discrete-math

[flat echo]

And I found this nice number theory book

[weary tiger]

i wish i was able to have the discipline to do that

lmao

[flat echo]

Haha

[sleek swallow]

consider two baskets of objects, yea?

Basket A contains m objects and basket B contains n objects

Now, you want to choose a total of k objects from both baskets

@flat echo so far so good?

[flat echo]

Yeah

[weary tiger]

Master guide me

[sleek swallow]

Alright, so if you want k objects from the m+n objects, then that's just $\binom{m+n}{k}$

[vital dew]

Abhijeet Vats:

[weary tiger]

actually dont help me yet ill attempt to do it by myself

[sleek swallow]

However, there are multiple ways for me to get k objects from both baskets. I could take 0 from basket A and k from basket B. I could take 1 from basket A and (k-1) from basket B. I could take 2 from basket A and (k-2) from basket B and so on

Yeap, attempt it yourself and struggle over it a bit

@flat echo

Do you get me so far?

[flat echo]

Yes, I I think so

[sleek swallow]

So, above 'multiple ways' is just encoded in the left-hand side of the equation you have

If i take i objects from basket A and k-i from basket B, then the total number of ways of doing so is:

$\binom{m}{i} \cdot \binom{n}{k-i}$

[vital dew]

Abhijeet Vats:

[sleek swallow]

And all i have to do is to vary i from 0 to k and that forms a sum that's equal to the right-hand side

[flat echo]

I was guessing so, but I need to prove it analytically and, I think, without using combinatorics

It's from a chapter that talks about the binomial theorem

And it doesn't require a background in combinatorics so far (it explains only the definition of a number of that kind)

[sleek swallow]

Consider $(1+x)^{m+n} = (1+x)^m \cdot (1+x)^n$ and look at the term with $x^k$ in it.

[vital dew]

Abhijeet Vats:

[sleek swallow]

If you compare those terms in both the representations of the same thing, you should be able to get the sum that you want

[flat echo]

I'm sorry, but I'm not getting it 😔

[sleek swallow]

Well:

$(1+x)^{m+n} = 1 + \binom{m+n}{1} x + \binom{m+n}{2} x^2 .....+ \binom{m+n}{k} x^k + ...+x^{m+n}$

[vital dew]

Abhijeet Vats:

[flat echo]

Ohh, yes, I get that

But I dont get it how to use it to prove the last thing

[sleek swallow]

Now, do the same for $(1+x)^m$ and $(1+x)^n$ and find all ways of multiplying them such that you get terms with $x^k$ in them

[vital dew]

Abhijeet Vats:

[weary tiger]

Abhijeet for structural induction i have to prove all base cases first right?

[sleek swallow]

Idk what structural induction really is but if it's the same thing as just normal induction, then you do have to start with the base cases

[weary tiger]

kk

[runic garnet]

nachens you can search up Vandermondes identity

[flat echo]

Im thinking haha

[sleek swallow]

After you've thought through this, do what Plum said.

[weary tiger]

im so confused, how do i prove the base cases

am i supposed to prove that 1 is divisible by 2

wat

for the base case im supposed to prove that lambda is divisible by 2, 0 is divisible by 2, and 1 is divisible by 2?

[flat echo]

Thanks

I have just sent it to some friends

It took me a good time to understand it

But thanks

[past dew]

I just want some reassurance for my lab. I don't know which is the better option so I can start writing my recursive equation. Basically the first Option#1 is just $100,000 and Option#2 is $1,073,741,824 because (1 x 2^year) ,but I'm confused on Option #3. Am I basically getting infinitely $1 each year, therefore Option#3 is better?

[weary tiger]

I need help finding the inductive step

[inner gyro]

where does the m-n come from

on both sides

[faint narwhal]

2m = (m + n) + (m - n)

[weary tiger]

@faint narwhal could u help me find the inductive step in my problem

or identify it

[analog dagger]

inductive step is something in your proof

[weary tiger]

I know that

@analog dagger with my specific question is what would i be trying to prove

i proved the base cases p(lambda), p(1), and p(0)

[analog dagger]

suppose p(k) is true <- induction hypothesis

then if you can prove p(k + 1) is true

that is the inductive step

[weary tiger]

https://cdn.discordapp.com/attachments/496785905474994186/682057322377248818/image0.png

In this case?

Inductive Step: Assume P(x) is true, Prove ???

im not sure about that

[analog dagger]

P(x + 1)

[weary tiger]

why P(x+1)?

[analog dagger]

thats induction

[weary tiger]

this is structural induction

im asking

not regular

[analog dagger]

no idea what "structural induction" is

[weary tiger]

ahh

alright

thanks anyways

[woeful galleon]

hey guys! can someone help me out here ?

[sleek swallow]

What do you need assistance with?

@woeful galleon

[woeful galleon]

dont really understand what i am supposed to @sleek swallow

[sleek swallow]

For the first part, write down a chain of equivalences

So begin with:

$x \in (A-B) \cup (A \cap B) \iff x \in A-B \lor x \in A \cap B$

[vital dew]

Abhijeet Vats:

[sleek swallow]

Then proceed from there

Of course, you should specify that x is an object contained in some universe of discourse, with A & B being subsets of that universe

[woeful galleon]

sorry for the stupid question but what does the "V" stand for in that case.

just wanna make sure

[sleek swallow]

Logical or

It's not a stupid question. You're learning a language and that takes time.

[woeful galleon]

aight, that is what i thought. had a hard time with my other classes since all these symbols are starting to mean different things lol

[sleek swallow]

Anyways, try to proceed from what I've written above

[woeful galleon]

so we can say that or = union

and "and = intersection"

[sleek swallow]

No, you may not. Logical operators are related to set operators but they're not the same thing

Let A and B be sets. Then:

$x \in A \cup B \iff x \in A \lor x \in B$

$x \in A \cap B \iff x \in A \land x \in B$

$x \in A - B \iff x \in A \land x \notin B$

[vital dew]

Abhijeet Vats:

[sleek swallow]

So, they're related in the sense that I've just described. But you can have logical operators act in a way that is independent of anything in set theory. There's no reason why a given proposition, for example, has to be set-theoretic

@woeful galleon

[woeful galleon]

aight, i am here. i believe u also helped me with my previous things lol

i highly appreciate the help

[sleek swallow]

You're welcome.

[woeful galleon]

(A-B) V (A and B) is this wrong so far

i got rid of the $x \in $ .. but it is still there

[vital dew]

Tareku:

[sleek swallow]

Well, without the $x \in$, what you've written has no meaning

[vital dew]

Abhijeet Vats:

[woeful galleon]

i am a lazy typer but i understand, where do i go from there ?
fixing it

[sleek swallow]

Well, okay. So:

$x \in (A-B) \cup (A \cap B) \iff x \in A-B \lor x \in A \cap B \iff (x \in A \land x \notin B) \lor (x \in A \land x \in B)$

[vital dew]

Abhijeet Vats:

[sleek swallow]

Oof

[woeful galleon]

lol

[sleek swallow]

But you should be able to parse through that

Understand so far?

