discrete-math

[last timber]

c part is permutations

[patent moat]

got another one im stuck

6 married couples sit in a 12 seats round table.
-boys sit together so does girls

i think this is 6! * 6!

and also one couple decides to break up and we want them to sit differently. rest 5 couples needs to sit together

this last one im confused.

[last timber]

cyclic permutations

wait lol how

oh i think i got it like

gbbg

well interesting

[patent moat]

yes it's cyclic permuitations

i sold most of the parts except the last one.

solved*

[lusty eagle]

ill try this q

order each of the couples in 2 ways, gvies 2^5 = 32

there are 5 couples and two individuals to seat around the table with the restriction that one couple does not sit next to each other

so

[patent moat]

no

those 2 individuals dont sit next to each other.

[lusty eagle]

ye

[patent moat]

ok yea sorry.

[lusty eagle]

2^5 x 1 x 3 x 5!

[patent moat]

2^5 for the 5 couples

[lusty eagle]

yuh

3 spots she or he can choose to not sit next to

[patent moat]

why 3?

[lusty eagle]

draw it

[runic cedar]

@last timber okay, I'll look into it. Thanks

[patent moat]

12 seats , 10 people sitting

2 people left , 2 seats left.

[lusty eagle]

group the couples

[patent moat]

5 couples 2 individuals

[lusty eagle]

arrange it so that one individual cannot sit that individual

oh its 5 spots mb

[patent moat]

It's 12 seats

the 10 are occupied

circled table.

[lusty eagle]

yes, but the couples have to sit next to each othrr

so you just count them as one

for simplification

[patent moat]

ok so

5 couples

that occupy 10 seats

2 seats left for 2 individuals

C C E C C E
E C C E C C
C E C C E C

there are more than 3 options

for them to sit

[lusty eagle]

so there are only 5 spots she can sit

[patent moat]

im burned.

[lusty eagle]

couples are meant to be together

[patent moat]

Yes the 5 couples

[lusty eagle]

you can group a couple

[patent moat]

i grouped them

it's 5 couples so 10 people

[lusty eagle]

so your groups are 5, 1, 1

[patent moat]

yep

[lusty eagle]

arrange them

[patent moat]

1,5,1

i mean if x = couple
1 x x x x 1 x
but it could be
x x 1 x 1 x x

[lusty eagle]

ye

so how many spots can you place the 1

[patent moat]

lots of them xD

[lusty eagle]

remember its a round table

[patent moat]

is it just 5?

[lusty eagle]

ye

[patent moat]

2^5 x 1 x 3 x 5!

the x1 and x3

how cme

2^5 for the 5 couples
5! for the empty seats

shouldnt be x2 cause those 5! seats they can sit the other way around as well

[lusty eagle]

i accounted for it in the 2^5

[patent moat]

for the couples yes

but for the individuals/

[lusty eagle]

its 2^5 x 1 x 5 x 5! btw

mb i couldnt count

no in round table you sacrifice one

[patent moat]

what's up with the x1?

[lusty eagle]

the sacrificed one

there is essentially only one way to seat thr first person, because until then, all seats are equivalent

[patent moat]

aight thanks

also

if they want the boys grouped + the girls grouped , is it
6! * 6! ?

[lusty eagle]

ye

my first time explaining lol

your first client

xd

[patent moat]

😄

i have one last left

which is 10 character long password

using only small letters + no repitition + lexigographical order

e.g bdgjkmpqr

So for the first we can pick out of 26 (small letters)

but the second depends on first.

[lusty eagle]

lexigoraphical order

[patent moat]

yes

[lusty eagle]

what that

[patent moat]

bdgjkmpqr

if u pick for first O

[last timber]

a > b > c > d

[patent moat]

then the second needs to be after O

[last timber]

oh correctly

a < b < c ... < z

[patent moat]

alphabetical order.

[lusty eagle]

ok

[patent moat]

bdgjkmpqr see this

[last timber]

i bet it is permutations

[patent moat]

every letter comes after the previews

yes it is.

[lusty eagle]

yes

[patent moat]

do u have any idea?

[lusty eagle]

uhh

mhmm

let me see

nah, cant come up one in 2 minutes

[patent moat]

😄

ok thans

[lusty eagle]

u got any clues?

[patent moat]

I can only imagine a more algorithmic way for this

because the second depends on first

[lusty eagle]

yh

but how

[patent moat]

if first is X out of the 26 letters of the alphabet)

the second one will be picked from 25- X

third will be 24-second

23-third etc

i think

[lusty eagle]

but thats dependable on the first

[patent moat]

but i can't get it into math because the next always depends on previews.

it can be

A then D so the third one

needs to be after D

it could W

then next shuld be aftyer W

someone said we have 16 letters since it's 10 char long

so

cause u cant have qrstuvwxyz

i mean u can have that

but u cant start from R

caus then there are only 8 chars left. for a 10 pass

[lusty eagle]

so 16 letters at the start then

[hollow olive]

I have no clue how to go about solving this

i feel dumb

can anyone help

[halcyon ledge]

Well you start with turning the sentences into math notation

$u \implies (r \lor c)$

[vital dew]

deekaan:

[hollow olive]

I have this

im just not sure how to do the actual proof

[halcyon ledge]

normally you would use truth tables

or transform the statements

$u \implies (r \lor c) \equiv u \lor \neg(r \lor c) \equiv u \lor (\neg r \land \neg c) \equiv (u \lor \neg r) \land (u \lor \neg c)$

[vital dew]

deekaan:

[halcyon ledge]

something along those lines

what you currently have doesn't proof anything

[sleek swallow]

@hollow olive lmao your second premise is literally:

$\lnot{c} \implies \lnot{r}$

[vital dew]

Abhijeet Vats:

[sleek swallow]

By the law of contrapositives, it follows that $r \implies c$

[vital dew]

Abhijeet Vats:

[proven girder]

When asked to define a graph that models all possibilities, would that mean to actually graph it?

Something like this:

[pallid lintel]

possibilities of what?

[dapper mason]

do you mean a complete graph?

in that case each vtx is adjacent to every other vtx so maybe that's what the question is implying?

yes, pairs of natural numbers

probably, they're asking for the equivalence classes after all

maybe there's some pattern in the relation that can make identifying solutions faster

lol gl

[hexed granite]

@craggy sentinel what do you mean by deal with it?

