( Sorry for not using tex )

I managed to prove the first part of the proof and that is to suppose f is surjective and prove that f-1[A] /= f-1[B], but I can't prove the other way around, I'm having trouble with that, can anyone point me to the right direction?

[untold raptor]

is it possable to get from [ p(y) → s(x) ] ∧ [ s(𝑥) → p(y) ] to [ p(y) → s(x) ]

[weary tiger]

this is completely wrong, right?

[clever sage]

well I mean by definition a-b should also be in the ring R if a,b are elements of the ring

[edgy vine]

A is an affine room. u vector and ub a point. What is the subtraction meant to be here?

Like I know affine rooms have defined addition: point+vector = point, so I would understand the subtraction to also be a point. And a norm of a point is just irritating me.

Or can I view ub as a point which is very close to the point the arrow would show, if it was set from the (0,0,0)+u

[plucky thunder]

a tautology means that it is always true right?

[fallen vigil]

yah

[unreal stump]

How do the games work

[quasi quest]

Is this question incomplete?

[quasi quest]

Thats all the info provided

[unreal stump]

Yea,feels incomplete

[flat harness]

I'm assuming you have a knowledge given any pair of them, who beat the other

and it need not be transitive

that's my guess, @quasi quest

then a cycle of length m would be a chain m-long such that person i bears person i+1, yet person m beats person 1

these are all of my guesses

so for all of them to form a cycle, means the existence of a cycle of length n

otherwise I think you can do some partitioning using cycles, but play around with it

[quasi quest]

Thanks!

[flat harness]

go ahead and ping me individually if you need more assistance

or helpers. either way I will get it

one thing to consider could be given some person, look at the set of participants reachable by a downward chain from that person

[mystic moss]

I think you can do a proof by contradiction

I.e. if there is no such partition, then there must be a cycle

[weary tiger]

S is not a ring in general

[still wagon]

it's okay because that's exactly asking S to be an additive subgroup of R ?

i'm saying that the condition you added

is exactly asking S to be an additive subgroup

[weary tiger]

this is from a book recommended by this server lol

[still wagon]

which one ? @weary tiger

[weary tiger]

Advanced Linear Algebra (Steve Roman)

[still wagon]

in which chapter ?

[median mica]

how come H is a forest here?

is it not connected?

[tawdry edge]

I don't understand the first question, are they asking the total combination of participants or the orientations of juniors and seniors allowed possible in a committee?

[median mica]

I believe total, so either only juniors or seniors, or both

[tawdry edge]

is this a n choose k problem or would it just be a permutation of n=10?

[wary scarab]

it's a combination problem

Sum (40 Choose K)(60 choose (10-k)) from k=0 to 10

[median mica]

any thoughts on this @wary scarab ?

[tawdry edge]

wait... what?

how did you come up with the that equation @wary scarab

[wary scarab]

A forest is an undirected graph in which any two vertices are connected by at most one path. H satisfies this property right?

[median mica]

yes it does

[wary scarab]

Hence it's a forest

[median mica]

it is a tree to be specific, but since all forest are trees, then it is a forest

as you said

cool

thank you

[wary scarab]

How would you solve it given it's a combination problem

[tawdry edge]

Well, why couldn't it be 100 choose10?

[wary scarab]

Oh yeah it is indeed that
I thought you were meant to use vandermonde's identity because of which i wrote it like that

[naive gulch]

Are questions ongoing or can I pop in?

[tawdry edge]

I don't believe we were ever taught that...

I just find the question to be equivocal

like the permutation of a committee with seniors and juniors? Or like out of the pool, how many combinations of the committee irrespective of class?

Seriously... fuck these problems

[wary scarab]

When the question says choose you don't need to think of permutations.

[tawdry edge]

There is no way this is true...

unless I'm misunderstanding the LHS expression

because for base case 1 you do not get 7 = 7

[coral raven]

why not

7 = -1 + 8

[tawdry edge]

woah... hold up

why are you insinuating that you don't need to apply the full LHS summation?

[coral raven]

it's ..., it's informal

[tawdry edge]

why do the 7 + 5 + 3 just disappear

[coral raven]

it's just understood that that's what meant

[tawdry edge]

...

