discrete-math

[gritty crescent]

Practice problems, ask around, spend some time thinking about problems, crying and coping, etc.

[naive marten]

my book has 2 practice problems a section its almost as if youtube is better

[gritty crescent]

At this stage, nothing really substitutes for proper practice.

You gotta get book and problems, and grind.

[naive marten]

i like the

crying and coping

one

[gritty crescent]

[naive marten]

No reason to lie : )

IM JK

[quaint star]

Say that again bitch

[gritty crescent]

[naive marten]

also uh

would it be safe to assume there are

|a| x |b| equivalence classes

[quaint star]

Idk what that means

[naive marten]

like

a = {1,2,3}

b = {1,2,3}

would it be safe to assume that

there are 3x3 equivalence classes

so 9

[quaint star]

For what relation? Or a relation on what?

[naive marten]

on a relation a + b = c + d for (a, b) R (c, d)

[quaint star]

Oh, no.

[naive marten]

:(

:(((((((((((((((((((((((((((((((((((((((((

:(((((((((((((((((((((((((((((((((((((((

[quaint star]

a x b has 9 elements. But not all of the elements have their own equivalence class.

Each equivalence class has multiple elements

It has at most 9 equivalence classes

But if every equivalence class has 3 elements, then there will be only 3 equivalence classes.

of equivalence classes * elements in equivalence class = 9

[naive marten]

i get

now

yay

[quick folio]

@naive marten send qtns

https://tenor.com/bl8K0.gif

[naive marten]

hello

i am out of questions

:(

maybe we can be besties instead?

@quick folio

[quick folio]

no time for besties

[naive marten]

:(

why is that

[quick folio]

only time for discrete math qtns

send qtns

[naive marten]

i send when i find one okay bestie

[quick folio]

okaye

[dawn junco]

could someone help me understand this non-linear reccurence relations problem?

the only similar problem on math exchange I found was this: https://mathematica.stackexchange.com/questions/197464/find-a-stable-3-cycle-of-the-sine-map/197467

and I kinda understand how cycles work etc and where to find it, but I don't know how to prove it in mathematica

[weary tiger]

Can someone help me understand how the second expression was derived from the first?

[weary tiger]

Nvm solved it using the convolution theorem

[fading spear]

could someone tell me what the dot between the X and Y represents?

[gritty crescent]

This is absurd, but I'm guessing it is some kind of composition, treating X and Y as functions.

[vital dew]

Manan.

[gritty crescent]

This is still a guess though, you should ask for clarification @fading spear .

[fading spear]

thanks a lot... so kind of like f dot g in that convention sense, as in f(g(x))

nice name btw

[gritty crescent]

Thanks.

[abstract otter]

can someone help me with this question? I have no idea how to start it

it seems like something to do with planar graphs having at most 3n-6 edges

but i dont see how it relates to forests

[weary tiger]

@abstract otter you are basically there

how many edges does a forest have?

[abstract otter]

|V| - |E|

[weary tiger]

whats E in this?

the number of components?

[abstract otter]

i think so

[weary tiger]

seems like weird notation but anyway

forests have less edges than a tree on the same number of vertices

trees have (n-1) edges

so G has <= 2(n-1) edges

[abstract otter]

ok so i calculate the amount of edges 2 trees could have

[weary tiger]

yeh so E_1,E_2<=(n-1)

and E <= E_1+E_2 <= 2(n-1)

(add some modulus signs)

[abstract otter]

which is under 3n-6

[weary tiger]

yeh exactly

[abstract otter]

ayt thanks !

[kind geode]

hey quick question for a relation to have property p (symmetric, transative, reflective, antisemmetric, etc) does each edge need to satisfy the condition.

an example if i have R={(1,1),(1,2),(1,4),(2,1),(2,2),(3,3),(4,4)} would this be symmetric?

or maybe to rephrase, would each node need to satisfy the given requirement for the relation to have property p

[naive marten]

does this break transitivity?

@quick folio bestie i have question for u ❤️ ^^

no right

it works right :/

[quick folio]

hello

[naive marten]

bestie

[quick folio]

yes this breaks transitivity

[naive marten]

it does? :(

[naive marten]

does there need to be

B to A

SIGH

wait why does b to b have to be a thing

and i guess the proof of it being not transitive would be smt like

wait why is that not transitive

[quick folio]

yes

there needs to be from b to a and a loop in both

[naive marten]

transitive is

aRb and bRc implies aRc right

so we have ARA and ARB here

so isnt ARB enough for transitivity?

[dusk stirrup]

Can someone simplify what the sentence "work backwards from the sequence of constant differences" means in my textbook? I'm stumped on this part:

It's just throwing me off and I can't seem to find formulas because I don't get that part.

Sorry if that is a dumb question.

Oh wait, ignore me. I think i figured it out - sorry.

[humble violet]

hi! this is my first question here (im sorry if my english is not the best), im helping a guy with a discret maths (but im really never do discret maths to computer science or logic exercises) so, the problem is:

found the set of the triples (a,b,c) that c is the remainder of $a= b\lfloor a/b \rfloor +c$

[vital dew]

Krayom

[kind geode]

hey my prof kinda went over this quicky but he siad these are the strongly connected components of the graph

but why??? lmaoo

i get the first part BCD but how are A and E strongly connected.

[quaint star]

Do you know what a strongly connected component is?

You have to be able to get from each vertex in a component to every other vertex in the component.

Consider E. The only vertex that you can reach from E is A, but you can't get from A back to E. So E is its own component.

Consider A. You cannot get from A to any other vertex. So A is its own component.

Consider B,C,D. You can get from any of them to any of the others. And although you can also go to A or E, you cannot go from A or E to one of B C or D. So B C and D is a component

@kind geode

[kind geode]

ohh ok so if a node has a single path to another node, but no reverse path, then the node itself is a strongly connected component. its just the definition is "if theres a path from a - b then theres a path from b -a" but in my mind, if A ISN'T able to get to and from another vertex, then why is it a component yk, but if the former statement is true then ill understand completely haha sorry

just the definition confuses me as it implies there needs to be two nodes, or at least i thought

[weary tiger]

Is there a typo?

I think it's missing a (-1)^n before the binomial coef in the sum but I'm not sure

[weary tiger]

I think it's correct.

