discrete-math

[wild flame]

whoops.

https://i.imgur.com/ah2wzBH.png

@sharp raptor

Lastly. Ifor the contrapositive and inverse.

Would it be okay if I used 'does not increase'? Or would the negation adequately be decrease.

[sharp raptor]

I think does not increase

Technically speaking it could remain the same so idk if decrease would be equivalent

Though idk how nitpicky about it your teachers would be

[wild flame]

Fair.

The last thing is this question that confused me in wording.

I guess they wanted me to identify the rules used to change the formulas around?

https://i.imgur.com/U9FkU9b.png

[sharp raptor]

Is the stuff below the (a) line written by you?

And yes, that is correct

[slow socket]

can someone explain how they got "~B". i dont understand it

[sharp raptor]

Does P equal something?

[slow socket]

no?

wat u mean?

hypothesis 1: A -> P
hypothesis 2: A and notB
conclusion: P and notB

im good so far up to "P"
but idk how they got notB

[quaint river]

since A /\ ~B is true

that implies A is true

and Not B is true as well

[slow socket]

thats for step 1?

thats how i got A

[quaint river]

its also how you get Not B

by the same logic

[slow socket]

would this be correct?
P: Pat did it
Q: Quincy did it
R: Quincy was reading
https://cdn.discordapp.com/attachments/238956921581862913/542191273424060426/a229b5bdf478df31c9e226532f0b9ab2.png

to start

[sour arrow]

Looks good!

Need the rest?

[slow socket]

nah i think im good

ty

[slow socket]

if i have (notp and q) using commutative laws can i do (q and notp)

[sharp raptor]

yes

[topaz parcel]

how to solve for three missing parts in a venn diagram?

[sharp raptor]

I'm sorry, I didn't understand the question. Not sure if it's lacking information or I'm just stupid

[analog sonnet]

Can you give an example @topaz parcel

[topaz parcel]

i managed to get it but idk how to solve it

i cant come up with algrebraic stuff to make it pop out
i sorta made assumptions and found a way out of it

[analog sonnet]

Usually with Venn diagrams, your intuition is correct

[topaz parcel]

so no need for solutions?

[analog sonnet]

Well you do need to write it down in a proper way

[topaz parcel]

i think so

[analog sonnet]

The most formal way is to use set theory notation

[topaz parcel]

ok imma search it up

oh my we havent learned this yet

[analog sonnet]

Then try to express it in words as clearly as possible

Might get a bit verbose, but it does the trick

[topaz parcel]

thanks man

[sharp raptor]

If you want an easy way to do it math-ly without set theory:
36-11-9-8 gives you 8 which is the amount which like only blueberries
then a+b+c = 54-36=18
if a = amount that like just apples, b amount that like just cranberries and c both
a+b+c = 18
a+c = 28-11-8
a+b = 31-9-11

And you end up with a simple 3 equation system

[topaz parcel]

thank youuuu

[verbal furnace]

👀.

[sharp raptor]

😍

[lean leaf]

Discrete Mathematics and Its Applications by Kenneth Rosen is pretty awesome too.

[topaz parcel]

guys i still dont know how to solve it so bad :(

could u guys help me out how to solve for intersections?

[weary tiger]

@topaz parcel split everything into disjoint sets

then it's just a matter of adding things together

[slow socket]

hey i got this question. part a i put "if x is an element of both A and B" and i got it right

but for part b i put "if x is in either A or B, not both" and i got it wrong. can someone help

[weary tiger]

$\mathbf 1_A + \mathbf 1_B - \mathbf 1_{A \cap B} = \mathbf 1_{A \cup B}$

[vital dew]

TendentiousTorturousTopics:

[whole marsh]

I need some quick help. I don't have a problem, persay, I just need someone to explain a concept to me.

Could someone explain this to me?

Just ping me if someone thinks they can help

[whole marsh]

Can anyone help me?

[thin kite]

it looks like you're meant to notice a pattern

that a_n = a_(n-m) + m(3)

[thin kite]

and then set m=n and you get an expression with only a_0 and n

[uncut haven]

heyo, for graph theory, how would you calculate centrality for a node in a graph?

specifically eigenvalue centrality

o i misinterpreted something

got it, thank you!

[paper thicket]

Hello, I need help with the following

For a) does ∀x ∈ Z (1/x) ∈ Q work?

Not sure if the notation is correct

[azure lichen]

I would add a : or so

but yes, that’s really all there is to it

[paper thicket]

Where would I add the : ?

[azure lichen]

$\forall x \in \mathbb{Z}: \frac{1}{x} \in \mathbb{Q}$

[vital dew]

Sascha Baer:

[paper thicket]

Oh I see

Would negation work on this?

Or would I have to make predicates that define what a reciprocal is

[sudden prawn]

Yes, I would recommend using negation for this. It should work better opposed to defining the reciprocal.

Do you know where to start if you were to negate it?

[slow socket]

Holy

[copper ore]

how to do?

i have to use direct proof

[sour arrow]

Write down your definitons.
Even number: Can be written as 2k for some natural number k

That is, x = 2n
And y = 2m

For some n, m

[copper ore]

yeah

i did that

[sour arrow]

So what have you done?

[copper ore]

that's what i did

stuck after that

i tried doing

2k + 2p
2(k+p)

but i dk

[sour arrow]

Nice! Now, k + p is a natural number.

So, that's 2k for some natural k. That is, that's an even number.

@copper ore

[wooden swift]

halp

[copper ore]

so i proved it by just doing this?

2k + 2p
2(k+p)

nothing more? @sour arrow

[sour arrow]

Yes indeed! You had to recognize that 2(k + p) is an even number. Do you understand why?

[copper ore]

no

@sour arrow

can i @ you?

or is it annoying

[sour arrow]

2m, where m is a natural number, is even. This is by definition. With me on that one?

[copper ore]

yes

[sour arrow]

Now, I didn't have to call it m. m is just a label. As long as the number can be written as 2(natural number), the number is even

[copper ore]

yeah

[sour arrow]

You wrote x + y as 2(k + p)

But k + p is a natural number

Again, 2(natural number) is an even number

[copper ore]

ok

but do i have to put a therefore statement at the end

or any kind of statement before i start the proof

[wooden swift]

halp

[copper ore]

yea so post what you need help with

[wooden swift]

it is in math help questions k

[copper ore]

and someone will help if they know what they're doing

[wooden swift]

i hope so

[weary tiger]

hllo

[slow socket]

hi

[acoustic valve]

hey guys

can i get some help with this problem

im rly bad at induction

i proved the base case already

but idk where to go from there

[quaint river]

$\sum_{i = 0}^{n} x^i= \frac{x^{n+1}-1}{x-1} $

[vital dew]

Victoria:

[quaint river]

so we assume that this is true

for some i

then, to prove next case, we have to add the next term

and we get

$\sum_{i = 0}^{n} x^i + x^{n+1}= \frac{x^{n+1}-1}{x-1} + x^{n+1}$

[vital dew]

