discrete-math

[weary tiger]

it has to add up to all natural numbers

$A \cup B = \mathbb N$ means there are exactly the same elements in $A \cup B $ as are in $\mathbb N$

[vital dew]

Godel:

[balmy nacelle]

wait

maybe im confused on natural numbers lol

are they not infinite?

[weary tiger]

yes they are

[balmy nacelle]

so then how would you make 2 sets that target infinity

infinite sets right?

[weary tiger]

yes, at least one has to be infinite

[balmy nacelle]

so something like

A={1,2} B={3, 4, ...}?

[weary tiger]

yes

or, A = even, B = odd numbers

[balmy nacelle]

wait is 0 a natural number?

[weary tiger]

yes

for me it is

[balmy nacelle]

so A={0, 1, 2}, B={3, 4, ...} is a valid solution?

[weary tiger]

or $A= \mathbb N , B = {1}$

[vital dew]

Godel:

[weary tiger]

yes it is

your solution and the ones I mntioned are all correct

try to verify that the ones I mentioned work

[balmy nacelle]

Give three sets A, B and C such that A ∪ B ∪ C is an infinite set and A ≠ B ≠ C.

A={1,2,3} B={4,5,6} C={7,8,...}?

does one infinite set make that work?

[weary tiger]

add 0 seomwhere

[balmy nacelle]

Since the set ℤ contains both positive and negative numbers, then the cardinality
of ℤ is greater than the cardinality of ℕ.

this is true, right?

[pale epoch]

no

[balmy nacelle]

why's that?

[pale epoch]

how would you define cardinality for infinite sets in the first place

well, essentially we say that 2 sets have the same cardinality if there is a one-to-one function between them

and there is such a function from N to Z

[balmy nacelle]

Ahhhhh that makes sense, I understand because infinite sets

what about this one, is this false?

Since the function 25n2 + 800n is in O(n2
) then 25n2 + 800n is also in O(n3
).

that pasted weirdly but it's n^2 and n^3 and whatnot

[shut fjord]

with this induction problem, how does the solution jump to k/(k+2) on the first line of the inductive step on the RHS?

The Inductive Hypothesis should be: 1/(k[k+1]) = k/(k+1)

wouldn't this be plugged into the k+1 induction setup?

There's some algebra going on that I can't see..

[shut fjord]

so its a typo

[obtuse lance]

pretty broad question

I guess main two are injective and surjective if that's what you're asking for

[zealous mantle]

you can always graph it and see

[weary tiger]

"A group is composed of 5 women and 8 men. How many committees of 5 people can be formed who have a male president and a female vice-president ?"

I did $\binom{8}{1} * \binom{5}{1} * 3!$

[vital dew]

total_sleezeball:

[weary tiger]

So I got 240 possible committees as an answer, is this correct ?

[robust mango]

5 women and 8 women

fix it pls 😄

@weary tiger It's 8c1 * 5c1 for choosing the president and vice-prez and then you need to choose 3 ppl from the remaining 11 people which is 11C3

You multiplied by 3! as if you're arranging 3 people but you only chose 2 at the start, this is wrong

[weary tiger]

Okay I gotcha, thanks for the help 🙂

[weary tiger]

@robust mango how about this one :

"How many trains of 8-bit must we generate to have atleast 4 whose first 3 bits are the same?"

By a train of bits I mean this: 101010101, for example.

I did (2^8 - 2^5) * 4 = 896

2^8 total number of bits, 2^5 number of trains whose first 3 bits are equal

[weary tiger]

nvm solved, it's 3 * 8 + 1

[lyric pumice]

@weary tiger Hello. Pardon the delay.

Suppose n distinct items have been arranged in a circle. Selecting k non-adjacent items renders n-k items non-selected. The non-selected items are separated into k parts by the selected items. Since there are n-k-1 spaces in between the non-selected items when they are arranged in a line, it follows that the number of ways to select k-1 of those spaces is n-k-1 choose k-1, which gives the number of ways to select k non-adjacent items from a line of n identical items so that the item at a particular end of the line is selected. If the items in a line are considered to be those items that have been arranged in the circle, then there are n ways to create a line so that the order of the items as they appear sequentially in the circle is retained since a sequence of items in the line depends upon the first item in the line and there are n different items that can appear first. Consequently, any k particular items can appear in k different orders for a given sequence of items in a line since the sequence of those k items depends upon the item that appears first among them in the line and there are k different items that can appear first. Therefore, the number of ways to create a line of items from the circle so that the order of the items as they appear sequentially in the circle is retained and then choose k items so that no two of them are adjacent and one of them is the item at the end of the line is (n-k-1 choose k-1)n/k, which gives the number of ways to select k non-adjacent items from n distinct items arranged in a circle.

[lyric pumice]

@weary tiger Consult me for further clarification.

[desert edge]

When you want to talk about the degree of a vertice, how would you write it?

If my vertice has a degree of 2, how would the notation look like?

it says d(V)

[pale epoch]

you just have a degree function from the set of vertices into the set of naturals

how you call that function and the name of the vertex are ultimately arbitrary

[desert edge]

How would the notation look?

[pale epoch]

if v is a vertex, d(v) is fine

or deg(v)

[desert edge]

so if the degree is 2, do I just write d(2)?

Im confused about the notation

[pale epoch]

no

if v is a vertex and the degree is 2

then deg(v) = 2

[desert edge]

ok so I just write d(V)=2

[pale epoch]

yes

[desert edge]

So if Im looking at lets say vertice 2 in my graph

[pale epoch]

that is the shorthand for "the vertex V has degree 2"

[desert edge]

i write d(V_2)=2

Ok another question I have. Whats the difference in notation with {} and ()

One indicates is directed right? ()

[pale epoch]

you have an example?

[desert edge]

Dont mind the danish

Im introducing the subject

So im starting from the basics

Just clearing up some confusion I have

[pale epoch]

{} is just a collection of things

order or multiplicity is not important

it's what is called a set

() is a tuple

so basically a list

order is important

and whether or not the same element appears multiple times

[desert edge]

Ok so its wrong to write it like that

Since its not directed

[pale epoch]

what is wrong?

[desert edge]

My graph is not directed, so the order is not important

Thats not how I wrote it before

[pale epoch]

oh, yeah

except

[desert edge]

This is how it looked

[pale epoch]

it should be {V_1, V_2}

elements in a set are separated by commas

and {V_1, V_2} = {V_2, V_1}

because order is not important

your graph only has 1 edge

[desert edge]

Hm

Its because the terminology pdf im using is saying

V_i and V_j is v_i v_j

didnt know you seperated them by commas

[pale epoch]

is this talking about edges or paths?

