discrete-math

[hollow bloom]

Have hard time wrapping my head around

[sleek swallow]

You have to just learn it and struggle a bit

[hollow bloom]

Gotcha

[sleek swallow]

Ask for help if you don't get it after a long period. But don't go around willfully memorizing these things

[hollow bloom]

Alright

Lol I remember

After class

Me and 2 friend were standing outside

Talking about it

Hold on

Lemme find what it wass

Ahhh its not important never mind

[sleek swallow]

Yo mamma

Prank

[hollow bloom]

lmao

Wait

I think

A sufficient condition for compoundXto boil is that its temperature be at least 150◦C.

Is it because the temperature

Is the q

A sufficient condition FOR compound to boil

okay

I understand now

q is the sufficient condition that it needs for p

please tell me I'm right

[sleek swallow]

'Is it because the temperature is the q' is just incorrect wording

Fix your wording, approach it logically

You're doing logic, so it makes sense that you should approach it logically

[hollow bloom]

Okay I got it

Basing on the list that you gave me regarding to the equivalence of p implies q, p is a sufficient condition for q. First the compound x to boil is p. Then temperature over 150 will be the q, in the context of the problem "A sufficient condition for compoundXto boil is that its temperature be at least 150◦C"
Then the temperature over 150,q, is the sufficient condition for p. Which means that this problem translate into q implies p. However, like you said in the list, p implies q is same as p is a sufficient condition for q, and not q is a sufficient condition for p.

Then borrowing your list of special conditionals, q implies p is the converse of p implies q which is false.

I finally understand

Why I messed them up

[hollow bloom]

Thank you so much! @sleek swallow , tomorrow Imma copy those lecture notes down

[sleek swallow]

Copy deez nuts down

[weary tiger]

Hey guys, does anyone have any good recommendations for a discrete math textbook? I'm looking for something that's a lot easier to parse than Knuth's "Concrete Mathematics."

[sleek swallow]

Logic & Discrete Mathematics by Goranko & Conradie

That book focuses quite a bit of logic but it leads well into other topics

[weary tiger]

Thanks I've just saved it on Amazon. Have you by any chance ever read "Book of Proof" or any of Richard Hammack's books? I'm reading that one right now. It seems like a very gentle bridge between the calculus series and other university level math. I'm pretty sure that was his goal when writing it.

[sleek swallow]

I tried reading it and it didn't work for me

[weary tiger]

Oh, did you think it was too simple? Like the pace was too slow?

[sleek swallow]

I didn't like the hand-waviness? I read Bernd Schroder's Fundamentals of Mathematics and I found that that was fantastic

It starts with Logic, then Axiomatic Set Theory (though not too deep cos it's one of only two chapters), then the construction of the naturals, integers, rationals and reals, before proving the unsolvability of quintic polynomials via radicals and some more set theory (haven't gotten here yet, i've putting it off till after i leave the army)

[weary tiger]

Oh, I'll check that one out too. Yeah, I'm looking for a book to tackle after I'm done with "Book of Proof." Not sure where to go next. "Logic & Discrete Mathematics" might be a good one.

I was also looking at "Introduction to Graph Theory" by Trudeau, and "How to Solve It" by Polya (I've been putting off reading this one for a long time).

[sleek swallow]

Not sure about Trudeau's book cos i've not used it. How to Solve It by Polya is really just a book that shows you various ways to approach math as a whole

It's nice for insights but doesn't give you the requisite logical/set theoretic language to approach the rest of math

[stray reef]

are the couples all meant to be man+woman couples

why didn't it say that explicitly >::|

[weary tiger]

@sleek swallow I feel like I need that really badly. I think I'm stuck in the habit of feeling confident because I can do the problems that come up in the Calculus series. But those problems are all basically pattern recognition then algebra. Whenever I see a problem that doesn't have a structure I've been walked through already I start struggling.

@iron crescent I know what you mean. Even some of the basic set problems in "Book of Proof" were unintuitive for me. I was pretty disappointed in myself.

[sleek swallow]

Try brilliant.org

It's pretty good

Polya's book is great but just remember, it's not gonna give you all your requisite set-theoretic/logical language for further use

Find the probability that neither of them are selected.

[lyric pumice]

@weary tiger Are you planning to major in math?

[sleek swallow]

1 from the men and 1 from the women?

Give a more detailed explanation

[weary tiger]

@iron crescent This is actually the same thing I was just thinking of when I said some of the problems were "unintuitive." So for your problem, I believe you'd find the probability that neither are selected, then subtract that from the "universal" set. (I think)

@lyric pumice No, I'm not. I just like Mathematics. I'm a computer science person, though.

[sleek swallow]

Okay see

[lyric pumice]

@weary tiger Are you planning to go to grad school for computer science?

[sleek swallow]

That really doesn't make much sense. 6c1 means you're choosing 1 individual from 6 people and there are 6 ways to do that

Whereas you really wanna just choose joanna

Now, you're also not accounting for the possibility that joanna is paired with someone else

Or that douglas is paired with someone else

[weary tiger]

@lyric pumice No, I'm not planning on ever going to grad school.

[sleek swallow]

Yea. Also, when you do the calculation like that, you're implicitly saying that one of them has to be a man and one of them has to be a man. That's not a requirement at all

[lyric pumice]

@weary tiger Then you probably won't be very concerned about pure math.

[weary tiger]

@lyric pumice Oh... I'm very concerned with pure math. lol

[lyric pumice]

@weary tiger Do you find it to be very useful in your other work?

[sleek swallow]

I think what they're concerned with is just making sure they learn math without half-assing it

I mean, Charles University forces its comp sci students to take quite a lot of classes in pure math

So maybe you want to understand what you're doing with math, as opposed to just using it for comp sci?

@weary tiger

[lyric pumice]

@sleek swallow Why, what for?

[sleek swallow]

Oh I'm not quite sure why but I think the university focuses quite a bit on getting you into academia

But that basically has allowed quite a few of their comp science students to move on and do masters degrees in math

[weary tiger]

I think @sleek swallow is correct. I don't know where I would use some of this stuff, or if I ever would. But you never know. I feel like I'm not getting enough out of the math side of things. And there are some areas that I'm just interested in for some reason (can't explain why). Stuff like linear algebra, functional analysis, etc. I'm really interested in how we can use the "abstraction" of R^n to "describe" more "abstract" vector spaces, etc. That stuff is really "cool" to me.

[faint narwhal]

(you should just do math instead)

[sleek swallow]

It doesn't really matter

[weary tiger]

@faint narwhal How dare you?

[lyric pumice]

In most employable contexts of computing, you will not need such math.

[sleek swallow]

If your degree allows you to do many math courses, it's just whatever lol

Just do the math courses you like

[lyric pumice]

You should not need any more than basic applied math.

In more theoretical contexts of computing, in which there are relatively very few positions, you should be comfortable with basic discrete math.

[sleek swallow]

Just study the day before the exam

[lyric pumice]

@weary tiger Then you might have some potential.

[weary tiger]

@iron crescent The best way I've found is basically to put in the work throughout the semester. Do homework every day, visit office hours, youtube videos, etc. Not very comforting, especially if you have an exam this week, I know.

[lyric pumice]

I believe that deep curiosity is indicative of underlying ability.

Testing is not the same as exploration.

[weary tiger]

@lyric pumice Thanks, I've had a few people try to steer me towards math. I'm very interested in lots of different areas, but once I get to a certain point, then I become less so. I don't think I've hit that point with math yet, though. But I have a strong feeling I would hit it long before I could complete a degree. lol

[sleek swallow]

Then just don't feel the pressure

[weary tiger]

@iron crescent Oh, like generalized anxiety? I have that, too. Usually when I'm headed to school. As soon as I'm in class it goes away. You could always talk to a therapist or psychiatrist! (Helped me).

