discrete-math

[coral parcel]

That's a cute theorem.

[sacred dune]

Friend and I have struggled on this one for a bit yet it seems so simple: show there are 2n^(n-3) spanning trees of K_n that contain a chosen edge e

I don’t want a solution just a hint in the right direction

[coral parcel]

Have you already proved the formula for the number of spanning trees without this restriction?

If so: what is the probability that a randomly picked spanning tree contains your chosen edge?

[true venture]

Hey does the way I've formatted this equation make sense? Reffering to the boxed one.

The summary currently adds many more terms than is needed but I'll try and get that lower later

[coral parcel]

It looks like what you really want is just to sum n-H(i) for those i where it is positive?

That would be clearer as $$\sum_{i=1}^{n} \max(0,n-H(i))$$ than with the absolute-value gymnastics.

[vital dew]

Troposphere

[true venture]

Ah thanks, Yes I only want the sum when it is positive

[sacred dune]

Hmm haven’t thought of going the probability way I’ll think about it thanks

Oh I think I might see what to do thank you

[stuck bone]

Hi. I don't really know if this counts as discrete math or not... It looks like a problem in discrete math, maybe... Say I have a list of numbers $\theta_1, \theta_2, ..., \theta_k \in [0, 2\pi)$ in (strictly) increasing order. I know that $0 = \sum_{j=1}^k e^{i \theta_j}$. I'd like to show that $e^{i \theta_{j+1}} = e^{i \theta_j}e^{i 2\pi/k}$ holds for all $1 \leq j \leq k$, where $\theta_{k+1}$ is to be taken as equal to $\theta_1$.

[vital dew]

phao

[stuck bone]

For k=2,3 I know how to do it... but what I did doesn't really generalize...

(btw... idk if this is true or not -- I think it is, but I'm not sure)

[coral parcel]

It is not necessarily true for k>3.

[stuck bone]

right

[coral parcel]

We can reformulate it as: Given a convex polygon where all sides have length 1, is it necessarily regular?

Yes if there are three sides, not if there are more.

[stuck bone]

Ok.

[coral parcel]

For example we could have theta1=0, theta2=10°, theta3=180°, theta4=190°.

[stuck bone]

Yep... when you said it

that wasn't true for k > 3

I thought of combining "two working cases for k=2"

tbh... idk if it would work, hehe, I'm just assuming that is what you did.

But thank you!!

[coral parcel]

Essentially that gives the same solution as I have. I was thinking of it more geometrically, though.

[stuck bone]

Right. Thank you very much. You really helped me here hehe

[vital dew]

jo_fish

[wide vine]

I am sorry to hear that

[craggy wharf]

I'm kind of new to sets, am i stupid or are none of these actually correct

if you do the math in A it'd be {3, 3, 3} and B would just be {2} right

[narrow trench]

sets don't have repetition so A={3} and B={2}

But yeah, something's off here

As it stands none are correct

[craggy wharf]

see that's what I'm saying

ok thanks for the confirmation though

[unreal stump]

If you were to be pedantic,A inter B is literally null

Well ok, Nothing matches even if you don't assume that

[burnt lynx]

can I get help with the inductive step

I have case 1 covered

but idk how to go about case 2 when b=0

[vale cairn]

You haven't defined what b is

[burnt lynx]

2a + 7b

[vale cairn]

Wdym by 'when b = 0' though - surely the point is you need to find such a and b

[earnest breach]

when b is 0, then a is at least 3, and u can replace a with (a-3) and b with (b+1) since 7 = 6+1

[burnt lynx]

ahh

can you guys check If I did the problem correct

[earnest breach]

why did u do a-10?

