discrete-math

[vital dew]

Deus_Vult

[prisma breach]

im going over some notes about cardinalities
now a theorem states that for some sets, A,B , |A| = |B| iff there exists an equivalence function f B^A

that would imply there exists a bijective function between the reals, and all subsets of all naturals
how would I go about implementing such a function ?
could I simply map any real number to some subset containing all of the real numbers decimals?
say, 3.00001 = > {30000,1}?
or 3.253 => {3,2,53}?

[pale epoch]

how is this a map?

[prisma breach]

from real numbers to subsets?

[pale epoch]

how is it well defined

i don't see how it is supposed to work

[prisma breach]

I... admittedly haven't thought this through

[pale epoch]

you also run into more problems

because decimal expansions of real numbers are not unique

[prisma breach]

unique in what way ?

[weary tiger]

0.99999.... = 1

[pale epoch]

like 1.000000 = 0.99999

i mean thats not that big of a problem

the bigger problem is how you decided that 3.253 => {3,2,53} and 3.00001 = > {30000,1}

you cant just take the decimal expansion and place commas arbitrarily to turn it into subsets of N

[weary tiger]

or like what would 0.9999... map to at all

[pale epoch]

i mean you can, but that doesnt show surjective

ah ye, thats also a problem

if you want to work towards a map that works, consider binary expansions of real numbers

[prisma breach]

Okay I see
so how else would you go about forming such a function

[pale epoch]

and think ||of 0/1 as true/false||

[coral parcel]

Beware that you still need to deal with double representations, since 0.1000...=0.0111... in binary.

(Ah, you don't want a bijection, that makes it easier).

[weary tiger]

but actually you dont need binary representation at all for just surjection

[coral parcel]

Right, there are other strategies than binary representations.

[weary tiger]

I mean you can work in base 10 and just use the numbers with 0s and 1s in their expansion

and map rest to say empty set

[coral parcel]

Hmm, fair. Yes, that is easier.

[prisma breach]

oh, did I state surjection?

oh.. I did mean bijection my bad

[weary tiger]

oh well then you might have to use binary expansion

[prisma breach]

by binary expansion I assume you mean expressing numbers uniquely through sums of powers of 2?

[weary tiger]

yeah

[coral parcel]

Honestly, I would go for injections R->P(N) and P(N)->R, and then appeal to Cantor/Bernstein/Schröder for gluing them together to a bijection. It's possible to write one explicitly, by taking explicit steps to move the double representations out of the way, but the result is not really enlightening...

[weary tiger]

what loch said

[prisma breach]

conisdering 0/1 as false true in this context is determining which powers of 2 are to be summed

.. ?

[coral parcel]

(Note that this is the same as making 0/1 the coefficients of the powers of 2).

[prisma breach]

yea ofc

[coral parcel]

(Except that would make 0/1 into false/true rather than true/false :-p )

[dark geode]

could someone help me write a recursive formula for part b of this problem
i have
(0,0), (1,1), (6,36), (35,1225), (204,41616)
and the problem is
Part (b): Find a recursive relationship expressing (xn,yn) in terms of (xn-1,yn-1) for any n≥1.
??

[tidal tulip]

for the statement "x is neither big nor fuzzy", could i express that in terms of P,Q defined as P= x is big, Q= x is fuzzy with the expression ~(P or Q)

and this would be equivalent to (~P and ~Q) which would read "x is not big and x is not fuzzy" -- is that correct?

[coral parcel]

Correct.

[tidal tulip]

awesome, thanks!

[prisma breach]

hmm...
well every yn term seems to be a square of the xn term

and the x terms seem to follow a certain pattern of multiplication and subtraction

can you see what it is?

[dark geode]

6* previous term - the term before that

but is there any mathematical way to reach that

x(n) = 6x(n-1)-x(n-2)

but idk abt the y

and plust

the question asks for it in terms of x(n-1)

[upbeat bough]

helo?

anyone can help with ∀x∈R ∃ y∈R [x ≠y → (x<y ∨ x > y)] ?

[pale epoch]

what about it?

[upbeat bough]

can you help me solve it

[olive condor]

I think the proof is pretty straightforward

What have you tried?

[upbeat bough]

ok nevermind why did i bother asking

[coral parcel]

It is quite unusual to have → directly under a ∃. Are you sure you don't mean ∀y?

[weary tiger]

could someone explain how they went from the first line ~p V (~q A ~r) V (P A q A ~r) to the next step?
I get it all except that one law they used there

how did they move the ~p and left the first term as is??

[hard citrus]

I believe it's just the associative property

[weary tiger]

but

~P OR ( P AND Q AND ~R)

NOT p or ( or or or )

associative law has to have the stuff the same connective no?

[hard citrus]

Oh there

[weary tiger]

would u be able

to wirte it down

on paint or something

LOL its so confusing

[hard citrus]

They distributed ~p to p and qAND~r

[weary tiger]

I get that the ~P moved to the other side

so comutative law

and dist?

they didnt use associative law then

[hard citrus]

Right

Not there

They used associative from 2nd to 3rd step

[weary tiger]

ok 1 sec

is this written correctly?

like notation wsie

@hard citrus

[hard citrus]

Yes

[weary tiger]

ok thanks

[hard citrus]

Just put a <=> inbetween them

[weary tiger]

ahh I see

[vital dew]

Yaluna

[weary tiger]

Would someone tell me how to validate this argument

[hard citrus]

I'd give a more descriptive answer as to how you concluded that q must be true and that r must be true since you avoided using rules of inference

Otherwise it seems valid to me

[weary tiger]

But how to use laws of logic

I understand the reasoning

[proven silo]

If n=k=0 is allowed for example floor is needed

[prisma breach]

n+1 > 2k then

[proven silo]

Oh right

Likely the answer is just why not?

Easier to see floor inequality holds

Than convince ceiling also works

[prisma breach]

actually
flooring down should've removed the equal sign

k is strictly bigger

[proven silo]

No, n=k=1 (and doesn’t matter if it was strict if they only need less or equal)

[prisma breach]

wait

n k are integers?

also yea you're right

but if they're integers wouldn't the inequality not hold if n and k are negative?

[twilit sierra]

 Consider the string w = 0^k1^k. Since w ∈ L, it must be accepted by M. We think of processing w by M as a traversal of the corresponding directed graph.

Since the number of symbols in w is the same as the number of transitions made by M upon processing w, and the number of states in M is less that |w|, some state qi must be revisited while processing w.```
can some1 explain to me why ```he number of states in M is less that |w|``` this is true?

[prisma breach]

-3<-2
-2>-4 ?

@bright shale I think the inequality is flawed unless absolute values are taken

'integers' implies Z... ?

[regal gate]

What is the number of permutations $\sigma$ of $[n]$ such that $\sigma(i)\neq i$ for each $1\leq i\leq n$?