[grand elm]

for k=1 it's correct, and then it's products of positive coefs, so they stay positive

[weary tiger]

Yup I got confused with the binomial coef for negative exponents

[scenic pecan]

hi, i need help with this question, more specifically, is R transitive?

i've figured out the rest using a number line

i've only gotten this far listing out the transitive property and got stuck trying to prove it on a number line

[unreal stump]

take x=4,y=5,z=9 xRy and yRz is true but xRz is not true

[scenic pecan]

alright

also, for antisymmetric, my answer is yes, because x=y

is that correct?

[unreal stump]

You get x<=y and y<=x

So x=y

[scenic pecan]

thanks bud!

[minor lake]

can anyone explain me what's wrong in my steps

[minor lake]

nvm

[weary tiger]

Set $a_{n} = b_{n} + \alpha \times n$

To get a homogenous linear recursion of order 1 in b_n

[vital dew]

Laïka

[minor lake]

If I had a recurrence like the following

$a_{n} = 2a_{n-1} + 2\cdot4^{n}$ for $n > 0$ and $a_{0} = 11$ would you change anything?

because the non-homogeneous terms more looks like ab^x instead of c^x

term*

@weary tiger

[weary tiger]

Cdot?

Could you retype that again pls

[minor lake]

yea

[vital dew]

Chad

[weary tiger]

Oh

You’d have to change the choice of the auxiliary sequence I believe

[minor lake]

because I tried $2^{2n + 1}$ but I get stuck

[vital dew]

Chad

[weary tiger]

So basically

In this case doing a substitution would get you stuck

The right way to go would be by solving the associated homogenous recurrence separately

[minor lake]

? welp I already tried to solve the homogeneous one

[weary tiger]

This is correct

Because the characteristic root is 2

You now have to find the particular solution

F(n) = 2x4^n

4 is not a root so the particular solution is of the form cx4^n

[minor lake]

hmmmmm how would I deal with the 4^(n-1)

or maybe

I can just factor it

[quick folio]

Whats the qtn

[minor lake]

$a_{n} = 2a{n-1} + 2\cdot4^{n}$ for $n > 0$ and $a_{0} = 11$

[vital dew]

Chad

[minor lake]

I already solved an example once with exponent by the use of substitution but apparently I can't use substitution for this one so idk what to do

maybe just turning everything in log

$a_{n} = 2_a{n-1} + 2\cdot4^{n}$ for $n > 0$ and $a_{0} = 11$

[quick folio]

why are u logging?

ill send my working out

sec ill work it out

[vital dew]

Chad

[minor lake]

just have no idea about how to handle that

[vital dew]

Meliodas

[minor lake]

oh

hmm

can you explain what happened here

i think i get the left side ok

let me think a bit hmmm

\frac{2(4^n)}{c\cdot4^{n}\cdot4^{n}} ?

$\frac{2(4^n)}{c\cdot4^{n}\cdot4^{n}}$

[vital dew]

Chad

[quick folio]

u have a particular homogen soln

yeah

[minor lake]

oh it's a division 4^n ok

i thought that some magic happened

ok makes sense

thanks

[valid wave]

hello my brain is expanding

so would strong induction only be used for like recursion proof questions

like im struggling to see how it would be used for something like this

when only 1 base case is needed

[valid wave]

<@&286206848099549185>

[keen girder]

you don't need strong induction here.

[tawny blade]

i need a short summary of how the answer of this question is 14 https://images-ext-2.discordapp.net/external/eLH0Po7Gabe9hVliekPv8pgOCSJemfQNZ8u7QfMqYzk/https/i.gyazo.com/thumb/1200/9fef529d54a3afb6413529b3cbacebb9-png.jpg

[weary tiger]

that is not discrete math

[kind geode]

@quaint star

sorry 😓

[quaint star]

Show me the definition you are talking about

[kind geode]

its uh here

[kind geode]

.

[quaint star]

I don't understand what's being defined there

Can you send me a picture of the defn from your notes or book?

[kind geode]

yea sure igmma sec

[quaint star]

This does not imply there needs to be two nodes

It says IF there are two nodes in the graph THEN there needs to be a path from the one to the other and vice versa

[kind geode]

ohhh so if theres only 1 then its just stongly connected if its within the graph

[quaint star]

Yes, a directed with 1 node is always strongly connected

But when you are talking about strongly connected components, you want to choose the biggest possible strongly connected subgraphs.

[kind geode]

ahhh okok i see now thank you so much

[quaint star]

Np

[neon frost]

Can someone explain to me why G-v would have a (k-1)-colouring?

[weary tiger]

@neon frost read the definition of k-colour critical

G-v is a subgraph hence it's chromatic number is less than k hence it has a k-1 colouring

[neon frost]

Oh damn it was staring me in the face

thank you

[mint bane]

can someone please help me with these, i have a hard time with proving injectivity and surjectivity

like i know and understand the formal definitions i just have a hard time applying them to actual proofs

[errant bear]

which one

[mint bane]

any of them, understanding one can probably help with the rest

[shadow egret]

hey

can someone help me with these?

[mint bane]

#rules

[shadow egret]

well you saw it

i removed it

but if you can dm me

[sand cipher]

need help going through this

already lost at the first, what is lm(f)

ik what im r is but is it same with im(f)?

[errant bear]

sure, look at problem a, with surjectivity to start off @mint bane

what does this mean, in words

[errant bear]

the image of the function

[sand cipher]

would that be xmod 7?

[errant bear]

yes

so what are the possible outputs

[mint bane]

surjectivity is tryna show that $$\forall a \in \mathbb{Z}, f(x) = a$$

[vital dew]

nitezba

[sand cipher]

only x=7?

[errant bear]

this is wrong

[mint bane]

in the reals, my b

[errant bear]

what is 2 mod 7

[mint bane]

that the range is equal to the codomain

[sand cipher]

o wait mb i thought divide smh

2mod7 is 2

[errant bear]

yes, so what are all possible values

[errant bear]

for all a in R, there exists an x st f(x)=a. is this true?

[mint bane]

intuition says yes...?

[sand cipher]

{1,2,3,4,5,6,8} but not {0,7}

[errant bear]

does -1 get reached? @mint bane

[errant bear]

you sure?

[mint bane]

oh im dumb

for any negative real number it just becomes 0

[errant bear]

yup

[sand cipher]

well 0 and 8 would be 0 won't it?

