discrete-math

[subtle charm]

Can someone explain to me why is this transitive?

I don't quite understand how a R a and a R b -> a R b is shown here. (second line) Since there isn't an arrow from a to b.

[last timber]

@subtle charm still need help?

[subtle charm]

@last timber Yes please!

[last timber]

ok so yes there is no arrow from a to b

but do both arrows from a to c and c to b exist?

or a to a and a to b

[subtle charm]

a to c doesn't exist

and a to b too

[last timber]

yes so a R a ^ a R b is false as well as a R c ^ b R c?

[subtle charm]

ok I think i get it. it is true by default right?

[last timber]

ye, vacuously true

since you get propositions F->whatever

[subtle charm]

yup, I got it. I kept forgetting about the vacuously true part of implication statements. Thanks for the help!

[last timber]

yw

[silk hull]

@snow sleet

Hey I can’t message u? Not sure if u could help with a few questions?

[subtle charm]

Can I check if x is an element of an equivalence class S, then S = [x]?

[last timber]

wym

[subtle charm]

Like if this is true?

[last timber]

i mean what is definition of equivalence class

[subtle charm]

[x]∼ ={y∈A: x∼y}.

[last timber]

so if S is equivalence class and x is in S then S contains elements equivalent to x

now show that if y~x then [y] = [x]

[subtle charm]

Is this due to the same component relation?

[last timber]

it is due to definition

[shut jacinth]

i removed vertices {c, d, h} which leaves 4 components left

and by a theorem which states that by removing k vertices s.t. the graph splits into at least k+1 components, you can prove the original graph does not contain a hamiltonian circuit

was wondering if my approach was correct here

[last timber]

hmm

looks like this graph is semihamiltonian tho

[shut jacinth]

how do u figure

[last timber]

like there is hamiltonian path (not cycle)

[shut jacinth]

oops

sorry i want to prove no hamiltonian circuit exists

i should've been more specific

[last timber]

g->e->c->a->b->d->j->h->f->i

ye

[last timber]

well i do not remember theorem

but i remember something similar to this

so looks like you are indeed on the right path

[shut jacinth]

sorry for the bad drawing (did it in mspaint)

but this is what G' would look like if i removed {c,d,h} right

which has 4 components

[last timber]

yep

actually

if you remove j

you see that graph immediately becomes hamiltonian

[shut jacinth]

true

[shut jacinth]

does this contradict what im trying to disprove then?

[last timber]

nah this just indicates what the problem with current graph

j is its bad vertex

but iirc there is no theorem on preserving hamiltonianness after adding removing vertices

[shut jacinth]

hm i see

i just dont know if this method is rigorous enough lol

i know there's 3 rules most textbooks follows to see if you can build a HC

[last timber]

well if this theorem is correct it is rigorous enough

[shut jacinth]

alright, thank you for your help

[last timber]

yw

[shut jacinth]

oh also

couldn't you just say that c-d-j is a subcircuit therefore a HC cannot exist

[last timber]

idunno

is there such theorem?

it feels wrong

this has subcircuits but is hamiltonian

i guess it just means that you cannot do path like a->b->a if you have c also as vertex

[shut jacinth]

oh i see

[patent terrace]

How to prove that $A(2,n)=2n+3$ for all $n \in \mathbb{N}$ (Ackermann)

[vital dew]

ksg

[patent terrace]

Hi, I've used induction but I'm stuck on the inductive step

[stray reef]

have you made any progress on this so far?

[glossy plover]

not really. not sure how to start

is a perfect square like two same integers multiplied

where it is possible to take the square root of that product

[stray reef]

is a perfect square like two same integers multiplied

if you insist on phrasing it that way, yes

here's a hint: how much do adjacent perfect squares differ by?

perhaps write out the first few to get an idea.

[fresh venture]

Can someone help me understand the topic Method of proof?

[fresh venture]

What kind of questions are related and etc.

[gritty crescent]

You should probably consider reading an introductory book on proofs, or discrete math. You can ask for help here on more specific problems if you're stuck somewhere.

[fresh venture]

I have a question

Proposition is a statement that can either be true or false right?
It can't be both at the same time

[gritty crescent]

Depending on your system, it likely should be

[last timber]

@gritty crescent abandon law of excluded third

imagine using it

it's so lame

[gritty crescent]

[weary tiger]

a ≡ −15 (mod 27) and −26 ≤ a ≤ 0.

Been stuck on this

m | (a - b )

27 | (a - (-15))

How is the answer a = -15

27 | ( -15 - (-15))

27 | 0

[stray reef]

yeah, 0 is a multiple of 27

@weary tiger

[weary tiger]

So when you do this problem you look at the multples of the mod?

a ≡ 24 (mod 31) and −15 ≤ a ≤ 15.

[stray reef]

you seem to have doubted the truth of the statement 27 | 0.

did i get that wrong?

[weary tiger]

isnt that undefined

[stray reef]

do not confuse the statement 27 | 0 with the expression 27/0

"27 | 0" means "27 divides 0", or "0 is divisible by 27"

[weary tiger]

0 / 27 = 0

im not sure how to get to a = 0

a ≡ 24 (mod 31) and −15 ≤ a ≤ 15.

how would you start a problem like this

(a - 24 ) / 31 = (needs to be an integer not a decimal )

Would guessing and checking through −15 ≤ a ≤ 15 be the wrong way to go about it

[stray reef]

im not sure how to get to a = 0
but who said anything about a=0...

also no you don't need to guess and check

just go down by 31 from 24

i.e. 24 - 31

[weary tiger]

−15 ≤ a ≤ 15 ?

it has to be between those

it's -7

-7 ≡ 24 (mod 31) and −15 ≤ a ≤ 15.

but thats just cause i guessed and checked

[stray reef]

i.e. 24 - 31

start with what's on the right hand side

if it's too big then subtract 31

if it's too small then add 31

[weary tiger]

to what

[stray reef]

......okay evidently i failed to explain it clearly

forget i said anything and let me try again

you want to find $x$ such that $x \equiv c \pmod{m}$ and $A \leq x \leq B$

[vital dew]

Ann

[stray reef]

(we assume the range from A to B has length m so that there's exactly one solution)

ok?

c, m, A and B are known

[weary tiger]

ok

[stray reef]

start with c

if c is between A and B then you're done and that's your x

if c is not between A and B then add or subtract a multiple of m to c to put it in your range

whether you add or subtract depends on which side of the range c is on

[weary tiger]

a ≡ 99 (mod 41) and 100 ≤ a ≤ 140

41 * 1 = 41

41 + 41 = 82

41 | ( 82 - 99 )

41 * 2 = 82

41 + 82 = 123

41 | (123 - 99)

[stray reef]

you're walking in circles here

[weary tiger]

yes

[stray reef]

evidently i've failed to make myself clear

despite my best efforte

efforts*

[weary tiger]

im trying to understand

what am i doing wrong

[stray reef]

everything

you have $a \equiv 99 \pmod{41}$ and you want $100 \leq a \leq 140$

[vital dew]

Ann

[stray reef]

is 99 between 100 and 140?