[sleek swallow]

Rupert, just literally do what you like for math

[weary tiger]

@iron crescent I think a degree of anxiety is normal. There is a lot of "pressure" to perform. But if you feel like you're "sick", then it might be more serious. Or if you feel like you're in a state of distress or something. Otherwise I'd say it's probably very common. @sleek swallow Thanks for the recommendations!

[sleek swallow]

Lmao

Literally the worst kind of pedantry

Your handwriting is pretty crappy, btw

But the 'if' has a very specific meaning

Typically, 'If....then' is substituted for implications

That's probably why they took away points

Also, you should've just written "It's a shoe brand."

[weary tiger]

Wait, would "If ... then" imply causality, whereas without it, it'd just be "correlation"? (Is this a dumb question?)

Or are they both just correlation.

[sleek swallow]

Logic has nothing to do with causality

Statements p & q don't have to be related to each other at all. We're only concerned with their truth values and the truth value of the resulting proposition

[lyric pumice]

I would disagree. Logic and causality are related.

[weary tiger]

Oh shit... didn't mean to start something here. Anyways, thanks for the advice. I gotta go!

[sleek swallow]

How are they related though?

I mean, okay yea, you ideally want the statements you're working with to be causally related. But the study of propositions doesn't require any sort of causual relation, no? We're not concerned with the specific nature of the propositions?

[lyric pumice]

Causality can be established using logic.

Suppose a chemical A manifests only if a chemical B is present. Then you can assert that B --> A.

[sleek swallow]

Sure but that still isn't really the concern of formal logic, though? Formal logic doesn't concern itself with specific statements.

[lyric pumice]

It is an application of formal logic.

You are suggesting that formal logic deals with more general notions, abstractions even.

[rare vault]

Assuming one starts at the left circle and draws a line through the figure, ending at the top right, without being allowed to trace over lines or travel left, how many unique paths can be taken?

[sleek swallow]

Yea i'm suggesting that formal logic deals with more general notions, as opposed to such specific things

[rare vault]

Here is an example using 3 arbitrarily chosen potential paths. Colored.

[dapper rose]

yo the witness

[rare vault]

YES ITS FROM THE WITNESS

My friend wanted to brute force a puzzle and we're both into mathematics

and we can't figure out a way

I tried giving each intersection a value, depending on how many decisions can be made (from 1 to 3) then drawing lines as if it were a weighted graph

even if this covers possible paths, it doesn't cover unique paths

one different turn and the entire thing falls apart

[obtuse lance]

since you can't go left, there are only 3 possibilities at each of the 4 vertical sections, so there are 3^4 possibilities

[rare vault]

Hm.. do you think you can provide a more detailed proof?

I dont think the first and last vertical sections hold the same reachability as the middle ones

[obtuse lance]

https://prnt.sc/r0lrja

start at the left, you have 3 choices of path to go right starting from the first column

once you end up in column 2, you again have 3 choices, there are no restrictions on where to go right

in all you have 4 choices

3*3*3*3

[rare vault]

Ah I miscounted the columns.

[obtuse lance]

obviously the last column there is no choice because you just are at a wall

cool makes sense?

[rare vault]

That makes sense. It almost seems too simple

[obtuse lance]

make a shorter version of it and brute force it and check that it works when you compute it the same method I showed you

if you're still in shock heh 😛

[lyric pumice]

@rare vault Try the problem in a board with m rows and n columns for a more interesting challenge.

[sleek swallow]

@iron crescent Well, what are the possibilities you have to account for?

Is 0 not an even number?

You’re welcome.

[lyric pumice]

@iron crescent Try a geometric approach.

Start by illustrating all of the possible outcomes.

[tardy laurel]

Hey I have a question, how do you prove that 2^(1/3) is irrational

[exotic sun]

@tardy laurel Take a look at the 2^1/2 proof

Is this the right channel for graph theory questions?

[pale epoch]

yes

[exotic sun]

Im gonna explain my thoughts on a problem, lemme know if there are logic issues

So I have an undirected graph that can have any kind of connections i.e. sparse subgraphs dense subgraphs some lone nodes, etc

I want to cut all edges such that i am left with the most one edge connected components

an example

my current thought process is to start with the vertex with the fewest neighbors and only keep the edge with the neighbor with fewest neighbors and repeat until until all extraneous edges are cut

not sure if this is sound tho

[lyric pumice]

@iron crescent Hi. The problem can be solved by constructing the outcome space as ordered pairs and computing the fraction of outcomes confined to a triangular portion of the outcome space.

The final result is the ratio of areas (1/2)(12-1)^2/(20*12).

@iron crescent What course level of probability are you taking?

What course level of probability have you taken before?

What kind of school/program are you attending?

[exotic sun]

yo if you take eecs 482 try and get peter chen

[lyric pumice]

@iron crescent Then expect to be challenged.

[exotic sun]

office hours

ah

rip

[lyric pumice]

@iron crescent You must practice for lots of hours.

[faint narwhal]

Read the definition, ask about what you're confused about

[lyric pumice]

@iron crescent You can find texts with exercises by searching online.

Otherwise, we can make up problems to solve.

The videos online are limited by time.

And it is difficult to find texts with a sufficient number of exercises.

Yes.

It is right.

Four of a kind means 4 of the same number or of the same letter letter.

No.

52 choose 4 is the number of combinations of 4 different cards.

Those 4 can be from different kinds.

[lyric pumice]

The probability is (13 choose 1)(4 choose 4)(12 choose 2)(4 choose 3)^2/(52 choose 10).

The number of kinds to choose from is (13 choose 1).

The number of ways to have 4 of a kind with 4 cards is (13 choose 1)(4 choose 4).

After choosing the first kind, there are (12 choose 2) possible combinations of 2 additional kinds.

The number of ways to have 3 of a kind with 4 cards is (4 choose 3).

So, the number of ways to have two instances of 3 of a kind with 6 cards is (12 choose 2)(4 choose 3)^2.

Then the number of ways to have two instances of 3 of a kind with 6 cards and 4 of a kind with 4 cards is (13 choose 1)(4 choose 4)(12 choose 2)(4 choose 3)^2.

[kind nest]

i'm sorry but i have a question about combinatorics

[lyric pumice]

Don't apologize.

[kind nest]

so, the question is: how many 4 letter words can be formed from 30 letters of the alphabet ( my language has 30 letters, 25 consonants and 5 vowels, which will be important) if every consonant can repeat only once.

Now to explain my thought process

the number of vowels is either 0,1,2,3 or 4 so there are 5 cases

so if all 4 letters are vowels, there are 5^4 possible permutations?, ways to arrange the words

next if there are 3 vowels
i choose 5^3 vowels and then 25C1 consonants and for that one consonant i have to choose its place, so
5^3*(25C1)*4

if there are 2 vowels then it is 5^2 * (25C2)

and now i'm stuck

how do i choose the places for those 2 consonants?

[eternal sparrow]

@fickle fable

[kind nest]

and later for 3 and 4

i tried 5^2 * (25C2) * 4^2

but that is wrong

would it be (4C2) as in from the 4 places in the word choose 2 places

[kind nest]

why 60?

oh yes

5 * 4 * 3 is 60

i am tired, thought it was bigger

i'm sorry man i can't thinkstraight

andi dont knowmuch about this so rather wait for someone smarter thanme

[lyric pumice]

@kind nest You posted a problem of a kind whose generalization is interesting and quite difficult. Are you familiar with the principle of inclusion-exclusion?

[kind nest]

Don't know, maybe

because i don't know the english terms of this stuff

[lyric pumice]

I prefer the following result. 30^4-30(25 choose 5)(4 choose 3)+3(25 choose 5)

a)-c) look good to me.

