discrete-math

[narrow agate]

If P is necessary for Q, shouldn't it occur before Q?

[vestal bronze]

it's all timeless

[narrow agate]

I don't get it.

[vestal bronze]

just true or false things

a number has to be even to be divisible by6

div by 6 → div by 2

[narrow agate]

ohhh

Thanks @vestal bronze!

[azure smelt]

Can someone pls solve

[glass leaf]

say where you're stuck

[storm violet]

$\exists !$ means ``there exists a unique,'' right?

[vital dew]

Altanis

[slim geode]

Yes

[kind reef]

i: Correct. Let y be x+1, so y=x+1 if x>0 , and let y be x-1 so y=x-1 if x<0. So y^2>x is always guaranteed. Because there is always an x-1 or x+1, in terms a |x+1| in R.
ii: Correct. if x^3=8, then x=third root of 8 =2. There is no -2, cause the grade is uneven with 3.
iii: Wrong. Let x+y=x be a thing. If x=/=0, u can say y=0 as the only answer, which is not "For all y€R". If x=0 it only works if y=x=0 too, so also not "for all y"
iv: Wrong: Let x=0. Then 0=2y+1 -> -1=2y -> -1/2=y and thats not from Z.
If u say x=0 is not from Z, try x=2: 2=2y+1 -> 1=2y -> 1/2=y --> Not from Z

[lofty cloud]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[charred tapir]

no

[south quarry]

someone save me

this is my last class for my degree and I am down bad

[lofty cloud]

Asking the actual question right away is more likely to get responses.

Asking "Can I ask...?" or "Does anyone know about...?" doesn't give people enough information to decide whether they can help, and answering can feel like a promise to help with the actual question, which they might find themselves unable to.

[south quarry]

something like this

[worn mulch]

do you know where to begin or...?

[south quarry]

I have no idea

[worn mulch]

Have you proved stuff in math before?

[south quarry]

like not gonna lie im just trying to get a good enough mark to pass the exams

[south quarry]

not necessarily

[worn mulch]

hmm alright

[sharp rose]

Is your course about proofs in discrete math

[south quarry]

im so desperate to learn how to finnesse the class that I searched up math discords to learn from people smarter than me

[sharp rose]

Maybe review some of the material

[south quarry]

its about discrete math

but this is the first unit

im currently going through the practice exam. and im struggling

it all looks so foreign

honestly if you know tips that could guide me through it that would be helpful

[worn mulch]

I don't think you'd have much problem if you took the time to understand this
how comfortable are you with the notation? cause if you're not comfortable we have to address that first

[south quarry]

scientific notation?

[worn mulch]

there's some youtube channel for discrete math I know, I can share that

[south quarry]

that would be helpful if you could share it!

[worn mulch]

no like x belongs to A can be written as x \epsilon A and and the subset, union, intersection notation

https://www.youtube.com/watch?v=rdXw7Ps9vxc&list=PLHXZ9OQGMqxersk8fUxiUMSIx0DBqsKZS

skip to the part with set theory

[south quarry]

geez...im going to be honest sounds like a foreign language when you say it like that versus what chatgpt says

[worn mulch]

https://www.youtube.com/@MichaelPennMath
this guy is also good, look for his proof videos

[south quarry]

ive been working with chatgpt for like the last week prompting it to teach me like im a 5 year old

[sharp rose]

Probably should read textbook or lecture notes rather than chatgpt

[worn mulch]

if you're more comfortable with books check out book of proof by richard hammack

(it's free online)

probably the best resource

[south quarry]

I should but I kinda of am just seeking the easiest way out to complete this course. I want to know just enough to pass

[sharp rose]

Chatgpt is not the easiest way out

[worn mulch]

don't seek the easiest way out

[sharp rose]

It will take more work to learn the same material

[south quarry]

maybe yall can give me advice about my thought process

i figured if I studied the practice exam and learned how to do each question I would be good enough to pass the exam?

obviously you guys dont know the cirriculum and such but thats the base line i put for myself

I just want to learn how to do specific questions from the exam

[fossil mural]

the obvious potential issue with that is that you'd end up learning the practice exam instead of the underlying skills, and then you'd be faced with different questions based on the same skills and not know what to do

[worn mulch]

yeah but there's still the possibility of failure, the more you do things the right way (as in understanding the material), the higher the chance you pass
even if your goal is just doing it for the sake of doing it it's still putting you at risk

[south quarry]

ya I was hopinh to gain enough knowledge from practicing the same question to finesse my way through the exam

[worn mulch]

u can't do that if you don't understand the notation as well 😭

[ember obsidian]

in your first proof based math class youre basically learning a new language. trying to do the bare minimum without getting familiar with the notation and definitions would be like being stranded in a foreign country without a translator

[south quarry]

aorts honestly shout out the youtube video you sent its a lot shorter and less intimidating considering how long the videos are. I might just binge his videos and try to learn from there

[worn mulch]

you probably could if you had some baseline to begin with, you're sort of skipping that baseline entirely by trying to find the easiest way out

[fossil mural]

i don't know exactly but i think there's a general tendency for "discrete maths" to be around the point in most people's experience with being taught maths where it switches from "you can just scan the question for keywords to figure out which of a fixed list of symbol manipulation procedures it wants you to execute" to "you are presented with a problem you have never seen before and have to apply generic problem-solving strategies and to some extent creativity to construct a solution"

[south quarry]

I see.

[worn mulch]

hmm yeah do try but refer to your class specific material as well!!

they're short because he teaches basically whatever the bare minimum is

[fossil mural]

so if you're used to questions that, once you read through the words, really just want you to set up a system of equations and then solve them or whatever, this is something completely different to that

[south quarry]

I feel like the youtube will give me enough foundational skills to complete the course...at least thats what im hoping

[worn mulch]

yeah that and the material u have

[south quarry]

thanks for all the advice! if I have more specific things ill reach out

[fossil mural]

also even if you have the underlying skills you'll need to know how your specific course wants you to do things, just in the sense that there are some ultimately arbitrary/surface-level choices that different people will set up differently even if they're talking about the same mathematics, and your course might expect you to know which specific choices they made

like, i could probably get through this course with very little effort, but even i wouldn't want to show up to the exam without having looked at any of the course's provided material, because they might have a question that asks for the fourth principle of induction and i'd just be sitting there going "huh?? what's the fourth one?" even though if the actual statement was presented i'd recognise it as obviously true

[south quarry]

ok ill keep that in mind as I get grades on previous things I submitted

[hidden totem]

i think everyone here is trying to softball the main point by offering valuable nuanced points, which i will contrast by directly hammering at you once

basically, dont half-ass your education. if you find yourself caring more about passing than genuinely learning, this mentality is only going to make things more difficult, not less

utilize all the resources at your disposal, your lecture notes, the textbook, youtube videos, guided help, everything. each resource serves some function that is slightly different from the others

finding a reason to care is genuinely one of the hardest things about learning, but is, at least i would argue, the most important thing

[south quarry]

completely understand. As this is my last class in exclusion to a captsone project I might just be making excuses for myself but I just want to get it over with

[hidden totem]

that is also understandable, burnout and stress are definitely real and not to be ignored

take care of yourself first and foremost

[fathom crystal]

hello

anybody know why the middle section is included in a symmetric difference of three sets? im thinking maybe its because it gets negated 3 times which ends up including it in the final solution? i hope someone doesnt mind clearing this up i appreciate it.

[quick folio]

it's just how the math works out when you take (AΔB)ΔC

[fathom crystal]

what if it was the overlapping section of a (AΔB)Δ(CΔD)? im just curious what that would look like now

[quick folio]

it is shaded if it is the overlap of an odd number of sets

[fathom crystal]

ohhhhh that makes sense

[quick folio]

it is unshaded if it is the overlap of an even number of sets

[fathom crystal]

okay that clears things up now thanks!

