discrete-math

[modest zealot]

ive just come to the realization

of

wat the fak is dis

[azure lichen]

basics

[stray reef]

^

[azure lichen]

is what this is

getting fluent with that will be useful later

[modest zealot]

hmmmmm

wait just one question

publication date, number of pages.
(a) What is a likely primary key for this relation?
(b) Under what conditions would (title, publication date) be a composite key?
(c) Under what conditions would (title, number of pages) be a composite key?```

what is a composite key

google says its something that requires 2 fields to identify a particular element

[stray reef]

2 or more, probably

and the key (pun not intended) thing is uniqueness

[azure lichen]

I assume it would mean that if you have that info it uniquely specifies the book?

but no guarantee that was right

[stray reef]

i.e. given just the key there's one and only one entry to be found under that key

[modest zealot]

wut does it mean by "under what condition"

wouldnt that be a terrible idea?

[stray reef]

so accordingly, what you don't want is for the same key to match two different items

[modest zealot]

could have same titles on the same date

[azure lichen]

that’s the condition

:P

well, the opposite of the condition, rather

[stray reef]

(title, date) would be a composite key unless you had two different books with the same title published in the same year which were nonetheless different

[modest zealot]

as long as there isn't two books with the same exact titles on the same exact date

then that would be perfectly fine

[azure lichen]

and yet, we use (author, year) usually

which fails a bunch of times

[stray reef]

it's common in math

(title, author) is more common in more quotidian usage

[azure lichen]

and then they have to do (author, year, index), where the index says the how-many-eth book of that author in that year it was

like idk, Dixon 2005b

[stray reef]

[modest zealot]

hmmm i c i c

[azure lichen]

at which point, honesly, the title would be easier :P

[modest zealot]

tru tru

wtf is question 4 asking

that shud be my last question

[stray reef]

the projection $P_{2,3,5}$ keeps the 2nd, 3rd and 5th coordinates and discards the rest

[vital dew]

Ann:

[stray reef]

by defn

[azure lichen]

and I think the table doesn’t have anything to do with it

[modest zealot]

o

wait rly

sooooo it would be like (b, c, e)

oh also question, can a set be both symmetric and antisymmetric?
or can a set be NOT symmetric and NOT antisymmetric?

or is symmetric/antisymmetric one of those things, where its one or the other

[stray reef]

can a relation be sym and antisym at the same time? yes

can a relation be neither sym nor antisym? hell yes

[modest zealot]

o

, the inverse of the relation R, be found from the matrix
representing R, when R is a relation on a finite set A?```

do we just hold all the i=j values

and then swap i and j

so for exaple a matrix of like

5 7 8
3 1 3
5 2 5

becomes

5 3 5
7 1 2
8 3 5

[azure lichen]

sym & antisym:
the relation = on any set is both

neither sym nor antisym:
just about any relation witout much structure

[modest zealot]

gotcha, that makes sense

[stray reef]

@modest zealot the matrices that represent relations can only have 0 and 1 as entries

[modest zealot]

oh fuck

mb

uhhh

UHHH

shiet

[azure lichen]

now, what’s the inverse relation?

[modest zealot]

dont u just flip the rows and columns

and leave the rows = columns alone

[stray reef]

||but yes the matrix for R^-1 can be found from that of R by taking the transpose||

[azure lichen]

that’s noth the answer to my question

[modest zealot]

uhhh

inverse relation

is basically going backwards

in a nutshell

rite

[stray reef]

the definition

give us the definition

[modest zealot]

hol on

lets say u have a relation

then an inverse

is backwards

IDK THE REAL DEIFNITION

[azure lichen]

try to write one, then

try to formalize your gut feeling

like, aR⁻¹b if and only if………

[modest zealot]

Assume you have a relation in set A, then an inverse is

for all relations, where some element a maps to b (a->b), then an inverse relation is when the element b maps to a (b->a)

o

idk that form

thats weird form above

[azure lichen]

do you know what aRb means

[modest zealot]

a relation b

OH SHIT

[azure lichen]

which means, in your notation?

[modest zealot]

aRb = bR^-1a

BOOm

ez

[azure lichen]

not =

well

[modest zealot]

oh shit

mb

uhh

[azure lichen]

I guess yes = but like

[modest zealot]

INVERSE

[azure lichen]

I would rather write ⇔ than =

or “if and only if”

but yes

[modest zealot]

where tf do u get weird symbols

nani

[azure lichen]

good keyboard

[modest zealot]

wo

where u get cool keyboard

i wanna e cool

[azure lichen]

neo-layout.org, it’s optimized for German though

has a qwertz-mode

but not qwerty-mode

it’s like, really good though

[modest zealot]

ahhh ic ic

dis shit

reflexive, antisymmetric, and transitive right

ez

[azure lichen]

correct

[modest zealot]

awesome

dis lil shit is only reflexive

[stray reef]

correct

[modest zealot]

bro im nailing this

i still gotta get back to ur hellish exercise after this

[azure lichen]

you don’t have to do all of them but honestly after doing 2-3 the rest will be easy

[stray reef]

it's 8 smaller exercises in a trenchcoat

[azure lichen]

