discrete-math

[tidal plinth]

That's not the question

How many places (vertices) are there in Kₙ where you can start a cycle from?

[sick fiber]

Well we can start the cycle from any vertex in complete graph

So n places/vertices

[tidal plinth]

@obtuse lance Please continue. I'm waiting to answer this with a slightly different approach after you're done xD

[It's going to be almost the same thing, except for the first step. But the non-trivial part of it will be the same so you'll have to do the hard work here]

[obtuse lance]

oh haha

ok so now imagine walking to another vertex, how many choices do you have to walk to @sick fiber ?

[sick fiber]

n - 1 ( since its complete)

[obtuse lance]

yup

so that means in our first step we have n(n-1) options of paths to take

[sick fiber]

I am taking a random example in my head. Like i started with v1 and then went to v2...

[obtuse lance]

perfect

[sick fiber]

Yeah yeah sure

U r ryt

[obtuse lance]

how many choices do we have when we take k-1 steps

[tidal plinth]

Oh, so you're doing the same thing. I thought you were going to count each step

[sick fiber]

n(n-1)(n-2)...(n - (k -1 ))

[tidal plinth]

↑ That is counting each step. So I'm safe, haha

[obtuse lance]

I guess quick sanity check, we will do a cycle of length 3 after taking 2 steps since we have no choice where to return

your answer gives n(n-1)(n-2)(n-3)

so I think you should really have n(n-1)(n-2)...(n-(k-1))

we wanna do one less than k, not one more than k

[tidal plinth]

Yeah you messed up the sign there

[sick fiber]

Ohk word

[obtuse lance]

now we've overcounted but we can save it

how many times have we overcounted our cycles?

[tidal plinth]

[You were thinking n - k + 1 [which is a very common simplification], but then you partially reasoned it out, so you knew it's n - (something), then mixed those up]

[sick fiber]

Yeah xD ryt.

[obtuse lance]

nice, yeah good point

[sick fiber]

@obtuse lance trying to see overcounting here.

[obtuse lance]

so let's maybe just imagine the case with a 3-cycle how many times would we have overcounted?

it's tricky

[sick fiber]

Ohk let me try..

[tidal plinth]

@obtuse lance I think it's better to use 4 vertices. S₃ = D₃ but S₄ ≠ D₄ (for the same reason that makes this example unsuitable)

[obtuse lance]

oh, I don't see that making a difference for how I'm thinking about it

but I see how that's a good strategy

[tidal plinth]

(All permutations vs. only certain permutations)

[obtuse lance]

well ok, I will say I am just thinking of them directly

well I'll wait to see what gurjit says

[tidal plinth]

Yeah

[sick fiber]

Ueff.. i am trying to just take an example and come up with some answerr

[obtuse lance]

well, how do you decide to start a cycle?

how many ways could you have done that?

[sick fiber]

Like i took 4 vertices v1,v2,v3,v4...
4 * 3 * 2 ( eg v1 --- v2 --- v3)

[obtuse lance]

ahh that was to make all our cycles

but now we're focused just on a single cycle

that we have traversed multiple times

[sick fiber]

So if from this example

I would say

V1 - v2 - v3

And v3 - v2 - v1

Is Counted twice

[tidal plinth]

https://discordapp.com/channels/268882317391429632/496785905474994186/659084513124548608
The question is: How many ways could you have chosen the start vertex so that after choosing the other vertices in the following steps (in some way), you'll finally get the same cycle?

For example if your graph as 5 vertices: v₁, v₂, v₃, v₄, v₅,
and your cycle is v₁v₂v₃v₄,
how many ways could you have chosen a starting vertex so that after carefully choosing the remaining ones, you'll get the same cycle v₁v₂v₃v₄?

[Or am I misunderstanding @obtuse lance's question?]

[obtuse lance]

well I guess I should say how I was thinking it

I imagine a single cycle as like a necklace with beads

and we counted a cycle for each starting bead

but not only that, we could have gone clockwise or counter clockwise

so we have over counted by 2k

[tidal plinth]

That's true. But is that all?

[obtuse lance]

for each cycle, so we divide n(n-1)(n-2)...(n-(k-1))/2k

I don't know, is this all?

[tidal plinth]

Which cycle is v₂v₃v₄v₁?

[obtuse lance]

sorry I didn't read what you wrote

[tidal plinth]

1234 is the same cycle as 4321, like you said

But what about 2341?

[obtuse lance]

I'm counting that I believe

[tidal plinth]

https://discordapp.com/channels/268882317391429632/496785905474994186/659086278330941490
Sorry. Rather, I didn't read what you wrote

[obtuse lance]

I count all k cycles 1234, 2341, 3412, 4123 as being all possible starting places

and then go around them backwards yeh

[tidal plinth]

I somehow missed this one: https://discordapp.com/channels/268882317391429632/496785905474994186/659085746581274635

[obtuse lance]

ok cool we're on the same page

[sick fiber]

I just counted it as twice actually

But I got that i am overcounting many cycles not alone 2 times

@obtuse lance yeah for 123

It should be same as

231

312

[obtuse lance]

ah not quite

when you came up with our way of counting

we allowed ourselves to start at any of the n vertices remember?

so that means 123 is the cycle we made from starting at 1

231 is the cycle we made starting at 2

312 is the cycle we made starting at 3

so we have 3 cycles

[sick fiber]

Ahh yeah yeah...

[obtuse lance]

but we could have also gone backwards along these like you noticed

so 2*3

[sick fiber]

Backward in the sense ?

[obtuse lance]

so we can start at 1

and then go to 2 or 3

there are two paths around a cycle starting from the same point depending on if you go clockwise or counterclockwise

[tidal plinth]

https://discordapp.com/channels/268882317391429632/496785905474994186/659088174848540700
E.g., 1→2→3(→1) or 1→3→2(→1)

[sick fiber]

Oo yeah right

[obtuse lance]

that was nice I didn't know this result before in general just the case with K_n with k=n

nice to see that it does simplify down to (n-1)!/2 like I was expecting

[tidal plinth]

Okay, so my turn now

[obtuse lance]

ok go for it 😛

[tidal plinth]

I'll start with the (slightly) harder part that you've already done

@sick fiber If I give you k vertices, in how many ways can you make a cycle out of them?
To make a cycle, this is all you need to do: Make a path, and then connect the first and last vertex.
So in how many ways can you make a path out of the k given vertices?

[sick fiber]

Alright... So v1, v2, v3 ,v4,..vk, v1...(k -1)!

Like i fixed v1 and v1 in the ends

And just checking permutations of the vertices in the middle

(just not sure is that correct approach)

[tidal plinth]

Yeah, the first vertex is fixed as being adjacent to the last vertex, but the last vertex is also counted among the k.
So in how many ways can you arrange k vertices into a path, that is, in a line/row?

In other words, how many permutations of k objects (the k vertices) are there?

[sick fiber]

K!

[tidal plinth]

Yes. And once you get a path, you make the first and last vertices adjacent, so there's no choice and hence nothing to count there

But now the same overcounting-logic from before applies

So there are 2k different permutations that give the same cycle by this process.
So the actual number of different cycles that we can form using the k given vertices is
k!/2k = (k - 1)!/2

Okay so far?