[desert edge]

Edges

actually wikipedia says youre correct

[pale epoch]

i mean, notation is not universal

[desert edge]

They are seperated by commas

[pale epoch]

the standard notation would be {v_i, v_j} is the edge connecting v_i and v_j

and E is then the set of all edges

but in theory your prof could define edges differently and say v_iv_j is the edge connecting v_i and v_j

but then E for your example would be {V_1V_2}

and honestly that would be dumb

[desert edge]

Yeah I can see that

Glad to have it clarified atleast

[weary tiger]

pigeonhole principle/ dirichlet's box principle

[lilac pivot]

can think of it as a string 7 long, with one part of the string = "111000" and the other 6 parts are either 0 or 1

[lyric pumice]

@iron crescent Using a certain formula, you get 26-14+1 choose 14, which is 0. Then 26 choose 14 is greater than 0.

@weary tiger

[near prawn]

use the definition

oh by 1:1 you meant one to one mapping?

well I'd check both conditions separately

the injectiviity part is straightforward

no

you can have g o f being surjective without f being surjective

[tired axle]

Could anyone help with a fucking difficult discrete geometry problem

[dawn loom]

Could anyone help with a not so difficult discrete math question involving permutations

[robust mango]

just ask guys

dont ask to ask, you're wasting your own time

post the question, someone will see it and reply

[dawn loom]

Alright I have 12 friends to invite to a BBQ, I can only bring 10 though

Two of them are married and must come together

How many ways can I invite 10

Im thinking 10 C 8

quite unsure though

[robust mango]

Yeah it's 10C8. You can also think of it as 2C2 * 10C8

[dawn loom]

But how about if I dont invite the two

at all

so its just 10 c 10

[robust mango]

Yes

[dawn loom]

Yeah but does this account for that?

[robust mango]

can u show the exact wording of the question

[weary tiger]

I have 6 shirts, 6 pairs of pants, and 6 hats. Each item comes in the same 6 colors (so that I have one item of each color). I refuse to wear an outfit in which all 3 items are the same color. How many choices for outfits do I have?

does anyone else think this question is extremely ambigous?

[dawn loom]

I have 12 friends I would like to invite to a barbaque but can only accommodate 10. How many ways can I invite ten if: Two of the friends are married so If I invite one I have to invite the other

[robust mango]

what's the other part?

[dawn loom]

No other part?

Just a. there are no restricitions

c. Two of my friends do not like each other so if I invite one, I cannot invite the other

Thats what im saying

im confused

Yeah

M ig its just 10 C 8

Now about this second part

Two of my friends do not like each other so if I invite one, I cannot invite the other

How do I go about this

without

no more married couple

they divorced

[robust mango]

you can invite A and leave B or invite B and leave A or leave them both

[dawn loom]

theyre the two friends that dont like each other

Sooo

is it

2 C 1

2 C 1 * 11 C 9

[robust mango]

There's 10ppl + 2(AB). To invite 10 ppl, we have 2C1 * 10C9

it's not 11C9 it's 10C9 because you removed AB from the 12

[dawn loom]

Oh

Thanks man

[robust mango]

you could leave them both

so now you have 10C10

[dawn loom]

so I need to account for that

?

[robust mango]

Yeah

[dawn loom]

So just add one?

Oh so for the previous problem the married couple, is it. (10!/(8!2!)) + 1

[robust mango]

for the married couple you said it was necessary to bring them?

so it's 2C2 * 10C8

[dawn loom]

No

[robust mango]

well if it's not necessary you coudl leave them

[dawn loom]

Yeah

[robust mango]

and add the other possibilities yes

[dawn loom]

Ok

so its just add 1 then

[robust mango]

Yeah

[dawn loom]

To the possibility that I did invite them

Alright awesome, thank you

[lyric pumice]

@weary tiger Figure out how many outfits are possible such that the items of clothing that you are wearing all have the same color.

[desert edge]

Can someone help me understand this.
If a town has an odd number of bridges, then there's an easy way to figure out whether you can make the journey or not: If the sum of the number of times each letter must appear is one more than the number of bridges (like the eight-letter solution we noted in the first paragraph of this section), you can make the journey. If the sum is greater than that, then you can't

Euler tried to make a journey across the bridges and he couldnt, cause the sum was 9 of the ltters and not 8

Ohh nvm.. Im stupid

I just realised

ffs. ill try spelling it out next time

[weary tiger]

can I say that?:

$A \subseteq A \cup B$

[vital dew]

nxue:

[stray reef]

you can say that, sure

an even stronger statement holds: you can say that and be right

[weary tiger]

lol

thanks

for K(10) and K(5, 10) are the chromatic numbers both 2?

also how do you determine chromatic number for C(20)

[pale epoch]

is C(20) supposed to be the cycle graph on 20 vertices?

also the answer to your first question is no both before and after you edited it

[weary tiger]

well isnt k(5,10) 2 for chromatic?

yes C(20) is cycle graph

[pale epoch]

a graph is bipartite iff it is 2-colorable

but that was not the question you asked

determining the chromatic number of a cycle graph should be easy, where is your problem?

[weary tiger]

its 2 right its the even odd rule?

3 if odd 2 if even

for the cycle graph

[pale epoch]

i don't know, is it?

[weary tiger]

yes

now

k(10) 🤔

[pale epoch]

i mean complete graphs are also very simple

just start coloring it and see how many you need

[weary tiger]

so the pattern is chromatic number = n so in this case 10 would be needed?

nice it is

ok thanks for the help!

[rigid tree]

@iron crescent well when you think of the operations you have on sets the problem can be simplified

can you add some values a + b where the result is not a number

[empty nest]

@iron crescent As I remember it from my class, countable union of countable sets is countable... so no, that's not possible

[vital dew]

N/𝔄:

[weary tiger]

@iron crescent

It should say surjection not injection mb

[rigid tree]

@weary tiger explain this what is this latex bot

[weary tiger]

if a company has 3 director roles: chairman, ceo, and manager, and 40 total staff, then how many ways can we fill those roles if john will not be director if josh is a director?

i am trying to find a solution by just reasoning and not too many formulas if possible

[stray reef]

uh

if john will not be director if josh is a director?
what's up with this condition

oh

[weary tiger]

idk

it just is

lol

[stray reef]

nevermind it's just phrased weird

so john and josh aren't allowed together on the board of directors

[weary tiger]

yeah but are individually

(or not at all)

[stray reef]

yes

without the john/josh restriction there are 40 * 39 * 38 possible ways to fill those positions

[weary tiger]

yup

[stray reef]

while forcing the two to be together makes 3 * 2 * 38 possibilities that are to be excluded

[weary tiger]

why 3 * 2 * 38?