I mean it actually is

[modest zealot]

oh god

[weary tiger]

lmao trenchcoat

[modest zealot]

ok might need a little help on something i never seen before

C L O S U R E S

[azure lichen]

in what context

there’s like five different kinds of closures at least

…I mean I know two but still

[modest zealot]

dis

[weary tiger]

kek

[modest zealot]

context

wdf r closures

[azure lichen]

okay, closure wrt …something basically means you make it bigger so it fulfills that property

but only as big as you have to

like, the closure of the natural numbers for - would be the integers

[modest zealot]

o like barely making the pass mark in schoooooool

[azure lichen]

cause you just add all the numbers you can get by doing - as well

but nothing more

[modest zealot]

so just kee padding shit til u meet the conditions

no more

[azure lichen]

aye

[modest zealot]

for the problem above

can we just do

a=b

or is that too much

overkill

[azure lichen]

well, it also has to continue fulfilling what was there before

[modest zealot]

well fak

uhhh

a+b = 1?

or a+b = something

[azure lichen]

…how do you even get that idea

[stray reef]

[modest zealot]

is that absolutely wrong

prob is

[azure lichen]

it’s about as wrong as it could get cause that’s neither reflexive nor does it extend the given relation

like, what you need is:

[modest zealot]

oh shit

i was thinking about symmetry

mb

so ur telling me

it has to map to itself

[azure lichen]

if a≠b, then aRb
and additionally, aRa for all a

[modest zealot]

but cant be equal to each other

bro

thats like looking left and right at the same time

[weary tiger]

i can do that

[azure lichen]

no, you’re misunderstanding

so the thing is

a relation is just a set of tuples, right?

[modest zealot]

ye]

[azure lichen]

like, aRb here just means that the tuple (a,b) is in the set R

[modest zealot]

yup

[azure lichen]

now, you want to just add more to R

so you keep all the ones in there

and then add whatever is necessary to make it reflexive

[modest zealot]

sooooooooo

a=/=b AAAAND a = a

boom easy?

[azure lichen]

OR

:P

[modest zealot]

wait wut

WAIT THATS THE ASNWER?!?!?!

[azure lichen]

well, it has to fulfill either of them

a≠b AND a=b is not fulfilled by anthing

[modest zealot]

all u can do is just slap

a= a

[azure lichen]

a=b

[modest zealot]

o

[azure lichen]

:P

[modest zealot]

o shit

u rite

o hfkffkfk

[azure lichen]

so, what is R now?

[modest zealot]

either its not equal or equal

has the original condition

but is also reflexive

[azure lichen]

yea but find a nice description of the relation

like, nicer than just “this or that”

[modest zealot]

wait was i not nice neugh

[weary tiger]

no

like what you said can be simplified further

[modest zealot]

do i have to pet it

wait

o shit

uhhh

{(a,b) | a =/= b or a = b}

there

[azure lichen]

that’s what you have right now

[modest zealot]

thats nice enugh

uh huh uh huh

[azure lichen]

that’s what you can make nicer though

[modest zealot]

how tf do u make it nicer!?!?

draw flowers n shit?

[weary tiger]

by simplifying it

[azure lichen]

well think about what it is

[weary tiger]

what are the possible choices for b?

[modest zealot]

waaaaaaiiittt

UNiverse?

cuz its either equal or not equal

[weary tiger]

and what is the universe here

[modest zealot]

set of integers.....

WAIT

JUST

WRITE
T

hE

WERID LOOKIN Z

[azure lichen]

well… almost

[weary tiger]

olmost

[modest zealot]

aww FUCK

[azure lichen]

getting there

[weary tiger]

well b is in Z as you said

what then are the choices for a

[modest zealot]

{(a,b) | Z{

ZxZ?

[weary tiger]

haha yes

[azure lichen]

there we go!

[modest zealot]

BOIZ

AAAYAYYy

[weary tiger]

ok.

[modest zealot]

bro u guys make math fun as hell

[azure lichen]

successes are fun

[weary tiger]

yeah but we lied
the real way you make it 'nicer' is by drawing flowers around it

[stray reef]

you might as well just write

[modest zealot]

there, thats some nice mfkin flowers

[weary tiger]

lol

[modest zealot]

boom done

[azure lichen]

those are leaves

[weary tiger]

clover

[modest zealot]

naw fok off

those r flowers

lfowers with aids

cuz they dont like math

idk

[azure lichen]

I can accept clovers

[stray reef]

might as well write "a R b no matter what a and b are"

or "everything is related to everything"

[modest zealot]

ZxZ

3 letters > however the fuck many letters is in "Everything is related to everything

[weary tiger]

lol

[opal crescent]

🌻 ZxZ🌻

[modest zealot]

^ this guy got it

😄

[azure lichen]

🥀 ℤ×ℤ 🥀 seems more like what they wrote tho

[weary tiger]

lol

[modest zealot]

10/10

just for clarification

wtf is this asking

[opal crescent]

🍀ZxZ🍀 BOOM

[azure lichen]

well, you should now know what the symmetric closure is

so you’re given a directed graph showing some relation

like the ones above

[modest zealot]

wuts directed graph for this relation

OH

So

all u do

[azure lichen]

and you wanna make it symmetric

[modest zealot]

is add an inverse thing right

if u got a->b

then u add b->a

[azure lichen]

ya

[modest zealot]

if u got nothing

dont do nothing

[azure lichen]

that’s it

[modest zealot]

or shits gonn ablow up

[weary tiger]

lol

[azure lichen]

(btw, while i’m sure they want the smallest one that has the property when they say closure, in one particular context, namely the algebraic closure, I don’t usually see that restriction)

[modest zealot]

o

so more like

as long as u satisfy the cnodition

dont gotta restrict urself

[azure lichen]

yea, but again, I think usually that restriction is a thing and then it’s usually unique too

but I think with algebraic closures you don’t have uniqueness so you might as well not bother with the “smallest”

[modest zealot]

ahhhhh gotcha

ok just for clarification

i answered all of them already

what happens if you have a a a

will that be true?

like if its reflexive, is it only true for two iterations?