It just means that your solution is periodic

Aka the eigenvalues are complex

Some cool stuff is that for a non-homogeneous equation if the real part of the eigenvalues is negative, then the long term behavior will solely depend on the particular solution

[crimson canopy]

is this the place for graph theory(introductory course)

[stray reef]

sure

[crimson canopy]

thank you. will ask here to at questions. have an exam like tomorrow, so hopefully i can clear all misconceptions by then

[obtuse lance]

maybe try to do a similar problem you can work out by hand with only 4 letters or so that way you can see

yeah

[crimson canopy]

how about finding A becomes M and B becomes W

then take A becomes M only minus the above?

idk

[halcyon ledge]

BCDA

CBDA

CDBA

BDCA

DBCA

DCBA

or do you mean its just one switch

[stable valve]

Hey can someone check if this is correct, I'm not sure. Especially with T6

[dense creek]

Can someone explain to me how 4 is congruent to -1 mod 5?

[weary tiger]

what's -1+5?

one way of thinking about it is integers that are congruent to -1 mod 5 are the integers which can be written as -1+5n for some integer n

or better yet, 4+5n for some integer n

or more properly you could verify it using the definition

https://i.imgur.com/kmTJ4Zh.png

since (4- (-1))/5 is an integer, the congruence is true

[weary tiger]

so a graph that has only one cycle subgraph

is basically a graph that is a square?

or one in which each of its vertices only have 2 degrees?

[ionic wind]

anyone here familiar with how to solve markov chain questions?

[patent rover]

Is an "element of a set" a special case of a "subset of a set"?

my thought was || no because say A = {1,2,3} and 1 is an element but not a set (because there is no set defined to be 1) and we then defined B = {1},
the set B would be a subset of A but the element 1 would still be contained in B
and we couldnt ever say 1 is a subset of A
because we would have to say B is a subset of A or that {1} was a subset of A which is not an element but {1} and B is not the same as the element
which means it is not a special case ||

[young ferry]

I would agree with that assessment @patent rover

[pallid lintel]

you put verticies instead of edges? the counterexample indeed shows there's no cycle from z that uses y. but e1Re2 and e2Re3 shows e1Re3. Your example doesnt show a common edge e2.

[crimson canopy]

to prove 2 graphs are isomorphic, do i simply compare if they have the exact same graphical sequence?

true

oh...adjacency matrix...but wouldn't the rows and columns of the numbers be different

lemme check out its definition again

my understanding of adjacency matrix is that it can be transformed

PA1P^-1 = A2 for some matrix P given 2 adjacency matrices A1 and A2 of graphs 1 and 2 respectively

can't find it in my notes

so the question is how do i find P

[cunning slate]

quick question

for this definition

c can equal a right?

c = a is possible?

so if aRb and bRa it means aRa or else it ain't transitive?

[stray reef]

if you have aRb and bRa but not aRa then your relation is not transitive, yes

[cunning slate]

Thanks Ann

Ann are you like universally an expert in all fields of math?

[stray reef]

uh

no

not by any means

math is huge and it's impossible for one person to be an expert in everything

[cunning slate]

Well you seem to be knowledgable in a lot of fields that is for sure.

More than the vast majority of people here

[stray reef]

i am knowledgeable in a number of things, i guess

[dapper rose]

she is also knowledgeable in numbers

[stray reef]

you're fine

[last timber]

well, just simple counting gives that
there are 120 outcomes with all dices not equal

out of 216 possible

does not look similar to 0.9 probabilty

well, yes you can just count

but i think what you should do is to use inclusion-exclusion principle and probably conditional probability

i mean look, you do not firstly care about the first roll, and count the probability that at least one of the remaining two will be the same as first

then you count the probability of that they are not equal to the first but equal between them and sum it together

[weary atlas]

any help with this?

[obtuse lance]

wrong channel this is #calculus or #advanced-analysis

[weary atlas]

oh gotcha. is this also #calculus ? @obtuse lance
https://i.imgur.com/Cz9KISW.png

[little ermine]

GRAPH THEORY: Prove that a graph with minimal degree 2 has at least 1 edge which isn't a bridge.

How do I prove that?

[weary tiger]

Sets and logic anyone?

<@&286206848099549185>

[last timber]

GRAPH THEORY: Prove that a graph with minimal degree 2 has at least 1 edge which isn't a bridge.
@little ermine just show that if you'll remove this edge graph still remains connected

i mean, construct firstly graph so that its minimal degree is 1 and in which each edge is bridge

and then show that adding (and then therefore removal) of an edge does not change anything

[weary tiger]

Can I get help on this? I even @ the helpers...

Here is my attempt don’t know if it’s right need confirmation

[last timber]

look

draw sets p q r

p -> q is the same as [ not p or q ]

[weary tiger]

I know the first 2 are correct

I’m just unsure about the third one

[last timber]

p or r?

[weary tiger]

Yes

[last timber]

it is union of the sets

not intersection

[weary tiger]

Would it be U?

so ~PUR?

[last timber]

yep

[weary tiger]

So all 3 of the sets are correct?

Like that?

[last timber]

looks so

but what ~ stands for?

[weary tiger]

Negation?

[dire orbit]

questions about basic enumerative combinatorics go here?

hi, i was reading a chapter about the use of generating functions in combinatorics, mostly for the purpose of finding a closed form equation for recursive formula of a series, and although i understand the mechanics and able to solve the basic exercises, i find the whole thing kind of magical. i mean, why suddenly using polynomials and basic calculus come into play, in a way other than just magically helping with the computation? why does going from and to geometrics series helps us solve these king of questions. sorry for being vague im a beginner...

[obtuse lance]

that's a good question I'd like a better answer to myself but I can at least give some partial answer

a lot of things we're taught is not really stressed what kind of combinatorial interpretation they have, for instance the binomial theorem

expanding out (x+y)^3 = xxx+xxy+xyx+yxx+yyx+yxy+xyy+yyy literally encodes the combinations of strings if we are careful when we expand

I think over all there are more things like this lurking around if you think deep enough about it, which might make things seem more natural

if you find an answer let me know

[dire orbit]

interesting, in fact some of the proofs in the chapter were exactly like this: showing that algebric/polynomial form is just another way of representing the combinatorical problem in question @obtuse lance

[obtuse lance]

since it's bipartite, all of the vertices in V_1 are not connected to each other

I mean, directly by an edge

they are only attached to vertices in V_2

because there are no edges connecting within the partitions, only across

we can count the degrees of the vertices of V_1 to get |E|

likewise for V_2

you can do it by induction I guess sure

but seems like a waste of time to do that

because it's bipartite, every edge has one foot in V_1 and one foot in V_2

the degree of each vertex just tells you how many edges it's connected to, so looking at all the vertices by adding them up tells you this

[obtuse lance]

assume there's a cycle of odd length

cycles have to end where they start right?

think about counting your steps and saying if your step is even or odd

and keep an eye on which of the two partitions you're in

yeah, try to think about what happens in general

yeah

[pallid lintel]

@last timber quite a few ways to prove this theorem. If you assume by contradiction that G has all bridges (its a forest), thats a nice way.

can also think about the longest path in G

[weary tiger]

Hello, can someone tell me why C is wrong

[pallid lintel]

https://en.wikipedia.org/wiki/If_and_only_if#Distinction_from_"if"_and_"only_if"

[weary tiger]

So do I just flip what's inside the ()??