[sharp rose]

(A \triangle B \triangle C \neq {x : x \text{ is in precisely one of } A,B,C})

[vital dew]

qiu

[fathom crystal]

Ty ty I get it now

[wicked holly]

@unkempt edge what is it brochacho

[high skiff]

are recurance relations questions allowed here?

how to get the exact solution of a divide and conquer recurrence? am not talking about an asymptotic, but the actual solution of the reccurance that gives the number of operations done by an algorithm exactly.

[vestal bronze]

wolfram can solve them

[past rivet]

<@&268886789983436800> spam

apologies for the triple ping

[fast vale]

i can click through channels too you know

[past rivet]

clearly not as fast as me

[fast vale]

always fun getting to mark all the channels as read when a bot comes

[flat eagle]

(graphs) what's a quasitree? my math teacher showed me a counting result concerning them that he was trying to prove but i can't really find a good explanation of what it is online

[main falcon]

you can mark a server as read yknow

[near seal]

I have the answer sheet to this but I don't really understand it. It starts of as:

Claim/Hypothesis?: i != j, W^i != W^j
for i >= 0
and j <= n-1

Proof by contradiction:
Assume that W^i = w^j

That means,
(w^i / w^j) = (w^j / w^j)
w^(i-j) = 1 
w^0 = 1

For some reason this contradicts the assumption that w is a prim. root of unity? idk why?

and

w^k != 1 ??? where did k come from  
and
0 < k <= n-1

Therefore w^i = w^j is not possible.

If someone could explain this like im a 5 year old i'd appreciate that. Also plz tag me so i get notified.

[winged delta]

For some reason this contradicts the assumption that w is a prim. root of unity? idk why?
It's just part of the definition of being a primitive root of unity. For n>1, the primitive roots of unity z have that 1 =/= z^k for k=1, 2, ... , n-1

[near seal]

Ok so checking if im getting it right:
a complex num w is an n-th root of unity if: $w^n = 1$
A complex num w is a prim. n-th root of unity if:

1. $w^n = 1$
2. n is the smallest possible integer such that $w^n = 1$
   meaning: $w^k =/= 1$ for every $1 <= k <= n-1$ (doesn't go to n-1 cuz for n -> $w^n = 1$)

[winged delta]

Mmhm!

[near seal]

but in the question, w^0 exists

wait is it because j <= n-1 ? is that part of the proof? like,
Proof by contradiction:

Assume that W^i = w^j
for i >= 0
and j <= n-1

[winged delta]

Yeah pick i and j to be among the list 1, ..., n-1

You already know that 1 is not equal to any of the w^1, w^2, etc.

So you need to show that the non-zero powers are not equal to each other

[vital dew]

gosha404

[near seal]

Is that accurate?

[little prism]

the key point that you somehow arent mentioning is that (wlog i>j) you have 0<i-j<n

[near seal]

What does wlog mean?

[little prism]

without loss of generality

by switching the roles of i and j we can just assume that i is the bigger of the two

[near seal]

that's why i have to mention 0 < i-j <n ?

[little prism]

if j>i then -n < i-j < 0

thats just bookkeeping at that point, it doesnt change much

the point is that you arent mentioning that k=i-j is the precise exponent which goes against the definition of primitive root because 0<k<n but omega^k=1

your proof currently just writes some things then writes down the definition of primitive root and then concludes a contradiction from nowhere

[near seal]

Do I have to mention that? If w^i = w^j ? So it doesn’t ever matter if i<j cuz they’ll be equal to each other.

[unborn bronze]

Need a mentor for this

[amber nexus]

The way I understood this is that, wlog, we can interpret this as (AΔB)ΔC, or

"In only one of AΔB or C"

A int B int C is definitely a part of C

but, beause A int B int C contains members that are part of both A and B, it cannot be in AΔB

Therefore, it is in C and not AΔB, therefore it satisfies AΔBΔC

i just saw that this is from the 24th

[spark field]

Use one of the methods of solving recurrences

The basic ones are forwards or backwards substitution

for example

Let $a_n = 3a_{n-1}$

Then $a_{n-1} = 3a_{n-2}$ by the same rule. Similarly, $a_{n-2} = 3a_{n-3}$ and so on.

Therefore
\begin{align*}
a_n &= 3a_{n-1} \
&= 3(3a_{n-2}) \
&= 3^2(3a_{n-3}) \
&\vdots \
&= 3^xa_{n-x}
\end{align*}

Suppose we have an initial condition $a_1 = 1$.
Then we set $n - x = 1$, i.e. $x = n - 1$, so $$a_n = 3^xa_{n-x} = 3^{n-1}a_1 = 3^{n-1}$ and we have solved the recurrence.

[vital dew]

Coolempire93
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[spark field]

someone in #combinatorial-structures may be able able to help

[flat eagle]

thx o7

[keen flower]

can someone help me understand the definition of prime and compositive?

[spark field]

The intuitive definition is

Prime - only divisible by 1 and itself
Composite - can be divided by some number > 1 (that isn't itself)

Using the definition of divisibility: b is divisible by a if and only if b is an integer multiple of a (that is, b/a is an integer)

An example would be: 17 is prime
You can try to divide 17 by any number other than 1 (or 17) and you always end up with a fraction [that can't be turned into an integer]

6 is composite
I can divide 6 by 3: 6/3 = 2 which is an integer, so 6 is divisible by 3, which is a number > 1

[keen flower]

i understand that but using the definition in the proof is what gets me

maybe im attacking this wrong

[spark field]

What type of questions do you have

I'll write an example proof for you to see

[keen flower]

okay

There exists an integer n such that 6n^2+27 is prime

[spark field]

Ah yes this requires a little more than just the definitions

[keen flower]

si si

[spark field]

Let me see if I can find a similar problem in my ENT book to write a proof of

[keen flower]

thank you as always

[grand ibex]

are you asked to prove the statement itself, or are you asked to prove true or false

[keen flower]

prove that its false

[spare slate]

orz u

[spark field]

ah false is a different story

[grand ibex]

think about what coolempire said earlier, and see if you can manipulate this term to always be a multiple of an integer greater than 1

[spark field]

(since not prime == ||composite||, so we need to show that ||something > 1 divides 6n^2 + 27||)

[grand ibex]

that would then fit the definition of composite numbers

[keen flower]

right

[keen flower]

im following

[spark field]

no u

[spark field]

Think about what you know about factoring
That definition I stated before, b/a = integer is usually written as b = ka, where k is an integer (same as me saying b is a multiple of a)

[grand ibex]

also for another hint, think about how you can tell if a number is even

this is a similar principle

[keen flower]

3(2n^2+9)

[grand ibex]

yes

[spark field]

🥳

(conclude: no matter what n is, 6n^2 + 27 is ||divisible by 3||, so it's always composite -> never prime)

[keen flower]

how do i write it under the definition of composite

hold on

[spark field]

Usually formally they just define composite as 'not prime', so the process entails pointing out that it fails the definition of prime, whichever definition it is that they gave you

or at least that's what they did for us 😆

[keen flower]

im horrible with writing proofs man

i need more practice

can you give me the full proof writing and ill ask questions or maybe the other way around?

[spark field]

That's what this class is for 👍

[thorn hemlock]

What are you using to practice proofs? A book?