[sick fiber]

Just a moment plz

2 was because of clockwise and anticlockwise

[tidal plinth]

And k because the k "rotations" of any cycle will give k different ways of writing the same cycle

1234 = 2341 = 3412 = 4123

[sick fiber]

Yeah yeah exact

And could we rephrase it in terms of objects here ?

So like relating it to the example of merosity of beads

[tidal plinth]

This part is exactly the same. So we're still asking: If you're given k different beads, how many (truly) different necklaces can you make with them?

That is: In how many ways can you arrange k objects in a circle (if all that matters is which object is next to which, and not on which side of an object is the other — if that mattered, then we wouldn't divide by 2)

[sick fiber]

Cool now makes much more sense to me.

🙂

[tidal plinth]

Okay, so the number of cycles we can form with k given vertices is (k - 1)!/2

But we're not given k vertices, we're given Kₙ.
So we must first choose k vertices out of the total n. We can choose any k, since there's nothing special about any vertex.
Also, two different choices of k vertices will never give the same cycle (no matter how we arrange them).
So the total number of k-cycles is:
#(Ways to choose k vertices out of n)×(k - 1)!/2

In how many ways can you choosek vertices out of n?

[sick fiber]

Nck

[tidal plinth]

Yes. So: C(n, k)×(k - 1)!/2

Which looks different, but can be easily simplified to what you got before

@obtuse lance And C(n, n)×(n - 1)!/2 = (n - 1)!/2 when k = n

[obtuse lance]

nice

[sick fiber]

@tidal plinth @obtuse lance thanks a ton guys

[obtuse lance]

yeah I guess I didn't even post the final form but

n!/(n-k!)2k

was what the original starting thing was, but I was afk I'll try to catch up to see how you did it

you did something with like, making a single loop first and then building up all the loops is that right?

@tidal plinth

[tidal plinth]

@obtuse lance I chose k vertices out of n (in C(n, k) ways) and formed all possible different cycles using these given k vertices (in (k - 1)!/2 ways → that's the number of the so-called circular permutations of k objects)

[obtuse lance]

ah I see, gotcha

[slender skiff]

Hello,
Can anyone explain to me why the 2nd one is the correct answer? couldn't the 3rd one be correct aswell? If P is true it makes R be true, and if they are both true then Q and S is also true, which makes the hole statement be true.

[tidal plinth]

But it's not given that they're both true, or even that one of them is true

[slender skiff]

oh so its just stating the fact that P implies R, not that any of them are tru (R is true if P is true doe)

[tidal plinth]

Exactly

[slender skiff]

ah but in the 2nd case If P is true, that means that Q is to, and if Q is true S is true,

ah

i see

Thanks!

[kind tapir]

wtf does $p \to q \to r$ mean? don't you need parantheses?

[vital dew]

Intel:

[dapper rose]

$(p\to q)\land(q\to r)$ perhaps

[vital dew]

EpicGuy4227:

[kind tapir]

i mean, this would agree with the intuition that you can sort begin with a proposition which you know is true and sort of flow in the direction of the arrows to get new true propositions

but it's not standard notation as far as i know

so i don't know what to make of i

[sleek swallow]

Lmao just check if (p => q) => r iff p => (q=>r)

If they’re equivalent, then associativity holds

[kind tapir]

plot twist: it doesn't

for example, in the case where p is false, and r is false.

but in the question above something of the form $p \to q \to r$ is written, and i'm trying to understand what the fuck is up with that

[vital dew]

Intel:

[dapper rose]

chaining implications is pretty common. here's an example

[kind tapir]

i think that arrow indicates inference rather than implication, but i might be wrong

anyway, i've never seen it in a pure propositional logic context. is it common?

[dapper rose]

well I never saw it during the 1 semester of logic I did

[kind tapir]

but it does unambigiously mean $(p \to q) \land (q \to r)$, right?

[vital dew]

Intel:

[sleek swallow]

‘The arrow indicates inference rather than implication’

What

[kind tapir]

"this, therefore that" rather than "if this than that"

and i meant in the context of the picture epicguy sent

like 3 dots arranged in a triangle pointing upwards

[sleek swallow]

$[(p \implies q) \land (q \implies r)] \iff (p \implies r)$

[vital dew]

Abhijeet Vats:

[kind tapir]

that's not in general true

[sleek swallow]

‘Inference’ has a very specific meaning though

No that is true. It’s a tautology. The transitive law

[kind tapir]

plug in p is false, q is true, r is false

[dapper rose]

the iff should be only if

[kind tapir]

exactly

[sleek swallow]

Oh right right fair enough

Fk i forgot my bad

[kind tapir]

then it's a tautological implication corresponding to hypothetical syllogism

[sleek swallow]

Okay yea but if you just wanna check if it unambiguously is the conjunction of two implications, then you just need to check if the equivalence holds

It comes down to the same type of problem

[kind tapir]

no

it's not a math problem

it's a matter of knowing the convention

the thing in question is what is meant by $p \to q \to r$.

[vital dew]

Intel:

[kind tapir]

it's not like this thing is defined a particular way and we're trying to figure out something about it

we're trying to figure out how it's defined

which isn't a matter of mathematics, but of convention

anyway, $(p \to q) \land (q \to r)$ does seem like the most reasonable interpretation so far

[vital dew]

Intel:

[dapper rose]

found this post https://math.stackexchange.com/questions/104143/in-what-order-are-chained-implications-evaluated-i-e-a-implies-b-implies-c

[sleek swallow]

Oof that’s cool, i didn’t know that

[dapper rose]

wtf from the post there's also $p\to(q\to r)$ and $(p\land q)\to r$ interpretations (these 2 are equivalent). rip.

[sleek swallow]

Might be best to consult a text on logic

[vital dew]

EpicGuy4227:

[sleek swallow]

It seems like there are deeper reasons for preferring one convention over the other

[dapper rose]

computer scientists ruining everything

[sleek swallow]

It’s pretty interesting tbh

[covert flume]

why is this correct

$\Gamma(\frac{n-1}{n})*\Gamma(\frac{1}{n}) = \frac{\pi}{sin(\frac{\pi}{n})}$

[vital dew]

Angelium:

[obtuse lance]

because it's the reflection formula with x=1/n plugged in

[covert flume]

oh

thanks

actually no

thats incorrect

wait no

[obtuse lance]

$\Gamma(1-x)\Gamma(x)=\frac{\pi}{\sin(\pi x)}$

[vital dew]

Merosity:

[obtuse lance]

plug in x=1/n

this doesn't really answer "why" though, just that in anchors it to a name so you can look up the proof of that formula yourself to read

[covert flume]

yes

[last sigil]

Post the full question @slender skiff picture is cropped

[errant bear]

no

[slender skiff]

Hello,

How does this answare become 40.320, I keep getting 336 when i do: 8 choose 5 * 6 choose 5, what am i missing, by doing that you get the different ways all the men and woman can be arranged.

[rugged sorrel]

Order matters.

You chose the people that will be matched, but not who will be matched with whom.

[weary tiger]

Actually, I don't get this one either. What would you have to do to get 40320? That's 8!

