discrete-math

[coral parcel]

Who put the checkmark there?

[wide vine]

the teacher I guess

I guess she believed the omitted 2 wasnt such a big deal

Im also confused on my teachers emails and really don't know what more to say

[coral parcel]

I'm a bit confused about what's confusing you, then. It sounded vaguely like you thought the comment in red was a criticism of the answer, but it looks to me like it's explaining how the answer is right.

[wide vine]

Yeah I wasn't sure if it was a comment or not

[sharp shell]

How do you solve 500n^2 < 2^n

[silent hinge]

hello friends is this a right translation

for number one \

[coral parcel]

How do you solve 500n^2 < 2^n
I iterate n --> ceil(log2(500n²)) until I reach a fixed point. It happens fairly quickly.

[olive condor]

Nope, maybe it helps if you read “everybody is an obnoxious elf” as “for all character x in a fantasy world, x is an elf and x is obnoxious”

And for number two, don’t omit the parentheses around the implication. It means two different things with and without parentheses in this case

[silent hinge]

Thanks man

[civic cedar]

Can someone help me? Not sure how to proceed from here

[floral abyss]

How would I go about proving the following
Suppose a,b,c are in R
Prove that a(b+c)=ab+ac

[stray reef]

you need a construction of R

because sometimes R is constructed axiomatically and this is just one of the axioms

otherwise it's probably constructed via cauchy sequences of rationals in which case the distributive law in R reduces to the same in Q

[floral abyss]

I think we’re constructing it from Q

[stray reef]

you are constructing it from Q but how exactly

dedekind cuts?

[floral abyss]

Honestly no idea they never said anything about that

[stray reef]

.......

[floral abyss]

I’m assuming the simplest way to construct it whatever that is

[stray reef]

again.

no construction

no proof

you need notes from your class

that say how R is constructed formally by your professor

because if you use a different construction than your professor, then your proof will be pointless

is this for an assignment?

[floral abyss]

It’s just for homework

Doesn’t count so I guess I’ll just go to his office hours and ask him

[stray reef]

a homework assignment

can you share it in its entirety

i have a feeling that there might be something else in there that helps clarify what's going on

[floral abyss]

There’s like 900 pages lol

Idk how I’d share that

I think it’s something to do with Cauchy sequences and converges but honestly not sure

[stray reef]

how does ONE homework assignment take up nine hundred pages

[floral abyss]

Wait no that’s the notes for the whole course

The questions are inside the notes

[stray reef]

well i'm asking you to share the questions that were given to you as homework

[floral abyss]

They’re scattered throughout the notes

I mean I can screenshot like 20 pages lol but idk if that’ll help

[stray reef]

well fine then i guess i'm not going to get any clarification

[pale epoch]

you already gave the correct answer, ann

which is "figure out what a real numbers is and how to add an multiply them"

[stray reef]

did you word it that way on purpose

[pale epoch]

i wanted to sum up what op is supposed to do and possibly come back with

in case they are still reading

[stray reef]

i mean "real numbers" instead of "real number"

[pale epoch]

i just cant type

[stray reef]

ah

[shadow geyser]

Hello!

Can anyone explain me understand how this was to be manipulated?

i'm just trying to understand why the second term was broken up into two of the same terms

[livid cobalt]

p + p <=> p

[shadow geyser]

what does <=> mean?

[livid cobalt]

logically equivalent

not sure how that's useful in that situation though 🤔

[shadow geyser]

not sure why that was still broken up though

you could just pull out the NotX in the first line

like NotX (NotY + Y) + XY

so that NotX + XY leading to (NotX + X)(NotX + Y) which cancels to NotX + Y

Still looking for help if anyone can btw ^

[pearl apex]

Hello guys

I had a question pls

how can those be symmetric ?

like for example if I have the pair (1,2)

how can this be symmetric ?

how is 1 = 2 !?

or am I assume that if 1 = 2 then 2 = 1 aswell ?

[floral abyss]

@stray reef prof said R is the completion of Q in which there are 2 possibilities of constructing R

1. Dedekind
2. Cauchy: in which every Cauchy sequence converges
   (with respect to | | )

We’re using the 2nd definition to answer the question

[cerulean wind]

(1,2) is only in R_6 for the symmetric relations

[pearl apex]

so how is R_3 and R_4 symmetric ?

does he mean like if (1,-1) works in R_3 then (-1,1) should work aswell?

and in R_4, if (a=2,b=2) work then (b=2,a=2) should work aswell ?

one more thing cuz it confuses me

here

why R1 is antisymmetric ?

[worthy mist]

Well does it fulfill the definition of antisymmetry?

[hard citrus]

$\bar{X}\bar{Y}+\bar{X}Y+XY = \bar{X}(\bar{Y}+Y)+XY = \bar{X}+XY = (\bar{X}+X)(\bar{X}+Y) = \bar{X}+Y$

[vital dew]

.itsjustnai

[hard citrus]

for the first one

[shadow geyser]

i understand that

but my TA wrote what i sent instead

im just trying to understand why that was written

[hard citrus]

Hold on, let me reread what you wrote

Oh

Basically what nerayo said

[livid cobalt]

if that's a minimization exercise, expanding like that seems completely useless

but it's still valid

[hard citrus]

When you have
x+x it is the same as having just x

[shadow geyser]

hm

okay

does that apply for any case within boolean algebra?

you can break up one thing into two of itself

[hard citrus]

I believe so

If you corelate it to logic, it be

$p \equiv p \lor p$

[vital dew]

.itsjustnai

[livid cobalt]

yes, for any proposition p, p + p = p
it's called the idempotent law

[shadow geyser]

idempotent law

i se

thank you

[pearl apex]

yes but how can it be symmetric and antisymmetric at the same tijme

[worthy mist]

it isnt? Your previous slide even says that R_1 is not symmetric

[lone raft]

We know (-1)^n gives us the sequence -1, 1, -1, 1,.... Is it possible to find something similar for 1, 1, -1, 1, 1, -1. Two positive then negative etc?

[pearl apex]

yea sry i meant R_4

I think I kinda understood it

its cuz (2,2) is symmetric cuz (2,2) would work aswel

and (2,2) in R_4 is anti cuz (2,2) and (2,2) works but ( 2 = 2 )

then it is isymmetric

[livid cobalt]

it exists on oeis, so i guess

[pale epoch]

something similar?
$$a_n = \begin{cases}
-1 &\text{ if $n \equiv 0 \pmod{3}$, }\
1 &\text{ otherwise.}
\end{cases}$$

[vital dew]

Lochverstärker

[stray reef]

well you can do $(-1)^{n(n+1)/2 + 1}$ i think

[vital dew]

Ann

[pearl apex]

is 2 a rational number in discrete mathematics ?

and is 1/3 and 1/4 rational numbers ?

[stray reef]

a rational number in discrete mathematics as opposed to a rational number in what?

do you think the answer to the question "is 2 a rational number?" depends on the phase of the moon or something?

if you know what a rational number is (i.e. that it is a number expressible as the ratio of two integers, the denominator being nonzero) then you should be able to answer yourself whether 2, 1/3 and 1/4 are rational.

[tidal tulip]

if i want to prove “1 divides any even number” would this be valid: the definition of even is n=2k, k in Z and the definition of a divides b is b=am for some integers b,a,m. from the definitions we see that 2k=1m, hence we see that 1 divides any even integer

[stray reef]

sure, though it's unnecessarily long

1 divides any integer, and in particular it divides any even integer

[tidal tulip]

how do you prove it divides any integer. like n=1k for some int k?

[stray reef]

n = 1*n

or if you really insist on introducing another variable, n = 1*k for some integer k, specifically for k = n.

[tidal tulip]

k, what you said makes sense

thx

[limpid fossil]

for a complete binary tree with 2^k - 1 nodes and no duplicate keys, how can i show the worst case number of nodes that must be examined to search for a node is the number of nodes?