[azure lichen]

(for reference, the algebraic closure is basically all the numbers so that every polynomial has a root)

[modest zealot]

x^2 + 1 = 0 hue hue

[azure lichen]

and yes, aaa is in there

and thus, the algebraic closure of ℤ contains i

[modest zealot]

so for reflexive shit, u cna have as many iterations to itsefl and it'll be true?

[azure lichen]

because x² + 1 has to have a root

[modest zealot]

icic

[azure lichen]

and yea I mean there’s no reason for you not to

[modest zealot]

oops discord bug

[opal crescent]

it's not a bug just anti spam ig

[modest zealot]

oh

a sequence of a a a a is tru then

[opal crescent]

wat

[azure lichen]

surprising

[modest zealot]

ur busted

[azure lichen]

actually

we’re probably exempt from spamfilters

aaaaaaaaaaaaaaaaaaaaaaaaaa

yes

[opal crescent]

yea that's what i thought

[azure lichen]

aight we should stop tho

:P

[opal crescent]

definitely

[modest zealot]

for the first part part a

a->c and c->a

do i leave those laone?

bRd and dRb

and it also seems like i cant do shit for part a

cuz it technically is transitive

[azure lichen]

what’s that algorithm you learned?

[modest zealot]

lmao idk, i was sleeping in class i think

[azure lichen]

it’s not transitive

you have (e,d) and (d,b)

[modest zealot]

lemme check

[azure lichen]

but no (e,b)

[modest zealot]

apparently its something like

first see if there r any of the transitive relations

like aRb and bRc

and then connect those first

aka first repeat

and then u generate new relations

and then using the original + new relations

draw more transitive shit

aka second repeat

[azure lichen]

okay, on infinite sets you might have to do that infinitely often :P

but on finite sets it’ll eventually terminate at least

[modest zealot]

yes

15a only new relation i got was (e,b)

[azure lichen]

ya I think that’s it

without spending too much time thinking

actually, (b,e) too, right?

or?

not sure

would have to think

don’t feel like doing so

[modest zealot]

naw dont think so

cuz (e,b) doesnt introduce new transitive

[azure lichen]

ye you right

[modest zealot]

damn this is annoying

this shit feels like ur trying to fix a bug in ur program

fix one, twenty pop up

[weary tiger]

lol

relatable

[modest zealot]

ok done iwth hmwrk, time to do this ridonculous challenge

https://cdn.discordapp.com/attachments/496785905474994186/561993611340808210/183668144404037632.png

[weary tiger]

lol

[modest zealot]

i=0 {(a,b) | a+1=b}
i=1 {(a,b) | a>b}
i=2 {(a,b) | a+b=1}
i=3 {(a,b) | a*b=1}
i=4 {(a,b) | ???}
i=5 {(a,b) | a>=b}
i=6 {(a,b) | ???}
i=7 {(a,b) | a=b}

this is the best i can do

[azure lichen]

okay let’s see

0 is correct

[modest zealot]

yaas

[azure lichen]

1 is correct

[modest zealot]

yaaaaaaas

[azure lichen]

2 is correct I think

3 is obviuosly not correct

[modest zealot]

boi im nailing this

o shit

yea

lmao idk

[azure lichen]

4 is wrong

wait

what does 4 even mean

[modest zealot]

idk

i just pulled it out of my ass

[azure lichen]

that would just be all of them

cause for any (a,b), a=a

so, wrong

[modest zealot]

ok thats clearly wrong

LOL

[azure lichen]

7 is correct

[modest zealot]

bro 50%

nailed it

[azure lichen]

aight but 3 was one I gave you initially so

[modest zealot]

that one is hard

[azure lichen]

and 5 is ez

[modest zealot]

lemme do 5 first

[azure lichen]

just take the reflexive closure of one you already have

[modest zealot]

damn this is tough

im starting to get confuzzled

[azure lichen]

5 has a standard example

in some sense

[modest zealot]

a >= b for

5

[azure lichen]

yes

[modest zealot]

ok now 3, 4, 6

left

which one shud i tackle?

[azure lichen]

do 3 first

for 4 there’s like, a really stupid way to do it

[modest zealot]

symmetric and transitive only

not reflexive, cant map to itself

lemme think

[azure lichen]

well, some numbers may

just not all

but yes

probably easiest to just make no number relate to itself

[modest zealot]

FUUUUUCK

[azure lichen]

sym, trans, not refl has a “one symbol” one too

[modest zealot]

a*b = 1????

[azure lichen]

…it’s not wrong

the only thing in it is (1,1)

[modest zealot]

wait wtf thats actually correct?