[burnt lynx]

just to keep the template of 2a

[earnest breach]

its possible for there to be a case where a is less than 10 though

for example the base case, where a = 3

just do n = 2(a-3) + 7(b+1)

[burnt lynx]

Oh yea, thx

[gilded axle]

I'm looking for an efficient algorithm to create cayley matrices, to be more precise:
I'm looking for a randomised algorithm that can efficiently colour every vertex of a dxd grid with d colours, such that every colour appears exactly once in every row and once in every column. kind of like a sudoku generator in more dimensions, but without the small box constraints.
I'm not sure what the best way is to go about this, since I want d to be very large and at the end I should just have d matrices of that size.

an additional constraint that isn't necessary, but one that I would possibly also like to have is that the colorings of the grid are symmetric with respect to transposition, but that is of lower priority. I can just symmetrize the matrices I got from before, worst case scenario

[gilded axle]

ok what I might just end up doing is shifting the identity matrix diagonal step after step, and then at the end just multiply the all matrices by the same random permutation matrix at the end

not sure if this generates all of them, but this should be good enough

[frosty dirge]

guys, in automata and formal language theory, what would be the result of this: {{a} U {b}}*

• for Kleene Closure

[narrow trench]

The set of all words on the alphabet {a,b}

I mean, by definition, the set of all words on {a,b} is {a,b}*

[frosty dirge]

would it include the empty word

[narrow trench]

Yes, as does the Kleene closure of anything

(even of the empty set)

[frosty dirge]

Thanks man

[earnest portal]

yo sweet this server does discrete too?

I might come here often this school year too

[weary tiger]

Is my understanding right?

If I can picture properly in my mind, the Cardinality of sets A, B and C (only respectively) remain the same no matter if we change the Cardinality of the intersections

[coral parcel]

The expression looks right.

I don't have enough context to comment about "the cardinality of A, B and C remains the same".

[weary tiger]

Cardinality of elements in A only, Cardinality of elements in B only, Cardinality of elements in C only remain the same (and are equal) even if we change the Cardinality of the intersections

[coral parcel]

It's certainly possible to change the sets such that the cardinalities of each set stays constant while the cardinality of the intersections change.
I don't think you have a described any particular assumptions about the (so far unspecified) change that would force that to happen.

[rough crystal]

im trying to define an equation and am using set builder notation for part of the definition, i want to say x is an element of a sequence, but im not sure if I'm allowed to use ∈

I said p ∈ w, and defined earlier that w is a sequence, is that sufficient?

[stray reef]

sounds a little dodgy

can you show the full definition you're trying to put into notation?

in plain english, if possible?

[rough crystal]

C sub i, where i is the index of a word, is equal to the tuple (p, x) where p is an element of the sequence w (where w is all words in the excerpt), and x is an element of the sequence i (i contains the index’s of all words in the sequence w), and if the cosine similarity of the embeddings of w sub i and p is greater than .8

Sorry if thats hard to follow

I guess i can assign w to another variable and u that as a set in this scenario

[coral parcel]

Hmm, I would expect "plain English" to imply "without assigning letters as names for everything".

[rough crystal]

It doesn’t necessarily need order

[rough crystal]

Ok yeah ur probably right

Need to create a sequence of sets, the set at each index is a word in an excerpt, containing the same index. Next get the word embedding of said word and treat it as an anchor word, then get the word embedding of all other words in the same excerpt and compare it against that anchor, if the cosine similarity is above .8 u add it to the set

[coral parcel]

What is a "word embedding"?

[rough crystal]

Basically a list of numbers that represent some features, in this instance it’s derived from the weights of a neural network called word2vec

It encodes features into those numbers, and lets u do traditional math on words basically. The traditional example is king - man would output queen

And u can go further and compare the 2 vectors to see how similar the 2 words are

[serene perch]

does anyone know how to prove this proposition "∀x, y ∈ R, if x is rational and y is irrational, then x + y is irrational." using the definition "∃a, b ∈ Z, a = bx and b /= 0"

[final cliff]

I'd try for a contradiction. Assume x+y is rational, see where that takes you by applying your definitions and stuff.

(See what happens if you solve for y.)

[outer rune]

Can anyone direct me to a resource to learn how to do problems like this

my professor explained it in such a bad way

im very confused

[serene perch]

oh wait if ur still here, is y supposed to contradict my assumption since I assumed that its irrational and I got it to be rational?

so then my assumption is wrong?

[final cliff]

Yeah basically. You assume x+y is rational, show y must be rational too. But you had already previously assumed y was irrational. So you found a contradiction.

So, by contradiction x+y can't be rational.