Victoria:

[quaint river]

now we just have to simplify the right side so that it is equal to $\frac{x^{n+2}-1}{x-1} $

[vital dew]

Victoria:

[quaint river]

@acoustic valve

[acoustic valve]

thanks, victoria

i ended up watching a youtube video and it made some sense there but with this you made me understand even more

:D

my brain is literally ginormous now 🤠

[quaint river]

a lot of induction is just done like this

you just add a new term

[crimson reef]

ok so question

if A n B do not intersect does that mean A n B = empty set

or A n B = 0

[quaint river]

empty set

[glad oriole]

0 is not a set

[crimson reef]

ty

ik 0 is not an empty set im just wondering if A n B does not intersect what does it mean

in the sense that its 0 or empty set

so thank you

[finite cipher]

Hi guys, I have a question that goes like this:

On the menu of a chinese restaurant there are 6 chicken dishes, 5 beef dishes, 9 seafood dishes and 10 vegetable dishes.
a)In how many ways can a family order if they choose exactly one dish of each kind?
b) In how many ways can they order if at most one dish of each kind is chosen?```

Can someone tell me whats the difference between a) and b)? they look the same to me

[tawny pawn]

Problem 35: Let Σ = {a, b} and Λ be the null string. Define A ⊆ Σ
∗ by the following rules.

1. Λ ∈ A.
2. If ω ∈ A, then aωb ∈ A.
3. If µ, ν ∈ A, then µν ∈ A.
4. Every ω ∈ A must come from a finite number of applications of rules
   1, 2, or 3.
   Prove by mathematical induction that, for every ω ∈ A, ω has equally
   many a’s and b’s

[azure lichen]

uh, when you say {a,b}, I take it you don’t mean the set that only contains the elements a and b, cause that would not make sense

did you mean {a,b}*?

[tawny pawn]

yea oops

[azure lichen]

yea so, I don’t really see the issue here, this seems pretty straightforward

you can induct over the number of rule applications

[sharp raptor]

how far are you getting?

Did you manage to get the base case?

[crimson reef]

anyone mind helping with this

like mind explaining what its saying when its definted under the power set

isnt the power set of N = { {empty set}, {1}, {2}, {1,2} , etc }

so 1 is not in any power set?

[weary tiger]

show ∃x(P(x)→Q(x)) and ∀xP(x)→∃xQ(x) is valid:

1) ∃x(P(x)→Q(x))
2) P(c)→Q(c)         Existential Instantiation

am i allowed to use Q(c) as a premise because they both imply →Q(x))?

also am i allowed to use conditional statement equivalences?
what do i do next?

[vocal sable]

@Thorng ok so the difference is a and b is that in a you have to order one of the dishes of the specific type but in b you can or cannot order one from that set , the maximum size limit is one and the minimum is zero.

@finite cipher^

[opal crescent]

@crimson reef that relation is indeed defined on P(N), ie X and Y are themselves elements of the power set of N

so they're just subsets of N

[azure lichen]

X and Y are subsets of N. you say that they are equivalent if 1 is both in X and in Y

[lethal solstice]

Does anyone know a method to draw a graph from a given degree sequence?

[verbal furnace]

Share more.

[lethal solstice]

you have been given a degree sequence like : G(3,3,5,5,5,5) and you need to draw a graph, here the number of vertices are 6,in which 2 vertices have degree 3 and 4 vertices have degree 5

By trial and error method it tedious,.. I looked on Google but was bombarded with theorems,

@verbal furnace

[verbal furnace]

Is the graph connected ?

[lethal solstice]

I have to draw it

[verbal furnace]

5, 5, 3, 3, 5, 5.

[lethal solstice]

check the exact question

[verbal furnace]

Yes. Do it.

Done ?

[lethal solstice]

yeah I have done it

but took me quite a few triala

trials*

[verbal furnace]

It seemed easy.

Train.

[lethal solstice]

🎊🎊

[lethal solstice]

figure 6 is bipartite but how?

[verbal furnace]

You can use 1 and 4 to connect with 2 and 3.

[lethal solstice]

but the definition of bipartite graph is that each vertex must join a vertex in set A and a vertex in set B, considering the two sets A and B

[late gust]

1 and 3, 2 and 4

[lethal solstice]

but figure 6 has 3 edges while, 1 and 3, 2 and 4 has only 2 edges

[late gust]

uhhh, I don't think I explained myself clearly

what I was trying to say is that

you number the vertices of graph 6 as 1, 2, 3, 4 from the top down

1 and 3 are a set, 2 and 4 are a set

because 1 and 3 are not connected to each other but are each connected to something from the other set

and similarly for 2 and 4

since they form the sets that way, 6 is bipartite

[lethal solstice]

So 1 connects with 2 and 3 connects with 4,right?

[late gust]

thats true

but the question you needed to answer was to show that graph 6 is bipartite correct?

because 3 is also connected to 2

[lethal solstice]

no, I had to find if it was bipartite or regular,

[late gust]

ah ok

its bipartite, because we can separate the groups in such a way that we have 2 sets of vertices

within each set of vertices, they arent connected

but each vertex is connected to a (at least one) vertex from the other set

[lethal solstice]

yes 3 is connected with 2 also

thank you very much

[lethal solstice]

is it possible to draw a cubic graph with 11 vertices

[azure lichen]

a cubic graph is one which can be interpreted as the vertices and edges of a hypercube, right? in that case, no

cause a hypercube has 2^n vertices, where n is the dimension

[lethal solstice]

okay

thxx

[weary tiger]

can somebody verify if im reading this correctly?

If the inbox is full, than all emails are lost 

where F(x) is inbox is full, where x is all emails
and L(x) is emails are lost, where x is all emails

F(x)  → ∀xL(x)   like this??

or is it 

∀x(F(x) →L(x))

<@&286206848099549185>

[azure lichen]

I don’t really see how those definitions for F(x) and L(x) make any sense

what is x

[weary tiger]

x is for all emails

[azure lichen]

how can a function take in an email and determine if the mailbox is full?