[pale epoch]

how does the search work?

[limpid fossil]

the pseudo code is like this

[pale epoch]

then it's not true? you discard half the (sub)tree at every call of that function

[limpid fossil]

oh.. does complete binary tree mean every node has two children

[pale epoch]

not necessarily

if you have an odd number of nodes, then not
also leaves do not have any children

otherwise yes

but thats just what binary tree means

complete means that every layer except the last is filled completely

and the last layer is filled from left to right so to say

[limpid fossil]

so.. the worse case would be the height?

[pale epoch]

yes

[limpid fossil]

but how do we prove that is true?

[pale epoch]

induction on the height works

[limpid fossil]

for the sake of the question, we were given 2^k - 1 nodes

so can we use the property that the height is log_2(# nodes)

[pale epoch]

sure but the height will still depend on k

[limpid fossil]

how would the inductive step work? for the base case on a height of one we only have to check the root, and that is height 1

[pale epoch]

you check the root and the immediately move into a subtree that has lower height

thats basically it

[limpid fossil]

so is the predicate forall n in N, if T is a complete binary tree with height n then the worst case nodes examined is height n?

[coral parcel]

Phrasing it as induction might be making it more cumbersome for yourself than it needs to be -- especially if you can get away with saying something like "each recursive call is for a node on a lower level, so we'll examine at most one node at each level, so in total we'll examine at most k nodes".

[limpid fossil]

right, but for the sake of this problem im trying to construct a formal proof and it sounds like it would be done by induction

[coral parcel]

"Formal proof" is a technical term from proof theory, and I somewhat doubt that technical concept is being used in the same breath as a property as intricate as binary search trees.

I believe even the definitions necessary for even stating the property you're proving in a completely formal notation would run to several pages.

[tidal tulip]

Let $P(f)$ be the following statement:
$\forall \epsilon > 0 , \exists \delta>0$ such that $\forall x \in (2-\delta, 2 + \delta)$, we have $L - \epsilon < f(x)$ and $f(x)< L + \epsilon$.

[vital dew]

strings

[tidal tulip]

Is the negation of $P(f)$: $\exists \epsilon \leq 0$ such that $\forall \delta \leq 0 $, $\exists x \in (2-\delta, 2 + \delta)$, we have we have $L - \epsilon \geq f(x)$ or $f(x)\geq L + \epsilon$?

[vital dew]

strings

[tidal tulip]

Is the negation of $P(f)$: $\exists \epsilon \leq 0$ such that $\forall \delta \leq 0 $, $\exists x \in (2-\delta, 2 + \delta)$, we have $L - \epsilon \geq f(x)$ or $f(x)\geq L + \epsilon$?

[vital dew]

strings

[ember obsidian]

dont blindly negate every symbol

$\forall\ep>0$ is shorthand for 'for all $\ep$ in the set of positive reals'

[vital dew]

RokabeJintaro

[tidal tulip]

ok

then $\exists \epsilon \leq 0$ means for all $\epsilon$ in the set of the negative reals or 0?

[vital dew]

strings

[ember obsidian]

u wrote exists

[tidal tulip]

oops

$\exists \epsilon \leq 0$ means there exist an $\epsilon$ which is either 0, or in the set of negative reals?

[vital dew]

strings

[ember obsidian]

yes, u can also say nonpositive reals

[tidal tulip]

ok

does the rest of that negation look ok

[ember obsidian]

my point is u dont blindly negate every symbol u see

negating a quantifier is just swapping forall by exists & vice versa

[tidal tulip]

yeah

[ember obsidian]

so 'for all ep in the set of positive reals' negates to 'exists ep in the set of positive reals'

or exists ep>0 for short

[tidal tulip]

i am confused, isn't $\forall \epsilon > 0$ negated: $\exists \epsilon \leq 0$

[vital dew]

strings

[tidal tulip]

i want to see whether P(f) negated is: $P(f)$: $\exists \epsilon \leq 0$ such that $\forall \delta \leq 0 $, $\exists x \in (2-\delta, 2 + \delta)$, we have $L - \epsilon \geq f(x)$ or $f(x)\geq L + \epsilon$

[vital dew]

strings

[ember obsidian]

dont blindly negate every symbol
'forall ep>0' is shorthand for 'for all ep in the set of positive reals'
negating a quantifier is just swapping forall by exists & vice versa
so 'for all ep in the set of positive reals' negates to 'exists ep in the set of positive reals'
or exists ep>0 for short

[tidal tulip]

what don't you negate the inequality there

[ember obsidian]

'forall ep>0' is shorthand for 'for all ep in the set of positive reals'
this is the key to the misconception

[tidal tulip]

ah, i see

let me re-work it then

[ember obsidian]

just having > does NOT make it a statement

its just a way to lazily write a quantifier

[tidal tulip]

ok i see

when you negate things you shouldnt negate the domain the elements reside in?

and thats what i did

incorrectly yeah?

[ember obsidian]

you basically replaced 'in' by 'not in'

[tidal tulip]

ok, that makes sense

[ember obsidian]

but the only replacement should be 'forall' to 'exists'

[tidal tulip]

i'll re-tex it

how is this

Negation of $P(f)$: $\exists \epsilon > 0$ such that $\forall \delta > 0 $, $\exists x \in (2-\delta, 2 + \delta)$, such that $L - \epsilon \geq f(x)$ or $f(x)\geq L + \epsilon$

[vital dew]

strings

[ember obsidian]

yes

[tidal tulip]

ok, what you said makes sense

[ember obsidian]

yeah so in the future look out for symbols used to denote the set a variable is in

[tidal tulip]

ok, good call. i'll do that

[floral field]

So, I played around with this a little bit, and I noticed that we are pretty much permuting the elements of [n] in each case. So, my claim is that there are n! ways to place n rooks. I think this is reasonable, so I was wondering if someone could point me towards a good direction of how I could argue this.

I feel like my instinct is to try induction, but I feel like there's a clever counting argument for this

[obtuse lance]

the way I think of it is when you place a rook on an n by n board, you have to remove the row and column that it's in, and you're left with an n-1 by n-1 board. There's n ways to place it in the first row, then n-1 for the second row, etc... so n! ways to place them, seems like what you're saying is fine

[floral field]

Would this be sufficient to prove my claim?

[obtuse lance]

it's sufficient to me, idk

[weary tiger]

Is this correct? Besides the typo

[tidal tulip]

@ember obsidian in regards to the negation problem yesterday it asked to show a graph that shows P(f) is false. is L here a constant or are they using L as a limit? assuming it’s a limit my attempt was lim x->2 f(x)=2 and f(x)=L+1=2+1 with a graph like this, does that work?

or no because f(x)=2 except when x=2, f(x)=3

and if not how do i fix it

[ember obsidian]

L is any constant

this isnt a limit. a limit at 2 further requires x!=2 in the x quantifier

[tidal tulip]

so f(x)=L+1 works

what do you mean a limit at 2 requires x!=2 in the x quantifier?

let L=2, then f(x)=L+1 would break the inequality making P(f) false

i got super confused thinking this was about a limit

if it isn’t about a limit then i won’t worry about it further too many other problems to do.

but now i’m curious. would this statement be true for limits?

[ember obsidian]

the def of limit at 2 instead has $\forall x\in(2-\delta,2+\delta)-\brc2$

[vital dew]

RokabeJintaro

[tidal tulip]

ok

so for this problem

f(x)=L+1 works?

f(x)=3

L=2

and graph f(x)=3

[weary tiger]

I have a question about this negation here. The statement we're trying to negative is $\forall \epsilon \in S, \exists \delta \in S, \forall x \in I, |f(x) - L| < \epsilon$.