[woeful galleon]

looking

[sleek swallow]

aight

[woeful galleon]

$(x \in A \land x \notin B)$

[vital dew]

Tareku:

[woeful galleon]

this is A - B

correct?

[sleek swallow]

Yeap

That just means that $x \in A-B$

[vital dew]

Abhijeet Vats:

[woeful galleon]

yep. just making sure of everything so i dont get lost

[sleek swallow]

Sure, take your time

[woeful galleon]

i follow so far

[sleek swallow]

Aight nice

So, now I'm going to use distributivity of logical operators and do the following:

$x \in (A-B) \cup (A \cap B) \iff [x \in A \land (x \notin B \lor x \in B)]$

Ah fuck

[vital dew]

Abhijeet Vats:

[woeful galleon]

just to have it near lol $(x \in A \land x \notin B) \lor (x \in A \land x \in B)$

[vital dew]

Tareku:

[woeful galleon]

aight i see it

[sleek swallow]

Yeap

and if you noticed, $x \notin B \lor x \in B$ is just a tautology and taking the conjunction of a proposition and a true statement just yields that proposition

[vital dew]

Abhijeet Vats:

[sleek swallow]

In other words,

$p \land TRUE \iff p$

[vital dew]

Abhijeet Vats:

[woeful galleon]

is it wrong to say that it is = 1

[sleek swallow]

Uh you mean $x \notin B \lor x \in B = 1$?

[vital dew]

Abhijeet Vats:

[woeful galleon]

yes

sorry for not explaining lol

[sleek swallow]

Uh i don't particularly like it but i know it's common to use the 1 and 0 thing in comp sci

So i guess you can

[woeful galleon]

aight yea lol. i like 0s and 1s

okay. so part 1 is done 😄

let me look at part 2

[sleek swallow]

Yeap

The point is to go back to the definitions

And work from there

[woeful galleon]

so cardinality is the set's size

correct ?

[sleek swallow]

The number of elements in the set

[woeful galleon]

so the first one is 0

[sleek swallow]

Are you asking me or telling me?

[woeful galleon]

both ?! i am telling you lol

the first set has 0 elements

second set has 1

and then i am kinda lost. i wanna say that 3 has 2 and 4 has 4 ?

[sleek swallow]

List out the elements of each set and then count them. Like, write them out separately so you're not confusing yourself

[woeful galleon]

were all my answer wrong

cuz i am pretty sure 1 and 2 are correct

[sleek swallow]

I didn't say that they were wrong. But you need to be sure that what you're writing makes sense. I won't tell you if they're right or wrong till you're more confident and can justify why you think you're right.

[woeful galleon]

aight, let me do it then

written i should say lol

well, let me do it slowly, the first one is ${empty}$

mh

[vital dew]

Tareku:

[sleek swallow]

Indeed, so $|\varnothing| = 0$

[vital dew]

Abhijeet Vats:

[woeful galleon]

second one is $|{\varnothing}| = 1$

[vital dew]

Tareku:

[sleek swallow]

Correct, because it has $\varnothing$ as its single element

[vital dew]

Abhijeet Vats:

[woeful galleon]

now for 3, same thing. $|{\varnothing, {\varnothing}}| = 2$

[vital dew]

Tareku:

[woeful galleon]

and for 4,$|{\varnothing, {\varnothing},{\varnothing, {\varnothing}}}| = 4$

[vital dew]

Tareku:

[sleek swallow]

Are you sure there are 4?

[woeful galleon]

before that, do we agree on 3

[sleek swallow]

Yeap that's correct

[woeful galleon]

i am not gonna lie, i am kinda "guessing" not really, but i am looking at it and doing it in my head. is there a better way of doing this ?

[sleek swallow]

You should go back to your notes and review set theory if you're unsure about this.

[woeful galleon]

hmm

i did just now lol

i did do something and it might be 2 ?

[sleek swallow]

No

Like i said, go and study them in detail

Then, do the problems

You're guessing your way to the answer and while guesses can be useful, this isn't the sort of thing you should be guessing

[stray reef]

what even is the problem

[sleek swallow]

https://media.discordapp.net/attachments/496785905474994186/682080026937917507/unknown.png

[stray reef]

and why exactly would one ever need to guess when it's a straightforward counting matter

[sleek swallow]

That's exactly what I'm saying. It's not the kind of thing that needs to be guessed.

[woeful galleon]

i am aware that it "straight forward counting matter" that is why i am here asking for help and studying instead of shooting random numbers until it looks right lol

[stray reef]

the elements of {∅, {∅}, {∅,{∅}} } are ∅, {∅} and {∅,{∅}}

there's three of 'em

[woeful galleon]

yea, i found a nice video explaining that lol

[sleek swallow]

Wtf

How tf...

How did you type that out without texit?

[stray reef]

autohotkey lol

[woeful galleon]

∅

cheater

[sleek swallow]

Fuckkkkkkkkkkkkkk

[woeful galleon]

i just finished watching and reading notes again and it makes a lot more sense lol. i was missing a few things xD

[weary tiger]

How would one mention a lemma in a prood

proof*

[sleek swallow]

Use the following sentence:

"Lemme use this lemma"

[weary tiger]

Awesome thank you!

man im dead as shit

ive been doing homework since 1 PM

and working on my discord bot since 9

[sleek swallow]

Honestly

[weary tiger]

life is hell

[sleek swallow]

I haven't had homework in so long

[weary tiger]

but this discord bot be bringing me stacks

[sleek swallow]

I'd actually like to have homework

[weary tiger]

why not self learn stuff

and assign ur own homework

lmao

[sleek swallow]

Lel yea

[weary tiger]

i made a discord server a week ago dedicated to cheating on homework

[sleek swallow]

:/

[weary tiger]

1700 members baby

[sleek swallow]

😦

[faint narwhal]

sad

[weary tiger]

lmao

[sleek swallow]

It's not a good thing to do

[weary tiger]

i dont do it

specifically its a chegg server

for free chegg unlocks

[sleek swallow]

Idk what chegg is

[faint narwhal]

oh, screw chegg, im good with this

[weary tiger]

but none of my homework is on chegg

thats why i go to my senpais

when ur professor creates his own assembly language just so students wont cheat

[near prawn]

damn

that is hardcore

[small kayak]

Do i have to write a binary string with length of 30 for representing set B and C ? Or there is other ways to write binary string representations for B, C ?

[sleek swallow]

?

$B = {6,12,18,24,30}$ and $C = {8,16,24}$. Then, write all of these elements in binary?