[errant bear]

yes

so your set should have a 0 not 8

[errant bear]

so this answers both injectivity and surjectivity for you

[mint bane]

ohhhh and to prove that those are false a counterexample will probably suffice

[errant bear]

sure

[sand cipher]

wait but 0 mod 7= 0 and 8 mod 7 = 1

[errant bear]

oh sorry, forgot you were mod 7

[sand cipher]

wasn't it saying like set N

[errant bear]

yes

[mint bane]

alright well how about one that is surjective and/or injective

appreciate you helping two people at once btw <3

i dont think b) is injective

[errant bear]

why not

[sand cipher]

how would i find if f is onto or not, i tried to do y=xmod7 and try to solve if x=N but i'm not sure how it will go

or what will mod= to

[errant bear]

mod7 gives the remainder when divided by 7

[sand cipher]

and what will mod turn to if it move to y

would it be y/7?

because if y/7 it will be xR(smthing) where R is remainder

[errant bear]

what

does every element of the set X get mapped to? thats what you want to answer

[sand cipher]

X= {1,2,3,4,5,6,8} this is the elements rite

[errant bear]

no

look at your question

[sand cipher]

wait this is the set {0,1,2,3,4,5,6,7,8}

but Im(f)={1,2,3,4,5,6,8}

is that right?

[quick folio]

no

8 wont exist

8 is 1 mod 7

[errant bear]

this one is wrong

[elfin flume]

Is the complete graph K_3 a cayley graph?

According to Sabidussi's theorem, a graph is a Cayley graph if its automorphism group acts simply transitive on G. I know K_3 is vertex-transitive, but how can I tell if it's free?

[quick folio]

b) is it surjective ? No not all of codomain is used up
c) 1-1 ? no because a lot of N can point to X if it helps
d) No because f-1 is not bijective

[sand cipher]

i thought 1 mod 7= 1 and that 8 mod 7 = 1

[sand cipher]

the whole thing or just the element 8?

[errant bear]

you dont have 0 in your imagw

but it is

[mint bane]

was this @ me?

[sand cipher]

1 / 7 = 0.142857
0 x 7 = 0
1 - 0 = 1
isn't this the mod for 1 mod 7
and won't this be
8 / 7 = 1.142857
1 x 7 = 7
8 - 7 = 1

[mint bane]

also i didnt think b was injective bc (-1,1) will give the same output as (1,-1)

[sand cipher]

wait but 0 mod 7 is 0 and 0 is not a natural number

[quick folio]

yes but mod 7s equivalent classes are 0-7

and ur partitioning them into 0-7

not 0-8

ur right that u can take a mod of a number from N

but the partitions are different here

and also

0 is a natural number in discrete

😳

thats how my course is structured at least

[sand cipher]

ye i heard different course have different natural number cuz my professor doesn't include 0 in it as he specifically stated "Zero is not in it(Natural number)"

[errant bear]

??

why is this an issue

it doesnt matter if 0 is in N or not, i never used it

[sand cipher]

"Let f be the function from the set ℕ into the set
X = {0,1,2,3,4,5,6,7,8}"

[errant bear]

yes

[sand cipher]

set N and isn't N natural number?

[errant bear]

yes

[sand cipher]

or is does that hae no relation at all

?

[errant bear]

what is 7 mod 7

[sand cipher]

0

[errant bear]

okay

so why is there an issue

[sand cipher]

well the thing is you said 0 is not in my image

but 0 mod 7 is 0

same like 7 isn't it?

so won't it be excluded?

[errant bear]

i said

it isnt in the image that you gave

but it is

in the actual image

meaning you were wrong

[sand cipher]

im (F)= {0,1,2,3,4,5,6}"

would this be it?

[errant bear]

yes

[sand cipher]

lol i'm just making it harder

....

[errant bear]

what

[sand cipher]

how would i find onto?

[errant bear]

you can just explicitly give examples in this case

[sand cipher]

ex: if it was y=3+x
y-3=x
then i can solve onto, but what about mod?

[errant bear]

weve already done like 4 of them

[sand cipher]

wait we have??

[errant bear]

1, 2, 7

okay 3 examples

[sand cipher]

would this mean it's not onto

because not all x =imf

[mint bane]

sorry to interrupt but was i write about that b not being injective

for convenience

[mint bane]

@errant bear 👉 👈

[elfin flume]

In graph theory, the definition of regular graph is a graph that is vertex-transitive and free, but I also saw a definition that says a graph is regular if each of its vertices have the same number of neighbors. Are these the same?

[tiny wigeon]

can i get some help??

how do i simply this [P V (Q Λ ¬P)] V ¬Q Λ P?

[pine iris]

Use distributive on the first side

[stray reef]

this expression is ambiguous as-is

[lilac pivot]

100% quicker kekg

[stray reef]

do you mean [P V (Q Λ ¬P)] V (¬Q Λ P) or ([P V (Q Λ ¬P)] V ¬Q) Λ P

in any case, a quick way to do it may be to use k-maps

[tiny wigeon]

i presume it like this [P V (Q Λ ¬P)] V (¬Q Λ P)

since that is the only given

[stray reef]

do you have to use boolean algebra laws to do this

[tiny wigeon]

yeah

[stray reef]

damn

sad

[tiny wigeon]

i get stuck here (pvq)^(-pvq)

then i dont know what is next

[pine iris]

Reverse distributive

That results from a distributive if you look closer

[tiny wigeon]

wdym by reverse distributive?

(pvq)^(-pvq) to -(pvq)??

[pine iris]

No

No that both ( ) have one letter in common

And the $\vee \wedge$ are also in the format

[vital dew]

Mns98

[tiny wigeon]

ohh right

they said the final asnwer would be P v Q

but i dont know how they reach to it

[pine iris]

You got it?

[tiny wigeon]

they did but i dont know how it got to P v Q

or is it correct or not

[pine iris]

What

[tiny wigeon]

they manage to get the answer P v Q

but i dont know how it got there

[pine iris]

$P \vee \neg Q$ or $P \vee Q$?

[vital dew]

Mns98

[pine iris]

This is the format?

[tiny wigeon]

$P \vee \neg Q$

[vital dew]

elxn

[tiny wigeon]

this onee

[pine iris]

Oh ok thats right thn

It doesnt have the second parentheses

On $\neg Q \wedge P$

[vital dew]

Mns98

[pine iris]

Therefore the question needs to be $[P \vee (Q \wedge \neg P)] \vee \neg Q \wedge P$

[tiny wigeon]

still the same

yeaah

[vital dew]

Mns98

[pine iris]

Well the result is true

[tiny wigeon]

ohh okay

thank youu

[pine iris]

I mean $P \vee \neg Q$ is the right one

[vital dew]

Mns98

[tiny wigeon]

okaayy

[weary tiger]

Would it be better to argue that this graph is not bipartite by saying that it has an odd cycle (coloured in red)?