[vital dew]

n/c

[weary tiger]

The negation would then be $\exists \epsilon \in S, \forall \delta \in S, \exists x \in I, |f(x) - L| \ge \epsilon$?

I didn't read their negation yet

But is this how you would do it?

[vital dew]

n/c

[tidal tulip]

i write my negation above i don’t have a lot of time to tex it got like 10 mins free but scroll up a bit

wrote

sorry super busy

[weary tiger]

Oh cool this

[tidal tulip]

i did it wrong initially

then yeah

that’s it

[weary tiger]

Okay

I think mine matches yours

Sorry for interrupting just wanted to practice lol

[tidal tulip]

np! glad you asked

[ember obsidian]

sure

[hearty elbow]

Why is it that when you

have an even number and you mod it by an even number

it gives you an even number

2k (mod 2n) = 2x

[pale epoch]

even number - even number = even number

[weary tiger]

Can we hide discrete in primes series ? Like hiding a markov chain in a prime series ?
I think it's a safe bet.
I think its a safe bet to say you can hide P(universe)!=1 in P(universe)=1

At least P(universe)!=1 is max n-1
How i know that ? Because to do retro analisys we used markov chain property. It don't think it's safe to use them to predict. But to retro analyze stuff, it's pretty safe actually.

I have others bets to talk about the wheeler experiment. First i will wait some feedback from you all.

[weary tiger]

ZFC is consistent
It means that you can say everything AND its opposite
ZFC is reducible to a number
This number is the number of space units available in ZFC

So if a model uses all the space units available in ZFC then although it cannot be proved, this number is contradictory.

What about QM ?

[olive condor]

3^2 - 1 = 8 = even? So when x = 3, R(x) => Q(x) is true

But this alone is not sufficient to prove the theorem

Since you have only proven it’s true when x = 3, while you need to prove it for all x

[olive condor]

Since the theorem claims that it’s true for k >= 2, you can “disprove” it with a single counter-example

You can show that when k = 2, the equality doesn’t hold

[weary tiger]

do u just plug in

k=2

thats it?

what did they sub k-2 for?

[olive condor]

Ya, just plug it in, do something like:

[olive condor]

Huh? k-2?

[weary tiger]

i will show you

@olive condor

[olive condor]

It seems overkill to me. This is to prove that for all k >= 2, k^2+k < 3k is false

[weary tiger]

hmmmmmm

[olive condor]

But one case of a counter-example is sufficient to show that the theorem is incorrect

Doesn’t it make sense? If the theorem claims that it’s true for all k >= 2, but you found out that it’s false for some value k = 2, then the theorem is false

Since it’s no longer true for all k >= 2

[olive condor]

But if your instructor insists you to prove this ⬆️, then maybe you should stick to it

Does it make sense to you?

[weary tiger]

yeah

what you're saying is does make sense

[weary tiger]

but whats

I don't get how the instructor went from l(k-2)>0

to saying well this means it's a contradiction

I dont get how is k^2+k >3k

[olive condor]

It adds 3k to both sides

[weary tiger]

oh wait so

ok now thats so clear

thank u so much

but I have a small quesiton

when we you move the 3k to other side

in step 1

shouldn't be elss 0

since its k^2+k- <3k

so k^2-2k<0 no?

like

why did we flip the inequality

[olive condor]

Wait, hold on, what are you saying?

So you’re asking why adding 3k to both sides doesn’t flip the inequality?

Erm, because adding and subtracting will never flip the inequality

You flip the inequality only when you are multiplying or dividing a negative number

It makes sense because the inequality will still hold after the operations:

[weary tiger]

no no I'm saying

[weary tiger]

I'm saying from

going from k^2+k<3k to k^2+k-3k>0

[sturdy walrus]

Hi I am having a hard time understanding this proposition. Could someone please take a moment and help me understand this?

[olive condor]

No, no, if you read the proof, there is no k^2+k<3k, only k(k-2)>=0

And you are going to prove or “disprove” k^2+k<3k, which is the goal of your proof

[olive condor]

If (x-1) is a factor of p(x), you can write p(x) = (x-1) * q(x). So when x=1, p(1) = 0*q(x) = 0. Since p(x) = ax^n + …, p(1) = a_n + a_(n-1) + … + a_0. Thus, a_n + a_(n-1) + … + a_0 = 0.

[weary tiger]

@olive condor

[olive condor]

Ya? So that’s the goal, you want to prove it to be true

But you are only given k is an integer and k >= 2

So a proof is a series of steps going from the givens to the goal

[weary tiger]

Ok

You moved 3k to the other side

3so it becomes -2k

K^2-2k<0?

But it says >0

[olive condor]

Wait, it seems that you are very confused. Read your given “proof” again.

So, you are only given k >= 2. From k >= 2, you get k(k-2) >= 0. Then, by adding 3k to both sides, you get k^2 + k >= 3k, which is the total opposite of your goal. So you conclude that the theorem is incorrect.

I won’t say it’s a contradiction 🤷‍♂️

[weary tiger]

What is non-attacking roots ?

[floral field]

So, if I have two rooks on the same column, they would attack eachother (regardless of side)

[weary tiger]

It's a markov chain property

[weary tiger]

So i ask my question again

I have some surface to say everything AND its opposite.
This surface is strictly divisible by n entities.
So I have a number of size n.
Can this number be said contradictory ?
Can we prove by this number that it is contradictory ?

[pale epoch]

surface?

[coral parcel]

What do you mean by a number being "contradictory"?

[versed sigil]

Why doesn't transitivity apply to d and e? Shouldn't they be pointing to g?

[stray reef]

do you have the original problem statement

[versed sigil]

Literally draw the directed graph from the hasse diagram

[stray reef]

right, so you're asked to draw all arrows for the order relation...

then yeah, d->g and e->g are missing.

[versed sigil]

Okay phew I thought I was losing my mind

[storm drum]

Hi, is the following phrase true?
if X⊆(A-B) ⇒ X⊆A∧X⊈B
in words: If X is a subset of the difference between A and B so , X is a subset of A and X is not a subset of B

[stray reef]

no, this is not true

[storm drum]

can you give me an example how to disprove it?

[stray reef]

the simplest counterexample would be X = empty set

[storm drum]

i am trying to understand the logic, i tried the empty set and ye it disproved it, but i don't really get it why its not true

[stray reef]

you have literally been presented with a counterexample

[storm drum]

ye ik but except of showing that its not true, why isn't it true

the idea behind

[stray reef]

???????????

"i know this shows why it isn't true, but why isn't it true?"

do you hear yourself

[storm drum]

ye i do, and i am trying to figure out why my intuition told me it is true

while it isn't

[stray reef]

you only brought up intuition now.

and honestly, there could be any number of reasons why your intuition misled you.

[storm drum]

Thanks

[heavy tinsel]

From practice test

[heavy tinsel]

I cant think of another way of solving this without brute forcing it, like dividing this into three cases: 1) H is at the end, 2) I is at the end, 3) E is at the end. Then in each of these cases, "slide" the P or G around to make it the third consonent

But that would take too long

[proven silo]

Start with fixing placement of 3rd letter, then fix placement of the rightmost letter then count ways to arrange the O’s

[heavy tinsel]

so like the "opposite" steps of what ive done?

[proven silo]

The idea is you fix the most “restrictive” thing first in these types of questions

Only 2 ways to have 3rd letter correct so start with that

[proven silo]

Just by following this it also prevents double counting in most questions

[heavy tinsel]

ok, i'll give that a try

But just to make sure I understand, there would be two partitions: 1) P is the third consonant, 2) G is the third consonant. Then there are sub-cases for the third condition. Is that correct?

[proven silo]

Right so ways to place 3rd letter is 2, so we have 2 * …

As the answer

[nova star]

T(n) = T(n-1) + T(n-2) +n

is this relation similar to lucas numbers?

the last +n is through me off

[stray reef]

similar in what sense?