(yes or no)

[weary tiger]

no

[stray reef]

correct

is 99 too big or too small?

[weary tiger]

small

ok so i messed up u add it to c i was adding it to m

[stray reef]

you were just generating multiples of m and that's it

[weary tiger]

i meant to add to 99

99 + 41

41 | (140-99)

gets u 1

sorry for being slow

[stray reef]

41 | (140-99)
gets u 1

no

41 | (140 - 99) is true by construction

you still seem to think m|n is an operation and not a relation

[weary tiger]

So, it means 41 is divisible by 41.

[stray reef]

yes

i mean

you got 140 by adding 41 to 99

so of course (41+99)-99 will be divisible by 41

[weary tiger]

ok

thank you for the help

[hallow blaze]

is there a set for any number >= 0

[sand cipher]

could someone help me with no 1

do i write it down like, if x = 1 then ..... and x = -1 then ....
if x = 2 then the result will be odd since x is even

[faint narwhal]

This isn't really going to work, are you going to write out x = 3 and then x = 5 and so on for all the odd numbers?

[sand cipher]

what if it's like if x + 1 is even there exist an integer n such that is x + 1 = 2n

[faint narwhal]

yeah thats towards the right idea

write out what it means for an integer to be odd or even

and then use those to show what you want

[sand cipher]

so can i just write ( Integers that are not divisible by 2 are odd number while integers that are divisible by 2 are even numbers, therefore if x + 1 is even then there exist an integer n such that is x + 1 = 2n)

and should i provide examples too?

[faint narwhal]

you don't need any examples

but what you've written isnt sufficient at all

[sand cipher]

what should i add more?

[faint narwhal]

It's not clear why if you take an integer thats not divisible by 2 and add 1 then that number will be divisible by 2

Also, iff means you have to show both directions

Also, I'm pretty doubtful that this is the definition of odd thats used in your book/class

[sand cipher]

yep just went to check the lecture again and it seems i need two proofs,
so 1 should be let x be an odd integer then x+1 is even
and
2. let x+1 be even then x is odd

[faint narwhal]

Thats right

[sand cipher]

ok so definition of odd integer is that if x is odd it can be expressed as x=2m + 1 for n is an integer
and we need to prove that x + 1 = 2n
Manipulate it: x+1 = 2m + 2
x+1 = 2(m+1) where m+1 = n
x+1 = 2n

since sum of 2 integer M + 1 = an integer (n)

[surreal crescent]

i dont even know how to start this problem/get anything that resembles the basic rules of inference

[surreal crescent]

i think i posted this in the wrong channel i just now saw the proofs and logic one haha

[lavish oar]

can someone explain this to me?

[gritty crescent]

@lavish oar What have you tried?

What would it mean for a subset B to contain {1,2}? What elements must B necessarily have? What additional elements, if any, can B possibly have?

[last timber]

@gritty crescent these questions are offensive since as mathematician i find it offensive not to get just the answer directly

[fresh venture]

In graph theory what are the leaves of tree?

[last timber]

nodes that have no childs

[fresh venture]

Are they those vertex with only 1 edge connected

[last timber]

e.g tree (a,b), (a,c), (c,d) -- b and d are leaves

[last timber]

yes, but except for the root

[fresh venture]

Ok

[last timber]

in tree (a,b) b is leaf and a is root

[fresh venture]

But that root one we can just assume any random, right?

[last timber]

in directed tree no

[fresh venture]

Like can't we say b is root and a is leaf

[last timber]

in undirected it actually does not matter much and you can choose any node to be root if needed

[fresh venture]

How does directed one looks like?

[last timber]

haven't you seen directed graphs?

[fresh venture]

I might have but idt it was specified

[last timber]

this is directed

[fresh venture]

Ahh ok ok

Got it

What does this mean?

[stray reef]

what does what mean?

is there a word you don't understand?

@fresh venture

[fresh venture]

The one in yellow box

[stray reef]

again, what exactly do you not understand?

is there a word you don't understand?

[fresh venture]

The whole sentence

[stray reef]

so you understand every word individually, but the sentence as a whole escapes you. did i get that right?

[fresh venture]

How does it become eulerian path with zero or 2 vertices with odd degree

[stray reef]

nothing "becomes" anything.

there exists an euler path in your graph <=> there are only zero or two vertices of odd degree

[fresh venture]

Can't be more than 2?

Or 1?

[pale epoch]

indeed

the idea is that you have to leave every vertex you enter (except the starting and end point)

[fresh venture]

Yeah

What if there is 1 vertex with odd degree and rest all are even?

I am unable to visualise

Ig it's either not possible to have only 1 vertex with odd

[pale epoch]

well, you will have to end somewhere

and it will have to be at a different vertex

but if that has even degree, you will have to leave it again

which you cant

[fresh venture]

I see

Thanks

Got it

How do we do this?

Perfect matching is when every vertex is joined with some other right?

[stray reef]

this cannot happen ever

the handshake lemma makes it impossible for such a graph to ever exist

[fresh venture]

Yeah I got it

[stray reef]

yes a perfect matching is a matching that involves all vertices

[fresh venture]

So all except 2nd graph is alright?

[stray reef]

parts of the graphs at the bottom are cut off.

[edgy vine]

What exactly is meant with the hint. Can I assume on a infinite walk that it will find v and look at every possible path without repetition? Ans what is meant by direction (one way only like in directed graph? )

[hallow blaze]

is there a set for any real numbers >= 0?

[gritty crescent]

You just described it yourself

Set of real numbers >=0

[hallow blaze]

oh

I was just wondering if there was a formal set for it

[gritty crescent]

You can denote it with $\bR_{\geq 0}$ or whatever if you like

[vital dew]

Manan.