[pallid lintel]

if and only if is the conjunction of two statements (if p then Q, and if Q then P)

[weary tiger]

Okay, got it. Makes sense now thanks

[dire void]

Hi #discrete-math !!!

I’m trying to understand what an intersection of an arbitrary collection of sets actually is, mainly what it means for an element to be in that intersection.

These are personal notes that I’ve written. I hope it is correct.

Mainly are 3 and 4 correct examples of the definition?

Maybe this belongs in #proofs-and-logic ?

[sleek swallow]

@dire void That's something called a generalized intersection/union. Essentially, it's a very neat generalization of the definition of union/intersection given for that of a finite number of sets.

In particular, the index set doesn't have to be finite. It can be countable or uncountable. The definition of a generalized intersection/union is practically the same as that of the regular union/intersection. Obviously, there are a few more details to be wary of.

Let $X$ be our set in question and denote $I$ as the index set. Then, an indexed family of subsets of $X$, denoted by ${S_i}_{i \in I}$ is a family of subsets of $X$, So, it contains elements from $P(X)$. Then, we define the generalized intersection and union of this family as follows:

$\cap_{i \in I} {S_i} = {x \in X: (\forall i \in I: x \in S_i) }$

$\cup_{i \in I} {S_i} = {x \in X: (\exists i \in I: x \in S_i) }$

[vital dew]

Abhijeet Vats:

[sleek swallow]

Notice that, if the index set is finite, then this definition I've put up just reduces to the regular definition for finite unions and intersections. That is what you'd expect, actually.

In your picture sent above, the point of examples 3 and 4 is to illustrate the difference between using a countable set as your index set and an uncountable set as your index set. In the case of using the natural numbers as the index set, it's very easy to characterize the union/intersection in terms of what has been stated. However, when using an uncountable set as the index set, this becomes relatively more difficult to do.

Ping me if you do have questions and I'll try my best to respond.

[dire void]

@sleek swallow maybe I’m getting confused by notation here:

Is it correct to write 1) $$\bigcap_{i \in I}S_i$$

or is it correct to write 2)
$$\bigcap_{i \in I}\left{ S_i \right} $$

[sleek swallow]

:?

[vital dew]

ninnymonger:

[sleek swallow]

I've seen both notations being used, though the former is far more common.

If you state that {S_i} is a family of subsets, then people are probably not going to confuse it with the set containing the element S_i

[solid forge]

does inferential statistics belong here?

[sleek swallow]

#probability-statistics

[solid forge]

thank you!

[pallid lintel]

yo, i want to prove that for 3 sets, which are pairwise intersections non empty, then i cannot have n/2 elements in each set without making them all interesect

for n elements that i wish to distribute

[wanton steppe]

can anyone help?

[umbral sky]

Hello everyone. Hopefully someone can help me with a short question on the transitive property on a relation. Let R = { (1, 2), (2, 1)} to I need to fulfill the transitive property just (1, 1) or also (2, 2)?

[pale epoch]

what do you think

[halcyon ledge]

Is R also symmetric?

[umbral sky]

R should be symmetric. I think, we have two connections in R. (1, 2) (2, 1) and (2, 1) (1, 2) so I think we need both (1, 1) AND (2, 2) to let R be transitive.

[pale epoch]

you are right

[umbral sky]

Great, thanks a lot.

[last timber]

You always can apply Warshall's algo to produce transitive closure

[weary atlas]

https://i.imgur.com/V3N1vKH.png

The answer to this is {(1,1),(2,2),(3,3)} right?

[stray reef]

there are many possible answers

but yours is one of them, yes

[weary atlas]

Thank you!

[versed flare]

you can try the first cases until you notice a pattern

[obtuse lance]

look at them mod 100

notice 99=-1 and 93=-7 mod 100

there's a bit of symmetry that cancels out

@neon thorn

because you only care about the last two digits

and mod 100 cuts off all higher digits you don't care about

when you carry after adding and multiplying, the digits go to the left, so they are just thrown away as irrelevant

[north jewel]

How can I solve the recurrence relation $a_n = 6a_{n - 1} - 8a_{n - 2}$ where $a_0 = 1, a_1 = 0$ for an explicit formula?

[vital dew]

Frank:

[north jewel]

I tried something of the form $a_n = \alpha r^n + \beta s^n$, but it isn't able to satisfy the initial conditions for any combination of $\alpha, \beta, r$ and $s$ (I think)

[vital dew]

Frank:

[obtuse lance]

no not quite

(-7^13) + (7^13)=0

but the other two (-1)^12 + 1^12 = 1+1=2

12 is even, 13 is odd

makes all the difference for the negative sign

@neon thorn

[daring harness]

does anyone remember what its called when there are two vertices on a graph with no path connecting them? its on the tip of my tongue and i cant remember

is there even a word for it, or are they just on separate subgraphs?

[last timber]

Well. they are just called disconnected vertices

and also it means that graph contains at least two (strongly) connected components

[daring harness]

@last timber thanks "connected components" is what i was looking for

[novel quartz]

I disagree with discrete math

[last timber]

And?

[weary tiger]

Hello people i am new

I have a question that i would really appreciate any help with in matrices

[vital dew]

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

[last timber]

loop always increases ii-th entry by two

undirected graph's adjacency matrix is always symmetric

but for directed $a_i_j$ not equal to zero means that there is an edge from ith vertex to jth

[vital dew]

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

[weary tiger]

Thanks ur a good person but i dont understand i need easy terms

I dont get discrete at all so the ith vertex i dont know where that is and same with jth infact i dont know whats going on at all

[last timber]

just as an example

adj matrix would be

0 1 0 1
1 0 1 0
0 1 0 1
1 0 1 0

so first row and second column is 1

it means that between vertex labelled 1 and vertex labelled 2 there is an edge

the same is true for other entries

for example 1 row 3 column is zero and there is no edge between vertices labelled 1 and 3

[weary tiger]

Bro thanks a lot even though i can recall some of this ima have to go back to basics its been a while since i touched the content

[last timber]

yes

[prisma arch]

need help with B. I know the answer because its in the back of the book but i dont know why it is 14.