[gritty agate]

Would anyone know how to simply this? The answer is supposed to be the complement of B. I simplified each side on its own and ended up with:

[nova star]

idk i just have to find upper and lower bounds but im not sure how to do it

[stray reef]

oh you've multiposted...

[craggy juniper]

multiposting can result in someone spending a lot of time helping you without them knowing that you’ve already solved the problem

so basically that person wasted their time

[heavy tinsel]

Can anyone give some general advice on proving identities by combinatorial arguments? Im kind of stuggling on this type of proof

Simpler ones I can do, but ones like these take a while

[stray reef]

nC3 is the number of ways to choose 3 numbers out of n

try looking at this in a different way

imagine that the numbers are sitting in a row in increasing order

[heavy tinsel]

ok, so 1, 2, ..., k-1, k, k+1, ..., n. So then (k-1) essentially takes 1 number before k, and n-k essentially takes 1 number after k. So we still take 3 numbers out of n.

What about the sigma part?

[weary tiger]

Space

[pale epoch]

that is even less descriptive

[weary tiger]

Can not be true at the same time.

[coral parcel]

What does it mean for a number to "be true at the same time"? Or even just "be true"?

[weary tiger]

Any metrics.

[stray reef]

congratulations @weary tiger you've somehow managed to be even more and more incomprehensible with every message

[stray reef]

that sums over all possible values of k, the middle number

[weary tiger]

I have the answer

[pale epoch]

glad we could help then

[weary tiger]

The point here is to compare what is in my mind is the same in yours about this problème.

I'm not a mathematician

[stray reef]

nobody has any idea what you're talking about, malebranche.

[pale epoch]

well, you use words that have some mathematical meaning (or at least connotations) but they dont really fit together

so you will have to be a lot more precise for anyone to even guess what you mean

[weary tiger]

So What is the meaning of 'markov chain' ?

[heavy tinsel]

ahh right that brings it together

its hard to piece that together sometimes

I guess i just need more practice

[pale epoch]

there is a lot of stuff going into the definition

[weary tiger]

Good thing.
I'm a specialist, an expert in définition

[pale epoch]

ok?

i don't think i am able to help you, so i am just going to step away

[weary tiger]

I will give u the result now. This for you to be able to tell me with maths, if it's the good result.

Some are saying the number is undefine.

There is 2 cases
P(universe)=1
P(universe)!=1

When it's well defined there is no doubt.

The space unit is the space needed to compare their lenght.
P(universe)!=1 is at max 'n-1'

I will stay here and wait for an answer.

Go on talking your stuffs guys.
It will take time to get my answer.

[soft bobcat]

Hey, I need a little bit of help getting to an answer on this homework assignment, I asked in the help channel but there's not much luck,

[weary tiger]

what did you try

and post a pic without the pencil in top left corner

[soft bobcat]

I don't even know where to start. I know I need to prove that R is antisymmetric, which isn't much of a problem. I'm just not sure what to start with this proof

[soft bobcat]

I just removed that because it has my real name

[weary tiger]

ok ok

Can you first state what it means for a relation to be antisymmetric?

[soft bobcat]

When a relation is antisymmetric, it the definition (in my eyes) is that if you have the point (a,b) and (a,b) is in a relation r, and (b,a) is in the relation r, then a = b

[weary tiger]

yeah

so if (a,b)R(x,y) and (x,y)R(a,b) then (a,b) = (x,y)

So you have kinda 2 choices

[soft bobcat]

Yep! That's the part that I understand, I'm just not sure how to define that relation

[weary tiger]

Because for elements in relation R you either have that a<x or the second thing

it cant be the first property of the relation

because that would imply a<x and x<a

right?

[soft bobcat]

Correct.

[weary tiger]

Okay so you need to look at the second condition

which means: a=x and b<=y and x=a and y<=b

so now I guess its easy to see (a,b)=(x,y)

[soft bobcat]

And that condition works perfectly, since the or condition only requires one of the conditions to be true

[weary tiger]

I also think its often beneficial to draw the relation on the plane in this case

for some numbers

to understand it more

[soft bobcat]

What do you mean?

[weary tiger]

like for example why (1,2)R(7,2)

or (7,7)R(7,123)

on R^2 plane

so if the first coordinate is smaller than the first coordinate of the other point, they are in the relation

but if you had lest say numbers (7,2) and (1,2) then its not true that (7,2)R(1,2)

and if the first coordinates are the same then you compare the second coordinate like normal numbers

this is the definition of this relation

[soft bobcat]

Alright, I understand. But I wouldn't be too sure how to go about drawing that

[weary tiger]

I guess can't really 'draw' it, just pick some random numbers and see if they are in the relation

for example, point (0,0) is in relation with all the numbers on the right of that point

but not vice versa

thats why its antisymmetric

[limpid fossil]

for a max heap, storing consecutive numbers from 1 to n, if you know the value of the second node, what is the possible values n can take?

[soft bobcat]

I see, I see. That does make sense. But how would I start a direct proof on this?

[limpid fossil]

i know it must be atleast 1 + the value of the node

but i dont know whawt it at most can be.

[weary tiger]

Well what I said at first is pretty much the proof: you assume aRb and bRa. From this you conclude the first coordinate has to be the same

Then you conclude the 2nd coordinate has to be the same

End of proof

[soft bobcat]

It's that simple?

[weary tiger]

Yeah

I guess fill in the details why you can conclude those things

like b <= y and y<=b implies b=y

[soft bobcat]

Alright great! I'm going to start writing this all out, can I tag you if I have any more questions?

[weary tiger]

sure idc

but might be not responsive soon

[soft bobcat]

That's fine, I'm much better off than I was

Thank you so much

[weary tiger]

All good, relations seemed pretty hard for me when I first learned them, its much easier when you first try to understand how it works (i.e. what numbers are in relation with each other)

(on a side note, this is often called 'an alphabetical order' for an obvious reason)

[soft bobcat]

Oh wait, you said assume that aRb and that bRa, could you go into more detail of that in the context of this question?

[weary tiger]

im just using definition of antisymmetric relation

a and b being points on R^2

[coral parcel]

It is more common to see this order called "lexicographic". Apparently "alphabetical" is too obvious a name. :p

[soft bobcat]

So is "a" a point on R^2 or is it just the x value of another point?

[coral parcel]

The a in the definition of "antisymmetric" is an entire point of R^2.

It makes it somewhat confusing that the a in the definition of your particular R is just one half of that point.

[soft bobcat]

Gotcha, so I need to assume that (a,b)R(x,y) and that (x,y)R(a,b) and then go from there?

[coral parcel]

Excactly.

[soft bobcat]

Gotcha, gotcha. This stuff is complicated lol

[weary tiger]

Wich one ? This point is half of sinus !

[soft bobcat]

Can I have an opinion on the proof I wrote?

@weary tiger

[weary tiger]

Its pretty wrong

Not 'we will use 2nd condition' - we dont have choice, its forced because you cannot have a<x and x<a

And later on I dont see why you can conclude such things - you didnt use the fact you assume two relations hold

'This shows R is anti...' no you didnt show it yet

[soft bobcat]

So what can I say to assume that fact?

[weary tiger]

youre ussing only the fact that xRy not yRx

You concluded b<=y rightuflly, but I dont see how you concluded the other direction

'therefore (x,y)R(a,b)...' you knew that at the start (thats what youre assuming)

[soft bobcat]

I see, I see. Let me try again. Is that all that's wrong?

[weary tiger]

https://electronics.stackexchange.com/questions/456257/ltspice-antisymmetric-sine-pulse

Just to show i'm a real expert

[soft bobcat]

This one should be much better

[dark geode]

i can’t really parse this

can anyone help?