[hallow blaze]

with a letter

oh

ok thanks

[gritty crescent]

Some conventions include 0 in R^+ but I generally avoid that

[last timber]

just use \overilne{R^+}

[gritty crescent]

[last timber]

closure of R^+

[gritty crescent]

[ruby cobalt]

how do u give a counterexample for this

[pale epoch]

not

[shadow shuttle]

it's true so there is no counterexamples 😄

[ruby cobalt]

what about this

[pale epoch]

have you tried anything?

[weary tiger]

I understand a and b

Confused on c)

What does the subscript n-1 mean

[stray reef]

it means the term at position n-1 in your sequence

nothing special

if you wish, $a_{n-1}$ is the term in the sequence right before $a_n$

[vital dew]

Ann

[weary tiger]

how did they go from row 2 to row 3

not sure how you change n-2 -> n-1

[elder berry]

6 = 3*2 so there is an extra 2 in the product that can be put in 2^(n-2)

[vital dew]

peaceGiant

[weary tiger]

ohh thank you

[elder berry]

yw

[weary tiger]

how did they go from step 1 to 2

the . is meant for multiplication

@elder berry

[elder berry]

8 . 4^{n-1} = 8 . 4 . 4^{n-2}

After that they factored the 4^{n-2}, it's a common factor for the two numbers

[slate burrow]

When we try to prove something in induction, is there a rule for how we should like prove k and k+1? Like i mean, are we suppose to derive k from k+1 or k+1 to k?

[weary tiger]

ty 🙏

[pine iris]

usually this is the way to go

[slate burrow]

we assume p(n) is true then we prove p(n++) so we go from p(n) to p(n+1)?

Having a hard time understanding their explanation

[pine iris]

yes, you assume p(n) and prove p(n+1)

[slate burrow]

Alrighty, thank you :)

[outer hinge]

I’m really confused. I have a problem that’s telling me to write the truth values for ~p —> q if the statement p —> q is false

[slate burrow]

Does this make sense?

Any criticism?

[weary tiger]

how did they get n-1 -> n

@elder berry

[elder berry]

Why am I getting pinged?

also, just distribute the negative sign

[deep portal]

its been a while since I took this class so i wanna make sure I remember. so statement

"if S is a square then s is a rectangle. "
negating this statement is: S is a square and s is not a rectangle

right?

[last flicker]

yes

[onyx zephyr]

When I get a question like this, do I make something like a diagram or write out all the 32 sets? Or is there an easier option?

"Let R be a relation on the set P([0,1,2,3,4]) so that A R B if A is a subset of B U [4]. Decide the properties of R."

Also what does B U [4] stand for in this example? I am so damn confused and stuck.

[stray reef]

[x] is the equivalence class of x

[weary tiger]

can someone help me in the #help-4 channel, a question regarding discrete math

[calm comet]

ive got no idea where to put this but i got an idea that id like to hammer out that has to do with the lucas numbers pascals triangle and f(x,y,n) = x^n + y^n

if anyone would like to discuss this new idea in mathematics voice, @ me
I'd love to iron it out with a second perspective

[shadow shuttle]

Do you guys know what is the best book for practicing discrete math ? , Tks

[desert edge]

can somebody explain to me how A = {a,b} is not a subset of p(A)?

says in my book A is a member of the powerset A

but they both contain the set {a,b}. doesnt that qualify A to be a subset of p(A)?

[faint narwhal]

A is an element of p(A), not a subset like they say

a and b are not elements of p(A), so you can't say that {a,b} is a subset of p(A)

[desert edge]

that hurts my brain

or wait it does not actually

ohh i get it

[mystic elbow]

Hii

<@&286206848099549185>

Need help with 5,6

Please helppp

Ping me too

[cerulean wind]

5. let D represent the number of mismatched pairs of cards, R represent the number of pairs of red cards and B represent the number of pairs of black cards.

there are a total of D red cards in the D pairs of mismatched cards and D black cards in the D pairs of mismatched cards.

there are a total of 2R red cards in the collection of R red pairs and 2B black cards in the collection of B black pairs.

when you add the D red cards to the rest of the 2R red cards, you get 26. make a similar statement about the black cards.

6), a/b is either 0, positive, or negative. if it’s zero, then ur done.

if a/b > 0, then we have either a,b > 0 or a,b < 0

if a/b < 0, then either a > 0 and b < 0 or a < 0 and b > 0

[exotic fog]

moved my question from #proofs-and-logic because it fits here more

https://imgur.com/PdiHyb7

does this translate into ¬s → w & d ?

[mystic elbow]

Thank you so much for u help but could u also elaborate q 5 I’m not quite understanding sorry

[cerulean wind]

what part r u confused on?

[mystic elbow]

The mismatched card part

[cerulean wind]

so it i have 5 pairs of mismatched cards, it looks something like this

rb
br
br
rb
rb

there are 5 reds and 5 blacks in that collection

[weary tiger]

Can someone explain me combination with repition

in voice

?

[mystic elbow]

Ok

But why do u have 3 RB

[cerulean wind]

it was just a random collection of 5 pairs of mismatched colors

the order doesn’t matter

[mystic elbow]

Ohhh ok

[cerulean wind]

D also has to be even.

[mystic elbow]

Ok

How bout the black cards

@cerulean wind

[cerulean wind]

u tell me (or at least any idea that you have)

[mystic elbow]

Ok wait

Same like what u said about red?

[cerulean wind]

yea. so 2B + D = 26

and 2R + D = 26

[mystic elbow]

Yes

Ok

[cerulean wind]

now what? (we’re super close)

[mystic elbow]

What

Hmmm

Do algebra and cancel stuff out?

[cerulean wind]

yea

[mystic elbow]

So what would the final value be

[desert edge]

since this is a proof by contradiction, do I try to prove that A ∩ B ⊄ ℤ?

