discrete-math

[errant bear]

part b

i thiught the probs were listed 1) 2) 3) not a) b) c)

so you are done w a)
now do b)

[weary tiger]

2.5

[errant bear]

what is the ordered pair

because the intersection is a set, so your final answer needs to be in the form {(x1, y1), (x2, y2) etc.}

[weary tiger]

there is 2 ordered

pairs

[errant bear]

what are they

[weary tiger]

(2.5, - 2.5), (2.5, 7.5)

[errant bear]

no

is 3 times 2.5 = -2.5?

[weary tiger]

7.5

[errant bear]

so why do you have (2.5, -2.5) as an element of the intersection

those elements in the intersection have to satisfy both properties

[weary tiger]

x - 5 = y

[errant bear]

and it has to satisfy y=3x

[weary tiger]

not sure what you mean

[errant bear]

you literally have the intersection of two linear lines

[weary tiger]

yes

[errant bear]

there can only be 1 point of int as is has to satisfy both requirements

[weary tiger]

2 lines and the formula for geting the intersection is make to make the two formulas equal to each other

[errant bear]

omg

yeah, however you always have to plug the x value you get into the equation being substitued in

listen, id strongly recommend doing a quick sketch on some paper of the graphs

[weary tiger]

(-2.5 , -7.5)

[errant bear]

ok

now do c

[weary tiger]

this is everything not in B or C so "not S" is everything in B or C

[errant bear]

yess

which is

[weary tiger]

is there an equation for this?

[errant bear]

you have 2 actually

[weary tiger]

yes

I have 3x

[errant bear]

cough B and C

[weary tiger]

and x - y =5

[errant bear]

its just asking you to define it

[weary tiger]

so do I just write {(x, y) | 3x and x - y = 5}

this seems wrong to me

I know it is both equation tho

[errant bear]

its or

[weary tiger]

since A U B is both set A and B

yes or and would be the middle only

[errant bear]

or

[weary tiger]

could I write it like this though

not S = B U C

[errant bear]

i mean yes, if thats S bar

yes

[rough shard]

How do I compare the cardinality of countable and uncountable sets?

For example, rational numbers between 1 and 2 and natural numbers

Is the cardinality of natural numbers less than that of all rationals?

[stray reef]

no. Q is countable.

[rough shard]

Oh i meant real numbers

sorry

[stray reef]

R is not countable. |R| > |N|.

countable < uncountable generally

[rough shard]

Ok perfect thank you

Is there something I can google to find a proof of that

Or a textbook article or something

[stray reef]

uh

any set theory book should have that

[rough shard]

Ok i'll look around in my textbook

thank you

thats not continuum is it?

No bc thats dealing with the powerset of naturals

[stray reef]

actually no |R| is continuum by definition

[lethal sequoia]

Hi guys, I have a question that I have almost solved but I know it’s wrong. Can you help me?

Question 3

[lethal sequoia]

This is what I did

By definition: $G(x) = \sum_{k \geq 0} a_{k}x^{k}$

[vital dew]

Mac:

[lethal sequoia]

$ = a_0 + a_1 + \sum_{k \geq 2} (a_{k-1}+6a_{k-2})x^{k}
\
= 5x + \sum_{k \geq 2} a_{k-1}x^{k} + \sum_{k \geq 2}6a_{k-2}x^{k}
\
= 5x + \sum_{k \geq 1} a_{k} x^{k+1} + \sum_{k \geq 0}6a_{k}x^{k+2}
\
= 5x + x \sum_{k \geq 1} a_{k} x^{k} + 6x \sum_{k \geq 0}a_{k}x^{k}
\
= 5x + x \sum_{k \geq 0} a_{k} x^{k} - xa_{0}x^{0} + 6x \sum_{k \geq 0}a_{k}x^{k}
\
G(x) = 5x + x G(x) + 6x^2 G(x)
\
-5x = 6x^2 G(x) + x G(x) - G(x)
\
-5x = G(x) (6x^2 +x-1)
\
G(x) = \frac{-5x}{6x^2 + x - 1}
$

[vital dew]

Mac:

[lethal sequoia]

Then I use partial fractions to find the formula

$ 6x^2 + x - 1 = 0
\
G(x) = \frac{-5x}{(x- \frac{1}{3})(x+ \frac{1}{2})} = \frac{A}{x- \frac{1}{3}} + \frac{B}{x+ \frac{1}{2}}
\
-5 = A(x+ \frac{1}{2}) + B(x= \frac{1}{3})
\
A = -2,B=-3
\
\frac{-2}{x- \frac{1}{3}} + \frac{-3}{x+ \frac{1}{2}}
\
\frac{-6}{x - 1} + \frac{-6}{x+ 1}
\
6(\frac{1}{1-x} + \frac{-1}{x -(-1)})
\
G(x) = 6 \sum_{k \geq 0} (-1 + 1) x^k
\
a_k = 6(-1+1)^k
$

[vital dew]

Mac:

[lethal sequoia]

The last few parts I'm sure I didn't do it right but not sure how to manipulate it properly

[random basin]

Hello everyone, I have an exercise I need help with.

Fermat's little theorem:

I don't see how to get from Fermat's Little Theorem to the congruence asked in the question.

[analog sonnet]

What have you tried?

[random basin]

One second, I will show you!

This is the whole exercise:

These are my answers to 1b):

I still don't see the connection between a) and b), although b) was easy to solve.

[analog sonnet]

,w 4^14 mod 15

[vital dew]

Display results online and refine query

Requested by Rijinaru#5601

[analog sonnet]

Yeah I don't really see the connection either

But what did you try for a?

[random basin]

Nothing yet, I thought I could prove the general case if I solved the examples.

[analog sonnet]

Also, you realize that 15 is not a prime number, right? :p

You were lucky that 4^14 is one more than a multiple of 15

In any case

I suggest that you try proving the converse "<=" first

[random basin]

Hmm you're right, totally overlooked that. Thanks for pointing it out.

[analog sonnet]

🦈

[random basin]

I think I got it:

@analog sonnet thanks for the help

[cerulean ore]

it looks wrong '='

Right to left, first place can be filled by 7 digits only, then 7, then 6....

[stray reef]

yeah this solution is fucked up actually

[cerulean ore]

Good, thank you!

C and R occupy end places means _ _ _ _ _ _ _ _ C and _ _ _ _ _ _ _ R or _ _ _ _ _ _ _C R and __......RC?