[vital dew]

Croqueta

[prisma breach]

isn't there a defined formula for that?

[regal gate]

that's what I'm asking.

I don't really know about this stuff

[prisma breach]

using the principle of inclusion exclusion

I think it's nothing more than formality

for every real number these equalities will hold no matter whether you use the flooring or ceiling function on the expression

nvm

[prisma breach]

the formula derived from said principle is n! * sum(0->k)(-1)^k / k!

[regal gate]

$n!\sum_{k=0}^n \frac{(-1)^k}{k!}$

?

[vital dew]

Croqueta

[prisma breach]

yea

[regal gate]

ok thank you

uhm

I didn't expect that tbh

because then its always even

[prisma breach]

..?

even?

[regal gate]

yea?

[prisma breach]

it's not even a natural number

[regal gate]

watf

???

[prisma breach]

actually, as n --> infinity the number can be estimated as n!/e

[regal gate]

i don't even know what you are talking about

[prisma breach]

.

[regal gate]

and you are saying its not a natural number? then what's the point of the formula?

[prisma breach]

you can estimate the number by flooring down the result

The sum in the formula is the definition of e as n approaches infinity

[regal gate]

oh so its not what I poste dthen

cuz what I posted is at least an integer

you didn't specify the upper limit

[prisma breach]

No no no , the formula doesn't require n to approach infinity

[regal gate]

so that's an integer

[prisma breach]

im saying the your sequence, the number of such permutations approaches n!/e

[regal gate]

oh

so hold on

wasn't my question clear?

[prisma breach]

you're summing ab alternating harmonic series yea

No your question is right

the result is not an integer

you'd have to floor it down

[regal gate]

what?

[regal gate]

dude what are you talking about

[carmine hornet]

whats this question asking exactly

[regal gate]

This is always an integer, because $n\geq k$, simple

[vital dew]

Croqueta

[prisma breach]

wha... how is that relevant?

k is just an index

[regal gate]

dude

is this trolling?

[prisma breach]

wh

[regal gate]

seriously I don't know what you are talking about

[prisma breach]

okay you're asking how many permuations exist such that pi(i)!=i right?

pi is the notation we use in my course for a permutation

[regal gate]

yes

[prisma breach]

okay great

[regal gate]

wait

[prisma breach]

first we start of with the total amount or permutations right?

[regal gate]

permutations of the list [n]

which is (1,...,n)

nothing approaches infinity

[prisma breach]

I never SAID

[regal gate]

n!

[prisma breach]

N APPROACHES INFINITY
I

[regal gate]

just in case

[prisma breach]

I JUST SAID IF IT DOES THEN YOU GET AN INTERESTING RESULT

[regal gate]

well you started talking about infinity and convergence stuff, which was totally unrelated to my question

dude

shut up

don't fucking scream in discord

[prisma breach]

I didn't intend to scream

sorry

was meant as emphasis

it's just the alternating sum in the formula equals e as n approaches infinity

[regal gate]

ok

[carmine hornet]

bruv?

[regal gate]

it doesn't

oh sorry

[prisma breach]

so as your sequence is bigger and bigger, number can be estimated as n!/e

[regal gate]

yeah, the alternating sum yeah, thought you said the formula

[prisma breach]

yea

[regal gate]

that's were I got confused, because I'm not really interested in that

[prisma breach]

I see

in either case that is the formula for computing it

which can be derived from the principle of inclusion exclusion

[carmine hornet]

am i not seeking for help properly or sum?

[regal gate]

no its not

its wrong

because its always even, and its not the case

[prisma breach]

https://en.wikipedia.org/wiki/Derangement

isn't this what you're talking about?

[regal gate]

ok yeah im an idiot

yeah thank you

[prisma breach]

sure thing

[regal gate]

for some reason I thought it would be even, since there was an n! in front lmao

[prisma breach]

oh umm sorry
you can ask help on one of the help channels as well

[prisma breach]

well yea you have a fair point
but the sum of an alternating harmonic series kinda ruins that lol

[carmine hornet]

its a discrete math question tho

[carmine hornet]

.\

[regal gate]

yeah it was not so smart. I should have taken your answer since the begining lmao. Sorry for wasting too much time

[prisma breach]

it's fine lol

sorry for capsing

[regal gate]

"A herd of 1000 cows of nonzero weight is given.
Prove that we can remove one cow such that the remaining 999 cows cannot be split into
two halves of equal weights."

This was all to solve this problem, and I was stressed because I don't know combinatorics

[severe swan]

How can I calculate the number of rows in a truth table? 5 variables would be 2^5 = 32? If there are 128 rows then to calculate the number of variables would be squareroot 128 = 11 rounded down?

[prisma breach]

im actually confused about this question..

how can you prove that?

[regal gate]

use a little of linear algebra and then the derangment formula will be useful

[prisma breach]

are you not given any details about their weight?

[regal gate]

it is positive

the cows set up is to confuse you probably

you can take the weight to be negative anyway

[tranquil vector]

How many ways are there to place n distinct people into four different rooms?

Would the answer by 4^n

[regal gate]

wanna see a proof?

[severe swan]

Am I understanding this correctly?

original implication
If it is not monday then I am happy.

Would the converse of this implication be

If I am happy then it is monday.

[regal gate]

I think so, because for each person there are 4 possibilities, and in total there are n persons.

[tranquil vector]

How many ways are there to place n distinct people into four different rooms so that there are at least two non-empty rooms?

[cerulean wind]

count how many ways there are to put n people into exactly 2 rooms, add that to how many ways there are to put n ppl into exactly 3 rooms, and ad that to how many ways there are to put n ppl into exactly 4 rooms

to get you started, there are 2^n - 2 ways to put n people into exactly two rooms.

this is related to counting surjections

[tranquil vector]

why the -1

[cerulean wind]

my bad

it should be -2. there are only two ways to put n people into two rooms with exactly one room empty

wait

i may have misunderstood the problem

[full tartan]

could someone help with trig ratios?