[keen flower]

yes

susanna book

thats what the class uses

and its a pdf too ugh

[spark field]

What's the definition of prime they gave you

I'll give you some similar examples, and then we'll look at how you write it

[spark field]

Ah good definition

Okay here are two related examples, and see if you can construct an answer based on how I write these

[keen flower]

its confusing broski

okay

[grand ibex]

idea of proof that i learned, for a direct proof, state definition and apply to the problem, and derive some result, then repeat

[keen flower]

yes pretty much

you need to state some " closed under blah blah"

[spark field]

• Proof or disprove that 6 is prime.
  I will show that 6 is not prime.
  \begin{proof} If 6 is prime, then if $6 = rs$, either $r$ or $s$ is 6, by definition. But $6 = 3 \cdot 2$, and neither $2$ nor 3 is 6, so 6 cannot be prime.
  \end{proof}

oops

I clicked enter

instead of backspace

[vital dew]

Coolempire93

[spark field]

Is the first example

Now comes the second, one involving a variable

[grand ibex]

i don't know about other places, but for the two weeks that i took discrete math last semester at my uni, i remember they were very strict about the most trivial of definitions lol

but it's good practice

[keen flower]

thats how my professor is too

[keen flower]

did u not complete the class?

[grand ibex]

cause in later classes when they loosen up a bit or the definitions are a lot more technical and/or way less trivial, it will feel more comfortable

[grand ibex]

i got bored and dropped the class 😭

[keen flower]

what major are you

[keen flower]

this is easy to follow

[grand ibex]

decided that i could learn basic proof writing much more efficiently on my own over break between semesters and then just cover the same requirement with combinatorics instead, which i like much more

[grand ibex]

computer science

but based on my course load, you would think i'm a math major

[keen flower]

u need discrete math for cs tho no?

[grand ibex]

math was a side quest that turned into the main quest

[keen flower]

so u will need to take it

ahahaha

thats how im feeling too

fellow cs major here

[grand ibex]

yes but combinatorics covers the same requirement

[keen flower]

ohh thats a seperate class

[grand ibex]

and i'm doing that now

[keen flower]

this class covers combinatorics too

[grand ibex]

yeah

yeah it covers some basics

[keen flower]

what is combinatorics?

[grand ibex]

i believe later you go through some basic graph theory

[keen flower]

yes that too

[grand ibex]

study of enumeration techniques

[keen flower]

ohh

very useful for algorithms it seems

[grand ibex]

definitely

[keen flower]

i was originally going to go for math major

but decided to change

i still have time to minor cs and major math

[spark field]

• Proof or disprove that there exists a positive integer $n$ such that $2n + 4$ is prime.

I will show that $2n + 4$ can never be prime.
\begin{proof}
Let $n$ be a positive integer. Then because integers are closed under addition, $n+2$ is an integer, and so we can write $2n + 4$ as a product of integers: $$2n +4 = 2(n+2).$$ By definition, if $2n + 4$ is prime, then any product of integers $2n + 4 = rs$ has either $r = 2n+4$ or $s = 2n+4$. We can verify that this can never be true here:
$$2n + 4 = 2 \implies 2n = -2 \implies n = -1$$ which is impossible because $n$ is positive.
Otherwise, $$2n + 4 = n + 2 \implies n = -2$$ which is again impossible because $n$ is positive.

Therefore $2n + 4 = 2(n+2)$ always violates the definition of prime, and thus $2n + 4$ can never be prime if $n$ is a positive integer.
\end{proof}

[vital dew]

Coolempire93

[grand ibex]

i'm the opposite, cs major and math minor

cs major is still good though cause my focus will likely be on graph algorithms

[spark field]

Indeed

[spark field]

https://tenor.com/view/emoji-sigh-sludge-demon-no-tears-no-crying-gif-10716016711387478941

[grand ibex]

aye it was for a good cause, took combinatorics instead

[keen flower]

by the time im done with cs, i will need 2 extra classes for math minor

[grand ibex]

i'll be honest

i would do a math major if it weren't for the language requirement for that department

[keen flower]

cs bs ms math?

hows that

[spark field]

Fair enough, you did succeed

[spark field]

crazy definition of combi

[grand ibex]

whether it's cs major + math minor or math major + cs minor though, pretty much the same for me

[grand ibex]

damn it was i wrong? 😭

shit i just spread mathematical blasphemy then

[grand ibex]

as in cs bachelors and masters in math?

[keen flower]

si

[grand ibex]

i'm doing something similar

except math phd instead of masters

or well

most likely math phd

depends on which department, not sure about all those specifics yet

[keen flower]

u tryna go into fintech or something

[grand ibex]

absolutely not

[keen flower]

lol

ahahah

then?

[grand ibex]

i'm trying to study graph theory and graph algorithms

[keen flower]

tell me more

what does that mean

[grand ibex]

well

i know too little to say much 😭

[spark field]

I mean that's like enumerative combi and a little analytic but graph theory is combi, combi has a bunch of extremal subfields, there's algebraic and geometric, etc.

[keen flower]

well you're curious about it

[spark field]

Not that the definition I usually give is particularly different

[keen flower]

if you havent had the chance contemplate

[grand ibex]

sort of

well

see it's hard for me to define concretely

[spark field]

Usually when people ask I say combi is 'the study of how many ways there are to do things and what the ways are in which you can do them'

[grand ibex]

but i know this is what i want to do

[spark field]

me wishing I had another year for chem minor and to strengthen my resume

[grand ibex]

damn, combinatorics and chem

interesting combination

chem is cool though

one of my favorites among all natural sciences

[spark field]

They do actually have quite a bit of interplay, although not necessarily that I would have gone into it if I could

[grand ibex]

permutations of atoms to discover new substances

[spark field]

Funny

[grand ibex]

(i'm speaking like an uneducated individual aren't i)

[spark field]

[keen flower]

lets do 4.3 together

holy cow i have no idea

[spark field]

What's your definition of rational

Btw I assume the theorems/exercises they are referring to you not being able to use are rationals closed under addition and multiplication, which we are basically going to prove the special case of here with r and s

[keen flower]

Proof: By the definition of rational where integers a and b such that r = a/b and b != 0

[spark field]

Oh I have a document that might be useful for you

if my google drive would load

I'm only 85% full it should still be working

[spark field]

Right, our definition of a rational number r is that it can be written as r = a/b, where a and b are integers and b != 0

[keen flower]

right

[spark field]

So begin there, rewrite r and s

[grand ibex]

construct 2r+3s in terms of r and s written as fractions

[keen flower]

Proof: By the definition of rational where integers a and b such that r = a/b and b != 0
So we can derive 2(a/b) + 3(c/d) by method of substitution

[spark field]

So now how can you rewrite $2\frac{a}{b}$

[vital dew]

Coolempire93

[grand ibex]

now you have to simplify that and state why the numerator and denominator are integers

[spark field]

Into something more useful

[grand ibex]

with denominator not equal to 0

also maybe restrict denominators to positive numbers

[keen flower]

2a/b and 2c/d

[spark field]

Good

And since integers are closed under multiplication, 2a is?

[keen flower]

integer

damm we have to include that as well

[spark field]

So $\frac{2a}{b}$ is an integer over an integer, with $b!=0$, so it must be a _ number

[grand ibex]

man this just reminded me of the triviality of what i did during the only discrete math homework i did 😭

[vital dew]

Coolempire93

[grand ibex]

was a pain in the ass

[keen flower]

rational

[spark field]

Good

See if you can do the same with the s term

[keen flower]

damm i feel insulted

[grand ibex]

no no i didn't mean that way

[keen flower]

https://tenor.com/view/dog-crying-meme-doggo-crys-megan-soo-crying-dog-gif-5276199764143986284

[grand ibex]

i don't mean to say it was easy

[spark field]

Actually you already rewrote it before

So continuing

[grand ibex]

i meant in terms of stating simple definitions like this

[spark field]

$\frac{2a}{b} + \frac{3c}{d} =$?