@small kayak

[vital dew]

Abhijeet Vats:

[small kayak]

No, write it as binary string. For example, U = { 1, 2, 3, 4, 5, 6, 7, 8}. Then subset of U which is A = {1, 3, 4, 6} is represented as the bit string 10110100

[sleek swallow]

Oooh interesting. Uh then i guess you have to do it by hand or get a computer to do it for you lmao

[small kayak]

I guess i have to write a binary string with length of 30 for representing B, C, union of B and C, intersection of B and C. But i don't know how to write binary string representations for complement of B.

[small kayak]

Is my solution correct ?

[pale epoch]

your argument for symmetric and antisymmetric is unclear

moreover it is wrong as R is not symmetric

the last point can be written more clearly

[small kayak]

Why the relation is not symmetric ? I think for every (a, b) in the relation there is (b, a) in the relation too, so it is symmetric, but here a is not different from b so the relation is also antisymmetric.

[pale epoch]

({1}, {1,2}) is in R but ({1,2}, {1}) is not

[small kayak]

Oh i forgot that. Tks a lot !

And the last point you meant transitive relation right ?

[pale epoch]

just the way you wrote it down is not so nice

there should be at least a comma between "(a, c)" and "R"

[small kayak]

Okay. Tks a lot for your help

[open narwhal]

I'm having a little issue, I have to write sentences symbolically using quantifiers, then write the negation until there are no negation symbols left, then i have to write the negation in an English sentance.

the first sentence: The is a real number whose square is nonzero.

this is what i put for the statement with quantifiers

[pale epoch]

is this supposed to be $\mathbb{R}$ or $\mathbb{Z}$

[vital dew]

Lochverstärker:

[open narwhal]

$\mathbb{z}$

the z, did that wrong ig

[pale epoch]

Z is the set of integers, R is the set of real numbers

[open narwhal]

okay so thats 1 thing i did incorrectly thank you

i'm not sure how i would right the negation symbolically for this at all i thought it was:

[sleek swallow]

$\lnot{(P \implies Q)} \iff \lnot{(\lnot{P} \lor Q)} \iff P \land \lnot{Q}$

[vital dew]

Abhijeet Vats:

[pale epoch]

negating that statement just gives you $\neg(\exists x \in \mathbb{R}: x^2 \neq 0)$

[vital dew]

Lochverstärker:

[open narwhal]

How would i keep that statement symbolically but not have any negation symbols

[pale epoch]

now you want to move the negation symbol "past" the 'exists'-symbol

more generally, you should know what $\neg(\exists x: P(x))$ is

[vital dew]

Lochverstärker:

[open narwhal]

That makes sense

i way over thought this.

this was what i needed

[pale epoch]

that is correct

[woeful galleon]

hey guys! i have been sick and i got a few questions due today lol.. if anyone could help with them, it would be highly appreciated. i can get u a cookie if u want

[near prawn]

what flavour tho

@woeful galleon this just a matter of checking whether the definitions hold or not

[woeful galleon]

Any flavour. I got some Italian cookies since I am Italian lol

But yea.. I struggle with this. I feel bad asking for a walktrhough for it

[near prawn]

whats the definition of reflexivity

[sleek swallow]

Do you have pizza flavored cookies?

Or cookie flavored pizza?

[woeful galleon]

that is not italian food sorry lol xD

[near prawn]

mama mia

[woeful galleon]

mamma* lol

a set b is reflexive if it relates every elements of itself to itself

[near prawn]

uhh

wow the wording

[woeful galleon]

lol

xD

dont make me cry

plz, i have cookies

[near prawn]

you want to check whether $\forall x \in \mathbb{Z}$ xRx is true or not

[vital dew]

the one n only:

[near prawn]

do you think its true

[woeful galleon]

i think it is true

but i know because of graphing powers and not cuz of that line u gave. can u explain how you would go at it ? i wish i had good teachers

[near prawn]

huh

whats graphing powers

[woeful galleon]

i graphed it in my head

xD

[near prawn]

ok lets break down what this is saying

its say if u pick any integer x

you have x times x >= 3

or just x^2__>__ 3

clearly this is false when x=1

[woeful galleon]

we talking about ur statement right ? or xy>=3

[near prawn]

which statement is mine

[woeful galleon]

you want to check whether $\forall x \in \mathbb{Z}$ xRx is true or not

[vital dew]

Tareku:

[near prawn]

thats just the definition of reflexivity

R here is a relation: xy>=3

[woeful galleon]

aight got it

[near prawn]

ok

so we good with the first one?

[woeful galleon]

kinda? sorry if i am a pain lol. how would i prove it tho ? u setting y = x right ?

[near prawn]

we just disproved it

xRx is not true for every integer x so the relation isnt relfexive

[woeful galleon]

got u! ty. yes we good for 1

[near prawn]

aight

now apply the definition of the next one and see if it holds

[woeful galleon]

aight

that is why i was confused. i confused reflexivity with symmetry

i cant upload img but i have the rule

aRb <-> bRa

where a and b are elements of set

[near prawn]

i did that once in a test too so dw about it

yes

[woeful galleon]

so yes, it is symmetric

a = x and b = y right ? (making sure)

[near prawn]

huh

its symmetric yes

whats a and b

[woeful galleon]

i thought they were the elements

[near prawn]

[woeful galleon]

my face lol

[near prawn]

ehh as long as it works for u tbh

[woeful galleon]

well, tell me what YOU do

like ur mental process lol

[open narwhal]

List all integers between -100 and 100 that are congruent to -3 mod 25

what is this asking for because i can only think that -75 would be the only answer

[faint narwhal]

-75 is -3 mod 25?

[open narwhal]

wait is it asking for a number divided by 25 with a remander of -3?

[faint narwhal]

That is what mod means yes

[open narwhal]

brain did a big dead, i apologize

[open narwhal]

@faint narwhal how would i go about doing something like -222 mod 87?

im very confused

[faint narwhal]

It's the same

What's the remainder of -222 after dividing by 87

[open narwhal]

39 gets left over if im doing math right

[faint narwhal]

Yes that's right

[spring latch]

you add 87 to -222 until the answer is on [0, 87)

[past dew]

Hi, I don't know how this discord channel works. I'm hoping to get help on creating a correct recursive algorithm that correctly computes n^2. I am given my base case as n=1 so I added that ,but I don't understand If I did my recursion right. I just feel like its going to infinitely call itself and never directly compute n^2

Also heres more context, as the question I am trying to get. By creating a recursion function. It just means to create code like my example above.

[shell ore]

recursive functions needa call themselves

int rec(int n)
{
return rec(n-1);
}

some shit like that

obv that runs infinitely bc no base case to stop it

also what happens if user inputs n = 0 for ur program

[weary tiger]

if the cardinality of {1, 1} is 1, would the cardinality of {1, {1}} also be 1?