[stray reef]

john can go in one of three positions

josh in either of the two that remain

and the last position is filled by any of the other 38 people

[weary tiger]

what about 1 * 1 * 38 * 3!, since 3 ways to "arrange" the roles

(i know it gives the same answer but i'm asking about the logic)

@stray reef also are there any ways to solve this using casework?

[stray reef]

since 3 ways to "arrange" the roles

well... ok i guess you could like. line up john and josh with another person and then line up the three roles in a row in one of 3! ways

anyway if you want casework ig you could do like

john only, josh only and neither

[weary tiger]

what would that look like

38 * 37 * 36 + 3 * 38 * 37 + 3 * 38 * 37 maybe?

i think this makes sense

the only part i'm unsure about is the 3 part

since we have 38, 37 choices of people for the other two multiplications

but we only have 1 choice of person

but we multiply by 3 anyway..

[stray reef]

3 for the number of positions said person could be in

[weary tiger]

yeah not sure about that

we don't do that for the 38 and 37

like they have 2 * 1 choices as well

[stray reef]

we don't bc we just fill the positions that remain one by one in order

[weary tiger]

how to change my late(ks) font

what do you mean by that

my eks key aint working

one by one in order

[stray reef]

i mean exactly that

there are two positions left and they must both be filled

[weary tiger]

yeah but there's 2 * 1 ways to fill them tho

[stray reef]

no

there are 38*37

and besides, does it matter if you first give martha the chair role and then give steve the ceo role, or give steve the ceo role and then give martha the chair role

[weary tiger]

no? but then why do we multiply by 3

the last person only has 1 choice after martha and steve have their roles

[stray reef]

the last person as in john or josh?

if you're assigning martha and steve their roles first then you have to choose which two of the three roles they're gonna get

you're overthinking it

[weary tiger]

idk i am confused by this simple thing 😦

@stray reef can you unconfuse me please oh god

[stray reef]

ok so say you're doing the case where you're going with john but not josh

start with all three roles being empty

give john one of the roles

in 3 ways

then of the two roles that remain

pick one of the 38 people remaining for it

and then for the last role

pick one of the 37 people remaining

[weary tiger]

but this doesn't address my concern of not multiplying the remaining people by 2 * 1

[stray reef]

why would you need to multiply by that

[weary tiger]

why not?

that's how to arrange the offices

or to assign roles

w/e

[stray reef]

the offices are not arranged

[weary tiger]

assign roles

[stray reef]

...

i think you're misunderstanding something here but i have zero idea how to get to you

[weary tiger]

😦

don't give up

please

i am hopeless so much anxiety from this one problem

if anyone else would like to give it a try...

[stray reef]

okay maybe calm yourself down then if you're anxious

[weary tiger]

i'm just upset that i'm not getting it

i can feel that it's a simple thing

like

idk why i'm not getting it

[faint narwhal]

That's not right because that doesn't contradict the statement

That's an example of the statement working

[weary tiger]

@iron crescent the issue is that idk why you don't multiply by 2 * 1

even but you multiply by 3 for the 3 choices for that 1 guy

lol

[weary tiger]

yes

i am pretty sure ann is a woman

maybe

can you send me that worksheet

[stray reef]

oh the combinatorics worksheet? one moment

i should recompile the answer key for that actually

or find where i've got it written down on paper

[robust mango]

@weary tiger After the base step and assuming that this hypothesis holds for some integer k where k>=1, we have sum of the first k cubes = k^2 (k+1)^2 / 4. We need to prove this holds for n=k+1. You can add the (k+1)th term on both sides, which will be (k+1)^3. LHS becomes the sum of the first (k+1) cubes which is what we needed. Simplify RHS.

[lean basin]

Can anyone help me with the following two problems:

[robust mango]

@weary tiger Yes but the third step is wrong, you have to prove it's true for n=k+1. If you put n=k+1, one should get (k+1)^2 (k+2)^2 / 4.

You have k^2 (k+1)^2 / 4 + (k+1)^3, you didn't even simplify it properly

[vital dew]

Sup?:

[robust mango]

\begin{align*} \sum_{n=1}^{k+1} n^{3} & = \frac{k^{2} \cdot (k+1)^{2}}{4} + (k+1)^{3} \\ & = \frac{k^{2} \cdot (k+1)^{2} + 4(k+1)^{3}}{4} \\ & = (k+1)^{2} \cdot \frac{k^{2} + 4(k+1)}{4} \\ &= (k+1)^{2} \cdot \frac{k^{2} + 4k + 4}{4} \\ & = (k+1)^{2} \cdot \frac{(k+2)^{2}}{4} \end{align*}

[vital dew]

Sup?:

[zealous summit]

@lean basin is "polynolial" a technical term, or...?

[brisk osprey]

I've been trying to understand this algorithm problem that asks about giving a theta notation for the # of times a statement is executed in an algorithm, but it's been giving me trouble

[pale epoch]

i mean you have to count how often x = x+1 is executed and its in 3 nested for loops

every time the inner for loop runs, it is executed once

it runs $\sum_{k=1}^{2j}1$ times

[vital dew]

Lochverstärker:

[pale epoch]

this for loop runs for every iteration of the middle for loop

and the same argument applies to the outer loop

which gets you the solution

so "x = x+1" get's executed $\sum_{i=1}^n(\sum_{j=1}^{10i^2}(\sum_{k=1}^{2j}1))$ times

[vital dew]

Lochverstärker:

[pale epoch]

@brisk osprey

[brisk osprey]

@pale epoch I appreciate it

[zinc pewter]

For this step I don't understand how my prof turned -20x to 20x

it's absolute property

but I don't see how it could happen when real numbers have to be greater or equal to 0

since only negative numbers could make it positive

any help would be appreciated thanks

[craggy gale]

when $x\geq 0$, you do have $-20x\leq 20x$

[vital dew]

Tuong:

[craggy gale]

there's nothing much to say about this

your prof could as well have written
$$3x^2-20x+100\leq 3x^2+100\quad\text{when }x\geq 0$$

[vital dew]

Tuong:

[lean basin]

@lean basin is "polynolial" a technical term, or...?
@zealous summit No, it is a typo, it's supposed to be polynomial

[little nacelle]

what the hell is this

[lyric pumice]

@little nacelle Prove Pascal's identity.