[stray reef]

no

it's C_________R or R_________C

[cerulean ore]

fml

That was a rookie mistake

[errant bear]

the number of possible paths from the origin to any fixed point in the first quadrant in R^2, if you are allowed to move up or down, but strictly right, is aleph naught, right

we did only up and right in class, and the answer was x+y choose x or y, just wondering about if you arent restricted

[faint narwhal]

yes

[errant bear]

ok makes sense

same for freedom in both horizontal and vertical, as it would still be the (infinitely) countable union of (infinitely) countable sets/families

?

[cerulean ore]

Permutation and Combination has never been intuitive for me

[cerulean ore]

Why we are multiplying 6! with 5!? 5 software engineers has to be seated in 6 seats.

Isn't that 6C5?

[stray reef]

looks like a typo

is this the same book as the previous problem

bc if so i wouldn't trust it

anyway the 5! is (6-1)!

[cerulean ore]

Same book

(6-1)! * 6C5? right?

[stray reef]

no

there are 6 software engineers, not 5.

wait

no

nvm

[cerulean ore]

5 SE and 6 EE

[stray reef]

hang on though

this screws up their entire reasoning

bc their fig 5.2 shows a table with 12 seats

as if there's 6 of each type of engineer

[cerulean ore]

[stray reef]

yeah their solution to this problem is just fucked up

as was the last one

the book's trash

though i guess that wasn't exactly shocking

[cerulean ore]

Lol, if you remember the books I usually use are trash.

[stray reef]

uhh

ok alright maybe this can be saved somehow

so since there's 11 engineers to be seated

[cerulean ore]

11?

[stray reef]

5 software, 6 electric?

[cerulean ore]

oh 6+5 (if you read trash, your mind becomes trash)

[stray reef]

and there's 5 software engineers who can't sit next to one another
the gaps between them in any seating arrangement will be of sizes 1, 1, 1, 1 and 2

so

ok lemme get a pic rq

they're always gonna sit like this

does this make sense

[cerulean ore]

Agreed!

Thank you so much for spending time^

[stray reef]

yeah so at this point it's kinda obvious that the orientation of the table is fixed and the number of ways the 11 engineers could sit is just 6!5!

coincidentally, arriving at the same answer as the book, but without the wonky reasoning!

[errant bear]

o

[cerulean ore]

How 6!5!?
Is it like to make electronic engineers sit we have (6-1)! ways, to sit 5 software engineers when available seats is 6, we have 6C5 * 5! = 6!?

So, total ways is 5!*6!

[stray reef]

no??

the electronic engineers can be seated in 6! ways

and the software engineers in 5! ways

[cerulean ore]

round table? (n-1)!

[stray reef]

no! the orientation is fixed!

you can't apply rotational symmetry anymore!

[cerulean ore]

Which book helped you to get better at P&C?

[stray reef]

none in particular ¯_(ツ)_/¯

idk

[cerulean ore]

Genius then

[stray reef]

eh

no

it was probably a collection of different resources that even i couldn't replicate if i tried

[cerulean ore]

Unfortunately, fucked up university hinders a lot of your learning.

[true crag]

So I thought of an interesting argument, but I can't think of a way to formalize it.

If you take a complete graph on n vertices, that's the net of an n-1 simplex

From that perspective, it makes a little bit of intuitive sense that k5 can't be drawn without crossing

It ostensibly means that there's no way to orient a 4-simplex such that all vertices are visible, looking from the top down

Can anybody think of a way to formalize this?

[analog sonnet]

You’d have to define visibility in 4D space

And orientation, and the notion of “top-down”

[true crag]

Maybe think of it like a spot light, shining on the object?
If we think 3 dimensionally, I can orient a tetrahedron so the casted shadow has all 4 points visible

So we don't need topdown persay

We can define orientation by any rotation of the object

[analog sonnet]

The cube is a planar graph

Yet we can’t see all vertices at the same time in the shape of a Platonic solid

[true crag]

As long as it isn't solid we can

(Wire frame)

The argument would ostensibly be about the projection of a 4-simplex into 2d space

[clever nacelle]

protocol, create a regular expression that finds ESTABLISHED and LISTENING
network connections on TCP port 502.```

Does anyone know how to approach this

[weary tiger]

can anyone help me on this question

[reef thistle]

@weary tiger Can you use the definition? Show your work.

[weary tiger]

I kind of understand it but I dont understand it fully

its not super neccasary though if you are busy I can just ask my TA

if you ar busy

[cerulean ore]

@reef thistle Did you recommend me a book on graph theory?

[reef thistle]

I don't remember if I did

[plucky wasp]

What is the number of sequences of six digits where the number of even digits is equal to the number of odd digits?

how to solve this???

[stray reef]

if there's as many even digits as odd digits in this sequence, and there's 6 digits in total, how many are there of each

[plucky wasp]

3

[stray reef]

ok

so there are three even digits and three odd digits

[plucky wasp]

sorry i think in there 5 even , and 5 odd

[stray reef]

in your string

[plucky wasp]

and there should be only 3 odd/even in two places like that i think

[stray reef]

your string is composed of three evens and three odds, potentially repeating since there's no specification that they don't repeat

now before we answer this

[stray reef]

it might make sense to ask yourself this simpler question

how many ways are there to arrange 3 E's and 3 O's in a row

[plucky wasp]

but what about the other two odds and evens left

[stray reef]

what do you mean, "the other two odds and evens"

you seem to be under the impression that the digits in your string can't repeat

when that is not the case at all

[plucky wasp]

like there should be 6digit number which should have 3o ,3e so the other 2e/2o we should use them right ?

[stray reef]

no?????????????????????

you're missing the point??????????????

are you even reading what i'm saying?????????????????

[plucky wasp]

no they can repeat because the digit sequence can occur

again

i mean with position change

[stray reef]

what if the question was asking for the number of strings of length 20 instead of length 6

then there would have to be 10 each of evens as odds

and in your words

what would we do with the -5 that remain

[plucky wasp]

what if for like 23 ? not 20 ,30

[stray reef]

23 isn't even, you can't have equal numbers of even and odd digits in a string of length 23

anyway you're really overthinking all this

[plucky wasp]

oh ok

We first select three positions for odd digits. For each of these three positions, we select one of five odd digits. For each of the remaining three positions, we select one of the five even digits. Overall, this gives 6c3 * 5^3 * 5^3

[stray reef]

well there you have it

[cerulean ore]

@reef thistle Can you please recommend?