[grand ibex]

as simple as definition of an even number

[spark field]

(write this as a single fraction)

[vital dew]

Coolempire93

[keen flower]

Proof: By the definition of rational where integers a and b such that r = a/b and b != 0.
So we can derive 2(a/b) + 3(c/d) by method of substitution.
By using algebra, we can derive 2a/b and 2c/d where 2a and 2c are both integers closed under multiplication.

[spark field]

That's the idea, now add them (turn into a single fraction)

[keen flower]

Furthermore, 2a/b + 3c/d = 2ad/bd + 3bc/bd which is (2ad + 3bc) / bd

[spark field]

Good ✅

[keen flower]

whats next

we r so close

[spark field]

And because integers are closed under addition and multiplication, the numerator and denominator are

[grand ibex]

oh shit i'm slow, why did i say that we needed to restrict denominator to being positive

my bad

i was thinking of this other problem

here it doesn't matter

[keen flower]

integers

[spark field]

You really should be able to see it from here (our goal was to show this sum is rational)

So we are showing it satisfies the definition of rational

[spark field]

Right, so we have a fraction of integer/integer

What's the last condition

for our fraction to be a rational number

[keen flower]

that denometer cant be 0

[spark field]

Right

[grand ibex]

well we know that

[spark field]

So since b isn't 0 and d isn't 0, we can conclude?

[grand ibex]

oh right yeah probably should state it too

[keen flower]

that 2r+ 3s is rational

[spark field]

We can conclude that bd isn't 0

[keen flower]

ok im getting the hang of it

[grand ibex]

maybe step back a bit

[keen flower]

ohh

ahahhah

[spark field]

and thus 2r + 3s is rational, yes

[keen flower]

ok i will write it on paper and show you guys

[grand ibex]

by the way, is your name a reference to luffy

https://tenor.com/view/luffy-gordo-luffy-gordinho-gif-25125796

[keen flower]

yes, ive had this name since joy boy was first mentioned in the manga

ahaha

[grand ibex]

peak

luffy is so whimsical

https://tenor.com/view/luffy-one-piece-gif-23767041

oh nah there's a gif where he is in gear 5 and is taunting by smacking his ass 😭

[keen flower]

This is my WhatsApp pfp 😭😭

[grand ibex]

i think zoro needs a buff to catch up to gear 5

[keen flower]

No zoro's strength comes from his dedication and loyalty towards luffy and the crew. As long as he becomes the world's strongest swordsman he doesn't need to overcome luffys strength. Beside I dont think oda would allow captain being weaker than right hand

Zoro my favorite character btw

He'll surpass g5 but by then luffy will too

theres a beautiful analysis of zoro character on yt

@spark field hopefully u can read my handwriting

[grand ibex]

i don't mean surpassing

i just mean getting as strong as or nearly as strong as

a true right hand man

[keen flower]

ohh i see

[grand ibex]

my favorite among one piece characters as well

[keen flower]

nah nah hes my favorite ahaha

[grand ibex]

i'm not super into one piece as much as dragon ball though

[keen flower]

really?

[grand ibex]

my favorite among all of anime is vegeta

[keen flower]

whos ur favorite dbz char

ofc ofc mine too

ahaha

[grand ibex]

vegeta is too peak

after that i like goku best

[keen flower]

You may have invaded my mind and my body, but there's one thing a Saiyan always keeps... his pride!"

listen to that shit when u deadlift ez extra 100lbs

[spark field]

It's not unruly 😆

[grand ibex]

yes

[spark field]

Should be 3(c/d)

[keen flower]

lemme know how it is

[spark field]

not 3(b/c)

[keen flower]

where

[spark field]

Should be 2ab + 3cb (or 3bc) in the top of the fraction

[grand ibex]

gave me a good idea

i've been struggling to get to 4 plates on deadlift

i need to listen to more vegeta audio

[keen flower]

ohh got it

[grand ibex]

"impossible to defeat you? don't make me laugh" 🗣️

[keen flower]

i recently did 4 plates leg press and 3 plates squats

i dont do deadlift i need to

[winged delta]

y'all. Take it to a general board :p

[grand ibex]

damn, 3 plate squat is good

[grand ibex]

my bad

@keen flower there's a training channel by the way

[keen flower]

oopsie we got too excited

[grand ibex]

#1213611686565777439

[winged delta]

I'm not a mod, so like, no worries XD

[spark field]

And usually in the line concluding the proof you recall the original point, so you might say ``$2r + 3s = \frac{2ad + 3bc}{bd}$ is rational by the definition"

[vital dew]

Coolempire93

[keen flower]

everything else looks good?

cuz i need to head to bed

4.4 tmr and monday test

[spark field]

Yeah I would say very acceptable for a first class in proofs

[keen flower]

such a backhand compliment ahahha

but ill take it

ahahaha

[spark field]

Haha I don't want to outright say "👏 wonderful" as if there's nothing to improve but I also want you to know that you did good

Hopefully that struck a good balance 😆

[keen flower]

it did

thanks amigo

[spark field]

No problem 🙂 good luck!

[robust thunder]

We just started mathematical induction in class and it's extremely confusing. Any tips on getting better at these proofs?
Would you recommend looking at all sorts of proofs and writing them down to try and understand.. or any other method?

[spare slate]

from friend

[storm violet]

well what's confusing about it

is it the actual setup or is it the proof

the setup goes as follows:

1. you choose a base case (for example, n = 1) and prove the hypothesis is true for that case
2. you assume it's true for a generic case n (for weak induction. for strong induction you may assume it's true for cases 1, 2, ..., n in this example).
3. you show it's true for case n + 1

if you're struggling with something besides that it's likely the actual proofwriting process and not induction

or are you struggling on identifying if a proof should be done by induction?

[dull briar]

Hello

I am currently "opening a new page" to discrete mathematics!

I just finished highschool and i ultimately decided id want to self study compsci and discrete mathematics to supplement my compsci journey as well

I wonder if there would be anyone here who is also self studying this subject as well, if there are, please give me some advice :3

The book I would be flipping through is "Discrete Mathematics and It's Applications by Sussana Epp"

It would be helpful for anyone to give their opinion and their experience for my book of choice here

Self studying can be quite challenging, so any information about this subject can be helpful :3

[spark field]

I think @keen flower is using that book, maybe he can comment on his thoughts on it

[dull briar]

Oo okie

[spark field]

I would say in general there are things in discrete math that will be hard without a teacher, but you can always ask here so same difference

[dull briar]

Thank youu

This channel would be helpful for me :3

[keen flower]

Whats the question @dull briar

[keen flower]

Read it thoroughly, take notes and understand concepts before moving on. Do the quiz and tests on it

Do both odd n even numbers

Do not check answer before writing ur own

[dull briar]

Sorry for the late response, but ty so much

[simple solstice]

I want to clarify some new intuition for strong induction because I'm trying to really break into math/cs. I'm making good progress in my review of last semester material!!

For some statement $P(x)$ that requires inductive reasoning, we adopt this approach--

We prove an $i$ amount of cases and essentially show that $P(0) \wedge .. \wedge P(i)$ holds. Boom, base case done.

Our inductive hypothesis is essentially treating $P(i) \wedge P(k)$ as an axiom such that $k \ge i$ and $0 \le i \le k$. When actually working on the inductive step, we just use the statements we previously proven to our best capability.

[vital dew]

ginja - ギンジヤ

[spare slate]

Your hypothesis is $P(0) \land P(1) \land \cdots \land P(k)$.

[vital dew]

Civil Service Pigeon

[simple solstice]

Yeah that's what I meant by that, thanks.

[spare slate]

[keen flower]

prove the statement, the differebce between the squares of any two consecutive integers is odd

[winged delta]

Call them n^2 and (n+1)^2, can you go from there?

[keen flower]

that, i got

so we plug in n^2 and (n+1)^2 into the definition of odd

2k+1?