[lyric pumice]

@minor dagger Chinese Remainder Theorem

[zenith saddle]

Could anyone help explain this:

[faint narwhal]

Do you know what the chinese remainder theorem says?

[zenith saddle]

yeah, if a and b are relatively prime then for all m,n:
x congruent m mod a
x congruent n mod b

[faint narwhal]

That's not a statement what

[zenith saddle]

my book has this:

[faint narwhal]

Yeah that's a statement

Well try to use this

you're proving that the function is a bijection

so show that it is injective and surjective

[zenith saddle]

ok, I think this notation is throwing me off:

I'm not understanding what exactly that means

[faint narwhal]

Yeah this is pretty poorly written

the first one is supposed to be 0,1,2,..., ab-1

like all the integers from 0 up to ab - 1

and the same for the others

[zenith saddle]

ok, that makes way more sense already

[faint narwhal]

it should be 0,1,2,...,a-1

[zenith saddle]

got it

then how is the [0,a) x [0,b) read? Does that mean the function implies you multiply each pair?

[faint narwhal]

Do you know what cartesian product of sets is

[zenith saddle]

kinda, thats like a cross product?

[stray reef]

no, it has absolutely positively nothing to do with the cross product from vector algebra

and there's no "kinda". you either do or you don't

[zenith saddle]

ok, then no I need to study that.

ok, thank you. After looking up cartesian product I think I understand all the notation now.

[charred cargo]

hey

mRn ↔ m|n)

This means that the relation will be true if n divides m evenly?

[pale epoch]

what

what does it mean for a relation to be true

sick leftrightarrow though

[charred cargo]

that the the m and n are related

i mean

@pale epoch

[pale epoch]

then you're right

wait no, it's the other way around

m|n is "m divides n"

[charred cargo]

@pale epoch

is this asking me to make the equivalence relation that includes this set

as a class

[pale epoch]

I don't know what that means

but also no

this wants you to craft an equivalence relation such that this is the set of the equivalence classes

look up what an equivalence class is and this should be easy

[covert plinth]

I think it may be x~y <=> x+y=5 or x=y

[obtuse minnow]

hmm but 5~5 is not 5 🙂

nor is 3~3

nvm I wasn't paying attention

[weary tiger]

True for $n=2$ trivially. Let $G_n$ be a graph of order $n$. Let $G_{n+1}=G_n+x$ Then $$d(G_{n+1}V_1\cup x)+d(G_{n+1}V_2 \cup x)=d(G_{n+1}(x))$$ W.L.O.G. $d(G_{n+1}V_1\cup x)\leq d(G_{n+1}V_2 \cup x)$ so $d(G_{n+1}V_1\cup x) \leq \frac{1}{2}d(G_{n+1}(x))$. Let $U_1=V_1\cupx$ and $U_2=V_2$ then $$e(G_{n+1}[U_1])+e(G_{n+1}[U_2])\leq e(G_n[V_1])+e(G_n[V_2])+\frac{d(G_{n+1}(x))}{2}\leq \frac{1}{2}e(G_n)+\frac{d(G_{n+1}(x))}{2}=\frac{1}{2}e(G_{n+1})$

[vital dew]

same:
Compile Error! Click the reaction for details. (You may edit your message)

[weary tiger]

whatever it doesnt format properly but is the idea correct?

like by induction then just adding the new vertex to the set with which it has less edges

[empty forum]

need help with some homework

im not really understanding relations

[gleaming zephyr]

what does a reflexive relation mean

[empty forum]

it means to be equal to itself

https://www.youtube.com/watch?v=q0xN_N7l_Kw&list=PLHXZ9OQGMqxersk8fUxiUMSIx0DBqsKZS

my professor linked me this video to watch

[gleaming zephyr]

a reflexive relation

[empty forum]

to understand it

[gleaming zephyr]

is a relation that satisfies aRa for all a in A

[empty forum]

he isnt lecturing anymore unfrotunately hes just linking us youtube video

[gleaming zephyr]

a relation is just a subset of a cartesian product

here the relation is on A

so the relation is a subset of AxA

we say aRb iff (a,b) is in R

( R is the relation )

now a reflexive relation is a relation that satisifes aRa for all a in A

understand?

[empty forum]

yes kind of

[gleaming zephyr]

okay cool

now that you understand

what are relations

try to do this problem

[empty forum]

were asked the smallest ordered pair i think

[gleaming zephyr]

you are asked to find the smallest set of ordered pairs

that makes R reflexive

okay so

[empty forum]

oh yes

[gleaming zephyr]

i gotta ask first

is R reflexive

here?

[empty forum]

no

[gleaming zephyr]

why

[empty forum]

i dont think it is at least

[gleaming zephyr]

return to your definition

what does a reflexive relation mean

after you do

does this relation satisify the definition

[empty forum]

equal to itself r = r?

[gleaming zephyr]

?

[empty forum]

am i interpretting this incorrect?

[gleaming zephyr]

yes

what is a relation?>

[empty forum]

it has do deal with the subset

uh

[gleaming zephyr]

can you define it for me

formally

[empty forum]

im not entirely sure how to define a relation but like

i think its like how two subsets can compare

[gleaming zephyr]

do you ahve a textbook

[empty forum]

yes

[gleaming zephyr]

can you return to yoru definitions

in the textbook

[empty forum]

yea one sec ill look throguh

[gleaming zephyr]

for you the solve a problem you gotta first know what these things mean

formally

and get used to this ig later in the course

u gotta do defintion ---> definition ---> definition ...

[empty forum]

yes its mostly english

[gleaming zephyr]

okay

[empty forum]

forgive me

its kind of a dense definition

if you're curious the textbook we use is a free resource

https://www.people.vcu.edu/~rhammack/BookOfProof/Main.pdf

[gleaming zephyr]

cool book

so understand this definition

[empty forum]

this is kind of similar to what you said earlier

[gleaming zephyr]

yea ig

now whats a reflexive relation

once you got that you got all u need

try to look at ur relation

is it reflexive

and if its not whats the samllest set

you can 'add' to it so it becomes a reflexive

relation

[pale epoch]

honestly, did you watch the vid provided

if the formal definition of what a relation is hard for you to parse at first, just sketch the given relation as in the video

see what is missing to make it reflexive in the image your drew

add that

and then think about what that means formally

[empty forum]

thank you ill take everything into account

thankfully i have time to think it over

i think its an empty set?

correct me if im wrong but because they are all in the set they're related to itself?

so what i mean to say is that its reflexive

[gleaming zephyr]

no

it is not

reflexive

[empty forum]

it is not?

