discrete-math

[stray reef]

does it?

[steel valve]

supposedly this statement represents "There is somebody whom no one loves"

why can't the quantifiers be reversed?

L(x,y) represents "x loves y" btw

where x and y consist of the entire planets' population of humans.

nvm i get it now

🤦

[tranquil cargo]

for all x there exists y such that x+y=8

there exists y such that for all x x+y =8

?

[steel valve]

is it a true or false statement i assume you're asking?

i think it's false cause all real numbers can't satisfy one existing condition

say the existing real number is just 1

7+1 may work

but 8+1 does not

not all numbers can satisfy the existing y value.

someone pls correct me if im wrong

thanks

[wide karma]

First intro to discrete math class. So i am not sure how to approach this assignment, any help?!

[steel valve]

tf you guys are so far ahead

[wide karma]

Really this is the first no grade hw he gave us

[steel valve]

lol

[wide karma]

More of a excersie

[steel valve]

well for my curriculum we started off with propositional logic

and built on top of that with symbolic reasoning

[wide karma]

We are doing sets

[steel valve]

yeah idk what you guys are doing

[wide karma]

Lol this professor is feared by all

[steel valve]

the way the curriculum is taught we're i'm at is about writing logical proofs.

and how to apply them in matrices and sets

etc

[wide karma]

That sounds hard

[steel valve]

idk tbh

i haven't read far ahead to tell how hard it is

[wide karma]

Writing proof is no joke tho

[steel valve]

i assume it should be ok

if it's anything like the proofs i've done so far for propositional logic

it shouldn't be too bad

just look at the rules chart and pray a solution comes

[wide karma]

Ohhh well u seem to get it

Is ur prof a hard ass

[steel valve]

nah he chill af

like a HS teacher

[wide karma]

Ur so lucky

[steel valve]

he does assigned hw reading

i always tell him that i'm late on the reading assigments cause i have trouble with the material and would like to practice more. gives me full points for them.

i am telling the truth because i have those detailed notes and practice problems.

really nice guy

[wide karma]

Lmao if i tell that to my prof by the vibe i got off of him he will drop me or fail me

[steel valve]

oof

that sucks

[wide karma]

I been trying to avoid him for 1 year. Friends told me he is crazy but alas i am here

[steel valve]

oof

best of luck to you

[wide karma]

U too man

I should post my question again

First intro to discrete math class. So i am not sure how to approach this assignment, any help?!

[analog sonnet]

Lots of typos

But I'd suggest just do those set operations on a couple of sets that are indicated and see if you understand what's going on

Start writing some down, it's mainly to get the hang of it

[loud ingot]

Have a question, I think the answer to it is yes

8. Can a multiple of ten be easily identified from its binary representation?

Does anyone agree/disagree the answer is yes?

[stray reef]

what's your reasoning behind it

[loud ingot]

@stray reef I think it would be because the last digit of a binary representation would be a 1, and the first digit would be a 0 for multiples of 10.

[stray reef]

well

it's true that IF your number is a multiple of 10, THEN its last digit in binary will be 0

doesn't quite work the other way around though

10000_2 isn't a multiple of 10 despite ending in a zero in binary

[loud ingot]

Hm, before my response above I was kind of leaning towards no, because I initially couldn’t find a specific pattern for identifying multiples of 10 reliably. Would you agree it’s correct to say no @stray reef

[stray reef]

well you CAN come up with a divisibility rule for 5 in binary

wouldn't call that "easily identified" tho

[loud ingot]

I see, why would it be difficult with the divisibility rule you mentioned? Couldn’t find a specific answer in regards to that online @stray reef

[stray reef]

why do you keep pinging me in every message

i'm here

[loud ingot]

sometimes people just up and go for hours, lol. If it annoys you I’ll stop

[stray reef]

anyway uh

yeah i mean i can give you what amounts to a generalized version of the divisibility rule for 11 in base 10

it's not something i'd consider allowing something to be "readily identified"

[loud ingot]

Sure anything helps. as of right now, not sure how I could explain my reasoning for 8. why it couldn’t be easily identified

[stray reef]

really the only ones i'd consider "easy to identify" in binary would be divisibility by powers of 2

[loud ingot]

I could see that. Thanks

[hidden crow]

Would [B U C’] U A

Just be C’

[crude dagger]

How does one solve a recurrence that cannot be solved with the master method? $T(n) = 4T(n/2) +n^2log(n)$ . I made a recurrence tree and found the that root node was $n^2log(n)$ obviously, and the leaf nodes "work" was $n^2log(1/2)$ which leads me to believe that the correct solution is $O(n^2log(n)$

[vital dew]

DangerousDan:

[crude dagger]

but log(1/2) would result in a negative amount of work which doesnt make sense

[craggy gale]

https://cdn.discordapp.com/attachments/488104216678760469/619599394433531914/94730921676115968.png

[crude dagger]

thanks

[craggy gale]

You're welcome

[static rapids]

this place... is going to used alot

[blazing nova]

Does anyone know how to do this problem/

?

[craggy gale]

Yes

You have to know stuff like power functions grow faster than logarithms

[blazing nova]

Let me show you what I have . Not sure if it's right.

Fb < fd < fc < fg < ff < fa < fh < fi

[craggy gale]

Where's f_e in this?

[blazing nova]

v?

oh lols

not sure where ot put it

[craggy gale]

Well, for all $n\geq 2$
$$\sum_{i=1}^n\sum_{j=i+1}^n\pi=\sum_{i=1}^n(n-i)\pi=\sum_{i=1}^ni\pi=\frac{n(n+1)}2\pi$$

[vital dew]

Tuong:

[craggy gale]

slightly faster than n²

[blazing nova]

wait how did you guys get from n-i pi -> ipi?

[azure lichen]

it’s not quite correct. if i started at 0, it would work though (and it could be salvaged without much issues I’m sure). if i starts at 0, then the first sum goes
5π + 4π + 3π + 2π + 1π + 0π
and the second goes
0π + 1π + … + 5π

[blazing nova]

okay so I'm guessing the line up is ff < fe < fa

[azure lichen]

because i starts at 1, it needs some shifts or sth

[blazing nova]

losl it's the same thing

since 0pi is 0

[azure lichen]

no, because the former doesn’t have 5π in it

[blazing nova]

I'm just wondering how they got this

[azure lichen]

when i starts at 1, (n-i) is never 5

[azure lichen]

yes I just explained that

it’s wrong, because of the indices

because the first sum here goes
0 + π + 2π + … + (n-1)π
and the second sum goes
π + … + (n-1)π + nπ

so the second sum should only go to n-1

and then they’d be the same

but other than that it’s literally just inverting the order of summation

[blazing nova]

I get that.

the second one is npi more

so hwo do I aply this to the problem..