L(x) is fine

but F(x) makes no sense to take in mails as an argument

however, if you redefine F to take in a sensible argument then F( ) ⇒ ∀x∈Mails: L(x) does make perfect sense

[weary tiger]

this is the full thing

Let M(x) be "email is sent"
Let S(x) be "saved in inbox"
Let F(x) be "inbox is full"

∃x(M(x) Λ ¬S(x))   (there exists* an email is sent, but not saved in the inbox)
∀x(S(x) v F(x))  (All emails* are (saved in the inbox, or the inbox is full))
F(x)  → ∀xL(x)   (If the inbox is full, than the emails are lost)

therefore: ∃xL(x)

There is an email that is sent but it is not saved in the inbox.
All emails are saved in the inbox or the inbox is full.
If the inbox is full, then all emails are lost.
Therefore, some email is lost.

oh okay

[azure lichen]

however, as this reads right now all emails are lost

including whatever had been received

if you don’t mean that, you have to specify further for which x L(x) really is true

[weary tiger]

okay let me think about it and get back to you

[azure lichen]

oh I see how F(x) is meant

it’s intended to be like “when trying to save x, the inbox was full”

[weary tiger]

oh does it still work then

ya

[azure lichen]

in which case, F(x) ⇒ L(x) is all you want

you don’t want all mails to be lost

just the ones that didn’t have space

so whenever the mailbox is full upon arrival, it is lost

[weary tiger]

i have to use the rules of inference to show that the premises have a conclusion that some mails are lost ya

[azure lichen]

also M(x) Λ ¬S(x) without any qualifiers would to me imply an implicit ∀x

but I think oyu meant ∃x based on the text

[weary tiger]

ya

i forgot to update my post

i fixied it before with Ex and Ax before the first two premises

im just kind of lost on the last one if i wrote that right before i begin trying to solve it

[azure lichen]

well, for every x, if the inbox is full (F(x)), then it is lost (L(x))

the way you wrote it, is “if the inbox is ever full, all mails are lost”

[weary tiger]

the question had the premise wrote like this which confuses me

"If the inbox is full, then all emails are lost."

that's why i wrote it like that i think is that still wrong

[azure lichen]

that’s not what you wrote in parens at least

[weary tiger]

ohh

so i wrote it wrong

like in predicate form

[azure lichen]

well idk

your predicate logic and your english in parentheses afterwards don’t say the same thing

also, then*

[weary tiger]

oh

how do i write it like the question asks?

the question asked it as like the parenthesis

which is how you reworded it? for every x, if the inbox is full F(x) then it is lost L(x)

[azure lichen]

yes

[weary tiger]

could you teach me how to read and re-word it?

[azure lichen]

no

[weary tiger]

i think that's a problem i have is trying to rewrite it

[azure lichen]

as in I actually don’t know how to teach that

[weary tiger]

okay

i know i did it wrong and it should be wrote like that, since i dont think i can solve it without having the ∀x encapsulating the F(x) and S(x) which is why i asked

thanks im gonna try solving this, then i'll just ask the pforessor for help after class on my homework

on how to re-word lol

are there any points that i can give you on this server? i just joined yesterday

[azure lichen]

convince a moderator that I am Honorable and should get a fancy yellow name
please do not actually do that

(no, there is no such thing)

[weary tiger]

lol i think i know how to think about it now

i just have to think about x is the mails

then the functions are what happens to it

and go from there maybe

[azure lichen]

yes

[weary tiger]

i see why it didn't make sense now for you

because if i set

let x be all emails

M(x) would be, x email is sent

S(x) would be, x emails are saved

and F(x) would be, x emails are full, which wouldn't make sense lol

i think

[azure lichen]

well, I now interpret F(x) as “when the mail arrived, the box was full”

[weary tiger]

okay

what do you study to know so much math

also im starting to understand the beauty of math and why everything in society is built on it's foundation or something lol

[azure lichen]

math?

I study math

it turns out, if you wanna know math, study it :P

I probably learned more in my first year of undergrad than in at least all of highschool combined, at least when it came to math

[pale berry]

hs is a bit of a meme

[azure lichen]

(I probably learned somewhat more history in hs though)

well mine was actually rather good

and, importantly, nonamerican

[pale berry]

Same

but it was dick

[azure lichen]

math was piss easy if you cared and impossible if you didn’t

[pale berry]

best hs in the region, even got their shitty little inspection rewards

[azure lichen]

is more or less how I experienced it

[pale berry]

One thing I found really enjoyable about studying logic

was solving predicates and problems pretty much algebraicly

I'd love to have more of that

[azure lichen]

I love math but in math class I sat last row (which had advantages such as: the teacher not seing what I am doing because it was like a mini-auditorium so he was below us; being in reach of our own wi-fi; more legroom)

and rarely if ever participated

it was just booring

[pale berry]

I just skipped class so I could actually learn

[azure lichen]

we had a number of classes we could skip per semester

without like, having to jump over a lot of hurdles to not get into trouble

this included missing because of sickness just as much as missing cause you felt like it

(sickness could get excused with a doctor’s note, but even then only if it was a regular occurence or lasted more than three days)

despite being one of the better students, I usually used at least 60% or so of the possible hours that I could miss

(I left some buffer in case of sudden sickness)

[pale berry]

probably makes you a better student even

I hate it when you get punished for inattendance

when the reason I want to stay home is to actually learn xd

[azure lichen]

I mean I more or less ignored most classes anyway

I got through on my +5 to fast learning perk mostly

I tend to pick up stuff rather quickly, it’s like the one thing I got going for me

[pale berry]

+5?

[azure lichen]

at least

[pale berry]

as in a passing grade at the very least you mean right

[azure lichen]

no, as in a DnD-like stat

[pale berry]

oh wait, you mean like an rpg skill

yeah xd

same, used to be extremely driven too, but energy has left my body recently

and details sometimes slip through my fingers

[azure lichen]

nah I was never driven back in high school

I was definitely one of those “naturally smart” kids or sth who never learned and got good grades. had to change that up in university

I had luckily heard enough stories about people like me failing hard in university

and decided to not let chance determine my fate :P

I wouldn’t call me diligent but I’m not lazy anymore either

just… scared of failure?

[pale berry]

wasn't talking about high school

High school sucked the life out of me, because I knew I could be doing so much more in the time I was forced to spend there xd

[azure lichen]

I… idk, I kinda enjoyed high school. I didn’t care much for it but in my last two years I had great classmates and was in a fun club

(robotics. actual robotics, not just that lego bullshit)

complete with deciphering data sheets. fun times

[weary tiger]

sascha if i had my 4 variables

x is the domain of all mails

M(x) - x mails were sent
S(x) - x mails were saved in the inbox
F(x) - when x mail arrives, the inbox was full
L(x) - x mails are lost

If i deduced a single variable like M(x)

does the "addition" rule for rules of inference mean i can  add
M(x)  with any existing variable I have right now? Like  state 

M(x) v F(x)

A mail x was sent, "or", when x mail arrived the inbox was full?

addition rule is

p

---

there fore p v q

[azure lichen]

what does “doing M(x)” mean to you

[weary tiger]

my mail was sent

[azure lichen]

M(x) is a value, either true or false

every mail has that value

it cannot be anything other than true or false, and it can never change

[pale berry]

M(x) is not a correct predicate

[weary tiger]

oh

[azure lichen]

I mean for any given x it is

it’s either “true” or “false”

[pale berry]

mhm

[azure lichen]

but in general you always have to specify what “x” is, which tends to mean either any x (in which case you use for all) or some specific existing one (in which case you use exists)

[pale berry]

it's not how i've seen predicates typically been used, but you could do it this way

[azure lichen]

the way you should think about this problem is, we’ve taken a snapshot in time

every mail has a determined state, which is either “not sent” or “sent”, and either “lost” or “not lost” and either “saved” or “not saved”

and either “inbox was full when it arrived” or the opposite

there is no “doing” anything

[weary tiger]

oh i made a type

typo

how exactly does the addition rule work from rules of inference

basically it says i have my predicate, then i can magically add another predicate using or connector

[azure lichen]

I can’t really give you good advice cause to me Logic is a thing of, well, logically makign conclusions, not blindly applying rules

[weary tiger]

so does that mean i can just take my predicate M(x) mail was sent (or not sent) and connect it with any other one?