[gleaming zephyr]

that means

there exists an elementr ,a, in A such that (a,a) iis not in R

[pale epoch]

did you sketch the relation as in the video?

[empty forum]

{0,1,2} is A correct

[gleaming zephyr]

yes

[empty forum]

well maybe im interpretting this completely wrong

but i tried relating it to the video

[gleaming zephyr]

try to watch the video as loch said

[empty forum]

alright ill draw it instead of interpretting it

[gleaming zephyr]

( try to also understand it formally cuz thats improtant )

[zinc pewter]

for questions like these are we allowed to do anything as long as it's > or = to the original equation

Cause for this example a whole part of the original statement is just removed

The way I would have approached this problem is making n >= 1 removing log(n) and that would give me -6n^2 would that be correct still?

[obsidian rivet]

Let A = {a, b, c} and B = {1, 2, 3, 4} be sets and R = {(a, 1),(a, 2),(b, 3)} be a relation. Is R a function? If not,
explain why.

Is R not a function because it doesn't have a rule for c?

[pale epoch]

you can say that, yes

[sour arrow]

@obsidian rivet
The function would have used each member of A exactly once. It uses "a" twice and never uses "c".

[obsidian rivet]

thank you :)

[gleaming zephyr]

what is xR1

maybe its a typo

pro beh emans xR1y iff x+y is evene

[sour arrow]

You'd agree that -55 and 1 are equivalent under this relation, right?

xRy is a statement "x is equivalent to y"

We don't mean "equivalent" to mean "equal"

We mean "they satisfy an equivalence relation"

In this case, 6R10 is true
Because 6 + 10 is even

We say that 6 and 10 are equivalent

Similarly, -55R1

Because (-55) + 1 is even

We don't actually care what x + y is here, or if it belongs to A. We only need that x + y is even

-54 is an even number, so -55 and 1 are equivalent

Alright, so if we're clear on how to tell two elements are equivalent, we can move on to equivalence classes.

The equivalence class of -55 is the set of all elements equivalent to -55

Note that 6 is not equivalent to -55, so 6 is not in the equivalence class of -55

6 is not equivalent to -55
Because -55 + 6 is not even

x doesn't have to be -55. x can be anything in A.

For another example, -7 and 11 are equivalent because -7 + 11 is even.

We write (-7)R11 as another way to state this

And can also write that 11 is a member of -7's equivalence class.

-55 + (-6) isn't even

So -55 is not equivalent to -6

And finally -6 is not in -55's equivalence class

So I can choose any two elements, and see if they're equivalent

Well, let's check a few. Is -55 equivalent to -2?

So -2 is not in -55's equivalence class

Now, there's a trick here. Another way to tell two elements are equivalent is if they are both even, or if they are both odd

-55 is equivalent to any other odd number
And is not equivalent to any even numbers

[white jetty]

How to do question A?

So I rearranged it to: ≡5 = {5 | (x - y)}

[faint narwhal]

how you rearranged doesn't really make sense

[sour arrow]

You wouldn't say =5 = 2

[bleak kite]

Hey, I need help with a combinatorics problem. How many strings can be formed using the letters AAABBCDE if no 2 A's are consecutive

[sour arrow]

We can run through an example to make sure you get it. 8 is equivalent to 3 under this relation, because 5 divides 8 - 3
@white jetty

[zenith saddle]

Consider some graph G that has 14 edges and 8 vertices. What is the maximum possible length of a longest path in this graph?
From my understanding a path never repeats a vertex. And length is counted by edges connecting each vertex, so longest possible path here should be 8-1 = 7. But I'm getting marked wrong for that answer.

Am I missing something?

[weary tiger]

it should be 7

unless im making the exact same mistake as you

but the definition in my notes definitely says distinct

@zenith saddle

[white jetty]

@sour arrow I got it now

Thank you so much

[lyric pumice]

@bleak kite How long have you been working on the problem?

The solution involves the use of letters as separators.

[zenith saddle]

cool thanks @weary tiger

[lyric pumice]

@bleak kite Keep trying.

[white jetty]

Could anyone give me an example of "An infinite relation on Z that is not reflexive and not anti-reflexive"?

Using set builder notation?

[faint narwhal]

What have you tried?

[white jetty]

A > B

[faint narwhal]

why is this not anti-reflexive?

[white jetty]

I got confused

so like by definition, being non reflexive means that you cannot have something like: R(1, 1)

and meanwhile, you cannot have anti-reflexive, which means that you cannot have something is a reflexive inside the set....

So I'm thinking that, does that mean the set have some reflexive members and anti-reflexive members at the same time?

but not all?

@faint narwhal do you have any idea?

[faint narwhal]

thats the right idea yeah

[white jetty]

So it's reflexive and anti-reflexive at the same time?

[weary tiger]

3 cards are chosen at random from a standard 52-card deck. What is the probability that they form a pair? (A 3-card hand is a 'pair' if two of the cards match in rank but the third card is different. For example, 668 is a pair, but 999 is not.)

what's wrong w/ this

52 ways to pick first card

3 ways to pick 2nd card

48 ways to pick last card

so we have

(52 * 3 * 48)/(52 * 51 * 50)

= 24/425

what am i undercounting?

[weary tiger]

<@&286206848099549185>

1 - (48/51)*(44/50) (cards are all different)- (3/51 * 2/50) (cards are all the same)

i think its that

complementary counting?

huh

so the probability of picking a pair is equal to 1 - P(all same cards) - P(all different cards)

right?

cuz the only outcomes are that all the cards are the same, only 2 of the cards are the same(pair) and none of the cards are the same

its easier to subtract the cards that are not pairs

which is easier to do

cuz like probability of all the cards being the same is if they are in the same suit

so 3/51 * 2/50

but i want to know why mine is wrong

ya lemme check

you didnt consider that the pair selected is the first and third card, second and third card or first and second card

if you were just accounting for pairs you would have to consider the order in which those cards could be selected

hi @obsidian tendon

@weary tiger wdym

here let me give you some intuition

in what ways would selecting 3 cards become a pair

like what are the orderings in which that could possibly happen

??

what does that mean

ok so like you are selecting cards one by one right

so the ways you get a pair only occurs if ...