It has to do with the pigeonhole principle

[sleek swallow]

Well, the goal is to get two socks of the same color, correct? Since he is picking the socks at random, it stands to reason that the firsts 12 socks he could pick could all be brown. So, picking 12 socks is clearly not enough to guarantee that you have at least two socks of the same color.

If you pick 13, then you are guaranteed to have at least one black sock. But you need 2, so you need to pick 14 socks.

[prisma arch]

So these are worst case scenario situations right?

[sleek swallow]

Yeap

[prisma arch]

Ah ok that makes a little more sense.

[sleek swallow]

You COULD pick out two black socks without picking out 14 socks

No one is denying that that's possible

However, you want to guarantee that you do have 2 black socks and picking 14 guarantees that outcome.

[prisma arch]

Right definitely makes more sense. This I will start my homework now that I get that much lol

[sleek swallow]

Yeap

[prisma arch]

legit was going to quit the class because its on zoom and hard to understand the prof and I couldnt understand the concept lol

[sleek swallow]

Same tbh

[pallid lintel]

r

im just wondering how this problem would be solved...if anyone out there is into ramsey theory. (im just getting into it today)

you would colour edges for 2 colours though right, one colour for a number that increases and one for a number that decreases?

for K6 we have either a blue triangle (increasing), or a red triangle (decreasing)

halp

oh, we sequence Kn vertices, and each next vertice in sequence either has to be increase, decrease or the same

[lime jolt]

Is there anyone that can help me with this problem?

Using algebra, show that C(n+1,r+1) - C(n, r+1) = C(n,r)

[robust mango]

@lime jolt By C(n+1,r+1), you mean $\binom{n+1}{r+1}$ right?

[vital dew]

Sup?:

[lime jolt]

yea

[robust mango]

shouldn't be too hard, did u try anything

[lime jolt]

I mean, i kind of looked up a solution, but I don't fully understand it

I missed the class my professor taught because I forgot to even look at the clock before i went on my walk in which class was in 5 minutes -.-

[robust mango]

Well you should know the binomial coefficient is defined as $\binom{n}{r} = \frac{n!}{r!(n-r)!}$

[vital dew]

Sup?:

[lime jolt]

Yea, i know that part

[robust mango]

Nice, so we're given $\frac{(n+1)!}{(r+1)!(n-r)!} - \frac{n!}{(r+1)![n-(r+1)!]}$

[vital dew]

Sup?:

[lime jolt]

Okay

[robust mango]

Now what we wanna do is make both the denominators same

So we can add these fractions

Are you familiar with the fact that n! = n(n-1)!

[lime jolt]

Yea, i kind of like factorials 😄

First time I saw it i was like O.O then i realized how awesome it is

[robust mango]

The denominator of the fraction on the left can be written as (r+1)! (n-r) (n-r-1)!

using the property stated above

Why do we do this? Because it resembles the denominator of the second fraction

So wo we have $\frac{(n+1)!}{(r+1)!(n-r)(n-r-1)!} - \frac{n!}{(r+1)!(n-r-1)!}$

[vital dew]

Sup?:

[robust mango]

we just distributed the negative sign there in the denominator of the second fraction

Now our denominators are almost the same

To make the second fraction's denominator the same as the denominator of the first fraction, what we do now is just multiply/divide the second term by (n-r)

$\frac{(n+1)!}{(r+1)!(n-r)(n-r-1)!} - \frac{n!(n-r)}{(r+1)!(n-r)(n-r-1)!}$

[vital dew]

Sup?:

[robust mango]

You can surely do the rest

[lime jolt]

Thanks a bunch 😄

Also, i did not know you could do that with TeXit

[robust mango]

😄

[lime jolt]

Which is pretty useful

[robust mango]

yes

[idle violet]

hep[o

hey ppl

[lyric pumice]

Hello,

[weary tiger]

Is every reflexive and symmetric relation is transitive?

[pale epoch]

no

[weary tiger]

in predicate logic how to write that two numbers are such that their sum is equal to their product

real numbers

[stray reef]

x+y=xy?

[weary tiger]

yess thanks

anyone good with counting principles?

How many subsets of A of cardinality 3 are there?

A is {1,2,3,4,5}

[near prawn]

what have u tried

[weary tiger]

I cannot figure it out

Im new to that

so any help would be appreciated

[halcyon ledge]

well the simplest way would be to count them

[near prawn]

have you covered combinations?

[halcyon ledge]

I mean it's basically just a question of possible permutations

[weary tiger]

how to do it by permutations?

[halcyon ledge]

well a set with cardinality 3 has 3 distinct elements

so just check

[whole shard]

What is the difference between subset and proper subset. Aren't they the same thing?

Or a set of A is a subset of set B, if both set have equal elements?

[gleaming zephyr]

proper subset means they cant be equal

like a<b and a<=b

[split python]

A being a subset of B just means all the elements of A are also contained in B. If the sets are equal then this is certainly the case, so A would still be considered a subset of B. Set A being a proper subset of B means all the elements of A are in B but there is at least one element of B not also in A.

[whole shard]

oh so a<b would be proper subset, and a<=b is a subset?

or a=b a subset?

[whole shard]

Idk the difference on these

{x:x+3∈N}.
{x∈N:x+3∈N}.

[gleaming zephyr]

first set is the set of all x such that x+3 is in N

the second set is all x in IN such that x+3 is in N

':' means such that

[whole shard]

The second statement is all x is natural numbers?

but first one includes x such as x=-3?

[pallid lintel]

what does a ramsey number of a clique of a line graph tell us about the helly property of a induced partial subhypergraph if anything?

[whole shard]

thanks mo2men and moonside, everything makes sense about sets now

[weary tiger]

you only need the first point right

actually no they're not

all of them contribute something

[quasi ferry]

Alright, first of all X_i \cap Y_j are disjoint with X_i \cap Y_k, sinc Y_j and Y_k are disjoint.

So basically, X_1 contains each element from Y_1, Y_2, Y_3 and Y_4, all distinct.

So basically, you can think of X_1 as coming from the first elements of Y_i, X_2 as coming from the second elements of Y_i, X_3 as coming from the third elements of Y_i, and X_4 as coming from the fourth element of Y_i.

i.e. X_i doesn't really add anything.

so you only have to count Y_i

sorry that's a bit of mess.

yeah, which is easy

If I didn't make sense, try it with X_1, X_2, Y_1, Y_2 with two elements each.

yeah, because Y_i are pairwise disjoint

blame it on Corona lol

maybe do an example, should make things clear.