But I don't understand why would 8! apply here?

I think it's binom(8,5) * binom(6,5) * 5! which does indeed give 40320, but I am not sure if I can reason the 5!. Hmm. I guess the first two binoms are the ways to pick 5 women and 5 men, and then 5! ways to arrange the couples? But the logic seems kind of crooked.

I think I thought of a good way: (8 * 7 * 6 * 5 * 4) * (6 * 5 * 4 * 3 * 2) ways to get the men and women, but then we divide by the number of ways we arrange the couples, which is 5!. Thus giving us 40320. Can anyone confirm if I'm correct?

<@&286206848099549185>

[stray reef]

yeah it's really just a coincidence that this happens to also be 8! @weary tiger

[weary tiger]

I know @stray reef But what do you think of my reasoning?

Is it valid or?

[stray reef]

yes your reasoning seems ok

[weary tiger]

How would you do it?

The problem I mean

[stray reef]

i would do the exact same calculation as you did

[weary tiger]

Thanks

[sick fiber]

https://cdn.discordapp.com/attachments/659064044866764837/660849974271279105/JPEG_20191229_195151.jpg

TIFR 2018.8

I just able to figure that vertex n should be of degree odd

[lean creek]

talk about a sus angle lol

[sick fiber]

@lean creek sup ?

<@&286206848099549185>

[young ferry]

I've been working on this problem over break and it's killing me. I feel I'm close. It's a very satisfying problem so I thought I'd share for y'all:

Given n red points and n blue points in the 2d plane where no three points are colinear, must there exist a set of line segments connecting one red dot to one blue dot, such that no two cross

[obtuse lance]

I think I have a solution

I'll do induction, for n=1 we have one pair and one line, easy

now for the nth case we assume is true and connected and we'd like to show the n+1 case is true

so I pick out two points where I want my red and blue dots to lie in this already nice looking connected picture

and I imagine taking a pair of red and blue dots already connected from far away and sliding it in towards their position

as I approach, I might have to split it

but when I do, I split and trade with that pair of vertices

[young ferry]

Who says that during the reconnecting process, a line might not be created that crosses another? This crossing will never trigger your splitting process

[obtuse lance]

and so I am always maintaing connections on its path to the final resting place

[young ferry]

That was my first idea too

[obtuse lance]

because when they come in, I can approach narrowly

[young ferry]

What do you mean?

[obtuse lance]

one sec

https://prnt.sc/qh6gpq

[young ferry]

Any kind of continuous transformation strategy I've tried has suffered from that problem

[obtuse lance]

once they're there, I can trade

[young ferry]

There might be so much noise surrounding the spots you want the dots to end up in, that the trading process just creates another crossing

And then you'd have to rectify that one, which makes more

[obtuse lance]

no since n is a finite number

[young ferry]

I'm not convinced a loop can't occur

The process of unhooking crossings could go on forever

[obtuse lance]

yeah you're right, this is not good enough

[young ferry]

My current strategy is to represent every red-blue pair as nodes in a graph which connects nodes when the pairs they represent don't share a node and don't cross

And then pigeonhole

I'm sure if I knew more graph theory or combinatorics it could work

[obtuse lance]

hmm

do the points partition the plane into little convex sections?

[young ferry]

of the representation graph, or of the original picture?

[obtuse lance]

just points in the plane with no edges drawn

[young ferry]

You could triangulate it, but that hasn't been fruitful for me

[obtuse lance]

I guess I'm just trying to understand what kind of barriers there are to this

after you triangulate it what do you do

like some kind of sperner lemma type thing?

not sure how that would help lol

I guess I'm just imagining maybe if it's triangulated you have some kind of paths to enter or exit this maze like thing

and try to imagine what that looks like

since we can define doors in the triangles which have all different colors at the edges where there are two different colors

[young ferry]

Maybe some useful information: it's trivially easy to do this for pretty much any configuration you draw. Part of why it's so frustrating to prove

[obtuse lance]

so each triangle with different colors has an in and out door

and is necessarily connected to anothe rwith an in/out door

[young ferry]

There are configurations that can't be triangulated without tons of triangles all with the same color

In your method would every triangle be 2-1?

[obtuse lance]

no I am saying just triangulate it however, doesn't matter

then look at the corners

let me draw a little pic

[young ferry]

Sure

[obtuse lance]

https://prnt.sc/qh6k1t

I drew points

did an arbitrary triangulation

then imagined that edges where there are different color vertices on each end are doors

so then you can walk through it

since by design every triangle will have an in and out door, as long as the colors are different

this is not the best example I should have triangulated so there are closed rooms

it doesn't matter, it's just some kind of idea that's probably not helpful to consider

[young ferry]

What about a huge configuration such that there are closed rooms that are completely bordered by closed rooms?

[obtuse lance]

that's possible

[young ferry]

My other biggest strategy has been trying to show that the unhooking process can't loop on itself

Idk, some food for thought. It's not very deep but it's been really fun to do

[obtuse lance]

good problem, nice and simple to say but annoyingly hard

[scenic shard]

Hi I have troubles understanding this question

I thought the induction step would be something like 'if P(k) holds, then P(k+1) also holds'

[stray reef]

this is a somewhat different kind of induction

it's not induction on natural numbers

[scenic shard]

but what kind of induction is it? I'm trying to understand it, but I don't see it. 🙂

[undone python]

MemyselfandI
I'm not convinced a loop can't occur
The process of unhooking crossings could go on forever

MerosityToday
yeah you're right, this is not good enough

Any pair of opposite edges in a rhombus is shorter in total length then the two diagonals, so the total length after "uncrossing" decreases. Since it's monotone, bounded below (by 0), and discrete, it must obtain a minimum, and that minimum must have no crossings.

[young ferry]

Holy shit you’re right

I knew the geometry had to get used at some point

[vital dew]

gaiameansearth:

[weary tiger]

hello folks, I'm learning an intro to discrete mathematics and I'm having trouble understanding why this set means the set {0,1,2,3,...} base on the book I'm reading. Currently, I'm interpreting the set given above as the set where x is an element of natural numbers such that it is all natural numbers plus 3 which should mean {3,4,5,6,...} right?

[last sigil]

@weary tiger

You have a mistake in your statement in "any natural number"

The first part means to consider the set of natural numbers

Now the : means such that, so you are now putting a rule on then elements of your set (naturals so far)

So if the element x is in the naturals and x + 3 is part of the naturals, x is in your set

So here 1 is natural and 1 + 3 is natural, so 1 is in your set

[weary tiger]

gotcha, so the x + 3 part just means I substitute x to natural numbers (as implied by the precondition) and that when x + 3 is processed it should also be an element in the natural numbers, correct?