[winged delta]

Hang on, do the subtraction

[keen flower]

wdym

[winged delta]

(n+1)^2 - n^2

Expand the first one

[keen flower]

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

ohh my mind is all over the place

[winged delta]

Yeah, so when you do the subtraction, then you can use this definition

[keen flower]

how do i formally say the proof now?

[keen flower]

theoritical cs sounds fantastic!

[winged delta]

I think you can dispose of it in one line:

Pick an integer n, and compute (n+1)^2 - n^2 = n^2 + 2n + 1 - n^2 = 2n+1, which is easily seen to be an odd number.

Some profs might want a tiny bit more detail, like

Recall that the odd numbers are exactly those that can be written as 2k + 1 for integer k. Pick an integer... ... = 2n+1, so this difference is odd, as desired.

[keen flower]

@winged delta is there anything particular you're focsed on in theoritical cs? i'd like to do that too

[winged delta]

Not to get too off topic from the board's subject, but I'm in computability theory, which is vaguely an intersection between logic and CS. I don't really do pure CS research but I do teach a foundational course (that could honestly be under either umbrella) and algorithms (which is firmly CS!)

[clever rose]

can someone explain how to tackle this problem?

do i need to know ring and group theroy to compute this exercise=?

[little prism]

you need to know what modulo 7 is

otherwise this is just normal polynomial division

just that instead of e.g. 2/3 you compute 2*3^-1 where 3^-1 is the inverse of 3 mod 7

ok well you dont actually have to divide anywhere so you dont even have to think about that

[keen flower]

how do i get started on 4.5?

[keen flower]

thats sick!

hints arent really hints lol

i could factor out the a(a^2-1)

which is

a(a-1)(a+1)

[haughty garden]

you are taking the product of three consecutive terms so one of them must be divisible by 3 (why?)

[keen flower]

because the difference between the third and 1st consecutive number is 3?

perhaps?

[haughty garden]

difference is 2 but that’s not quite the reasoning

it might help to consider what the remainder of a could be when you divide it by 3

[keen flower]

thats a good question

i do not know

either 0 1 2?

mod 3?

[haughty garden]

Yup

Now consider each of these cases individually

[keen flower]

okay

[haughty garden]

If the remainder is zero, what does that imply

[keen flower]

that it is divisible by 3

[haughty garden]

yup

so what does that mean for a(a - 1)(a + 1)

[keen flower]

that it is being multiplied by a 3

?

one of the number has to be 3

[haughty garden]

well if a is divisible by 3, then the product must also be divisible

[keen flower]

right

[haughty garden]

now you can do something similar for the other two cases

[keen flower]

what do you mean by that?

i appreciate you being patient while guiding me 🙂

[haughty garden]

we only showed that if a is divisible by 3, then the product is divisible by 3

[keen flower]

i was sick for a couple of days and i feel like im behind a year

[haughty garden]

but a could have a remainder of 1 or 2

so you should try to argue that in these other cases, 3 is divisible by some other term

(and hence, the product is divisible by 3)

[keen flower]

im lost now can you explain it to me like im 15?

[keen flower]

i followed up till here i believe

[haughty garden]

so we know that a itself could be 0, 1 or 2 (mod 3)

[keen flower]

right

[haughty garden]

we don’t know which one a is right now but if we can show that, in any of these cases, the result we want to prove is true, then we would be done with the proof

so far, we only argued what happens when a is 0 (mod 3)

[keen flower]

okay

im following

[haughty garden]

so now we need to see what happens when a is 1 (mod 3)

[keen flower]

ohh okay

and then when a is 2 (mod 3)

[haughty garden]

Yup

[keen flower]

but if we already know that one of the number is divisible by 3, the product of 3 consecutive number is also divisible by 3 no?

or am i stating the obvious or am i being stupid

[haughty garden]

this is correct

[keen flower]

ohh ok

[haughty garden]

so we need to show that one of the terms is divisible by 3 in the case where a is 1 (mod 3) and 2 (mod 3)

[keen flower]

ohh okay

im following

[haughty garden]

so if a is 1 (mod 3), then which term is divisible by 3

[keen flower]

a-1?

[haughty garden]

why do you say that

[keen flower]

intuition

lol

[haughty garden]

can you write a small line of working

[keen flower]

because the reminder is 1 and we need that reminde gone

[haughty garden]

if you know that a is 1 (mod 3), then write a = 3k +1 for some integer k

so now what can you say about a - 1

[keen flower]

3k+2?

[haughty garden]

try again

[keen flower]

3k

[haughty garden]

yup

[keen flower]

wait how

a = 3k+1
a-1 = 3k+1
move 1 to the right side, 3k+2

[haughty garden]

a = 3k + 1 so a - 1 = (3k + 1) - 1 = 3k

[keen flower]

im doing something wrong fundamentally

[haughty garden]

you just have to substitute a for 3k + 1

[keen flower]

ohhh

i seeeeee

[haughty garden]

I think you’re trying to solve for a but that’s not what we are trying to do

[keen flower]

makes sense

ok everything so far is clear

[haughty garden]

ok so we established that if a is 1 (mod 3), then a - 1 is divisible by 3 which means the product is divisible by 3

now can you do something similar for the case where a is 2 (mod 3)

[keen flower]

how so?

3k+1 + 2

3k+3 = 3(k+1)

[haughty garden]

idk what you’re doing here

[keen flower]

ok i apologiz

[haughty garden]

try to write out your reasoning

if you know that a is 2 (mod 3), then you can write a explicitly

[keen flower]

1 minute please, brb

[keen flower]

how so?

[haughty garden]

do something similar to this^

[keen flower]

3k+2

[haughty garden]

what is 3k + 2

[keen flower]

2 (mod 3)

[haughty garden]

uh huh so which term is 3k + 2

[keen flower]

3rd?

[haughty garden]

ok so let’s take a step back

[keen flower]

sure

[haughty garden]

we are in the case that a is 2 (mod 3)

If a = 2 (mod 3), then what does this mean

[keen flower]

when a is divided by 3, reminder is 2

[haughty garden]

yes

so you can write a as a = 3m + 2 for some integer m

[keen flower]

isnt that what this means?

ohhhhh

right right

we need to switch the variables

[haughty garden]

Better to do so because you’ve already defined k in the previous setting

[keen flower]

ight right

[haughty garden]

but the idea is that you need to explicitly state that a is 3… + 2

[keen flower]

got it

[haughty garden]

once you know this, what term is divisible by 3 and why

[keen flower]

give me a second im thinking

whichever has a reminder of 0

so a?

ugh i dont think this is right

[haughty garden]

we said that a has remainder 2

hint: what if you just add 1 to a

[keen flower]

a+1

but

[haughty garden]

what does a + 1 look like

replace the a with 3m + 2 and see what happens

[keen flower]

a + 1= 3m+3

i feel my brain going numb lol

[haughty garden]

is 3m + 3 divisible by 3

[keen flower]

yes

3(m+1)

[haughty garden]

yup

so what does that mean for the product

[keen flower]

it is divisible by 3

[haughty garden]

yup

so now let’s tie everything together

[keen flower]

i got confused here cuz i thought we said originally a is 0 mod 3 which means it is divibislbe by 3

[haughty garden]

That’s only one of the cases

we still needed to worry about what happens if a is 1 (mod 3) since we don’t know that a is 0 (mod 3)

[keen flower]

ohh i see

ok lets put it all together

[haughty garden]

So we showed that if a is 0 (mod 3), then a is divisible by 3 and hence, the product is divisible by 3

If a is 1 (mod 3), then a - 1 is divisible by 3 and hence, the product is divisible by 3

If a is 2 (mod 3), then a + 1 is divisible by 3 and hence, the product is divisible by 3

And we know one of these cases must happen so a(a - 1)(a + 1) is divisible by 3

[keen flower]

ugh idk how to formally write this down

[haughty garden]

the way we did it before is fine as a formal proof

Suppose that a = 3k + 1. Then a - 1 = (3k + 1) - 1 = 3k so a - 1 is divisible by 3

Do something similar for the 2 (mod 3) setting

[keen flower]

im gonna write it down and send it here

can you let me know how to revise it please

[haughty garden]

sure

[keen flower]

we went straight to mod 1, what about mod 0?