[sleek swallow]

No. The cardinality of {1,1} is 1 because {1,1} = {1}. The latter set is not the same.

There is a stark difference between the set containing the element 1 and the set containing a set that contains the element 1

[woeful galleon]

hey boyz, almsot done lol. can anyone give me a hand(y) ?

(i remember those problems lol, similar to mine)

[sleek swallow]

? Which one?

[woeful galleon]

all of those D:

i am trying to do the first summation rn

[sleek swallow]

Okay so for the first one, you realize you can simply treat it as the difference of 2 geometric sums?

So, there's a way to evaluate those

@woeful galleon

The second sum telescopes

[woeful galleon]

yep! got that one done, lookin at the second sum ❤️

[sleek swallow]

Write out the terms of the sum explicitly

@neon thorn the 5 dashes and 8 dots are the same. Like, you can't distinguish between the dashes and you can't distinguish between the dots.

Consider another way of looking at the problem

Suppose you have n objects

And you want to mark k of those objects with a red mark, leaving n-k objects unmarked

[woeful galleon]

(got all the sum) moving to induction lol
(ignore me pls for now)

[sleek swallow]

In other words, you need to pick k objects from n objects

Now, there are n ways to choose the first object, n-1 ways to choose the second object ...... (n-k+1) ways to choose the last object. So, the total number of ways would seemingly be:

$\frac{n!}{(n-k)!}$

[vital dew]

Abhijeet Vats:

[sleek swallow]

However, you have to consider the fact that there are k! ways to arrange these k objects

And the order you pick them in doesn't matter.

So you need to get rid of all those permutations. Hence, the total number of ways of choosing k objects from n objects is:

$\frac{n!}{k!(n-k)!}$

[vital dew]

Abhijeet Vats:

[sleek swallow]

(n-0)(n-1)(n-2)(n-3)....(n-(k-1))

Notice that this product has k factors. This is in line with the fact that there were k objects chosen

Your question doesn't make much sense. 'If i wanted to choose 3 objects from a set of 8 objects to see how many permutations would be possible' is all over the place

That's a different thing entirely, though? Like, it doesn't have much to do with the initial question you had. You'd have to pick 3 objects from the 8 objects and then permute them. So, it's the number of ways you can pick 3 objects from 8 objects multiplied with the number of arrangements of the 3 objects.

Yea

I urge you to learn the logic behind the derivation of the formulas, as opposed to just learning the formulas

How can 5 dashes have a length of 13

13-3 = 10

I explained how to form that denominator above. Look at the derivation and try to understand that first.

The question is: How many ways are there to arrange 5 dashes and 8 dots?

The equivalent question is: How many ways are there for me to choose 5 objects, amongst 13, to become dashes, with the rest becoming dots?

My advice is to work out the derivation on paper. Like, use pen & paper to write each step and internalize everything. After that, this problem shouldn't be too much of an issue

[mint horizon]

how do I prove that |L =<| = |L|

i know that they're equal

[minor zephyr]

I might just not know what this is, but something tells me that notation is non-standard? What are you trying to prove exactly, in words?

[mint horizon]

that a set of linked lists sorted in non-decreasing order is equal to a the set of linked lists

idk how to use the formatting tool, which is why i frequently type in an amended form of cs script.

[small kayak]

I have a question, for closure of a relation which is both symmetric and transitive, i think i can take union of reflexive and transitive closure of a relation. Am i correct ?

[night otter]

I have a question on finding the greatest common divisor using prime factorizations. Question is: a = 2^2 * 3^3 * 5^5 * 7^7, b = 2^7 * 3^5 * 5^3 * 7^2

[faint narwhal]

and what are you confused about?

[reef thistle]

@night otter how was the greatest common divisor defined?

@small kayak symmetric closure is adding (a, b) when you have (b, a), reflexive closure is add (a, a), transitive closure is add (a, c) for (a, b) and (b, c)... what sort of closure are you trying to find?

[dense creek]

can someone help me with 29a?

I think the statement says "There is a rectangle that is a square."

[dense creek]

<@&286206848099549185>

[tough locust]

@dense creek what's your doubt?

[dense creek]

is the statement that i came up correct?

[tough locust]

Why do you think it is or isn't?

[sour arrow]

More strictly, it says
"There exists a geometric figure such that the geometric figure is a square, and the geometric figure is a rectangle"

[tough locust]

When they say "rewrite the statement", do they mean in english?

[dense creek]

yeah

this was similar to the format that the example answers said

[tough locust]

Then it's weird to ask you not to use quantifiers

Can i see an example answer?

[dense creek]

ok

[tough locust]

"There is a" is just English for the existential quantifier

[dense creek]

That is for 28d

[dry patrol]

Hey guys I need help with a concept, counting.

I dont think this is correct at all

Can someone help me out?

s is the 2d sequence

it is a sequence of (sequence of T)

is this a correct statement (referring to out := ...)

another way I can do this is by doing
\forall x : Nat | x \in [0..|s| - 1] : \forall y : Nat | 0 < y < |s[0]| : s[x][y] = t

i hope that makes sense

[reef thistle]

what sort of notation is this?

@dry patrol

[dry patrol]

Wait sorry what do u mean?

I am defining it through quantifiers

in this case the + means summation

another way I can do this is by doing
\forall x : Nat | x \in [0..|s| - 1] : \forall y : Nat | 0 < y < |s[0]| : s[x][y] = t
@dry patrol this notation is just discrete math lol

[reef thistle]

$\forall x : Nat | x \in [0..|s| - 1] : \forall y : Nat | 0 < y < |s[0]| : s[x][y] = t$

[vital dew]

Element118:

[reef thistle]

@dry patrol I mean the notation $out := (+point:T|(point\in s)\land (point=t):1)$

[vital dew]

Element118:

[short juniper]

Hi, i have a question

I am trying to show that $\neg{A-\neg{B}} \cap \neg{(A-C)} = \neg{A}\cup(C-B)$
I understand how to do this up yo the point it involves the distrutive law.

[vital dew]

joseph2531:

[short juniper]

<@&286206848099549185>

[short juniper]

sorry, i figured it out

[dry patrol]

what does this mean?

I am trying to create a fucntion which will output the maximum value of a 2d sequence

but i dont think this is the right way

[pale epoch]

there is a +k added at the beginning

so you have to subtract it again

[weary tiger]

can someone give me an example that shows that the hypothesis in Euclid's lemma is necessary. That is, find integers a,b,c such that a|(bc) but a∤b and a∤c.

[faint narwhal]

What have you tried?

[weary tiger]

couldnt find any hence why im here and confused about how it works

[faint narwhal]

You know what the hypothesis says

so you have to do something that doesn't satisfy the hypothesis

[weary tiger]

so itd just be the opposite?