[last sigil]

Better way to state it is the set of x where x is an element of the natural numbers such that x+3 is also an element of the natural numbers

[sleek swallow]

The set of all elements x belonging to the natural numbers such that x+3 is also an element of the natural numbers

[last sigil]

^ much nicer

[sleek swallow]

Anyways if it were me, i'd be anal about it and write:

${ x \in \bN : (x \in \bN) \land ([x+3] \in \bN) }$

[vital dew]

Abhijeet Vats:

[last sigil]

[glossy adder]

ew

[last sigil]

Still kinda confused why 0 is there, but I guess the book is including it

[glossy adder]

$0\in\bN$

[vital dew]

CaptainLightning:

[glossy adder]

this is not arguable

[sleek swallow]

Well, some books define 0 as part of the naturals

Lol

[weary tiger]

thank you @last sigil && @sleek swallow

[sleek swallow]

Sure

[dapper rose]

shit now I can't argue with that @glossy adder

[small kayak]

can anyone give me some instruction on this problem ?

[slender skiff]

Hello,

How is that ~P is true? i mean woudn't it make more sence if it was the first one? Since it would make P and Q be true, and therefor the first implication true, and since P is true R is true aswell.

[sleek swallow]

P & Q wouldn't be true necessarily with the first option, btw

P would definitely have to be true

but either of Q or R could be true

Anyways, if both Q & R are true, then the conclusion is true. That's pretty easy to see.

If Q is false and R is true, then you have the entire conclusion being true by ex falso quod libet.

If Q is true and R is false, notice that your entire antecedent is true but your consequent is false. That is, P -> R is false but the antecedent is true by virtue of P and Q being true.

Now, suppose that not(P) is true. Then, P is false.

If P is false, then P and Q is false. If P is false, P -> Q is automatically true by ex falso quod libet. So, the truth value of the antecedent, whose truth value, mind you, only depends on not(R) and Q, is inconsequential to the truth value of the entire implication. Hence, the conclusion is always true when not(P) is true.

@slender skiff Ask questions if I haven't been crystal clear about what I've said.

[slender skiff]

Im not sure how you got to the conclusion that Q is false if P is false?

there is no implications for that?

[sleek swallow]

Huh where?

[slender skiff]

"If P is false, then P and Q is false"

[sleek swallow]

Yea that's correct. A conjunction is true only when both propositions are true.

Since P is false, P and Q is false. Q can be either true or false on its own.

[slender skiff]

Oh

so with not(P)

its true by default?

because the first implication is false

[sleek swallow]

Yeap. Look at the argument I've made.

[slender skiff]

idk what "falso quod libet" is

XD

guessing true by default

[sleek swallow]

So, not(P) is true. Hence, P is false.

If P is false, then P and Q must be false.

Hence, the truth value of the antecedent is only dependent on that of not(R) and Q

Now, if P is false, then P -> Q is true, regardless of the truth value of Q. So, the consequent is true. You have an undetermined antecedent and a true consequent. Hence, the undetermined antecedent does NOT affect the truth value of the entire compound proposition.

ex falso quod libet is the principle that you can prove that any given proposition is true from a false premise.

[slender skiff]

ah

[sleek swallow]

Something that's 'true by default' would be more appropriately described by the word 'tautology'. So, a tautology is a proposition that is true regardless of the truth values of the elementary propositions that form it.

Understand everything?

[slender skiff]

Yes

Thanks you so much!

[sleek swallow]

🙂

With logic, always try to break things out into smaller arguments. That will help you quite a lot. Actually, that's a strategy that works for math in general as well.

[slender skiff]

Hello, how is nr 3 the answare? dosn't c have 3 degrees? my definition of degrees says "Equals the number of edges that are incident on v, with an edge that is loop counted twice."

and im reading that "A simple path is a path with no repeated vertices." and "A path in a graph is a sequence of vertices connected by edges, with no repeated edges."

so if a simple path is a path with no repeated vertices how is the 3 answare true? since the only way to get a length of 4 from a to e is to take the loop on a, which means that it is repeating a.

[undone python]

Let's go through each of the statements, from "truest" to "falsest":

D is obviously true, there is a bd edge.

B is also certainly true, but you have to remember a loop counts for 2 in the degree computation.

A is true, here is one such path: a - a - a - d - e

C is not true. One way to see this quickly is that there are only 4 vertices in the connected component containing a and e, so the longest possible simple path in that component is length 3. So there are no simple paths of length 4, much less one that starts at a and ends at e.

[slender skiff]

I see

@undone python Thanks! 😄

idk how i messed that up lol

[noble forum]

What does E notation mean?

[sleek swallow]

?

$\exists$?

[vital dew]

Abhijeet Vats:

[stray reef]

"E notation"?

[noble forum]

Im a noob at descrete, im just starting to learn it and idk what EURO symbol means

[sleek swallow]

$\in$?

[vital dew]

Abhijeet Vats:

[sleek swallow]

Is that what you're referring to?

[noble forum]

ye

[sleek swallow]

That's the 'belonging' symbol

[noble forum]

please elaborate

[sleek swallow]

So, for example, if $x$ is an object and $S$ is a set, then we say that:

$x \in S$ if x belongs to S.

[vital dew]

Abhijeet Vats:

[noble forum]

so x is in S

[sleek swallow]

Yes

It's an element of S.

[noble forum]

so like an array

[sleek swallow]

Uh if you list the elements in S, then yea.

Though

[noble forum]

ye

[sleek swallow]

Array kind of implies that they're arranged in an order. Sets do not have to be ordered.

[noble forum]

why is it useful and why do people use it.

true ^^

[sleek swallow]

Well, it's used quite a bit in set theory because you're always talking about objects and sets and objects belonging to sets.

[stray reef]

yeah, $\in$ can be read as "is in"\ or "belongs to"

[vital dew]

Ann:

[noble forum]

thanks

[sleek swallow]

It's useful as a notational shorthand, i suppose

[noble forum]

Yes, but why is it useful information, couldnt u just say its some variable

[sleek swallow]

You really wouldn't wanna write 'is in' or 'belongs to'

What do you mean?

[noble forum]

me?

[stray reef]

yes you

[noble forum]

oh

[sleek swallow]

Lmao i can't be asking Ann right?

[stray reef]

Yes, but why is it useful information, couldnt u just say its some variable
what are you talking about here

[sleek swallow]

What do you mean by 'couldn't you just say its some variable'?

[noble forum]

Why use the notation, it hasnt been defined, so couldnt u appose is as some variable such as, x,y,z

[glossy adder]

huh

[stray reef]

uh

what

[sleek swallow]

Well, you're not wrong

[noble forum]

lol

[sleek swallow]

The notion of belonging technically is primitive in set theory

[noble forum]

Im difficult

[stray reef]

set membership is the most basic relationship out there

[noble forum]

Ye

[stray reef]

like

"this object is in this set"

is used overwhelmingly often in math

often enough to warrant its own symbol

[noble forum]

Give me a simple example please using the notation and i will explain more intuitively

[stray reef]

ok suppose A is the set of all even integers

then 18 ∈ A

[noble forum]

Yeah, simply COULD u say, A is all even integers

[stray reef]

A is the set of all even integers.