1. the first and the second card are a pair and not the third
2. the second and the third card are a pair and not the first
3. the first and the third card are a pair and not the second

in otherwords you cant just assume there is one way of picking the cards

but why does the order matter here

cuz there are the 3 ways you can get the desired results

intuitively it does depending on how you think about it

because the chance given by each card isnt constant for all 3 of those results

to fix this you either calculate one of those and multiply by 3 because the probability is the same for each

[minor zephyr]

So i found this answer for part a

the \leq 12n comparisons part makes no sense to me...

like what is being compared exactly??

what does it mean by "3 names with a majority in each list?" How can more than 1 name have a majority??

[halcyon sierra]

not sure if this is the section I should ask, but could someone help me understand the discrete nodal domain theorem and how to determine one from a graph?

[zealous summit]

!15m

oh bot's down

anyway, please read #❓how-to-get-help

[pale epoch]

this server is not for you to cheat on a quiz

[sleek swallow]

<@&268886789983436800> please kick this cheater out, thanks

[weary tiger]

@distant lagoon this is not permissible

😮 he got banned by someone

[sleek swallow]

Oh shit he got rekt

[halcyon sierra]

does someone know how to solve my problem? i can find the eigenvalues and eigenvectors of a graph. im just unsure how to assign signs to vertices of a graph.

$\begin{bmatrix} -1\ 0 \ 1 \end{bmatrix}$

For example, if this is an eigenvector for my graph's Laplacian, do the components' signs give the sign to the vertices? Vertex1 would be negative, Vertex2 would be 0, Vertex3 would be positive?

[vital dew]

TheRad1:

[stray reef]

3^n < n! so 3^(n+1) < (n+1)!

you're kind of using the very thing you're supposed to prove

as if you've already proven it

which you have not

[vital blade]

I really hate proof, you're absolutely right. Maybe I need sleep afterall.

Thanks.

I'll come back after more sleep and another attempt along with a visit back to my textbook. Hopefully I can gather a little bit of my pride back along my way and remember to actually prove something this time.

[quaint lichen]

How many multiples of 2,3,5,7 are there in 1<n<1000?

trivial application of inclusion exclusion and I got the right answer, but are there any tricks/shortcuts without access to calculator? (I only used it for division)

[weary tiger]

theres nothing that hard to do without a calculator there

[quaint lichen]

999/21

I need to brush up on some basics XD

perhaps not hard, but tedious

[weary tiger]

i mean

its really not lol

u know its about 50

50*21=1050 minus a few 21s

and ur there

(its about 50 because 1000/20=50 lol)

[quaint lichen]

ahhh you're right lol

[boreal beacon]

for this question

Suppose we have a path P. This path has two leaves, v1 and vk. By adding an edge to v1 or vk, we remove that leaf and introduce a new leaf in it's place. This is still a path because we can get to v1 to vk without having to repeat vertices. However adding an edge to any other vertex which is not a leaf introduces a new leaf while not removing the leaves v1, vk. This also wouldnt be a path because we cant get from v1 to vk without repeating a vertex.

^^^ is that a good answer ^^^^

[reef thistle]

@weary tiger can you show for n=1?

Then next just follow induction proof...

[weary tiger]

Carson flips over the cards of a standard 52-card deck one at a time. What is the probability that he flips over the ace of spades before any face card (jack, queen or king)?

i think it is just 1/12?

because

1 way to get ace of spades

but

$\binom{12}{1}$ ways to get face card

[vital dew]

polynomial:

[weary tiger]

[weary tiger]

<@&286206848099549185>

[fossil pewter]

1/13

You also have to include that ace too

[weary tiger]

oh

but is 1/13 correct? hmm

[fossil pewter]

I think so I did it with a long way to

$ \sum_{r=1}^{40} \dfrac{ \binom{39}{r-1}}{r \cdot \binom{52}{r}} $

[vital dew]

Krishna:

[fossil pewter]

This also gave me 1/13

[weary tiger]

thanks @fossil pewter

[pale wing]

**The girth of a graph G is defined to be the length of the shortest cycle. Prove that if G= (V, E) is a planar graph with girth g≥3, then|E|≤ g(|V|−2)/ (g−2) **

can someone help me with this

I do understand I have to use the v-e+f=2 to reach somewhere since they are planar graphs but I have no clue how to continue

[weary tiger]

@pale wing

[reef thistle]

ah yeah that makes sense dual graph it

[stray reef]

so what was the statement again

"let f: A -> B, g: B -> C be functions, prove g.f is a function from A to C"?

as stated this seems like,,, definitionally trivial

[mint tiger]

yes

[reef thistle]

yeah, but we need the definition of function and composition

if there is something to prove

[stray reef]

my point exactly

so.....

[stray reef]

okay, and the definition of composition for relations?

[mint tiger]

So I would have to prove that the composition of gof satisfies both those conditions for it to be a function

[stray reef]

uh huh

great

alright

so we have two functions f: A -> B and g: B -> C

their composition g.f is a relation from A to C

so now we need to prove two things

$(\forall a \in A)(\exists c \in C)[(a, c) \in g \circ f]$

[vital dew]

Ann:

[stray reef]

this is point i

$(\forall a \in A)(\forall c_1, c_2 \in C)[(a, c_1) \in g \circ f \land (a, c_2) \in g \circ f \implies c_1 = c_2]$

[vital dew]

Ann:

[stray reef]

and this is point ii

[mint tiger]

yes

[stray reef]

do these translations into logical notation make sense to you

[mint tiger]

yes

[stray reef]

ok then go forth

write up a proof and ping me when you're done

[mint tiger]

ohh

okay cool

[mint tiger]

@stray reef I've done it lol. Sorry for the wait. I had lecture to attend

i hope you can read it because it's a lot to type.

[stray reef]

the domain of g o f is in A

that bolded "in" should go

[mint tiger]

oh okay

[stray reef]

otherwise this is fine.

[mint tiger]

omg okay thank you so much

[hollow narwhal]

Before i post my question, I just wanna make sure this is the right place. Does this chatroom include combination/permutation/counting method topics?