[dapper rose]

No you assume that A ∩ B ⊄ ℤ

[desert edge]

can you help me through the steps

i am super new to proofs

but thats what i meant. sorry the terminology is not totally set

[dapper rose]

Assume for a contradiction that A ∩ B ⊄ ℤ

[vital dew]

@dapper rose

[dapper rose]

Oh wait

[vital dew]

@dapper rose

[desert edge]

no proper subset

which is the right one i assume

[dapper rose]

Okay

[vital dew]

@dapper rose

[desert edge]

wish i could tell you

[dapper rose]

can you see how it follows

[desert edge]

well yea both set a and b make up for the integer set

is that what you meant?

wait i meant they make up for the positive integers. Which means they are a proper subset

as they dont cover the negatives, right

[dapper rose]

yeah that's right, just they want you to use contradiction for some reason

[desert edge]

hm

i dont know enough properties to do anything

ive done 2 proof by contradictions so far but they werent like this

[weary tiger]

Consider the function f : Z × Z → Z where f(m, n) = |n|. Is this function onto?

why is it NOT onto

f(m,n) = y

m=9 n = -y

|-y|

y = y

[desert edge]

I would still love some help on this if possible

[vale cairn]

Suppose the intersection of A and B is not contained in Z

Then there is an element of A and B which is not an integer

Can you show that is a contradiction

[weary tiger]

-2 = |n|

[desert edge]

is he talking to you or me

[weary tiger]

not sure

[desert edge]

think its me, as mine has to do with proof by contradiction

since its a proof by contradiction, I assume that A ∩ B ⊄ ℤ

then there is an element of A and B which is not an integer hm

∃x A ∩ B ⊈ ℤ

i dont really know what im doing. can you guide me please

[vale cairn]

Talking to tugge

Well I assume they mean proper subset by that symvol given the hint. suppose intersection of A and B is not a proper Z. Then either A cap B is Z, or A cap B contains an element which is not an integer

[desert edge]

A cap B meaning?

[cloud cargo]

its the intersection symbol

A ∩ B ⊄ ℤ <=> {} ⊄ ℤ which is a contradiction?

[fallow sable]

So on my practice test I have this question, and the given answer is C i just dont really understand why

it help for me to translate this into words: so what I was thinking for the original is "For all X there exists a Y that makes P(x,y) true"

But doesn't c say that "For all X there exists a Y that makes P(X,Y) untrue"?

wouldn't you just need to find one outlier Y that makes P(x,y) untrue

[calm comet]

i think its asking you to provide a negative of the given statement

which C certainly is

[pale epoch]

(c) is not the negation of $\forall x \exists y P(x, y)$ ...

[vital dew]

Lochverstärker

[pale epoch]

is there more to this question @fallow sable ?

[fallow sable]

No, that is it

Nevermind i see why B doesn't make sense

[pale epoch]

why?

[fallow sable]

Well hold on , i guess i need to understand the original statement more. Does it mean that for all X's , there is a Y that makes P(x,y) true, so for any X that you plug in, there is a corresponding Y ?

[pale epoch]

yes

if i give you an x, you can find a y that makes P(x, y) true

[fallow sable]

So wouldn't that mean we need to find an outlier to negate that? So there exists a Y in X that makes P(x,y) false

[pale epoch]

a Y in X?

[fallow sable]

what im trying to translate into math is that there is at least one Y for the X's that negates P(x,y)

if that makes sense

[pale epoch]

ok so the way i would think about this is the following:

your original statement starts with "for all x ..."

so to disprove this, we would have to find a counterexample (this is what you mean by outlier)

[fallow sable]

yes

[pale epoch]

so we would have to find an x, such that P(x, y) is always wrong

so whatever y you plug in, this will always be wrong

this is (b)

does this make sense to you?

[fallow sable]

So B doesn't mean there exists an X for every Y? it means that there exists an X for some Y?

[pale epoch]

the order of quantification matters

i think of it like a game

so b) here means that i can give you a (specific) x

and no matter which y you try, you will never make P(x, y) true

i.e. for this specific x there does not exist any y that makes P(x, y) true, so this x is a counterexample to the original statement

[fallow sable]

okay now i see

that makes sense

it was very confusing cause C is the correct answer on the key

i guess the key is wrong

[pale epoch]

the key is wrong, yes

i suggest telling this to your professor

also if you want to think about this more concretely try P(x, y) as "x > y" and "y < x"

also also i get the wish to translate this into "natural language" but just mechanically negating statements is also super easy and quick

i think of it as pulling the negation symbol inside and every quantifier that it touches gets "flipped"

$\neg(\forall x \exists y P(x, y) \equiv \exists x \neg(\exists y P(x, y)) \equiv \exists x \forall y \neg P(x, y)$

[vital dew]

Lochverstärker

[shadow grove]

I have to prove this.

I tried using De Morgan's Law

Such that

(leftside):

A ^ (-A)^(-B)

Can I use some law of idempotency to turn A AND (-A) into just A?

[vale cairn]

wdym by 'A AND (-A)' ?

[shadow grove]

-A = Neg(A)

[vale cairn]

I don't see what negation has to do with this

[shadow grove]

I turned them into logic

[vale cairn]

the notation means to remove elements of B from A

[shadow grove]

al expressions.

Yes but

A-B = A^(neg(B))

[vale cairn]

er okay i see what you mean, not standard notation though

You could prove this by showing double containment and arguing that way - I'm not sure how formal you want it

But assuming by -A you mean the complement A^c of A, it should be clear that the intersection of A and A^c is empty

the set of things that are and are not in A

[shadow grove]

I have to do it pretty formally

I migh tjust do a truth table

[vale cairn]

Eh there is a way to do it more directly with De Morgan's law

[shadow grove]

unless, once again, there is a law, such that I can turn either left side into the right side or vise versa.

[vale cairn]

Well there is

[shadow grove]

Which is?

[vale cairn]

Well could you show me your working thus far?

You seem to have made a slight mistake applying de morgan's law

[shadow grove]

It is on paper

but I will try to write it again

[vale cairn]

ok fair enough

But yes, writing the left hand side as A cap (A cap B)^c should get you started

(cap = intersection)

[shadow grove]

Nooo

because I don't know the logical equivalent of complement :/

[vale cairn]

You seem to have used that already using 'negation' right

But I also think it's very important to actually learn the set theory instead of having to port stuff over to logic

i originally thought of it in boolean algebra or equivalent too but it's helpful to do it more directly

[shadow grove]

I know the set theory, but I have chosen tod o it this way

and I have 2 hours 'till delivery, so I am far beyond point of no return 😄

[vale cairn]

What does that notation in the first line mean? as that negation seems misplaced

[shadow grove]

I know if

[vale cairn]

$a \land \neg(a \land b)$ ig you mean

[vital dew]

potato

[shadow grove]

oh yeah, oops

[vale cairn]

well anyway, that is not how demorgan's law works like

that is equivalent to $a \land (\neg a \lor \neg b)$

[vital dew]

potato

[vale cairn]

now it's very close to what we wanted

[shadow grove]

oh

I see

yes omfg

Or, not where to go, but that, that is how you use De Morgan's law

I can use distributivity now, right?

And then I will (I assume) get something ala

(A\wedge\negA)?

[vale cairn]

yes

[shadow grove]

Thank you!