[tranquil vector]

Would it be 4^n +3^n + 2^n

or is that too easy

[tidal tulip]

is it correct to say that the following statement is true “if n is a multiple of 6, then n is even”. this is true because 6 is an even number and even * even = even and even * odd = even. so no matter whether the other multiple or n is even or odd the product of n will be even.

even * odd = even proof
let k, m be in Z then let x be even and y be odd where x=km and y=2m+1

x * y = (2k)(2m+1) = 4km+2k = 2(2km+k). let t=2km+k in Z, then we see x*y is of the form 2(t). hence even * odd = even

case even * even =even

let x=2k, y=2m for k,m in Z

xy=(2k)(2m) = 4km = 2(2km) and let t=2km in Z then we see xy=2*t, hence even * even = even

[cerulean wind]

there are 4 C 2 = 6 ways to have two non-empty rooms. you need to multiply that by the number of ways there are to send n people into two rooms so that none of the rooms are empty. in that sense, there are 6(2^n - 2) ways to do this.

similarly, there are 4 C 3 = 4ways to have three non-empty rooms. you need to multiply that by the number of ways there are to send n people into three rooms so that none of the rooms are empty.

there are 4 C 4 = 1 way to have 4 non-empty rooms. you need to multiply this by the number of ways to send n people into four rooms so that none of the rooms are empty

add them all together and you will get your answer

it should be 6*2!*{n, 2} + 4*3!*{n, 3} + 4!*{n, 4}
where {n,k} is the stirling number of the second kind

[tidal tulip]

could someone do a proof check for my problem

[cerulean wind]

yea this is fine, but why not just say that n = 6k = 2(3k) is even

[sage mulch]

u could also say n = 6k = 2k + 2k + 2k 😆

theres many ways to prove it

[tidal tulip]

if you say 6k = 2(3k) is even then you have to show that m * 2(3k) is even and show that’s true whether m is even or odd no?

and thanks for checking my proof

[severe swan]

Can somone explain how to find the answer to this question?

When is the implication !q --> ! p false

Do I switch p and q to positive and negatives and figure out which one leads to all Fs? I did that and I didn't get all Fs. I am not sure which one it would be below.

when p is true and q is true
!q --> p
T
F
T
T

when p is true and q is false
!q --> p
T
T
T
F

when p is false and q is false
!q --> !q
T
T
F
T

when p is false and q is true
q --> !p
F
T
T
T

[tidal tulip]

@cerulean wind : if you say 6k = 2(3k) is even then you have to show that m * 2(3k) is even and show that’s true whether m is even or odd no?

[sage mulch]

@severe swan do the truth tables for each of the componets of !q and !p and then do the if statement table

[tidal tulip]

and thanks for checking my proof!

[severe swan]

Ok thanks

[sage mulch]

if u want to be extra cautious, you can set a varialbe like a = 3k and then itll be 2a, which is even

[cerulean wind]

a number n is even if it can be written as 2m for some integer m

3k is an integer

2(3k) = 6k = n is even

[tidal tulip]

i see 6 is even but i don’t see how m * 2(3k) is even unless you break m up into cases for m being odd or m being even

[sage mulch]

didnt u say how odd * even = even

[tidal tulip]

yeah

but i had to prove it

[sage mulch]

u could use that proof to indicate that m * 2(3k) is even

[tidal tulip]

m * 2(3k) would require you to prove that product is even where m is odd or m is even

oh i see you’re breaking 6 down

[sage mulch]

yes

[cerulean wind]

what m are you talking about?

[tidal tulip]

and using the proof i did to make the claim m * 2(3k) is even

[sage mulch]

👍

thats one way to go about it

though i bet theres a faster manner

[tidal tulip]

ok i see

what’s the faster way

[cerulean wind]

you want to show that if n is a multiple of 6, then n is even.

n is even provided that there exists an integer m such that n = 2m

if n is a multiple of 6, then n = 6k for some integer k.

n = 6k = (2 * 3)k = 2(3k) is by definition even

[tidal tulip]

i’m moving around i’ll read this and respond in a few

ok so you’re saying n is a multiple of 6 which means n = 6k and then you’re rewriting 6 in terms of its own multiple 6=2(3k), but then you notice that n can be written entirely as n=2(3k) because 2(3k) can get you any product that 6 is a multiple of

[cerulean wind]

everything is fine until the last line of reasoning. i am merely writing n as a multiple of 2

thats all the problem is asking you to do

[sage mulch]

@tidal tulip m = 3k here

if that helps

[tidal tulip]

i’m kind of confused because “what happened to the 6” when you’re writing n that way

[cerulean wind]

6 got turned into 2 * 3

[tidal tulip]

ok

how would this proof go:

n = 6k = 2(3k) where 3k is in Z, let t=3k then we see n=2t, thus we see n is even

that side stepped all my case proofs which is great

it’s clever too

basically you’re rewriting 6k

and then when you rewrite 6k you get n in the form of what it means to be even

for any k in Z

is that the right understanding yeahv

?

[cerulean wind]

yea

[waxen spire]

does anyone have good resources for combinational proofs?

[tidal tulip]

awesome thanks for the help

[severe swan]

I got this incorrect on a test. Can somone explain the correct answer to me?

Which of the following is the converse of the implication "If it is not Monday, then I am happy"

A) If it is not Monday, then I am happy

B) If I am not happy, then it is Monday

C) If it is not Monday, then I am not happy

D) If I am happy, then it is Monday (my answer)

E) If I am happy then it is not Monday

F) If it is Monday, then I am not happy

G) None of the answers are correct

[cerulean wind]

E

converse of p --> q is q --> p

[severe swan]

Thank you

[wide vine]

can you get R^3 from a cartesian product?

like if you did R x R you would get a set with elements that are a tuples with 2 entries

I assume R^3 = {(x,y,z) : x,y,z ∈ R}

something like that?

[coral parcel]

(RxR)xR is close enough for many purposes.

[whole hamlet]

If a divides c and b divides c can the product ab divide c? I got to the proof c^2(jk)=ab where jk is an integer. Does that mean ab can divide c or not because it’s squared? Or did I just get the proof wrong ?

[cerulean wind]

2 divides 100 and 100 divides 100 but 200 does not divide 100

[floral field]

What would be the optimal way to prove surjectivity injectivity here? I don’t quite see what algebra to do to deduce (a,b)=(c,d)

[cerulean wind]

you mean injectivity?

[floral field]

Oh yes I do

Oops

Yup, I was proving injectivity first

[cerulean wind]

try showing that for every natural number n, there exists unique m,j such that n = (m C 2) + j with 0 <= j <= m

[floral field]

How does (mC2)+j help here?

[cerulean wind]

its basically the inverse function

[cerulean wind]

hint: ||take m to be the largest m such that T_m = (m C 2) <= n, where T_m is the m-th triangle number. this is unique since it is the maximal element of some set of bounded above natural numbers.||

[distant bobcat]

Whats the procedure for checking if a graph is a bipartite graph? Do I check for each vertex in the graph all of its neighbors and see if I can construct a disjoint set? Is that the idea?

[pale epoch]

isnt bipartite iff 2-colorable true

and a 2-coloring (of a connected component) is uniquely determined once you color a single vertex

so this can be done checked in linear time

[distant bobcat]

Yes, this is true . Its bipartite if there exists a two coloring s.t its vertices can be colored in for example black and white in such a way that each edge joins a black vertex and a white vertex.
Im trying check if any of the platonic graphs are bipartite and was wondering if there was a convenient method of verifying this?