[coral raven]

like technically you could go all sigma

but i don't think ppl like sigma, easier to see when it's term 1, term 2, term 3, dot dot dot

even if it's not gonna have a term 2 or 3 or so on if it's only n = 1

[tawdry edge]

ellipsis also means "something happens between , or a continuation of the pattern before"

so I don't understand how you can derive a universal interpretation of this problem

it is equivocal

[coral raven]

sure, but i don't think it's easily misunderstood, perhaps i'm wrong

[tawdry edge]

Well, to me it is misunderstood

[coral raven]

i think most people here would understand it fine but again it's subjective

[tawdry edge]

because when you define a recursive expression... sometimes you use ellipsis to denote that you start at some number then summation happens then you end up at nk or whatever

so it is stupid

[coral raven]

it's a little stupid, but it's understandable so idrc

it's a little stupid, but it's understandable so idrc

[tawdry edge]

I guess... maybe I'm pedantic but this shit makes me unwilling to solve the problem

Like why even do this

[coral raven]

bruh

it makes it easier for you to see the pattern

would you actually prefer the sigma

i know i wouldn't

[wary scarab]

For the base case n=1 so the lhs is 9-2=7 so it's just the first term

[naive gulch]

Regarding probability, shouldn't p(E) + p(F) for separate events = 1?

[tawdry edge]

You don't need sigma G

[naive gulch]

I understand that if set E has 3 elements and set F has 1, then p(E) = 3/4, and p(F) = 1/4, from the examples I've seen

[coral raven]

so what would you do

[tawdry edge]

Actually, if this is a summation of the expression (9-2n) through n - 1

then it is even more stupid

that they didn't use SIgma

[coral raven]

?

[tawdry edge]

FUck this problme

[coral raven]

it's i from 1 to n

i don't understand

[tawdry edge]

exactly!

[coral raven]

it's just notation

[tawdry edge]

It is the most stupid thing ever

[coral raven]

it's not

[tawdry edge]

how am I supposed to derive that

[coral raven]

i don't understand

derive??

[tawdry edge]

derive, yes...

[coral raven]

i just completely do not understand why you have a problem with this

i do not get it

[tawdry edge]

given a signal / sign, you must interpret and derive meaning

this expression on the LHS IS EQUIVOCAL

[coral raven]

oh my gods

[wary scarab]

Iif E=complement of F or vice versa

[naive gulch]

Ah hm

[coral raven]

i just don't see the point in getting worked up over an expression being not a hundred percent rigorous if the meaning is clear

means to an end

[tawdry edge]

Homie... it is obviously an intentional equivocation on behalf of my professor is induce confusion that is unneeded.

It is one of those, "tricks" used in pedagogy that serve absolutely no pragmatic utility besides creating confusion for the sake of complexity.

If you don't have an issue with that, you allow pedagogy to be a dissipated waste of time

[wary scarab]

Do you understand how to do this?

[naive gulch]

Yikes, lowkey sounds like you're starting to pull out words out of thesaurus

[naive gulch]

Sorta, trying to work out the logic rn

[wary scarab]

Ok

[tawdry edge]

@naive gulch Ehhh, sorry my diction isn't the same as yours

[coral raven]

i have to say that i think you are objectively wrong if you think that anyone did this to cause confusion on purpose. i believe that it would be perfectly clear to most people and that you simply hadn't run into this before

i don't think talking about this further will be helpful especially given that ppl are trying to do maths here

[naive gulch]

For the latter portion, since 0.8 * 0.6 = 0.48 where p(e)p(f) = p(e intersection f), that satisfies the p(E intersection F) >= 0.4 right?

[tawdry edge]

Yeah, I changed it to summation and now I can solve the problem... what a wondrous thing

[wary scarab]

The formula you wrote is valid only if e and f are independent

They could be dependent

[naive gulch]

Ah true shoot I forgot that only applied for independent events

[wary scarab]

P(E intersection F)=1-P((E intersection F)^c)
Try to use de Morgan's law now

c is the complement

[tawdry edge]

Does anyone know the rule for sigma summation ?