[blazing nova]

so I got that f is theta of e

sorry

fe(n) = theta of ff(n)

but it doens't pass the ratio test

like ratios go to pi

lim of f(n)/g(n) -> pi as n -> infiniti

[reef thistle]

hmm

so, that means they are of the same order

[cerulean ore]

what is the meaning of this symbol over here?

that q divides c and p?

[cerulean ore]

nvm,

Thank you!

Shouldn't underlined part be d | b

2 is gcd(2,4)
therefore, 2 | 2,4

d = gcd(a,b,c)
and that guy is writing (a,b) | d

The concrete mathematics book says:

[sonic ferry]

Hello i need basic discrete math help

Wouldnt it be false because the set {a, b} is not inside the set {a, b, c, {a, b, c}}

or is my logic wrong

[ripe ferry]

youre right

[sonic ferry]

Thank you! my friend keep saying im wrong and that its True because a,b is inside the {a , b , c , { a ,b ,c }} @ripe ferry

but the question is an element question so its : Is there an element set {a, b} inside the set {a , b , c , { a, b , c }}

[stray reef]

"is inside" is ambiguous

{a,b} is not an element of {a, b, c, {a,b,c}}

[sonic ferry]

Thank you @stray reef

[cerulean ore]

Guys, I am working upon a programming problem.
Suppose there are N integers, then we have to count the number of subsequences whose linear combination gives us the given constant K.

[reef thistle]

linear combination???

so you are adding termwise?

[cerulean ore]

Yes like if the array is 2, 4 then subsequences are (2), (2,4), (4)

[reef thistle]

Okay, so

[cerulean ore]

Now, what I have done yet is that I have learnt that gcd(a,b,c) = gcd((a,b),c)

[reef thistle]

okay you have subsequences, and what do you mean by "linear combination gives us the given constant K"

[reef thistle]

hmm okay

[cerulean ore]

gcd(a,b) = m(a)+n(b)

[reef thistle]

okay sure so you need to make sure gcd of the entire subsequence divides k

How big is N, K?

@cerulean ore

How big are the integers in your sequence?

[reef thistle]

hmm okay

subsequences are like subsets here

since order is not really doing anything much

so I'm really tempted to sort the array

[cerulean ore]

order should be maintained

[reef thistle]

nah, it wouldn't change the number of subsequences

[cerulean ore]

https://www.codechef.com/SEPT19B/problems/FUZZYLIN

My mistake

it should be contiguous subsequence

[reef thistle]

ARGH

okay no wonder

that makes it easier on that other side

I'm still thinking of factoring K

then we find the number of subarrays with GCD = x

for some x

Let's call them subarrays

[cerulean ore]

I constructed a GCD matrix as GCD will never change for the subsequences

[reef thistle]

GCD matrix?

[cerulean ore]

Let me show you

If the elements are 2, 4, 8, 24 then this matrix is what I am making

[reef thistle]

That seems too big

N is 10^5

you can't store it all

[cerulean ore]

Yeah that's the problem

[reef thistle]

I think I have the solution

[cerulean ore]

I need a hint

[reef thistle]

Divide and conquer

My solution should run in N polylog N

[cerulean ore]

With that matrix, do you think I should get the answers if the test cases are smaller?

[reef thistle]

idk what that matrix would do

Are you trying to precompute all the subarray GCDs

[cerulean ore]

yes

[reef thistle]

if so, it's definitely the wrong approach

Here's my solution

Divide the array into 2 halves

and consider subarrays that pass through the middle

@cerulean ore Can you work on that?

[cerulean ore]

1. How is my approach wrong?
2. And what does that "subarrays that pass through the middle" means?

[reef thistle]

1. Too slow

[reef thistle]

2. Just find the middle of the array, and count the number of subarrays that begin in the left side and end in the right side

[cerulean ore]

Yeah, it is slow af

[reef thistle]

So if your array is 5 4 2 | 8 2 10 5, where I split at the bar, then I want to deal with subarrays like
5 4 [2 | 8 2 10] 5
but not
5 4 2 | [8 2 10] 5 (fully on the right)
or
[5] 4 2 | 8 2 10 5 (fully on the left)

if you can do this, you are more or less done

[cerulean ore]

If the given array is: 2 4 | 8 24
2 [4 | 8] 24
2 [4 |8 24]
[2 4 | 8] 24
[2 4| 8 24]

How does that helps?

[reef thistle]

Notice if a subarray can, then subarray containing it also can

[cerulean ore]

fuck

woah

[reef thistle]

So, you want to find the maximal subarrays that cannot

So, you can find the maximal subarrays starting at each element that cannot

at some point you must make sure you don't compute too many gcds

you can do that with a sparse table or "segment tree"

[cerulean ore]

That's a new term for me

You never leave a chance to amuse me.

[reef thistle]

well, the problem with "segment tree" is that we found that it's not well defined in the literature

https://www.geeksforgeeks.org/segment-tree-set-1-sum-of-given-range/

https://www.geeksforgeeks.org/sparse-table/ this is the sparse table

[cerulean ore]

Did it take a lot of coding to get to the solution of this problem or is it about more Mathematics?

[reef thistle]

It's more about algorithm design

and proving your algorithm works

[cerulean ore]

and that comes with practice?

[reef thistle]

maybe experience too

and knowing what you can try

you learn as many things as possible that you can try

and then you take them out and throw them one at a time at the problem

[cerulean ore]

aah, so it is like you've gathered so many tools

[reef thistle]

with experience you can see "my, this screwdriver here can't help with nailing that nail" while you search for a better hammer, so you spend more time on more promising ideas

[cerulean ore]

Our batch coordinator wants me to represent the university next year in ACM-ICPC

[reef thistle]

oh no

[cerulean ore]

and you know my level of math

[reef thistle]

says me who hasn't represented a university in ACM-ICPC

and potentially is going to

[cerulean ore]

You don't need ACM-ICPC

Go for bigger ;-;

[reef thistle]

what's bigger?