[stray reef]

platonic graphs?

like, the graphs representing each of the platonic solids?

if so then only the cube is bipartite, all the others have odd cycles (the faces)

[distant bobcat]

yes, I found the cube to be bipartite as well through trial and error. How do odd cycles discount the bipartite graphs?

Its easy to see for a regular graph like k3, but how do I show that for all odd cycles its impossible to construct two disjoint sets?

Nevermind, I found an answer

https://math.stackexchange.com/questions/311665/proof-a-graph-is-bipartite-if-and-only-if-it-contains-no-odd-cycles

[stray reef]

odd cycle implies lack of 2-colorability very trivially

try coloring just the vertices that are part of your odd cycle

youll run into two that have to be the same color

[distant bobcat]

yep, I saw this for smaller examples while doing earlier exercises, but didnt know that it held in general. Thanks.

[twilit sierra]

is that considered discrete?

or where should i ask my question

[stray reef]

yes this channel is ok for that

[prisma breach]

Let R be a relation on $\mathbb{N} \cross P(\mathbb{N})$ such that
for every $(m,A),(n,B) \in \mathbb{N} \cross P(\mathbb{N})$
$(m,A)R(n,B) \iff m-n$ is \ divisible by 3 and $A \cup B$ is a finite set
Is R an equivalence relation on $\mathbb{N} \cross P(\mathbb{N})$ ?

[vital dew]

Deus_Vult

[prisma breach]

small question on this one, the answer was no because R is not reflexive
and I sort of fail to see why ?
since m-m = 0 would count as being divisible by 3
does the check fail because A U A is not necessarily a finite set ?

[pale epoch]

yes

take (1, N)

[prisma breach]

oh

right

that's trivial
thanks

[severe swan]

Can someone explain this to me? I got it wrong on my test and have no idea how to figure it out.

Without making a truth table, what truth values need to be given to variables p,q,r,s,u,w to make the following proposition evaluate to True

!(p&q) & (u --> p) & q & (w <--> u) & (!w --> s) & (r V u V w)

what variables should be true and what variables should be false

[prisma breach]

hmm.. maybe start with the most obvious, and see how you can advance from there

what was your answer?

[livid cobalt]

since all your subformulas are part of a conjunction, you know they all must evaluate to true then

[valid wave]

how does this represent addition

when the symbol is Z4 X Z4

g(i,j)

[pale epoch]

its addition modulo 4

[valid wave]

i get that

but

Z cross Z

[pale epoch]

and g(i, j) = i + j

[valid wave]

!= multiplication table?

[valid wave]

so then what represents multiplication in a function

[pale epoch]

g is just some function that takes two inputs in Z_4 and spits out a number in Z_4

[valid wave]

wait ohhh

since Z4 is a set

[pale epoch]

multiplication also is a function $Z_4 \times Z_4 \to Z_4$

[vital dew]

Lochverstärker

[valid wave]

thats the exact same as addition function though?

[pale epoch]

there are many functions like that

[valid wave]

weird

so how would one know the distinction

[pale epoch]

by being told

[valid wave]

interesting

[pale epoch]

the text tells you that g is supposed to be addition

there is also a function g(i, j) = 0 that sends everything to 0

[hard citrus]

This is just the domain and range of the function here

[pale epoch]

this does not have a special name

[valid wave]

so Z4 cross Z4 doesnt necessarily mean multiplication, means the pair of elements from the set Z4

[pale epoch]

(constant 0 function maybe)

oh

[hard citrus]

You also need a rule, in your example thats addition modulo 4, it could be anuthing else really

[pale epoch]

this is the cartesian product

its the set of all tuples (i, j) with i, j in Z_4

[valid wave]

makes sense

but if its addition

this Z4 cross Z4 means

2 memebers fro mZ4

map to something on Z4

[pale epoch]

yes

[valid wave]

when g(i,j)

[hard citrus]

Exactly

[valid wave]

gotcha

🤯

but just making sure im understanding

g(i,j) could mean anything

have to have context

[pale epoch]

yes

[hard citrus]

Look more into cartesian product

[valid wave]

got it, thank you!

[pale epoch]

it can be any element in Z_4 in this case

in this case there is a "rule" g(i, j) = i+j (mod 4)

but you can also arbitrarily choose a result for each (i, j) without any rule

say f(0, 1) = 0, f(1, 0) = 1 and f(i, j) = 2 for everything else

thats also a function Z_4 x Z_4 -> Z_4

albeit not a very interesting one

[valid wave]

got it

Z_4 x Z_4 != multiplication rather the idea of some g(i,j) mapping onto Z_4. A pair, cartesian product, of values from Z_4

😄

thanks again!

[valid wave]

even this just looks like random stuff just written down

[wide vine]

now this is a bit pedantic but how do I list elements of a set?

I wrote this

but I think

it is just you tuple, tuple, tuple, .... $\in$ A x B x C

[vital dew]

Brandon7716

[wide vine]

because I believe I listed the set rather than stating the elements of the set

[pale epoch]

what exactly is the question

i only see your solution

[wide vine]

that is the qusetion

1

oops

[pale epoch]

ah ok

what is (2)?

[wide vine]

that shouldnt have an \in

well I was trying to show my steps

and Im not sure a good way to indicate that

[pale epoch]

ok

[wide vine]

that is why I did (3)

[pale epoch]

you forgot a close curly brace and open curly brace in (2) then

the way its written rn does not make sense

(3) is the set $(A \times B) \times C$

[vital dew]

Lochverstärker

[pale epoch]

so that is correct

depending on pedantism and how you defined cartesian products

[wide vine]

yes

[pale epoch]

generally you would list the elements as (a, b, c), not ((a, b), c)

and you are correct that you listed the set technically and not the elements

[wide vine]

yeah I also wasn't sure about ((a,b), c)

because I asked early about how R^3 is defined

and didn't know if people just do RxRxR

[pale epoch]

well

[wide vine]

or

{(x,y,z) : x,y,z in R}

[pale epoch]

the issue is that you dont want an extra definition for R^n for every n, so you define it inductively as R^{n-1} x R

but then you get typographical nightmares like ((((a, b), c), d, e)

and you technically get $(R \times R) \times R \neq R \times (R \times R)$

[vital dew]

Lochverstärker

[pale epoch]

so we generally just agree to write tuples as (a, b, ..., x)

[wide vine]

ahh

[pale epoch]

i dont know your professor, i would mark both as correct

but yours isnt as pretty to look at as the alternative

(same for the set vs its elements distinction, i would mark both correct)

[wide vine]

It seems I must have closed off that part

glossed*

idk why I have been doing the ((a,b), c)

well this will clear up my messy stuff

ahh I think I knew why I did that

I thought of (a,b) as 1 element

so are both accepted?