[azure lichen]

but it’s not wrong

[modest zealot]

jesus fuck

[azure lichen]

I would’ve gone with a≠b though

[modest zealot]

but not necessarily tho right?

a=/= b and b=/= c doesnt imply a =/=c

not always tru tho

[azure lichen]

…you’re right

my example wasn’t an example

[modest zealot]

LOL

hmmm

[azure lichen]

good thing to keep in mind too, that ≠ isn’t transitive

[modest zealot]

yes yes

ok now two more, reflexivie, not symmetrical and not transitive

[azure lichen]

yea the best one I can think of there is stupid af

[modest zealot]

ok im leaving 4 alone for now

[azure lichen]

wanna hear mine?

[modest zealot]

reflexive, symmetrical, but not transitive

ye sure

[azure lichen]

so you take all the (a,a)

to make it reflexive

and then just add like (1,2) to it

and nothing else

that breaks the symmetry

oh and it also has to be intransitive

[modest zealot]

that works oh my

[azure lichen]

so also add like, (2,3)

[modest zealot]

but its odd LOL

[azure lichen]

and there, done

[modest zealot]

its like u have everything that obeys a lot of rules

and then add one lemenet

that breaks everything

[azure lichen]

yea

[modest zealot]

icic

[azure lichen]

well, two elements

[modest zealot]

how about #6

reflexive, symmetric, but not transitive

a=b or a=-b ?

a = |b|

lol

idfk

[azure lichen]

-b is not in ℕ

[modest zealot]

well fok

ok i quit

too much brainpower

welp thanks a BUNCH for the help on discretem athematics

learned more in this session than i did in a week of class

appreciate the help

[azure lichen]

you did the thinking

that’s why :)

[modest zealot]

well not all of it

[azure lichen]

we just told you what to think

[modest zealot]

exaaaactly

aka not me thinking all of it

[azure lichen]

you don’t need 100% original ideas

getting guided is fine as long as you do the vital steps yourself

and learn from it how to get to them

[modest zealot]

ye ye agreed

[copper ore]

(3/𝑛)+11 is O(𝑛)

how do i do this?

"Determine whether or not the following are true"

[loud copper]

O pi

[weary tiger]

@copper ore definition of function being O(another function)

[tranquil iris]

Could someone guide me to resources on this type of propositional logic? Not sure if I'm googling the right thing as all I get are the basics of doing up truth tables etc.

[weary tiger]

how would i go about soving this using the contradiction method?

if a is an integer even number and b is an odd integer number, then their product, ab is even

[topaz parcel]

"7 is a prime number if and only if all dogs are black"

T⇔F = T right?

wait is this the real table? i think our prof taught wrong

[sonic narwhal]

@weary tiger well let a=2n and b=2n+1

Multiply them

And you should get a product of 2n therefore the number is even

[stray reef]

T <=> F = T

[weary tiger]

what do you do when you have a system of congruences and the gcd between 2 of them is not 1?

[analog sonnet]

Just use the lcm

[cyan socket]

I have a test coming up and I can not find any practice problems to help me study. Does anyone know where I can find similar problems like this?

[copper ore]

so i have to show if n^3 is big omega(n^3) or not

[weary tiger]

use the definition

[copper ore]

omega sorry

[weary tiger]

oh

use the definition

[copper ore]

lol

yeah

but

am i allowed to add terms in f(x)?

f(x) = n^3 btw

so like can i do n^3 + x >= x^3

@weary tiger

[weary tiger]

what

you are trying to find a number n_0 and a constant k > 0 such that for all n > n_0,
n³ >= k * n³

so what do you want to choose for k, first of all

[copper ore]

idk

what should i choose?

1?

[weary tiger]

yeah that works

[copper ore]

if choose 10

itd be n^3 is not big omega?

[weary tiger]

no

just because you chose the wrong constant doesn't mean there is no constant that will satisfy the inequality

[copper ore]

oh

so 1<=n<=9?

[acoustic valve]

How many ternary sequences of digits chosen from {0, 1, 2} of length twelve have exactly three 1's and two 0's?

I can't seem to figure it out.

I thought it would be 12 choose 3 * 10 choose 2, but it wasn't, apparently.

any ideas?

[stray reef]

(12 choose 3) * (9 choose 2).

[acoustic valve]

wait, why?

[stray reef]

...okay wait what was your reasoning behind your answer?

[acoustic valve]

so you have

length 12, right?

so 12 things to pick

bc its ternary

u have 3 things to pick from

so 12 choose 3 for 3 ones, then from the remaining

OH

OHHHHH.

thanks fam

[stray reef]

yeah, there's nine symbols left to pick from rather than ten 😛

[acoustic valve]

how do i buy someone cake

over the internet

🍰 i'll give you this emoji

digital marie antoinette

thank you a million

i was having so much trouble with that problem im sleepy its late

[iron basin]

Hey friends! Quick question about Theoretical CS:

if L := {a, b, c}
is L^2 then {(ab),(ac),(ba),(bc),(ca),(ac),(aa),(bb),(cc)}?

as in, words with length 2 that you can make with that language?

[stray reef]

you forgot (cc)

[weary tiger]

i mean it's not an infinite set

[stray reef]

and also that

[iron basin]

I know, I was just too lazy to continue

sorry

[stray reef]

abbreviating one element into an ellipsis

[iron basin]

ouh

I didnt check

damn it was only one element missing

HOWEVER

lemme fix i

there u go

is that right though?

Then L^3 would just be words with 3 letters with that language

and is the empty word in those ?