[sleek swallow]

154(-2)+[497+154(-3)] (9) = 154(-2)+497(9)+154(-27) = 497(9)+154(-29)

@weary tiger Consider a = 6, b = 4 and c = 3

[faint narwhal]

bezout's

you've definitely learned this

This is exactly the statement of bezout's, maybe they didn't name it but, they're just using this statement here

it doesn't indicate that

[stray reef]

$abx_0 + ady_0 = a(bx_0+dy_0) = a \cdot 1$

[vital dew]

Ann:

[stray reef]

@iron crescent

you should pay closer attention to what is said

| isn't an operation

...

[faint narwhal]

yeah but its not

It's just the thing you're assuming from the start

[faint narwhal]

by the recursive definition of F_{k+2}

how is the sequence F_k defined?

That's not a definition of F_k

There are a lot of sequences that satisfy this property

[stray reef]

no

how is the sequence F_k defined?

you have still yet to answer this question

[faint narwhal]

What number is F_0

[stray reef]

......... do you not know what "define" means

[faint narwhal]

What about F_1

And F_2 and F_3?

By knowing what this sequence actually is?

It should say it somewhere in your notes

Probably earlier in your course

This is the fibonnaci sequence

yes

Look at the picture you just sent

and think about it

[potent vortex]

So i need help with a problem and I can't seem to wrap my head around it

[stray reef]

it continues with 19 from here on out

[faint narwhal]

Think about the differences between each term of the sequence

Yes

[stray reef]

this is literally a textbook example of a recursive definition

[potent vortex]

The last two numbers in that one add up to make the next

[faint narwhal]

what do you know about f_k+1 to start?

[stray reef]

$F_{k+2} = F_{(k+2)-1} + F_{(k+2)-2}$

[vital dew]

Ann:

[potent vortex]

That's called learning

Mine is still giving me trouble.

[faint narwhal]

"Think about the differences between each term of the sequence"

[potent vortex]

I must be real smooth brained but all i see is that it goes up by 3 and down 1 after each one.

[faint narwhal]

That's just the pattern

[potent vortex]

I'm not very good at this yet. i'm still trying to solve it

[stray reef]

this is not an instance of a commutative law

are you allowed truth tables

well then

no, that's and

yes

[lyric pumice]

@iron crescent Find the geometry of the restrictions and compute through it.

[stray reef]

[lyric pumice]

TTTTT...T ----> H_H_H_H_H...H

[H...H]T[H...H]T[H...H]....T[H...H]

Choose where the tails go between heads.

There are 46 spots where the tails can go between heads because there are 44 spots between the sequence of heads and 2 spots at the ends of the sequence of heads.

Every tail has to go between heads or next to 1 head otherwise consecutive tails occur.

Each head has a left side and a right side.

Starting from the left, first count 1 space, then count 1 head.

Continuing on, you eventually count 45 spaces and 45 heads.

But there is one more space on the right side of the 45th head.

That makes 46 spaces.

[near prawn]

hi

does anyone know the generall approach to working out the number of circular permutations when you n1 type 1 objects, n2 type 2 objects n3 type 3 objects....etc

where a permutation is invariant under rotation

[faint narwhal]

Burnsides lemma probably

[hardy viper]

mayyybe it's https://en.wikipedia.org/wiki/Necklace_(combinatorics)

[near prawn]

ye that keeps popping up zoph but like

looks kinda

lemme check out that necklace link

[faint narwhal]

It's not too bad to understand

If you know some basic group theory

[near prawn]

i think this is it

[hardy viper]

phi god phi

[lyric pumice]

Anything scary is probably useful.

[sour arrow]

"Every step below is true"
vs
"The last step below is true"

@iron crescent

[stray reef]

did i share the ladder analogy for induction before @iron crescent

in weak (normal) induction, to prove you can climb a rung, you assume you've climbed the rung immediately below it. in strong induction, you assume you've climbed up to every rung below it.

[stray reef]

5! divided by 5 for rotational symmetry

bc once you have seated the men there is no longer any symmetry

[empty grove]

6!/(2! 2!). Can you reason why it should be that?

oh I didnt read that

lemme see

[reef thistle]

hmm then looks like it's the 180 is for no restrictions

[robust mango]

That is odd, I'm getting 36 as well

[reef thistle]

and provided answer is wrong

[empty grove]

Yeah 180 should be for no restrictions

[reef thistle]

yeah 36

[solemn cipher]

stuck on part b

part a is 10, since for the first fixed vertex (i.e. the first vertex to obtain a permanent value) can only have 0 temp values). the second fixed vertex can have 1 temp value (which is the weight of the direct connection from the first fixed vertex), the third fixed value can have 2 temp values and so on... which is 0+1+2+3+4=10

i got 10 using this thinking, but the answer is 6

[weary tiger]

In how many ways can you write numbers from 0 to 9 in one row such that 0 isnt next to 1 and 8 isnt next to 9?

[robust mango]

@weary tiger Find total arrangements without any restrictions and subtract the no. of arrangements where 0, 1 and 8,9 are together. I'm getting 10! - ( 8! * 2! * 2! )

[weary tiger]

ok I tried different method but Im not sure if my final answer is correct, I'll try to calculate and compare it to your 10!-(8!*4)

[robust mango]

okay

[weary tiger]

ok yeah gives the same answer. I checked like 3 different cases though

[robust mango]

idk how you did it with any other method but gratz

[weary tiger]

where did u get 8!*2!*2! from

[robust mango]

8 blocks, 6 of them containing 234567, with the other two containing 01 and 89

2! * 2! is for internal arrangement of 01 and 89

[weary tiger]

8 blocks?

ok yeah

thx

[robust mango]

np!

[weary tiger]

@robust mango wait actually I dont get it, why 8 blocks?

wdym by blocks

[robust mango]

It's just something called block approach my teacher taught me

01, 89 need to be together

So you put em in one block

[01] [89] [2] [3] [4] [5] [6] [7]

This is what you get

Now we have 8! for their arrangements

With 2! * 2! for the internal arrangements of 01 and 89

[weary tiger]

hmm ok I thought about it differently, thanks, worth knowing this one

[robust mango]

np

may i know how you did it

@weary tiger don't leave me

[weary tiger]

ok its kinda comp;licated but I can draw in paint

1 min

[robust mango]

sure

[weary tiger]

@robust mango

the grey curly brackets are possible 0,1 pairs

3rd case is wrong, I just realized, you get the picture though

should be 6 pairs left for (8,9)

[robust mango]

what about 8,9 same thing?

i see

[weary tiger]

For the first case we assume 0,1 are first or last, blue are possible (8,9) pairs which is 5, times 2 cuz swapping

times 6! remaining etc... I added it up and your answer and both work. Yours cleaner tho lol

[robust mango]

thing is this method is too much thinking for me that's why i always go for shortcut, it's awesome you could do this

[weary tiger]

thats not correct I think

[near prawn]

u mean plus

[weary tiger]

yeye

[near prawn]

u need to find 2 different equivalent ways of choosing r objects from a set of n+1 elements

[weary tiger]

lol I think he knows what he needs to do, just cant find the 2nd way

[stray reef]

consider a bowl with n+1 apples, n of them green and 1 red

consider a class of n+1 people, with 1 person in there called jim

you wanna make a committee of size r

you have two options

either you include jim, then you have r-1 more people to pick from the other n people in the class

or you exclude jim, then you have r people to pick from the other n

(n+1)Cr is when you don't give jim special consideration

nCr + nC(r-1) is when you do

[near prawn]

np

[weary tiger]

How many nonincreasing functions f:{1,2,....,k} -> {1,2,...,n} are there?