I was thinking about doing a colouring of the vertices and say that we'd necessarily end up with monochromatic edges

[stray reef]

better than what

[weary tiger]

But my concern is that graph colouring is misleading

Unless we prove that the colouring is "forced"

[stray reef]

oh, i mean... if your teacher allows you to use the bipartite-iff-2-colorable theorem then that's ok

bipartite iff no odd cycles is also ok

[weary tiger]

But doesn't that require more explanation? If you didn't find a 2 colouring then there are chances you've done it the wrong way and a 2-colouring might exist. So if you want to justify a graph isn't bipartite by simply colouring the vertices in some way and showing that you end up with monochromatic edges isn't enough. You really have to try all possibilities or show the colouring is forced (doing a BFS colouring)

[stray reef]

for a 2-coloring if a vertex is red then all its neighbors have to be blue

[weary tiger]

I have a question regarding 2-regular graphs, if anyone is keen on helping me understand this. As far as tours are concerned, we can always find an Euler Tour in a connected graph that has all vertices of even degree. Such proof goes by finding consecutive tours until covering all edges of the graph exactly once and then joining them to get an Euler tour. This procedure uses the Tour Lemma which assures the existence of a tour in a graph that has all vertices of even degree and at least one vertex of degree bigger than zero. Now in 2-regular graph, all vertices have degree 2>0, so a tour exists (by the Tour Lemma). I want to be able to say that we can cover all edges of a 2 regular graph with edge-disjoint tours without necessarily using the connectivity argument.

[stray reef]

are 2-regular graphs necessarily connected tho

just take two disjoint cycles of any length

make that a graph

[weary tiger]

Is it because we can apply the tour lemma to every vertex of the 2-regular graph ? (every vertex has necessarily degree bigger than 0)

[stray reef]

you don't have an euler tour in that do you

or a tour at all

[weary tiger]

Ok! The definition wikipedia gives has some additional property for a regular graph to be connected

[stray reef]

then a 2-regular graph is just a cycle

[quick folio]

Ann ur actually so mathematically gifted 😮

[weary tiger]

I agree

[stray reef]

gifted ≠ experienced...

[quick folio]

😳 OK BOTH

[stray reef]

meh

i kinda. dont like the "gifted" label

[weary tiger]

So yeah I just wanted to make sure that we can apply the tour lemma repeatedly on a given graph and get a collection of edge-disjoint tours (that we don't necessarily have to join into an euler tour, this happens tho if the graph is connected) but such that they cover all the edges of the graph. In the case of 2-regular graph we can do that because every vertex has degree >0. Having done that, we can infer that every 2 regular graph is a collection of disjoint cycles.

[dusk stirrup]

Does 9 choose 2 include doubles? so if I have 9 types of things to be selected by two people - both being able to select the same type of item - is that included in the answer?

So say I have A, B, C, D and want to select two.. so is 4 choose 2 AB, AC, AD, BC, BD, CD or is it AA, BB, CC, DD, AB, AC, AD, ... ?

[stray reef]

no, nCk does not count doubles

[weary tiger]

9 choose 2 is the number of subsets of size 2 of a set of size 9 so no

[dusk stirrup]

THank you

[thorn axle]

hi! i followed almost all of this calculation until the stage here

i'm not sure why it is n-1 on the bottom

i can send the full question if you want context

[weary tiger]

hi

what would be the remainder/quotient of -37 is divided by 16?

do you do 16*-3 + (48-37), so the quotient is -3 and r = 11?

or is it 0 and 19?

or is it -2 and -5?

i thinking the first, but i'm not sure

[thorn axle]

yeah you're right i think- i.e. the first

you just want to rewrite your number -37 as -37=q*16+r where q and r are integers

[weary tiger]

okok, ty

[thorn axle]

<@&286206848099549185>

[grand elm]

the product is n·(n+1)···(n+k-1), which is (n+k-1)!/(n-1)!, and together with the k! in front you get the binomial

[thorn axle]

perfect, tysm!

[grand elm]

:)

[weary tiger]

How should I go about doing this?

[weary tiger]

tbh it's stupid to do it by induction, you can prove something more general, in a simpler way without it

but should be straightforward, have you ever done induction before?

(note to self don't call things stupid again, it sounds pretentious, people have different opinions)

[dry reef]

hey any chance someone could help me with this?

[naive gulch]

This question looks horrifying, I don't even know where to begin here

[weary tiger]

Well, how did you define E(X)?

@naive gulch

and why is this in this channel lol

[naive gulch]

This is Discrete math is it not haha

E(x) is the expected value right

[weary tiger]

yea, how did you define it

that is probability but ok idrc

[naive gulch]

oh, I just get these questions from my discrete class textbook so I expected it'd have relevancy here too

[weary tiger]

oh you expected

[naive gulch]

I'm not sure what you mean by how did I define it, we have a summation from 1 to infinity of the probability that X(s) >=k?

I'm not sure if I'm reading the question right though! Any clarifications would be helpful to get started on this one

[weary tiger]

they want you to show that E(X) is equal to that thing

so they must have defined E(X) before

[naive gulch]

Unfortunately it's an independent question it seems

[weary tiger]

So you have that $\mathbb{E}[X]=\sum_x x\mathbb{P}(X=x)$ so in this case $\mathbb{E}[X]=\sum_{x=1}^\infty x\mathbb{P}(X=x)$

[vital dew]

same

[weary tiger]

Can you show that $\sum_{k=1}^\infty P(X \geq k) = \sum_{k=1}^\infty kP(X=k)$

[vital dew]

same

[naive gulch]

ooh boy, interesting

Hmmm

I can try, I think

[kind geode]

and sorry P(E1 union E2) would be 1/2 times 0.5

aorrow is misleading

[fallen vigil]

I’m in discrete math still but isn’t it 4! x 3! x 6!

[twin crater]

why are you using conditional probability

[weary tiger]

The question asks you to count all possible arrangements of fruits, why do you use probabilities at all?

[kind geode]

yea soryr it was the wrong quesiton i sent lmao

and i got it sorry ahahahah

[short steeple]

hi, i have a bonus problem on my hw i'm struggling to get around:

Let the sets $A_1,A_2,\ldots,A_m$ partition the set ${1,2,\ldots,mn}$, with $|A_i|=n$ for all $i$. Let $B_1,B_2,\ldots,B_m$ also partition ${1,2,\ldots,mn}$, with $|B_i|=n$ for all $i$. Prove that there is some permutation $j_1,j_2,\ldots,j_m$ of $1,2,\ldots,m$ for which no intersection $A_i\cap B_{j_i}$, with $1\leq i\leq m$, is empty.