[azure lichen]

but this does not seem to fit into the sort of questions that are posed here about logic

there are all these fancy words like “syllogism” and “rules of inference” but it’s all just common sense written down

of course if P is true, then P∨Q is true, look at a damn truth table for ∨

[weary tiger]

oh okay

[azure lichen]

or like… p⇒q and q⇒r
yes, of course p⇒r, what do you think “implies” means? why does this need its own name and deserves a rule worth memorizing?

(sorry, just ranting)

about the only nontrivial thing in all these logic rules is contraposition

that is (Q ⇒ P) ⇔ (¬P ⇒ ¬Q)

and even that is readily verified from a truth table

[weary tiger]

do you know if im allowed to use equivalences while solving with rules of inference?

[azure lichen]

no of course I don’t know that

I’m literally ranting about how senseless I find the concept of having a fixed set of rules like that in a field called logic

[weary tiger]

ya

me too

it makes me more confused

but it did teach me some new things but overall i think its confusing me more lol

omg

@azure lichen i did it

i did the question!

i liked the way you wrote the last premise

[azure lichen]

yay

[weary tiger]

for every x, if the mail box is full, then x mail is lost

thats what you did and i like it because

the textbook question is stupid i think

it just literally says "If the inbox is full, then all emails are lost." but you wrote it so much more beautifully with logic behind it

by telling me that if i wrote it the way i did, it means i lose every sing mail from before too lol

sascha want to read some of my statements and see if you can guess what i'm trying to say?

[azure lichen]

I’m kinda doing stuff, sry

[weary tiger]

oh its okay

[thin hinge]

Is the answer R={(1,3), (1,5), (2,4), (3,5)}

[azure lichen]

that looks right

[thin hinge]

or should I include stuff like (5,1)

[azure lichen]

oh uh

yes

you should

[thin hinge]

hm okay

[azure lichen]

ordered pairs

[thin hinge]

ah okay

[azure lichen]

it’s an equivalence relation so it’s symmetric, but not all relations are

[thin hinge]

okay that makes sense

[azure lichen]

e.g. if R was ≤ then you wouldn’t have (5,1) in it but you would have (1,5)

[thin hinge]

okay thanks

R={(1,3), (1,5), (2,4), (3,1), (3,5), (5,3), (5,1)}?

[azure lichen]

actually missing a few still

come to think of it

[thin hinge]

oh I see

I missed (2,4)

[azure lichen]

no you didn’t

you missed (4,2)

[thin hinge]

wait

[azure lichen]

but that’s stil lnot all

[thin hinge]

I mean (4,2) lol

[azure lichen]

you’re missing five more elements

[thin hinge]

like (1,1), (2,2) etc?

[azure lichen]

yep

[thin hinge]

okay

[lethal solstice]

Guys what is the difference between Eulerian and semi-Eulerian graph?

The definition in the book is thick

[analog sonnet]

If my memory serves, a graph is semi-eulerian by definition if there is an Euler trail, and eulerian by definition if there is an Euler tour (i.e. closed Euler trail)

So every eulerian graph is also semi-eulerian but not the other way around

[heady sedge]

can i get some help with this

[azure lichen]

what have you been able to show, where are you stuck?

is there anything confusing in the definitions or quesitons?

I don’t know what A_n is in the last part though, that must rely on some definition from elsewhere

but the first two seem pretty straightforward proofs by contradiction

[analog sonnet]

Tanaka from Tanaka is Always Listless

[mint harbor]

i came across this on mathstackexchange, how do you even derive $$Y_n \leq Y_{n-1} + 1$$ to begin with? and i have no idea how you get from that to $$Y_n \leq n-1$$

[vital dew]

manzu:

[mint harbor]

to give some context: this problem is about recurrence relations for the tower of hanoi game with 4 towers/pegs instead of 3

<@&286206848099549185>

[radiant bough]

I'm not sure of the problem, but I know you can derive the second statement from induction and using the first statement

you know Y1 = 0 \leq 1

so Y2 \leq 0 + 1 = 1

so Y3 \leq 1 + 1 = 2

[mint harbor]

hmm...

[radiant bough]

So Y4 \leq 2 + 1 = 3

(and so on)

[mint harbor]

thats neat...

okay i think i get that

still confused about how they'd derive $Y_n \leq Y_{n-1} + 1$ though

[vital dew]

manzu:

[radiant bough]

I mean, I think you just do it

and what I mean by that is: you have an expression for Yn

[mint harbor]

i mean intuitively yeah its true that the inequality holds

[radiant bough]

and you have a relationship between some Ws

[mint harbor]

but i wouldnt have thought of using Y_n-1 and add 1 to it

[radiant bough]

no you don't do it that way

[mint harbor]

hmm?

[radiant bough]

you don't start by thinking "oh yeah you know what's cool? Y_n-1 + 1 is pretty cool, so let's put that on the other side"

you start with Yn = the thing with the W

then you just use the inequality given for the Ws

and then you should literally find that once you simplify that, Y_n-1 + 1 pops out

[mint harbor]

holy crap youre right

its like magic ! 🤔

thank you ! @radiant bough

[radiant bough]

haha I'm not sure what else you expected to happen!

In a problem like that, there's really only one thing you can do

so you just have to do it

and hope it works out

[mint harbor]

i dunno, i thought the RHS of the inequality felt like a completely arbitrary choice

so i had no idea how it came about

[radiant bough]

I mean you're not really choosing it

it's just what comes out of the computation

like, you know you want some inequality with the Ys. And you are given some inequality with the Ws

so all you can really do is try to use the W one to get the Y one

[mint harbor]

hmm well, i also wouldnt have thought of substituting an expression Y = something with W

[radiant bough]

how else could you ever prove anything about Y?

[mint harbor]

initially i proceeded with brute force through mathematical induction

[radiant bough]

You know nothing about Y except how it's defined in terms of W

[mint harbor]

well

the Y was introduced as a way to solve the problem

originally this is how the problem looks like

[radiant bough]

I see, I thought you were starting from the Ys

[mint harbor]

so my first approach was doing mathematical induction aka brute force lol

i would never have thought of introducing a Y

[radiant bough]

yeah I will give you that introducing a Y is not obvious at all

but in any case the first step would be to prove the inequality between the Ws

[mint harbor]

yep i see how it all comes together now

thank you!