Like when you have

n + 1 for your "top part"

so you can separate out the +1 and just get n sigma plus something?

[naive gulch]

Ah ok thanks let me try that!

[wary scarab]

n sigma plus the (n+1)th term

[naive gulch]

I mean, isn't it kind of ironic you're annoyed at the unneeded complexity of the problem when you could've used like five words to say the above message

[tawdry edge]

Well, if you wish the speak for me, could you tell me what those 5 words are and how they would be equivalent to what I said in my message? @naive gulch

Unlike Math, rhetoric is relative and more abstract, sorry

[dense sapphire]

How is this done

[neon frost]

Hello! I have a problem. We were given that the number of cycles of length 4 in an n-dimentional hypercube is n(n-1)2^(n-3) where n is greater than or equal to 2 and we must prove this by induction.

I've tried writing out the adjacency matrices for Q2 and Q3 as their entries can be used to determine the number of paths of length 4, but I'm not sure how to relate paths to cycles so I gave up on that.

Then I found out that the number of edges in a n-dimentional hypercube is equal to the sum of the entries in the nth row of Pascal's triangle which is crazy but I haven't been able to do anything other than find where the n(n-1)2^(n-3) formula comes from.

The only answer I've found online (here: https://math.stackexchange.com/questions/431472/number-of-cycles-of-length-4-in-the-n-cube) requires splitting up the hypercube into smaller cubes? No idea what this means or how to do it. If someone could either help me find another way or explain what it means to break a hypercube into other cubes I would be very grateful.

[proven garnet]

Consider the $3$ -cube. We can represent it with vertices $(x_1, x_2, x_3)$ where $x_i \in {0,1 }$. A way to split this cube into two $2$ - cubes would be to consider the cube with vertices where $x_3=0$ and the one with $x_3=1$.

You can do this for any $n$-cube by considering sub-cubes where the last co-ordinate is $0$ and the one with last co-ordinate $1$.

[vital dew]

andreO

[fallen vigil]

is this a good proof for part b?

[wraith parcel]

Saw the question up there earlier and wanted to check with yall if my proof by induction is correct.

[proven garnet]

You're not really making use of the induction hypothesis

For $n = a+1$, you could do 
$$\sum_{k=1}^{a+1}(9-2k) = \sum_{k=1}^{a}(9-2k) + (9-2a-2)$$
$$= -a^2 + 8a + (9 - 2a -2)  \text{ (using the induction hypothesis) }$$
$$= -a^2 -2a-1  + 8(a+1)  $$
$$= -(a+1)^2 + 8(a+1) $$

[vital dew]

andreO

[wraith parcel]

ahhh

so its better to rephrase it sort of in terms of p(a) + another part

[proven garnet]

yup

[wraith parcel]

then use the induction hypothesis for p(a)

i get it now, thanks alot

[proven garnet]

np

[remote cape]

I was wondering how FFT is used in the Schoenhage strassen algorithm, does anyone have an idea on how that works?

[remote cape]

https://en.wikipedia.org/wiki/Schönhage–Strassen_algorithm

[jolly ridge]

But isn't radix a sorting method and not a searching?

You have two lists A and B.
A is an unsorted list, whereas, B is sorted.
Given ′𝑋′ a number input by the user, you need to determine whether or not 𝑋 is present in A and B.

In how many ways can I do this?

How is radix searchdifferent from binary search?

@stray reef

[stray reef]

i don't have the energy to explain it right now, you can look it up

[jolly ridge]

Just one question, isn't radix a sorting method and not a searching one

[stray reef]

you can adapt it to searching

do not ask me anything else about this

[jolly ridge]

Other than Binary search, which else method can you use for searching a sorted list?

[weary tiger]

there are infinitely many ways to search do it. most of them are extremely inneficient tho.