[vale cairn]

$(a \land \neg a) \lor (a \land \neg b)$

[shadow grove]

Saved my life 😄

[vital dew]

potato

[vale cairn]

np

[shadow grove]

oh wait...

Doesn't this mean, that it will always be correct...?

[vale cairn]

wdym

no it doesn't

[shadow grove]

oh nvm

it will always be wrong

oops xD

[vale cairn]

it won't always be false either

but a \land \neg a is a contradiction yes

[shadow grove]

Yes, exactly waht I meant.

But nonethless, thank yoU 🙂

[shadow grove]

quick question, how do I do xor in formal logic...?

@vale cairn

We haven't covered this yet, but I feel like I can effectivize some if it by using it

Nvmmm

[shadow grove]

Is this logically correct?

I.e If the person saying the statement is a liar, then the true statement is the negation of that said statement?

Or am I committing one of the dread fallacies? Can't quite think of which, but something feels off...

But on the other hand, analogously it feels like algebra, where I "multiply by neg" to both sides.

Please tag me, if you have an answer 🙂

<@&286206848099549185>

🙂

[cerulean wind]

there is no “final value”. you get that
2R + D = 26 = 2B + D
so you have to have R = B. which is what we wanted to show

[rich siren]

Can anyone explain what this is asking

I don't understand

For any equivalence relation R on a set T, the fact that for all x in set T, x is an element of the equivalence class of x, because R is symmetric

Is how I translate it in my head, which still makes no sense

[pale epoch]

which part do you not get?

do you know what an equivalence relation is? what the equivalence class of an element is?

[rich siren]

[x] = all elements that relate to x in a set right?

[shadow grove]

Could someone also help me? 🙂

[pale epoch]

yes

so [x] is some set

and for an equivalence relation we have $ x \in [x]$

why?

[rich siren]

Hm. So is it saying for all equivalence relations on set T, all x in that relation will also be in the equivalence class of [x]?

[pale epoch]

what does "x in that relation" mean?

[rich siren]

x in the equivalence relations

of set T

will also be in the equivalence class of [x]

[pale epoch]

that makes no sense

[rich siren]

I agree

[pale epoch]

x is an element of T (which is just a set)

R is an equivalence relation

(its a subset of T x T)

[x] is a subset of T

and x is in [x] (this is true)

this questions essentially asks why its true

(and gives symmetry of the relation as a reason)

[rich siren]

Ah, ah, ah. Okay I get it I think

so the answer is false

but if it said reflexive, it'd be true

Yes? 🙏

[pale epoch]

sounds good

[rich siren]

Lets goooo

Thanks :)

[shadow grove]

Could you help me with my problem now?

[pale epoch]

you can just ask, someone will (probably) help you...

[shadow grove]

I did

Further up in the chat

[pale epoch]

oh that

its a weird question honestly

if "anna is a liar" means lying in that what she said above, its fine

[shadow grove]

what..

[pale epoch]

I.e If the person saying the statement is a liar, then the true statement is the negation of that said statement?
this is true if you model your statements with FOL

a statement is false if, and only if, the negation is true

[sharp basin]

I just have a question on how to formalize a statement. Here is the given statement "for all integers n, n2 − n + 3 is odd". Is it ok if I restate the statement as shown in the attached image or is it incorrect to say it as stated in the image? Thanks for the help.

But also if my version is incorrect how would I restate it.

[faint narwhal]

latex

https://detexify.kirelabs.org/classify.html

[quaint seal]

could someone please explain this to me?

im beyond confused

[dry raven]

Im having trouble realizing why this is true....

x and y are the set of real numbers

[amber prism]

y:=1-x

[dry raven]

so couldnt I make x any negative and y 0 though?

[amber prism]

corrected it

$y:=1-x\implies x+y=1\geq 0$

[vital dew]

Mosh

[sharp basin]

Can anybody help me with my question up top

[amber prism]

a y just needs to exist

so even if you pick a y that doesnt work, you can just pick a y that does work

like I demonstrated

[dry raven]

ohhhh that makes sense then

[sharp basin]

Latex is not to bad you could learn it in a day here is a good tutorial on it https://www.youtube.com/watch?v=Jp0lPj2-DQA&t=394s

[amber prism]

in words what it says is

"For any real number x, there is a real number y such that x+y isnt negative"

[weary tiger]

hi

[amber prism]

(in fact there are infinitely many y per x but I digress)

[dry raven]

ya these problems somewhat confuse me because when I switch the existential and universal then it makes that false.... which makes no sense to me

[amber prism]

it.. shouldnt

since addition is commutative in R

$\forall y, \exists x (x+y\geq 0)$

[vital dew]

Mosh

[dry raven]

second one was aparently correct

this is why im confused haha

because whats the difference x and y are just variables and are interchangeable in this example which is why I dont understand the logic behind this

[weary tiger]

https://tenor.com/view/calculation-math-hangover-allen-zach-galifianakis-gif-6219070

[river harness]

hello?

can anyone help with discrete math hw?

[amber prism]

https://dontasktoask.com/

[river harness]

I have learned new etiquette 😅

[mystic elbow]

Yes got it that you also for another question no explanation needed?

[cerulean wind]

do u mind rephrasing?

that was hard to understand

[mystic elbow]

There were 2 questions I sent yesterday right

So another one doesn’t needed any explanation(proving)

@cerulean wind

[cerulean wind]

yes, 5) and 6)

[mystic elbow]

So for 6 u sent some values so that’s the answer?

That’s it nothing else to add up ya?

[cerulean wind]

yea. 2R + D = 26 = 2B + D

so you get that D is even and B = R

[mystic elbow]

Ok got it thank you

[mystic elbow]

I have one question if u could help

[cerulean wind]

ok

[mystic elbow]

8,9

[cerulean wind]

its pretty late where i am. thought they would be a bit less involved. somebody else should be able to help later tho

[river harness]

Hi everyone, I'm trying to figure out this homework problem where it asks to show that if p and q are distinct rational numbers with p<q then there is a rational number r such that p<r<q

Any thoughts?

[pale epoch]

think about what you would do geometrically (on the number line)

[river harness]

So if i wanted to find a number between p and q i could do (q-p)/2

Or (p+q)/2 and that would give me r

Is that what you meant?

[pale epoch]

not quite, but that's what i mean, yes

[mystic elbow]

Can anyone please help <@&286206848099549185>

[cerulean wind]

this right here is perfect

[river harness]

Thank you! ☺️

[tacit pumice]

Can anyone please help me on predicate logic?😩
There exist an x for all values of y such that x > y for the domain x and y in the negative
G(x, y): x > y
∃x∀yG(x, y)

But I have no idea how to do it for the domain of all real numbers.