[reef thistle]

idk I don't have a graph theory textbook

that I use

[weary tiger]

Hey i am new to proofing and Discrete Maths in general, i am a bit unsure if my proof is "Correct"and i would appreciate it if one of you could take a look 🙂

[stray reef]

fourth line fucked up

[weary tiger]

Doesn't the \mid sign mean that the second number can be described by the first number times some other number q?

[stray reef]

that isn't the issue here

you're using q for two different things here

[weary tiger]

so exists q,r in N would fix it?

[stray reef]

with proper placement of quantifiers, yes, there is a chance that the proof might be fixed

but just one question

are you explicitly required to write the proof in this clunky formal manner

[weary tiger]

I am required to proof the statement a => b by showing not b => not a

[stray reef]

that is not what i am asking you

are you required to shit quantifiers all over the fucking page

or are you allowed to use words like a human being

[weary tiger]

I thought it would be normal in a proof 😄

Thank you for helping 🙂

[toxic fjord]

Recursively define the set of bit strings that have more zeros than ones.

how do i do this..

[obtuse lance]

not entirely sure what you're being asked to do, but I'd probably imagine I have a set of bit strings of length n (and less), then try to describe how to construct the strings of length n+1

[toxic fjord]

yea i think that's what it means while fulfilling the condition "have more 0 than 1"

[obtuse lance]

I guess there are two routes, take all the length n strings, and then put a 0 on the end of them, and then the other route is to make the "maximal" type ones which we don't get from this way

kind of hand wavy, might need to be fleshed out a bit more than that, like where you add 0 to the left or right might make duplicates and when you add "maximal" ones it might depend on if the string is even or odd, I'm not thinking too much here but hopefully that helps

[toxic fjord]

btw this is the answer i got, can't really understand what it means

[obtuse lance]

interesting, I see what they're doing

try to describe what you do understand or what you're thinking about it

[toxic fjord]

at basis we starting with a bit string with length 1, so obviously string "0" it is.
in recursive step,
that's where i stuck

does it mean that s and t is just "0"?

[obtuse lance]

yeah when you first start all you have is "0"

so s="0" and t="0" so you can make st = "00"

now S has "0" and "00" in it

so you keep building in this way

recursively

[toxic fjord]

aah

now i see it

[obtuse lance]

you could also do 1st with s=0 and t=0 to make 100

yup

[toxic fjord]

damn big thanks for clearing this up

[obtuse lance]

yeah cool trick they have, I think the way I was doing it was not so elegant but I was more planning on creating a recursive formula so I could count them in mind so I don' tfeel too bad haha

[slender skiff]

I'm trying to construct a phrase-structure grammar based on some sets, but how do I know if the start state S can transition to the empty string?

[toxic fjord]

anyone know how to solve this?
i believe it is a popular question though

[slender skiff]

@toxic fjord Pretty sure its the pigeonhole principle you have to look into

[toxic fjord]

yea it is a pigeonhole principle problem

just kinda lost with the explanation

[runic mauve]

can someone help me with this problem?

Suppose p and p + 2 are both prime numbers with p > 3, show that p(p+2)+1 is divisible by 36

i tried to do it with induction but i don't think thats the pathway

if someone could just give me a little nudge in the right direction that would be all i need

[faint narwhal]

You want to show that p(p+2) is -1 mod 4 and -1 mod 9

[runic mauve]

could i say it's -1(mod2) and -1(mod 3)?

instead of that i mean

[faint narwhal]

No

Because that's not enough to conclude that p(p+2) + 1 is divisible by 36

[runic mauve]

yeah i figured that after i sent it

thanks

@faint narwhal would i use the chinese remainder theorem for this btw?

[faint narwhal]

No you don't need it

[runic mauve]

hmm

oh ok im getting it more now

actually this is still kinda confusing me, is there some elementary way to work around mods that I haven't learned or smth? I'm trying to make an argument mostly with english but i'm having trouble coming up with stuff

[faint narwhal]

Nope, you know everything you need to know here

Think about what p can be

mod 4

[runic mauve]

well it has to be a multiple of 3 but not of 6

[faint narwhal]

uh what?

p is a multiple of 3?

[runic mauve]

or no thats not what i meant lmao

i mean it has to be one less than an even multiple of 3

[faint narwhal]

Why

[runic mauve]

because since they're both primes then they cant be even and for any given 3 consecutive numbers one of them must be a multiple of 3

which means one of p, p+1, and p+2 must be a multiple of 3

since p and p+2 are prime then it must be p+1

[faint narwhal]

Ah sure

Sure

[runic mauve]

and if p+1 was an odd multiple of 3 then p would be even

yea

[faint narwhal]

My original question was

What can p be mod 4

Using the same type of logic

[runic mauve]

well it can be any number one less than 4?

if you're only considering the mod 4 part of it

[faint narwhal]

And why?

[runic mauve]

well just because of the mod part

but idrk what to do with the prime aspect of that you know

like with the multiple of 3 it was pretty easy to say that p+1 has to be a multiple of 3 and 2 but -1(mod 4) makes it odd already

[faint narwhal]

No, you're not thinking in the right logical direction

You're trying to show that p(p+2) is -1 mod 4 and -1 mod 9

This is what you're trying to show, not what you know

[runic mauve]

ok

but im supposed to use the fact that it's prime at some point though right?

[faint narwhal]

Of course

so what can p be mod 4

[runic mauve]

what do you mean when you say what can it be mod 4?

like 0, 1, 2, 3?

[faint narwhal]

can p be 0 mod 4

[runic mauve]

no

nor can it be 2 mod 4

[faint narwhal]

So you have two cases

[runic mauve]

so i have to show that it cant be 1 mod 4 and then do the same thing with 9

[faint narwhal]

No

think again

about what you're trying to show

[runic mauve]

oh ok so i have to take the two cases where p is 1 mod 4 and where its 3 mod 4 and show that that number times p+2 is always -1 mod 4 and -1 mod 9?

[faint narwhal]

well, do the first one first

but yes

[runic mauve]

ok

[sonic ferry]

Can someone help me with this question

If my answer is correct

[faint narwhal]

this is

a terrible question

[sonic ferry]

@faint narwhal the question or my answer being incorrect

[faint narwhal]

the question is terrible

[sonic ferry]

ahhhh

[faint narwhal]

The relation they give

Isn't an equivalence relation

So it really makes no sense to talk about equivalence classes

Or at least, not the equivalence classes people usually talk about

[sonic ferry]

i tried looking around but all examples i see is with divides @faint narwhal

[sour arrow]

There's a reason for that lol

The question is broken. They're assuming a result that isn't true

[sonic ferry]

i dont see any similar question like this with inequality @sour arrow

D:

[faint narwhal]

It's just terrible

Talk to your teacher or professor

[sonic ferry]

yea ill ask her

thanks!