[stray reef]

well, does the empty word have length 2? or length 3?

[iron basin]

I don't know, since it could be whitespace 3 times but then whitespace probably would have to be in the language?

I'm in my first week so it all is very new to me

[stray reef]

no, the empty word definitionally has length zero

[iron basin]

okay, so the empty word would only be in
L^0?

[stray reef]

it is a special object unto itself

if you want to define L^0 as consisting of just the empty word, go ahead

nobody is stopping you

[iron basin]

oh okay!

So L^n (n€N) would be what if I had to write down the words?
Does that mean any length or infinitly long?

I think it's the first one, makes more sense

[weary tiger]

that means the set of all strings of length n

[iron basin]

so it would be {(a), (ab), (aba), (ababa) ... (cacacacbabac) ... } ?

literally everything that is any combination of the letters in the language?

of any length

[weary tiger]

of length n

[iron basin]

oh

oops

I did a fuckety

[weary tiger]

all strings would be like

$\bigcup_{n \in \bN} L^n$

[iron basin]

nah I get it, my brain went poof for a second there, thanks D:

[weary tiger]

:D

[iron basin]

What is L*?
I know Sigma* is all words of any length but I don't really know what the * means and google isnt much help

[weary tiger]

try kleene star

[iron basin]

oh nice thanks!

How many words can I build from a language?
I've done some testing and saw that with L={a,b}
you can build 2^n words for L^n

But it doesn't seem to check out if L = {a,b,c}

ye I did that

[weary tiger]

oh

[iron basin]

either I totally miscalculated the words in L^3 for L= {a,b,c}

[weary tiger]

what did you get

[iron basin]

21

[weary tiger]

3³ is 27

[iron basin]

L {(aaa),(aab)(aba)(baa)(abb)(bab)(bba)(bbb)(aac)(aca)(caa)(acc)(cac)(cca)(ccc)(bbc)(bcb)(cbb)(bcc)(cbc)(ccb)}

those are 21

[weary tiger]

ew ew ur listing them out

[iron basin]

sorry

:

[weary tiger]

lol

where's (abc) etc

[iron basin]

oh.

welp.

what is love

[weary tiger]

baby don't hurt me

[iron basin]

dont hurt me

[weary tiger]

btw

are you on homework help

[iron basin]

no

I'm in university

[weary tiger]

i mean

the discord server

[iron basin]

oh uh

no

[weary tiger]

ah ok

lol

[iron basin]

I am in the coding den, this one and [glitched]

guess my bachelor

[weary tiger]

cs

[glossy adder]

english

[iron basin]

both are actually correct

[weary tiger]

lmao

[glossy adder]

as planned

[iron basin]

I studied english / math before this

[weary tiger]

why did you quit math

[iron basin]

I cant do math

[weary tiger]

D:

[iron basin]

:)

nah it was way too much workload for me to handle

and no fun

[glossy adder]

[iron basin]

:

[weary tiger]

idk, being shown a trivial counterexample to your claims can be fun

[iron basin]

I enjoy math, I don't enjoy having 20 new pages of material every week

[weary tiger]

hahahahaha

[iron basin]

and despite putting like 3 hours of work into math every day having the backlog getting bigger n bigger

no fun

[weary tiger]

those are rookie numbers

[iron basin]

remember I had english as well

so that was 5 hours of work I put in every day + 5 hours of university / day (average)

that is a 55 hour work week + I had a part time job to keep my appartment

which adds up to a 70 hour work week

no fun

[weary tiger]

mmmm

[iron basin]

So L* means all words that can be formed with the alphabet of the language, including the empty word?

which means that if
L0 = {a,b} and L1 = {11, 22}
then the intersection of L0* and L1* would be the empty word and not { } ?

[stray reef]

yes, the intersection of L0* and L1* is { "" }

[iron basin]

nice, thanks

[pastel temple]

any help with this question?

P(n) is a simplified version of the pattern from S(n). I notice that for increment of n+1 , the two values from the multiplication in the denominator is increased by 2

[quaint river]

looks like partial fractions

then you get telescoping series?

What are you even supposed to find

[iron basin]

I dont understand the question

honestly

[quaint river]

like does it want you to simplify or something

[iron basin]

^

[quaint river]

or find the value for specific N

or find a partial sums expression using telescoping series

[pastel temple]

Do you see a generarl pattern for the simplified value of S(n) ? Write your guess to complete the proposition P(n) below.

[quaint river]

Then do partial fractions and it becomes a telescoping series

[slow socket]

i have the value set {0,1,2} and P(X=0) = P(X=1) = P(Y=0) = P(Y=1) = 1/4
and = P(X=2) = P(Y=2) = 1/2

and have to find P(X=0 or Y=2)

do i just add 1/4 + 1/2?

[stray reef]

are X and Y known to be independent?

[slow socket]

Yes

[stray reef]

then no

[slow socket]

Wat do i do

[stray reef]

i mean there's at least two ways you could do this

one would be to write P(A or B) as 1 - P(not A)P(not B)

the other would be to make a table of all possible outcomes

[blissful imp]

im in highschool calc and im having trouble understanding how 3 coplanar normals relate to 3 planes intersecting at a line, the lesson was the intersection of 3 planes

(sorry if this is the wrong channel)

[slow socket]

@blissful imp #calculus

[ashen magnet]

whadup.
Say i had a row with 13 slots and and i had 4 objects that can be assigned to slots.
2 of them can be in slots 1 - 6 and 2 of them can be in slots 7 - 13.
I already calculated the number of possible ways the left side can be arranged as well as the right side. to get the total ways these 4 objects can be arranged,
do i multiply or add their number of possibilities.