[cerulean ore]

You must know

BTW, let K = 8
2 [4 | 8] 24 , since gcd(4,8) = 4 | 8 so, we have 3 pairs over here {4,8, {4,8}}
2 [4 |8 24], since gcd(4,8,24) = 4 | 8 so, we have pairs over here {4, 8, {4,8}, {4,8,24}}
[2 4 | 8] 24]
[2 4| 8 24]

[reef thistle]

@cerulean ore Nope I don't know

[cerulean ore]

Even I don't

[reef thistle]

erm I don't know what you are doing

[cerulean ore]

If people like you are coming to the contest then it is better for me to be at home

[reef thistle]

Well, if you have stronger people on your team

you can distribute the problems so that the stronger people do the harder problems

[cerulean ore]

They're choosing me for Math

[reef thistle]

2 4 [8] 24
2 [4] 8 24
[2] 4 8 24
These are what the subarrays need to contain

[stray reef]

can i get the original problem statement please

[reef thistle]

https://www.codechef.com/SEPT19B/problems/FUZZYLIN

[cerulean ore]

These are the subarrays which doesn't contribute to the pair, right?

[reef thistle]

The solution is probably divide and conquer

there may be other solutions involving data structures that help compute gcd of a subarray quickly

[cerulean ore]

If K = 8 then 8 is the solution

[reef thistle]

??

[cerulean ore]

(0 indexing) (2,2)

[reef thistle]

hmm, wait we need to support Q queries of K?

[cerulean ore]

Yep

[reef thistle]

I guess would still be possible

[cerulean ore]

my gcd matrix computes all the GCDs at first step ;3

[reef thistle]

Hmm my solution is still too slow

You can use a sparse table or "segment tree" to get GCDs quickly

nah divide and conquer not working fast enough

[cerulean ore]

(0th indexing) (arr is 2 4 8 24)
So, how I have constructed the matrix is:
gcdMatrix[i][0] is the element from which we're currently counting our subsequence
That's why gcdMatrix[0][0] = 2, gcdMatrix[1][0] = 4, gcdMatrix[2][0] = 8...

I am using this gcd(a1,a2,a3,....,a(n-1), an) = gcd((a1,a2,a3,...,a(n-1),an)

So, at each step for gcdMatrix[i][j] where j>0, I am taking the previous term gcdMatrix[i][j-1] and my current term from the array and finding their gcd out

So, row 1 is :
2 | p = gcd(2,4) | q = gcd(p,8) | r = gcd(q,24)

@reef thistle What do you say?

[reef thistle]

I say spend more time coming up with the algorithm

because in ICPC it's either solve or not

wait until you actually have an algorithm to solve the problem

then code it

[cerulean ore]

is 9 the correct answer if K = 8

[reef thistle]

there are 9 subarrays here yes

but what you need to try to do is try to find an algorithm that works in general

[cerulean ore]

It must work in general

But, this as submission gives wrong answer ;_;

Where do you think this will fall?

Previously I was making N^2 matrix but, then I started dynamic array

[cerulean ore]

update: the wrong answer might be due to the slow computation of GCD

update2: what I was implementing is called DP ;-;

[severe path]

What’s the difference between $\forall a\forall b$ and $\forall a,b$?

[vital dew]

QuAnTuM:

[sharp mauve]

Is there a context? Seems the same for me

[distant hamlet]

Is there a quick way on solving large numbered factorials such as 680!/620!60! ?

[upbeat minnow]

Anyone got a really good source that explains how quantifiers and logical connectors work in equations?

I'm 4 weeks fresh in uni and totally friggin' lost with maths.

[sharp mauve]

Maybe try to find a book about how to write in math?

[upbeat minnow]

Somebody recommended "A mind for numbers" to me. I'm like near-dyscalculia levels bad in maths.

[faint narwhal]

Read How to Prove it by Velleman

[upbeat minnow]

Alright I'll put it on the list. Cheers.

[distant hamlet]

I’m guessing no one can help me out?

[west hedge]

you want to find the value of 680!/620!60!?

[distant hamlet]

Yes

And I want to know how

[west hedge]

well do you know how to cancel the 620! in the denominator?

[distant hamlet]

Yes

But I don’t know how these numbers won’t overflow the calculator

[west hedge]

wolfram is able to handle the product from 621 to 680

https://www.wolframalpha.com/input/?i=product_{n%3D621+to+680}+n

[distant hamlet]

I hope my professor lets me use that on the test lmao

[west hedge]

well I think if you have 680!/620!60! there is no way in hell you're expected to write it out in decimal form lol

[weary tiger]

733538391785775455343676371444874620941478691823229721225727410087430498353530681156000

If i did everything right

Or 2^5×3^3×5^3×7×13^2×17×23×37×61×67×71×73×79×83×89×97×107×109×113×127×131×157×163×167×211×223×311×313×317×331×337×631×641×643×647×653×659×661×673×677

@west hedge

[cerulean ore]

@upbeat minnow I was also struggling with something similar and "How to prove it" was recommended to me as well

I vouch for the book.

[stray reef]

who wants to try their hand at a silly proof problem i just came up with

Let $X$ be a set and let $A$ and $B$ be two nonempty disjoint subsets of $X$. Let $R$ be a relation on $X$ such that $R \subseteq A \times B$. Show that $R$ is transitive and irreflexive.

[vital dew]

Ann:

[faint narwhal]

@cerulean ore that was also me lmao

[reef thistle]

hehe, follows pretty easily

||Note if aRb, then a in A and b in B. Transitive because we can't have both aRb and bRc, then b in A and B, which is not possible as A, B are disjoint, so transitive. If aRa, then a in A and B, not possible as A, B disjoint, so irreflexive.||

[cerulean ore]

@faint narwhal It was you or @reef thistle ?

[reef thistle]

@cerulean ore What are you asking?

[cerulean ore]

Who recommended me "How to prove it By Velleman"

[reef thistle]

https://discordapp.com/channels/268882317391429632/496785905474994186/621036111342338050

I don't know that book so

[cerulean ore]

Do you remember last day's question?

Sequence of integers means eg 1,2,3..?

for [3 5 7] and K = 7
Possible arrays: [3] [3 5] [3 5 7] [5] [5 7] [ 7]
[3 5] is possible combination 34 -5
[3 5 7] as 30 + 5*0 + 7
[5 7]
[7]

[reef thistle]

yeah

[cerulean ore]

Now for, [2 3 5 7] my program is generating 7

which is the correct number

K=7^

this time I am not computing GCDmatrix

[cerulean ore]

@reef thistle Do you think the first answer will help?

https://math.stackexchange.com/questions/85830/how-to-use-the-extended-euclidean-algorithm-manually/85841#85841

[reef thistle]

probably not

because you don't need to calculate the coefficients

[cerulean ore]

There is no faster way

even after using two theorems the question still remains unsolved in lesser time

[reef thistle]

the problem is you need to solve it quickly each case

I think it's a good idea to sort the K

somehow

well, we want to find number of subarrays where gcd = x

[cerulean ore]

x%gcd == 0

if we sort then won't we lose the contiguous subsequences?