[faint narwhal]

Ask her why the relation is an equivalence relation

[sour arrow]

10N5 is true
But 5N10 is false

[faint narwhal]

I mean yeah, it's not symmetric either

[sour arrow]

We don't expect that to happen with an equivalence relation

Where's the question from lol?

[sonic ferry]

exam question @sour arrow

i got it wrong, im trying to fix it. to prepare for the finals

[sour arrow]

Whoa. They actually gave you a completely broken question for an exam. I'm so sorry

[sonic ferry]

yea this question deducted 5 points from me

yikes

[sour arrow]

Complain. And get a teacher who has at least seen set theory before

[faint narwhal]

I mean

is there a definition of equivalence class somewhere in your notes

maybe they use this term a bit differently than what's standard

[sonic ferry]

no we base it on the book

Discrete Mathematics with Applications 4th

edition.

ill check

[sour arrow]

Susanna Epp?

[sonic ferry]

yep

[sour arrow]

That definition is standard

[sonic ferry]

yea ill definitely ask tomorrow

Does this mean my answer shouldve been: T0 = {0,0,0}, T1 = {0,1,-2} and so on? @faint narwhal @sour arrow

Wouldnt index for this mean T0UT1UT2UT3?

[sour arrow]

Derp. Yes, that's what T0, T1, ... are

However, you're taking the union of 4 different sets

[sonic ferry]

Oh

I went over

[sour arrow]

That returns a set that has everything all four sets have

The answer should be
{0,1,2,3,-2,-4,-6}

[sonic ferry]

Hence why it says all together

Ahhh gotcha

[sour arrow]

U means "squash into one set"

[sonic ferry]

Is there a difference if the question was [0, i, -2i]? Instead of {} @sour arrow

[sour arrow]

That wouldn't make sense

{} denote sets

[sonic ferry]

ah okay, because the book had something like this which is what im practicing on. @sour arrow

[sour arrow]

Okay. [a,b] means the set of all real numbers between a and b

a and b are in the set as well.

They define it with the set builder notation to the left

[sonic ferry]

oh okay i see

[quaint pike]

I need some quick help regarding congruence relations

I have the problem:

[8] + [6] in Z base12

And the answer sheet says [2]

I'm wondering what steps I need to take to achieve the answer [2]

[ivory badge]

what's 8 + 6?

[quaint pike]

14

14 is 2 higher than 12

so the answer is 2

[ivory badge]

yes

[quaint pike]

Got it thanks

[ivory badge]

but like
what do you mean 22

[quaint pike]

accient

[ivory badge]

the answer is 2

[quaint pike]

accident

[ivory badge]

aight ye

mod 12 means you do things based on how far they are from 12

[quaint pike]

yeah

i understand now

thx

okay

what about multiplication

[5] times [12] in Z base 8

5 times 12 is 60

the answer sheet says [4]

[ivory badge]

what's 60 mod 8

[quaint pike]

uhm

[ivory badge]

60 = 7*8 + 4

[quaint pike]

60 / 8 = 7.5

that .5 represents 4

since 4 is half of eight, 4 is the remainder

i see

[ivory badge]

aight

[quaint pike]

and what about [9]^7

to the power of 7 in Z base 7

[ivory badge]

Well, that's gonna end up being 9 iirc

but uh
what's 9^7

[quaint pike]

it is 4782969

[sour arrow]

Are you asking about 9^7 (mod 7)?

[ivory badge]

it's gonna be 9 because of a fermat theorem

which is 2 btw since [9]=[2]

[quaint pike]

let me check

[runic mauve]

i've got a similar problem in #help-6 if 1 of u guys knows about that one, no rush tho sry for interrupting

[runic mauve]

Hey i'm posting the same problem in here again because the guy who helped me got stuck but i figured i would try again if someone could figure it out? I'm just gonna skip the problem for now:

"find the last 3 digits of 17^2019"

I've gotten to 17^19 = 17^19(mod 1000) but idk how to get past that

my professor did something where he used phi(phi(phi(1000))) (or maybe more phi's after that idr) to find the answer when it gets small enough to not reduce but also still not be computable and somehow was able to set that up to work but i've forgotten the method exactly

[weary tiger]

What is 1 + 1

Our teacher didnt tell us that

they say its a mystery of the world

[stray reef]

you're not being serious are you

[weary tiger]

jk

[woven olive]

🤨

[runic mauve]

i guess ill just leave that problem blank lmao

i dont think i actually need it to keep 100% on homework so no worries

[weary tiger]

heheh

[woven olive]

Honestly a pretty interesting problem. If/when you get a solution set, would you mind posting it?

[faint narwhal]

@runic mauve I woke up if your homeworks not due yet

[runic mauve]

It’s not but maybe I can @ you tomorrow? It’s 1:43 am here lmao

[faint narwhal]

Sure

[runic mauve]

It’s due Thursday so Wednesday is all good for me

[weary tiger]

@runic mauve you essentially did it

Using Eulers, we get 17^19 mod 1000

Then you can just break it up

Agh fuck this is bigger than i thought ahahaha

(17^3)^3 * (17^3)^3 * 17

(913)^3 * 913^3 * 17

11^6 x 17 x 83^6 mod 1000

So powers of 11

[faint narwhal]

There are easier ways

[hardy yoke]

nCr = !n / r! * (n-r)!. 52 playing cards. What's the probability of picking out 3 cards and have at least one spade? I've gotten to : (13C1 * 51C2 + 13C2 * 50C1 + 13C3)/52C3 but the probability is pretty high here - 93ish%. And logically should be somewhere under 75%.

[analog sonnet]

I think it's easier to calculate the complement probability: how likely is it that you draw 3 cards, but none of them are spades?

Once you have that, subtract it from 1 to get your desired outcome

[hardy yoke]

1-(39/52x39/52x39/52) that would get me at 57,8125‬%

[analog sonnet]

Looks right

Wait

No

First of all, those two numbers aren't even equal

Second of all, the expression you wrote down doesn't represent the probability that you have to calculate

[hardy yoke]

39/52 is probability of drawing none of the 13 spades.

[analog sonnet]

Only on your first draw

Remember that when you draw your second card, the probability will change

[hardy yoke]

That would create multiple possibilities

I can't just do 39/51

Or can I?

[analog sonnet]

Why 39/51?