I feel like multiply would make more sense because for each position on the left side, the right side can be in one of its many positions

[stray reef]

that is exactly why you multiply.

@ashen magnet

[acoustic valve]

[twilit wadi]

show that the formula Q V (P∧¬Q)V(¬P∧¬Q) is tatutology

can anybody tell how to solve this problem

[stray reef]

what tools do you have available?

[twilit wadi]

@stray reef do you mean by tools

[stray reef]

did you accidentally a "what"?

anyway, i'm asking for some context. what class is this for?

what methods do you have?

[wispy gull]

What rules have you learned (ie De Morgan's)

[stray reef]

like, depending on the class, a simple application of a bunch of Boolean algebra laws may or may not be appropriate

[twilit wadi]

double negation

associative law

distributive law

[wispy gull]

So you have equivalence and inference laws

[twilit wadi]

I am using p ---> q equivalent to ¬P V Q

[wispy gull]

If you can slim it down to conditional sets you can use conditional proof and reductio ad absurdum to find your answer

[stray reef]

distributive law, great

okay that is enough context for me

[twilit wadi]

can you explain what he has done here

[stray reef]

he = who?

[twilit wadi]

it is written in the book

Q V (P∧¬Q)V(¬P∧¬Q) = Q V ((P V ¬P) ∧ ¬Q)

[stray reef]

that's an application of the distributive law.

$Q \lor [(P \land \neg Q) \lor (\neg P \land \neg Q)] \ = Q \lor [(P \lor \neg P) \land Q]$

[vital dew]

Ann:

[twilit wadi]

@stray reef thanks

[modest zealot]

question, how am i supposed to show transitivity of matrices

partial orders = antisymmetric, transitive, and reflexive

or i should reword, how tf do i figure out if a matrix is transitive

[slow socket]

@modest zealot look at the first matrix, u can see is 3x3

[modest zealot]

ye

[slow socket]

u can see row 1, column 1 is 1 right?

think of it like this

3 rows, 3 columns

[modest zealot]

oh

wait

i get it now

its the thing

where its like

(1,2) and (2,3) implies (1,3) is in there

GOTCHA

[slow socket]

idk wat u mean

but look'

row 1, column 1 in the first matrix

what number is there

or another of saying, what number is at 1 - 1

[modest zealot]

i think i got it figured out

sry LOL

[slow socket]

k

[weary tiger]

whats the difference between a circuit and eularean circuit

[verbal furnace]

It has to do with edges. Do you use multiple times the same edges? Do you use all edges?

[twilit wadi]

obtain PDNF of P-->Q

=(¬P∧T) V(T∧Q)
=(¬P∧(QV¬Q)) V((PV¬P)∧Q)
=((¬P∧Q)V(¬P∧¬Q)) V((PV∧Q)V(¬P∧Q))
=((¬P∧Q)V(¬P∧¬Q)) V(PV∧Q)```

can any one tell how they got the last line

[stray reef]

PDNF?

also, PV∧Q?

[weary tiger]

principal disjunctive normal form, probably

[twilit wadi]

sorry

the correct one is

=(¬P∧T) V(T∧Q)
=(¬P∧(QV¬Q)) V((PV¬P)∧Q)
=((¬P∧Q)V(¬P∧¬Q)) V((P∧Q)V(¬P∧Q))
=((¬P∧Q)V(¬P∧¬Q)) V(P∧Q)```

how they got the last line

[stray reef]

you have (¬P∧Q) occurring twice in a disjunction

(¬P∧Q) v (¬P∧¬Q) v (P∧Q) v (¬P∧Q)

[twilit wadi]

so

it is made into one
right

[stray reef]

idempotence

[twilit wadi]

super

thanks

[wild harness]

Hi!

“The relation R ⊆ Z x Z given by (a,b) ∈ R ↔ 3|b-a is symmetric, transitive and reflexive”

Basically have to prove all three separate

[stray reef]

and what's holding you up?

[wild harness]

i don’t know how to start it period

my textbook doesn’t have examples with a divisibility for it

[stray reef]

do you know the definitions of "symmetric", "transitive" and "reflexive"?

and of "divides"?

[wild harness]

yeah i have them

[stray reef]

you can't not know how to start if you know the definitions

[wild harness]

well we never did a example on it yet

not good with proofs at all

[stray reef]

have you ever proved any of those three properties for a relation before

[wild harness]

no, only wrote the definition lol

[stray reef]

okay, so like

let's do reflexivity. its definition is the simplest.

we want to show that R is reflexive

this means that we want to show that for every a ∈ Z, we have (a, a) ∈ R.

do you follow so far?

[wild harness]

yeah

[stray reef]

according to the definition of R, (a,a) ∈ R means 3 | a-a.

i.e. (a,a) ∈ R iff 3 | 0.

and 3 does divide 0, since 0 = 3 * 0 and so 0 is a multiple of 3.