[faint narwhal]

I'm very lost on what you're even saying

store 100 what

[tardy laurel]

hey anyone willing to help me

Show that if |A| = |B| and |B| = |C|, then |A| = |C|

[obtuse lance]

uhhh |A| = |B| = |C|

[tardy laurel]

With bijections?

I kinda don't know how to prove bijections

[faint narwhal]

What does |A| = |B| mean

in terms of bijections

[tardy laurel]

not gonna lie i need help with that

on what a bijection is

so its like every input has an output and every output has an input right

[faint narwhal]

Do you know what a function is?

[sour arrow]

Don't multipost Arc

[reef thistle]

@neon thorn what sort of calculations do you need to do

any restrictions on how we can use the device?

[reef thistle]

what happens if the result is > 100?

what do you mean "store"?

do you have paper?

and pencil

why not do it on paper and pencil

that is strange

what's the original problem formulation

exact phrasing please

picts if easier

what happens if we divide?

say 5/10, what's the result on this calculator

I have a solution that doesn't even need CRT.

if the calculator can store 5/10 = 0.5

that is a ****** assignment

are you sure you can do mod on your calculator?

not sure if you can even input it

I still don't get the point

Do we need to calculate the answer in base 10?

Honestly the only useful moduli seem to be 11, 9, 8, 7, 5 assuming we have to do everything with the calculator and that the calculator rounds down for division.
How are you going to even calculate other moduli on your calculator?

You said your calculator has no modulus

how do you calculate that

OR... can we use negative numbers on the calcualtor?

If we can use negative numbers we just might have enough bases

I'm going to assume larger than 100 includes smaller than -100

using mod 19, 17, 16, 13, 11, 9, 7, 5 gives you mod 232792560

which ARGH IT'S JUST NOT ENOUGH

It seems we can't use any bigger mods. Try calculating big number mod 21 on your calculator.

We can't input the entire big number because the calculator cannot store anything > 100

So, how would we calculate big number mod 21?

what moduli can you store?

what do you mean

can you calculate a number mod 21?

if you can't reduce number mod 21, then you can't even store it in the first place

but how do you use the calculator to calculate big number mod 21?

(EDIT: 21 is probably possible, do 23)

Are you going to subtract 23 from it until it's negative?

oh no

we can't input the big number to subtract

so wait, you are saying that there's a separate calculator that we can do moduli on?

then I don't see any way to take mod 23

calculate 184629 mod 23 on your calculator

yeah, so if you want to represent it with mod 23, you need to calculate 184629 mod 23

well, what mods do you want to use?

if you use mod 100, how do you add 87+76 mod 100?

*you're

nervously looking right, just in case something is happening there

So, do you agree the only mods that are worth using are 19, 17, 16, 13, 11, 9, 7, 5?

their product is 232792560

which is too small

if you use mod 23, how do you calculate mod 23?

184629 mod 23 is?

yeah, but how do you get that on that calculator

unless you are saying P is initially stored with your method

can you multiply them though?

how do you multiply something 18 mod 19 with something 18 mod 19

What I do is store 18 as -1, then it's just -1 * -1 = 1 (mod 19)

honestly afterwards any solution will degenerate into just grade school multiplication

yeah, you store negatives

but you can't get above mod 19 because 11*11 mod 23

[weary tiger]

anyone up?

[stray reef]

i'm sure at least one person is awake at this moment

[reef thistle]

I don't think Euclidean Algorithm is the right method

because Euclidean Algorithm is GCD

Just think about what each digit of 123 in base 8 means.

@iron crescent

[stray reef]

can you hammer a nail with a microscope

[stray reef]

a relation from A to B is simply a subset of AxB

consider that each such relation is uniquely determined by which of the 7 elements of AxB \ {(1,5),(1,2)} are in it

the whole point is that these have essentially already added in for you and you have no control over them.

the restriction is that they are included.

ALWAYS.

read carefully what i said

a relation from A to B is simply a subset of AxB

you want the number of subsets of AxB that contain two particular elements

this is the same as the number of subsets of the set that is AxB without those two particular elements

it makes PERFECT sense. one can pass from one class of sets to the other by adding or throwing out these two particular elements from each set.

[reef thistle]

y=-7x, z=10x

@iron crescent

[potent vortex]

I need help wrapping my head around a certain problem

So to simplify things I'll use the lottery as an example. You get 6 numbers to pick from and 50 choices so the answer to that is c(50,6) = 15,890,700

but

To calculate the chance of winning if you bought like 12 tickets would this be correct c(15890700, 12)?

[stray reef]

no

[potent vortex]

Would it be 12/15890700?

[stray reef]

that should be it

[stray reef]

y=-7x, z=10x

,calc 7 * (-7 * 11) + 5 * (10 * 11)

[vital dew]

Result:

11

[stray reef]

@iron crescent

[thin badge]

I passed in my discrete maths class 😩 🙌

[weary tiger]

Anyone got a good recurrence relations resource ?

My textbook goes from: "solve a(n) = a(n - 1) + 3"

to

"solve f(n) = 3(f(n-1)) + 2" without any explanation

the first one is easy, but I can't get the second one simplified

I'm stuck at this step,

I know that the right-hand side of the + is a geometric series

But I can't simplify it to the final answer :/

[weary tiger]

$f(n) = 3^n \cdot g(n)$

[vital dew]

tet:

[weary tiger]

the answer is supposed to be $f(n) = (3^n-1)/2$

[vital dew]

total_sleezeball:

[sour arrow]

Base case. Can you prove P(0)? By that I mean, can you prove that the formula agrees with the recursive definition for a0?

@iron crescent

The recursive definition depends on a0 and a1 to start it off, it needs two values. They are both the base case

Do those agree with the formula on the bottom?

So you tried it?

Are we both looking at the same question you posted a picture of?