[radiant bough]

np

[harsh warren]

can anyone explain to me why when dealing with combinations containing a certain suit in a deck of cards (a hand containing 5 spades for example), we use C(13,5) instead of C(52,5)? we’re still drawing cards from a deck of 52 cards isn’t it

[earnest oxide]

http://prntscr.com/mjt6vw

so for the first part would it be {2,3},{2,6},{3,6}

?

[modest zealot]

what is symbolic derivation

also what is this question asking

[craggy gale]

$f(\bbZ) = {f(x)\ |\ x\in\bbZ}$

[vital dew]

Tuong:

[craggy gale]

it's asking to describe these sets

(the thing I wrote was just in case you weren't familiar with the notation f(Z))

[modest zealot]

describe these sets/

?

how tf would u describe them?

[craggy gale]

idk something like

$f(\bbR) =\bbR$

[vital dew]

Tuong:

[modest zealot]

like describe the resulting domain when the numbers r plugged into f(x) = 2x?

i mean range

[weary tiger]

@craggy gale choose any non-real number x, for every NON-integer n, true or false, X^n > 0

[craggy gale]

What

If you go non-real, how do you still dare talking about the usual strict order relation?

[weary tiger]

i dont know my question asked me

to negate that sentance and i got that

and i have no idea if that's true or false lol

how do i type notation?

hold on i'll just draw it

@craggy gale https://i.imgur.com/wm2VzJ4.png

the negation of this, and see if it is true or false

[craggy gale]

$\forall x\in\bbR,\exists n\in\bbZ, x^n\leq 0$

[vital dew]

Tuong:

[weary tiger]

ya but the negation of that

[craggy gale]

$\exists x\in\bbR, \forall n\in\bbN, x^n>0$

[vital dew]

Tuong:

[weary tiger]

ya

i need to learn how to type that lol

that's what i tried telling you in words xD

[craggy gale]

Ah xD

[weary tiger]

except you have to negate the E

so (not) in R and (not) in N

or, not in Z*

$\exists x\not in\bbR, \forall n\not in\bbZ, x^n>0$

[vital dew]

nini:

[weary tiger]

how come mine is black?

[craggy gale]

Uhh there was a command to change that

[weary tiger]

how do i type "not in" or is that the same as negation of episolone

[craggy gale]

$\notin$

[vital dew]

Tuong:

[weary tiger]

oh okay thanks

$\exists x\notin\bbR, \forall n\notin\bbZ, x^n>0$

[vital dew]

nini:

[weary tiger]

so basically i got this, from the negation

is that true or false.. i havn't learned complex numbers yet

[craggy gale]

you don't even have the usual order relation in C

meaning for x in C\R, this "xⁿ>0" does not make any sense

I think you got the negation wrong

[weary tiger]

oh

well that was the question given to me

let me copy and paste

negate this: ( ∀𝑥 ∈ 𝑆,∃𝑛 ∈ 𝑎, 𝑥 𝑛 ≤ 0 )

the S is a R

wait let me just screen cap it lol

[craggy gale]

half the characters don't render on my phone lmao

[weary tiger]

https://i.imgur.com/9w6xWO5.png

[craggy gale]

so

I'm claiming the negation is

$\exists x\in\bbR, \forall n\in\bbZ, x^n>0$

[weary tiger]

oh... am i not suppoed to negate the epsilone?

[vital dew]

Tuong:

[weary tiger]

so the "in" doesn't get negated?

[craggy gale]

it doesn't

[weary tiger]

WAIT REALLy

omg. i'll have to redo another question after this

[craggy gale]

it depends on the context

[weary tiger]

oh

the context is this, Write down the negations of the following expressions, so that the ¬ symbol
does not arise to the left of any quantifier. Indicate whether the negated statement is true.

[craggy gale]

"not (for all x, P(x))" is the same as "there exists x such that (not P(x))"

[weary tiger]

ya you are right

lol

so for all in R negated would be there exists some in R...

not there exists some NOT in R LOL

okay back to re-doing the question before that

i have another one i did wrong then let me brb

wait but @craggy gale

for this question then...

https://i.imgur.com/xPi23H6.png

does that mean i just ignore the negation of the term in the middle?

[craggy gale]

It's different here

[weary tiger]

$exists/in/bbN/, x^2/in/bbN/or/1/2x/notin/bbN$

[vital dew]

nini:

[weary tiger]

is it because there's a comma....

oh...

[craggy gale]

antislash

[weary tiger]

its because there wasnt any quantifier

before the middle term?

so i just negate the epsilone

$exists x\in\bbN,\x^2\in\bbN\or\1/2x\in\bbN$

[vital dew]

nini:
Compile Error! Click the reaction for details. (You may edit your message)

[weary tiger]

lol i suck at this

[craggy gale]

$\lnot(\forall x\in\bbN, x^2\in\bbN\land\frac 1{2x}\notin\bbN)\
=\exists x\in\bbN, \lnot(x^2\in\bbN\land\frac 1{2x}\notin\bbN)\
=\exists x\in\bbN, \lnot(x^2\in\bbN)\lor\lnot(\frac 1{2x}\notin\bbN)\
=\exists x\in\bbN, x^2\notin\bbN\lor\frac 1{2x}\in\bbN$

[vital dew]

Tuong:

[craggy gale]

there

[weary tiger]

ya that's what i got

[craggy gale]

with steps

[weary tiger]

for the first one at least

excpet i negated the episole for the 1st term.. oops xD

okay.. so i get the rule now

if there's no quantifier i negate the episolone

if there's a quantifier i just move onto the next term

[craggy gale]

$\lnot(\forall x, P(x)) = \exists x, \lnot P(x) $

[vital dew]

Tuong:

[weary tiger]

ya i know the demorgan's law

i just thought that E N was a predicate lol.

$\in\bbN$

[vital dew]

nini:

[weary tiger]

i thought this was a P(x) LOL

[craggy gale]

x')

[weary tiger]

oaky let me try solving this myslef then ill ask you for help if im unsure

one day im going to be a one strong tuong apprentice!

[craggy gale]

haha <3

[weary tiger]

@craggy gale is this correct for how i solved this one?

https://i.imgur.com/VICtAVO.png
Let S(x) be x solution for x^2-x-1=0
Let N(x) be x in all the natural numbers
∀x(N(x)→¬S(x))

[craggy gale]

I'm not sure I understand what your S(x) and N(x) are supposed to be

[weary tiger]

oh

Let x be numbers
Let N(x) be x in the domain of natural numbers
Let S(x) be x numbers in the equation x^2-x-1=0

maybe i should just say numbers

[craggy gale]

Is this just a quantifier translation question?