[serene perch]

Ohh ok perfect and also I set y = sqrt(2) but for x I just subbed it in as a = bx. Is that fine

so then I made m = ny which is m = sqrt(2)n

just double checking to see if im supposed to do that in my proof

[final cliff]

Why do you think y=sqrt(2)?

What if y is some other irrational number?

[serene perch]

oh yeah true, idk I just assumed it to be sqrt(2)

but ig I could leave it as y and not specify it

[final cliff]

Well the proof won't be correct if you only do it for y=sqrt(2) because the thing you are trying to prove needs to hold for any irrational y.

[wide vine]

Suppose you have k length list of objects to choose to permutate from n objects. Not all these objects are distinct. By this I mean we can partition these n objects into m distinct "types" (1<=m<=n). How many ways are you able to permute a list k long out of n objects with there being m distinct "types" in the set of n objects

To add to this I will supply this partition set function that tells you the amount of each type are made up in this n long list.

Basically this question stems from me wanting to general an example such as

AABBBCCD and permute/ take combinations of length 2

if m =1 then all the objects in the n long list are the same, if m=n then all the objects in the n long list are unique (idk how I handle the cases in between) (plz ping me if you know how to do this in general or if you have some resources that says you need to just brute force it and there is no nice math formula)

[wide vine]

<@&286206848099549185>

[surreal crescent]

Hello. What does practical Graph Theory look like? What are the methods?

For example, in Category Theory (at my level) the time is split more or less evenly between trying to imagine and type check complicated definitions, assembling diagrams from given building blocks, and drafting equational proofs about said diagrams. All of these kinds of work are productive and apply everywhere.

In Graph Theory, many problems and definitions are formulated in terms of graph homomorphisms. But finding graph homomorphisms is a hard problem. For example, a core of a graph is a smallest retract. Am I supposed, given a graph, to find and test all of its retracts? Then, most of the work must be done on a fast computer? Or maybe I could look my graph up in a library of graphs on the Internet? This is made even worse by how all the nodes are equally boring. In Category Theory we commonly have a few vertices half of which are somehow special and recognizable. Not so in Graph Theory.

I want to study big random graphs, but I cannot figure out where to start. I have been opening some course books about Graph Theory — they are easy to understand, but the problems they offer are toy problems — curious little theorems and cute little examples.

Eventually I want to connect the study of graphs with the study of simplicial complices and topology. Are there any common methods to these areas? At first they look distant. I know there is the cycle and cut space, but I never figured out how to use these tools for any practical purpose.

Another way to look at graphs is of course as at homogeneous binary relations, but commonly seen relations are boring — for example, the most intricate equivalence is simply a bunch of cliques. So it does not seem to be immediately fruitful to integrate these two areas.

TL;DR: I want to do something with big random graphs, what are my options?

[obtuse lance]

based on your example AABBBCCD and permute/ take combinations of length 2 you can write a generating function for it by combining generating functions for the letters. For example B appears 3 times so we would make $1+x+x^2+x^3 = \frac{x^4-1}{x-1}$ for that, the exponents represent the number of times you can pick that letter. So when we multiply them all together, the exponents will add, so in this case you'd have $$f(x)=\frac{x^3-1}{x-1}\frac{x^4-1}{x-1}\frac{x^3-1}{x-1}\frac{x^2-1}{x-1}$$ and the coefficient on $f(x)$ of $x^2$ is the number of combinations of strings you can make of length $2$.

[vital dew]

Merosity

[obtuse lance]

I realize that's kinda gross but it's better than nothing

[wide vine]

ngl I don't really follow where the 1+x+x^2+x^3 came from

Im not familiar with this

and I don't really see how I extract information to solve the problem from this :v

[obtuse lance]

might help to look at a smaller example or work it out explicitly by expanding $(1 + x + x^2)(1 + x + x^2 + x^3)(1 + x + x^2)(1+x)$ and look at the $x^2$ term

[vital dew]

Merosity

[obtuse lance]

at least to see how it works

[wide vine]

I am more so confused on the polynomial stuff

we never did anything like this in my prob/stats class

[obtuse lance]

ok so simpler example, let's say we have these two polynomials (ax^2+bx+c) and (Ax^2+Bx+C) and I multiply them together, what will be the coefficient on x^2?

the method is what matters not the final answer here

[wide vine]

Merosity I am honestly confused where the polynomials are tying into this is my problem.