[slow jewel]

Any ideas guys

[mystic moss]

Check each element 100 times to make sure the array isn’t trolling you 😌

[mystic moss]

Also you can start with the subgraph of just the neighborhood of the largest degree vertex and see what happens as you add vertices to complete the tree

[quick folio]

can someone teach me how to do this question

after diracs qtn is answered ofc

🙂

[slow jewel]

I used the idea that if its a tree it must have n-1 edges. Starting with the degree(n-1) we get n-1 leafs and decreasing the degree each time for v_1 say, decreases the number of edges connected to v_1. Since the number of edges it fixed (n-1) and the fact the T cannot form a cycle. Then this produces another leaf. Hence L<=D

@mystic moss

[mystic moss]

Wdym “produces another leaf”? It’s possible that it just replaces another leaf

Otherwise the number of leaves would stay n-1 no matter the tree

[slow jewel]

i see

so it remains constant

[mint shore]

ok so i'm having a bit of a brain fart right now

A shop sells 5 types of sodas. If one was to buy 8 cans of soda, how many possible purchases would there be?

it's probably with C(n,k)

but man, brain stuck

[wary scarab]

Every can has 5 soda types to choose from

[mint shore]

oh

so 5 * C(5,1)?

[quick folio]

umm

[wary scarab]

5^8

[quick folio]

ummmm no?

[wary scarab]

what is it then?

[quick folio]

8^5

[mint shore]

seems a little high

[quick folio]

do u have the answer?

[mint shore]

nope

[quick folio]

i dont think its either

tbh

[mint shore]

i have these formulae though

[quick folio]

yep

so look

8 cans of soda

first can can be any 5

second can can be any 5

....

[mint shore]

so 8 times the same thing

you pick 1 out of 5

[quick folio]

last can can be any 5

8^5

[mint shore]

so C(5, 1) makes sense

[quick folio]

no

[wary scarab]

You should multiply 5 8 times duh

[quick folio]

5C1 means that you are not allowed repetitions

[mint shore]

true

[quick folio]

its 3 am here

oop

mb

yeah

[mint shore]

but wait

[quick folio]

its just function m -> nxnxnxnxnxnxn tuple

so 5^8

lol

sorry @wary scarab

[mint shore]

if it's 8^5 for 8 cans, for 1 can it would be 1^5 which would be 1

[quick folio]

pls forgive me

[mint shore]

yea 5^8 makes more sense

[quick folio]

no no

5^8

yessir

[mint shore]

nice

[quick folio]

can u help me with this qtn now

[mint shore]

ty lots

i'll try

[quick folio]

also @mint shore

send more qtns

for perms and combs

[mint shore]

wait

what does it mean by containment relation?

(english isn't my first lang)

[wary scarab]

yeah what's containment relation

[mint shore]

what's the relation between the 2 sets?

[quick folio]

so like

subset of

yep

[mint shore]

oh ok

[quick folio]

or equivalence too

[mint shore]

ok so

think about i)

what does that set have in it?

not talking about ranges

what do you apply U on

you have f(a) U f(b) right

on the one side you combine f(a) with f(b), on the other you combine a with b

so you have to think about what those Us will return to you

think functions

[quick folio]

they will return sets?

[mint shore]

yes

but what do those sets have in them

let me help a bit

f(a) holds the output of f for every value in a

whereas a alone is the set that goes as input in f

[quick folio]

f(a) or f(b) will return a collection of sets
f(a or b ) will return same collection of sets
let x in a and y in b, f(a) = a1 and f(b) = b1
f(a or b) = {a1,b1}
f(a) or f(b) = {a1,b1}

so its equivalent

is that right approach

[mint shore]

i'm not quite sure tbh

let me try something

[quick folio]

sure bro

or sis

or anything

random person on internet

[mint shore]

ok so suppose

f's domain is {1,2} and range is {3,4}

if i wrote that correctly

so let A = {1} and B = {2}

this isn't a solution, just for testing to get a feel for what we're being asked to prove

ok so with some mock values it is correct

f(a or b) = {a1,b1}
@quick folio i think this part of your solution is what you need to explain a bit

f(a) or f(b) = {a1,b1}
this works

because of your hypothesis

now if you can get the other equation using some smart moves, you're set

[quick folio]

uhh

a is an arbitrary element in A

which maps to a1

same with b

like how u said dom is {1,2}

say a = 1 and b = 2

[mint shore]

come to think of it my little thought is a little wrong

because it has to do with what info you're given regarding f

[quick folio]

like what

if A and B are disjoint sets

f(a) or f(b) will give out lets say image S

a and b are like elements of A and B

f(a or b)

will spit out S as well?

because the subset a or b is mapped to S by f as well?