[vital dew]

Snodlop

[short steeple]

i have the idea to create a bipartite graph with the sets A and B on either side of the partition, and as of right now i'm drawing an edge between A_i and B_j if their intersection is not empty (so each vertex is degree n)

i know that an n-regular bipartite graph has a perfect matching, and i'm sure that applies to this problem since this has been what we've been learning about thus far, but i'm not sure how to apply it to this problem or where to go from here

any help would be appreciated, thank you!

[short steeple]

<@&286206848099549185>

[short steeple]

never mind, i think i was able to work it out to the best of my ability

[frigid abyss]

can someone help me out with this. I know its pigeonhole but idk how to do it

[quick folio]

coul you partition the initial set such that you have n odd terms first and n even terms next

if u take n terms from either partition ur bound 2 get 2 from the even

@frigid abyss where is this from

[kind geode]

hey quick qeustion

for euler circuts

a circuit will have a euler circuit iff it has verticeis of even degree but is this for all vertices?

[quick folio]

ye[

@kind geode where do u do problems from for discrete

[kind geode]

?

wdym?

[quick folio]

like uk for revision problems

for hw and stuff

[kind geode]

a lot of its just school stuff

and tb has a lot of good stuff

[quick folio]

what tb do u use

[kind geode]

some green one fomr oscar smth

lmao

dik sepecific name

[frigid abyss]

im still confused lol. could u use an example

well i kinda see what youre saying. like if n=5, then we get 1,3,5,7,9,2,4,6,8,10. and if we chose 7 numbers then two would have to be next to each other.

would would the number of pigeonholes be in this case when n=5 tho

[golden widget]

Im pretty sure for the Hamilton circuit one, if you can make a line touching all vertex ones, and returning back to the starting point it is one.

[dawn robin]

ok, so I have what I consider to be a very hard question and am not sure where to post it. But I would reeally like some help figuring it out. This seems like the closest place i can think of. I also dont know the exact language but here goes

You are given a set of points which must be fully linked.
Each point is given a random set of possible directions from which they may link expressed as a set x = {up, down, left, right} in which 1 means a connection is able to be made, and 0 means no connections can be made.

For example, if given the set of points A = {1,0,0,0}, B = {0,0,1,1}, C = {0,1,1,0}
Point A would not be able to form a link to Point B, but it would be able to form a link to Point C.

Likewise Point B would be able to form a link with Point C, but only from Point B’s right side to Point C’s left side.

Every set of points will contain a starting point and at least 1 point E = {0,1,0,0} which cannot be linked to the starting point.

The starting point of any given set of points will always have a 0 in the down space. {x,0,x,x}

No points may consist of {0,0,0,0} and no two connecting points may fully terminate.
The list of points may contain duplicates, and can consist of any number of points and no two points may share the same connection.
Connections between points may not intersect, and all consist of the same length

With these rules in mind, what constraints would any given set of points require to ensure that the list can be fully linked, and are there any set of constraints which would ensure that the list can be fully linked even if the two points being linked are chosen arbitrarily from the initial set of points?

image for clarification

if you think there is a better place for this question, please let me know

[obtuse minnow]

@dawn robin still there? tag me if you are.

[unreal stump]

Number of left subgroups will be |G|/|U|

So,It depends on |U|

[deft olive]

I don't quite understand

[unreal stump]

Is |U| given?

You can't say otherwise

[deft olive]

wait, shit. It formatted wrongly

[deft olive]

|U| is not given

[unreal stump]

So, Should be can't say

[stray reef]

can you show the whole problem?

[deft olive]

What I have shown you is all I am presented with...

[stray reef]

show us the entire text of the proble

problem*

even if it's in another language

[dawn robin]

@obtuse minnow srry, kinda knocked out just before you messaged. Question still unsolved though :D

[gritty crescent]

This is correct.

[weary tiger]

okay thank you!

[merry dirge]

I have to answer whether these are true or false

I got True, False, False, True

Can someone verify if thats correct?

g-j?

can you @ me if you do end up helping me please.

[weary tiger]

do you know what functionally complete means?

[vast python]

anyone know what the curly brackets mean

[keen ingot]

It’s just set notation, you have the source and sink set

So all those nodes are on one side of the cut, and the rest on the others

So try to draw your cut on the diagram so that it does divide it into those two sets

[merry dirge]

im not sure

[weary tiger]

Why are you trying to answer questions before knowing what they mean?

[agile kernel]

Is it appropriate to use the "is an element of" symbol when assigning variables their definitions at the beginning of a proof, or is there a separate "let (var) be an element of (set)" symbol?

[weary tiger]

Let (var) $\in$ (set) is ok

[vital dew]

Carla_

[agile kernel]

Perfect thank you so much

[limber loom]

does anyone know how to do this problrm?

[dreamy solar]

anyone know how to solve this?

i cant remember the steps

[mortal nexus]

My book just used this but i ve never seen it before, anybody knows where i can read about it??

and @dreamy solar sorry i sent my message right after yours. i can try to help

so. consider only age now

if there are at least 5 people, at least 2 will have the same age, right?

and if there are 9 people, then at least 3 will have the same age! which is what you want

but there's colors too

for example, with 9 people, 3 of them will have the same age, but they could still all be wearing different colors

try to work from that?

[dreamy solar]

Gotcha

would the right answer be 49?

because there are 16 possibile combinations of colored shirts and ages

and then 3 students have to have the same so we do 16 * 3

+1

would that be right?

[mortal nexus]

I GENUINELY DONT KNOW

Im trying to draw it

I was thinking maybe 33

Im trying to imagine a worst case scenario where they're all different?

And im waiting for a moment where if i add a student, then three will necessarily all match

Im Positive it's 33 because, as you said, there are 16 different combinations, so if you do 16*2 +1, then at least 3 have to match

[merry dirge]

For this would I do 2^5

or 2^32

32 comes from 2^5

[versed kernel]

2^(2^5), yes

[merry dirge]

ah

[merry dirge]

wait but its not asking how many different right

is this a trick question

if it was different wouldnt it be 2^32

[dawn robin]

reposing this as I am still seeking help on it. I have made some minor progress and added some clarifications to the question, but am still working on finding the less obvious restraints.
As always if this belongs somewhere else let me know.
Instead of fully re-posting I will simply be using a reply to link to this message unless major changes need to be made to the description from here on.