[stray reef]

"x is negative" can be written as G(0,x) by the way

$\exists x[G(0,x) \land \forall y (G(0,y) \to x=y \lor G(x,y)) ]$

[vital dew]

Ann

[tacit pumice]

That made sense, I'm 99% sure that the answer is correct. But could you pls simplify or elaborate more? I'm slow...

[tacit pumice]

Can I answer it this way? ∃x∀y(G(0,x) ∧ ((G(0,y)→G(x, y))) or do I have to state that x=y v G(x, y)?

[stray reef]

well you need to put the forall y

there exists a number x such that:

• x is negative
• for every negative number y, either y equals x or y is less than x

[orchid idol]

just found out answer but i will post here again, i was so dumb all this time:
if each column and row has the sum x,
by summing all 5 columns we have total sum of 5x,
by summing all 4 rows we have total sum of 4x
and since we are summing all the squares the same way, we have 5x = 4x, meaning x = 0, which is impossible

[shadow grove]

I am doing an assignment on logical statements.

The logical statement in natural language is at the top, no more info is given.

I have modelled it, in the case Anne is lying and Anne is telling the truth.

Have i modelled it correct? And am I correct in assuming, that before we can say anything about their truth/liar relation, Anne must be telling the truth. Otherwise, we can say nothing?

[orchid idol]

we are summing the same number

[fresh grove]

A = {1, 2, 3, 4}
P(A x A) = P({(1, 1), (1, 2), (1, 3), (1, 4), (2, 1)... (4, 4)})
= {{}, {(1, 1)}, {(1, 2)}, ... {(4, 4)}, {{(1, 1)}{(1 , 2)}}, ...}

assuming each relation in P(Ax A) is equally likely to be chosen, the qn is what's the probability that a randomly chosen relation is reflective?

here, i got 4 relations to be reflective, (1, 1), (2, 2), (3, 3), (4, 4)

and the total number of relations in P(Ax A) = 2^16

but my answer isnt one of the options

[celest hazel]

Hey I was kinda wondering is there any reason we can’t use the size of sets as the definition of the natural numbers?

Like you define 0 as the size of the empty set. Then you define 1 as the size of a set S such that for all x in S, x=x1 and the size of S is not 0.

And so on

[vale cairn]

that's remarkably similar to the construction used in ZF for example

[celest hazel]

Kinda? But there’s several key differences. For example in ZF 4 is an element of 5 whereas this isn’t the case in this method necessarily

[vale cairn]

agreed

[pale epoch]

what is the "size of a set" if you don't have access to natural numbers

[celest hazel]

Well you define size recursively

The way I started doing above

[pale epoch]

"0 is the size of the empty set"?

what kind of object is this

[celest hazel]

The empty set is a set such that for all x, x is not an element of S

[pale epoch]

i know what the empty set is

what is it's "size"

[celest hazel]

Well you define 0 as being it’s size

[pale epoch]

what does size mean

if you define 0 as something else you have to know what that something else is

[celest hazel]

Can’t I define it as it’s own entity?

[pale epoch]

no?

[vale cairn]

I'm also struggling to see how you'd define, say, addition with this

[pale epoch]

this is akin to defining 0 as the soul of the empty set

what does this mean

[vale cairn]

Other than just making it symbolic and saying 1 + 1 = 2 etc in which case calling it the size of the empty set seems sorta irrelevant

[celest hazel]

You could define addition of natural numbers by saying that if A and B share no elements, then the size of the union of A and B is equal to the size of A + the size of B

[pale epoch]

then you have to know what size is again ...

like, we say that a set A has cardinality n in N iff there is a bijection between A and n

[cerulean wind]

0 is defined as a set, as is everything in ZF. is your size a set? it doesn’t seem like it

[pale epoch]

this notion requires the existence of natural numbers

[vale cairn]

I suppose this answers your question though lol, it has clear disadvantages ig

[celest hazel]

I mean ZF definition of numbers is really weird tho. Like 4 is an element of 5 in ZF

But I feel like that should be wrong

[vale cairn]

why? it allows you to naturally define an order too for example

[celest hazel]

You can define order in a different way other than saying that 4 is an element of 5

[cerulean wind]

example?

[celest hazel]

I could say that |A| “comes right before” |B| iff there exists some set C such that |C|=1, A and C share no elements, and A union C is equal to B.

Where A, B, and C are all sets.

Or smth like that

[pale epoch]

the order with respect to cardinality of sets is defined by injections/surjections between the sets

they were talking about the natural numbers

[celest hazel]

Plus a really convenient thing you can do instead is say that |A| is less than or equal to |B| if there exists some C such that the union of A and C equals B.

[pale epoch]

which in your definition are not sets but "sizes" of sets

[celest hazel]

Ye my “size of sets” are not themselves sets

[pale epoch]

for sets its really clear, we have |A| <= |B| iff A injects into B

[celest hazel]

“Injects” seems a bit more complicated then you’re letting on

Injection is like this massive recursive element you need to do in this case where you need to do stuff like check the elements of your elements

Whereas just defining natural numbers in terms of sizes seems more natural to me? Plus it seems like a convenient extension of the idea of natural numbers as “counting things” where numbers wind up essentially being defined as how many elements are in a set. It’s a cardinal definition essentially

[vale cairn]

i don't see anything easier about this compared to saying there is an injection from A to B for example

[celest hazel]

Wait one of my definitions earlier was a bit wrong. Its actually the case that |A|<=|B| iff there exists C such that |A union C| = |B|.

[vale cairn]

and how are you defining equality of set sizes?

[celest hazel]

Well if |A|=n and |B|=n then |A|=|B|.

And I can define what that n is recursively using the way I’ve defined sizes of sets. So like I could then define 2 as being the size of any set S such that if x is an element of S, then x=x1 or x=x2 and x1=/=x2 and the size of S is not 0 or 1.

[celest hazel]

Ig what’s really going on is that the ZF definition of numbers is defining them as a thing, whereas my definition is kinda closer to describing numbers as a quality.

[pale epoch]

nothing of this works, because you can't define what the size of a set is

[celest hazel]

The size is essentially being defined as how many elements are in the set

[pale epoch]

and how do you define this in the language of sets

and more importantly, without natural numbers

you wanted to define 0 as the size of the empty set, because it has 0 elements?

that is very circular

[celest hazel]

You don’t need natural numbers to define how many. I can say that a set S has a size of 1 if there exists an x such that x is in S and for all y if y is in S then x=y.