[wild harness]

ok makes sense for that

Is that all for reflexive?

or there’s more to it

[stray reef]

yes, that is it

[wild harness]

now for symmetric and transitive, how does it go?

cause for transitive there’s a a, b and c

[weary tiger]

try to prove symmetric first

i.e, if 3 | (a-b), then 3 | (b-a)

[stray reef]

it's all a matter of unpacking the definitions, really...

[wild harness]

so how would you wrap it up after the i.e, if 3 | (a-b), then 3 | (b-a)

[weary tiger]

first you prove that 3 | (a-b) implies 3 | (b-a)
then you have basically proved that (a,b) in R => (b,a) in R

so you are done

[wild harness]

hate to ask so much, but how do you prove that 3|a-b implies 3|b-a

[weary tiger]

b - a = -(a-b)

[stray reef]

if a-b is a multiple of 3, then by definition there exists an integer k such that a-b = 3k

[wild harness]

how about transitive

that’s my last question sorry

[weary tiger]

now you are trying to prove 3 | (a-b) and 3 | (b-c) => 3 | (a-c)

the trick is to use the fact:
if k | m and k | n, then k | (mx + ny) for integers m,n

[wild harness]

how would you apply those integers to it

[weary tiger]

well

you are trying to go from (a-b) and (b-c) to (a-c)

the most natural choice for m and n would be 1 for both

[wild harness]

what would k be? 3?

[weary tiger]

yes

[wild harness]

would I write out that fact first and then apply it?

cause if I’m using 1 for both integers, it would be written out as 3|1 and 3|1

[weary tiger]

oh sorry

i meant

if k | m and k | n, then k | (mx + ny) for integers x,y

and the most natural choice for x and y would be 1 for both

so in this case, it would result in:

3 | (a-b) and 3 | (b-c) => 3 | ( (a-b) + (b-c) )

now simplify the parentheses

[modest zealot]

how does b have a degree of 6

[neon pasture]

1 + 1 + 1 + 1 + 2 = 6

[modest zealot]

i count 5

[cyan talon]

Isn't the nod on b considered as two because it's not oriented

Like taking it both ways

?

@modest zealot

[modest zealot]

oh undirected graphs imply two way

or are degrees just the number of lines?

[cyan talon]

So it's 6

[modest zealot]

this logic feels weird ngl

[cyan talon]

I mean yeh you can consider too the arrow from b to c and from c to b

This will count as two

But it may also count as one because you are interested in one way only

Idk

[modest zealot]

asme

same

no fucking clue

[cyan talon]

@modest zealot

[modest zealot]

oHHH

[modest zealot]

anyone got any clue wtf this is asking

[sour arrow]

For which values of n are these graphs bipartite?
(a) Kn
(b) Cn
(c) Wn
(d) Qn

Also is the 8th question, and is worth 20 points.

[modest zealot]

our notes have nothing about bipartite and what Kn, Cn, Wn, and Qn are

:/

[sour arrow]

"bipartite" means you can colour the nodes two different colors, such that no two nodes of the same color are neighbors

[modest zealot]

oooohhhhh ic ic

wut is Kn, Cn, Wn, and Qn, are those special types of graphs or something?

[sour arrow]

I don't remember them by heart. But google will know!

,w Kn graph

[vital dew]

Display results online and refine query

Requested by Kaynex#7520

[sour arrow]

Uh huh.

[modest zealot]

uhhhh

5 euros?

[sour arrow]

Kn is the graph with n nodes, such that every single edge is between them

"The complete graph"

I can already see that K3 isn't bipartite and it gets worse from there

[modest zealot]

so Kn is the graph where every single node is connected with every single node

is there anywhere where i can search wtf these graphs r? its so annoying none of the notes or lectures explain this

:/

[sour arrow]

Google?

I just punched "Kn graph" in and found it

[modest zealot]

ok yea im doing that

but i can do it for other things like C and W and Q

ok found C

ok just to make sure

wheel graphs can NEVER be bipartitie

nvm

last question, is there a trick to draw a 2-tuple hasse diagram

part c

[echo lintel]

https://gyazo.com/749e21b692938a36585aa0975826cffa

im not understandning how part 2 turned into part 3

[modest zealot]

line 2 to line 3?

[echo lintel]

yes

[modest zealot]

basic algebra manipulation

[echo lintel]

yeah im kinda of missing what's happening

[modest zealot]

do u understand the first term?

the ( (k(k+1))/2 )^2?

[echo lintel]

in line 1

?

[modest zealot]

line 2

ur having problems from line 2 to line 3 right?

[echo lintel]

there's not /2 in line 2

OH

wait

line 3 to line 4

sorry

https://gyazo.com/11e6beae9a83a4b2d9de64f7a7832b1a

[modest zealot]

oh line 3 to line 4

what they did was factor out the (k+1)^2 term

thats about it

[echo lintel]

ohh i see it now

that brings the k^2 on the left side to the right inside the parentheses

that was what was confusing me

[modest zealot]

u okay now?

[echo lintel]

i believe so

and then they distribute 4 to k and +1

[modest zealot]

and then factor ye ye u got the idea

[echo lintel]

thanks ^_^

@modest zealot sorry t o ping but i have one more question

https://gyazo.com/fb264927f6f950b4ee5be63c4291b19b

[modest zealot]

ye

[echo lintel]

why did they make the common dom 4 instead of 2

[modest zealot]

so you can combine the fractions

[echo lintel]

why not 2 tho

[modest zealot]

its 2^2, not 2

in line 2

[echo lintel]

OH

ahh

[modest zealot]

ye ye

[echo lintel]

yeah u right thanks im tired hahaha

[modest zealot]

happens to all of us, np

[echo lintel]

https://cdn.discordapp.com/attachments/238956921581862913/564282176007569416/8745c40e63a23c83bd4e1a9770c1c758.png

is 3 correct?