@iron crescent Perhaps, you should multiply by 4.

[proud mason]

Is there a way to find the cardinality of a finite set other than counting how many elements it has?

Or is this just a vague question?

[weary tiger]

Hello

What algorithm would you guys use to factor 3127?

Into 2 primes: p and q

Intro to discrete math class

and has to be done on paper

[weary tiger]

I did it with fermat's factorization method

but why do we start by assume a is the sqrt(n) ?

a = ceil(sqrt(n))

and then b^2 = a^2 - n

[exotic sun]

Im gonna explain my thoughts on a problem, lemme know if there are logic issues
So I have an undirected graph that can have any kind of connections i.e. sparse subgraphs dense subgraphs some lone nodes, etc
I want to cut all edges such that i am left with the most one edge connected components
https://cdn.discordapp.com/attachments/496785905474994186/676900131387211788/unknown.png
an example
my current thought process is to start with the vertex with the fewest neighbors and only keep the edge with the neighbor with fewest neighbors and repeat until until all extraneous edges are cut
not sure if this is sound tho

[obtuse lance]

hmm well since we can treat the separate disconnected components independently, we can just focus on a single connected graph

so then one thing I figured out is that for a connected graph is that you will end up with at most half the number of vertices of connected edges left

so in your example with 6 vertices, you ended up with 3 edges which is maximal

I don't know if it's possible to always achieve this to within 1 (depending on if it's even or odd) but I haven't thought about either

but it gives you something to check for if you do achieve it, here's the derivation of the result:

$\sum_{v \in V} \deg(v) = 2|E|$

[vital dew]

Merosity:

[obtuse lance]

we start here, and recognize at the end of cutting the degree of the vertices is either 0 or 1, so at most they're all 1

$\sum_{v \in V} 1 \ge 2|E'|$

[vital dew]

Merosity:

[obtuse lance]

in other words,

$|E'| \le \frac{|V|}{2}$

[vital dew]

Merosity:

[obtuse lance]

my next step might be to try to show the max can always be obtained (hopefully) on an arbitrary tree since it depends purely on the vertices which the tree has in common. Probably unlikely but if it's true then it means you could take any spanning tree and that the algorithm is not too sensitive. Just an idea, I'm sure someone else knows better

[exotic sun]

so if im understanding you correctly, the max would be 3 here but really we only get 1

[obtuse lance]

yeah exactly

[exotic sun]

so the max always obtained assumption doesnt hold

[obtuse lance]

yep

[exotic sun]

so my question is what order/logic can you use to ensure youre always cutting the next optimal edge?

so here

the top would be an incorrect solution

and my current logic is that i would start at the right vertex since has lowest degree

[obtuse lance]

I guess we can try to prove this never causes a problem

or try to find some counterexample where this breaks

[exotic sun]

yeah ive been trying to think of a counterexample

but cant think of one

[obtuse lance]

I could believe that making a cut this way would then end up cutting something worse later but idk

[exotic sun]

and not really sure how to start with proving

[exotic sun]

#probability-statistics would prob be better id think

[lyric pumice]

@iron crescent Hello. For f), the answer should be the following.

[lyric pumice]

There are (4 choose 2)=6 ways in which 2 can appear twice in 4 rolls.

Since 2 is required to appear exactly 2 times in 4 rolls, there are 4 choose 2 different pairs of rolls that result in a 2 twice.

Yes.

The 2s can occur according to the following ordered pairs of rolls: (1st, 2nd), (1st, 3rd), (1st, 4th), (2nd, 3rd), (2nd, 4th), (3rd, 4th).

Combinatorics and probability are challenging.

What is your major?

What other classes are you taking?

Welcome to math.

I recommend spending at least 40 hours, including class time, per week on math to become good at it.

We should have the server admins put together a comprehensive list of practice exercises in a channel.

I will put in a request.

[pale epoch]

sounds like work lol

honestly publishing answerkeys is a bad idea in general

you should find your own arguments convincing enough that you don't need one

and it just enables you to lie to yourself

look up the answer and think "yeah, i could've solved that"

there is no answerkey for real life problems

and in math we are fortunate enough to have definite answers

you should be able to judge your own answers

and know if they are correct

at the very least after talking with peers

[lyric pumice]

Answer keys should be provided with special permission upon request.

At some point, students will need answers.

[pale epoch]

i mean, how do you know some random answerkey is correct

you should convince yourself with proof

if you are unsure in your answer, your answer is probably bad

similar thing goes for "not enough exercises"

there are dozens of books you can use

and at some point in your mathematical career, there wont be any practice exercises

you just have to learn to ask your own questions

[lyric pumice]

I would generally agree.

[pale epoch]

that being said, exercise sheets are a good idea

but it's a double edged sword, you have to know if they guy using it has the knowledge to solve them

do they not?

did you not define probability?

so if you only deal with those, you can "check" your answers

[lyric pumice]

Yes, there are good reasons why compiled math exercises are difficult to find.

[pale epoch]

you literally only work with the definition and in every step only apply theorems you know to be valid

then your argument is correct

[lyric pumice]

There is only so much that a mathematician is willing to put together on a particular topic.

[pale epoch]

i mean, it sounds simpler than it is

the same way that whoever wrote the hypothetical answer key knows it

then i would not trust them

[lyric pumice]

In general, you should at least see a proof of every formula.

[stray reef]

double, triple, quadruple check your work

at every step if you doubt it even slightly then subject it to scrutiny

[pale epoch]

the general idea is just divide cardinality of your event by cardinality of sample space

at least in theory, you should be able to list every element in both

bcs it's tedious you can apply theorems/reasoning to get there "quicker"

if you can't trust your own argument, it's a bad argument

[lyric pumice]

There is no "magical author of correct math exercises for every kind of problem in topic X."

[pale epoch]

yeah, it takes time

and especially when starting with a topic, you need a good teacher

to tell you what you are doing wrong early

why does everyone go to shit unis

things are often hard to explain in text imo

like, if you don't have enough mathematical maturity it is hard to self teach imo

that's why you need teachers, ideally in person

to stop you from doing beginner mistakes and give you intuition mostly

[lyric pumice]

It is hard to find young people who are interested in joining a serious math chat on Discord at this current time.

And Discord mostly has young people under 30.

[pale epoch]

actively teaching someone is work

and especially if you are in a class with a random prof, i.e. i don't make the curriculum, i will not spend hours teaching you something

it's just not worth it

[lyric pumice]

The older people who could help are not here on Discord yet.

Discord is still new. The server is very young.

[pale epoch]

does your uni not have tutorials or office hours

honestly, my uni pays people to teach things to students

sick

[faint narwhal]

probably

[pale epoch]

i guess, but it's an investment

the alternative is hoping it works itself out

which might happen honestly

a day is a pretty short timespan

depends on your definition of trivial

[faint narwhal]

Non-trivial changes the more you know

Like you're basically always going to be solving "trivial" problems

[pale epoch]

personally i have been having more fun continuously over my "career"

i always feel like i know very little though and am just starting

[faint narwhal]

I mean, you obviously have to think about these problems

Otherwise you would solving them easily

The other point is that putnam problems are very different from upper level math problems

Upper level math exercises are all "easy" if you understand the material

[pale epoch]

your prof will never give you non-trivial problems in a normal setting

[faint narwhal]

i.e., they're mostly there to test that you actually understand the material and know how to apply it, there's really nothing super crazy about them

[pale epoch]

they are designed to be solved in a week or so after all

[faint narwhal]

I mean okay, when you get to research this changes of course but

Learning math

lmao

[pale epoch]

the upper level math

[faint narwhal]

all of it honestly

[pale epoch]

i'm only taking classes i enjoy

[faint narwhal]

Upper level math, even what you're studying now, is so interconnected that everything affects everything else and so the more you learn about anything, the more your overall understand of math grows

That said, I mostly prefer algebra and number theory things

[lyric pumice]

Some Putnam problems are "easy" if you don't have a time limit.