Can we use LCM*HCF = a*b?

[stray reef]

what are you trying to prove rn

@cerulean ore

[cerulean ore]

Not proving, I am trying to solve this question

@stray reef https://www.codechef.com/SEPT19B/problems/FUZZYLIN

[stray reef]

ok i'm out

[cerulean ore]

@reef thistle sparse tablemay work

as nlogn is construction time

[daring blade]

I was wondering if anyone can explain this to me

I'm not sure what's going on

[reef thistle]

@daring blade So do you understand how f is a subset of A x B?

[daring blade]

No I am unsure about that

[reef thistle]

Let's revise that

A x B is the set of ordered pairs where the first element comes from A and the second element comes from B

f is a subset of A x B, and it contains (a, b) where f(a)=b

@daring blade get it so far?

[daring blade]

Yes I understand that so far

So AXB in this case would be
(a,x),(b,y),(c,z)?

[reef thistle]

no

A x B = {(a, x), (a, y), (a, z), (b, x), (b, y), (b, z), (c, x), (c, y), (c, z)}

@daring blade

[daring blade]

Oh

Okay so it's like foil

I understand that so far

[reef thistle]

So, f is a subset of this

and if (a, b) is in the set f, then f(a)=b

[daring blade]

That makes sense, since it's a subset of the main set

[reef thistle]

So, enough for you to figure out what's going on?

[daring blade]

I'm not entirely sure what second coordinates are sorry to be a bother

From what I know it's the range

[reef thistle]

If the pair (1, 2) is in f, then f(1)=2

that you get?

[daring blade]

Yes

If I understand it properly:
im(f) is [x,y] since the y's are x,y
same goes with im(g)

But why is (gof) = [s,t]?

[reef thistle]

it's not

$\operatorname{Im}(g\circ f)={s, t}$

[vital dew]

Element118:

[distant hamlet]

I know this is easy, but I can't simply think right now. What's the set builder notation for {5,12,21,28,40,...} and {3,7,13,21,31,43,57,...}?

[stray reef]

what are those even meant to be

[copper ore]

sets lol

but what set is that even

[reef thistle]

@distant hamlet depends on how you define the elements in those sets

[stray reef]

well then

{5, 12, 19, 21, 28, 40} and {3, 7, 13, 19, 21, 31, 43, 57}

[stiff arrow]

Is there a good YT Channel to watch to better understanding of proofs and simple set functions

[faint narwhal]

Or how about you read a book

[stiff arrow]

I said better understanding

I have a book

[faint narwhal]

I'm not sure what you're looking to get out of a video that you can't get out of a book

[plain imp]

why isn't a single vertex graph eulorian ? The walk would be of length 0, beginning and ending on that vertex, making it a Eulerian tour. Since there are no edges to traverse, the condition "traverses every edge exactly once" would be vacuously true. What am I missing?

[faint narwhal]

What says that it's not eulerian?

[plain imp]

oh wait it says "at most one" nontrivial component, I read it as at least one 🤦🏼

[patent notch]

How would one reduce the statement A delta A in set theory? Or essentially finding the symmetric difference of a singular set?

[stray reef]

the symmetric difference of a set with itself?

that's empty @patent notch

[stray reef]

what are m, Δx and t

[stray reef]

the units don't check out at all

and your line of reasoning makes no sense at all

and this is also not the right channel

[marble zealot]

She wasn't being a jerk though, she was just pointing out the fact that there is a flaw in your reasoning.

[reef thistle]

Ann is quite direct to point out mistakes instead of sugar-coating them.

[stray reef]

@marble zealot i'm a she.

[marble zealot]

@stray reef sorry, my apologies

[faint narwhal]

It's just nonsense

what does the universe having absolutely nothing even mean

In strict mathematical terms?

And how would this imply that time is irrelevant?

What is time? And how does the universe not having anything (whatever that means) imply that time is irrelevant?

And as Ann said, your equation still makes no sense

The units don't even match up correctly

Is that enough for you @weary tiger

[faint narwhal]

This is still rambling nonsense

How do you know that if the stage has no actors then it may or may not exist at all

That makes 0 sense

The Vienna state opera house is empty right now, that doesn't mean we don't know for sure the building isn't still there

Also you're just using an analogy

How do you know that the universe actually works like a play

Just stop typing

You have no idea what you're talking about

Honestly don't bother

You pretty clearly have no education in either math or physics

You don't know how either of these fields work

Or how to rigorously make arguments that are actually testable in the real world

You don't even know how to check that your equations have matching units on both sides

You're just typing up ideas that make "intuitive" sense to you, by way of analogy

Without any rigorous modelling going on

[stray reef]

they left

[stray reef]

oh you're back

you're not really making any more sense

all of these points still stand

you were never on any track to begin with

[faint narwhal]

That's not how argument works

Hey

There's a teapot behind the sun right now

Prove me wrong

[stray reef]

left again

[stray reef]

oh you keep coming back don't you

[faint narwhal]

Behind as in from our perspective

[stray reef]

quit the theatrics

[faint narwhal]

So you can't prove me wrong right?

Then I might be right?

That this is exactly what you're doing

You're providing some statement

That would be completely impossible for us to test

Because you haven't given it a rigorous basis for us to test anything on

So it'd be impossible for us to prove you wrong

[stray reef]

inb4 he actually comes back in a few minutes

[reef thistle]

@shut basin If you think V makes sense, then can you express it rigourously?

This started with e/Δχ=t

I think the problem here is that V didn't add a proportionality constant

No, I'm a mere mortal

[mild flicker]

i'm having difficulty in proving hte logical equivalence of this

[ripe ferry]

the right side should probably be ( p and q ) implies r

not p and ( q implies r)

[mild flicker]

?

[mild flicker]

so that's wehre i'm a bit confused
The problem just gave p and q implies r
after I apply implication does it just remain p and !q or r without any paranthesis?

[stray reef]

@weary tiger ok so what was your question again

uh huh

why do you think 8C3 isn't right?

well, ok, let's put it that way: what is 8C3 supposed to represent?

how did you arrive at that number in the first place?

uh huh

yes

there are 8C3 ways to do that

and that's exactly right

it is

once you've picked 3 more people to join steve & danny's team, the other 5 automatically go in the other team

does it matter though

we have two teams sure

S&D's team

and the other team

otherwise they're indistinct

well idk

because if you swap the two teams' places, it doesn't really make for a different division

because who gets the ball first is decided upon right at the beginning of the basketball game itself

[weary tiger]

its 8C3

ann say 8C3 is the correct answer

[stray reef]

@weary tiger you're not helping.