[whole shard]

How are there 8 functions here

[stray reef]

{1 ↦ a, 2 ↦ a, 3 ↦ a}
{1 ↦ a, 2 ↦ a, 3 ↦ b}
{1 ↦ a, 2 ↦ b, 3 ↦ a}
{1 ↦ a, 2 ↦ b, 3 ↦ b}
{1 ↦ b, 2 ↦ a, 3 ↦ a}
{1 ↦ b, 2 ↦ a, 3 ↦ b}
{1 ↦ b, 2 ↦ b, 3 ↦ a}
{1 ↦ b, 2 ↦ b, 3 ↦ b}

[whole shard]

is there a shorcut i need to learn?

[last sigil]

Think about possible outputs for each input

[stray reef]

for every input in {1,2,3} there are two options for what output it maps to

and you make this choice independently for every input

therefore

2 * 2 * 2

or 8

[whole shard]

wait it looks like a truth table?

Ann how about counting the injective/subjective. Do I have to write it down?

[stray reef]

surjective*

[whole shard]

Nevermind, wriiting it down like u did helped.

[stray reef]

and uh yeah ig you'll have to write it down

counting surjective functions is kinda hard

[tight shard]

I don’t get the bottom box

Like answering on how it says b)

[stray reef]

well you could write y-7.50x+300=0 i guess

[tight shard]

In the form of ax + by + c?

[tame gale]

Can anyone help me asap pls?

[whole shard]

I can try 😛

[tame gale]

How can i get those two to be the same?

[whole shard]

oh dis is logic

[tame gale]

christ

[whole shard]

inference i believe

Something about De Morgans law

[tame gale]

i have like an hour or so trying to do this mdfkr

[whole shard]

it becomes (~p^~(p^q))

(~p^~(~p^q))

[weary tiger]

@whole shard it can be proven relatively easily that there are $n^m$ functions that can be made from a set to with $m$ elements to a set with $n$ elements

[vital dew]

Botnuke:

[weary tiger]

for your shortcut you wanted

[whole shard]

n=domain m=codomain?

i mean number of elements??

[weary tiger]

what?

[whole shard]

i dont know where you got n and m? is this about the function i asked earlier?

[weary tiger]

the $n$ and $m$ are arbitrary

[vital dew]

Botnuke:

[whole shard]

@tame gale (~p^(pV~q))

[tame gale]

aha

[weary tiger]

stated differently - let $X$ be a set with $m$ elements and $Y$ be a set with $n$ elements. Then there are $n^m$ functions from $X$ to $Y$.

[vital dew]

Botnuke:

[weary tiger]

you were considering number of functions from ${1,2,3}$ to ${a,b}$

[vital dew]

Botnuke:

[weary tiger]

so I claim that since these are 3 and 2 elements sets

[whole shard]

Oh okay I understand clearly now. so 3^2?

[weary tiger]

no, other way around

there are 2^3 functions

[weary tiger]

send that in #prealg-and-algebra or any of the questions channels

[whole shard]

@tame gale then you can do this (~p^(p->~q))

(-p^-q)

how was it

im not sure how you write one line down in chat

botnuke so you're saying # codomain ^# domain

[weary tiger]

yes

but...

well nvm I guess the language is fine

[viral hinge]

how was it
@whole shard you are a G.O.D ❤️

[tame gale]

Why the conditional there @whole shard ? im lost

[whole shard]

I think something like this, im not sure

[tame gale]

@tame gale (~p^(pV~q))
@whole shard

But here the one with the ~ is the q

so is it still possible?

[whole shard]

oh so like (-p ^ (-p -> - (-q)))

[tame gale]

What the actually fuck

[whole shard]

(-p ^ (-p -> q) )

(-p ^ (p V q) )

Im stuck now

[tame gale]

well....

[whole shard]

oh man it would have been easy if it was just q

[tame gale]

its ok

But thank you anyways

[whole shard]

oops its pdf @tame gale

@tame gale is this better

[tame gale]

Bro

I love you

no homo

[whole shard]

im not sure if its right cuz i'm still learning discrete math

im about to begin chapter 4

[tame gale]

its ok thank u

[whole shard]

ok

[whole shard]

Hello, so for inverse functions, you just swap the set from both side?

Like X -> Y becomes Y -> X

[weary tiger]

what?

you need to swap the components in the pairs that define the function

what you are suggesting is more like..

a prerequisite for functions to be inverses of each other

or something

[whole shard]

what i understand is if set X:{1,2,3} and Y:{a,b}. the inverse would be X:{a,b} and Y:{1,2,3}

[weary tiger]

no, I don't think you understand

well

do you know what a function is

[whole shard]

oh yes.

im reading it by the book, thats what i'm trying to understand. The book is makes it really weird

[weary tiger]

ok, you should really understand the statement of the definition of a function before pushing forward into other things

do you know what a relation is?

[whole shard]

HUH

i dont think i have that

i mean in syllabus

[weary tiger]

how does you book define functions without touching relations at all?

[whole shard]

this is their definition

[weary tiger]

I am talking about functions, not inverses

[whole shard]

oh uh, functions is rule i know that

[weary tiger]

a function is not simply "a rule"

a function is a set of ordered pairs

the high school definition of functions as "rules" totally ruins functions for everyone

[whole shard]

I guess I have to read the whole chapter again, because the invese thing made me lose it all

[weary tiger]

yea I mean, you aren't going to understand function inverses at a rigorous level if you don't understand functions at a rigorous level

which is what you seem to be trying to do

[whole shard]

I got the onto and one-to-one

could you recommend a book.

[weary tiger]

hmmm

I can recommend -

https://brilliant.org/wiki/relations/

start here and understand everything on this page

https://www.tutorialspoint.com/discrete_mathematics/discrete_mathematics_relations.htm this is good as well

I am personally of the opinion that relations should be learned before functions, but that's just me

functions build from relations rather easily, once you understand them

[whole shard]

i never heard of this

[weary tiger]

functions can be defined as relations that satisfy one extra condition that a relation doesn't need to have

so you should learn what a relation really actually is at a rigorous level, if you want to really understand what a function from set X to set Y is

because saying a function is "a rule" is really useless if you want to understand or prove things related to functions at an abstract level

[whole shard]

alright, let me just start over this chapter. I might be able to change my understanding

cuz now ur saying functions are "sets"

[weary tiger]

yes, a function from set X to set Y is a subset of the cartesian product of X and Y (which is a set of ordered pairs) satisfying the condition that each x in X associates with exactly one y in Y

you should understand that statement fully before trying to understand function inverses and other things at a rigorous level

functions are, literally, sets of ordered pairs

not rules

function "rules" describe the ordered pairs

but they don't exactly define the function

they tell you how to find the ordered pairs, or what they look like

or however you want to think of it

[whole shard]

okay so I'm looking at this

A function assigns to each element of a first set exactly one element of a second set, where the two sets are not necessarily distinct

Is this what you're trying to say?