[pale epoch]

(a, b) is one element in A x B

[wide vine]

yeah

[pale epoch]

so if you did it in the steps you listed, it makes sense

generally we agree that ((a, b), c) = (a, b, c) = (a, (b, c))

[wide vine]

Okay

[pale epoch]

"we" i mean mathematicians

i suggest to use (a, b, c)

[random stirrup]

Hi can I ask a question?

[valid wave]

hi im trying to solve this question

im not at all seeing how induction can be shown for this

could someone point me in the right direction

i tried simplifying that r + 1/r

into (r^2 + 1) / r

and it doesnt seem to even help my base case

so for instance

id have

((r^2n) + 1) / r ^n

looking at the base case of n = 1

((r ^ 2) + 1)/ r

this changes nothing for me

wait omg, i can set this equal to the orignal r + 1 / r since thats an integer

🤯

[cunning girder]

Hey folks

Could someone explain why this statement is true?

∃x(x is even → x is odd) (assuming x is an integer)

As it stands, I figured it meant as follows:

there exists an integer x that is odd if it's even

Which should logically be false, since integers can't be even & odd simultaneously

Am I misinterpreting the statement?

[pale epoch]

check how implication works

if x is odd, then (x is even -> x is odd) is true

[cunning girder]

How come, though?

I'll look into implication shortly, haven't done this stuff in a while

but I thought it worked like an if statement?

so if x is even then x is odd

i.e., if p then q

[pale epoch]

this question comes up a lot

someone else recently explained it way better than i ever could, let me see if i can find it

[cunning girder]

Sure, that'd be great, thanks

[valid wave]

crap

am i approaching this right still.

it SEEMS like it would be right

since i need to basically prove that r6n + 1/r^n would be a multiple of r + 1/r

+ask

oof guess thats not a command here

https://dontasktoask.com/

[pale epoch]

the explanation starts here: #proofs-and-logic message

my explanation is right above and consists of the principle of explosion

but i think troposphere explained it a lot better, if you care to read it

[weary tiger]

my answer is:

[obtuse lance]

I'd use the binomial expansion of $(r+r^{-1})^n$ to prove that it's an integer

[weary tiger]

Mero

[vital dew]

Merosity

[cunning girder]

Right

So in our case

We can look at it as p being a subset of q

x is even -> x is odd is the same as (x is even) is a subset of (x is odd)

so by definition

since p is a subset of q

[valid wave]

can you elaborate please :

[cunning girder]

q is forcibly true

right?

So something like this

even though this wouldn't make sense in the real world, we have to assume that the statement is true?

so we assume that the premise (x is even) is false

[valid wave]

is this even in the form of binomail expansion

the (r^n) +1/r^n

[obtuse lance]

no

[valid wave]

for instance i have the base case of 1

but that n is confusing me

[obtuse lance]

(r+1/r)^n will be a bunch of terms paired up, the binomial coefficients being symmetric gives r^n+1/r^n + binomial coefficients* (r^k + 1/r^k) for k<n

[valid wave]

it LOOKS like it would be the right path

if i can get that n out

to show that the left is a multiple of the right maybe

[obtuse lance]

so subtracting an integer from an integer leaves an integer, (r+1/r)^n- binomial coefficients* (r^k + 1/r^k) for k<n = r^n + 1/r^n

[valid wave]

all of that went over my head

jesus im pea brain

[obtuse lance]

it's slightly annoying cause even vs odd terms will be slightly different

idk maybe someone else can help I'm about to make dinner GL

[valid wave]

thanks 😄

[wide vine]

is this a good enough explanation

I think it is correct

[pale epoch]

i am not sure if this is a useful perspective (im also not sure i get your argument)
im also getting tired, sorry i cant help further

[severe swan]

Can someone explain how to do this? I need to assign true or false to a, b, c:

give truth values to a, b, c which shows
(b->c)->a
is not equivalent to
b->(c->a)

a = T/F
b=T/F
c=T/F

[livid cobalt]

one way is to build a truth table and find the interpretations of a, b, and c for which [(b -> c) -> a] <=> [b -> (c -> a)] is false; another is to think in terms of what makes a conditional true or false - i.e., if we want (b -> c) -> a to be false, then it needs to take the form T -> F, meaning (b -> c) <=> T and a <=> F. here we find that b and c can still form multiple combinations ((F, F), (F, T), (T, T)), so maybe it's time to turn to the other formula - i.e., b -> (c -> a), which must be true. the only form it cannot take, then, is T -> F. we know from our previous assumption that a <=> F, so let's plug that in: b -> (c -> F). once again: since we want this formula to be true, we cannot have the form T -> F. the only pair (T, c) which we found while analyzing the other formula was (T, T), but this interpretation will make this second formula false, which is not what we want. so (T, T) won't help us here. what remains is (F, F) and (F, T), and both of these will make the second formula true and the first formula false (and you will also find these combinations in your truth table)

so (b -> c) -> a and b -> (c -> a) are nonequivalent for the combinations (read as (a, b, c)): (F, F, F) and (F, F, T)

[severe swan]

Thank you very much for the detailed explanation.

[distant bobcat]

Why do we have for a tournament (a complete directed graph where any two vertices are joined by exactly one edge) we have that the sum of the in-degree equals the sum of the out-degree which I understand, but I also found the following property I didn't understand:
(x_i + y_i) = n-1 where x_i is the in-degree, y_i is the out-degree of the graph. If we sum over the in and out degrees do we not just get twice the number of edges of the graph?

[gilded axle]

I think in the second paragraph x_i and y_i is supposed to represent the in- and outdegrees of a single vertex i, not the entire graph, which would make a lot more sense as this equality would then also be correct

[prisma breach]

hey , just a small sanity check

$$A={\emptyset,1} \ B={{\emptyset},1}$$
$$A\neq B$$
$$\emptyset \in A \ \emptyset\subset A \ \emptyset \subset B \ \emptyset\notin B$$

[vital dew]

Deus_Vult

[prisma breach]

is this correct?

[pale epoch]

yes

[stray reef]

\\ for newlines btw

[prisma breach]

oh thanks

[prisma breach]

Great, thank you

[distant bobcat]

@gilded axle This does indeed make sense. Thanks

[weary tiger]

soemone here got a spare minute to help?

In set theory

[hard citrus]

[weary tiger]

how do I proof stuff with induction on sets

AA A x i i n x A 1 2 '' ' " !n i ^ ^ ^ 1,2, , odd

[livid cobalt]

[weary tiger]

how do I proof that with induction

[wide vine]

So my prof released the answer and somehow (e) is not true?

$A \cap C \subseteq B$

[vital dew]

Brandon7716

[weary tiger]

Lol I guess the prof meant there is a difference between $\subset$ and $\subseteq$?

[vital dew]

catfan1337

[wide vine]

Im not sure

I guess?