[weary tiger]

sorry

[dire spruce]

Hello guys

I am new here

I have a lot of question about math is anyone want to be my firend's?

[reef thistle]

the math questions

[dire spruce]

:)

Those are personal questions dude actually i wan to asking in the tex

If i made a mistake inside of sentence sorry about that.. Also im studying for english. ✋

[olive dragon]

Hello any online resource for Discrete Math? like calculator solvers or something

[upbeat minnow]

I've tried to find some too but resorted to drawing a shit ton of truth tables

no wait thats logic not discrete

my bad

[faint narwhal]

Just read a textbook

[olive dragon]

I never seen math at this phase, it has been a nightmare so far

[atomic sapphire]

Hi could anyone help me with reflective, symmetric, and transitive properties? I still don't know how to prove them properly

[faint narwhal]

[atomic sapphire]

When I write it out in math terms would it be written this way?

[faint narwhal]

Uh no

[atomic sapphire]

How do I write it correctly?

[faint narwhal]

Like what's p, what's q

They've already written it correctly

[atomic sapphire]

All real numbers?

Oh I guess I can start proving something now

I know it has to do something with (p,p) but I don't know what to do after that

[faint narwhal]

What?

Why are you talking about real numbers

It doesn't mention real numbers anywhere

[atomic sapphire]

What does double Z x double Z mean?

Actually, I need a complete refresher what does the whole thing mean?

[faint narwhal]

You really should look in a textbook for this

It'd be pretty hard for me to teach you a whole lesson on set notation

[atomic sapphire]

Ok I'll need help while reading the book, so the C with the line through it means the relationship between an element and a set

[faint narwhal]

$\in$

[vital dew]

Zopherus:

[faint narwhal]

If you mean this, then kind of

[atomic sapphire]

Yeah that

[faint narwhal]

You really need to be specific when you say relationship between an element and a set

[atomic sapphire]

I'd assume the set would be on the left side and the element would be on the right side

Since a set is a collection of objects inside an element

[faint narwhal]

That is not what a set is

[atomic sapphire]

Wait a collection of objects is not the same as a collection of elements?

[faint narwhal]

Really depends what objects and elements mean

[atomic sapphire]

Ok so the left is what's inside the right of that set right? I'll give an example,
A = {9,3,8,7}
7 (c with a line) A

Basically, it means 7 is a number in the set of A?

[faint narwhal]

subset?

[atomic sapphire]

Oops I mean set

[faint narwhal]

7 is an element of the set A yes

[atomic sapphire]

So this type of notation is always the form element (backwards E) set?

[faint narwhal]

Yes

[atomic sapphire]

Ok I understand that part now

[faint narwhal]

vaguely

[atomic sapphire]

Ok I have an idea for how to prove that the relation R is reflexive

Back to this pic

So we pick numbers that would make this true which is (20,5)
It would be
(20-5)/(5)

15/5 = 3

And it would mean (20,20) would always be in set of all of the integers

[faint narwhal]

What do you mean by that last sentence?

[atomic sapphire]

I had used the definition of the book that the two x (in this case p) values are in the set of Z x Z

Here I found a better ss of the page

[faint narwhal]

What does (20,20) always being in the set of all integers mean

It's true that $(20,20) \in \mathbb{Z} \times \mathbb{Z}$

[vital dew]

Zopherus:

[faint narwhal]

If this is what you're trying to say

[atomic sapphire]

Yeah

I need to work on explaining proofs clearly

Ok so for Symmetric, if I flip p and a around and make it (5-20)/(5) it'll be -3 which is still true
Because (20,5) and (5,20) is also in the set of all integers

[faint narwhal]

Stop saying set of all integers

Set of all integers just means the set of integers

Which is not what you really mean

But yes, this would be how you show symmetry

You'd have to do it more generally, but

[atomic sapphire]

I'm sorry idk why I keep tacking on the word all

[faint narwhal]

Even without the all

It's not correct

(20,20) is not an integer

[atomic sapphire]

Wait let's go back a few steps

Double Z x Double Z means all of the coordinates right?

[faint narwhal]

That's one way to think about it

[atomic sapphire]

So just one Double Z mean integers?

[faint narwhal]

Yep

[atomic sapphire]

Ohhh so it wouldn't be symmetry because it's not inside the set of the Cartesian plane (I'm not sure if I'm using the correct terms)

[faint narwhal]

No I have no idea what you're saying

[atomic sapphire]

Dang now I'm confusing my self one sec I'll rethink this answer

Okay, this is what I'm saying

Ok (20,5) implies that (5,20) $ \in \mathbb{Z} \times \mathbb{Z}$

[vital dew]

Rising_Lamp:

[faint narwhal]

You should write

$(20,5) \in \mathbb{Z} \times \mathbb{Z}$ implies that $(5,20) \in \mathbb{Z} \times \mathbb{Z}$

[vital dew]

Zopherus:

[faint narwhal]

But yes

[atomic sapphire]

Oh yea your answer sounds more specific than mines

[faint narwhal]

Just saying (20,5) doesn't really mean anything

[atomic sapphire]

Yea I just realized that now

Ok for transitive, I'm completely lost

I don't even know where to start

[faint narwhal]

You haven't really shown symmetry yet

At least not fully

[reef thistle]

that's just one example

[atomic sapphire]

Wait we need more than one example?

[faint narwhal]

To show symmetry

You need to show it for all pairs

That if $(m,n) \in \mathbb{Z} \times \mathbb{Z}$, then $(n,m) \in \mathbb{Z} \times \mathbb{Z}$

[vital dew]

Zopherus:

[atomic sapphire]

So would I use variables or numbers to show the proof?

[faint narwhal]

You can't really just use numbers

There are infinitely many pairs in Z x Z

[atomic sapphire]

So the correct way to prove this is use what you used?

[faint narwhal]

Yes

You have to do it symbolically

[atomic sapphire]

You mean solving out (m-n)/(5) and (n-m)/(5) with algebra?

[vital dew]

euler/fermat:

[reef thistle]

It overcounts

because the 5th kid

if a boy, would overcount ways to pick boys

a boy could be picked by the 2 chosen, or the 5th kid

and the way a boy was picked would matter

Just add the 2 boy 3 girls and 3 boy 2 girls case together

[pale epoch]

in your suggested solution, the order of picking people matters

like

imagine you first pick boys A and B, whatever girls and pick boy C as fifth kid

this is counted as a separate case from picking A and C first, same girls as before and picking B as fifth kid

even though the resulting teams are the same

@weary tiger

you would need to divide by 3, yes

there are 2 cases

first case is you picked 3 boys

let's say A, B and C

you counted "A, B first, then C", "A, C first, then B" and "B, C first, then A" as separate cases

same for the case of having picked 3 girls

to account for this overcounting, you have to divide by 3

you're welcome

no

why?