[mint shore]

disjoint = no common element?

[quick folio]

yep

holy

godel is here to help

pog

[weary tiger]

I was here to ask about my question lol.

[quick folio]

v_v

what is it

send

[weary tiger]

What is your problem about.

[quick folio]

set theory

i cant imagine it

or prove it idk why reee

@weary tiger

whats ur qtn

ill tru urs

[weary tiger]

Well f(A and B) is in f(A) and f(B)

[quick folio]

yeah how is that the case

[weary tiger]

Actually no, let me think.

I think it may be neither.

[shadow pond]

Hey guys, could you help me with this problem: https://gyazo.com/b7a1b0ef2d0a53c65c824f134828b13b

I was thinking of using induction, but I didn't really get to the answer

[quick folio]

can u send the whole qtn

[shadow pond]

oh that is the whole question

the others are just other parts

i was thinknig maybe inclusion exclusion

but that only applies for union not intersection

[quick folio]

it is the inclusion exclusion

[shadow pond]

how so?

I don't see the application

|A ∪ B| = |A| + |B| – |A ∩ B|

isn't this inclusion exclusion ^

[quick folio]

yep

only sets in existence when m = 2 is X1 and X2

[weary tiger]

@quick folio Okay right one is in left one.

[quick folio]

Could you explain sur

ur procedure

ur brain works and ur mathematical preamble

[weary tiger]

$x \in f(A \cap B) \implies f^{-1}(x) \in A \cap B \implies f^{-1}(x) \in A $ and $ f^{-1}(x) \in B \implies x \in f(A) $and $x \in f(B)$

[vital dew]

Godel

[quick folio]

wtf

Godel

u genious

[shadow pond]

oh i see

wait so meliodas

[quick folio]

yessir

[shadow pond]

for the next part, isn't it the same as the first

or is it different

[quick folio]

Kinda

[shadow pond]

cuz inc exclusion changes w/ num of sets

but how do we quantify that

[quick folio]

u need to do the inclusion and exclusion on a bigger set

sec

thats why i said

send the whole qtn

.>

[shadow pond]

okay ya

https://gyazo.com/9a52720e838cd1a6eb365650b7cdefba

[quick folio]

have you considered the case when m = 3

[shadow pond]

yeah

should be like 3k - 2n?

or was i wrong in deriving it

[quick folio]

-2n??

shouldnt it be 3k-N

?

cause when m = 3 ur universal set cardinality is N

so >= 3k-N

so its mk-N

[shadow pond]

ooh yeah

[quick folio]

for induction u need to use A and B

so yeahh

[shadow pond]

i see

[mint shore]

does anyone know any type of solver for mod equations?

this type

i just want to find some way to verify whether my solution is right or not

[errant bear]

just....do it lol

or literally just google

[mint shore]

yea i just found one

man im dum

ignore

[shadow pond]

for part b, with three sets I end up getting n >= 3k + |x1 ∩x2∩x3| - |x2∩x3| - |x1∩x2| - |x1∩x3|

but now i'm stuck on making this expression mk-n

because how do i deal w/ the other three interseciotns

[valid wave]

Could someone explain what 2^ A intersection B means

Like what is the 2

In this

[pale epoch]

powerset most likely

so $2^U$ is the powerset of $U$

[vital dew]

Lochverstärker

[valid wave]

thank you

wait wouldnt most of these be true

oh wait

im confused

so were taking the power set of a the intersection of b

then comparing it to the power set of the intersection of a and b

arent these the same things in all cases

[pale epoch]

that's the question lmao

[valid wave]

So all of this is true

I’m saying isn’t this a dumb question then

It’s just rewording it

Unless I’m missing something

[pale epoch]

no

it's not just rewording

[valid wave]

could you elaborate

[pale epoch]

also not all of them are true

[coral raven]

no