[weary tiger]

"where the two sets are not necessarily distinct" what's necessary about this remark?

[whole shard]

I couldnt understand what it says

[weary tiger]

they surely don't need to be distinct

[whole shard]

two set are not the same

[weary tiger]

yea but you don't need to mention that

[whole shard]

OOH okay so function here is assigned from first set to second set?

now I see what ur saying. I need to read more

[weary tiger]

basically what I'm saying is, you should replace your "rule" intuition with "sets of ordered pairs"

[whole shard]

reading different websites about discrete math is really a bad idea. I should stick closely with the book

[weary tiger]

umm

functions should be pretty consistent across the board

[whole shard]

Maybe im reading these wrong

[weary tiger]

what book is it

[whole shard]

http://discrete.openmathbooks.org/dmoi3/sec_intro-functions.html

my classmate gave me that website, i thought it was easier to understand

[weary tiger]

it's easier to understand because it matches the intuitive notion of functions that high school gives

[whole shard]

damn, so I'm learning at a high school level by reading that website

unacceptable

[weary tiger]

I mean, it's almost there

but I'd have to say there are plenty of better things out there

[whole shard]

Lol I'm taking that off my bookmark list

one more thing, ordered and unordered, is like organized and not organized?

my vocabulary is bad

[weary tiger]

nah

unordered doesn't seem to come up much but it would mean

[whole shard]

something like {1,2,3} is ordered and {3,2,2,2,3,1} unordered

[weary tiger]

(a,b) is the same unordered pair as (b,a) even if a and b are not the same

where as with ordered pairs, (a,b) and (b,a) are only the same if a=b

[whole shard]

uhm like (fruits,vegetables) is unordered. then (fruits,fruits) is ordered?

[weary tiger]

no

if we were working with ordered pairs, then we would consider (fruits,vegetables) and (vegetables,fruits) to be 2 separate pairs

but if we were working with unordered pairs, then (fruits,vegetables) and (vegetables,fruits) would be the same object

[whole shard]

AHhhh

[weary tiger]

basically, the order in which we write them matters, in ordered pairs

while it doesn't for unordered

[whole shard]

Ah so what that mean is the ordered pairs are strictly (a,b)..

where unordered is (a,b) but can be (b,a)?

[weary tiger]

I don't know exactly what you mean

if (a,b) is an ordered pair, then (b,a) is a different pair, unless a=b

[whole shard]

I'm beginning to understand thank you

[clever patrol]

Hi, not sure where to put this but i think this channel is alright? Anyway, can anyone explain why the answer for part IV is simply {a,b,c,d}^2?

I'm normally used to adding new pairs to make the relation an equivalence relation (adding new elements by reflexitivity/symmetry etc). Thanks

[weary tiger]

https://i.imgur.com/to57Qiw.png hey @whole shard here is a good example

see how the function is defined by a set of ordered pairs?

a associates with 2, b associates with 4, and c associates with 2

that's the function

a set of 3 ordered pairs

[whole shard]

oh this is something different, i know function is like f(x)=2x

maybe thats why im confusd

[weary tiger]

well, yea, but that's not a very precise way to define a function

[whole shard]

so this way, function is a set of ordered pairs

because I see a set with 3 elements

[weary tiger]

what that function actually is

is a set of ordered pairs

for example, (5,10) is one ordered pair that defines the function

so is (1,2), and (20,40), etc

that's what defines the function

[whole shard]

oh ok, but for discrete math, we have finite variables right

most of the times

finite domain i mean

[weary tiger]

yea but I'm just trying to change your perception on functions lol

even the high school functions from real numbers to real numbers are still sets of ordered pairs

they just never get presented that way

which is a shame imo

makes functions very hard to understand rigorously later

[whole shard]

yes I can see that

im trying to adjust

now I'm beginning to understand what the book is saying about students enrolled in school and students enrolled in discrete math

the function is students enrolled in school who's taking discrete math

[tiny axle]

what's the logical connectives for but? like
A but not B

[stray reef]

$A \land \neg B$

[vital dew]

Ann:

[sterile yacht]

if i flip a coin and you win $1 if its Tails and you get $0 if its Heads, but if its heads 3 times you will 100% win on the 4th flip. What is your win chance? and how do i calculate that?

[sacred furnace]

the floor of x < x for all real numbers, that would be false right?

like if x=2

the floor of 2 < 2, would be false?

[obtuse lance]

are you sure?

[sacred furnace]

i think so, for the expression, the floor of x < x, it would have to be lessthan or equal too

for it to be true

[obtuse lance]

yeah

I guess so

[obtuse lance]

yeah

[cunning slate]

so 2 = 12 (mod 10)

2 = 22 (mod 10)

[stray reef]

those statements are both true.

[cunning slate]

I see,

I know modular arithmetic is not supposed to be this hard but I am confused.

is there a distinction between congruence and the equal sign for modular arithmetic?

Sorry to bother but 22 = 12 (mod 10) also true?

[sacred furnace]

Is this one to one?

[split python]

No

Where did you get your last line from?

[sacred furnace]

Myself

[split python]

To show one to one you want to show same output implies same input

[sacred furnace]

Would line one be correct?

R_1 + 2r_2 = s_1 + 2s_2 ?

[split python]

Sure that's the defn of the J's being equal

Your second line is also correct

[sacred furnace]

What would I plug in?

[split python]

But can you conclude from that the inputs have to be equal?

Can you think of any two pairs of inputs that give the same output?

[sacred furnace]

If r1 and s1 = 1 and r2 and s2 = 2

[cunning slate]

Can someone please explain modular arithmetic to me?

I understand why 37 = 1 (mod 12)

But I dont understand how 2 = 22 (mod 10)

[split python]

What's the remainder of 22 divided by 10?

[cunning slate]

*2

[split python]

Yes

So 2 == 12 mod 10

[cunning slate]

well you cant apply that logic to 37 = 1 (mod 12)

[split python]

Yes you can

[cunning slate]

what's the remainder of 1 divided by 12?

[split python]

1

[cunning slate]

This is why it is confusing to me

yes 1,

[split python]

12 goes into 1 0 times with a remainder of 1

[cunning slate]

but that doesnt explain how 37 = 1 (mod 12) (Which I understand through a different method)

[split python]

12 goes into 37 3 times with a remainder of 1

[cunning slate]

Oh ok I see

[split python]

So they both have remainder 1 mod 12

[cunning slate]

so its like a clock

[sacred furnace]

So do J(r1,r2) and J(s1,s2) both have to be the same ordered pair i.e (2,3) and (2,3) ?