[faint narwhal]

I mean, yeah you can bash out dumb solutions

[lyric pumice]

They are not like research problems that can't be solved in any reasonable amount of time.

[faint narwhal]

I mean, I don't think you should be doing math for the fame

[lyric pumice]

Sometimes an interesting problem has a mildly famous solution.

[pale epoch]

honestly it's more of a group effort

we are all just working together to create more mathematics

[lyric pumice]

I'm more interested in "hopeless" combinatorial problems.

[faint narwhal]

Proving anything at all is an amazing feeling, I don't really think the riemann hypothesis is special in that sense

[lyric pumice]

They are frequent in combinatorial game theory.

I like to try to be on the cutting-edge of knowledge.

The time varies.

[faint narwhal]

As long as I feel like I'm still being productive and doing good work

[lyric pumice]

For some people, any free time is math time.

[faint narwhal]

Maybe sticking to a set schedule every day would help with being productive, but I enjoy being flexible with my time

I'm not really sure you should be doing math if you don't enjoy it honestly. At least in the sense that you might need a break from it

This happens to me sometimes, that doing math becomes a chore but I don't really know what else to do

And I usually just take a break and study things that I find more interesting until I rediscover my passion for doing math

[lyric pumice]

For me, doing math is just part of life.

[pale epoch]

also note that higher level math is inherently creative

[lyric pumice]

I don't really think about the hours much.

[pale epoch]

like, even time spent not actively doing math is actually you doing math

[faint narwhal]

I mean yeah same, I've been doing math everyday for like the past three years, its just habit to do math at this point

True, I find solutions to research problems in the shower more often than you might expect

[pale epoch]

and as creative jobs go, time spent varies

sometimes you are on a run and work like 10 hours a day

[lyric pumice]

Yes, there is such a thing as subconscious solving.

[pale epoch]

sometimes you aren't and do little

[faint narwhal]

Why?

I'm not saying its wrong, lots of people feel the same way

And I've done a lot of comp math in the past too but

[lyric pumice]

I think that most subjects are more entertaining when performed outside of school.

[faint narwhal]

I'm going to say that learning higher math is kind of the same

[lyric pumice]

I am more of a problem-solver-->theory-builder.

I enjoy generalizing.

But I prefer concreteness over abstraction.

[faint narwhal]

Like if I asked you to find all integer solutions to y^2 = x^3 - 5, this probably isn't something you could do with your current knowledge. But learning number theory allows you to be able to solve something like this. And the ideas are something that you could have come up with yourself, but would have taken you like a hundred years to come up with

Yeah I mean, its a bit weird, but with that description, it seems that something like combinatorics and discrete math would be more up your alley

Because those topics are less about knowing everything that there is to be known, but more about solving problems with some basic tools

[lyric pumice]

Combinatorics often involves having the proper perspective.

It is very accessible but most complex.

Combinatorics and number theory are probably the most complex of the mathematical pillars.

Analysis and geometry are probably the most rigorous.

Algebra and topology are probably the most abstract.

[faint narwhal]

Hm, I have a couple disagreements with those statements but its not important

[lyric pumice]

Which?

[faint narwhal]

Well with the overall sentiment of trying to rank math fields like this, I'm just not so sure its super helpful.

Then, maybe I don't know your definition of complex, but some people would count abstractness as complexity

[lyric pumice]

Well, the study of computational complexity is largely concerned with discrete math problems.

[faint narwhal]

Sure, then how is number theory complex from that definition

[lyric pumice]

A suitable example is factorization.

[faint narwhal]

Most, almost all, of number theory doesn't really deal with computational questions like that

[lyric pumice]

Also, linear algebra could fit the description.

I would imagine that a topic that involves computations that are not "nice" or not "smooth" can fall under complexity.

Discrete structures lend themselves to such a nature.

Yes, most of number theory is concerned with abstraction.

[woeful galleon]

can someone explain me this please ?

[sleek swallow]

What is confusing you?

[woeful galleon]

well, lets start with the answer. i said false and true

is that correct

[sleek swallow]

Nope, let’s begin with your justification. Why do you think the first is false and why do you think the second is true?

[woeful galleon]

i am not really sure how to explain it honestly. lets say that the V symbol means for all and E means for some right ?

[sleek swallow]

$\forall$ means for all.

$\exists$ means there exists.

[vital dew]

Abhijeet Vats:

[woeful galleon]

alright, so my logic says that the first one doesnt always work.

u can choose and x value and then a y values that makes the function fail

[sleek swallow]

predicate

But yes

[woeful galleon]

lol sorry. so that is why i said false

from what i know the first one must work for all numbers , which is doesnt

[sleek swallow]

Yeap, that’s correct. So, there exist positive integers x & y such that x+y<6

Now, for the second one?

[woeful galleon]

wait confused

so the first one DOES work ? it obviously works if you choose specific numbers

but doesnt "for all" mean that it has to work in all cases or am i missing the point

[sleek swallow]

No it doesn’t. There exist positive integers x & y such that x + y < 6

[woeful galleon]

ohh

i didnt see the <

[sleek swallow]

Now, the second one.

[woeful galleon]

for the second one, same logic. but it sayins that there is a solution for all x with specific y ? (sorry for my wording, still new to this)

[sleek swallow]

The second one simply reads:

For all positive integers x, there exists a positive integer y such that x + y >= 6.

That’s what it reads.

So, is that true or not? If it is, why? If it isn’t, why?

[woeful galleon]

it is, you can choose any x and "compensate" with a bigger y

[sleek swallow]

Suppose that x >= 6. Then, you’re done. Any choice of y will do.

Now, suppose that x < 6. Then, for x = 1,2,3,4,5, you can clearly choose positive integers y such that P(x,y) is true.

Hence, the proposition is true

[woeful galleon]

alright ! i understand, you writing the propositions out helped a lot

[sleek swallow]

Yeap

It takes time to get used to a new language. So, take your time and do it properly.

[woeful galleon]

thank you sir. i appreciate it

[tranquil jewel]

Can someone tell me if my (c) is right? and I’m kinda confused on how to do the (d)

[stray reef]

no, your c is not right

[tranquil jewel]

Okay, for the complement of A is it (-6, +infinity).?

[sonic herald]

is the negation of a set the set containing all other elements in the universe of discourse besides those in the original?

[stray reef]

@tranquil jewel no.

[tranquil jewel]

How would I determine the complement of A? I thought it’s everything that isn’t it in A

[stray reef]

what about -6

is it in A all of a sudden

or is it not in A

[tranquil jewel]

Oh

Is it

[stray reef]

or is it somehow not in A and not not in A at the same time

[tranquil jewel]

[-6,+infinity)

[stray reef]

there we go. yes that's A^c

[tranquil jewel]

Okay so,

BnC is (-4,-pi)?

[stray reef]

yes

[tranquil jewel]

and then,

Is that right then?

Oh wait

[sleek swallow]

@sonic herald It makes no sense to talk about the negation of a set. You can negate propositions, not sets.

[tranquil jewel]

One sec let me change that

[sonic herald]

you're right, I misread the notation @sleek swallow , I assumed the line over the variable was a negation, apparently it means compliment

[sleek swallow]

Complement*

Compliments are what you give to nice people

[tranquil jewel]

Is that right Ann?