The Rules

You are given a set of points which must be connected by edges in such a way that there exists a path from the given starting point, to every other point in the set.

Each point is given a random set with four elements such that we'll call its first member up, second down, third left, fourth left, such that all its elements are either 0 or 1

Edges may only be formed between two points where both the exit and entering side of the 2 points are 1.

For example, if given the set Points = {A, B, C} where A = {1,0,0,0}, B = {0,0,1,1}, C = {0,1,1,0}
Point A would not be able to form a link to Point B, but it would be able to form a link to Point C.
Likewise Point B would be able to form a link with Point C, but only from Point B’s right side to Point C’s left side.

Every set of points will contain a starting point and a point E such that E = {0,1,0,0} which cannot be linked to the starting point.

The starting point of any given set of points will always have a 0 in the down element. {x,0,x,x}

No points may consist of {0,0,0,0}
The list of points may contain duplicates, and can consist of any number of points.
Connections between points may not intersect, and all consist of the same length

Finally, no two edges may connect to the same point by way of the same element.
For example, If given the set Points = {A,B,C} where A = {1,0,0,0}, B = {1,01,0}, C = {0,1,0,1}
Both, A and B could not connect to Point C through an edge from C's down element to both A and B's up elements. However, if A and C form an edge through C's down element to A's up element, B may still form an edge from its left element to C's right element.

With these rules in mind, what restraints would any given set of points require to ensure that there exists a path from the given starting point, to every other point in the set, and are there any set of constraints which would ensure that such a path will always be made; even if the two points being linked are chosen arbitrarily from the initial set of points?

The attached image is for a bit of clarification on some parts that I didn't quite know how to word.

I have worked out at least two restraints so far.
1: The number of Points must be greater than 2.
My thinking - In order for there to be only 2 points in the set it would have to consist of only the start and end points which cannot connect to each other. For any less than that a rule would have to be broken.

2:The total number of possible connections between any two points must be greater than 2
My thinking - 2 is the lowest possible total number of connections between 2 points in which an edge can be formed. However this would always result in a separation from the path as no edges can be formed back to the path.

[merry dirge]

Can someone check my work and see if its correct?

[shadow pond]

Hello, can someone help me with this question.

There's a bag of 10 cards. There are 5 cards of type A, 3 cards of type B, 2 cards of type C. How many different orderings are there if you pull all 10 cards from the bag?

[stray mural]

Orderings in the sense, number of ways??

[shadow pond]

I think so?

The answer is 2520

10 ! / 5! * 3! * 2!

[stray mural]

You are asking??

[stray mural]

Seems right to me

[shadow pond]

That's the answer

How do I get to it

[stray mural]

Oh this is the sense

Its like this will give the number of ways when no repetition

Ig

[urban saddle]

hey, I'm trying to show that $11^(n+1) + 12^(2*n-1)$ is divisible by 133

the case n = 1 works, but unless I fucking it up somwhere, n = 2 gives me 1475, which isnt divisible by 133

[quick folio]

btw shouldnt this be [5]

[gritty crescent]

,w (11^3+12^5)/133

[vital dew]

Results provided by WolframAlpha

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

Requested by Manan#0640

[gritty crescent]

So yeah, it breaks down.

[urban saddle]

@gritty crescent ups I fucked up its 2*n-1 not n+1

[gritty crescent]

Oh

[urban saddle]

but still doesnt work

[gritty crescent]

Hmmm?

[quaint star]

@quick folio Numbering questions with greek letters. Hmmm

[quick folio]

whats up

😳

[quaint star]

Why do you think it's just {5}? It's written like this btw

[quick folio]

because if u input both into the function

your image is {5}

?

oh idk why i put it like that

[quaint star]

Okay, but what about the points in between

[quick folio]

wait

🤦‍♂️

alright

i was thinking

that was a type setting thing

[-2,5] is actually the interval

OOP

ty ty

[quaint star]

Np

[quick folio]

so in this graph

to prove that its non planar

using kuratowskis theorem

we could knock out edge cf

and have that c remaining

the solutions were saying that c is "absorbed" into bd

is that legal??

i thought u could either a) remove edges

b) remove vertices, if that happens all edges incident to it are removed too

so if we knocked out c bc and bc edge will be gone too

also out of curiosity

when does e <= 3v-6 or eulers fail for planarity for graphs

[weary tiger]

Given a class of a function,
$$f(x) = f[\alpha (x)]:::::::::::::(A)$$
A particular solution to this is-
$$f(x) = f\left(\frac{x}{x-1}\right):::::::::(B)$$
where the a general solution for equation (A) with $\alpha (x)$ as an involution is given by -
$$f(x) = \tau[x, \alpha(x)]:::::::::::::::(C)$$
I am tasked with expressing the following solution of (A) in the form of the equation (C)
$$f(x) = \cos \log(x -1)$$

[vital dew]

xi64

[weary tiger]

an involution is a function which spits out the given variable after recursive operations

is this pretty much guess work or is there is something to it ?

[weary tiger]

oh and the involution is defined to be symmetric

this kind of problem seems familiar but i don't understand how you phrased it

i used the exact terminology used in the book, i am not sure if those terms are used globally. In case something isn't clear i can try answering it.

the question is posed in a paragraph so i didn't bother uploading the full texxt, if you say so i can upload the few pages

@weary tiger

you want to find all \alpha such that it solves f(x) = f(\alpha(x))?

i want to find <a function named tau> such that if i give it some other function alpha(x) and x , it gives out "x" whatever it maybe . Here the alpha(x) or x can be the cos log(x - 1) function

[vital dew]

xi64

[weary tiger]

where

it is actually $\phi^{-1}(\phi[\tau(x,\alpha(x))])$

at least that is how i am making sense of the problem

[vital dew]

xi64

[urban saddle]

I have this equation: 4x = 10x mod 27, what do I do with the x on the right side?

[unreal stump]

Rewrite that as 6x=0 mod 27

Which is same as saying 3x=0 mod 27 (since 2 is coprime with 27)

[urban saddle]

can you explain what you did there

[unreal stump]

Subtract 4x on both sides

[urban saddle]

so you ended with 0 = 6x mod 27 and then passed the 6x to the other side

[unreal stump]

Yes

[urban saddle]

alright, thank you

[stray reef]

bad latex

there are dedicated ways to label equations in latex without resorting to this shit with dozens of backslash-colons

[errant bear]

ok and

[merry dirge]

Could someone check my work and see if its correct?

mostly part b and c, cuz i think i did a correctly

[boreal lark]

That all looks right

@merry dirge

[merry dirge]

thanks!