[weary tiger]

ya

well it just says try writing the following sentence in predicate logic

does my updated one make more sense?

i'm still learning how to write more beautifully like you xD

i think my math grammar is so bad i write like a hobo

[craggy gale]

I still can't make any sense of it xd

[weary tiger]

aw okay

[craggy gale]

$\forall x\in\bbR, x^2-x-1=0\implies x\notin \bbN$

[vital dew]

Tuong:

[weary tiger]

ya i want to write it like that but i have to use predicates

so i just wrote it like ∀x(N(x)→¬S(x))

lol...

but i dont know how to define my predicates any well

$\forall x\in\bbN\implies\negation x^2-x-1$

[craggy gale]

maybe with S = {x in R | x²-x-1=0}
∀x in S, x not in N

[vital dew]

nini:
Compile Error! Click the reaction for details. (You may edit your message)

[weary tiger]

okay

[craggy gale]

$\lnot$

[vital dew]

Tuong:

[craggy gale]

Also there's this

$P\implies Q$ is the same as $\lnot P\lor Q$

[vital dew]

Tuong:

[weary tiger]

okay! i remember that one

im good at equating equivalences i think

[craggy gale]

I'll go to sleep now, good luck with your exercises!

[uncut isle]

https://media.discordapp.net/attachments/521095928443306019/544745734256459797/unknown.png

the domain would be p and q, and the range would be f(p,q)? I'm not exactly sure what the codomain would be

[sour arrow]

The domain is sometimes written B², for two boolean variables

Codomain is B

Range? I'm not sure give me a sec @uncut isle

Well, B since it can be 0 or 1

[uncut isle]

oh wait so it isn't p or q?

I know that like A --> B, A maps to B
Then domain is A, codomain is B, and range is f(A), but the mapping arrow isn't the same as the implication one, i dont think

not sure if it's supposed to be like outputting truth values or whatever

[sour arrow]

f(x) = x²
The domain and codomain isn't x

f(x, y) = p → q

Takes in two boolean values, and gives one back
f: B² → B
f: [0, 1]² → [0, 1]

[uncut isle]

ohh okay, domain would be B^2, codomain would be B and range is f(x,y)?

since my book i think defines it as "all the possible points that can be inputted"

also i just got done with an exam, this was one of the questions

a lot of my classmates just did sum(a) + sum(b), which i didn't think to do cause i havent used series in a long time

i did this instead

and someone told me i couldn't do that? if I can't, why can't i?

[vocal sable]

I guess there's a factorisation mistake

$4k^4+4k^3+1 =/= 4k^2(k^2+k+1)$

[vital dew]

DEAD.:

[uncut isle]

oh rip my bad

but im allowed to do like the first line right?

so many stupid mistakes lmfao

i just wanna know if im allowed to do 4k^2((k(k+1))/(k(k+1)))

$ \dfrac{1}{k(k+1)}+\dfrac{4k^2(k(k+1))}{k(k+1)}$

[vital dew]

Majesticat:

[uncut isle]

that ^^ instead of doing

$\sum_{k=1}^{n} \frac{1}{k(k+1)}+ \sum_{k=1}^{n} 4k^2$

[vital dew]

Majesticat:

[vocal sable]

Hmm.

[modest zealot]

what is the logic equivalence of "is"

n^2 + n + 5 is odd

implication?

[uncut isle]

im noob so im not sure but dont you have to use propositional logic?
like for all x such that x = n^2+n+5, then theres a y such that x=2y+1
2y+1 being the odd integer.

$\forall{x:x=n^2+n+5}, \exists{y:x=2y+1}$

[vital dew]

Majesticat:

[uncut isle]

that's logically equivalent to saying n^2+n+5 is odd

?? probably lmfao :')

[sharp ravine]

I have Epp's hardcover book to self studying Discrete math. I wonder if you guys read a Discrete Math book (Epp or Rosen) from cover-to-cover or no?

I also have Book of Proof by Hammack

[uncut isle]

i have one im trying to go through but i haven't gotten far sadly

its the velleman one

[sharp ravine]

I see. I am still in Chapter 1 (Rules of inference)

[uncut isle]

oh wait i have rosen

that's the textbook right?

[sharp ravine]

Yes

I also read that one and I am surprised that they have different rules of inference.

[pale berry]

so for recurrence relations

Are there recurrence relations we can't solve through diagonalizing matrices?

[sour arrow]

Non-linear ones

[craggy gale]

when the relation isn't "linear", you can't even have matrices

[pale berry]

makes sense

[sour arrow]

Why are we diagonalizing matricies? What's the method?

[pale berry]

does that render them unsolvable then?

I know you can derive an explicit formula for fibonacci through diagonalizing for example

Can we do non-homogenous ones too in this way?

also, how about the instances where we don't have enough eigenvectors to diagonalize?

[craggy gale]

You diagonalize to find, in the end, the powers of a matrix right?

[pale berry]

yes

taking the 1000th power of the diagonalized fibonacci matrix nets you the 1002th term if you transform (1 1) in this way

the vector (1 1)

[sour arrow]

Oh cool, I see what you're up to

[craggy gale]

Well if you can't diagonalize, there are other means to calculate these powers

[pale berry]

I know, but the linear algebra way seems the most intuitive to me

[craggy gale]

by other means, I mean still lin alg related :p

[pale berry]

Actually derived it this way myself, as it was way more straight forward than the linear combination proofs i've seen for these

owo

complex eigenvalues 🤔

[craggy gale]

Dunford decomposition maybe

[pale berry]

t!wiki dunford decomposition

[uncut grove]

📖 | ** https://en.wikipedia.org/wiki/Jordan–Chevalley_decomposition **

[sour arrow]

You can educated guess a function for fibbonacci

[pale berry]

and prove it through induction

not needed though

[sour arrow]

But that lin alg method is lit

[pale berry]

it is uwu

I wonder if this would work for non-homogenous recurrence relations too 🤔

i've noticed that linear homogenous differential equations of the second order had a similar looking explicit form to that of recurrence relations

What was the explicit form 2nd ordered sequences that only had 1 solution to the characteristic equation again?

(A+Bn)g^n = a(n) or something?🤔

Suppose we had 1 line spanned by eigenvectors i.e. 1 eigenvalue

if we restrict ourselves to 2 dimensions

[lean leaf]

Determine whether f is a function from the set of all bit strings to the set of integers if f(S) is the position of a zero bit in S.

Could someone help me with this? I know what a bit string is and I understand that f maps a bit string to an integer but I don't know how to form a proper answer.

[pale berry]

functions have 2 properties

for all x in the domain f(x) = y where y is in the co domain

and

for all x,y,z if f(x)=y ^ f(x)=z then y=z

[lean leaf]

Mhm...

[pale berry]

that's all you have to prove

of "a" zero bit in S

it's not a function

because S can contain multiple 0 bits

[lean leaf]

Oh, so it would map to several results?

[pale berry]

Well, your codomain is integers, not sets of integers

we don't call the circle equation x^2 + y^2 = 1 a function as an output for an input x can have multiple solutions for the same reason

the definition for proper functions becomes more intriguing when dealing with recursive formulae

as recursive steps can get you situations where f(5) =/= f(5)

[lean leaf]

oh lol

[pale berry]

owo

[lean leaf]

That's awkward.