[sonic herald]

Yeah, thanks

[stray reef]

no

$[-6, -4] \cup [-\pi, +\infty)$

[vital dew]

Ann:

[shut fjord]

can someone explain to me this arithmetic:

Here's the full page:

but this step has me lost

[pale epoch]

it's just geometric series

and well, n-n=0

[pale epoch]

yes, exactly

[shut fjord]

the 1 cancels the -1 at the end

[pale epoch]

a=1, r=3 and you only sum to n-1 instead of n

[shut fjord]

making it n okay

thank you

I was looking at this for like 20 minutes

and I thought there's some holes in my knowledge

[pale epoch]

apparently the hole in your knowledge is not recognizing a geometric series

[tranquil jewel]

Wait I have to include the U in that?

[wide birch]

Hey guys, I was recommended to repost my question in this chat

What are some good tools for finding solutions to a system of constraints? I'd like to iterate over all solutions for a through h which are positive integers and obey these constraints:

e + h    < N (for some positive integer constant N)
a^2 + bc = e
b(a + d) = f
c(a + d) = g
d^2 + bc = h```

I think that it's integer programming, but most information I can find is for linear setups, and I'm not sure if those squared terms throw that off. I have a little experience with SageMath, but wouldn't be opposed at all to learning anything new if you guys think it'd be faster.

[shut fjord]

I have two problems in regards to summations

For part a I want to apply the following summation

This probably isn't right

I was thinking this would be more appropriate

but can n represent infinity?

and if so, can i shift this gemoetric summation up by 1 to fit problem a?

[analog sonnet]

@neon thorn it's a fact you have to use quite late on in the proof

What you basically have to do is prove something stronger: namely that the differences between subsequent terms obey the chain of inequalities as well

[analog sonnet]

I proved it differently (credits to @eternal patrol)

Ok alright so

The base case 6 is easy to prove

64 < 720 < 729

So we can assume $\left(\frac{n}3\right)^n<n!<\left(\frac{n}2\right)^n$ for some $n\ge6$

[vital dew]

Rijinaru:

[analog sonnet]

Ok

So now we want to prove that $\left(\frac{n+1}3\right)^{n+1}<(n+1)!<\left(\frac{n+1}2\right)^{n+1}$, right?

[vital dew]

Rijinaru:

[analog sonnet]

Yes

Now here's the first tricky bit: you have to realize that it is sufficient to prove that $\left(\frac{n+1}3\right)^{n+1}-\left(\frac{n}3\right)^n<(n+1)!-n!<\left(\frac{n+1}2\right)^{n+1}-\left(\frac{n}2\right)^n$

[vital dew]

Rijinaru:

[analog sonnet]

Do you see why?

Ok, so bear with me:

I'll introduce a quite elementary fact: if a < b and c < d, then a + c < b + d

where a,b,c,d are any real numbers

Do you agree with this fact?

Ok

So if we know $\left(\frac{n+1}3\right)^{n+1}-\left(\frac{n}3\right)^n<(n+1)!-n!<\left(\frac{n+1}2\right)^{n+1}-\left(\frac{n}2\right)^n$ and also $\left(\frac{n}3\right)^n<n!<\left(\frac{n}2\right)^n$ (by induction hypothesis), we can conclude that, by adding each part (left, middle and right) of the inequality together, that $\left(\frac{n+1}3\right)^{n+1}<(n+1)!<\left(\frac{n+1}2\right)^{n+1}$

[vital dew]

Rijinaru:

[analog sonnet]

Not quite

Where I take the difference technically only serves a part of the induction step

But you've got the right idea

I've noticed that I got ahead of myself a little bit

I wanted to work with not the original chain of inequalities, but the version where every part has been put into a natural log

I'm sorry for the confusion

I'll go to bed soon but I'll try to do a writeup at some point

[sturdy sorrel]

Someone figure this out

[weary tiger]

Any discrete math tutors available for hiring a couple of hours here?

[stray reef]

i can try ig? i'm not a "discrete math tutor" specifically, but i do math tutoring and i know my way around what is normally called discrete math.

though i'm not available at this exact moment. i'll be free about 3 hours from now, if that's ok

@weary tiger

[weary tiger]

I saw your message now, I have exam around 20th this month. I am going through the lecture notes at this moment, we can continue to talk on DM
@stray reef

[stray reef]

no, but i can attempt to throw one together when i come home. how many problems do you want

50 questions sound good?

[weary tiger]

yes, same for me please

exactly what I am working on right now

[stray reef]

i'll just throw it in here once done ig

[nimble plover]

https://gyazo.com/93eb794eb17ba066c385fd25029c7201

can someone explain what does Q^x mean ?

[pale epoch]

group of units i guess

[nimble plover]

@pale epoch i'm still confused

[pale epoch]

by what?

[nimble plover]

i dont know what this notation means

how to apply it

i need to determine whether H is a subgroup of G

[pale epoch]

do you know what a unit is?

[nimble plover]

not in this context

[pale epoch]

units have multiplicative inverses

so Q^x is just Q\{0}

[nimble plover]

oooh

[pale epoch]

so this tells you what the group operation in those examples is as well

(it's multiplication)

[nimble plover]

wait, why is it Q\{0} in this case ?

[pale epoch]

every element except 0 has a multiplicative inverse in Q

[nimble plover]

so these are fractions, and the inverse would be 1/x

oh and 1/0 is undefined

[pale epoch]

0*x = 1 has no solution

in Q or in C

[nimble plover]

thanks

then iv) is a subgroup right ?

H is a subgroup of G

[pale epoch]

you tell me

[nimble plover]

because for example you can have 3/5 since a and b are both odd,

0 is even, so it won't be allowed here

everything in H is in G already

that's my logic

[pale epoch]

i mean yeah H is a subset of G

but that is not enough

[nimble plover]

how come ?

[pale epoch]

because that is not enough to satisfy the definition of subgroup

you also need that H is a group in the first place

and also that H has the same identity as G

[nimble plover]

oh shit, i forgot about that

[pale epoch]

there are multiple equivalent ways actually

but being a subset is definitely not enough

[nimble plover]

@pale epoch althought the question doesn't specify 'prove' it just says 'determine'

[pale epoch]

i would assume that this is synonymous with prove

[nimble plover]

@pale epoch when proving for an inverse, do i use additive or multiplicative inverse ?

[pale epoch]

the group operation is multiplication

[stray reef]

@iron crescent you asked for the combinatorics & probability problem set, here it is

if you want to check your work for any of these feel free to dm me. but do be warned i can and will ask you to show your work, so keep that in mind.

this applies to anyone else who wishes to use my problem set to practice.

[shut fjord]

Give an example of two uncountable sets Aand B such that A−B is a ) finite. b) countably infinite. c) uncountable.

[stray reef]

what is giving you trouble here?

[shut fjord]

A and B is A n B

[stray reef]

no

[shut fjord]

its just stating the two sets

[stray reef]

you shouldn't blindly replace all instances of "and" with the intersection symbol

[shut fjord]

okay

[stray reef]

you need to give two sets, one of them called A and the other B, such that they're both uncountable and A-B has the cardinality listed in each subproblem

on that note

which of (a), (b) and (c) do you need help with

[shut fjord]

b

countably infinite

this is a bijection where the domain maps to the positive integer co-domain?

[stray reef]

what

uh

[shut fjord]

A->B

[stray reef]

no

you need to give two sets, A and B, such that A-B is countably infinite.

and you should have some intuition for which sets are countable and which sets are not.

[shut fjord]

okay so that's everything in A only

[stray reef]

no

A - B is the set of all points that are in A but not in B.

[shut fjord]

so visually, this is what I'm working with..

Well if both A and B are sets in the real number domain, that makes them both uncountable.

[stray reef]

no, just because a set is a subset of the real number line does not mean it is uncountable.

also, your venn diagram has a very tiny sliver for the intersection of A and B, to the point where it is very hard to tell if it's meant to be included or not in the shading

[shut fjord]

the intersection is not included

only A part should be highlighted

[stray reef]

you're missing my point

and the visual isn't that helpful here anyway

[shut fjord]

not every subset of the real number is uncountable

because I can state a set like {1,2,3,4,5}

[stray reef]

this contradicts your earlier statement:

Well if both A and B are sets in the real number domain, that makes them both uncountable.

and your last statement is a restatement of what i said in return.