[weary tiger]

yes

just post it

lol

@hollow narwhal

what with

[hollow narwhal]

Part C on this problem

not quite sure if all the cards are dealt at the same time or if they're dealt one at a time etc

[weary tiger]

doesnt matter

[hollow narwhal]

mmk

[weary tiger]

@hollow narwhal

like

you dont even need to consider the cards charlie doesnt get

[hollow narwhal]

hm

so does his first card have a different probability than his second?

like

[weary tiger]

so for a flush he needs 5 of the suit right

[hollow narwhal]

assuming he does or doesn't draw a card of the same suit

ya

[weary tiger]

so needs 2 of the same suit in hand\

and 3 in the middle

[hollow narwhal]

or 1 in the hand, and 4 in the middle

[weary tiger]

(or 4 in the middle)

doesnt it say at the top

u use 3 from the middle

yeh

oh wait nvm

the games changed

ur right

[hollow narwhal]

but that also includes 2 of the same suit in hand and 4 in the middle

ya

[weary tiger]

so 1 in hand or 4 in the middle

but first

can you work out the chance he has 2 of the same suit in his hand

and has 3 out of 4 cards the same suit in the middle as the suit in his hand

[hollow narwhal]

lets see

[weary tiger]

(or 4 of the same suit in the middle)

[hollow narwhal]

so would it be something like 4 * 13c5

or

[weary tiger]

so from a probability standpoint not total hands

the probability he has 2 cards of the same suit in his hand is 12/51 because it has to be the same suit as the previous one

[hollow narwhal]

why out of 51

[weary tiger]

because he already has a card

[hollow narwhal]

oh

[weary tiger]

it doesnt matter what suit the first card is the second one just has to match it

and theres 12 cards which match it

[hollow narwhal]

so it would be 12/52 + 11/51?

if theyre the same suit

[weary tiger]

I'm just looking at the second card

the probability of his hand being the same suit is 12/51

[hollow narwhal]

mmk

[weary tiger]

which should match 4*13C2/52C2

[hollow narwhal]

so that is just the probability of his hand

[weary tiger]

yeh then

for the middle

[hollow narwhal]

so would the probability of the middle be 4*13C3/52C2

[weary tiger]

close but

we know 2 cards already

[hollow narwhal]

oh, so it would be 11C3 instead?

[weary tiger]

and 50

[hollow narwhal]

and 50C3 on the bottom

ya

my bad

[weary tiger]

50C4

[hollow narwhal]

woops

ya

forgot it changed lol

[weary tiger]

also the 4 was because theres 4 suits

[hollow narwhal]

ya

[weary tiger]

but now we need to match to a specific suit

[hollow narwhal]

ya so those two calculations could result in 2 different suits

right?

[weary tiger]

yeh

[hollow narwhal]

gotcha

[weary tiger]

also in hindsight ive told you a v bad way

we should just consider all 6 cards

at the same time

[hollow narwhal]

mmk

[weary tiger]

so we need the probability 6 are the same suit + probability 5 are same suit and 1 different

whats the first probability

using your C notation

[hollow narwhal]

so does it matter if they're dealt at the same time vs if they're dealt differently if we're considering all 6 rather than partitioning

[weary tiger]

no it'll get the same result

but will be fewer calcs with doing 6 at a time

[hollow narwhal]

gotcha

[weary tiger]

so whats the probability 6 are the same suit

[hollow narwhal]

that would be 13C6 / 52C6

[weary tiger]

yeh exactly

[hollow narwhal]

for 1 suit

[weary tiger]

now whats the probability that 5 are 1 suit and 1 another

remember you can have

SSSSSD, SSSSDS, SSSDSS, SSDSSS, SDSSSS, DSSSSS

where S are the same suit and D is a different suit

[hollow narwhal]

mmk

uh

hmm not exactly sure on that

[weary tiger]

so the way you can pick 5 cards from 1 suit is 13C5

[hollow narwhal]

probability/counting was just introduced this week

gotcha

[weary tiger]

and the way you can pick 6 cards from 52 is 52C6 right

then what is the number of ways we can pick the last card?

how many possible cards are there

(it's a different suit)

[hollow narwhal]

last card would be 47C1

?

[weary tiger]

I mean how many possible cards can be the last 1 of the 6

if we know that we have 5 of one suit and 1 of the others

[hollow narwhal]

47 cards are left so last card could be any one of them

[weary tiger]

no because it needs to be a different suit

[hollow narwhal]

oh

i see

so then 39C1

[weary tiger]

yeh

but then because ordering matters

and there's 6 orders

it's 6*39*13C5/52C6

and then add the two results and u get the answer

[hollow narwhal]

mmk so that part is for if you get 5/6 are of the same suit

and the other one is 6/6 of the same suit

[weary tiger]

yeh

[hollow narwhal]

sheesh man

this thing was mind boggling me the whole day

I feel like counting isn't that hard, the only thing thats confusing is keeping track of which parts you need to calculate/take out if they intersect

thank you so much

[weary tiger]

no worries

[hollow narwhal]

hmm

would that whole thing be multiplied by 4

because there are 4 suits or

because thats the probability of those happening for 1 suit

[weary tiger]

yeh

wait no

[hollow narwhal]

how come?

[lyric pumice]

Counting is hard.

[hollow narwhal]

^

[weary tiger]

Does Euler's Formula imply the existence of the graph?

From this we have that $e(G)=\frac{2n-2}{2}=n-1$ and then $2-n+(n-1)=1$ so it has $1$ face so it's a tree

[vital dew]

same:

[woeful galleon]

hey guys, can anyone help me out here ?

[main marlin]

@woeful galleon i can help you in #help-4

[lean basin]

Can anyone check my solution for #7?

[reef thistle]

@lean basin what solution

bonus is it can't be the chromatic polynomial of any graph (substitute k=1)

[lean basin]

For the Bonus Question, I thought you have to substitute k=4 since it is a planar graph and every planar graph is 4-colorable?

My (unchecked) solution for #7:

[reef thistle]

bonus sub k=1 what does it say about the graph

there are multiple ways to do the same question

[lean basin]

If we substitute k=1 we get 3 colorings?

Don't we have to use the 4 color theorem?

[reef thistle]

But think about it, if there are 3 1-colourings, then...?

what does it say about the graph

@lean basin

[lean basin]

A null graph?

[reef thistle]

yeah, it's a graph without edges, so how many ways should there be to colour with 2 colours?