[hardy yoke]

One less card in the deck

I'm not sure about that.

[analog sonnet]

Let's take a step back

How many ways are there to pick 3 cards out of the entire deck?

[hardy yoke]

(13C1 * 51C2 + 13C2 * 50C1 + 13C3)/52C3

So 52C3

22100 ways

[analog sonnet]

Ok, so far so good

How many ways are there to pick 3 cards out of the deck without the spades?

[hardy yoke]

39C3

[analog sonnet]

Awesome

So there you have it

[hardy yoke]

Why would this not work?(13C1 * 51C2 + 13C2 * 50C1 + 13C3)/52C3 Creating all the possibilities of (1spade 2 random cards + 2spade 1 random + 3 spades) / all possibilities

[analog sonnet]

The probability that you draw 3 cards, and none of them are spades, equals (39 choose 3)/(52 choose 3)

[weary tiger]

@faint narwhal oh with CRT

[analog sonnet]

@hardy yoke You're over-counting

For example, that 51 choose 2 includes every card except in the deck except the one you just picked

In particular, it contains the spades

[hardy yoke]

Ah, right

[analog sonnet]

So with your method, the correct calculation would have been something along the lines of (13C1 * 39C2 + 13C2 * 39C1 + 13C3)/52C3

[hardy yoke]

Yes, probably, thanks for the help.

[analog sonnet]

🦈

np

[runic mauve]

@faint narwhal yeah I think that I GOT the answer but it’s just that I’m fairly certain he doesn’t want us to just go calculating that huge number

Which is essentially what I did

I’m interested to see what you would do

[faint narwhal]

Think about chinese remainder theorem

[runic mauve]

Would i break it up and use Chinese?

[faint narwhal]

Yes

[runic mauve]

Is the answer you got 153?

Or did u not find one

[faint narwhal]

I didn't try it

I just know this will work

[runic mauve]

Ok

[weary tiger]

Its still not a nice solution

Even with Chinese

Noting 17 = 3*5 + 2

[coarse gyro]

How to prove?

[faint narwhal]

are you sure they're true?

[coarse gyro]

Pretty sure

[faint narwhal]

And why?

[bleak vector]

If you have an injection from set A to B does all of A get mapped to B?

I always thought the whole domain of A didnt havent o be used to still be injective

Just that the points were 1 to 1

[last sigil]

https://math.stackexchange.com/questions/2345629/in-an-injection-from-a-to-b-do-all-elements-from-a-have-to-be-used

[coarse gyro]

I made up some sets and tested it @faint narwhal

[faint narwhal]

Okay well how do you show two sets are equal

What does it mean for two sets to be equal

[errant bear]

subset has entered the chat

[rapid herald]

hey guys, im doing an internal examination high school paper on Dijkstra's algorithm, im not very aware of the notations and only really spend time udnerstanding the algorithm

so could someone have a look at my paper to see if everything checks in with the notations and stuff discrete math has?

ty

[sonic ferry]

is it an equivalence relation if every test is not symmetric, not transitive, and not reflexive?

[cerulean ore]

Can a multigraph have a loop?

My teacher has written in the definition of Multigraph(General graph) that
A graph which ahve either loop or parallel edge or both the edge is called general or multigraph.

[stray reef]

is that verbatim what your teacher wrote

what is it with the proliferation of broken english i've been seeing lately

[cerulean ore]

,rotate

[cerulean ore]

from my teacher's notes:

Teacher seems to be wrong ;-;

[stray reef]

hhhhhhh

this is all just meaningless musings

[cerulean ore]

[cerulean sentinel]

Find the number of sequences, containing $a_1$ ones, $a_2$ twos, $\ldots$ , $a_n$ numbers $n$, such that
for each number 2 there is a 1 to the left of it, $\ldots$ , for each $n$ there is an $(n-1)$ to the left of it.

[vital dew]

bilyan:

bilyan:

[cerulean sentinel]

<@&286206848099549185>

[obtuse minnow]

hmm I am going to have a go at it, but did you come up with anything ... partial discoveries?

@cerulean sentinel

[cerulean sentinel]

We've come up with a sort of algorithm to build them (not sure if it's even correct), but no idea about a formula

Another way would be to get all numbers from them and order them ascendingly (while keeping the condition satisfied)

[obtuse minnow]

hmm let's do 112233 as an example

123123 works!

actually 12XXXX always works which is nice.

[cerulean sentinel]

Yup

[obtuse minnow]

112XXX also always works.

1122XX also always works but has only one choice for the right side: 112233

oops

double counted

[cerulean sentinel]

Yeah 😄

[obtuse minnow]

Are there other patterns than 12XXXX and 112XXX?

for 112233

[cerulean sentinel]

No

I've written two programs, in python and cpp, the python one tries to count them on input but misses some as of right now, actually both

Currently I'm using a way of 'bubbling' each first occurence of n until it reaches the first occurence of n-1

[obtuse minnow]

so this is a very interesting problem but it's 8 AM where I am! Sleep calls 🙂

[cerulean sentinel]

No prob, thanks for taking a look

[obtuse minnow]

sure is this due within 24 hours?

[cerulean sentinel]

less than that 😄

we've been at it for a few days now

[obtuse minnow]

doh okay. If I don't make it back, good luck!

But I do think there is a lot to be said for this (just a thought I wanted to test but haven't):

All numbers must have:
1[]2[]3[] where neither 2 nor 3 cannot be to the left of the first 2, and 3 cannot be to the left of the first 3.

[cerulean sentinel]

Yeah, that's kinda what I'm doing rn

and I'm summing the product of the permutations of each []

https://gist.github.com/bilyanhadzhi/c49848ccd25b611e5bafd2a431b5a46c

If someone wants to run this and fix it cause it's not counting some rn

[obtuse minnow]

for 112233 ... 1[]2[]3[] --> 1[1*]2[2*]3[3*] where N* denotes the number of digits that can be there.

I dunno it's very half-baked.

1[*,1]2[*,1,2]3[*,1,2,3] would list the choices ... yeah that's all i got!

okay yeah very very half-baked. Anyway good luck.

thank you for sharing the problem though.

[vagrant goblet]

Hey I have a question about recurrence relations, can anyone help?

[rigid oriole]

If you ask it, perhaps.

Otherwise, no.