[shut fjord]

so I need to find a specific set in the R that is uncountable, not just any

[stray reef]

you need to find TWO sets

BOTH of them uncountable

such that their difference is countably infinite.

you are not required to specifically focus on the real number line.

though it is a convenient starting point.

[shut fjord]

we learned that (0,1) is uncountable since theres no way every number in N can be mapped to all the fractions in between 0 and 1

[stray reef]

careful with the wording there.

the set of FRACTIONS in (0,1), i.e. of RATIONAL numbers between 0 and 1, is countable.

(0,1) itself, the set of all REAL numbers between 0 and 1, is not.

you must also have learned, then, that R, the entire number line, is also uncountable.

[shut fjord]

well if (0,1) is uncountable considering all the real numbers

(0,1) should be contained in R

and R is definitely uncountable if (0,1) is uncountable

(if I'm considering the set of real numbers)

[stray reef]

i didn't exactly ask you to say any of that but whatever.

[shut fjord]

I think this would work

[stray reef]

it does.

[shut fjord]

it makes more sense just to talk it through sometimes

it helps me

thank you for the additional information

[stray reef]

you pinged the wrong person lul

i'm ann [aka discount tuong], not tuong himself

[shut fjord]

Prove or Disprove that all checkerboards of this shape can be completely covered using right triominoes whenever n is a positive integer.
a) 3×2

Here's my work:

[stray reef]

3 by 2 or 3 by 2^n?

[shut fjord]

I probably should replace 3 x 2^(k+1) with 3 x 2^(2)

my base case is 3 by 2

so n = 1 in the basis

which is the image on the right hand side

[stray reef]

Prove or Disprove that all checkerboards of this shape can be completely covered using right triominoes whenever n is a positive integer.
a) 3×2

[shut fjord]

oh its

[stray reef]

did you mean 3 by 2**^n** in the problem statement

[shut fjord]

yes

[stray reef]

okay

[shut fjord]

yeah

[stray reef]

i see a lot of words

but it seems like your basic idea is "to tile a 3 by 2^(k+1) board, cut it in half and tile each half"

[shut fjord]

yeah

[stray reef]

so i'm gonna say your thing is okay

[shut fjord]

n will always just double the checker board

thus we can divide it by n

or

we can basically get it back down to 3 x 2

which we know by the basis case to be true

[weary tiger]

Hi

I can not find conjunctive simplification

My teacher simplifies step 2 using conjunctive simplification, to me it looked similar to hypothetical syllogism but they are not the same

Thanks In advance

[stray reef]

conjunctive simplification is taking two things connected by an AND and discarding one of them

[weary tiger]

oh is that everything 😛

I think it was weird that was allowed

Why can we just discard one of them, can we discard the one to the right instead of the one to the left, or does it have to be the one on the left

I will continue where I left off tomorrow

[stray reef]

Why can we just discard one of them, can we discard the one to the right instead of the one to the left, or does it have to be the one on the left
it doesn't matter which one

and it should make intuitive sense why we can discard one of them

if i introduce you to a person named sarah, and then say "Sarah is a rock climber and plays guitar"

then surely the statement "Sarah plays guitar" will also be true

[hollow ice]

I am not very good at this sort of thing, but I have been wracking my brain trying to figure out how to go about starting to prove this:
"Prove that for all real numbers c, if c is a root of a polynomial with rational coefficients, then c is a root of a polynomial with integer coefficients."

[faint narwhal]

what have you tried?

[hollow ice]

I've gotten about as far as rewriting it as ∀c∈R, p(c) -> q(c)

[faint narwhal]

uh

what are p(c) and q(c) here

[hollow ice]

So far I'm used to proving statements that intuitively I know are true or false.. but I have no idea whether such a statement is true or not.
p(c) := c is a root of a polynomial with rational coefficients
q(c) := " ... integer coefficients

[faint narwhal]

maybe write out what p(c) means

so if p(c), then you have some polynomial f(x) = a_nx^n + ... + a_1x + a_0 where all the a_i are rational

and f(c) = 0

[hollow ice]

yea sure I can write a formal definition of a polynomial, and then what it means for the coefficients to be rational.. and for c to be a root, and I will surely need to, I just don't see where that gets me

[faint narwhal]

okay, maybe lets write it out more explicitly

We can write $f(x) = \frac{p_n}{q_n} x^n + \cdots + \frac{p_1}{q_1} x + \frac{p_0}{q_0}$ for some integers $p_i, q_i$ and $q_i \neq 0$

[vital dew]

Zopherus:

[faint narwhal]

and we know that $\frac{p_n}{q_n} c^n + \cdots + \frac{p_1}{q_1} c + \frac{p_0}{q_0} = 0$

[vital dew]

Zopherus:

[hollow ice]

intuitively I would assume there is a solution to this where q_i is not equal to 1 (making the coefficients integers)

[faint narwhal]

Well, do you see a way to get from this polynomial to a polynomial with just integer coefficients?

[hollow ice]

to zero you mean?

[faint narwhal]

awkward wording, reread my message

[hollow ice]

not elegantly, I only see stating that q0 through qn are equal to 1, or equal to the same thing and factoring them out

[faint narwhal]

what if we multiply everything by q_n? what happens

is c still a root of this new polynomial?

[hollow ice]

I honestly don't know, without seeing numbers

[faint narwhal]

Well you know that $\frac{p_n}{q_n} c^n + \cdots + \frac{p_1}{q_1} c + \frac{p_0}{q_0} = 0$

[vital dew]

Zopherus:

[faint narwhal]

is it true that $p_n c^n + \cdots + \frac{p_1 q_n}{q_1} c + \frac{p_0q_n}{q_0} = 0$?

[vital dew]

Zopherus:

[hollow ice]

seeing it outside of my head doesn't help sorry

I don't know if it changes the sum

[faint narwhal]

To get from the top to the bottom, we multiplied the left side by q_n

We multiply both sides of the first equation by q_n

so what happens to the right side

[hollow ice]

well put like that it doesn't make sense for it to modify the result I guess

[exotic sun]

yo is there a place i can ask a disc math question besides this channel?

[faint narwhal]

We took the top equation

and multiplied both sides by q_n

what happens to the right side of the equation when we multiply by q_n

[hollow ice]

I guess all the terms are scaled by the same amount, therefore the negatives should still be equal to the positives and the sum therefore still be zero

that only removes one q though

[faint narwhal]

I really don't think you understand

multiplying both sides of the equation by some number still keeps the equation true

we multiply both sides of the equation by q_n

and the right side is 0 * q_n which is 0

[hollow ice]

the end result is going to look like..
$p_n(q_0 q_1 ... q_n-1) c^n + \cdots + p_1 (q_n ... q_2 q_0) c + p_0 (q_n ... q_1) = 0$?

[vital dew]

D.B. Cooper:

[faint narwhal]

Something like that sure

[hollow ice]

No I think I understand fine, like I said since the sum is 0 the sum will remain zero regardless of how we multiply, so long as we multiply each of the terms

[faint narwhal]

Yeah but its much simpler than you're making it out to be

[hollow ice]

If it wasn't zero, we'd no longer know what the sum was though

and that was where I was getting caught up earlier

As far as I can tell though, this doesn't prove that the polynomial has integer coefficients though

ohh wait..

I see the proof now

It wasn't asking to prove that the polynomial with rational coefficients was actually a polynomial with integer coefficients. It was a proof about c being the root for two polynomials

[faint narwhal]

yes

[hollow ice]

I might have had better luck if I understood that

thank you for the help understanding this question

[shut fjord]

Is {2,3} x N countably infinite?