[weary tiger]

which is stupid

[wide vine]

She realized the solution via grading one of the hw's that was turned in

pretty dumb if that is false because of that

[weary tiger]

very dumb

check your notes to see how you defined those symbols

probably 'subset' vs 'proper subset' , but proper subset implies subset, so this is still true

unless subseteq means 'not proper but equal', but then why would you ever use this symbol lol

[wide vine]

If A ⊆ B and there is an element of B that is not an element of A (i.e., A ≠ B), then A is a proper subset of B, denoted as A ⊂ B. Venn diagrams are particularly useful for visualizing subset relationships between sets.

but based on the book questions

they don't have an assue

$D \subseteq C$

[weary tiger]

So as you can see A \subset B means "A \subseteq B AND exists b in B such that b isnt in A". Because this is AND, its definitely also A ]subseteq B

[vital dew]

Brandon7716

[wide vine]

plus this is also an exercise question

[weary tiger]

If you care about the points or whatever you should email your prof saying that its correct because proper subset implies subset

[wide vine]

it it was hw and optional and I believe everyone got full points for completing it

but I have a quiz soon and that would be pretty dumb to get something wrong because of that

[weary tiger]

Its better to answer the way you did and then argue imo, you shouldnt learn bad habits

[wide vine]

Yeah, I might email her to see if it was a mistake or something

[silent hinge]

Hello friends

Can I get a hint

For this question

Like for #2 now can I work that one out?

If the domain I have correspond to x

[hard citrus]

number 2 is just the negation of the statement in number 1

the domain is the same

[remote dock]

since the empty set is subset of every set , can say its equals ? ^

[hard citrus]

It's not the same

Here you treat the empty set as an element of A

In the second example it is both a subset and an element of A
In the first one it is only a subset of A

[remote dock]

oh i see

[coral parcel]

In the absence of contrary indications I would assume it's just sloppy editing of the question and when they wrote "the domain of x" they really meant "the domain of the quantifiers".

Or alternatively, that they were talking of the domain of the propositional function p, and it's a coincidence that the letter x they first wrote in "p(x)" is the same as the quantified variable in the first subquestion. (This also amounts to y ranging over the same set).

[remote dock]

I or ii is true?

[hard citrus]

This is how I usually encounter it, never really gave it a thought before

[hard citrus]

Well A={N} is a set which contains a set. Which one do you think is right, i) or ii)?

[remote dock]

ii

[hard citrus]

Yes

[remote dock]

thanks

[silent hinge]

Thank you guys

[hard citrus]

If you're just looking for a y/n answer, then yes. It can be simplified down to p*(r+q')

[vapid tinsel]

(~Q ∨ R) => ((Q | Q) (Q|Q)) | (R | R))

Is that correct? Or is there a simpler form?

[livid cobalt]

just to clarify: you're suppose to rewrite it using only nand gates, correct?

if so, it's right, assuming you just forgot to put the | between (Q | Q) and (Q | Q) (very important, since this absence could lead someone to think you're using an AND gate there)

[smoky glacier]

What is P(a,b)? A set?

[polar ferry]

Yes

[smoky glacier]

P(1,2) for example is a set, okay, but I don't know what P(a,b) is

[polar ferry]

what are a and b?

[smoky glacier]

yeah I'm struggling at that, is it a set of all sets made up of different a and b's or is it only one set with all elements of all possible a and b's

[smoky glacier]

so are the elements of P(a,b) like P(1,2), P(3,5) or points like (1,0,z)

real numbers

a,b are elements of Real numbers set

a bit confusing since the image says p_(a,b) = {(x,y,z...

but it makes more sense than a set like x e R^3

[polar ferry]

I made a mistake

It would be just a set

Not a set of sets

[digital nest]

wdym?

[polar ferry]

different question

[digital nest]

ohh

[smoky glacier]

okay, thank you

so P_(a,b) = {(x, y, z) e R^3)

[polar ferry]

Its a set of points (x,y,z) such that ax+by=0

so really z can be anything

as long as gcd(a,b)=0

[smoky glacier]

but what does P(a,b) mean

wouldn't it mean everything

like okay each P(a, b) where a and b are constants is a set of points with a line for x y. and all z values

[polar ferry]

Do an example

[smoky glacier]

for example P(1,2)
means all x and y values for y = -x/2

[smoky glacier]

so points will be (x, -x/2, z) for all possible x and z values

[craggy juniper]

P_(a,b) is the set of all (x,y,z) that satisfy the equation ax+by = 0

notice that z can take on any value

[smoky glacier]

I did

my question is that P(1,2) is a set with points

[smoky glacier]

like this

but if a and b are not specified, what is p(a,b)

[craggy juniper]

.

[polar ferry]

^

[smoky glacier]

I think I got it, thanks

[silent hinge]

for number one is this the right way of disliking joffrey?

im trying to watch the lecture again but i cant recall neglecting quantifiers in class. please provide a hint. thanks in advance

[last flicker]

No. You haven't even referenced jofferey or anya in your statement. what you hve written basically reads as " There exists someone who likes noone else"

[silent hinge]

assuming that arya = x and y = joff

[last flicker]

Then they wouldn't be variables....

[silent hinge]

how do i say x does not like y

[last flicker]

so the quantifier is meaningless

[last flicker]

not p(x,y)

You don't need variables or quantifiers for 1, just the objects of arya and jofferey

(and p and the not operator)

[silent hinge]

dang it

i get it know

~p(arya, joff) does the trick

[last flicker]

yeah it does

[silent hinge]

when ETH hits 10,000 ill buy you a ps5

[last flicker]

i already invested all my money in bingustoken I know its going to the moon anyway

/s

[wide vine]

how would I optimize this solution

i did

$X= (\bar{A} B)+(A\bar{B})$

$\not$

[vital dew]

Brandon7716

[wide vine]

$\bar{a}$

[vital dew]

Brandon7716

Brandon7716

[wide vine]

I think ideally in this solution we want to use nands whenever possible

[wide vine]

someone in #chill helped so no worries

[wide vine]

other than for the empty set, why would we ever need $\subseteq$

[vital dew]

Brandon7716

[wide vine]

Like If we are given sets A and B, when would we want/need to say $\subseteq$ instead of just using $\subset$ or $=$

[vital dew]

Brandon7716

[wide vine]

I can think of one condition being $\O \subseteq \O$

[vital dew]

Brandon7716

[wide vine]

but then I guess you could just use $\O = \O$

[vital dew]

Brandon7716

[stray reef]

@wide vine use \emptyset or \varnothing for the empty set

also im not sure what your point is

you want to require that all instances of $A \subseteq B$ be written as ``$A \subset B$ or $A = B$''...?

[vital dew]

Ann

[ashen cloak]

I need help understanding this.

suppose a|b and b|c.
I must show that a|c.
i get theres an x for a|b and y for b|c.
how does xa=b and yb=c.