[pale epoch]

ok sure

and then you define 1 as what?

the number that is

[vale cairn]

feels kinda ontologically lacking to me compared to the normal definition too

hm

[celest hazel]

That’s how you define 1. It’s the size of a set with the property given above

[pale epoch]

and what object is that

you didnt define what size means

only what it means for a set to have size 1

[celest hazel]

Well that’s bc we’re kinda defining numbers as a quality here not a thing in it of themselves essentially

[pale epoch]

that just means you arent defining numbers at all

[vale cairn]

I guess like, what is the real difference between this and just saying

a set S has 5 elements if you can count its elements to be 5

or smth

Like you're only sort of defining it in terms of sets

[celest hazel]

I mean that’s kinda what I’m saying. Numbers as qualities rather than objects.

[pale epoch]

whats a quality

and how do i manipulate it

but ok, the cardinality stuff works for finite sets

[celest hazel]

Ye and I’m just using this to define natural numbers

[pale epoch]

(and you can define one infinite set)

no, you dont

you only defined what it means for a set to have cardinality n

not what natural numbers are

[celest hazel]

I guess? Seems like defining “what they are” isn’t as useful then. Since I can just define most things using this idea of “how many”. Since for the most part, numbers are adjectives which is why you can describe so many systems with them

[pale epoch]

i mean i need numbers for a few things

i want to define a theory of arithmetic on them

which requires addition and multiplication

and i want induction

you can build addition and multiplication maybe, but you will always define it in terms of sets that have cardinality of a natural number

so you are again working with sets, which is essentially what the von neumann construction is

and induction will have the same issue

every time you will talk about a number n you will have to refer to some set of cardinality n

and if you like build the set of natural numbers you will have to consider all sets of finite cardinality and quotient by some equivalence relation

at which point you can just choose unique representatives of each natural number, which is exactly what the von neumann construction is

[celest hazel]

I mean in reality most of mathematics is adjectives without the noun. Like numbers are adjectives that can be given physical forms if you want to

[pale epoch]

no

when you use the word number it is very clear what you are referring to

usually some element of a skew field

[celest hazel]

Maybe I just need to learn more idk

[hallow horizon]

Same

[celest hazel]

It just seems that whenever we actually try to think about numbers in any way we turn them into adjectives describing something

[pale epoch]

i think of numbers as very concrete objects

[celest hazel]

And then the number is just an abstract way of describing the adjective allowing us to describe all possible things that this number could describe

[pale epoch]

that sounds like philosophy, not mathematics

[celest hazel]

I guess?

But I mean numbers are not concrete things by definition they are abstract

[pale epoch]

they are very concrete

[celest hazel]

What do you mean by concrete?

[pale epoch]

like, i know exactly what a real number is

its a very concrete object

[celest hazel]

Okay what is 1 then? How would you describe 1 to me?

[pale epoch]

its an element of the natural numbers that behaves in a certain way

[celest hazel]

What’s a natural number?

[pale epoch]

an element of a set, the natural numbers

which are the intersection of all inductive sets

or we can concretely define them via the peano axioms in first order logic, model them in ZFC, ...

[celest hazel]

I think it’s more so that the natural numbers are better as being considered to be homomorphic to the things called natural numbers in ZF

[pale epoch]

what does that mean

[mild oak]

help

[pale epoch]

help what

[stray reef]

"solve this for me please i will actively refuse to put in effort to learn"

[pale epoch]

ah ok

in this case: no

[pale epoch]

this seems very course specific

also don't ping helpers immediately, you are not that important (see #❓how-to-get-help)

[tawdry spindle]

guys I have a simple question

is 3<=5?

true or false?

[pale epoch]

[velvet pine]

I think it depends on what you set 3 and 5 to be

By building all the natural integers, you would start with the first iteration being 0=Card(∅)

Then 1=Card({a})

2=Card({a,b})

...

In that sense, then 3<=5

[foggy cloud]

what is a finite set?

I know something like the set A = {r | r is a real number and is less than 2} is infinite but what is a finite set??

thats not like randomly defined like A = {1,2,3}

[velvet pine]

If A is a discrete finite set, then Card(A)=/=Card(A without {a})

Wait

[pale epoch]

its a set that has either cardinality 0 or cardinality 1 or ...

alternatively they are characterized by the fact that injective maps are also surjective

maps from the set into itself

[foggy cloud]

Are there any examples where its not just a randomly defined set?

[pale epoch]

randomly defined?

[foggy cloud]

i just saw this and i dont know when I would ever use it

I've never seen a 'finite set' in my class yet

[pale epoch]

are you taking an analysis class?

[foggy cloud]

Its a mix between calculus and that

[pale epoch]

well, finite sets from the pov of analysis are very boring

since you can't really do interesting analysis on them

[foggy cloud]

oh ok

[pale epoch]

but in other fields of math, finite sets are studied a lot

[velvet pine]

A finite set is just a set that has finite number of elements in it

So it's a discrete set if you would prefer

Not a continuous one

[foggy cloud]

I know but I just haven't seen any non trivial examples lol my bad terrible wording

only time I saw it was in discrete math with some sets like A = {1,2,3,4,5,6} for some discrete math question

[velvet pine]

Okay so for example

[pale epoch]

the thing is

if you e.g. consider convergent sequences on finite sets

they are all eventually constant

if you consider finite subsets of real numbers, all the points are isolated

[foggy cloud]

yea you're right

[velvet pine]

The set {1/n | n in N*} is infinite

[foggy cloud]

im dumb thanks

[pale epoch]

so you cannot do any analysis on them

[velvet pine]

Since N* is infinite

[pale epoch]

nah, you aren't dumb

it's a good observation

[velvet pine]

Concepts in mathematics are always hard to get from scratch

Relatable

Hey a non-trivial finite set would be something like

{x in [0,2pi] | sin(x)=1}

Is a finite set

[pale epoch]

if you ever take an abstract algebra class, you will see a lot of more interesting finite sets (finite groups)

[velvet pine]

Hell yeah

[foggy cloud]

Oh am taking that next semester

Will it be in artin’s algebra, does it cover the topic you mentioned?

[pale epoch]

so yeah, a large part of that class will deal with understanding finite groups (and "almost finite" ones)

[foggy cloud]

(Book used in the class)

[pale epoch]

yes, artin talks about that

things like symmetry groups of geometric objects (cubes etc) are finite

and have interesting structure worth studying

[shadow grove]

Term vs variable:

A term can essentially be the specific value of the variable right?