[noble forum]

so yes

[sleek swallow]

Okay, see, you could just say that 18 is an even integer in that situation

[stray reef]

what's your point

[glossy adder]

[sleek swallow]

However, you could not properly talk about the existence of, say, the empty set without the 'belonging' relation

[noble forum]

I havent got a point, its my interpretation of understanding the symbol

[stray reef]

you sounded like you were arguing against something

[noble forum]

soz

im not

[stray reef]

∈ is in some sense similar to = or < or > as a symbol

it's a relation symbol

meaning it connects two objects

and the notation as a whole, like "x ∈ S", is a statement

which can be true or false

[sleek swallow]

Let A be any given set. Then, the empty set is just:

$\phi = { x \in A: \lnot{x \in A} }$

[vital dew]

Abhijeet Vats:

[stray reef]

uh

abhi that's

WAY overcomplicating things

[sleek swallow]

^That's something you can only get if you introduce the 'element' symbol

[stray reef]

also phi is not the symbol for the empty set

[sleek swallow]

Sorry, was just trying to give an example

[stray reef]

it has its own symbol

[glossy adder]

"\phi"

[noble forum]

So, would contain consecutives and sequences , but in it couldnt contain a random integer

[sleek swallow]

Lol fuck it

[glossy adder]

oh my god

[stray reef]

So, would contain consecutives and sequences , but in it couldnt contain a random integer
what?

no

no like look

vladislav

a set is just a collection of things

like

without getting into all the formalisms

[noble forum]

yes, im overcomplicating things.

[stray reef]

that's what a set is

please don't overcomplicate things.

[noble forum]

(laugh)

[stray reef]

the idea is so general it can take a while to grapple with

but like

the only thing that defines a set is what's in it

[noble forum]

I understand it, i just want to understand, now, what u can and cannot within the set

[stray reef]

with what

uh

"the" set? what?

[noble forum]

E

[stray reef]

??

[noble forum]

bruhh, im to ignorant

[sleek swallow]

You mean determine when an object belongs to a set?

[stray reef]

no you're just

not making yourself clear at all

[noble forum]

no @sleek swallow

[sleek swallow]

Then?

[noble forum]

You said that within the set, there can contain : all even integers including "18", or, all prime etc.

[stray reef]

bruh get a discrete math book and give the set theory section a read

no

vladislav

there is no such thing as "the set"

[noble forum]

wtf am i on about then

[stray reef]

YOU TELL ME BRUH

[noble forum]

please

[sleek swallow]

To be fair

It's not an easy thing to understand

I get why you're confused

[stray reef]

it's quite an abstract concept yes

[noble forum]

Ive read one chapter on this lol

[sleek swallow]

Just go back to your textbook and read it again

[stray reef]

give that chapter another read

[sleek swallow]

Bring out a particular section you're confused on

That will be more helpful

[stray reef]

^

[sleek swallow]

I could give you examples of sets and ann could do that

But that won't help

[noble forum]

define set(s)

[sleek swallow]

Informally, a set is just a collection of objects. The objects are the elements of the set.

Very naively, that's what you can use as a basic definition

Let x be an object and let S be a set. When x is an element of S, then we write $x \in S$.

[vital dew]

Abhijeet Vats:

[sleek swallow]

That's all there is to it.

[noble forum]

yes, so how was i wrongly using the word set

[stray reef]

you were talking about "the set" as if there's one single mathematical object out there named The Set™️

[sleek swallow]

Well

Ok nvm i'll stay hushed

[noble forum]

So, E is "belongs to/ apart of"

[sleek swallow]

A part of*

But yes

That's what that symbol means

[noble forum]

lol

ok good - progress !

[sleek swallow]

Give that chapter another go. Then, ask questions if you're still confused.

[noble forum]

Lol @stray reef

No i was talking generally with any set/ my set examples

[sleek swallow]

Then have your use of phi as the symbol for the empty set be made fun of

[stray reef]

then your use of the definite article "the" was inappropriate

[noble forum]

phi

how do u use the symbol on discord

[stray reef]

...

[sleek swallow]

Anyways, it's fine.

$x \in S$

[vital dew]

Abhijeet Vats:

[stray reef]

ugh

[sleek swallow]

x is an object, S is a set

[stray reef]

_<

[sleek swallow]

x belongs to S

[noble forum]

nonononono

[sleek swallow]

That is a proposition that literally reads 'x belongs to S'

[stray reef]

_<<

[noble forum]

how do u wrote it like that ^^^

[stray reef]

there's a bot here

[sleek swallow]

Oh the element symbol? \in

[noble forum]

thanks

[stray reef]

which uses a certain markup language called LaTeX

in which there are commands for various mathematical symbols

and other things

it'd take too long to explain

[noble forum]

No, its fine, I understand now. Thank you @stray reef @sleek swallow

[sleek swallow]

Sure

[noble forum]

I understand, to an extent lol

[stray reef]

.-.

[sleek swallow]

I don't blame you, it's not easy stuff.

[noble forum]

I dont study this stuff at school yet, but I just want to know it

[sleek swallow]

Certainly not if you don't have some propositional logic to assist you with learning it.

You learn it on your own?

[noble forum]

ye

[sleek swallow]

That's wonderful!

[noble forum]

Ive just decided to work on it today

[sleek swallow]

Continue doing it, it's very cool and interesting stuff

[noble forum]

Recommend any website?

[sleek swallow]

I'd recommend working on propositional logic a bit before you go to set theory.

[noble forum]

I mean I use a few, but im naive

[sleek swallow]

Introduction to Mathematical Thinking by Stanford on Coursera

Try that, just to get yourself introduced to basic propositional logic. Then, set theory will become easy to grasp.

[noble forum]

in propositional logic, do they teach u notations and uses?

to then apply

[sleek swallow]

Certainly. You'll get a good grasp of what a proposition is, what a quantifier is, what a logical connective is and what the laws of propositional logic are.

[noble forum]

^^^ with Keith Devlin?

[sleek swallow]

Yes.

[noble forum]

Thanks

[sleek swallow]

I generally dislike content made by american creators but this is one that i can recommend, because it's good.

[noble forum]

lol, i dont usually watch math lectures but ill give it a go!

[sleek swallow]

I'm not in a university so I can't speak of math lectures in general. These ones don't have a lot of material packed into a 50 minute class.

[noble forum]

so it begins...

[sleek swallow]

Rather, they are just trying to introduce you to basic logic. Basic logic is actually very easy.

[noble forum]

Is it a good connection to other maths?

[sleek swallow]

I'm probably gonna get pranked for this but i feel learning propositonal logic properly will lead to a better understanding of other things in math.

[noble forum]

Thanks.

Wait, ur not at uni? @sleek swallow

[sleek swallow]

Nope.

[noble forum]

Havent started or not going?

[sleek swallow]

I have not yet started. I will be going next year.

[noble forum]

so u doing A levels? Or already finished

[sleek swallow]

If Ann has better advice on how you should approach your basics or if anyone else has any other advice, then you should listen to them. They are in university right now.

I finished IB in 2017. I'm in the military at the moment but i'll be getting released soon.

[noble forum]

Why u in the army?

Even better, why did u join?

[sleek swallow]

Compulsory military service

[noble forum]

Ah, a mathematician whoms in the military

never heard of it lol

[sleek swallow]

I'm not even close to becoming a mathematician.

I'm actually very crappy but I'm trying to get better.

[noble forum]

Thats good

[sleek swallow]

That's why I told you to listen to the advice of the uni students if they're gonna give any. I can give my 2 cents but that isn't worth anything, coming from someone like me.