Anyway, I think I'm starting to understand this.

[pale berry]

Pay close attention to the definition of equivalence relations

reflexivity is one that cucked me out of a 10

[lean leaf]

@pale berry Okay. Thanks!

[harsh warren]

some help with a question please!
How many permutations of the letters ABCDEF contain the substring DEF?

[viral stump]

4!

[harsh warren]

could you explain how you got it?

[viral stump]

DEF has to stay as a block

[harsh warren]

what does substring mean?

[viral stump]

so we have 4 blocks

A B C DEF

we can permute these in any way

for example A B C DEF or A B DEF C

etc

[harsh warren]

ah i see

shouldn’t there be more than 4 then?

ABCDEF
ABDEFC
ADEFBC
DEFABC
DEFBAC
etc..

or is there no difference between the last 2

[viral stump]

4! isnt 4

[harsh warren]

can i ask what were your steps to arrive at 4! ?

i used 6P3 and still arrived at the same answer but i’m just wondering how you did it

oh wait never mind, i understand now

thanks!

[viral stump]

nice

[earnest oxide]

can anyone explain this? not solve it but help me decipher it

[earnest oxide]

@violet heath ?

[glad oriole]

when you mutiply two powers with the same base you add the exponents

so

[vital dew]

rockpaperscissors:

[patent prairie]

I second that

tbh though deciphering is hard to explain

[copper ore]

Using only the rules of inference and the logical equivalences listed on the last page of this quiz, show that the following argument is a contradiction by reducing it to a value of "False". You may assume that all the premises given are true. Make sure that you include both the rule and the line number(s) to which that rule is applied. 
𝑎→𝑏 
¬𝑏∧𝑐 
¬𝑎→𝑑 
𝑑→¬𝑒 
𝑒∧𝑓

how do i do this?

[azure lichen]

without listing that list of rules you are allowed to use, no one will be able to help you

also, what have you tried yourself?

[lean leaf]

Let f(x) = 2x. What is f(R)?

I know the answer is R, but I don't understand why.

[proven shard]

f(R) = 2R

[lean leaf]

Yes, I get that.

But f(N) for example is all even natural numbers

Same with f(Z)

[proven shard]

If R denoted the set of reals then show f is a bijection from R to itself

[lean leaf]

Ok, so to be a bijection it needs to be injective and onto, right?

[proven shard]

Yes

Honestly, you only need to show f is onto

[lean leaf]

Hm, I understand why it is injective

but I don't understand why it's onto

like

2R doesn't map to every element in R, does it?

[proven shard]

It's onto because you can divide by 2 in R.

It's not on to in Z or N because you can't divide by 2 in Z or N

[lean leaf]

why not?

f(Z) = {..., -2, 0, 2, 4, ...} (example)
I can divide all this elements by 2, can't I?

[quaint river]

Pick any element a in R. Then, f(a/2)=a

a/2 exists in R

[copper ore]

it's just the list of inference and logic @azure lichen it says it in the Q

[quaint river]

But the integers are not closed under division

So t does not work there

@lean leaf

[azure lichen]

there is no "the list of inference and logic" @copper ore

any statement you could prove (including this exercise) could be on such a list

[quaint river]

Rules of inference are rules like modus ponens, disjunctive syllogism, etc

@copper ore use e AND f implies e is true by simplification

Then d -> NOT e is equal to NOT d or NOT e

By disjunctive syllogism NOT d has to be true so d is false

Repeat with the other statements and you get that b is true

Thus, NOT b and C is never true

[paper thicket]

How can I go about proving that ¬(P v Q) = ¬P & ¬Q ?

Im starting with ¬P & ¬Q as a premise and deducing ¬P and ¬Q

And then im trying to join things with ¬P by adding or but idk where to go with that

[paper thicket]

Nvm I got it

I started with ¬P & ¬Q

Did a double negation

Then carried one negate using demorgans law which flips the v to &

[fast night]

Can somebody help me understand the meaning of what's being said in this text?

[patent prairie]

which part

@fast night

like, which part are you confused by?

@fast night are you available

[fast night]

All the text in general, I mean I understand the idea of a boolean function, what it is but i don't understand , for example, where that

2^{2^n}

comes from

[patent prairie]

darn it, I was just going to say I had to go soon

but let's see if I can crank out an explanation rn

the n represents how many inputs there are

2 ^ n is how many inputs can be inputed

i.e. for 2 inputs

there are 4 possible inputs total

00, 01, 10, 11

then 2 ^ that

is just how many binary outputs can be

I hope that helps

[fast night]

oh

thanks, but what's the reasoning behind that? I mean that number of outputs?

[patent prairie]

okay, I'm back

had to deal with something

so, imagine you can input 0, 1, 2, 3 into a function

and it can output either 0 or 1

actually, let's say you can only input 0

then there are only two possible functions

f(x) = 0 and f(x) = 1

if you can input 0 or 1 there are twice as many functions

0 -> 0; 1 -> 0

0 -> 1; 1 -> 0

0 -> 0; 1 -> 1

0 -> 1; 1 -> 1

this is because the choice of what f(1) is is independent of f(0)

each has 2 choices

2 * 2 = 4

what do you think of that explanation, I can be more specific if you are still confused

[fast night]

sorry, in some parts I think I start getting it and then I get confused. What do you mean by "actually, let's say you can only input 0
then there are only two possible functions
f(x) = 0 and f(x) = 1"?

[patent prairie]

what I meant was that

if you have f(x) which has domain {0}

and range {0, 1}

how many unique functions are like this?

@fast night sorry for not getting back to you sooner, again, doing stuff

actually this is the best part of the above explanation

each has 2 choices
2 * 2 = 4```

that's for domain {0, 1} and range {0, 1}

the chart in the first picture also exempifies this

where the domain is {00, 01, 10, 11}

they have a row for each possible function

[fast night]

I think I get it now, thanks for your dedication, and don't worry for not answering sooner, we all have stuff to do

[patent prairie]

np glad I could help

[fallow zealot]

Could anyone explain what is a linear in homogeneous recurrence relation?

[viral stump]

https://math.berkeley.edu/~arash/55/8_2.pdf

[sand vault]

Anyone available to help with some modular arithmetic?

@sudden knot

[sudden knot]

hi

[sand vault]

Are there any videos or anything that helps explain modular arithmetic that you know of?

[tranquil roost]

out of 10 how difficult would you rate this question

https://cdn.discordapp.com/attachments/355096113339957248/546002498553511936/Bildschirmfoto_2019-02-15_um_18.14.11.png

[azure lichen]

trivial

[tranquil roost]

97% of surveyed high school students in my country could not answer this correctly

[azure lichen]

were these high school students familiar with set builder notation and what an intersection of sets is

cause obviously if you can’t parse the notation it would be very hard

(also the notation is bad)

[tranquil roost]

They know set builder notations and know what the operators are

how is the notation bad, if you don't mind explaining?