[ember obsidian]

@ashen cloak whats ur confusion

[stray reef]

@ashen cloak are you still here

[floral abyss]

Bit of a dumb question but how would I show associativity. For all a,b,c in Z show that a(bc)=(ab)c

Like I get it’s pretty simple but I’m not sure how to write it out formally

[stray reef]

what is your definition of Z?

if you want a formal proof you probably want to appeal to whatever construction your class is using for Z

[weary tiger]

so all you have to do is to prove that | is transitive

lets suppose you A,B,C groups

so we suppose that a|b = a|c

[floral abyss]

Just an integer

[stray reef]

so your class did not construct Z formally from N?

[weary tiger]

that means that there is an a in a|b and also in a|c

[floral abyss]

Honestly I have no idea

[stray reef]

@weary tiger aren't you overcomplicating this for FAST_TRACK? and also not making any sense while you do so

a|b = a|c is either nonsensical or doesn't mean what you intend it to

[weary tiger]

what do you mean by that, the proof is just that | is transitivita

that's a proof rule you got to follow

to formalize an answer

how am I overcomplicating things, let me write the full proof for a sex

sec*

[stray reef]

what "proof rule" are you saying that one "has to" follow?

FAST_TRACK seems to have gotten themself confused at the definition of |, and you're going to shower them with formalism that explains nothing?

[weary tiger]

the proof rule of transitive, we assume them both, then proof it

I thought he didn't know how to proof it

[stray reef]

there is no such thing as "proof rule of transitive"

[weary tiger]

because he didn\t write I don\t understand the definition

[weary tiger]

not in a formal way but there is

the minute you learn to proof

you follow a structure

[stray reef]

in proving any statement of the form "if P then Q" you assume P and prove Q from it

proving the transitivity of a relation

isn't special in that regard at all

[weary tiger]

the structure to proof transitive is

[stray reef]

and it's pointless to pretend that every single little thing has a separate proof technique you have to memorize like a bag of tricks

[weary tiger]

we assume a = b, a = c and need to proof b = c.

[stray reef]

you have that the wrong way around

that's not the definition of transitivity and generally not equivalent to it

transitivity is aRb & bRc => aRc

[weary tiger]

I'm notr going by def, that is how you proof it

[stray reef]

???????

[weary tiger]

there is a book

how to proof it

suggest you read it

[stray reef]

what, velleman?

i am pretty sure velleman would not commit such a massive blunder in disguising what you wrote as transitivity

[weary tiger]

that is actually what you wrote

[stray reef]

also if you're not using the definition of transitivity to prove transitivity then you are almost definitely doing something disastrously wrong

[weary tiger]

I wrote it without R

[stray reef]

no

you wrote something different

[weary tiger]

because = is a relation

[stray reef]

your variables are the wrong way around

[floral abyss]

I don’t think so. We did something about equivalence relation on N*N in which the equivalence class of it is integers

[stray reef]

i dont care what symbol you use to denote your relation

you had aRb & aRc => bRc

i had aRb & bRc => aRc

these are not the same

would rather you didn't insist they were

[weary tiger]

you know that it doesn't matter, right? I can change them you know?

[stray reef]

it does matter

[ember obsidian]

w/e equivalence it is, thats the construction ann referred to

[weary tiger]

lets assume R is symetric

[stray reef]

OH

[weary tiger]

yeah

[stray reef]

so now

[weary tiger]

got you there

[stray reef]

we're assuming R is symmetric

where did you pull that from?

and going back to the original question

guess what

[weary tiger]

because in the question

[stray reef]

divisibility ISNT symmetric!

[floral abyss]

Hmm okay does this make much difference? Honestly I’m so lost in this course

[weary tiger]

I'm talking about the = sign

not divisible

[stray reef]

OP's question was about |

and you tried to pull the same shit on that

[weary tiger]

no

[stray reef]

yes you did

heres a screenshot of what you said

you started with what was probably intended as "a|b & a|c"

[weary tiger]

oh shit

U right

[stray reef]

which for divisibility is not the right way to prove transitivity

and again

[weary tiger]

yeah yeah

[stray reef]

if youre not using the definition

then you are liable to make mistakes like this

and making mistakes is not something you want to do

[weary tiger]

I took the wrong variables

[stray reef]

yes

thats what ive been telling you all this time

[weary tiger]

the thing is I got confused with thje equallity

nvm

[ember obsidian]

ur proof depends on ur def of Z which itself depends on ur equivalence

[weary tiger]

sup: a|b , b|c
ntp: a|c

[floral abyss]

Ok how would we start

I honestly thought this was gonna be a simple question 🤦‍♂️

[stray reef]

if you're proving something as fundamental as associativity about Z, then it's not something to take for granted

therefore you need to know exactly how Z was constructed from N in order to appeal to that construction in your proofs

[weary tiger]

sup: a|b , b|c
ntp: a|c
so there is a x and y such that:
xa = b and yb = c(that's because a is proportional to b and b is proportional to c, so there are these two integers that are not zero that satisfies that proportion)

[stray reef]

no

[floral abyss]

Idk this. We went over something about equivalence relation on N*N of which integers is it’s equivalence class

[stray reef]

"proportional" is the wrong word

[ember obsidian]

@weary tiger its their task to prove this so lets not write the full proof

[stray reef]

and youre missing key details such as that x and y have to be integers

[stray reef]

no construction no proof

[weary tiger]

for understanding that's the perfect word

you can go by def

but to get intuition

that's the exact thinking U need to have

[floral abyss]

The set of Z of integers is the set of equivalence classes of N*N under the following equivalence relation.
Let ~ be the relation on N x N defined as follows. For k,l,m,n in N (k,l)~(m,n) if k+n=l+m. Then ~ is an equivalence relation

That’s all it says in the notes

[stray reef]

okay, so that is the construction

it should be possible to reduce associativity of multiplication on Z to the same on N once the operation * on Z is properly defined

gonna say in advance that the formal proof is very symbol-pushy and very unenlightening

[remote dock]

both are true , right?

[livid cobalt]

what's P(A)?

[remote dock]

say A = {1,2,3}

Power set of A

[livid cobalt]

yep, then both are true

[remote dock]

ty

[arctic tundra]

i proved that log(1*2*3*...*(2n-1)) is O(nlogn), i dont know where to start with proving nlogn is O(log(1*2*3*...*(2n-1))). Any help so i can start the proof?

[weary tiger]

did you proof it in induction?

let n =1

then log(1) = log(1)

assumeto n

then prove to n+1

[arctic tundra]

thank you

[tranquil vector]

There are 9 horses, I want to ride horses 30 times but I never want to ride two identical horses in a row, how any possibilities are there?