[rapid herald]

hey, can somebody help me with a visual proof of the dijkstra's algorithm, like why it works

[weary tiger]

For 21, what even is

K, C, Q

What do those even stand for

What are they

[rigid oriole]

Types of graph I believe

rest - try google

K graph
C graph
Q graph

[worthy grotto]

K_n is the complete graph

C_n is the graph with a cycle

[rigid oriole]

oops

[worthy grotto]

K_m,n is the bipartite graph

i have no idea what Q_n is

[weary tiger]

Hmm gotcha ty

[indigo spear]

small question about paths in graph theory, only walks have alternating pattern of vertices and edges or no

[pale epoch]

there are very little standard definitions in graph theory

[sonic ferry]

I just wanna say thank you to this server. I have passed my discrete math class

[faint narwhal]

grats

[little pike]

https://cdn.discordapp.com/attachments/268882317391429632/652284637858496522/unknown.png

[obsidian tendon]

well, ok

we are told to use induction

right

[little pike]

yes

[obsidian tendon]

so in problems like this we should always start with the base case

especially because observing behavior of a structure in its simplest form

can often make it easier to think about more complicated cases

so lets say n = 1

this means that we have a function

$f \colon A \mapsto B$

[vital dew]

hegel:

[obsidian tendon]

there is only 1 element in A, lets say a

and B has m elements

how many distinct functions

can there be?

[little pike]

m?

[obsidian tendon]

right!

and m^1 = m

so our base case checks out, right?

[little pike]

yep

[obsidian tendon]

so now lets say we have a function f

with n elements

in A

and m in B

so by induction lets assume that there are m^n distinct functions from A to B

what happens when you add 1 more element to A?

since every element in A has to map to an element in B

how many elements in B can this new element in A map to?

[little pike]

just from the one element?

m

[obsidian tendon]

right

so we have n+1 elements in A

we assumed by induction that there were m^n functions distinct functions mapping n of our elements to m elements in B

and now for each of those functions, there are an associated m possibilities

for which the last element can map to

so how many functions do we have in total?

[little pike]

m^(n+1)

[obsidian tendon]

right

so now we've proved that if we have $f \colon A \mapsto B$ with $|A| = n $ and $|B| = m$

[vital dew]

hegel:

[obsidian tendon]

and if we assume that there are $m^n$ distinct functions from A to B, if we add 1 element to A and thus have $|A| = n + 1$, there are $m^{n+1}$ functions from A to B

[vital dew]

hegel:

[obsidian tendon]

i.e

by induction on n

our proposition is true

its true for n = 1, and if it is true for some n, its also true for n + 1

[little pike]

oh, so its mainly the wording

[obsidian tendon]

ya

dont get caught up in thinking "wow i cant figure out how this could work" if ur told to use induction

the problem is basically telling u what to do

you just gotta trust it for a little bit

and prove the base case

and that n implies n + 1 is true

[little pike]

I see. So for a concise answer, what key parts do you need to include for proving for it for n.

[obsidian tendon]

do you want me to write up a proof?

[little pike]

No no just the key parts you need to mention

[obsidian tendon]

well, obviously you need to prove the base case

[little pike]

ye

[obsidian tendon]

then you'd need to demonstrate what happens when you add 1 extra element to A

[little pike]

ye thats were I was sort of confused about this because you can just get away with writing the anwser without doing some mathematical working to prove how you got there

[obsidian tendon]

i.e show that if you have n elements in A, for each function f from A to B, there is now an extra m functions associated with it

and thus for each f you have m functions, with m^n f in the set of functions from A to B and thus m + ... + m m^n times, i.e m^n * m = m^(n+1)

and then ur done

[little pike]

Ok I see, thanks a lot for your help!

[errant bear]

when proving "if n^3 is odd, then n is odd" how would you do this without contradiction

with contradiction its like 2 lines but i feel like u should be able to directly do it

[faint narwhal]

You probably can, it's probably not nice

[errant bear]

hmm

[faint narwhal]

The contrapositive proof is probably what you want or are talking about

There's really no reason to use contradiction here

[errant bear]

oh yeah, contrapositive you dont use p at all until the end right

[faint narwhal]

uh whats p

[errant bear]

n cubed is odd

[faint narwhal]

p \implies q is equivalent to (not q) implies (not p)

That second statement is the contrapositive

[errant bear]

hmm

so saying: let n be even, then showing n cubed is even is the contrapositive

[faint narwhal]

Yes

And that of course, is straightforward

[errant bear]

yeah, so i wouldnt put the contradiction sign in this "proof"

?

[faint narwhal]

Not really

You would just say, we will prove the contrapositive

[errant bear]

i see

that makes sense, since they are logically eq

got it

[little pike]

when stating a range, how do you say natural numbers greater than 2

[errant bear]

n in N st n>2?

[little pike]

ah nvm

ye

[little pike]

im confused about each of these parts. how do you combine f and g and how do you get the inverse of f if it isnt in terms of x

[errant bear]

what does g dot f mean

[obsidian tendon]

composition

[errant bear]

im asking him

like does he understand it

[obsidian tendon]

o

lol

[little pike]

dw I solved it

[errant bear]

...ok

[versed granite]

hey guys. i gotta proof following 2 tasks.

i tried to solve this via using the definition of x|y. which means that xn=y and xn=z -> x*n =y+z

but i am not sure, how to continue

[stray reef]

no

you used the same letter (n) for three different things

[versed granite]

n1 n2 n3

[stray reef]

ok, so what you have is that $y = n_1x$ and $z = n_2x$ for some $n_1, n_2 \in \bZ$

[vital dew]

Ann:

[stray reef]

and what you want is to show that y+z can be written as an integer times x

$y + z = n_1x + n_2x = ; ??$

[vital dew]

Ann:

[versed granite]

but with your first tep i know that this is ture isnt it?

[stray reef]

what?

[versed granite]

step 1: y=n1x and z=n2x then i know that y+z =n1x+n2x : They left side is only true , if both terms are true

[stray reef]

what's your point

[versed granite]

if we assume that y=n1x and z=n2x is true

the right part is true

[stray reef]

of course we're assuming that. that's our premise!

what's the issue here?!