If so why?

[stray reef]

what do you think @shut fjord

[shut fjord]

I said because f:{2,3} -> N

It is uncountable infinite

[stray reef]

that doesn't make any sense

what do you even mean by "because f: {2,3} -> N"? what is f

[shut fjord]

It’s being crossed with all natural numbers

[stray reef]

what's being crossed with the natural numbers

[shut fjord]

{2,3}

[stray reef]

your set is the direct product of {2,3} with N yes

[shut fjord]

So you would get { {2,1}, {2,2},.....}

[stray reef]

no

{2,1} isn't an element of {2,3} x N

(2,1) is

[shut fjord]

oh right tuples are ( )

[stray reef]

{(2,1), (2,2), (2,3), ..., (3,1), (3,2), (3,3), ...}

[shut fjord]

not {}

[stray reef]

so why do you think this isn't countable?

[shut fjord]

I think it’s countably infinite

[stray reef]

oh now you're saying something different

[shut fjord]

I said uncountable infinite before

[stray reef]

three minutes ago you thought it was uncountable but now you think it's countable?

[shut fjord]

My mistake

Yeah I actually wrote it’s countably infinite

[stray reef]

okay so can you write out a bijection between this set and N then

or some other set known to be countable instead of N if that's more convenient

[shut fjord]

The cross is a bit confusing...

Is my domain strictly {2,3} ?

[stray reef]

no your domain is {2,3} x N

simply speaking you need to put the elements of {2,3} x N on a list

[shut fjord]

So I need to find a bijective function such that: f{2,3} x N —> N

[stray reef]

a bijective function f: {2,3}xN -> N

[shut fjord]

Yes

Thanks for the insight

[stray reef]

did i provide any?

all i did was repeatedly state the obvious but you're welcome

[weary tiger]

Can anoyone tell me what this plus sign means?

[robust mango]

Positive.

[weary tiger]

oh

is that all

[robust mango]

n belongs/is in the set of positive integers.

Yeah.

[weary tiger]

Hi

[proud mason]

Consider the constants $a, c \in \mathbb{N}^{+}$. To prove that ${ab \in \mathbb{N}^{+} \mid ab < c} = {ab \in \mathbb{N}^{+} \mid b < \frac{c}{a}}$, would $ab < c \implies b < \frac{c}{a}$ be enough?

[vital dew]

Noctilucente:

[weary tiger]

I have an issue understanding this problem:

https://gyazo.com/bff0683b6268569fb1617234b9512f3e

any help ?

How many ways can the letters MATEMATIKK permutate. So, 2 equal letters is not next to each other.
I understand this problem when it only involves one letter like A. But I am not sure how I should go at it when its multiple letters.

I am pretty sure it is easier then I think

[fresh flame]

Can you define predicates in terms of other predicates? E.g.: P(T(x), S(x))

[raw birch]

@weary tiger I have an idea that might work. Calculate the permutations when 2 equal letters are next to each other and then do n! minus that. So you divide by cases: When M is next to M (you multiply by 3), when M is next to M and T is next to T (you multiply by 3, and when M is next to M, T is next to T and K is next K. I think it is a lot of work but it might be a bit more organized

M is next to M. Decide position for M, we have 3 cases. M the first, the last or in the middle. In the first and last cases the second M has only one possible place, in the middle case it has two, and then the following are 8!... And so on...

[austere ridge]

@weary tiger Total ways of counting: $$ \frac{10!}{2! 2! 2!} $$.

[vital dew]

ilikenike:

[austere ridge]

Now we can just replace the repeated characters by a different character

$$ 7! $$

[vital dew]

ilikenike:

[austere ridge]

Because the repeated characters when joined can occur anywhere.

Just take their difference

[lyric pumice]

@weary tiger Hello. It is likely that the most convenient method of enumeration under such restrictions is an inclusion-exclusion sequence of multinomials whose poset is ordered vertically by the number of violations of repeated types in a permutation and horizontally by the distribution among repeated types of a particular number of violations within a permutation.

[weary tiger]

Hm

I heard about inclusions exclusion but didn’t read the lecture

Didn’t know it correlates to this

[lyric pumice]

If you want letters of the same type not to be adjacent, then you should apply the inclusion-exclusion principle. Otherwise, the solution to the problem is a single multinomial coefficient.

[weary tiger]

Alright, that’s what I want. I will take a look at the inclusion exclusion principle tomorrow

Does it have to do with sets?

[lyric pumice]

Indirectly, yes.

[sleek swallow]

@fresh flame I don’t see why not

[orchid fulcrum]

does the i_1, i_2 ... i_j mean that for each index of j, it does a summation from 1 to j?

haven't seen that notation before where the top of the summation has nothing on it

i also don't really get how to read that in the context of the intersections

[pale epoch]

it's a summation over every subset of {1, ..., n} with cardinality j

tbh this notation is bad

like, in the first step j=1 and you take all 1-element subsets of {1,...,n}

and because (-1)^(j-1) = 1 in this case

you add the cardinality of A_1, A_2, ..., A_n

then for j = 2 you look at all 2-element subsets of {1, ..., n}

so {1, 2}, {1, 3}, ... {1, n}, {2, 3}, {2, 4}, ..., {2, n}, ... {n-1, n}

and you subtract the cardinality of the intersction of those

[orchid fulcrum]

and at j=2 would you be adding the cardinalities of A_1_2, A_2_2, ... A_n_2

[pale epoch]

so -|A_1 intersect A_2| - |A_1 intersect A_2| - ...

[orchid fulcrum]

ahhh ok

[pale epoch]

just look at the formula for n=3

$|A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |A \cap C| - |B \cap C| + |A \cap B \cap C|$

[vital dew]

Lochverstärker:

[weary tiger]

Hi

If I want to solve this congruence

42x = 12 (mod 90)

Is it right to say that my first solution is

x = 12 (mod 90) ... x = 3 (mod 30) ... x = 1 (mod 10) ... first solution is 11

[stray reef]

[tacit willow]

could someone help me understand this question properly? ```1. Determine whether each of the following relations is a function on the set {1, 2, 3, 4}. Justify your answers.
f = {(1,2),(2,2),(3,1),(4,3),(2,3)}

I don

[raw birch]

f is not a function because it assigns to different values to 2

f(2) can't be 2 and 3 at the same time

[tacit willow]

hmm

oooh

So f = {(1,2),(3,3),(4,2)} would be yes, or no because there is no value for 2?

[raw birch]

It is not. Every element in the domain of a function must have an image

f(2) must exist

That is a function on the set {1,3,4} not the set {1,2,3,4}

[tacit willow]

aha

such easy to understand answers

thanks man

[raw birch]

No problem

[tacit willow]

    Is a function. 
    
    f = {(1,1),(2,2),(3,1),(4,2)}
    Yes

    f = {(1,1),(2,2),(3,4),(3,1),(2,1)}
    Not a function. 4 is missing a value and 3, 2 has two different values

So these answers are not completely out of touch?

[raw birch]

I don't know what you mean with out of touch

[tacit willow]

they are not all stupid

[raw birch]

They are correct

[tacit willow]

aha, good

[weary tiger]

hi

how do i approach this problem

x+y+z=10

x,y,z belongs to the natural numbers

how many possible ways do the numbers equal ten?

im not sure what channel to use

its a combinatoric problem, but its in the discrete math book lol

[stray reef]

do your natural numbers include 0 or not

[weary tiger]

it does not

[stray reef]

okay

then this can be reduced to the equation x' + y' + z' = 7, where x'=x-1, y'=y-1, z'=z-1 and x', y', z' are nonnegative

and that can be reduced to an arrangement problem

[weary tiger]

i see

[stray reef]

specifically, the number of solutions to x'+y'+z'=7 in nonnegative integers is equal to the number of ways to arrange 7 white and 2 red counters in a row

[weary tiger]

ok

cool.