[vital dew]

euler/fermat:

[gentle nebula]

Try with like 3 people in total, ABC, 2 together(AB), it’s obvious that the solution would be 2(if we ignore rotations):
ABC
BAC
But for your reasoning there there would be
ABC
BAC
CAB
CBC
then divide by 3, so 4/3

(it isnt a coincidence)
One way to count number of cases where you can rotate is just fix one person, so you don’t need to care about counting duplicates

Yes that's ok

Like you basically need now is 3!•number of permutations for 6 people in a round table

(uhhh idk any actly probably should ask someone else)

Maybe could try like math olympiad books on like combinatorics/counting, ik they do quite a bit of that

[gentle nebula]

uhhh

im not sure tbh

[steel valve]

simple question

the contrapositive of a+b is just a - b?

nvm

🤦

that makes no sense

contrapositive only applies to statements that are if/then etc.

[gray agate]

anyone know?

[faint narwhal]

Know what?

[wispy flicker]

I can't seem to find a server fully dedicated to logic proofs so I figure discrete math might be next best thing?

I ask because I'm trying to prove the following using just rules of inference

Can't use substitution. (So no De Morgans Law)

I got it halfway and can prove that (p and q) implies ~(p implies ~q)

but struggle to do so the other way around without De Morgans Law

[sour arrow]

Here's a trick. Try typing # and the the channels should appear. You can do a letter search. You can quickly find #proofs-and-logic this way.

But this channel also makes sense.

"Can't use substitution" does not imply you can't use demorgan's? Unless I'm misunderstanding. If there's another way to solve this, then you would also have to prove Demorgan's

And I'm pretty sure you can't prove Demorgan's without an argument on what NOT really is

[gentle nebula]

@gray agate idt theres a like purely cryptography channel here but basically the vulnurability usually lies in repeating the IV and the attacker being able to passively predict the IV

[gray agate]

@gentle nebula tnx but which channel?

[gentle nebula]

for like block/stream cipher kinda thing i think its closer to like discrete but it like its hard to really find a place where it fits

[wispy flicker]

@sour arrow sorry, I missed that channel.

In any case I managed to figure it out. But thanks.

[upbeat minnow]

Hi - I have a slight problem with very basic set theory. If A={1,2}, C={{1,2}}

is A=C ?

I'd say no but I'm not 100% confident

[faint narwhal]

What are the elements of A?

What are the elements of C?

[upbeat minnow]

There's, uh, a set of numbers containing 1 and 2 in A

And empty set and another set containing 1 and 2 in B

You can just remove the outermost {]'s right?

[frank robin]

@upbeat minnow you can’t

[upbeat minnow]

Oh

[frank robin]

@upbeat minnow in your case A contains two numbers and C contains 1 set

[upbeat minnow]

So they are different

[frank robin]

@upbeat minnow obviously

[upbeat minnow]

Alright thanks. I've just started and I'm really insecure about this stuff.

[frank robin]

@upbeat minnow two sets are equal if and only if for any x (x in A) <=> (x in C)

[stray reef]

@weary tiger it is perfectly possible to understand "logically" every use of the choose function

and exactly why nCk is equal to n!/(k!(n-k)!)

refusal to use nCk to save time and steps where it helps is just unnecessary obtusity

[steel valve]

@wispy flicker

I think this is how its done if you look at the logical equivalence table

Ping me if im wrong or so

[stray reef]

let the edge length of the cube be 1 for simplicity

consider what edge lengths your triangle may have

[reef merlin]

by SSS, triangles can be congruent

[stray reef]

1, sqrt(2), what else

why sqrt(3)/2

so what will it be then

the what

vertices are points, they can't "go through" anywhere

the space diagonal is sqrt(3).

anyway

1, 1, sqrt(2)
1, sqrt(2), sqrt(3)
sqrt(2), sqrt(2), sqrt(2)

that's it

this is 1, sqrt(2), sqrt(3).

sure of what

[void valve]

Hey all, could someone explain to me the difference between stirling numbers of the second kind and the stars and bars method?

[stray reef]

the triangle is a right triangle

i'm in class, so i can't watch this video

[stray reef]

the number of English professors is 2 iff the number of math professors is also 2, so moot point

[stray reef]

it's going to be the exact same calculation...

[upbeat minnow]

If I have A ∩ B where I have two complements, one above B and one above the entire thing, how does that work?

Like this

Oops had intersection instead of union but anyway

[stray reef]

wdym "how does that work"

you take the compliment of B, union that with A, then take the complement of what you get

[upbeat minnow]

I mean what's the order of operations

But yeah thanks you just answered :D

[stray reef]

the order of operations is completely unambiguous here

[upbeat minnow]

I tried going through my lecture notes and the online material we were distributed and didn't find it

[stray reef]

didn't find what

[upbeat minnow]

The order of operations

[stray reef]

there is simply no ambiguity here so no special conventions need to be specified

[upbeat minnow]

I'm like 3 weeks fresh into university math and 100% out of my head

[pale epoch]

what would your alternative order even be

[upbeat minnow]

I have no idea this is like the first time I came across complements in homework

And I second guess myself a lot because I'm super insecure about math in general

[pale epoch]

just always refer to definitions

you surely defined what a complement is somewhere

or a union/intersection

[upbeat minnow]

Yes

[pale epoch]

those definitions tell you exactly what you're working with

[upbeat minnow]

x∈C complement ⇔ x∉C in this case right?

[pale epoch]

what's C

[upbeat minnow]

whatever we're complementing in this case

I'm sorry I have to translate from Finnish to English with really unfamiliar terms and concepts

[pale epoch]

ok, yeah

[upbeat minnow]

Like I said I'm about 3 weeks into this stuff, haven't done any maths since '06 and super confused about most of it so if I babble nonsensical stuff I'll blame it on that

[pale epoch]

just in the case of your image you are looking at the complement of $A \cup \bar{B}$

[vital dew]

Lochverstärker:

[pale epoch]

then you can apply the definition of union, which tells you what that is

and apply definition of complement again, i guess

well, this is really basic set theory, i dont think it requires any special prior knowledge

you just have to adjust to doing math in general, which takes time

[upbeat minnow]

Yyyup

I've always been really bad at it too so it's been extra difficult for me

Applied for computer sciences and we have some obligatory math courses I'm hoping I can manage to get through

[pale epoch]

mathematics is extremely important for (academic) computer science

[upbeat minnow]

So I've been told and it's not helped with my self-confidence.