Other than that, it can be a variable itself.

But a variable cannot have a numeric value...?

Or should it analogously be understood as:

variable is a subset of a term...?

Specifically, I am trying to understand A[t/x].

[pale epoch]

terms and variables are part of FOL, numerical values do not play any role here

every variable is a term

terms are inductively built from variables, constant symbols and function symbols

[celest plover]

Trying to prove that m(n+k) = (mn) + (mk) holds for the natural numbers.
I know I need a base case and an inductive step.
I've done the base case:
m(n + 0) = mn = mn + 0
And I'm a little stuck on the inductive step. Pretty sure I have to show that m(n + k + 1) = (mn) + (m(k + 1)). This is what I have so far:
m(n + k + 1) = m(n + S(k)) = m(S(n + k))

[pale epoch]

apply the definition of multiplication?

[celest plover]

oh

so i can just

multiply normally?

[pale epoch]

you should have a recursive definition of multiplication

like m*0 = 0 and m*S(k) = m + m*k

[celest plover]

i see

ok ty

[celest plover]

wait, sorry for the ping but maybe i actually dont understand

this is what im doing but it i dont think this has to do with what you said

[vital dew]

ComicallyBad

[solemn fossil]

is a|b "a divided by b" or "b divided by a"

so is the bottom line correct? this is from my professor

also may i get help on b? (variables are integers)

[meager prairie]

what have you tried with b?

[solemn fossil]

i straight up dont know where to start

[amber prism]

were you able to do a?

[solemn fossil]

yup

[amber prism]

apply similar logic

[meager prairie]

@solemn fossil use a direct proof

since you know a doesnt divide b, and a doesnt divide c, write out what that means

its nearly identical, like mosh said

well in structure, at least

[amber prism]

a|b iff b=an, n in Z (assuming I have it right way round)

[meager prairie]

ill get you started even, $a \nmid b \to b = \alpha a + \beta$, with $b \neq 0$

[solemn fossil]

This was my proof for a

[meager prairie]

is the brown part of the proof?

its not especially clear

[solemn fossil]

the blue is the proof

[amber prism]

it's iffy but yes

you have a|b <=> b=aj, now what's a does not divide b?

[solemn fossil]

what happened to the texit bot stuff?

[solemn fossil]

uhhhh

[amber prism]

dont overthink it

[solemn fossil]

b=/= aj?

[amber prism]

yes

[solemn fossil]

ohh

[amber prism]

then it's literally repeat the proof, just with !=

[solemn fossil]

Ohhh

Thanks

[solemn fossil]

also what the hell is this

[boreal lark]

can someone help me with this?

i need to solve this using only basic proofs that have already been covered

[meager prairie]

negation, you gotta learn that if you wanna wrangle definitions 😄

google it

or if you have a textbook

[shut jacinth]

me?

[meager prairie]

no

the panda

[exotic fog]

anyone knows?

[pale epoch]

continue at what you had last:
m(S(n + k)) = m + m(n+k), then apply inductive hypothesis and the result is almost what you want

[mystic elbow]

Hi can anyone please <@&286206848099549185> ping me please

[gritty crescent]

@mystic elbow Wait for 15 minutes before pinging helpers, read #❓how-to-get-help . Also, what have you tried/thought so far?

[mystic elbow]

Ohh ok I’m not very sure how to do it

[gritty crescent]

For 7.2, try a few rational/irrational values

Same with 7.3 sorta

If I say anything more I'd virtually give away the answer

[mystic elbow]

Ohh ok

[fresh grove]

for q14

is it 2^12/2^16

[fresh grove]

<@&286206848099549185>

[exotic fog]

<@&286206848099549185> what does "must" mean in logic specifically in this question

[quaint star]

No

That would mean if you are not a subscriber then you have to pay the daily fee

Whereas you could just opt not to watch it

You basically have: If A is true then B must be true. So it just means the same as A implies B.

In this case I would write it as:
w -> d or s.

But if you want to start with the s you could say:
(not s) -> (not w) or d

[stray reef]

@weary tiger do you still need help with this

consider the game for small values of N

N=2, N=3, etc.

keep going. keep extrapolating

make a table

for each value of N specify whether you win or lose

see if you spot any patterns

that... looks wrong to me

one moment

oh nevermind

yes this is correct it looks like

wait but

3 is a winning position isn't it

you give 2 and now your opponent has 2 and he loses

and you win

pen and paper lol

the losing positions form a sequence $L_k$ that obeys the recurrence $$L_{k+1}=2L_k+1$$

[vital dew]

Ann

[stray reef]

you seem to have identified that

but the first losing position is 2 not 3

3 is a win for you

2 is a loss for you

the winning positions are all those that aren't losing positions

the losing positions are 2, 5, 11, 23, 47, ...

you do not

the opponent will give you 11 and you will lose

from 11, whatever your move is, your opponent will be able to cut it down to 5

the winning strategy is to place your opponent in the closest losing position below the current

i'm here but also kinda not here

i think you misunderstand me

you claim that if N=15, we play against each other, and i go first, then you will win.

is that the case?

yes or no

then what were you saying

ah.

okay, and what's your opening move?

the piece is 15 and it's you to move.

consider me as the opponent.

okay, so now it's 8 with me to move?

i split it into 5 and 3 and hand you the 5.

your move?

i split into 2 and 1, you get 2.

no, i get 1 and i win.

the person who receives the 1 wins.

the losing positions are 3*2^k - 1 for k = 0, 1, 2, ...

when not on a losing position, the winning move is to cut down to the biggest losing position possible.

when on a losing position you are screwed no matter what you do.

[stray reef]

it's the solution of this recurrence with L_0 = 2

...idk honestly

[stray reef]

here

i described the winning strat here

i have class now

this tells us how to win when we can win

the assumption is that it's us to move

[celest plover]

k think i got it now, thanks!

[stray reef]

yes that was exactly my point

[tribal zenith]

did i draw my hasse diagram correctly?

this is for the divisibility relation, for some reason my notes say this diagram shape is not allowed, so i added the numbers to show a counterexample

[stray reef]

@tribal zenith you missed 6

[tribal zenith]

But what if I define it to be {1, 2, 3, 4, 12}

[stray reef]

ah well then it looks ok

mmmm

maybe??

i'm in class rn.

so i can't fully concentrate

[toxic ermine]

Hey guys, can anyone help me with a little proof with identities?