Yes lol. That's much better

I also see
a(n+1) = 3a(n) + 4a(n-1)

So if you've confirmed the formula does agree that a0 = 3, a1 = 7, you've proven the base case

We know that P(0) and P(1) are both true

When I say P(n), I mean "the proposition holds for the number n"

A hint that we need two values for the base case is that the recursive definition needs two values

Basically, those two values follow no "rule", they were just given to you. Induction can't say anything about them

So they fit well as base cases

I'm going to assume that P(n) and P(n-1) are both true. That is:
a(n) = 2(4)ⁿ + (-1)ⁿ

a(n-1) = 2(4)^(n-1) + (-1)^(n-1)
a(n-1) = 1/2 (4)ⁿ - (-1)ⁿ

I would like to repeat, we are ASSUMING the above two facts. In our current world, they're just true. Now we want to prove P(n+1). We want to prove that the two ways to compute it agree.

Recursive definition:
a(n+1) = 3a(n) + 4a(n-1)
a(n+1) = 3[2(4)ⁿ + (-1)ⁿ] + 4[1/2 (4)ⁿ - (-1)ⁿ]
a(n+1) = 8(4)ⁿ - (-1)ⁿ

Plugging in the formula:
a(n+1) = 2(4)^(n+1) + (-1)^(n+1)
a(n+1) = 8(4)ⁿ - (-1)ⁿ

@iron crescent
For any two consecutive numbers where the formula holds, the next number does as well. Since you proved P(0) and P(1), the formula is true for all positive integers.

[weary tiger]

is this induction?

whats the question?

[sour arrow]

Induction with two assumptions

[weary tiger]

right

i havent done this in a while so idk, but ill see if i get it

[sour arrow]

@iron crescent
Induction always has the same structure, you'll want to be familiar with it. I like to think of it in three steps:
-Base case
-Assumptions
-Inductive step

The assumptions HAVE to be used to prove the inductive step. You owe it to yourself to clearly state what assumptions you're making

Classical example, prove that the sum of the first n natural numbers is n(n+1)/2

[weary tiger]

iirc for strong induction, i usually tried to prove equivalence algebraically, then find the all the other cases

[sour arrow]

You can assume more than that

[weary tiger]

ok i think i got it

[sour arrow]

@iron crescent
You can make cases where you assume 3 or more

[weary tiger]

hmm somethings off tho

[sour arrow]

Inducting downwards?

What's the question?

[weary tiger]

i think im missing a step, tho

i thought were proving the two expression are equivalent

lol

[sour arrow]

We are, but it's not all algebra

I'd suggest looking up the structure of an induction proof

[weary tiger]

idk which step im missing, but i think im supposed to pull out the necessary cases after this to properly prove it

[sour arrow]

OH I see what you did, you started with the assumptions and proved the last step

[weary tiger]

yea, but im still missing several steps of the full proof

[sour arrow]

Don't start by setting them equal. I like to call it a "t-table" proof. Put a line between them and prove they work out to the same thing

Putting "=" the entire way down is confusing. You don't know that they're equal

But yes, good job, you proved the inductive step

[weary tiger]

yeet, but i haven't shown the first few cases yet for it to be a proper proof

i'll use the t-table line thing next time tho

[sour arrow]

You need to prove "if and only if". Note that iff proofs are really two proofs in one.

If you prove what you put in paint, you'll have proven one of the directions. You'll still have the other direction to do.

What's a)? Lol

Is that the "if" part?

You want to prove:
"A + B = ∅" → "A = B"

By contraposative, you could prove instead:
"A ≠ B" → "A + B ≠ ∅"

Yeah okay lol

Two cases.

• There's an element in A that is not in B
  OR:
• There's an element in B that if not in A

You can do a wlog and make A the set with the extra element. That works because A ≠ B is the same as B ≠ A

"wlog" is "without loss of generality"
I assumed something about A and B that doesn't change that A and B can be any set.

That's not a contradiction! That's the result you want

If A has an element that B doesn't have, then A+B has at least that element. Ergo, A+B ≠ ∅

Don't be afraid to write out sentences in your proof btw. Proofs are almost always 95% words

You are trying to prove:
"A ≠ B" → "A + B ≠ ∅"

That means you start by assuming "A ≠ B" is a true fact, and follow logical conclusions to show that "A + B ≠ ∅" must also be true.

Note that ∅ is the empty set symbol, and is not phi φ.

I see the confusion though

Those are pretty similar

[obtuse lance]

yep

[desert edge]

Whats a good site for drawing graphs

Or a program

[last sigil]

Check #resources and scroll up a bit to horse kun's post

[desert edge]

ty @last sigil

[faint narwhal]

what are things you know about the gcd?

Sure, any other properties of the gcd you know?

you know the euclidean algorithm?

Uh, that's not the euclidean algorithm

sure, but what does the euclidean algorithm do

so what happens if you do the euclidean algorithm for this

I mean you're trying to find the gcd

Uh what are you dividing n + 3000 by?

and what's the remainder when you divide by n?

can you have a remainder of n when you divide something by n?

Really, you should think about why the euclidean algorithm is a valid way to find the gcd

Like why does doing this keep your gcd the same?

this is important in math, you should seek to not just know what is true

but why things are true

[pale wing]

Think about d= gcd(n, n+3000)

If d|n and d|n+3000, d|(n+3000)-n

[spiral cradle]

anyone know how to find N[a] and N[c]?

[pale epoch]

the definition is right above it

[spiral cradle]

yea it is but im a bit confused on what {a} would be

[pale epoch]

the set with one element a

[spiral cradle]

so it would be {b}, {e}, {a, b}, {a, e}?

[pale epoch]

what

[spiral cradle]

for N[a]

[pale epoch]

no

what do you have for N(a)

[spiral cradle]

b and e

[pale epoch]

{b, e}

N(a) is a set

[spiral cradle]

oh ok

[pale epoch]

N[a] is just the union of N(a) and {a}

so the union of {b, e} and {a}

which is {b, e, a}

go review your definitions

[spiral cradle]

ok thanks

i will

[sleek swallow]

Loch

[elder oak]

does anyone knows this?

Let S be a set of of propositional logic sentences and A a
sentence that does not belong to S. Prove that if S ╞ A and S ╞ ¬A, then
S is unsatisfactory.

[sleek swallow]

Unsatisfactory? What does that mean? Not sound and complete?

Also, if you look closely, your set of propositional sentences derive A and not(A). In other words, they've proved a contradiction. I'm assuming that you know that and still don't think it's enough to prove that it's unsatisfactory?

[fading widget]

How would I start proving or disproving this?

[fading widget]

<@&286206848099549185>

[lyric pumice]

@fading widget Try a proof by contradiction.