Is the dual of this just simply x + 0 = x?

[boreal lark]

I don't recall what dual is but I believe so after researching that

[merry dirge]

okay thanks!

[weary tiger]

yeah i figured, but figuring out numbering as latex excahnge said wasn't usable here. Mind showing how to do it ?

(plus the main point was the question , you see. LaTeX is auxillary, any help with that? )

[stray reef]

the question itself is worded too poorly to comment

[weary tiger]

well that is how it is in the book

¯_(ツ)_/¯

[stray reef]

exactly as written?

[weary tiger]

i wrote a paraphrased version

[stray reef]

see

don't paraphrase

[weary tiger]

yeah i can upload the pages

[stray reef]

your paraphrase is likely what's making it bad

[weary tiger]

no the paraphrase is a separate line, not in the latex

the latex one is not paraphrased

[weary tiger]

here is the paraphrased one

[stray reef]

can you send the picture from the book

[weary tiger]

yeah sure

equation 1.12 @stray reef

[gentle hedge]

Where to learn discrete mathematics !?

[weary tiger]

knuth's book is a good start i presume

[gentle hedge]

Any YouTube channel that you recommend ?!

[stray reef]

so your latex thing is NOT an exact copy of the book, xi64...

[weary tiger]

ok

well kinda expected

anyways any help now ?

[stray reef]

so you want to express cos(log(x-1)) as tau(x, alpha(x))

[weary tiger]

yeah.....

[stray reef]

where alpha is an involution on (1,+infty)

[weary tiger]

also , since it says that the tau function is symmetric does that imply that it has to be positive ?

[stray reef]

no

i think i got it. take alpha(x) = x/(x-1).

then your thing is cos(log(x) - log(alpha(x))

so tau(u,v) = cos(log(u)-log(v))

[weary tiger]

genius

thanks

[gentle hedge]

??

[weary tiger]

dn't think so,no

[quick folio]

could someone give me some pointers for b

what i am thinking since the cardinality of S is 91, we will have 91 (i,j) pairs. if we divide them into 45 partitions we have at least 3 such that they have same modular residue

is that good>

[unreal stump]

Ye,should work

[long bronze]

Hi can someone explain to me why I got this question wrong

Here is my working

[a-z] = 26 letters
[A-Z] = 26 letters
[0-9] = 10 numbers
Total = 62
No. of possible passwords = 62^6
No. of bits needed = (62^6) * 10 bits
                   = 5.6800235584 × 10^11 (Ans)

[stray reef]

what you calculated is the worst case scenario

you want the average, which is half of that

[long bronze]

thank you😄

[urban saddle]

is A - B the same as A/B?

[fair cedar]

yeah assuming you mean set difference by A/B

and set difference is usually backslash instead

$A \setminus B$

[vital dew]

Moldilocks

[urban saddle]

yep, true, thanks

[quick folio]

could someone please help me with this question

[river lion]

take any two vertices and look at their neighbourhood

[quick folio]

yeahhhh

i am stuck there LOL

so if they are not adjacent

What should i do

btw we just learning graph theory so any pointers will be nice

[river lion]

i have no idea how to give any pointers without giving away the answer

i guess you could work backwards? try to see, if they are not adjacent, what would need to happen for diam(G) <= 2 to be true

[weary tiger]

@balmy hornet fix the DE and the A so you are essentially picking strings of length 4 using BCFG so there is 24

[balmy hornet]

@weary tiger (sorry for the ping)

[weary tiger]

you are looking at strings of the form DE----A

where the remaining 4 letters go in the 4 slots

there are 4*3*2*1=4!= 24 ways to do this

for example you missed DEFCGBA (and 15 others)

[quick folio]

so when it says strings

could it not also be concatenated from the middle

alsooo

question never said reptitions in the middle not allowed

[weary tiger]

yeh i just assumed based on what they said

[balmy hornet]

yeah the question that i have does not have that said so i assumed there would be more. time to figure out 16 more

[weary tiger]

it says how many are there

you don't need to find them

[balmy hornet]

the question after this is to list all of then ;-;

[weary tiger]

oh lol

well think about how i argued that it is 24

you should be able to systematically find them all

using that as a hint

[balmy hornet]

yes thank you! i should be able to find all of them

@weary tiger what happens if theres no repetition?

[weary tiger]

then you have 7 choices for each of the 4 slots

so theres 7^4 combinations

(if you are asked to write them out it definitely doesn't want repetition)

thats not what u said before

then its a different answer

lol

your strings are length 7 not 5

[balmy hornet]

Im sorry i realize im confusing myself

i know that P(7,5) = 7 * 6 * 5 = 210 ( thats how many string we should have if i believe im right)

[sharp sonnet]

I am reading a book, and one of the examples given is finding how many possible combinations of passwords you can choose. The rules for the password are: Has length between 6-8 characters, a character consists of either a digit or an uppercase letter, at least one of the password characters must be a digit. The solution they give for when the password is of length 6 is 36^6 - 26^6, but I am confused, as to me (although I am incorrect), it makes more sense for it to be 36^5 * 10. As any of the 5/6 characters can be a digit or character (hence the 36), but as there has to be a digit of which there are 10, it must be multiplied by 10.

[river lion]

it would have to be 36^5 * 10 * 6 since you can still choose, which of the 6 characters is the digit but even that's wrong

you're still double counting, as choosing the 1st digit to be a 4 and choosing the rest to be 2ABCD gives the same answer as choosing the 2nd digit to be a 2 and the rest to be 4XABCD, namely 42ABCD

[languid blaze]

it's 10(36^5 + 36^6 + 36^7)

or wait

yea

[stiff comet]

is anyone good w/ counting and probability

[valid wave]

hi im a little confused on how p would divide this

could someone elaborate please

[civic horizon]

the fact that p divides a^2 or the fact that p divides a?

[valid wave]

both really

i can see how p would divide a if the first one is proved tho

[civic horizon]

what is the definition of disibility

[valid wave]

so really just the first case

[civic horizon]

if you have two integers m and n, then n divides m if and only if what?