None of my permuations or combinations dealt with this

[obtuse lance]

we're wanting to get combinatorial information out of it so we are thinking of choosing a term from one polynomial and term from the other

[obtuse lance]

right, that's what I'm trying to explain lol

[wide vine]

okay

so what example are we working with now

[obtuse lance]

an x^2 term can come from choosing an x^2 term from one and x^0 term from the other, or by choosing an x^1 term from one and x^1 from the other

the exponents are telling us the amount of times a letter appears, so maybe I should go more concrete than this

[wide vine]

I did a problem like this (a+b+c)(g+d+f) and we were asked how many terms are in the expression and it would be (3 c 1) * (3 c 1)

but again it didn't have repeats and stuff so

like the process of expansion is analogous to computing all the combinations

for the factors being multiplied

anyways idk if anything I said was related to what you were saying

[obtuse lance]

sort of

it might help to look at a separate example altogether to learn some of the basics

[wide vine]

Okay why are we using x^2 what example are we working with that we are rewrittting in this polynomial

[obtuse lance]

like the binomial coefficients

if you expand $(x+y)^3$ while keeping the order and not using exponents you end up with $xxx+xxy+xyx+yxx+yyx+yxy+xyy+yyy$ which is literally every string of length 3 with 2 letters

[vital dew]

Merosity

[wide vine]

okay

(x+y)(x+y)(x+y)

[obtuse lance]

the coefficient on x^ay^b will be exactly the number of ways to make a string of length a+b with a xs and b ys

[wide vine]

you have 2 choose 1

each time

but I guess you are taking order into account

[obtuse lance]

nope no order

[wide vine]

oh

[weary tiger]

hello

[obtuse lance]

the order was just to explicitly show the strings it's counting that it simplifies down

at each multiplication you're making a choice of x or y in (x+y) to pick in contributing towards computing (x+y)^n

[wide vine]

okay

[weary tiger]

i did this generalization intuitionally by thinking how to generalize these 2 theorems: (k₁!)(k₂!)...(kᵣ!) | (k₁+k₂+...+kᵣ)! and (k₁!ᵃa!)(k₂!ᵇb!)...(kᵣ!ᶜc!) | (k₁a+k₂b+...+kᵣc)!
is my generalization true? (couldn't prove it)

[obtuse lance]

so you can think of the coefficient on xy^2 in (x+y)(x+y)(x+y) as having come from picking x from the first (x+y) then y from the second two (x+y) when it distributes

or the other 2 ways of doing it,

which gets 3

[wide vine]

okay

[obtuse lance]

the main two things I'm trying to get at is we're going backwards from the exponent to where it came from to determine its coefficient

and that comes from looking at all the separate products and picking one term from those

[wide vine]

okay so you are saying something like

aab

would be like

choose 2

(a^2+b)^2

or is it

[obtuse lance]

might be easier to explain this in voice chat with me drawing it out

[wide vine]

(a + a + b)^2

maybe

[obtuse lance]

cause that's not right at all

[wide vine]

idk I am trying to figure out how the duplicates fit into this

[obtuse lance]

so let's say we have 2 As and 1 B what would I make as my generating function?

(1+x+x^2)(1+x)

now the first one is for A, the second is for B, so I think of it like this:

[wide vine]

I don't get where that comes from

[obtuse lance]

I'm showing you right now

the exponents are the frequency of that letter, so we can pick A either 0, 1, or 2 times

we can pick B 0 or 1 times

[wide vine]

why do you x in both cases?

why don't you use y for b

[obtuse lance]

so when we look at where the $x^2$ term comes from in the product we get $x^21 + x^1x^1 = 2x^2$ since there are only 2 ways to make a string of length 2 from these options

[vital dew]

Merosity

[obtuse lance]

the first term $x^2x^0$ means 2 of A, 0 of B, and the second term $x^1x^1$ means 1 of A and 1 or B, which are the only ways of picking two letters from this

[vital dew]

Merosity

[obtuse lance]