[cunning slate]

yep I understand now

thanks moonside

[split python]

Exactly like a clock

[cunning slate]

this has been bugging me for ages

[split python]

Np

[cunning slate]

So that is why people use the congruence

symbol instead of the equal

[split python]

Yes

It's a different 'type' of equals

[cunning slate]

yeah, I see

thanks

[split python]

No prob

[lunar flint]

moonside

[sacred furnace]

Sorry if I came off rude at all moonside

[lunar flint]

not sure where my question falls under but can someone help me?

[cunning slate]

Sorry another question, So I know that 2 = 72 (mod 10)

But is 12 = 72 (mod 10) , and 102 etc?

[sacred furnace]

Post it

[cunning slate]

a,b

e

oh wait

lmao

i got rekt

a, and e

[last sigil]

@lunar flint #geometry-and-trigonometry

[cunning slate]

not b

[split python]

What's up

[sacred furnace]

So do J(r1,r2) and J(s1,s2) both have to be the same ordered pair i.e (2,3) and (2,3) ?

[split python]

Huh that doesn't even make sense chondro

[sacred furnace]

I'm just confused on what to sub for those to show 0=0

[split python]

You want to check if J(s,r) = J(x,y) then you HAVE to have s=x and r=y

If you can find two different values of s,r

That give the SAME output J

Then J is not 1-1

Pick an output value and try to find two different inputs that give that value

Yes tom

2 = (any number with remainder 2 after division by 10) mod 10

[cunning slate]

thx again moonside

and so is 22 = (any number with remainder 2 after division by 10) mod 10

[split python]

Yes

It's exactly like a clock you're just changing how many hours in the day you have (well, ignoring am/pm)

[sacred furnace]

Does my spanning tree for dijkstrstras look correct?

[pale epoch]

this is not a spanning tree?

it doesn't span the graph and also isn't a tree

[sacred furnace]

oops

well i guess i mean to ask, does the pathing look correct

or edges

[pale epoch]

the shortest path from a to z is correct

[sacred furnace]

i guess im unsure of D to C

i want to say after the fourth edge has been added (A to D to F to C)

the next shortest distance would be A to D to C with a value of 8 (5th edge)

and then the final edge would be A to D to F to Z with a value of 9 (6th edge)

actually swap the 4th and 5th edge

So it might be F to C im unsure of

also maybe B to C is include where its value of 10 is less than the value of A to Z at 11

[pale epoch]

wait what

[sacred furnace]

well the way i was taught was to add the cheapest edge then relabel distances

so it would start from A to B value=1

Then A to D value 3

Then D to F value 6

Then D to C value 8

Then F to C value 9

Then B to C value 10

Then F to Z value 11

all values are from A

[pale epoch]

ok, let me think

you always choose shortest distance to an unvisited node

why would you visit C twice

once via D and once via F

you actually visit it a third time

via B

[sacred furnace]

Ahh unvisited node being the key word

[pale epoch]

also

you seem to have ignored E entirely

[sacred furnace]

Yes , so E has 2 paths that are both 11

A,d,e And a,b,c,e

Which one would I take?

Bm

Nwvermind

The shortest would be a,d,c,e

[pale epoch]

yes

in general if there are two paths of same lengths, you choose whichever

[sacred furnace]

Alright, that helps

I think this looks correct now

[pale epoch]

yes

[sacred furnace]

Cool thanks a bunch

[weary tiger]

hey guys, in regards to set algebra, how do you deal with A ∩ bar A

ik what to do if it's union but not intersection

[sacred furnace]

Intersection is what both sets share

[weary tiger]

right

but in terms of simplifying I learned it by rules

which idk the one for this case

[sacred furnace]

Do you have more information? What is the question asking?

[weary tiger]

Sorry for the late reply, this is the question

@sacred furnace

I was simplifying that expression and ended up with that case

[sacred furnace]

Is BarA the same as A^c ?

Complement = c

[weary tiger]

barA is not A

so everything not in set A

I've never seen that notation

hmm

[sacred furnace]

Ya kinda weird simplification, not sure what the circuit would look like

Maybe a node with a loop to itself

[sacred furnace]

Or two nodes that point at each other

[weary tiger]

So far this is what I have in terms of simplifying one side

[sacred furnace]

Making me read sideways and shit lol XD

[weary tiger]

Lol sorry I'm on mobile

[sacred furnace]

It's all good

That's still like every element of every set I think

I could be wrong

[weary tiger]

Ah ok

Np

[weary tiger]

I was wondering if anyone here knows truth tables, cuz I'm unsure if my work is correct

thought it is completely solved

[halcyon ledge]

show tables

[weary tiger]

ok one sec

[halcyon ledge]

looks fine

[weary tiger]

it's neither a tautology or contradiction right?

not sure how that works with biconditional statements but it's not all true and not all false

[halcyon ledge]

well tautology means always true and contradiction always false

so you take a good look at your result and you have your answer

[distant bay]

hey all

I was wondering which is the best book / online course for a someone self studying discrete maths

for context I've been covered algebra / trig pretty throughly

and plan on going through precalculus too

[spare jolt]

Knuth had a book called concrete maths which I think is pre good.

Is it correct that expanding:
(3-x)^-7 would be the same as expanding (3-x)^7, except I would just ^-1, each coefficient?

E.g
(3-x)^-7 = 1/(3-x)^7

= 1/(-x^7 + 21x^6 - 189x^5 + 945x^4 - 2835x^3 + 5103x^2 - 5103x + 2187)

= -x^-7 + 21x^-6 - 189x^-5 + 945x^-4 - 2835x^-3 + 5103x^-2 - 5103x^-1 + 1/2187

[halcyon ledge]

$\frac{1}{a + b} \neq \frac{1}{a} + \frac{1}{b}$

[vital dew]

deekaan:

[spare jolt]

rip

How would I expand it then?

Since binomial theorem uses combinations which I don't think support negative numbers.

[halcyon ledge]

$x^{-3} = (x^{3})^{-1} = \frac{1}{x^{3}}$

[vital dew]

deekaan:

[halcyon ledge]

you just don't do that weird last step you did

[spare jolt]

Right, how do I get rid of the ^-1 though?

Is there a way to remove the brackets?

[halcyon ledge]

what do you want to do exactly?

[spare jolt]

I want to get it in an expanded form, without brackets. Apparently the coefficients change?