[noble forum]

I mean, im still in secondary school. But i want to pursue physics or math. Not sure yet, both very interesting.

Thanks for the talk @sleek swallow

[sleek swallow]

It's admirable that you're learning this on your own. Keep your interest alive and pursue what you want in uni.

Not a problem at all.

[obtuse lance]

Does this have a closed form? $\sum_{k=0}^{n-1}\frac{(n+k)!}{k!}x^k$

[vital dew]

Merosity:

[stray reef]

[obtuse lance]

oh nevermind I think I got it, I must be blind

[ivory venture]

https://en.wikipedia.org/wiki/Menger's_theorem#Short_proof When they define what an AB-connector is, it should say internally vertex-disjoint instead of vertex-disjoint, right?

[prime steeple]

@obtuse lance What was it? The open upper bound makes it complicated

[weary tiger]

[obtuse lance]

no I didn't get it, but it turns out it doesn't matter for what I was doing since I left out terms in a preceding recurrence relation that led to those coefficients

[hardy yoke]

Combinatorics: 46 cards numbered from 1 to 46. You have to choose 11 cards. What's the chance of having number "1" in one of those 11 cards. 10C45/11C46

Would that be the right answer?

[stray reef]

no because that's 0/0 lol

you might have been looking for 45C10/46C11

...which should simplify to just 11/46

[hardy yoke]

Right

That's right though?

45C10/46C11

[stray reef]

yes

er

no you fucked up again

45C10/46C11

if you're using the choose function, most of the time your second argument should be smaller

[hardy yoke]

Yep, thanks.

And another thing related to combinatorics. Have to choose a number from 0-98 (including 0 and 98). Event A = "Divisable by 3", Event B = "Divisable by 2". Proof formula for independent events is P(A) = P(A|B) => P(A) = P(A and B)/P(B). Are the events independent? ```P(A) = 32/99 = 0,323(23).
|A and B| = 16, 98/6 = 16.333
|B| = 49, 98/2 = 49
P(A|B) = 16/49 = 0.32653061
Since 0.326 /= 0.323(23) events are related.

Would that be right?

[stray reef]

= 0,323(23).
ouch please don't write that

generally don't turn fractions into decimals at all unless you need to round something

anyway

besides that

you are correct, A and B are not independent here.

though "proof formula" is a wording i perceive as sort of weird

[hardy yoke]

If P(A) = 0.6, P(B) = 0.2, P(C) = 0.5. Chance of all events happening, chance of at least 1 event happening., chance of 1 event happening. All events are independent```
I'd assume chance of all is 0.60.20.2
Chance of at least 1 = (1-0.6)(1-0.2)(1-0.5)
Chance of only 1 happening I have no idea.
Perhaps,
P(A)(1-P(B))(1-P(C))+
P(B)(1-P(A))(1-P(C))+
P(C)(1-P(B))(1-P(A)) = 0.46?

[robust mango]

For all events, it should be 0.6 * 0.2 * 0.5. For at least 1 event, what you did was find the probability that neither A, B or C happen. So you need to subtract that from 1 and get 0.84 which is the probability of at least 1 event.

[hardy yoke]

Ah, right typo on the first one. You're right about the second one.

But only 1 happening?

[robust mango]

Lemme think 😛

I think you're right for part C, i was busy with my own thinking until I saw how you did it

[stray reef]

P(A)*(1-P(B))*(1-P(C))+
P(B)*(1-P(A))*(1-P(C))+
P(C)*(1-P(B))*(1-P(A)) = 0.46?

yup

[versed flare]

whatsup

can someone help me, im trying to prove that every 3 numbers, n, n+1, n+2 there must be one number divisible by 3 aka 3k

[soft thorn]

hmm

can't you consider this mod 3, and consider the cases

[ivory badge]

^

[versed flare]

they are asking me for doing it direclty

[sour arrow]

Oh, directly. How silly of us to not consider a direct method

[obtuse lance]

how often do numbers divisible by 3 occur when counting anyways

[versed flare]

1/3

[obtuse lance]

weird, so like if you were to count 3 numbers in a row, one of them would be divisible by 3 I guess

[sour arrow]

That's exactly it though. Consider the case where n and n - 1 are not divisible by 3. Then, n + 2 must be

[dapper rose]

n+1 not n - 1

[versed flare]

my intuition tells me that, but they are asking for some crazy formulation and shit

[sour arrow]

Oop yes that's what I meant

[ivory badge]

watch this: assume n is not divisible by three, then n = 3s + t by good old division with remainder

clearly t is 1 or 2

and because of this, either n+1 or n+2 will cause t to be a multiple of 3

qed

this is just mod with extra words tbh

but division with remainder is pretty direct and should be usable

what was it, euclidean algorithm?

[versed flare]

thanks you @ivory badge thats a pretty complete answer

[tawny hollow]

hey guys how would i go about converting ab* l b*a into an fsa?

[obtuse lance]

@tawny hollow what have you tried?

[tawny hollow]

@obtuse lance

[obtuse lance]

close

this is ab* | ba*

see why? try to just construct b*a by itself

[tawny hollow]

oh shit

i am trying to achieve this @obtuse lance

[obtuse lance]

yes I know

to construct ab* | b*a it's easier if you can construct ab* and b*a separately, then join them together

wait

what you wrote is different

[tawny hollow]

[obtuse lance]

write what "all the as appearing before all the bs" looks like

the regex I mean

take it one step at a time

[tawny hollow]

so in my regex i should have all a's or all b's?

[obtuse lance]

yeah

so for instance you should include aaab, aaaabb

also aaaaa or bbbb would also be accepted

[tawny hollow]

ok

thanks

that definitely helped

aabb or bbaa ? @obtuse lance

[obtuse lance]

write the regex for "all the as appearing before all the bs"

if that's what you're answering

[tawny hollow]

no

thats the string

[obtuse lance]

those are two examples of accepted strings

but there are infinitely many strings your fsa will accept

[tawny hollow]

ah ok

you know my first attempt should i discard that ? @obtuse lance

[obtuse lance]

your first attempt was what?

ab* | b*a

this?

[tawny hollow]

yes

[obtuse lance]

tell me why it's wrong

give an example of a string accepted by:

https://cdn.discordapp.com/attachments/496785905474994186/662635870049927178/unknown.png

that would not be accepted by ab* | b*a

[tawny hollow]

babab as the rejected string

abb or bba

[obtuse lance]

babab would be rejected by both

[tawny hollow]

if i had done a*b l b * a

aaaaaab

baaaaaaaa

[obtuse lance]

now you're on the opposite end haha, those would both be accepted by both

make one that's only accepted by
https://cdn.discordapp.com/attachments/496785905474994186/662635870049927178/unknown.png

but is NOT accepted by a*b | b*a

maybe the concept of what an FSA does is not clear

imagine it's a little machine, you give it a string, and it tells you "yes" or "no"

[tawny hollow]

can you do VC?