[haughty garden]

0 (mod 3) is already handled

If a = 0 (mod 3), then a is divisible by 3

You can write an extra line of working if you want

[keen flower]

right

okay

[keen flower]

lemme know how this is

i think its missing sauce that he normally has in his proofs

the professor

[winged delta]

Looks correct to me

The professor method would be something like

a(a^2 - 1) = a(a+1)(a-1), which are three adjacent numbers. No multiples of 3 are 4 apart, so any three adjacent integers contains a multiple of 3, and so their product is divisible by 3.

(Source: am professor)

[keen flower]

Do you recommend i take logic class after this? It's a philosophy class tho

[winged delta]

Ew, logic in the philosophy department

Not inherently evil, but often

I would have to see the syllabus

[keen flower]

There's a class called advanced math smth

Lemme see

[winged delta]

The sorts of things you'll see in discrete are often the same things you'd see in "intro logic" - proof techniques, truth tables, quantifiers. Certainly my discrete class has that overlap (partly because I use Levin, but partly because I am a logician). You have to pick up those things somewhere along the way as a math undergrad.

More advanced logic would be things like model theory/completeness/compactness, which of course I adore but are not always the most relevant to the working mathematician

Yes, take 307, those are the more intermediate/advanced notions that get left out of Discrete

[spark field]

That's an almost verbatim description of our Foundations of Advanced Mathematics class 😆 how funny

[keen flower]

You guys are very inspiring

Im beginning to love math again because of yah

[spark field]

weird that we have calc 2 as a prereq for this

[winged delta]

I am currently deleting Calc II as a prereq for Calc III, it shouldn't be a prereq for anything

[spark field]

Is this a /j or

[winged delta]

But the idea is that having both I and II means you have a higher level of "mathematical maturity"

[spark field]

Oh you mean leaving calc I as a prereq still

Okay yeah that's agreeable

[keen flower]

Whats calc 3 like

I heard calc 2 is just trig

[winged delta]

Genuinely not kidding, I use absolutely none of the integration techniques from II when I teach III - literally just e^x and 1/x. The problems are hard enough without requiring students to dust off all their complicated integration formulae! I want them to focus on the multivariable setting

[spark field]

Yeah

Typical

The only thing from calc II that came up was trig integration
But even then no techniques

[keen flower]

David, can I add you?

[spark field]

given cylindrical and spherical

[winged delta]

Sure :)

[keen flower]

Im struggling with trig rn ahahah

[spark field]

I just realized David was a name

[keen flower]

Its not letting me add you

[winged delta]

Fixed

[limber briar]

is the disquisitiones worth reading for a first introduction to some more advanced ENT topics like quadratic reciprocity

[winter briar]

What is discrete math even about bruh

[weary tiger]

Imagine math but it doesn’t have a progression

Kinda like logic? I think

I’m still confused too

[spark field]

See channel description

There's a lot of 'math' that comes under a typical discrete math class

[willow owl]

It’s discontinuous mathematical structures like sets of numbers between continuity

[spark field]

Discontinuous thats the word I was looking for

[frail sage]

My favorite math course so far :)) graph theory, combinatorics, logic, parts of basic group theory? (Although maybe the last one was only added as prep for actual group theory to our course)

[grand crown]

we learned some group and field theory in my discrete class too

[spark field]

in our discrete class for computer science majors we do things like strings and algorithms

In our discrete for math majors it's start of algebra

[robust monolith]

Discretion is advised

Mine's was very diluted

[frail sage]

yeahh im doing both math and cs and the cs discrete for us is literally the very basics of graphs and some graph algorithms + how to specifically prove combinatorial identities by induction

[spark field]

Interesting

Our cs discrete is still quite mathematical

[frail sage]

yeah unfortunately the cs department at my uni wants to not scare off too many students by making the bachelors "too mathy" which is why the only math courses that there are are very easy/shallow (yet most students still manage to find a way to complain)

[spark field]

real

Ours is the opposite, we have too many cs students so they continue making the curriculum stronger

We're slowly becoming a 'good' school for cs

[frail sage]

ahh yeah no for us theres 550 spots and they cant even manage to fill half of that 💀

last year people got in who didnt even attend the entrance exam since they just wanted people to come

[spark field]

We have ~120 CS students but the whole university is ~2500

So 120 is already too many

[frail sage]

oh thats the size of my cs course right now i think

what about math?

[spark field]

math, maybe 40 total

All classes <= 8 students

[spark field]

My largest was 36 iirc

[frail sage]

idk about other places but in my (pretty big) uni we started with 60 at the beginning of y1 (september) but by now more than half already dropped out

[spark field]

the professors really don't like it since there are only 4 of them

[spark field]

damn

[frail sage]

thats a smallll team

[frail sage]

wait sorry so like when you go to your LA class there's only 8 people and the prof or?

[spark field]

haha yeah

for math only 12 professors (since they have to serve the whole uni)

but pure math classes are taught by 2 mainly

[spark field]

LA is a class that a lot of majors have to take, so no

But LA2 there were 6 in mine

[frail sage]

ahh true true LA is like everyone

so something like group theory would be a math-only class right

[spark field]

Yeah

abstract algebra had 7

[frail sage]

damn thats cozy

[spark field]

Definitely a certain type of vibe

[frail sage]

how many people end up graduating?

[spark field]

In math? everyone

Everyone here for math is on scholarship

And they call a meeting every year to bring any new math majors into a scholarship

Once the students sign the contract they get serious ig

[frail sage]

oh woww okay

[spark field]

I've seen one student transfer schools, but no one drop out

[frail sage]

thats an interesting concept

[spark field]

In CS well my cohort started with like 40 but I think maybe ~16 will graduate now, at least 10 are still retaking courses, idk about the rest

[sand thorn]

I'm kind of not sure where to put this cause it's an algorithm course but the algorithm we're doing is essentially discrete math

[spark field]

Here is fine

[sand thorn]

Basically I have an algorithm for finding the minimum "longest distance" point relative to a point you select on the y axis (which is optimal) amongst a set of point

idk if that made sense

I had to derive it geometrically

because super strict plagiarism policy

and anti-ai

There's 1 step i am not sure

in terms of mathematical rigor

So basically my algorithm goes as follows:

Take 2 points. Look for each if their direct distance (x-axis) with the y-axis is bigger than the hypothenuse of the rectangular triangle formed with the other point.

If it is bigger, then you just pick that point's y coordinate as your optimal solution in this.

Otherwise, you pick the point on the y axis such that it forms an isoceles triangle with your 2 other points.

Now, when you add another point...

[spark field]

@wicked bolt You may like

[sand thorn]

You repeat the same thing. You look at whether it surpasses the "threshold" by which the hypothenuse is bigger than the direct (x-axis) distance to the y-axis

[wicked bolt]

hello

[sand thorn]

If it does not surpass, then you don't care about that third point

If it does surpass, then this is the step I am not sure about, but I am guessing the optimal "new solution" is to compute the midpoint between the first 2 points and you use that for computing the new isoceles triangle with the y axis?