[halcyon ledge]

$\frac{1}{(x + 1)^{3}} = \frac{1}{(x^3 + 3x^2 + 3x + 1)}$

[vital dew]

deekaan:

[halcyon ledge]

that?

[spare jolt]

Well I was hoping to get rid of the 1/

So like in a more: x^3 + 3x^2 + 3x + 1 like form

[halcyon ledge]

it's in that form, just the inverse

[spare jolt]

Can I get rid of the inverse?

[halcyon ledge]

no?

$(x^3 + 3x^2 + 3x + 1)^{-1}$

[vital dew]

deekaan:

[spare jolt]

Hmm, apparently there are different powers when I expanded it like that.

As in

I had a question and it was asking me what the coefficient of x^12 is, for (1-x)^-7

[halcyon ledge]

0

[spare jolt]

$$(1-x)^7 = -x^7+7x^6-21x^5+35x^4-35x^3+21x^2-7x+1$$

[vital dew]

Soap:

[spare jolt]

x12 seems non-existent.

A trick question maybe?

[halcyon ledge]

well you can add $0 \cdot x^{12}$

[vital dew]

deekaan:

[spare jolt]

Yeah I guess. But that feels like trivial answer.

0 seems most likely answer tho

Thanks

[spare jolt]

Ok apparently not a trick question.

$$(1+x)^{-n} = \sum_{k=0}^{\infty} (-1)^k \binom{n+k-1}{k} x^k$$

[vital dew]

Soap:

[spare jolt]

I don't get how this is allowed or even makes mathematical sense?

(1+x)^-n = 1/(1+x)

For the case: (1+x)^-4, x^69 exists?

Apparently in some bizzaro world: 59640x^69 is part of the expansion (1+x)^-4?

$$\binom{-n}{k} = (-1)^k \binom{n+k-1}{k}$$

[vital dew]

Soap:

[spare jolt]

Because of this identity, this whole wacky thing exists apparently.

[spark oyster]

Hmm, I imagine this binomial coefficient for negative n was defined in exactly such a way that it makes this identity true

since a priori, with the usual definition, the binomial coefficient only accepts nonnegative integers as arguments

And then, of course, you need to specify for what x you want the above expansion to hold. I imagine you want |x| < 1 or so

And for small x, this doesn't look super implausible at least

[spare jolt]

Now I need to figure out how it works for (1 - x)^-n

Cuz apparently it exists

Feels very wrong to me though

Algebraically doesn't make sense imo

[spark oyster]

What about it, though? It's unusual in the sense that it's a "binomial" expansion by an infinite series, which you don't have for positive n, but if you're comfy with infinite series then it should be okay

[spare jolt]

I guess so, since most things can be expanded into some infinite series.

[spark oyster]

https://www.wolframalpha.com/input/?i=sum+k%3D0+to+infinity+of+(-1)^k+(binom(2+%2B+k+-1%2C+k))+(1%2F2)^k for an explicit example, this checks out

[spare jolt]

Yea, this makes sense, https://mathworld.wolfram.com/NegativeBinomialSeries.html

Still weird tho

[spark oyster]

I imagine you can get this series fairly easily by just using the geometric series $(1 + x)^{-1} = \sum_{k=0}^\infty (-1)^k x^k$ and just using a Cauchy product formula or so

[vital dew]

Lartomato:

[spark oyster]

If that's anything you're familiar with

If it's not, you will 100% encounter it on your mathematical journeys

[spare jolt]

Probably, havent seen any Cauchy stuff, but will probably have to deal with it later on

Btw would it be the same expansion for (1 - x)?

But just negative of the coefficient?

[spark oyster]

Yeah, I imagine so. By the formula for negative binomial coefficients that you showed earlier, the expansion would look exactly the same for positive exponentials

[spare jolt]

Like since this one is true:
$$(1+x)^{-n} = \sum_{k=0}^{\infty} (-1)^k \binom{n+k-1}{k} x^k$$
Then I maybe I can assume:
$$(1-x)^{-n} = \sum_{k=0}^{\infty} (-1)^k \binom{n+k-1}{k} (-x)^k$$

[vital dew]

Soap:

[spark oyster]

Oh, yeah, sure. If you know that your previous formula holds for all |x| < 1, then the second one also holds for all |x| < 1

[spare jolt]

Aight cool, thanks!

[weary tiger]

hey guys

may someone pls explain what they did here in induction

like ok, they replaced k with k + 1 but how did the 10 get there

[sleek swallow]

10 = 3+7

Lul

[weary tiger]

yes i know but if so, why is the 3 still there

3k + 10

wouldnt it be k + 10

[sleek swallow]

3(k+1)+7

3k+3+7 = 3k+10

[weary tiger]

oh distributive

shit sorry lol

thanks

[sleek swallow]

Lololol

You're welcome

[weary tiger]

online classes be too stressful

[sleek swallow]

At least you have online classes lol

I'm gonna be waiting for 3 months before i go to uni

Well, a bit less than that

[weary tiger]

damn

good luck

[distant bay]

ok so I had a look into books and it seems like

http://discrete.openmathbooks.org/dmoi3/dmoi.html

This one is a good intro to discrete maths

but I can't find any information about a prerequisite

has anyone read this and knows (Discrete Mathematics: An Open Introduction, 3rd Edition)

[spare jolt]

Nope

But apparently its for math undergrads

This text aims to give an introduction to select topics in discrete mathematics at a level appropriate for first or second year undergraduate math majors, especially those who intend to teach middle and high school mathematics.

[distant bay]

Yeah isn't particularly clear for someone like me whose self studying

I have zero context what first / second year maths majors study

[spare jolt]

I assume its calc 1, calc 2, linalg

What are you self-studying for?

If its for compsci, I hear concrete maths is good

[distant bay]

yeah it's for computer science

[spare jolt]

concrete maths is written by knuth

who u mightve herd of

i think hes gonna die soon

hasnt finished his seminal work yet

[distant bay]

I having a look at that book now

and it doesn't have a prerequisite either

[spare jolt]

well i dont think most books have a page called prerequisites

[distant bay]

they should at least have a paragraph saying you'll need this much maths to understand this book

hell both of the graphics programming books I looked at have that

[spare jolt]

so apparently basic calculus, (everyone likes spivak fr some reason), geometry, trigonjometry, algebra,

"graphics programming"

generally if your good with your fundamentals which are geometry, trigonometry, algebra and calculus

you should be good with most of discrete mathematics

if your good with calc 1 and calc 2 and linear algebra (which are the uni-level fundamentals i guess)

you should be good with most other subjects

i assume thats why most people dont outline "prerequisites"

[distant bay]

it's still a bad practice