[alpine phoenix]

In this case, you might be able to get the answer graphically

if you know that x+y+z=10, then z is determined by x and y

[weary tiger]

yupp

[alpine phoenix]

then you can instead ask "what natural numbers x and y are such that 10-x-y is also a natural number?"

This approach might not scale well, but it's another way to look at this

[weary tiger]

okok

[alpine phoenix]

This picture illustrating 5 choose 3 may be inspirational https://upload.wikimedia.org/wikipedia/commons/thumb/6/65/Combinations_without_repetition%3B_5_choose_3.svg/800px-Combinations_without_repetition%3B_5_choose_3.svg.png

[tacit willow]

    f = {(1,1),(2,3),(3,2),(4,4),(5,5)}
    G =  {(1, 1), (3, 2), (2, 3), (4, 4), (5, 5)}
    f = {(1,5),(2,4),(3,2),(4,1),(5,4)}
    G = {(5,1),(4,2),(2,3),(1,4),(4,5)}

    f = {(2,1),(3,4),(1,3),(4,1),(5,2)}
    G = {(1,2),(4,3),(3,1),(1,4),(2,5)}

Does this seem correct? G = is what i answered

[verbal furnace]

It doesn't seem right for the second. Why?

[tacit willow]

hmmm, now im puzzled

2. Let A = {1, 2, 3, 4, 5}. Find the inverse of the following functions f : A → A .
    f = {(1,1),(2,3),(3,2),(4,4),(5,5)}
    G =  {(1,1),(3,2),(2,3),(4,4),(5,5)}

    f = {(1,5),(2,4),(3,2),(4,1),(5,4)}
    G = {(5,1),(4,2),(2,3),(1,4),(4,5)}

    f = {(2,1),(3,4),(1,3),(4,1),(5,2)}
    G = {(1,2),(4,3),(3,1),(1,4),(2,5)}

Better format

[verbal furnace]

The second G isn't right, right?

[tacit willow]

Well, i thought it was. But now i don

don't know 🙂

[verbal furnace]

You could draw to visualize the function.

[tacit willow]

damn

i see it now

lol

was very helpful to use mspaint. But how would i answer it correctley then? Just delete the (4,5)?

[verbal furnace]

You can't, describe more precisely your assignment.

[tacit willow]

well

the whole thing is copy pasted

that's litteraly it

[verbal furnace]

Then you can't give an answer, f needs to be at least injective to find an inverse.

[tacit willow]

So is this an error?

[verbal furnace]

The assignment no.

[tacit willow]

hm

Should be sufficient in this case then

[verbal furnace]

Explained then yes.

[tacit willow]

Thanks alot for the help. Feel like an idiot now

😄

[verbal furnace]

You are welcome.

[lyric pumice]

@weary tiger Count the number of integer compositions of 10 that contain 3 parts.

Prove that the result is 9 choose 2 = 36.

[shut fjord]

I'm a bit confused about this recurrence relation

If an = an−1 − n and a0= 4, what is a20?

the answer is way bigger than what I’m expecting

answer is -206

is my implementation wrong?

I should be getting -35

[wintry rock]

I'm a bit new to countermodels and predicate logic, what's the difference between the left side and the right side? 😖

[tacit willow]

    AD
    AB
    DA
    BD
``` Hmmm, am i wrong to assume that none of these are defined ? 🤔

[robust mango]

AD is not defined, AB is defined, DA is defined, BD is not defined.

just look up multiplication rules for matrices, you'll understand

[stray reef]

put the orders side by side

you're looking at the following maybe-defined products:

(6×4)(6×6)
(6×4)(4×4)
(6×6)(6×4)
(4×4)(6×6)

the inner ones have to match for the product to be defined.

[tacit willow]

aha, right because in order to multiply, number of row 1 of matrix a has to match the number of column 1 of matrix b. correct?

[weary tiger]

Anyone who are familiar with tasks like this?

I understand the concept, but I get slightly wrong answer, starting to wonder if its me or the teacher. But I guess its on my end

[tacit willow]

Does this seem correct?

[stray reef]

(c) is a function but isn't surjective; everything else is ok

[tacit willow]

Aha

Is it just a general function then?

Is a general function. Neither surjective or injectivebetter?

[stray reef]

remove the "general"

[tacit willow]

Hehe, ok 😄

Thanks

Scratched my head trying to figure out what kind of function C was

[tacit willow]

f(x) = y such that y=√xwould this be a function?

[sleek swallow]

It doesn't quite make sense to talk about a function without saying something about the domain & codomain

@tacit willow

[tacit willow]

hmm, tbh the whole task had me confused, but i'm just trying to understand it instead of asking for an answer

    f(x) =(13-x)/√(x^2-1)
    f(x) = y such that   y=√x
    f(x)= x^5-7

[sleek swallow]

Well, what has you confused?

When they're asking you to check if each is a function or not, what are the two things you need to check?

[tacit willow]

for domain and codomain?

[sleek swallow]

They're definitely related. What does it mean for $f:\bR \to \bR$ to be a function?

[vital dew]

Abhijeet Vats:

[tacit willow]

real number as input, real number as output?

[sleek swallow]

In fact, every element in the domain must be assigned a unique element from the codomain. That's what it means for this to be a function. So, there are two things you need to check:

1. Is f(x) defined for all real values of x?
2. For each real x, is there only one real value f(x) assigned to it?

[tacit willow]

hmm, so they are all functions?

Or am i misunderstanding this information?

[sleek swallow]

Well, let's take a look at part b), yes? Consider $f(x) = \sqrt{x}$.

Ask yourself the following question: is there any value of x in the real numbers such that f(x) does not belong to the real numbers? In other words, are there any real numbers x for which f(x) is undefined?

[vital dew]

Abhijeet Vats:

[tacit willow]

well, could be 0?

or negative?

[sleek swallow]

Indeed, $f(x) = \sqrt{x}$ is not defined when $x < 0$.

[vital dew]

Abhijeet Vats:

[sleek swallow]

So is it a function?

[tacit willow]

no

alright, i think i get it now

Thanks alot !

[sleek swallow]

You're welcome.

[weary tiger]

can someone show me an example of how to show this?

[tough locust]

@weary tiger is it true or false, do you think?

Hint: think about what operations you are allowed to do to both sides of an inequality. It's more limited than equality but there are useful things you can do

[weary tiger]

i understand when its some x, but havent seen some x, y

should we assume when x=1, y = 1

[tough locust]

Same idea, it's like saying

There exists a pair (x,y)

x cannot equal 1, why not?

[weary tiger]

if x is 1, then left side of the and is false, therefore whole thing false

x can ony be 4

not 3, since its not >=

[tough locust]

Okay sure

Or 5

Or 6,...

[weary tiger]

if thats the case then, the right side would be 4^2 + y = 16 + y

what about the y?

[tough locust]

I gave you a hint and you ignored it

[weary tiger]

if y is 0 then 16 is bigger than 5 and so on

you can multiply? and add, if you multiply by - you have to flip? is that the hint?

[tough locust]

Multiplying is allowed by a positive number

[weary tiger]

i am more confused about the y, since I am used to see, some x

[tough locust]

In fact if 0<x<y and 0<a<b

[weary tiger]

if you not flip the equality signs, right

[tough locust]

Then 0<ax<by

I ask because

You're given information about x

But what you need is information about x², not x

[weary tiger]

saying x is 3 is false right?

i believe the blueprint said 3

[tough locust]

I dunno, can x be three?

[weary tiger]

no, then 3 is not bigger than 3

[tough locust]

Right

My point is still