[azure lichen]

would imo be more elegant to write it as $${ x \in \mathbb{R} \mid -3 \leq x < 9 }$$

[vital dew]

Sascha Baer:

[azure lichen]

two issues:

why is the x bold

and why does the lefthand side not indicate what set the x is taken from

usually set builder notation goes {x ∈ larger set | attributes that x fulfills}

but honestly the bold annoys me almost more, cause imo bold x and regular x are just different entities

[tranquil roost]

Ah, thanks for the feedback, I'll let the lecturer know about that actually 😄

[azure lichen]

also, I assume most people who got it wrong went with 1 or 3?

(either misreading the < as ≤, or misinterpreting the intersect as a union?)

(though imo 0 is not in Z⁺ anyway)

(but these notations are always awfully ambiguous)

would be interested to see the stats

[tranquil roost]

a mixture of interpreting the operators incorrect - most were just flabbergasted because the notation was different to the standard notation they wer eused to

I'll see if I can upload them here

[azure lichen]

what would the standard be?

I find it weird that so many got it wrong. like that means most people were misled somehow, and weren’t guessing by chance

like if it was 75% who got it wrong then people just didn’t get it and guessed wrong

[mint harbor]

i can prove this algebraically, but im just curious as to how you can prove it using a counting argument

the left hand side counts the number of r-permutations of an n-element set, and we can count this by first obtaining an (r-1)-permutation of the same set. there are P(n, r-1) such permutations, and for the r-th element of such permutation, we have (n-r+1) remaining possible elements to add to the permutation

how can i proceed from here to show that the number of r-permutations is indeed (n-r+1) times P(n, r-1) ?

<@&286206848099549185>

[late gust]

Rigorous counting argument or intuition?

Because intuition wise, just think like this

If we have all of the r-1 permutations, we can construct all the r permutations by considering all the different possible endpoints

[mint harbor]

hmm, mind giving both the intuition and a rigorous counting argument? 😃

[late gust]

I'm not that good at rigour so you might need someone else

But here goes my attempt at intuition

[mint harbor]

hmm alright

[late gust]

So has what I've said made sense so far?

[mint harbor]

i dont really understand why its enough to consider 'all different possible endpoints'

so, there are n-r+1 possible remaining elements we have not added to the permutation

[late gust]

Yeah

[mint harbor]

we choose one out of these n-r+1 elements, but it doesnt mean that we simply append it to the end of the permutation

[late gust]

It's enough to think about just the endpoints, because if you envision putting another remaining element somewhere else, that permutation minus the endpoint already exists

Because you're still using elements of n

[mint harbor]

so its not exactly the 'endpoint' 🤔

[late gust]

It's enough to consider one position, the endpoint is just easiest to envision because it leads nicely from just thinking about counting permutations as a factorial

Let's use an example to make sure the thinking makes sense

If we've got numbers up to 10, let's start with permuting just 2 of them

One of these permutations will be 12

We can just consider adding something to the end of 12, and to the end of all the other permutations

If you think about adding something to another position in that permutation, say making it 132, 13 already existed as a permutation of length 2, so that's already covered

[mint harbor]

ehhh but you choose 1 and 2, the n-r+1 remaining elements include 3 (i.e 3 is not included as one of the r-1 initial choices)

so how could 13 have already existed?

[late gust]

The n-r+1 remaining elements won't always be the same n-r+1 remaining elements

[mint harbor]

😮

[late gust]

Because our already existing permutations use different elements

If we're considering 12, we have 3-10 to consider

If we're considering 13, we have 2, and then 4-10 to consider

So the number of elements will remain the same, but the exact elements we're considering won't

Which is why permutations initially work

[mint harbor]

oh... okay

thats... confusing

[late gust]

If we're doing a 3 letter permutation of the alphabet, we have 262524

oh crap

asterisks don't work on discord

uh

26 x 25 x 24

It doesn't matter which of the 26 we choose in that first one, only that there are 26 possible choices, and after that, there are 25 possible choices

If the first one is A, or B, or C, there are still 25 possible choices

[mint harbor]

okay i see the problem with fixing n-r+1 elements to be the same every time

[late gust]

Yeah combinatorics can be super counterintuitive, despite the fact that it's foundation is just clever ways of counting things

[mint harbor]

so by considering all the different possible endpoints

there are n-r+1 possible endpoints for each permutation

and P(n,r-1) permutations

so we get (n-r+1) times P(n,r-1)

[late gust]

Yup

For me, when combinatorics gets confusing, examples help as a sanity check

So think again about our 10 numbers

If we're doing up to 4 of them, we start with 10 x 9 x 8 x 7

Now if we extend to 5, which is our r, let's multiply by (10 - 5 + 1)

Which is 6

And it makes sense, if we're permuting 6 of them, we have 10 x 9 x 8 x 7 x 6

[mint harbor]

man i wish i thought of that example

i was thinking in terms of a set with n elements and forming r-permutations from this set

[late gust]

Keeping your examples more practical and less abstract tends to help, I find

Don't think elements, think numbers, or letters, or cards

[mint harbor]

but lets say we want to prove pascal's identity via a counting argument i.e. C(n,r) = C(n-1,r-1) + C(n-1,r)

[late gust]

Uh

A little over my head to do that

[mint harbor]

i would have immediately thought of using elements lol

[blissful basalt]

Any good resources on on Proofs? (direct argument, contrapositive argument and proof by contradiction), need to learn this stuff for compsci

[dense thicket]

How do I check given a relation on two sets whether they are isomporhic?

[dense thicket]

Like if one is dense the other one has to also be dense, any others? if ones well ordered, linear second one also has to be for them to be isomporphic?

[ripe cave]

These are more like ways to disprove two ordered sets being isomorphic

If one ordering is linear and the other one not then the sets are definitely not isomorphic

Same for well-order and dense order

But to show that two sets are isomorphic I think usually you can just find an isomorphism

[dense thicket]

@ripe cave k thanks : ) !

[ripe cave]

[copper ore]

The ceiling of a real number x, denoted by ⌈𝑥⌉, is the unique integer that satisfies the
inequality
⌈𝑥⌉ − 1 < 𝑥 ≤ ⌈𝑥⌉.

Prove or disprove that, for any real number 𝑥…
if ⌈𝑥⌉ − 𝑥 ≥ 1/2 then ⌈2𝑥⌉ = 2⌈𝑥⌉ − 1.

this is what i have so far and i’m stuck

[weary tiger]

can't you try induction on reals @copper ore

choose some value of x and try for any k, x + k

is that even discrete math

maybe try putting it in #geometry-and-trigonometry ?

[copper ore]

it is discrete math @weary tiger

[weary tiger]

@copper ore what @polar pulsar put was not