[valid wave]

if n is a multiple of m

[civic horizon]

okay

so is a^2 a multiple of p

and if it is, then why

[valid wave]

hmm

well since b^2p = a^2

p would have to be?

i guess im just not understanding that part

[civic horizon]

okay

so ill rephrase the defn of divisibility

n divides m if and only there exists an integer k such that nk = m

does this match up with your intuition

[valid wave]

somewhat

oh

yeah

crap

cause b^2 would be my k

[civic horizon]

yeah

[valid wave]

ah ok

that makes sense

thank you

[civic horizon]

cool

np

[valid wave]

wait

so for the p | k^2

p | k

is that proof complicated?

whats the idea behind that?

[violet kayak]

it's wrong

take a=6 (a²=36) and p=9

and b=2 (b²=4)

[valid wave]

wait thenn whaaaa

wouldnt that make this whole proof invalid then?

[violet kayak]

what came before that?

could you show me what they want to proof

and what the whole proof is

maybe they have some extra information on p, besides it being an integer

[violet kayak]

ah it's prime

then it's fine

you can use prime factorisation

to prove p|a² => p|a

[valid wave]

oh ok

so

p | k^2
p | k

when k is prime

and also

[violet kayak]

p prime

[elder scroll]

No when p is prime

[valid wave]

oh oof

ok thanks

but also

prime factorization is the

ability to show a number like 27

as

3 * 3 * 3

[violet kayak]

yep

[quaint star]

The idea is to consider the prime factorization of k. When you square a number, you don't add any new prime factors, you just multiply all the exponents of the prime factors by 2. So if p divides k^2, it also has to divide k. @valid wave

[civic horizon]

you can also just use euclids lemma

if a prime p divides a product ab, then p divides a or p divides b

then just pick a and b to be equal

[valid wave]

gotcha

thanks

sorry i have 1 more noob questions

for this

i understand that this means 12 | 3x - 9

so 12k = 3x - 9

3x = 12k + 9

so x = 4k + 3

but when its asked

how many solutions are in Z12

shouldnt it be infinite?

since theres an infiinte amount of numbers that are divisible by 12

[slow jewel]

I THINK IT JUST REPEATS SOLUTIONS

[clever sage]

z12 I believe is just the set of numbers 0 to 11

so in that range

what is a solution

because after that range it repeats

like Paul said

[errant bear]

what its really asking for is how many "classes" of solutions there are

[mint bane]

is the relation $\subseteq$ on the power set of integers transitive?

[vital dew]

nitezba

[mint bane]

i have a hunch that it isnt but idk

[still wagon]

it is

If $A, B, C$ are sets of integers, and $A \subseteq B$ and $B \subseteq C$, then $A \subseteq C$.\
To show this, show that if you take $a \in A$, then $a \in C$

@mint bane

[vital dew]

Shika-Blyat

[stray reef]

$\subseteq$ is transitive no matter what kinds of elements are in the sets you're considering, just sayin'.

[vital dew]

Ann

[still wagon]

Oh and yeah this ^, I should have said it

[languid blaze]

How many ways can you order 6 donuts from a shop that has 4 types A, B, C, D if
you can only order at most 3 donuts of type A.

i need to end a debate

49 or 74

or maybe 69 lmao

anyone up

[stray reef]

none of the above, if i understand correctly?

how were these numbers obtained

@languid blaze

[languid blaze]

through combinations with repetitions

b/c total number of ways is 84 no?

so i had multiple students subtract it different ways from total number of ways from having at most 3 or some stuff like that

idk people are at each others throat since its 50/50 on 49 and 74 lmfao

our final exam deadline just passed so everyone is talking about it

and no one knows

[stray reef]

hold on

okay i fucked up

theres more than one way of ordering ≥4 A donuts

without the at most 3 A restriction it's 84 yeah

[languid blaze]

right thats the total

i ended up getting 69 but i know thats wrong 😂

[stray reef]

so ok lets try to count the forbidden orders

an order is forbidden iff it has at least four A's so let's set those aside

we have two donuts left to pick

and they can be any type

4C2 + 4? that makes 10

so 74 is correct

[languid blaze]

the debate is over, one side has been granted victory through explanation

lol thanks, appreciate it

that answer makes sense

[quick folio]

i just did my discrete finals 😌

felt so goooood

[mint shore]

gj meliodas

anyways, can someone help me with a small brain far

i'm doing linear non homogenous recurrence relations with char eq

and i'm looking at an example off my powerpoints

basically

same polynomial

and different cases for F(n)

for the first one

F(n) = 3^n

you get your char eq

double root 3

all fair and good

here

what the fuck is p_0

n^2 is because of the double root and the s of s^n in the char eq being equal to one root

but what the fuck is p_0

it's a polynomial coefficient

as seen whenever you have other F(n)s

but the hell is its value?

[outer mirage]

Shoot just started this topic

[outer mirage]

What your language

[civic horizon]

Its determined by the initial conditions of your recurrence

[quick folio]

wait

greek be making it hard to read LOL

i think its just us trying out

different polunomials

cause its our geuss right

[mint bane]

can anyone help me on this please

[weary tiger]

whats your difficulty?

[mint bane]

well firstly im not even sure i get what a zero divisor is

but in general modular arithmetic has been messing me up

like i know what mod is ive had more than enough programming to understand it

but in this context it's a bit weird ig

[weary tiger]

well for example

2*2=0 mod 4. but 2 is not = 0 mod 4. so 2 is a zero divisor in integers mod 4

[mint bane]

hmmm

alright

but it doesnt have to be for all integers to be a zero divisor?

cuz 2*3 = 2 mod 4

[weary tiger]

no

else only 0 would ever be a zero divisor cuz in any ring 1*a=a*1 =a

[mint bane]

then multiples of 2 are zero divisors in mod 4

not worth saying multiples of 4 right cuz that's just redundant?

[weary tiger]

every number is either 0, 1, 2 or 3 in mod 4

2=6 mod 4

etc

so only need to say 0 and 2

you could say the classes of congruence of 0 and 2 mod 4

[mint bane]

damn see my prof skipped the number theory chapter and jumped to equivalence classes :|

would it still be safe to say the equivalence classes of 0 and 2 mod 4...?

[weary tiger]

if you show that the equiv classes of 0 and 3 are non zero divisors yea

[mint bane]

ay caramba

[weary tiger]

ayy lmao

[mint bane]

ooh wee

how might i go about proving that

cases of x?

[weary tiger]

you could go by cases theres easier way tho

[mint bane]

oh word

can u give me at least a hint pls

[weary tiger]

hint: the easier solution spoils (b)

tho there arent many cases anyway

you might want to check its just 3 cases

[mint bane]

that's funny cuz i was abt to just skip and go to b

but even the forwards direction of b idk

like ok let m be a prime number