[weary tiger]

what is your question?

how to solve it

or what is the right approach

[dark geode]

!!

and if i further need help

i can hopefully ask for that

hwo would i go about it

[weary tiger]

first thing that pops to my head, induction

strong induction thou

[weary tiger]

Uniform numbers or repunits are composed only of the digit 1 (concatenation of the digits 1).

Their multiples, all made of the same digit, are the repdigits.

In base decimal, a repunit is: 999... / 9 = (10k - 1) / (10 - 1)

In base b, a generalized repunit is worth: (bk - 1) / (b - 1)

n-1

Bye all

[dark geode]

??

wait

could u explain

[coral parcel]

I don't think it was intended as a reply to you.

[weary tiger]

you can include P(univers)!=1 into P(universe)=1

You can use the space unit of p(u)=1 to catch the period of p(u)!=1

You can hide any markov chain into primes.

Repunit is an example of that.

[dark geode]

oh

[heavy tinsel]

why is the 600 here the pigeonholes? Isnt that number just the sum of the elements of the largest subset? Like 10 ppl who are 60?

[distant bobcat]

If I have a weighted directed graph with positive weights how do calculate the probability for moving from one vertex to the next? This is in relation to random walks on a graph. I found the formula for directed graphs where its 1/deg(v_J), where v_j is vertex and 0 else, but not for weighted (positive) directed graphs.

nevermind I found it

[coral parcel]

Yes, so there are at most 600 different possible sums any nonempty subset can have.

One of these sums is 600 itself. There are 599 other possibilities.

[heavy tinsel]

yeah, that just clicked for me, we can construct a bijection using the possible sums of the non empty subsets and the numbers 1 - 600.

thanks for confirming it

[coral parcel]

Well, the possible sums for nonempty subsets are the numbers 1 through 600.

[heavy tinsel]

yeah... that makes sense.

[valid wave]

Couldn’t I say sqrt 3 is irrational so there exist no integers that makes this true

[obtuse lance]

is the stuff written related to the problem being asked or you just are using it as a white paper to write on

[valid wave]

Yes I’m solving num 4

WAIT OOPA

no I’m solving

[obtuse lance]

I know I saw you on that other problem the other day, I thought you were getting incredibly creative with your efforts haha

[weary tiger]

haha

lol

[valid wave]

I’m thinking because z/sqrt 3

This means no integer can make this

[weary tiger]

okay, try to assume by contradiction

[obtuse lance]

I don't think you can bring in sqrt(3) like that

I'd try reducing mod 3, one way is probably the method of descent

[valid wave]

Why mod 3

[coral parcel]

You've just concluded that a²+b² == 0 (mod 3)

[obtuse lance]

well you know z=0 mod 3 so it's divisible by 3, substitute in z=3c and then divide through and repeat

[valid wave]

Can you tell me more of your thought process

[valid wave]

Mod 3 because it’s divisible by 3?

[coral parcel]

Yes.

Also, the only squares modulo 3 are 0 and 1, so the only way to have a²+b²==0 (mod 3) is if a² and b² are each multiples of 3.

[valid wave]

this is going over my head

so mod 3 since because everything is divisble by 3

but then

that jump of a^2+b^2 = 0

im not seeing

this whole question just had me stumped

[coral parcel]

"P | Q" is the same as "Q == 0 (mod P)".

Set P=3 and Q=a²+b².

[weary tiger]

lets assume that x or y or z not equaling to zero
x != 0
x = sqrt((z^2-3y^2)/3)
we need x to be an integer so (z^2-3y^2 )/3needs to be a power, so we have three options for each one that y = 0 and one is z = o and one is they are both integers. this is the structure, do it and I think it'll be fine

[weary tiger]

also sorry if I have English mistakes, I learn math in another language

[obtuse lance]

when you look at x^2+y^2=0 mod 3, it helps to know what the squares are mod 3. So just square everything, 0^2=0, 1^2=1, and 2^2=1. In other words, we're looking for ways to make 0 mod 3 by adding only 0 or 1 together since 2 is not a square mod 3. we have 0+0=0, 0+1=1 and 1+1=2 so we're stuck to assuming x^2=0 and y^2=0 mod 3

so that's where I plug in x=3a and y=3b and then simplify my equation down to 3(a^2+b^2)=c^2 and we're back where we started but a,b,c are all a third of what they were. that's what infinite descent is about, cause we could keep repeating this forever with a solution. This means the only possibility is x=y=z=0

[valid wave]

im somewhat following

"what the squares are mod 3"

can you elaborate on this portion

[coral parcel]

(clarifying question here: are you familiar with modular arithmetic already, or just trying to follow along by looking up definitions?)

[valid wave]

somewhat familiar

not to the point that I can do proofs with it though

[valid wave]

i understand the first section after rereading

but

we have 0+0=0, 0+1=1 and 1+1=2 so we're stuck to assuming x^2=0 and y^2=0 mod 3

how are we stuck to make this assumptions

0+0=0, 0+1=1 and 1+1=2

are you saying these are the possible options

mod3

in our case

[coral parcel]

Yes, those are the only ways to add two numbers that are each either 0 or 1.

(Unless you count 1+0=1 as a fourth option, but that doesn't change the conclusion).

[valid wave]

hm interesting

but

"we're stuck to assuming x^2=0 and y^2=0 mod 3"

how are we stuck to assume this

and thanks for all the help

[valid wave]

when we have other "options" if im understanding correctly

waitttt

i think its clicking

we have 0+0=0, 0+1=1 and 1+1=2 so we're stuck to assuming x^2=0 and y^2=0 mod 3

this is like the logic table of all the possible options for us to end up with 0 mod 3

wouldnt we end this here

why is

this second paragraph needed

also it seems like a weird formal proof to just basically say that we are dealing with mod 3

in these possible casses mod 3 = 0 doesnt happen

thus statement true

[obtuse lance]

well we went from 3(x^2+y^2)=z^2 down to 3(a^2+b^2)=c^2

we assumed we had a solution, then showed that if we divided everything by 3, we'd get another solution

and we can do this infinitely

but dividing an integer other than 0 by 3 arbitrarily many times is going to end up giving us something that's not an integer anymore, which is the interesting part

[hearty elbow]

If I want to show equation A is equal to equation B

do I need to show how I can go from A to B

and from B to A?

or just A to B?

[dire obsidian]

Did I mess up anywhere?

[coral parcel]

If you want to show they're equivalent (that is, satisfied by exactly the same numbers) then you need both directions.

[coral parcel]

The beginning and end are equivalent all right. Is there anything in the middle you doubt in particular?

[dire obsidian]

I'm kind of new to logic laws

idk if i misapplied any of them

@coral parcel

[manic zinc]

Is this accurate big o by definition ?

For c*gx I picked 12

<@&286206848099549185>

[worldly knot]

need help with proofs #help-27

[candid harbor]

uwuquote 938564076534136920

oh rip

lemme post a message link instead

@rich fossil from #help-14 message

is there a formal way to prove
What did you not find formal enough about this proof?
Consider n = 2a books in a library, of which one is your favorite. The left sum represents the number of ways to choose an even number of books, the right sum represents the number of ways to choose an odd number of books. Make unique pairs between an even number of books and an odd number of books by unselecting your favorite book if you have selected your favorite book and vice versa. By bijection this means the right sum equals the left sum. But since these sums added together gives 2^n, we have proven the identity.

[muted patio]

sorta. I think you need an index capital N for which f > g for all n > N

[dire obsidian]

idk anymore

[coral parcel]

Send De Morgan to rehab, it doesn't work well on drugs. (You forgot to flip OR to AND in the last conjunct).

[dire obsidian]

Can you give me the line number? I'm really new to this and am having a headache

[coral parcel]

There are no line numers in the image I can see.

[dire obsidian]

there's a conjunction law?

[coral parcel]

I don't understand that question.

[dire obsidian]

Ok which law did i declare that I forgot to flip?