@lean basin

[faint narwhal]

think about the contrapositive statement

[stray reef]

no

as in no, the answer is not 13C6

$\binom{13}{3} \times \binom{39}{3} + \binom{13}{4} \times \binom{39}{2} + \binom{13}{5} \times \binom{39}{1} + \binom{13}{6}$

[vital dew]

Ann:

[white sundial]

Is it possible that you can help me on this problem please?

https://imgur.com/a/uTuMQ7R

[quartz gull]

@stray reef is (13c 3 ) (49c3)

so you are choosing 3 spades from 13 spades

then you put 10 spades back into deck of card

now we have 49 cards in total and then you choose 3 random cards

[gilded finch]

why don't they use sum notation for that question jesus

[stray reef]

@quartz gull no? i'm pretty sure that overcounts things to some extent

[quartz gull]

okay makes sense

thanks

[robust mango]

@white sundial Have you tried anything?

[white sundial]

Yes

[robust mango]

So did u do it

[white sundial]

No, I am still stuck.

[sudden knot]

tell us what you tried

yea it says in the pic that it's an induction problem lol @cursive elk

[white sundial]

Well, I tried removing 3 from 1/4

yes

[fleet lily]

in graph theory

is there a theorm that says two graphs are isomorphic if all their vertices have the same degrees, that is for each v in V, deg(v) = k, where k is an integer

[pale epoch]

no

k-regular graphs are not all isomorphic

at least not in general

[obtuse lance]

combinations because order in the set doesn't matter

[white sundial]

I am stuck on every problem here except a.

[weary tiger]

@fleet lilytwo graphs having the same degree sequence doesn't imply they are isomorphic

which is what I think you're asking

[fleet lily]

@sand musk no i know that, i said all vertices must have the same degree

[obtuse minnow]

yay isomers!

[fleet lily]

i think this is a good counter example

[stray reef]

here is fine

[ancient basin]

I'm currently stuck trying to do this question. Did I do things correctly so far? Any hints about what to do next?

[hybrid snow]

@ancient basin

What you're trying to prove in your inductive step is that for an integer $N, 2^N > N^3$, there's an integer $n = N+1$ such that $2^n > n^3$ holds. Can you guarantee for every integer $N$ that this is possible?

[vital dew]

dk:

[hybrid snow]

You can see that it falls apart, say we try N = 1. Then you're trying to prove n = N + 1 = 2 also holds, which you've shown it doesn't.

hmmm, the best hint I can give is try to find a way to increase n > N (by SOME increment, right? It doesn't have to be +1 or +3, *5 etc, as long as it's greater )(e.g. your "P(n+1)" step) so that it involves the inductive hypothesis you declared, but also guarantees that 2^n > n^3 actually holds.

How do you make something you know greater with absolute fact? err my wording is a bit off, but hopefully it helps.

[ancient basin]

I'm a bit confused at how we can use mathematical induction, when we can just prove that when N=0, the statement is also true with n = 1 by just using arithmetic and showing the end result.

Like at N = 0: 1>0
n = 1: 2>1

[hybrid snow]

mathematical induction is needed to prove that for any integer N you pick, not just 0, that it holds. N = 1 is just one example of the statement

Essentially, if you do it by example, then you have to prove every single integer N where that condition holds, you have to find a number n > N such that it also holds for n

which is physically impossible, since the integers are infinite

so we use induction to show that, if we pick ANY integer N, even something in the 10^999999 (assuming it holds), there's some integer greater than that for which the condition holds

sorta get it or nah?

let me know what part is funky

[ancient basin]

I understand the need for induction since integers are infinite, but how can I deal with the fact that the statement doesn't hold for N = 1?

[hybrid snow]

it does hold though, right?

remember, it's not saying n = N + 1

just some integer greater than N

ex:

N = 1, so 2 > 1 ✅

N = 2, 4 isn't greater than 8,

but we don't have to go incrementally

okay how about n = 10?

2^10 = 1024 > 10^3 = 1000

awesome, the condition holds for N = 1, since we found n = 10.

hmmmm

what ya think?

proving by induction doesn't mean you always have to show n => n + 1

[ancient basin]

Makes sense! So does this mean that we'd need to prove p(N+9) is true (as part as proving by induction) if we choose N=1?

[hybrid snow]

you don't have to

so I think the base case of N = 0, and n = 1 covers the starting point

we can prove for N = 1 as well by just saying: Let $N \geq 1$

[vital dew]

dk:

[hybrid snow]

when we do the inductive step ^.

so we're proving N = 1 in the inductive step, along with N =2 , N = 3 ... and so forth to infinity. (assuming they hold etc though - I don't think 2 or 3 hold lol)

ping me if you get stuck/have any more questions

[ancient basin]

Alright, thanks! I'll try to figure it for a bit to see if I can solve it

[hybrid snow]

okie

[feral rapids]

Question regarding sampling.

Say I have a sampling rate of 40001hz. Any perfectly aliased signals below 20000hz would be captured and reproduced properly by using sinc filter right? Or is there any additional condition on the sampling instant/phase.

What if the initial zero crossing of the sample doesnt align with the sampling points? Or rewriting the same. What would happen if all the points I'm supposed to sample were shifted by 1/2, 1/4 or any fraction of the sample period.

The samples otherwise follow the band limitations but are shifted by non integral multiple of the sampling period.

Currently looking at this article : https://royalsocietypublishing.org/doi/10.1098/rsos.140550

I was reading something about sub Nyquist aberrations but I'm kind of confused.

And what does this mean? It doesn't seem to correlate well. Either the above is wrong, or below is misinterpreted for what they mean by timing precision.

[dense creek]

I am trying to prove this by dividing it using cases and then use induction for that

However, when trying to prove P(k+1) when k is odd, I don't know how to advance because of 4f((k+1)/2)

[dense creek]

<@&286206848099549185>

[silk grove]

@dense creek not sure if I can help

But I don’t think the P(k) should be the proposition you’re assuming

Assume separate cases for odd n and even n

Hint: how do you write odd n and even n?

[dense creek]

But don't I need to prove using induction?

[silk grove]

Okay. I would think that

[pale epoch]

where exactly is your problem?

[silk grove]

You need to prove P(0), P(1), P(2) first

[dense creek]

I already have those base cases proven

[silk grove]

Then you assume some odd form of k is true - then do it again from the next consecutive odd

[dense creek]

@pale epoch I do not know how to prove (if n is even)

[silk grove]

Okay

[pale epoch]

you have $f(k+1) = 4f(\frac{k+1}{2})$ if $k$ is odd

[vital dew]

Lochverstärker:

[dense creek]

you mean even?

[silk grove]

How do you write an odd number in terms of k, if n is a whole number

[pale epoch]

no

[dense creek]

do you mean 2k+1?

[silk grove]

Yeah