[versed granite]

sry i was standing on a water tube 😄

thx for your help

[stray reef]

you were what

[versed granite]

something we say in germany if you dont know the answer despite the fact the answer is obvious 😉

[sleek swallow]

That'd be pretty useful for me lol

[rapid herald]

hey guys, i cant seem to understand what the heuristic value means in the A* search, can anyone explain>

please dumb the explanation down a bit for me ty

[hardy yoke]

P(A|B) = P(A) is for proving if events are independent of each other. If i have dice 1-6 and P(A) = "Roll 1", P(B) = "Roll even". P(A|B) = ((A and B) / B) = 0/3 . P(A) = 3/6. 0/3 != 3/6 so that means events are dependent...?

according to that formula?

but they're not

[stray reef]

P(A) = "Roll 1", P(B) = "Roll even".

no

P(A) is a number, as is P(B)

writing P(A) = "Roll 1" is incorrect

did you mean A = "Roll 1"?

but also, yes, "roll 1" and "roll even" are dependent

even though you calculated P(A) incorrectly as 1/2 when it should be 1/6

@hardy yoke

[hardy yoke]

Yes

@stray reef If I have to roll on first try 1-6 and on second try 1 then they're independent

How can i prove that with P(A|B) = P(A)?

[stray reef]

ok wait

what are your A and B now

[hardy yoke]

A = On first try roll 1-6

[stray reef]

and are you doing two dice now

[hardy yoke]

B = On second try roll 1

One dice

[stray reef]

well no

your experiment ought to consist of two tries

[hardy yoke]

I can use one dice though...

[stray reef]

_<

your sample space will consist of 36 pairs of numbers from 1 to 6

[hardy yoke]

Yes

[stray reef]

your A includes them all

P(A) = 1

[hardy yoke]

Yes

[stray reef]

your B includes six of them... P(B) = 1/6

and AB = A

so like

[hardy yoke]

AB = A?

[stray reef]

P(A|B) = P(AB)/P(B) = P(B)/P(B) = 1 = P(A)

yes typo sorry

[hardy yoke]

That would be 1 = 1/6

Or this is wrong way to do this, hmm.. probably wrong way

[stray reef]

no!

P(A) = 1 !

P(A|B) = 1 !

[hardy yoke]

P(A) = 1/6 chance

[stray reef]

no!!!!!!!

[hardy yoke]

What

[stray reef]

A = On first try roll 1-6

this is your message

[hardy yoke]

Ah, right.

[stray reef]

A happens ALWAYS

[hardy yoke]

Why did i want to compare it to B..

Lets say A = " roll 1st dice 1" and B = "roll 2nd dice 2".

P(A|B) now equals to 0

0 = 1/6 according to P(A|B) = P(A)

[rigid oriole]

uh

i dont have context

but are you just rolling 2 dice

because p a given b is 1/6 not 0

[hardy yoke]

Yes, i'd like to prove if they're independent or depended using formula

[rigid oriole]

ok and how is p a given b 0?

[hardy yoke]

P(A|B) = (A and B)/B

[rigid oriole]

yes

and also how is p(a) 1

[hardy yoke]

1/6*

[rigid oriole]

ok

so you have that formula for p a given b

how do you get 0 from that

I can roll a 1 on one dice and 2 on the other

[hardy yoke]

Well, A and B have 0 common outcomes.

Set A = {1} set B = {2} no?

[rigid oriole]

uh

think about it in real life

youre saying its impossible to roll 2 dice

and have one land on a 1 and the other a 2?

[hardy yoke]

I know that's possible, but I need to use this formula

Or some kind of proof

Using numbers

[rigid oriole]

what exactly is your formula for P(A and B) happening?

[hardy yoke]

Ah

Wait

Hmm, it'll be 36 total

1/36 chance

[stray reef]

Lets say A = " roll 1st dice 1" and B = "roll 2nd dice 2".(edited)
P(A|B) now equals to 0
0 = 1/6 according to P(A|B) = P(A)

no

P(A|B) is 1/6

P(AB) = 1/36

P(A) = 1/6

(1/36)/(1/6) is not 0

[rigid oriole]

also not 2 in 36

the dice are labelled

[hardy yoke]

it's 1/36 okay

How is AB = 1/36 😇 ?

[stray reef]

the event AB is "roll 1 on the first die and 2 on the second"

[hardy yoke]

One outcome from 36 possibilities?

[stray reef]

which describes precisely one of your 36 equiprobable possibilities

[rigid oriole]

please write and

[stray reef]

my probability prof uses $AB$ for $A \cap B$ and i've adopted this convention

[vital dew]

Ann:

[rigid oriole]

or is that common notation?

Feels like it could get confusing

with distributions you have like XY

meaning multiply the result

[stray reef]

yes but in something like P(AB) one udnerstands A and B to be events

you can also use commas for and

[rigid oriole]

yeah I guess...

[stray reef]

like P(X = 4, Y < 7)

[rigid oriole]

anyways

[hardy yoke]

Okay, I think i got this 🤷

[stray reef]

sounds like you didn't

[hardy yoke]

I just had this (AB) confused

Okay, just to confirm now. Single dice 6 sides 1-6. A="3" , B = "1". ```
P(A|B) = 2
P(A = 1/6)
2 != 1/6 => not independent

[stray reef]

aaaaaaaaaaaaaaaaa

your notation

is

fucking abhorrent

to the point where it's actually incomprehensible

[hardy yoke]

There, fixed notation, i think

[stray reef]

ok first off

P(A|B) = 2

this makes no sense

absolutely no sense

[hardy yoke]

A and B have 2 common outcomes

[stray reef]

whatsoever

[hardy yoke]

And B has only 1

So.. 2/1 = 2

[stray reef]

blink blink

how is it not a red flag for you that a PROBABILITY ended up being GREATER THAN 1

[hardy yoke]

Hmm...

[stray reef]

how is it not a red flag for you that a PROBABILITY ended up being GREATER THAN 1

[rigid oriole]

🙄

[hardy yoke]

Okay, right..

[stray reef]

what does A = "3", B = "1" even MEAN

[hardy yoke]

I just assumed it would be understood as rolling 3 or 1

[stray reef]

does it mean A = "roll 3", B = "roll 1"

then their intersection is EMPTY

hJEHLQRTGJHJKHGAKJHG

ARSGHKADFG

SADGAGSAG

NO

YOU

DO

NOT

ASSUME

THAT

INTERSECTION

IS

THE

SAME

AS

UNION

BECAUSE

THAT'S

NEVER

FUCKING

TRUE

[hardy yoke]

Okay

[stray reef]

it is IMPOSSIBLE to roll 3 AND 1 at once

the intersection is EMPTY

P(AB) is ZERO

"roll 3" and "roll 1" are NOT independent

[sour arrow]

AB for intersection is beautiful

It should have been how set theory was done

[rigid oriole]

And this wont be confused with cartesian product?

[stray reef]

no it won't

[rigid oriole]

If you say so.

[stray reef]

cartesian product is always denoted with ×

[rigid oriole]

R^2

[sleek swallow]

I mean, I see no problem with the symbol for intersections. It's pretty clear as it is.