[obtuse lance]

depending on if it matches one of the possibilities

sorry can't, I'm going to be going soon

[tawny hollow]

oh ok no worries

[obtuse lance]

the answer should be pretty easy if you understand it

keep trying

[tawny hollow]

will do

thanks for your guidance @obtuse lance

[hardy yoke]

11 questions. Student knows only 2. Stundent has to randomly pick 3 questions. Probabliliy of knowing atleast 1 question? ((2*1*9)+(2*10*9))/11C3

[stray reef]

no, you've just divided by 0

3C11 is 0

[hardy yoke]

So with 11C3 the answer should be right?

[robust mango]

The denominator will be 11C3, because that's the total number of ways for choosing any 3 questions from 11.

Think about the numerator now

[hardy yoke]

So that's wrong then, eh..

(2*1*9) Would be. 2ways to pick known * 1 way to pick known * 9 ways to pick the unknown.

(2*10*9) 2Ways to pick known * 10 ways to pick unknown * 9 ways to pick unknown.

That sounds like variations.

Shloud ve been 11A3?

[tranquil cargo]

rosen

is classic

never rread it b4

[stray reef]

how to prove it by velleman

that one's specifically for proofs

[sleek swallow]

Uh try Fundamentals of Mathematics by Bernd Schroder

[slender skiff]

Hey! Any tips on what i should do know? i dont see how i can get a function with the same variable to be a function with 2 different variables.

[sleek swallow]

uh

[weary tiger]

what on earth is that

[sleek swallow]

I can see a quantifier in there, some letters that, presumably, are variables and the predicate R

[weary tiger]

lol

[sleek swallow]

no u

[weary tiger]

no u

[delicate ridge]

is that jape?

[sleek swallow]

no u

[prime steeple]

@slender skiff But that one with the two variables is only the premise.
If you can prove A, then you can also classically prove B->A for any B.

[slender skiff]

Hello, Im sitting with this problem, I know that I need to use conditional probability, but I really don't see what A/B is and P(A)/P(B) is. How would i go about determining this?

[primal sun]

@slender skiff have you heard of the bayes theorem?

[reef thistle]

@slender skiff Try starting with something concrete, like a filesystem with 100000 files.

and then calculate the number of each type

[primal sun]

I'm pretty sure this is straightforward application of the Bayes Theorem

[reef thistle]

yes it is

but intuition for Bayes may be easier explaining through an actual population

@primal sun

[slender skiff]

@reef thistle using that way of thinking made it so much easier

[primal sun]

@slender skiff have you got it?

I usually do a branches diagram and that is more often than not sufficient to have a grasp of what's going on

but if it helps, do it!

[cerulean sentinel]

We have to switch the red and blue wagons using the yellow locomotive

(from my Graphs hw)

The locomotive can pull and push the wagons

At the top is a bridge, the locomotive can go under it but not the wagons

And at the last state we want the wagons to be switched and the locomotive to be back to its starting position

I can't come up with a solution, don't know what the graph is

[stray reef]

well there seem to be only four key points here

the current locations of the wagons and locomotive, and the intersection point of the circle rail and the straight rail

[cerulean sentinel]

Yup, they have defined three areas of the track: east (red), west (blue), and south (yellow)

[stray reef]

you also need the intersection as a separate point

so let's call that the center

so we have four points: C, W, E and S, with the locomotive at S, the blue wagon at W and the red at E

[cerulean sentinel]

Yes but the author says 'We want a solution with fewest moves, where a move is every time the locomotive goes from one section of the track to another' (i.e. the 3 sections)

We shall use State spaces, also

Here each state is a situtation of where the locomotive and wagons are, and when more than one are in the same section then their order also matters

And the graph will 'draw' the state space: the edges are the states, and the vertices are the moves between them

[scenic shard]

how can I find the DNF of (¬p ∧ ¬q) ∨ (¬r ∨ ¬q) ?

[sleek swallow]

DNF?

[scenic shard]

Disjunctive Normal Form\

[sleek swallow]

I assume that that's just you writing that as a disjunction of conjunctions?

[scenic shard]

yes exactly

[sleek swallow]

Okay, so what have you tried?

[scenic shard]

I know I can write it as (¬p ∧ ¬q) ∨ ¬r ∨ ¬q

[sleek swallow]

Okay, then?

[scenic shard]

then i get stuck

[sleek swallow]

Well, first of all, how would we typically get the DNF?

[scenic shard]

well we can use the rules of inference?

[sleek swallow]

How would that help?

[scenic shard]

to rewrite the statement

[sleek swallow]

I think it might be more fruitful to consider the combinations of the truth values of p,q and r where the entire statement is true

That's, after all, the basis of the proof that any logical statement can be written as a disjunction of conjunctions

[weary tiger]

A box has 52 balls. 47 are red, 5 are blue. 4 balls are removed from the box. Calculate the probability all 4 red. It's just Binom(47,4)/Binom(52,4), right?

@stray reef Can you confirm that I'm not being stupid?

[stray reef]

yea

that's correct

[weary tiger]

If there is a probability of 0.075 that something is broken independent, then the probability that 8 of those things being broken from a 110 total is (1-0.075)^102 * (0.075)^8, right?

@stray reef ?

[stray reef]

no, you forgot to multiply by 110C8

also i assume you meant exactly 8

[weary tiger]

Oh right

Oops

$(1-0.075)^{102}\cdot(0.075)^8\cdot\binom{110}{8}$

This is correct, yes?

[vital dew]

defective:

[weary tiger]

How do I calculate the average if at least like 9 are broken and 140 are made?

I don't want to do casework for that many.

[stray reef]

well you can calculate the probabilities for 0, 1, ..., 8 being broken and subtract the total from 1

[weary tiger]

Lol

[stray reef]

otherwise there's kinda no way around doing the sum for values "in the middle" if you want an exact answer

[weary tiger]

Any faster way

To get just the average?

[stray reef]

wdym the average

do you want to calculate the expected number of defects given that each item is broken with probability 0.075

[weary tiger]

110 balls are made. A ball, independent from the others, has a 0.075 probability of being broken.

If more than 9 balls turn out bad, what is the average number of balls that turn out bad if 140 balls are made instead?

Yes

But at least 9 are bad

[stray reef]

uhhh

hm.

i mean just going off of intuition it's got to be 9 + 131*0.075 but i may very well be wrong and it's close to midnight

[weary tiger]

Why 131 * 0.075?

[slender skiff]

How is the GCD 5, isn't this person using the algorithm in a wrong way? for step 2 arent u supposed to say 309 = 20 * 15 + 9, why did he say 55 = 3 * 15 + 10?

This is what i do:

a = b * c + d

a = b
c = d

And this method works for everything else,

[stray reef]

expected number of defects in the 131 that aren't known to be broken @weary tiger

[weary tiger]

Okay

@stray reef You smart

Why am I so dumb?

[primal sun]

@slender skiff you want the GCD between 55 and 17010, not 309 and 17010, which is what you'd be doing at step 2

[slender skiff]

okay got it, turns out i was doing it right before, but just got confused for some reason now

@primal sun thanks

[primal sun]

alright!