[obtuse lance]

try working out a smaller simpler example that you can write out all the possibilities to to try to figure out what the pattern is

like, 3 horses and riding 4 times

[tranquil vector]

Am I correct: in a case with 3 horses and riding 4 times: There are 16 correct posibilities (and 68 incorrect posibilities)

oh wait no is it 3^4

[obtuse lance]

yeah, 3^4 if you don't care about riding two identical horses in a row

but since we do care, the number will be smaller since we're restricting ourselves more

[tranquil vector]

3^4 - 4^3?

3^4 - (4^3-1)

[obtuse lance]

seems like you're just guessing

[tranquil vector]

something like that right

well I have 16 correct and 65 incorect (for total of 81)

[obtuse lance]

try this, write it out like a little diagram

[tranquil vector]

65 looks like 64+1 (4^3+1)

[obtuse lance]

make your horses A,B,C,D

[tranquil vector]

yeah that's what I did

[obtuse lance]

and write out what you can pick for your next choice

oh send a pic of your work

or explain it more, I can't help unless I see your reasoning

[tranquil vector]

Then I do that 4 timees

so I get 16 out of 81

[obtuse lance]

what's the black and red mean here

[tranquil vector]

red is correct

sorry incorrect

black is correct

thats for 1/4 of possiblities

you would do that 4 times

ie instead abab you do baba

I might be missing some posibilities tho

[obtuse lance]

I see

[tranquil vector]

oh

you do it 3 times

wait nvm

[obtuse lance]

I would take it one step at a time, as in

[tranquil vector]

i think i see my mistake

[obtuse lance]

how many choices do you have for the first horse?

then how many choices do you have for the second horse?

[tranquil vector]

they would be the same right

oh

ok i think i see it

its like 433*3

4 x 3 x 3 x 3

[obtuse lance]

yes exactly

[tranquil vector]

sorry no

3 x 2 x 2 x 2?

[obtuse lance]

oh right

I was thinking there were 4 horses not 3 for a moment too haha

[tranquil vector]

ok thank you

[obtuse lance]

yeah you're welcome

[valid wave]

Could someone look at this proof

Did I approach this correctly. Feel like I am missing something

Maybe should be

Should actually be this at the bottom I think

[heavy tinsel]

can anyone explain why those were chosen as the pigeonholes?

Why is it that for example, 2 and 3 dont form a pigenhole?

[coral parcel]

Because 2+3 is not a multiple of 8.

[heavy tinsel]

why do we want them to be a multiple of 8 anyway?

[coral parcel]

Because that's what we're trying to prove: that two of the numbers sum to something that's a multiple of 8.

[heavy tinsel]

I'll think about it more, thanks

[heavy tinsel]

I think I kinda get it

Still hard to put it into words, but Im convinced more

[ashen cloak]

Okay where do I go from there?

[ashen cloak]

I am just fell asleep.

[weary tiger]

Ann wrote it

the def on transitive

[ashen cloak]

I'm sort of understanding it better, just a bit confused
a|b means a divided by b correct?
b division symbol a correct?
And c would be on top?
So bc = a

[wide vine]

idk if my teacher is using a different definition of subset and subset equal than the book and I

can someone explain how this is true

The prof is saying that $A \cap C \subseteq B$ is false

[vital dew]

Brandon7716

[wide vine]

which doesn't seem to go well with the book defintion of subset

[craggy juniper]

i mean {2} and {6} aren’t even elements of Z

even though it claims they are

wouldn’t be surprised if there were more mistakes

[wide vine]

Which says if all the elements of A are in B then $A \subseteq B$

[vital dew]

Brandon7716

[wide vine]

yeah

2 and 6 are \in Z

[craggy juniper]

that symbol in B means much greater than right?

[wide vine]

not {2} , {6}

no

[craggy juniper]

oh yeah

greater than or equal to

[wide vine]

Yes

[craggy juniper]

weird looking

[pale epoch]

this is the most horrible thing i have read in a long time

[wide vine]

😦

[pale epoch]

please ask your professor what $\subseteq$ is supposed to mean

[hard citrus]

I feel like I forgot everything I ever learnt from this

[vital dew]

Lochverstärker

[wide vine]

I am kinda scared for the quiz if this is the case

[pale epoch]

when $A \subset B$ does not imply $A \subseteq B$

[vital dew]

Lochverstärker

[craggy juniper]

not like there’s a subjective meaning of subseteq

[pale epoch]

then is \subseteq just equality??

[wide vine]

this is the comment they added with 8(e)

im so confused . This conflicts with even the problem sets from thebook

[pale epoch]

your teacher is wrong

please ask them to clarify the difference between $\subseteq$ and $=$

[vital dew]

Lochverstärker

[hard citrus]

Is this a uni course?

[wide vine]

Yes

[pale epoch]

one could say that \subseteq "isn't precise enough", when actually \subset is true, but ^ shows they have no idea what they are talking about

[wide vine]

even has a Ph.D but I guess their main field was stats / teaching

:v

[hard citrus]

Regardless of the field, this is basic set stuff

[wide vine]

:/ yeah

[vivid bison]

hi can i ask question related to binary strings in this channel or is it wrong channel

[wide vine]

im not really sure what to ask them

[pale epoch]

ask them to clarify the difference between \subset, \subseteq and =

more specifically ask how $X \subset Y$ does not imply $X \subseteq Y$

[vital dew]

Lochverstärker

[wide vine]

IT would be like saying 8<=8 is false and that it is really 8=8 or that 7 <=8 is wrong it is is really only 7<8

[pale epoch]

indeed

[wide vine]

Funny thing is, this was an exacty question from the book

[pale epoch]

do you have a provided answer?

[wide vine]

Yeah

Let me find it

[pale epoch]

great

[wide vine]

but this also feeds back on their definitions which was

If every element in A is also an element of B, then A is a subset of B, denoted as A ⊆ B.

[pale epoch]

quote this and ask how $A \cap C \subseteq B$ is false but $A\cap C \subset B$ is true

[vital dew]

Lochverstärker

[obtuse lance]

that's awkward

it's just c^n right

[tranquil vector]

yeah

[obtuse lance]

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

[vital dew]

Merosity

[obtuse lance]

just a silly example of the binomial theorem

but I assume you're wanting to count it literally how they do with those terms and numbers

[tranquil vector]

we just have to prove c^n is equal to that weird summation

i dont understand what ur middle step is tho

[obtuse lance]

right, I just proved it by using the binomial expansion

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

[vital dew]

Merosity

[obtuse lance]

x=1, y=c-1

[tranquil vector]

ah I see

thank you

[obtuse lance]

you're welcome

[wide vine]

can someone explain this comment to me

isnt that what they did?

other than forgetting the 2

[coral parcel]

Hmm, you might need to explain the comment first. What are we looking at? A set of model solutions with comments telling the grader which features to look for in actual handed-in answers? Who wrote which parts of the screenshot, and in which order?

[wide vine]

sorry for late reply. This is a graded solution to someones HW with the the profs commetting on it

the comments are in RED