[coral parcel]

The one annotated "De Morgan's laws on drugs".

[dire obsidian]

sht

I messed up big

thanks a lot

can I use commutative law even if the operators are different from one another? (and/or) @coral parcel

[coral parcel]

The commutative law works only on one operator at a time, so I'm not sure what you mean there.

[dire obsidian]

wait mb

I meant associatative

wait sorry let me rephrase

I need to write stuff dow

n

[coral parcel]

No, (a AND b) OR c is not the same as a AND (b OR c).

[dire obsidian]

so the operators have to be the same

[coral parcel]

Yes. If they are different, you might want distributive instead.

[dire obsidian]

can you explain why this works?

[coral parcel]

A OR (B AND C) <=> (A OR B) AND (A OR C)
with A = ~p, B = p, C = q AND ~r

[dire obsidian]

$\neg p \lor (p \land (q \land \neg r) \Leftrightarrow [(\neg p \lor p) \land (\neg p \lor (q \land \neg r)]$

does latex bot not work in this channel

[vital dew]

Salt

[dire obsidian]

what we never learned this law

nvm

this is so weird

ok I see

wow

@coral parcel wait why can you just randomly add brackets like that

from $\neg p \lor (p \land q \land \neg r)$ to $\neg p \lor (p \land (q \land \neg r))$

is it still associatative?

[vital dew]

Salt

[dire obsidian]

you can add as many wacky brackets as you want as long as same operator?

[coral parcel]

Yes, the fact that \land is "associative" means that you get to bracket a chain of \land's however you want.

[dire obsidian]

so if you have \lor you can also chain as many as you want?

since duals

part of same associative law

[coral parcel]

Yes, just like (as a more familiar example) addition. If we write simply "1+2+3+4", we can choose to interpret that as (1+2)+(3+4) or (1+(2+3))+4 or 1+(2+(3+4)), etc. etc. etc., and all these have the same result because addition is associative.

[dire obsidian]

I see thanks a bunch
logic questions has just been made a lot easier for me thanks to that revelation

frick did i @ the right person

@unkempt fiber do you have any tips or tricks to solving logic proposition problems?

[weary tiger]

I'm not Ansh but I learned by just doing a lot of problems

[dire obsidian]

I'm starting to feel like

you just group like variables together

and hope for the best

since most of the cancellation comes from 2 same variable

negation law is wack

without negation logic problems would be a lot easier

[unkempt fiber]

and just follow the laws right ✓

n try to bring the same variables together

[dire obsidian]

ah so I was on the right track

[shadow geyser]

hello there!

I have a question

[dire obsidian]

Are there any issues with how I applied distributive?

@unkempt fiber

[unkempt fiber]

Also, you could've just used the identity again instead of distributing

[dire obsidian]

yeah that's what I ended up doing

[dire obsidian]

so with distributive law the operation symbols don't have to be opposite (and/or)

[unkempt fiber]

with distributive, you have (p V q) V r = (p V r) V (q V r)

[wide vine]

is there a symbol for exclusive or?

[dire obsidian]

yeah

[dire obsidian]

I see

[weary tiger]

I'm having trouble approaching this problem: Prove if A is a countable set, then cardinality of the power set of A is <= the cardinality of the real numbers. A hint would be greatly appreciated.

[wide vine]

is it $\directsum$

[vital dew]

Brandon7716
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[wide vine]

:/

i think it has multiple uses?

[dire obsidian]

I kinda lied to you srry $\underline{\vee}$

[vital dew]

Salt

[dire obsidian]

very scuffed but works

[wide vine]

is ⊕ also commonly used?

[dire obsidian]

yeah

i dont use it tho

[wide vine]

Although my previous unit used it for symmetric difference

for sets

[dire obsidian]

dang

[wide vine]

such as A ⊕ B = (A - B )U (B -A)

which I assume is also another use case?

[dire obsidian]

bruh I am stuck again

how do you go from first line to second line

im so confused

[wide vine]

is that an r?

is it p,q,r?

[dire obsidian]

yeah

you know how to do it?

idk what logic law is applied here

[wide vine]

well I just started logic so 🙂

[dire obsidian]

ah dang

[wide vine]

but maybe demorgan or something idk

let me write this out and see if I can get anywhere :V

[dire obsidian]

look at #help-10 ansh answered

it's backwards distributive

dang

[wide vine]

well have yet to learn about any of that but im sure I will

at least "formally"

[dire obsidian]

hf in logic unit 😢

haha

[wide vine]

will do 😉

[tired sable]

~~ Anyone here know how to do Logical implication and rules of inference? Thanks.~~ solved

[dire obsidian]

@unkempt fiber thanks a lot for all the help. Sorry for the late post

[unkempt fiber]

np

[weary tiger]

anyone recommend any problem set(s) for DFAs, NDFAs, moore mealy machines, their interconversions ?

[whole hamlet]

The following algorithm defines a sum. Following through, find the values for the lower bound a, the upper bound b, and the formula being multiplied by.

stop = 11

index = 4

result = 1

while index < stop:

result = result + (6
⋅
index+2)

index += 1

Can someone help me understand what the lower bound limit would be?

I understand that the upperbound limit is 4 and the function of the sum is 6x+2 but I dont understand hte lowerboud

[narrow robin]

How to convert a number from base x to base y without passing by base 10?

[weary tiger]

If you are an adult, you must understand first what a base is.
When you will get it, you will know the answer.

[narrow robin]

Bro this isn't helping

😂 😂

If you have no intention to help please don't answer

[weary tiger]

https://www.purplemath.com/modules/numbbase.htm

I built one, i call it base(p). Base primes.

[narrow robin]

This has no examples of how to change bases without passing through base 10

I know how to do it passing through base 10

[stray reef]

you would do the same thing as for converting base 10 into base y, but you would just do all arithmetic in base x

this includes having a multplication table or something for base x

[weary tiger]

This base is dissociative.

[weary tiger]

Exactly, and we get 'n-1' and repunit example.

I know where this base primes come from.

It is from radius.

That's why i'm here.
You mathematicians made this unit become in your order a derived unit.
You must correct that.
https://en.m.wikipedia.org/wiki/Radian

[narrow robin]

Ok thanks

Can you do me a simple example?

Makes my comprehension easier

[valid wave]

hello back again, you said that we can do this process infinitely. Isnt this just rewritting varaible names

when you look at x^2+y^2=0 mod 3, it helps to know what the squares are mod 3. So just square everything, 0^2=0, 1^2=1, and 2^2=1. In other words, we're looking for ways to make 0 mod 3 by adding only 0 or 1 together since 2 is not a square mod 3. we have 0+0=0, 0+1=1 and 1+1=2 so we're stuck to assuming x^2=0 and y^2=0 mod 3

[obtuse lance]

no because (x,y,z) = 3 * (a,b,c)

it's not just relabelling

[valid wave]

this by itself seems like proof in itself enough to prove num 3 right?

[valid wave]

so where does the fact that x^2 = 0 mod 3 come into this

[obtuse lance]

x^2=0 mod 3 means x=0 mod 3

so that's how I know to substitute x=3a

[valid wave]

ah so 3a + 3b = z

[obtuse lance]

squares

(3a)^2+(3b)^2=3c^2

[valid wave]

so where does this leave us though

[obtuse lance]

we came directly from x^2+y^2=3c^2

it leaves us with 3(a^2+b^2)=c^2

[valid wave]

god i have a pea brain

because a^2+b^2 have to be 0

because the mod arthemtic we were doing earlier

[obtuse lance]

I think you're mixing steps up

[valid wave]

if you haave time

[obtuse lance]

which is understandable, yesterday there were like 3 people trying to talk lol

[valid wave]

could you go over it all 1 more time

🙏

[obtuse lance]

wanna go into vc I can just write it out in order

[valid wave]

ok 😄

[obtuse lance]

k one sec

[valid wave]

yes

[valid wave]

thanks again @obtuse lance

[devout vault]

Could someone please help me understand how to draw a covering graph

The idea of a covering makes sense

Is it just taking the verticies from the covering and making a new graph with them?