I'm pretty decent at coding stuff in general but maths has always been a wrench in the gears / to the face

[pale epoch]

well, you will probably encounter different kind of maths than in school now

[upbeat minnow]

Definitely

Also I've doubled the age and hopefully motivation to learn too so maybe it's better now that I'm older and hopefully wiser

Anyway thanks for the help.

[steel valve]

also helps to not take a crazy load

taking 6 classes of math and science is adding a lot of stress and sleepless nights

lol

and i still have work unfinished

😰

[hollow osprey]

hey could anyone tell me how this is or is not transitive?

i dont understand the book, it says that it has 3 variables but this has only 2 variables

so does that mean it's not transitive?

[west hedge]

it says that it has 3 variables but this has only 2 variables

what do you mean @hollow osprey?

what has 3 variables?

[hollow osprey]

the (p,q)

[west hedge]

by having 2 variables, you mean that R consists of pairs?

it says that it has 3 variables but this has only 2 variables
I'm still trying to understand what you mean by this

[hollow osprey]

yes

[west hedge]

ok

what in the book suggests that R should have 3 variables?

[hollow osprey]

the x,y,z in it

[west hedge]

you have to give less cryptic answers

[hollow osprey]

its specifically this part of the book

[west hedge]

right that's the definition of what it means for a relation R (which consists of pairs (a,b)) to be transitive

[hollow osprey]

maybe im just not reading the book correctly

[west hedge]

but I don't see anything that suggests R should "have 3 variables"

so I'm not exactly sure what's confusing you

if you understand this definition then we can move on to deciding whether the particular relation R (given by (a,b) in R if a-b is a multiple of 5) is transitive

[hollow osprey]

i understand that part

[west hedge]

well in your initial post you said R is not transitive

or you suggested that it may not be, so it seems like you were unsure

so I guess you'll have to say which part is still confusing you, if any

[hollow osprey]

ok its this part where it is (p,q), where in the book, i see (x,y) and (y,z) and (x,z)

but i dont know where the z came from

[west hedge]

but it's just another letter

if you understand what it means for (p,q) to be in R, surely you understand what it means for (y,z) to be in R

[hollow osprey]

so this means that the letter z is made by the person and not by the book?

[west hedge]

ok how about this
let's look at a concrete example

I think you will understand it better

if x < y and y < z, is it true that x < z?

[hollow osprey]

yes

[west hedge]

right

are you confused in any way about the way that I used x, y, and z just now?

[hollow osprey]

im good with it now

[west hedge]

so you understand the use of x, y, and z in the definition of transitivity for any R

[hollow osprey]

the definition yes, the proof probably not

[west hedge]

ok

[hollow osprey]

sorry i mean not for the proof part

[west hedge]

so now we should decide whether R is transitive, where R = {(x,y) | x, y are integers with x - y a multiple of 5}

this means (x,y) is in R if there is an integer m with x - y = 5m

[hollow osprey]

could i plug in numbers for (x,y)?

[west hedge]

they are numbers, yeah

[hollow osprey]

when i prove stuff are the actual proofs in numbers or just variables?

[west hedge]

well if you want to prove that R is transitive you need to show that (x,y), (y,z) in R implies (x,z) in R for every x, y, z

if you do it for just one choice of x, y, z you aren't proving very much

[hollow osprey]

ohh because x,y,z are infinite numbers and i cant use all of them using just numbers

[west hedge]

it's not going to be any easier to prove if you just look at one case at a time

so assume x - y = 5m and y - z = 5k
and why don't you think about whether this implies that (x,z) is in R

[hollow osprey]

it has to be in R because if you switch it backwards it'll be a negative number but still a multiple of 5

[west hedge]

I don't think you understood the question
but what you just said sounds like a good proof that R is symmetric

[hollow osprey]

i only understand reflexive and symmetric but i don't quite fully understand transitive yet

[west hedge]

ok let me make this as straightforward as possible

assume that x - y = 5m and y - z = 5k

is x - z a multiple of 5?

[hollow osprey]

yes because it has 5 in it

[west hedge]

x - z?

[hollow osprey]

no not that, its where x - y = 5m and y - z = 5k so we would have to assume x - z is a multiple of 5 because both x and z is inside the two equations that has a multiple of 5 in it, idk i explained this correctly

[west hedge]

that's not very convincing

if x - z is a multiple of 5 then you should be able to write down an equation that shows me that it is the case

[hollow osprey]

ok im starting off with this, (x-z) = (x - y) + (y - z)

im not sure how to do the rest of it

[west hedge]

well we are trying to prove R is transitive
to do that we assumed two things (x,y) in R and (y,z) in R

you haven't used those assumptions yet

[hollow osprey]

how do i make that c with a line through it?

the set symbol

[west hedge]

"in"

[hollow osprey]

i think i should skip this for now and move on something that i could possibly answer

[west hedge]

@hollow osprey there is no reason you can't answer this, so like you say maybe you just need to come back to it later and look at it again
feel free to ping me if you make progress or continue to get stuck

[cerulean ore]

Started graph today

Looks good and heavy

[stray reef]

you're overcounting the cases where you have no bad coins

each one gets counted thrice, once for each bad coin that you happened to not choose

||11C3 (no bad coins) + 3C1 * 11C2 (exactly 1 bad coin)||

[stray reef]

wdym

is that so?

oh yeah

yes

of course

we're choosing 6 coins

i mean

you corrected it

i was under the impression that we were done

[vital dew]

euler/fermat:

[stray reef]

...ok but why add them

yeah no this is just complete bogus i'm afraid

and i just know you're going to ask me to specify exactly why it's bogus and i'm going to preemptively say i cannot do that

[pale epoch]

lol

and i just know you're going to ask me to specify exactly why it's bogus and i'm going to preemptively say i cannot do that

[steel valve]

@west hedge do a direct proof

Nvm ignore me lol

[sweet eagle]

can anyone point me into the right direction?

i'm not q uite sure on how to start a proof with arbitrary intersections.

<@&286206848099549185>

[modest zealot]

is the weird upside down u a negation?

or big upside down u

[long matrix]

is this the right channel to discuss about graph theory?