And then you repeat for "n" points

let me send pictures

Setup 1 where point 2 is too small to be considered so p1 wins

[wicked bolt]

that does look like a slayla problem but i will have to eat my breakfast first and also get some clarifications on the problem statement

[sand thorn]

Scenario 2 where the hypothenuse “surpasses” the direct distance so you take the isoceles distance which is optimal for choosing your ideal point on y axis

Scenario where you add a third point that surpasses them in terms of hypothenuse (and they surpass it in terms of their hypothenuse with respect to P3’s direct distance)

So here you take the new isoceles triangle for the optimal point using a “midpoint”

Excuse my complete lack of math vocabulary and rigor, I work with drawings more than anything

[sand thorn]

Very simply: find the optimal “y axis point” such that out of all your n points you get the optimal “farthest point distance” from that y axis point.

[wicked bolt]

we are given a collection of n points. find y such that the point (0,y) is optimal for… what exactly?

[sand thorn]

So every point has a distance with that optimal (0, y) point right?

[wicked bolt]

yep

[sand thorn]

You want that the point which has the biggest distance is also the minimum it can ever be

and that's the criteria you use

to find "y"

[wicked bolt]

ok i understand i think

[sand thorn]

Do you understand the procedure for 2 points which I explained above, I can reexplain maybe

[wicked bolt]

if we name our points $a_1, ..., a_n$, we want the $y$ that minimizes $$\max_{i}(\operatorname{dist}(a_i,(0,y))$$

[sand thorn]

exactly

[vital dew]

slayla

[sand thorn]

which for 2 points, the minimum distance is either directly the x-axis distance of one point with y

[wicked bolt]

i will go through your writing now

[sand thorn]

or it is the distance when you form an isoceles triangle

[sand thorn]

the drawings are easier to understand

[sand thorn]

wait what if it's not the midway point but it's the isoceles triangle with respect to P2 and P3?

And you pick p2 because it is farther from P3 than P1

Cause with the midway point, P2 is actually a bigger distance from the y-axis point than all of them, so it's not optimal. But if we set P3 and P2 to be isoceles, because they have same distance and P1 is between them then they have the biggest distance which is consistent with what we want to do.

And I don't think P1 can ever change that. If it does, then its x-axis distance would be optimal in the first place compared to P2, so you would never reach that scenario when you add P3.

Updated model

[wicked bolt]

can i just check something with you. is the idea of the algorithm like.. we start with 2 points, find the optimal y. then add another point in and find the new optimal y for these 3 points. and so on

[sand thorn]

exactly

which is only O(n)

we also get graded on time complexity

if it's too long I get 0 marks

[sand thorn]

I am using a divide and conquer algorithm. Technically if I use recursion which I think you can use it here, it might be O(log(n))

[sand thorn]

@wicked bolt I tested it and it works

[sand thorn]

no it actually doesnt always work there are edge cases and it has to do with selecting the correct third point unfortunately, which can make this O(n²) if we have to loop through every point in each iteration

[spark field]

hmmm I'm looking back at the problem now

Wait I thought you wanted optimal y-value why are you doing x-values

I wonder how well gradient descent performs here and what the time complexity would be

Probably not well I think there are local minima

this is a nasty computational problem xd

I can think of ways to improve the average case but worst case is still bad >.<

I think you can cut it down to a problem (or multiple problems) with just two points in log of something

with some arguments about convexity

I just can't do the geometry

[sand thorn]

yeah it ended up being x i misread the assignment

[sand thorn]

i think it works still if i sort the input set at the start

in ascending order of y coordinates

[sand thorn]

our prof gave us 3 problems like this for an undergrad level course and we're supposed to be able to complete it in a reasonable time

expecting us not to use AI

the real solution is ternary search, but I really wanted to derive the solution on my own

[boreal crescent]

can anyone help me with rules of inferences when applied to nested quantifiers?

how do I solve these?

[amber nexus]

Okay, so:

IF i use the whirlpool THEN i am sore

What can we infer from NOT sore?

[keen flower]

On a Monday a friend says he will meet you again in 30 days. What day of the week will that be?

30 = dq+r

30 = 4 weeks x 7 days = 28 + 2

2 reminder

so tuesday and wednesday ( monday + 2 days) = wednesday

did i get it

yeah i think i got it

can someone help me do 19

[keen flower]

for even 2k = 5(2k)+7(2j)?

so 10k+14j

2(5k+7j) even by definition

do the same for odd?

for odd 2k+1 so 5(2k+1)+7(2j+1)

i just had a problem with the word parity

i didnt undertand what they meant by m and n have same parity

[keen flower]

(10k+5)+(14j+7)

10k+14j+12

2(5k+7j+6)

even by definition

i have a hard time writing formal proof, how would i state this in an exam?

the result i got

( if its correct)

[keen flower]

What does parity mean

Sorry for the late reply im in bed

Ohh I see

[keen flower]

just ask

[fringe bluff]

Starting with a graph that has no cut vertex, if we perform DFS at any given vertex is the resulting tree a path? or is it only true that the root vertex has degree one?

[boreal crescent]

if I didn't used the whirlpool then I'll not be sore?

[amber nexus]

yep! That's the contrapositive
Now,
IF i play hockey THEN I am sore

I am not sore

What does that tell us?

You'll probably have to write out the line of thought using the formal syllogism name (I think modus tollens?) but that's more or less how you'd approach these, working backward. I find it useful to turn them into simple statements but you can do whatever you like, so long as it works for you

[hidden totem]

does anyone have an intuition for why dedekind cuts of the rationals are uncountable? i understand that they describe the reals and the reals can be proven to be uncountable, but im looking for direct connection without invoking the reals, only really using the fact that Q is a dense linear order

[wicked bolt]

this isn't what it says

🎶's statement is the contrapositive of what was written in the image, but not the contrapositive of what you wrote

[spark field]

How does one define the size of a dedekind cut

[grand crown]

i think a binary tree sort of idea would work:

for simplicity imagine we're just looking at two fixed points in the order a and b. from density you can always find an element between any two given elements, so you can create a tree that branches to narrower and narrower subintervals in [a, b], and you can think of each infinite path down the tree as a threshold for some dedekind cut. each path will correspond to a unique cut, and there are 2^N such paths

[hidden totem]

oooh thanks, this is really good

i really like this explanation

[dull briar]

hello

i would like to ask how this would make sense

the question ask to use set roster notation to indicate the elements in each following sets

from my understanding, d would not have any elements

probably thre must be some kind of switch up

but if that would be the case, how does c. U has every integer in the set?

the set would be U = {...-2, 2...}

if there would be a logical answer for this would be much appreciated

but this could be some sort of editor error

[cloud rampart]

it seems correct to me?

[dull briar]

really?

set U would not have integer -1, 0, 1 right?

[cloud rampart]

set U from c) is empty

like the answer key says

[dull briar]

oh

[dull briar]

i still do not understand

sorry

[cloud rampart]

is it possible for an integer to be both less than -2 and greater than 2?

[dull briar]

hm good point loll

yeah i get it now

i misundrstood the notation

tyy

[keen flower]

Hello hello

[boreal crescent]

Could you elaborate?

[boreal crescent]

Like in symbolic form?

[lunar cave]

Is taking an 8 week discrete math class this summer fully online a bad idea?

I've been programming for like 4 solid years and I know most of the basic DS&A's but 8 weeks feels intense

I would also be taking 2 somewhat easy classes along side it...

Would appreciate any insight

My school estimates it should take up 11-16 hours/week and that's with no prior exposure to the material so is it reasonable to assume the lower end of that time with my experience?