[sour arrow]

Nop. That's strong induction

[echo lintel]

ahh damn

i got wrong

https://gyazo.com/e3f995e59af9d064f9a10a5763873969

@sour arrow does this seem right?

for proving the basecase

[stray reef]

g_0

7^(0+2)

and you'll need two base cases

[echo lintel]

ahhh fuck too late

i only did0

[heavy dove]

Hey guys

this is a big question, but this one questions has 6+ subquestions which is why it might look a little big

please, if i could get some help with this, im unsure if this is a function

Muncipality lillestrand has eight locations. Four of them lays on the islands "Storøy" and "lilleøy". The islands has none bridgeconnectons but there is ferryroutes from Oddeneset-Lillegrend-Storøyhavn-oddeneset.
Map over the municipality.
http://prntscr.com/n8n86i
We can consider the road network as a relation between locations. that means road⊆location×location where
location= {Lillegrend, Yttervika, Berg, Storøyhavn, Oddeneset, Skipperhavn, Solvik, Dal}

Specifically we let the relation be represented like this:
Road = {(Skipperhavn, Dal), (Skipperhavn, Solvik), (Skipperhavn, Oddeneset), (Dal, Solvik), (Yttervika, Berg), (Yttervika, Storøyhavn)}

Moreover, the ferry route gives rise to a similar relationship Ferry⊆location×location. Specifically:
Ferry = {(Oddeneset, Lillegrend), (Lillegrend, Storøyhavn), (Storøyhavn, Oddeneset)
What is Road◦Ferry

if anyone can explain before giving answers so i understand it would be very nice :)

[sudden knot]

check the definition of a function

[heavy dove]

<@&286206848099549185> Please help:)

i just need help understanding what they mean

[weary tiger]

hey guys I have a little question regarding product notation in the context of Cartesian products

as shown here

https://wikimedia.org/api/rest_v1/media/math/render/svg/0813abf3bbeca6c3f8343424d52e4d054acfe5e5

https://en.wikipedia.org/wiki/Function_(mathematics)

it relates to the Cartesian product paragraph on this wiki page

could somebody please explain said notation to me because I'm not quite sure I understand it properly

is this just building a set from every possible combinations of X and Y ?

[sour arrow]

@weary tiger
Still looking for it?

[weary tiger]

yup!

[sour arrow]

@weary tiger
First one has to talk about relations in general. Let's say we have two sets
A = {a, b, c}
B = {1, 2, 3}

And a relation between them:
R = {(a, 3), (b, 1)}

That set encodes the arrows of the relation. "There's an arrow from a to 3, and an arrow from b to 1"

[weary tiger]

yup, I'm familiar with relations

[sour arrow]

Perfect! Notice that R is a subset of A×B and will always have to be

So we define a relation that way

[weary tiger]

yup!

[sour arrow]

A function is just a special type of relation, so all functions from A to B will also be subsets of A×B

OH you're just not sure of A×B, is that it?

A×B is indeed just every combination of A paired with B

[echo lintel]

https://gyazo.com/7b7e773129e8fc122779f23e11492e3b

where does the negative in 21 - go?

[sour arrow]

-(a - b) = -a + b

Remember to distribute your negatives

@echo lintel

[echo lintel]

ah thank you

i was looking too deep haha

[stray reef]

what do you mean by "both x and y"

the objects being compared here are ordered pairs

not individual real numbers

[slow socket]

i have P(A) =0.3 , P(B) = 0.4, P(notA) = 0.7, P(AB) = 0.1
how do i find P(notA and B)

[worthy thistle]

So how would I compare ordered pairs; because for reflexivity isn’t < for anything already not reflexive

[twilit wadi]

can anyone explain this problem

inference

[stray reef]

what is "rule P" and what is "rule T"

[twilit wadi]

even I do not know

[twilit wadi]

rules of inference

[quaint river]

Since P is true, P->Q implies Q is true by modus ponens, and since Q is true, Q->R implies R is true from modus ponens again

Rule P is just your starting premises which are given

[twilit wadi]

thanks

[wary salmon]

hey guys

can i get some help on part c of this problem

[weary tiger]

@stray reef senpai pls

[stray reef]

well what are you having trouble with ig

[weary tiger]

i have to determine whether the structures build semigroups, monoids, groups

,$ (\bbQ, o) x\ o\ y = x + 2y

[vital dew]

yami:

[weary tiger]

how to understando

[weary tiger]

@stray reef ok i got this but im in actual trouble now XD

[stray reef]

i mean it's all just a matter of checking the axioms

no?

[weary tiger]

except i dont know about this one

let A be a set. $ X = A^A = {f : A \rightarrow A} $ the set of all functions of $ f : A \rightarrow A $ with the opration $ f \ o \ g $ (sequential execution of functions)

[vital dew]

yami:

[weary tiger]

i read something about this being a monoid

ye why not

function composition is associative

and ofc there is a function f : A -> A such that f(a) = a for all a in A

but its not commutative, is it?