[rigid oriole]

Anyways

what exactly was the issue?

I couldve sworn the original problem involved rolling 2 dice

oh nvm there was another problem

[hardy yoke]

Okay A = "Rolled number is even" B = "Rolled number is 4". ```
P(A|B) = 1/1
P(A) = 3/6

Is this right?

[rigid oriole]

looking better

well

hm

[stray reef]

yes

[rigid oriole]

I dont see the point in writing 1/1

as opposed to either

1

or

(1/6)/(1/6)

[ornate geyser]

Hi! So I have a problem and I just need to know what direction to take when solving it, because it feels we've tried everything at this point.

So, we have n^4 + 4 = p, where n is a natural numbers that is not divisible by 5 and p is a prime number. We need to solve this equation. Any pointers are welcome. Cheers

[rigid oriole]

oh dear.

What solutions have you got so far

[ornate geyser]

n = 1 where p = 5 :p

But brute forcing wasn't an acceptable method

[rigid oriole]

of course not

always good to look for some

Have you tried taking the equation in various bases

You're expecting for there to be no other solutions, right?

[ornate geyser]

Yes, we concluded that n >= 1, but given that it's a natural number to begin with, that didn't really get us very far

Kind of, yes

[rigid oriole]

n greater than 1??

[ornate geyser]

or equal to

[rigid oriole]

sure u cant get any other info

i mean bases as in modulo arithmetic

eg. I can see the prime has to be 1 mod 4

[obtuse lance]

prime also has to be 1 mod 5 by flt

so p = 1+20k for some k

[rigid oriole]

i dont think so

[obtuse lance]

you're right scratch that last line about the 20

[rigid oriole]

However

[obtuse lance]

but the 1 mod 5 part is correct, idk how I butchered chinese remainder theorem so hard lol

[rigid oriole]

taking it mod 5 gives you good info

im pretty sure

[ornate geyser]

I guess we'll start there and see what happens

[obtuse lance]

it tells you the prime is 1 mod 5

[rigid oriole]

@obtuse lance um how

[obtuse lance]

by fermat's little theorem

[rigid oriole]

i cant remember it but surely not

take n to be 1 mod 5

[obtuse lance]

oh sorry 0\

wow I'm high

[rigid oriole]

well that did the q

[opal crescent]

it tells you p-4 is 1 mod 5 yeah

[obtuse lance]

$n^4+4 \equiv p \mod 5 $
$n^{5-1} - 1 \equiv p \mod 5$
$ 0 \equiv p \mod 5$

[vital dew]

Merosity:

[rigid oriole]

oh wow

[opal crescent]

double dollars

[obtuse lance]

jeeze what's it need

meh figures

well you can read it fine

[opal crescent]

no

[rigid oriole]

lmao i was brute forcing mod 5 in my head

[opal crescent]

$$n^4+4 \equiv p \mod 5 $$
$$n^{5-1} - 1 \equiv p \mod 5$$
$$ 0 \equiv p \mod 5$$

[vital dew]

emeric75:

[obtuse lance]

$n^4+4=5$

[vital dew]

Merosity:

[obtuse lance]

n=1 is the only solution

[ornate geyser]

Oh, wait. Yeah, that makes a lot of sense

Thanks a lot!

[south snow]

i is also a solution

[pastel wolf]

Let $${A_i}_{i\in I}$$ denote a partition of A. Define a relation R on A as $$\forall x, y \in A, (x, y) \in R \iff \exists i\in I (x, y \in A_i)$$ Prove that R is an equivalence relation.

[vital dew]

SpiderString:

[pastel wolf]

Can someone check my work with this?

For some reason I temporarily forgot that "there exists" is a backwards E and instead wrote an E so if you see that please forgive me. XD

I thought that looked weird

[pastel wolf]

<@&286206848099549185> can some check over my work for this discrete problem?

[weary tiger]

can anyone explain to me why this is true

[sleek swallow]

Well, let’s see

Let x be an element of the empty set.

Then, the empty set is a subset of the set containing the empty set iff all elements of the empty set also belong to the set containing the empty set.

In other words, the statement above A is:

A <=> [(x E phi) => ( x E {phi} )

Since there are no elements in phi, the antecedent is false and so, the conditional is true. Since the biconditional is true, it implies that A must be true as well.

^Of course, make sure you’re writing all this down very formally, yea?

But it’s the general idea

[sleek swallow]

Huh. If you wrote P({phi}), where P is supposed to represent the power set, wouldn’t you get { phi, {phi}}?

Besides, it can be proved that phi is a subset of any set, no?

Unless I’m getting this terribly wrong.

[stray reef]

please don't use "phi" for the empty set

[vital dew]

Ann:

[sleek swallow]

I mean, it doesn’t really matter. But yea, sure.

[pastel wolf]

Honestly I think I had a stroke, I was just working through why it made intuitive sense and then I decided to throw all that out in favor of some bs argument. I blame (even worse) sleep deprivation due to dead week.

[sleek swallow]

Huh wot

Are you referring to the argument you gave earlier?

[pastel wolf]

Yes, just ignore me.

[sleek swallow]

Ah don't be too hard on yourself. Making wrong arguments and realizing the mistake in them is a natural part of learning

[pastel wolf]

Also doesn't it follow, then, that ø={ø} since both contain the exact same elements

Which is what I was originally thinking but then I decided that was too reasonable, I guess.

[sleek swallow]

Uh, no?

The set containing the empty set is not the same as the set containing the empty set. The former has a cardinality of 0 and the latter has a cardinality of 1.

The empty set is empty cos there are no elements within it. The set containing the empty set has one element (the empty set).

[pastel wolf]

Aren't all sets subsets of themselves?

Ugh actually yeah I'm gonna just stop before I say something even stupider

[sleek swallow]

Yes. A given set is a subset of itself because x E A => x E A, for any set A, is a tautology.

However, that just having one set be a subset of another set isn't enough to guarantee equality.

In order for a set A to be equal to the set B, for all elements belong to A, x E A <=> x E B. That's the Axiom of Extension.

[pastel wolf]

Idk, doesn't make sense to me that you can say x is a subset of {x}

[sleek swallow]

Then, you can prove that that's basically equivalent to mutual containment.

[pastel wolf]

Because x is not a set, but just some element of a set

[sleek swallow]

Huh, of course not. The set of subsets of {x} is given by {phi, {x}}

[pastel wolf]

Then why can you say ø is a subset of {ø}

Whenever that's effectively the same argument