[spark field]

I would say no for a typical discrete class

overlap with DSA comes later in the course if ever

The beginning is typically focused on topics that have little to no overlap

Although if you're familiar with boolean algebra the first part shouldn't be particularly difficult

Depending on how the class is structured the proofs section would be all new and that's usually what people have the most difficulty with

Regardless as to whether it's a bad idea idk

I personally would be fine with it but I don't know if I would recommend it

[lunar cave]

Yeah no proof experience so that could be tough... The class follows the Susanna Epp book which I hear is really good for self teaching so that's a plus. I had a good experience in the past self teaching myself algebra and pre calc by reading/working a textbook which is the only reason I'm even considering this. I know I can sit down and self teach for long periods but are you saying the proof concepts might require a lot more than 11-16 hours per week to click?

[spark field]

hmmm

That depends a lot on how they present the content

for me if I spent 11 hours on a subject per week I would greatly outpace the class

I mean if you have time before it starts you could just start reading the book early

and that would hopefully help deal

[lunar cave]

Yeah I've never even done a fully online class so not sure what to expect in terms of presentation

Ideally they let you just read the book on your own and then do the homework but idk if cengage has some sort of interactive textbook

that would be a nightmare

Something makes me think they would hand hold you through the reading just to know you did it

I shouldn't have put off reading the Velleman proof book😭

[spark field]

tbf our DM book was interactive as well

but doing exercises online is just like doing homework

problems are problems

[lunar cave]

I mean I'm fine with that if it's a 1:1 with the example exercises in the book

Like is it just the same examples as the section just greyed out with an input field?

I'm used to just covering the example with a sticky note on a hard copy so that would be the same thing...

[spark field]

I don't remember how our cengage book was

I think it was almost like a separate thingy

[lunar cave]

😭

If you work through the whole section of the hard copy you should be able to blow through that right?

[spark field]

In general no

If the exercises are good

But that's the case for any math book

Exercises will be challenging and require you to develop ideas

Unlike for example a physics class

you apply principles you learned or plug things into equations

That is not the case for proofs

But once you build the skills/familiarity then yes

It's mostly an art of gaining intuition behind definitions, and knowing when they are used and how

[lunar cave]

Most math homework pulls stuff from the end of each section and then for the case of an interactive lesson I would assume most of the problems would be equal to like the section examples right?

[spark field]

That's the difference between what you are used to as math

[spark field]

e.g.

compared to proof-based math

Which is a different form of math

For example

Actually rather than make an example I will take one from a textbook I have on hand

1. Use a direct proof to show that the sum of two odd integers is even.
   This is the first exercise in the first section where proofs are introduced

At this point you would have this definition and these two examples

So it's not a case of 'same question, different numbers'

It's a different question from anything that you've been shown

Your work is now to adapt what you see in example 1 to fit the scenario from exercise 1

in this case it's fairly easy

It (ideally) gets more complicated as you go

But since it's an online class maybe they will focus less on this route and maybe they'll do a simpler route like ours did "fill in the missing steps of the proof"

Which typically will match the examples from the book

since it's just click and drag in order

Ours we had to write by hand on tests only

Nice but doesn't really develop the student imo

So you'll probably get more of that

Now that I think more clearly about it

Regardless what I can say for certain is that this

the end of each section
will not be true

In general

Since it's a breadth course it's not like you're working towards a final goal and each chapter works towards a final thing for you to apply

[spark field]

You often hearken back to early definitions like this

definition 1 more like thing #1 to record

[lunar cave]

Is that example from the epp book?

If so what section?

[spark field]

No this is an example from Rosen Discrete Math and its Applications, I just happened to have this book on hand

[lunar cave]

I see in the epp book she has "Test yourself" at the end of each section which is more of what I was expecting lol

[spark field]

I've seen some students working out of her book and it seemed fairly typical

So I would be surprised if exercises were solely limited to content from the end of sections

Although they do typically appear as a list at the end of each section

Which is normal

[lunar cave]

Yeah I mean that example you sent is intimidating in a vacuum but if it builds up to that naturally I would hope it would be doable when it's finally put infront of you

[spark field]

I wouldn't say it builds up naturally but you will understand what's being asked of you

Even if you don't know how to actually execut

Which is usually the problem 😆

All the tools are there but the route to the end is foggy

[lunar cave]

Yeah I've seen videos of people doing simple proofs and it reminds me of the early days of programming being super lost but following just enough to keep going lol

Do you think programming gives you an edge in proofs? The raw logic of it all?

[spark field]

hm

[spark field]

I guess to expound more on this the first lesson in discrete math is a formal introduction to logic

So depending on how adept you are at programming, you will have an edge

Proofs are then built upon formal logic (specifically, they must have a valid logical structure)

for example:

if p:
  q = True

p = True

We can determine that if we run both blocks that q is True

But you can also tell by that example that it's a little bit different

[spark field]

Regardless yes, this

[boreal crescent]

can anyone suggest a good YouTube channel/textbook that can help me understand Discrete Mathematics better? I'm really bad at it and I think I'm gonna fail my exams 🫠

[lunar cave]

Yeah a lot of the stuff in the epp book looks like just the mathmatical version of stuff I've seen in programming

Besides the proof

But even the proofs reminds me of attacking a problem in programming

I don't really know what I should do😭 You think I should just follow through and do the 8 week class? @spark field

I need someone to either push me off the edge or tell me not to jump💀

[nocturne yew]

Er... given the meaning that that last sentence usually has, I'm going to opt for saying "Don't jump"

[lunar cave]

Jeez that is grim... I was thinking of a metaphorical plank lol

[dire pine]

Discrete math helps a lot in programming

Also if you're interested in the formal study of algorithms

[lunar cave]

Yeah I see a ton of value in discrete math for CS which is why I'm worried an 8 week class is too short to fully soak it up

I mean it's a 3 credit class over 8 weeks which should only take 11-16 hours per week which I can handle

[dire pine]

That's pretty standard

[lunar cave]

You took it over a similar time scale?

[dire pine]

Mine was during a 3.5 month semester

but it was 4 hours a week

[keen flower]

8 weeks course vs 3.5 months is usually the same 3k minutes

U just meet more often

Usually 75 min 5 days a week

[dire pine]

That much of anything would be the end of me

[keen flower]

I'd suggest you go over the book cuz u still have time and then take it

[lunar cave]

I just checked and last summer I took Calc 1 (4cr over 12 weeks) which projects the same level of work (11-16hr/wk) and that wasn't bad at all

I did go in person for that

[lunar cave]

I'm taking DE rn so don't have much time for more math atm

[keen flower]

Ohh damm

[lunar cave]

Honestly just the fact that the time/week of a 3 credit 8 week class is = time/week of a 4 credit 12 week makes me feel a lot better

Appreciate the input everyone I think I'm going to see it through this summer

[lunar cave]

Back when I self taught pre calc I would do like 40 hour weeks of nothing but pre calc but that was unhealthy lol

[dire pine]

Well if the pace suited you, it was probably good

[lunar cave]

It wasn't so much pace but rather spending way too much time on exercises lol

Basically going overkill to get over my math anxiety

[keen flower]

would #6 just be K?

number 7, k+1?

[dense sable]

yes its the smallest integer greater than or equal to k

[keen flower]

but it asks why

[dense sable]

like you need to define the ceiling function?

[keen flower]

yeah i think so

n-1 < x <= n

i dont understand it ngl

[dense sable]

yeah im not sure what it wants you to say since thats just how the ceiling function is defined, its not like theres anything to prove

[keen flower]

ok i think i got it

im confused with number 8 now lol

we'd use floor because celing would mean a completed unit

right?

[dense sable]

yes thats right

[keen flower]

if you have enough material for 5.8 units, you can only do 5

nice

[dense sable]

so you could write it has floor of n/7

[keen flower]

right

what about 9?

the opposite i asssume

[dense sable]

similar logic yeah

[keen flower]

noicee

[keen flower]

can someone help me with 4.7

[spark field]

Did you work with the problem a bit