[whole cobalt]

yo

i need help with repeated squaring

i have to calc 3^383 in my head

google doesnt help

[sleek swallow]

[whole cobalt]

[stray reef]

i am pretty sure that it would be impossible to do such a thing mentally

do you need 3^383 itself or 3^383 mod something

[whole cobalt]

ah fuck my bad

mod 7

[stray reef]

now THAT'S a different story.

ok, what have you tried

[whole cobalt]

i know to divide the exponent by 2 then floor to get my calc exponents

then i just tried calculating mod 7 on that

[stray reef]

you... what?

[analog sonnet]

Are you allowed to use Euler's theorem

[stray reef]

"divide the exponent by 2 then floor" what

[whole cobalt]

idk i didnt understand it in the lecture itself

so some guy said to do that

[stray reef]

uhh

that's

gotta be THE worst way to describe repeated squaring

why not write 383 in binary, compute 3^(2^k) mod 7 for as many k as needed, and then multiply the required ones together to produce the final answer?!

[whole cobalt]

i have no idea how to do that

[stray reef]

which part do you have no idea how to do

[whole cobalt]

if i had to calc 383 in my head into binary

also what do you mean "required ones"

[stray reef]

if i had to calc 383 in my head into binary
i mean obviously i'm not expecting you to do that mentally

are you being held at gunpoint and being yelled at to compute the value of 3^383 mod 7 in your head? or what

[whole cobalt]

its an exam without help

[stray reef]

but surely you have some paper with you

right

converting 383 to binary isn't an insurmountable task as far as hand calculations go

[tawny hollow]

has anyone done finite state automata ?

[unreal berry]

a) and the solution to it

no

just no

is there someway to get out of this course during this lifetime?

[glass badge]

huh?

[unreal berry]

(b,b) doesn't even exist as a set in R

I can only assume they meant (d,d)

[stray reef]

that's what they said though. (b, b) ∉ R.

what's the issue?

[unreal berry]

the fact that I don't even know why they are stating the obvious

which is also irrelevant

[stray reef]

they're showing a counterexample to transitivity.

it is 100% relevant.

IF your relation was transitive, THEN from (b,c) ∈ R and (c,b) ∈ R you could conclude (b,b) ∈ R.

do you agree?

[glass badge]

not really grallak

[stray reef]

@unreal berry

[unreal berry]

could also conclude (c,c)

[stray reef]

ok yes sure

you could also say "(c,b) ∈ R and (b,c) ∈ R but (c,c) ∉ R, therefore R fails to be transitive"

but do you or do you not agree with what i said above

IF your relation was transitive, THEN from (b,c) ∈ R and (c,b) ∈ R you could conclude (b,b) ∈ R.

do you or do you not agree with that

[unreal berry]

is a function not transitive if I can find a loop or just a parallell?

[stray reef]

FUNCTIONS can't be transitive. RELATIONS can.

and what is this talk of loops or parallels in the first place??

ANSWER MY QUESTION.

[glass badge]

lol

[stray reef]

DO YOU OR DO YOU NOT AGREE WITH THE FOLLOWING STATEMENT:

IF your relation was transitive, THEN from (b,c) ∈ R and (c,b) ∈ R you could conclude (b,b) ∈ R.

[unreal berry]

because b,c c,b looks like a parallel

[stray reef]

DO YOU OR DO YOU NOT AGREE WITH THE FOLLOWING STATEMENT:

IF your relation was transitive, THEN from (b,c) ∈ R and (c,b) ∈ R you could conclude (b,b) ∈ R.

[glass badge]

noo

(b,c) and (c,b) r different

[stray reef]

@unreal berry answer my question.

[unreal berry]

I see the second half, but I do not see why the first part is true

[stray reef]

what second half

what first half

this is an IF-THEN statement

[unreal berry]

IF your relation was transitive based on something which I am not sure can prove something is transitive

a transitive function is 1 goes to 2, 2 goes to 3, 3 goes to 1 so 1 goes to 3*

[stray reef]

NO

NO!

[glass badge]

this is not a function man

[stray reef]

NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO! NO!

[glass badge]

its a Relation

[stray reef]

YOU DON'T EVEN UNDERSTAND WHAT A TRANSITIVE RELATION IS, THEN!

NO WONDER YOU'RE CONFUSED!

YOU DON'T KNOW WHAT IS BEING TALKED ABOUT, DESPITE HAVING CONVINCED YOURSELF THAT YOU DO!

[unreal berry]

I never said I was convinced of anything

[glass badge]

(a,b) means a is related to b

[unreal berry]

if I was, then I wouldn't be here

[glass badge]

u know this right

[stray reef]

ok

so

grallak

do you understand

[glossy adder]

[stray reef]

that IF a relation called R is transitive

AND (b,c) ∈ R, AND (c,b) ∈ R

THEN (b,b) ∈ R

do you understand this or not

yes or no

give me one word

do you understand this, yes or no

don't ghost me now, i can clearly see you're online

[unreal berry]

...

[stray reef]

if you need some time then say so explicitly

[unreal berry]

then no

[stray reef]

sigh

ok

we call a relation transitive if and only if, whenever (x,y) and (y,z) are both in R, (x,z) is also in R.

this is the definition of a transitive relation.

do you understand this. yes or no.

[unreal berry]

so if both xy yx are in R it's transitive?

[stray reef]

NO!

read the damn definition!

ugh

[unreal berry]

(x,y) and (y,z) are both in R, (x,z) is also in R.

[stray reef]

NO!

you've

[glass badge]

dont miss the last part

[stray reef]

managed to OMIT THE KEY FUCKING WORD

CONGRATU

FUCKING

LATIONS

YOU'VE CONTEXTOMIZED THE FUCK OUT OF THE THING I WROTE

JESUS FUCK

YOU MUST'VE TRIED REALLY HARD TO DO THAT

[glass badge]

chill ann 😂

[stray reef]

IT TAKES EFFORT TO PULL SHIT LIKE THIS

jeez no i'm really fucking HURT now

[unreal berry]

I just want to get through this course

I think. . .

[gleaming zephyr]

whats the problem

[unreal berry]

the problem is the problem

as it's not clear what it is, how it is solved, and why it matters

[gleaming zephyr]

whats the problem

[glass badge]

lol

[unreal berry]

I suppose the correct definition of transitivity in relations

I think

in layman terms

[gleaming zephyr]

'we call a relation transitive if and only if, whenever (x,y) and (y,z) are both in R, (x,z) is also in R.
'

[glass badge]

go ahead

[gleaming zephyr]

if you have a relation R such that:

(x,y) is in R and (y,z) is in R implie that (x,z) is in R

we call R transitive

thats it

[unreal berry]

yeah

[stray reef]

he doesn't understand the word "implies"

[unreal berry]

makes sense

[stray reef]

i tried that already mo2men

WOW

[gleaming zephyr]

XD

[stray reef]

WOWOWWOWOWOWOWPOWO

WOWOW

DGI

ASJAWOA

FSDA

ASD

I CAN'T BELIEVE IT

[glass badge]

and grall just had to check if the following property hold on that set

[unreal berry]

you can just cut out the middle man

[gleaming zephyr]

yea maybe

if you can

then its transitive

A implies B

means if A then B

[unreal berry]

not really sure why one value should be able to give two different results

but sure why not