[sweet eagle]

@modest zealot no that is the an arbitrary intersection

hm, maybe this isn't exactly discrete mathematics

but i'm not sure what other channel i could put this on

[hollow osprey]

hi could anyone help me understand the solution please? i know how to get the equation but i dont understand whats after the equation

should i assume that its the last order without a number in front that is the answer for every problem like this? (the (3n^2)log(n))

wait nvm i know how the middle is calculated now, its by taking ratios

[sour arrow]

@sweet eagle
There's something extremely weird about the notation. You can't really take the arbitrary intersection of a single set. Maybe if A and B were a family of sets?

Anyway, remember the definition of equal sets, you have to show they're each a subset of eachother. Start with an element x that is in the left side, show that it's also in the left. Then, do the opposite direction

[sweet eagle]

@sour arrow yeah i figured it out

its exactly what us aid

but yah

big intersection notation means arbitrary intersection

like if A = {{a}, {a, b}, {a, b, c}}

the Arbitrary intersection of A would be {a}

but yah thanks i wish i read ur suggestion sooner

[static rapids]

hey guys im having trouble completing this question

[faint narwhal]

What have you tried?

[static rapids]

i started off with

there is an x such that (-2 >= x >=3)

but that doesnt make sense

im thinking of single inequalities

how do i wrap my head around these ones?

[faint narwhal]

You should think about what the negation of a statement means

[static rapids]

so the original statement is... for every x, x is between -2 and 3

and a negation of this is... there is an x such that x is not in between -2 and 3?

does that make sense?

[long matrix]

i think you have to do it with quantification

[static rapids]

yea i would the not symbol and the "there is an x such that symbol" but i cant rn

[long matrix]

rn?

[static rapids]

i mean im on a laptop

dont have those symbols...

[long matrix]

ah

but you dont need not

am not sure its a long time i did this kind of stuff

but you jsut need the E quantificaiton

[static rapids]

im just confused on what the inside of the bracket will look like

yea the backwards E

but rn i dont understand what to do in the brackets with the inequalities and numbers

[long matrix]

you either change the numver 3 and -2 or do the oppisite of the bigger smaller symbol

[static rapids]

i did the 2nd option but it doesnt make sense

-2>= x >= 3

there are no numbers like that

[long matrix]

you are right

[static rapids]

do i change the sign of the numbers and the big/small symbols?

[long matrix]

maybe you can just assume that (x>=3)

i hope someone can help with my problem

do i have to take the longest augmented path when i try to find a matching? or is it up to me which AP i take?

[static rapids]

so i can do Ex(x>=3) or (x=<-2)?

[long matrix]

no i would take or

[static rapids]

that good?

[long matrix]

like i said am not rly sure

[static rapids]

thats fine

im still trying to figure this out

@faint narwhal i tried to think about what the negation means

was i valid or no?

[faint narwhal]

Yeah that's right

For negations, only one of the statements can be true

If you look at the statement you wrote down and the one you started with, you can see that they're oppposites of each other

[static rapids]

so does opposite = false in terms of this question

or do i have to find an untrue statement for the original question

[long matrix]

i wouldnt say false

[weary tiger]

im sorry idk where this fits http://prntscr.com/p8dax8

[faint narwhal]

It means only one of them can be true at a time

[weary tiger]

but can someone explain this to me

[faint narwhal]

If the original statement is true, then the negation can't be true

And vice-versa, if the negation is true, then the original statement can't be true

[static rapids]

but the original statement is true

so the negation is false but idk how you would put that in terms of symbol form

and without the negation symbol

[long matrix]

Ex((x>=3) v (x=<-2))

[static rapids]

would that be the negation?

[long matrix]

i would do it like that

[static rapids]

why the or symbol

[long matrix]

it cant be and symbol

x cant be bigger then 3 and smaller then -2 at the same time

[static rapids]

ooh

@faint narwhal do you agree with him?

[faint narwhal]

Yeah he's right? You wrote or earlier

why would you be questionable

[static rapids]

cause my prof was stating something about how you needed and's

thats why i was conflicted

[long matrix]

just do waht you think is right and explain it to him. if hes a good guy he will explain it to you

[static rapids]

alright thanks guys

ok i have a couple more questions that i didnt complete from my sets

I understand a. just not b.

i need to find a value of y that is also compatible with x in order to counter this argument

thats what i found out so far

[lethal sequoia]

@static rapids Try solving for y in terms of x and see if there are any solutions

[static rapids]

ok thank you

ok so i got y< 10 +sqr x

what do i do now

[lethal sequoia]

rememeber $\sqrt{100+x} \neq

$\sqrt{100+x} \neq \sqrt{100} + \sqrt{x}$

[vital dew]

Kei:

[static rapids]

ah ic.

so can i just leave it as sqrt(100 + x)

[lethal sequoia]

yup so can you find some value of x where there is no solution for y in the integers?

[static rapids]

negative integers?

[lethal sequoia]

yup can you find a specific negative integer?

[static rapids]

numbers larger than 100

does that mean i choose any or..

[lethal sequoia]

yeah can choose any x larger than 100

less than -100*

[static rapids]

wait less than -100?

oh yea negative wise

so after i choose the number can i just state y<sqrt(100+(-101)) as my answer?

as a counterexample

@lethal sequoia

[lethal sequoia]

yup

may want to explain why it has no y solutions in the integers

depending on how its marked

[static rapids]

oh yea

thank you

[lethal sequoia]

no problem!

[lethal sequoia]

You could try h = x^2

can you find a f,g?

[faint narwhal]

@weary tiger What have you tried?

[weary tiger]

I got it now thanks 🙂

[faint narwhal]

Which one are you confused about?

[static rapids]

oh i wanted to verify if my answers are correct

cause im not sure about e)

[faint narwhal]

d isn't correct either

[static rapids]

oh yea

its vx

whats wrong with e)

[faint narwhal]

Either wasn't really correct

e is correct

Do you understand why?

[static rapids]

yea cause if you want to negate that original statement it now has to change its domain

but for d) idk why

its wrong

[faint narwhal]

No you fixed it

It's just \forall x

[static rapids]

oh, thank you for checking

[sand moth]

I'm a n00b

List all the subsets of A
𝐴 = {𝑥|𝑥, 𝑘 ∈ ℕ, 𝑥 = 𝑘^3, 𝑘 < 4}

Would this be {}, {0}, {1}, {0,1}?

[faint narwhal]

Nope

[static rapids]

im so lost during proofs of these questions

[faint narwhal]

What are you confused about?

[static rapids]

the concept of onto and one to one with combination function

im thinking of adv func

which isnt helping

[faint narwhal]

I'm not sure what combination function means

Or adv func