<@&286206848099549185>

[spring surge]

https://cdn.discordapp.com/attachments/418981682117345288/938926742494662696/unknown.png

It generates elements like (c 1) (d 1) (c 2)...
How to modify it so that there are no repeating elements? Like we have (c 1) so no elements (c ...) or (.. 1) ?

[inner lava]

Does graph theory come under discrete mathematics

I have a vehicle routing problem and I'm struggling to categorise what VRP it actually is.

[tidal tulip]

does my proof work: I want to prove if a,b,c in Z and a|b, b|c, then a|c. Proof: Suppose a,b,c in Z and a|b and b|c. Then there exist k in Z st b=ak and there exist m in Z s.t. c=bm. So we have c=(ak)m -> c=a(km) [since mult is assoc] and therefore we see a|c

[craggy juniper]

yes that works

@tidal tulip

[tidal tulip]

@craggy juniper awesome thanks!

[wicked basin]

do i consider the cases where
x is even and y is even,
x is even and y is odd,
x is odd and y is even,
x is odd and y is odd?

[native zodiac]

Yes, basically. When doing addition and substraction the parity behaves as follows:
even ± even = even
even ± odd = odd
odd ± odd = even

So you apply the pigeonhole principle on the set elements and prove it

[olive condor]

Huh? “There exists” means you only need to prove it for one case

[coral parcel]

They have to be elements of S, though.

And you can't pick any one of the cases and be sure it occurs in S.

[craggy juniper]

we don’t know which case it is

[inner lava]

Is there a version of the multi commodity flow problem where the construction of the network is being optimized rather than the flow within it?

[olive condor]

Ohhh, okay 😯

[wicked basin]

idk how to do the 3rd paragraph

i know i have to apply the binomial theorem

but idk how

[coral parcel]

I don't see any obvious need for the binomial theorem here.

[wicked basin]

o

then what do i do

[coral parcel]

Follow the explicit hint?

Let the three possible remainders be pigeonholes.

[wicked basin]

o i mean i dont know how to do the "Generalize the previous two parts"

[coral parcel]

That's the fourth paragraph as far as I can count.

[wicked basin]

o ya

[coral parcel]

There's a pretty clear pattern already. "Difference divisible by 2" requires 3 numbers. "Difference divisible by 3" requires 4 numbers ...

[wicked basin]

o lit

so U must contain n+1 integers?

[coral parcel]

At least that's a fair guess. Try to prove it, or find a counterexample.

[wicked basin]

what do u mean for

[coral parcel]

Sorry, typo fixed.

[floral field]

So, we did this example in class, and we stopped a little early. How would I count the set of integers coprime to both 2 and 3?

[wicked basin]

ahhhhh so now i use binomial thm

[wide vine]

can a partition of a set be 1 set?

In others set you have some Set A

can you have a partition that is one set B

where B = A

I guess this has to be possibly

true*

let A = empty set

wait now im confused

nvm

If $\emptyset$ then the partition is ${{ }}$

[vital dew]

Brandon7716

[wide vine]

right?

[amber prism]

it'd be the set containing emptyset

[wide vine]

okay

[amber prism]

since ${\emptyset}\not\subseteq\emptyset$

[wide vine]

well like when you define a partition

is it considered a set?

like

A= {1,2,3}

[vital dew]

Mosh

[wide vine]

(Partition of A) = {{1,2}, {3}}

is that how you write out partitions?

[amber prism]

I'm not sure on the exact notation, however that feels fine as long as you define it like that

[wide vine]

Yeah im not sure either on it

[amber prism]

when it doubt define your own notation

[wide vine]

For any non-empty proper subset A of a set U, the set A together with its complement form a partition of U, namely, { A, U ∖ A }. what does the {A, U \ A} mean

a set defined with sets A given U is the complement of A? NVM figured it out

[inner lava]

Does anyone know if there is a name for a version of the vehicle routing problem where the demand and supply is being generated in real time while the deliveries are happening? This effectively means that a longer delivery route will deliver less as a backlog of deliveries is being generated the longer its out and the vehicles will be a constant looping closed path inside the network to keep the flow going.

I am unsure if this problem is actually the vehicle routing problem or a subset of the network flow problem

[amber prism]

Yeah, A and the complement of A form a partition

(provided A non-empty or A=U)

[wide vine]

yeah im just not familiar with the A\ B notation

[amber prism]

$A\setminus B :={a\in A|a\notin B}$

[vital dew]

Mosh

[amber prism]

ie take A and literally remove everything from B that's in A

[wide vine]

and this is called a relative complement?

[amber prism]

A\B is set difference

[wide vine]

oh

well how do you define a complement

with an unknown U

and why would they not just say A-B?

[amber prism]

A-B is equivalent

$A^c:=U\setminus A$

[vital dew]

Mosh

[amber prism]

U is usually implied from A

[wide vine]

okay

[balmy ivy]

for a question like that would it be mathematically rigorous enough if i use a direct proof involving euclid's algo?

[obtuse lance]

yeah that's preferred

[floral field]

So, I played around with some examples and claimed that this is the number of ways. Below it, I wrote down part of an argument, but I’m not sure how to finish it

In particular, I'm not sure what considering (n-1)x(n-1) matrix instead of nxn has anything to do with the sums/columns being even

[vital dew]

beeswax

[stray reef]

@floral field consider constructing a matrix of the kind they ask you for as follows:
fill the top left (n-1)×(n-1) submatrix arbitrarily. (this means everything except the last row & the last column)
then there will be exactly one way to fill the last row and the last column in such a way that each row and col has an even sum

[floral field]

So, you're basically saying that once we fill out the matrix except the last row and column, our hand is foced for those?

[stray reef]

yes

and of course the number of ways to fill the top left (n-1)×(n-1) submatrix is 2^{(n-1)^2}

[floral field]

Ahh, I understand. Thank u

[floral field]

Sorry, another counting question. I claimed this for this exercise, but I’m not quite sure how to argue this.

[stray reef]

each quadruplet of points uniquely determines an intersection of diagonals

[floral field]

Wait, what does this mean? Like for n=5, I have this figure:

[stray reef]

nC4 by defn counts the number of quadruplets that can be made from n points

each diagonal intersection point can be mapped to a quadruplet by looking at the four endpoints of the two diagonals that intersect at it

i guess that's going the other way around

to go from a quadruplet of points to a diagonal intersection, consider the quadrilateral formed by the four points in the quadruplet and look at the intersection of its two diagonals

[floral field]

Ahh

[weary tiger]

sosososo

like im watching this utube series n topic is relations now

i didnt understand this one

si

so*

is the formula for no of possibilties of reflexive relations

2^(n^2)-n

or

2^(n^2) - 2^(n^2)-n

????

where set is

like

a = {1,2,3}

a x a

set

[spring surge]

I want to define a set which says { a and all x's which are bigger than a }

{ 1 3 5 6 7} { 2 4 5 6}

How to do it in set builder notation? Am I allowed to self reference a set in its definition?

Example. C = { x | x=3 -> x not belongs in C } ?

[olive condor]

Erm, what is “a”?

Don’t really get what you meant

You can do something like let A = {1,3,5,6,7}, B = {2,4,5,6}, then $C = {a \in A | \forall b \in B (a > b)}$

[vital dew]

zacque

[olive condor]

It says C is the set of elements in A such that each of them is larger than all elements in B. In this case, C = {7}.

Self-referencing is not allowed as far as I know 🤔 Because if the set is not yet defined, how can you refer to it?

[spring surge]

That's nice clarity in logic. Sometimes I forget those pure sane questions to ask.