[near prawn]

erm

does it mean no 2 vowels are adjacent

or that not all of them at once are

[fathom garden]

@obtuse lance can i say this $n\in\mathbb{N}\quad y=x^n\quad\Delta^ny_1=n!$

[vital dew]

Jeffrey:

[fathom garden]

for the problem i came up with? i’ve poked around a bit and wanted to know if the notation is correct

sorry for kinda randomly mentioning you too, know you’re probably busy, but i had to go do work when we’re talking about this last

and never got a chance to think about it

y sub x =

oops

[weary tiger]

poops

[obtuse lance]

hey, yeah I mentioned I could show you a proof of it by induction if you wanted @fathom garden

[weary tiger]

can someone help me decompose this solution?

why do the current guests go to room 3n?

[hardy viper]

there's 3 infinities of people

[weary tiger]

why 3?

oh, 2 buses + current guests?

[hardy viper]

yea the hotel is filled already

[weary tiger]

can someone help me find an example of a function for which the domain is the integers, the codomain is the real numbers, and the image is not equal to the codomain.

[sleek swallow]

Sure. Let $f:\bZ \to \bR$ be the function such that $f(n) = n$ for all $n \in \bZ$. So, $f(Z) \neq \bR$.

[vital dew]

Abhijeet Vats:

[sleek swallow]

@weary tiger

[weary tiger]

tysm!

[sleek swallow]

You're welcome.

[weary tiger]

@sleek swallow im kinda confused how u got that still

[sleek swallow]

Oh it was nothing, it was just a little bit of this and a little bit of that

[weary tiger]

what does that even mean?

[sleek swallow]

I mean, it was just the simplest function that came to my mind. I could've chosen anything else, really.

[weary tiger]

well alright thank you

[weary tiger]

@stray reef i suck tho not early uni lol

[stray reef]

look this is combinatorics so it belongs here

okay so

recap

the original problem: bob has 24 cookies to share between himself, bill and bart. the latter two must get at least 2 cookies each. in how many ways can this be done?

the equational reformulation: count the integer solutions to x+y+z=24 with constraints x>=0 and y,z>=2

the substitution: count the integer solutions to x+y'+z'=20 with constraints x,y',z'>=0

[weary tiger]

yes

but i don't know how to do that

or am i dumb

well i am dumb but

lmao

[stray reef]

i was going to go over that

i'd appreciate not having to sit through self deprecative remarks from you

[weary tiger]

sry

[stray reef]

alright

so

there's a one to one correspondence between the solutions of our equation and the arrangements of 20 white and 2 red cards in a row

[weary tiger]

what is a one to one correspondence

[stray reef]

....oh god

okay, let me rephrase what i was saying

the number of solutions to our equation is the same as the number of arrangements in the scenario i laid out, because you can construct a solution from an arrangement, and vice versa, in a way that doesn't assign the same arrangement to two different solutions, or the same solution to two different arrangements

[weary tiger]

do you mean bijection btw

[stray reef]

yes i do mean bijection

i was under the impression you didn't know that word given that you said you didn't know what a one to one correspondence was

but yes a bijection is precisely what i was and am talking about

the point is that instead of counting solutions we can count arrangements

[weary tiger]

so what do you suggest now

[stray reef]

can you count the number of ways to arrange 20 white and 2 red cards in a row?

[weary tiger]

binom(20,2) if that's what you mean

[stray reef]

no

binom(22,2)

[weary tiger]

oh

[stray reef]

there aren't 20 cards in total that you're arranging, there are 22

[weary tiger]

hmm

idk about this

[stray reef]

the point is that binom(22,2) is the answer to the original problem as well as to all the intermediate reformulations thereof.

the goal of all these transformations and rephrasings was to reduce the problem to one whose solution is known

[weary tiger]

why are there 22?

we need to give 2 to each person

or?

[stray reef]

no

[weary tiger]

i mean

at least 2

[stray reef]

there are twenty white cards and two red cards

there are twenty-two cards in total that we are arranging

[weary tiger]

why do you separate them though

[stray reef]

what do you mean

[weary tiger]

like the cookies are the same

but your 22 white cards and 2 red cards are not the same

as in

you have 22 white cards and 2 red cards

not 24 just cards

[stray reef]

no

NO! I DID NOT SAY 22 WHITE CARDS!

[weary tiger]

sorry my bad

[stray reef]

there's a bijection between the solutions of our equation and the arrangements of 20 white and 2 red cards in a row

[weary tiger]

20 white cards

2 red cards

but still

in terms of cookies

i don't get it

[stray reef]

there are no cookies anymore

[weary tiger]

because you have to give 2 to each person

[stray reef]

no that's already been taken care of.

that's what the substitution was for.

[weary tiger]

and what if we wanted each of the three people to have at least 2 cookies then

[stray reef]

then we would have 18, and not 20, cookies to share freely

[weary tiger]

instead of just the two friends and not the person w/ the cookies

[stray reef]

then the x>=0 constraint would change to x>=2 and an additional substitution, x=x'+2, would have been made

leading to x'+y'+z'=18

[weary tiger]

tbh i am a bit confused with the cards

how would you use the cards here

[stray reef]

look, if you weren't clear on some of the intermediate steps, you should have told me

[weary tiger]

i thought i was

[stray reef]

but you didn't, leading me to assume that you were following along, but then it turns out you didn't

[weary tiger]

but now i realized i am not

when the conditions changed

[stray reef]

okay so let me try to sidestep the whole equation thing

let's start over

okay?

[weary tiger]

ok

[stray reef]

so bob has 24 cookies

him, bill and bart are to split these among themselves

are we requiring bob to keep at least 2 to himself or not?

[weary tiger]

yeah

[stray reef]

okay

so everyone takes 2 cookies

[weary tiger]

yes 18 left

to go around

[stray reef]

this leaves 18 in the pile

now let's take these 18 cookies and 2 separators

and arrange these 18+2=20 objects in a row

[weary tiger]

what is a separator

[stray reef]

a separator is a separator

it can be anything

we're considering it as the same kind of object as a cookie

[weary tiger]

where did this come from

[stray reef]

bear with me for a moment. i was just about to explain

[weary tiger]

yes

[stray reef]

a separator is an object distinct from a cookie. it doesn't really matter what it is

the point is

[weary tiger]

ok, but why do we need two of them and why do we need them at all

[stray reef]

an arrangement of cookies and separators corresponds to a way to share the cookies

and vice versa

to go from an arrangement to a sharing plan, give bob all the cookies to the left of the left separator, give bill all the cookies between the separators, and then give bart all the cookies to the right of the right separator

to go from a sharing plan to an arrangement, place, going left to right: bob's cookies, a separator, bill's cookies, a separator, and bart's cookies

[weary tiger]

wait, but why do we need them?

i don't understand

[stray reef]

wdym

their point is that in the arrangement they separate the cookies between the three friends

*******|*****|******

• for cookie, | for separator

this arrangement corresponds to giving bob 7 cookies, bill 5 and bart 6

[weary tiger]

oh ok? but why do we need to involve this thing

[stray reef]

wdym

what's wrong with it