[spring surge]

a belong to a set which has not element smaller than a. Is it correct?

[olive condor]

No, no, “a belongs to a set” means $a \in A$ for some set A

[vital dew]

zacque

[olive condor]

“A set which has no element smaller than a” means ${x | \neg (x < a) }$

So, all in all, it’s $a \in {x | \neg (x < a) }$

[vital dew]

zacque

zacque

[olive condor]

Not sure where your {1..10} fits in though

[mortal hemlock]

agghhh

[tidal tulip]

does this proof work: prove that the statement “if p is a prime number then p+1 is not a prime number” is false

proof by contradiction: assume p is a prime number then p+1 is not a prime number. then let’s consider 2 which is prime. then this means that 3 is not a prime by our assumption, but this is a contradiction since 3 is prime. thus our assumption that if p is prime then p+1 is not a prime is false. so we see if p is prime then p+1 may also be prime

i’m shaky on the last statement “if p is prime then p+1 may also be prime” line in my proof because it isn’t always the case if p is prime then p+1 is prime but the statement itself that ‘if p is prime then p+1 is not prime’ is false in certain cases so i wasn’t sure how to word the conclusion there

[mortal hemlock]

@tidal tulip do you know what quantifiers are?

or just basic propositional/predicate logic in general

[tidal tulip]

yes

[mortal hemlock]

okay great this will be much easier

[tidal tulip]

k

[mortal hemlock]

so you can try to formulate the proposition they gave with quantifiers

[tidal tulip]

ok

[mortal hemlock]

like this:
$\forall p \in \mathbb{N},\ p \text{ prime} \Longrightarrow p+1 \text{ not prime}$

[vital dew]

Loxvan

[tidal tulip]

that makes sense!

[mortal hemlock]

and then you find a counterexample

that was 2 and 3

[tidal tulip]

okay makes sense since

T -> F

is F

[mortal hemlock]

yes exactly

[tidal tulip]

how would i reword my proof then

[mortal hemlock]

so now you have proven that the statement is false, so its negation is true

to reword it, just write the negation using quantifiers and word it in plain english

[tidal tulip]

would the negation be ~(p->q) = ~(~ p or q) = ~~p and ~q= p and ~q

but does that imply if p is prime then p+1 is always prime?

[mortal hemlock]

correct, but you negate the entire statement, with the quantifier

[mortal hemlock]

no, because you're missing a quantifier

[tidal tulip]

which one

[mortal hemlock]

the negation is this:
$\exists p \in \mathbb{N},\ p \text{ prime}\ \wedge \ p+1 \text{ prime}$

[vital dew]

Loxvan

[mortal hemlock]

for all becomes there exists

[tidal tulip]

ah i see!

i’ll rewrite in a bit and tag you but have to go right now and i’d like to see if my revision id correct. i’ll tex it up with quantifiers and add more detail than necessary to verify my understanding

thank you

[mortal hemlock]

you're welcome

the key thing here is to always negate the whole statement including the quantifiers

[tidal tulip]

k

[mortal hemlock]

good luck!

alright now i finally got that tex to format right so i can ask my question
_ _
_ _
_ _

Anyone know some good (free) sources with relatively advanced analytical proofs for some identities/results in combinatorics (counting)?
I don't have time to work them all out by myself and sometimes I get stuck, and I haven't found much on YouTube. Most of the videos establish very basics results like nCr = nCn-r; I want something more elaborate and advanced, for example this:

\[
\text{Proof that } \binom{n}{0} - \binom{n}{1} + \binom{n}{2} + \dots + (-1)^n \binom{n}{n} = 0\]
\[\text{Let } S = \binom{n}{0} - \binom{n}{1} + \binom{n}{2} + \dots + (-1)^n \binom{n}{n}\]
\[S = \sum_{k=0}^{n}\binom{n}{k} (-1)^k = \sum_{k=0}^{n}\binom{n}{k} (-1)^k1^{n-k}\]
Using the binomial theorem
\[S = (-1+1)^n = 0^n = 0
\]

[vital dew]

Loxvan

[mortal hemlock]

What I mean by analytical is, proving it using the formulae and working it out merely by calculation, not the combinatorial argument. It's kind of frustrating being stuck on one for a while and then finding that it's just proven by induction and that there aren't any other simple methods.
It doesn't matter what the type of the source is, whether it's a video (doesn't have to be on YouTube), an article on a website, an ebook/paper, as long as it's not behind a paywall I'll take it.
Also I'm not looking for just one or two, preferably many of them.

[weary tiger]

hello, since 15 min have passed I'll ask my q:

so, I'm reviewing some graph theory and I came across Euler's theorem; and wanted to ask if someone could confirm my understanding of it. First we define the eulerian circuit and eulerian path, which means that we use each of the edges only once for both, however, for the eulerian circuit we start and end on some vertex X as well as fulfilling the aforementioned condition. Now comes the part I'm not so sure about, to my understanding - euler's theorem states, that, if a graph has a vertex with an odd degree, then no eulerian circuit is possible. However, a eulerian path exists if all but two vertices have an even degree, is my understanding correct? what if we have only one vertex of odd degree?

[pale epoch]

what if we have only one vertex of odd degree?
then you won't have an eulerian path

just look into the proof

[weary tiger]

ah

I see I see

[pale epoch]

basically you have to leave a vertex as often as you enter it

this gives you a circuit if every degree is even

[weary tiger]

so to clarify then, my understanding is correct insofar as: if we have two vertices with an odd degree, we can form a eulerian path however a eulerian circuit is impossible

[pale epoch]

but if you only need a path you can leave the starting vertex once more than you enter it

and enter the ending vertex once more than you leave it

allowing for exactly two odd degree vertices

ye, everything you said is correct

[weary tiger]

I see I see, thank you then

so if we have one graph of which one vertex has an odd degree, neither the eulerian path nor the circuit are possible

got it

[pale epoch]

👍

[cold hound]

If a connected regular bipartite graph has a Hamilton cycle then it has an Euler tour.

is this statement true?

[coral parcel]

You can't have a (finte) simple graph where exactly one vertex has odd degree.

The sum of the degrees is always even.

(Because it is twice the number of edges).

[stray reef]

this is vacuously true

but emphasis on vacuously

[weary tiger]

I am not a big brain and hence will search up the definition of vacuously

be right back

[coral parcel]

She means it is true for all such graphs that they have no eulerian path, because there are no such graphs.

It's equally true for all such graphs that they do have an eulerian path.

[spring surge]

I want to describe a set like where all elements are the pairs of numbers which represent same fractions. Am I doing it right? Is there simpler ways to represent relations between the elements in a set in set builder notation?

[coral parcel]

Hmm, can you give a concerete example of one of the things you want to be an element of your set?

What you've written so far is just a roundabout way to write the set of all pairs of numbers (which can be divided), which I don't think is what you indend.

[spring surge]

{ (1 2) (2 4) (3 6) (4 8) ... } or { (4 1) (16 4) (20 5)... }

[coral parcel]

Do you want each of those sets to be an element of the set you're trying to define?

[spring surge]

No.

[ember obsidian]

it seems each set should be the result of fixing the value of the fraction

to get { (1 2) (2 4) (3 6) (4 8) ... }, u write smth like

[spring surge]

I want that my set would define both of those if its possible

[ember obsidian]

$\brc{(a,b)\in\bN^2:\frac ab=2}$

[vital dew]

RokabeJintaro

[ember obsidian]

u can replace 2 by any natural n to generalize

$\brc{(a,b)\in\bN^2:\frac ab=n}$

[vital dew]

RokabeJintaro

[spring surge]

that means that every pair represents same n?

[ember obsidian]

this set is defined as the set of pairs (a,b) satisfying a/b=n

[spring surge]

thanks.

I want to build a set where all elements has R relation between themselves. Is it correct?