I'm not even sure if this is what you want because I didn't understand the original question but we got side tracked into discussing how generating functions work rather than whether this generating function could work for you

[wide vine]

basically my question is this

https://math.stackexchange.com/questions/2372/how-to-find-the-number-of-k-permutations-of-n-objects-with-x-types-and-r

[obtuse lance]

the second answer there is basically how I would fix up my solution

that's an exponential generating function

I guess I am taking generating functions for granted here, I'd recommend looking at some of this, it's pretty fun https://www2.math.upenn.edu/~wilf/gfology2.pdf

[wide vine]

okay well I think I have a resource. I'll look into this generating function thing

so are the solutions to these even easy to get

Like You get some generating function but then you still have to I assume find for where x^100

[obtuse lance]

depends, sometimes there are tricks, other times no

the most straight forward way to get the x^100 coefficient is differentiate 100 times, evaluate at 0, and divide by 100!

so practically speaking you probably won't be doing this by hand

the generating function is considered the solution, and you can throw the thing in a computer to get the answer out of it

basically, as math gets harder getting a nice simple solution is just not always possible, but generating functions are a way of nicely packaging the information

although they can be used for other things as well even when you do know a closed form solution

[weary tiger]

can someone help?

[serene marlin]

why can we not intesect A = B both sides by C
and have them become equal
the answer says its false

[final cliff]

Trying to come up with concrete counterexamples helps.

So for example pick C to be the emptyset

Pick A=C and B={1}.

What happens?

[final cliff]

Fwiw to find a counterexample you would want to start with two sets A and B that are not equal.

but when intersected with C give equal sets.

Rather than starting from A=B.

[stray reef]

you're asked to tell if it's possible to go the OTHER way

[serene marlin]

But how do you know when to look for a counter example?

do you just automatical;ly assume its false?

[final cliff]

For me I just periodically check

Like, if I keep failing to prove something on this type of q

Eventually I wonder if finding a counterexample is easier so I give it a shot and see what happens.

[serene marlin]

oh i see, if u struggle you just try finding a ce

that makes sense

[final cliff]

Yeah. Idk if there's a perfect answer to that kind of question.

[serene marlin]

honestly I just dont have this math vision like for this example it took me so long to realised to union both sides by a intercet c

like how does one even spot that

aka i get its practice

but still lol

[final cliff]

Over time it can get a lot easier.

You get used to seeing certain tricks and stuff.

Also with those kinds of questions you can usually prove them multiple ways.

[serene marlin]

ya i know alot of times my method varies but i still get it right

[final cliff]

It's a good exercise to try to prove this by element chasing too.

[serene marlin]

i have never even heard of that

[final cliff]

Using the definition of set equality.

A=B if and only if A is a subset of B and B is a subset of A.

Is that sounding familiar?

A is a subset of B if and only if x being an element of A implies x is an element of B.

[serene marlin]

ohh ya aka not a proper subset

[final cliff]

So a common way to prove A=B is to use this

To show A is a subset of B, then to show B is a subset of A. Then you can conclude A=B.

Sometimes people call that kind of thing element chasing is all.

[final cliff]

Like for this.

Assume A-B=B-A.

Assume x is an element of A.

If x is not an element of B, then x is an element of A-B, so x is an element of B-A too. But then x would end up being an element of B, which is impossible.

So, x must also be an element of B.

So, A is a subset of B.

Similar reasoning will show B is a subset of A.

So, A=B.

[serene marlin]

aghh its too late for this shit hahah

but thanks for all the help

really did clear som,e shit up as far as methods go

thank!

[weary tiger]

…

[weary tiger]

[serene marlin]

idk how you guys can study that late

[stray reef]

"late" is relative, because timezones.

[serene marlin]

past 12

[coral parcel]

Every time is past 12; it's just a question of which 12.

[pastel hollow]

silly tropo, if x and y both equal 12, then x = 12 = y. so there's only one 12

[vale cairn]

I reject the transitivity of equality

I mean, just because carrots and cucumbers are both orange doesn't mean they're the same thing

[wide python]

Hi, What is formula for generalized counting and permutation?

n! / k!(n-k)!

is this formula for generalized counting?

n+r-1 chose r
and is this formula same as above? kindly explain difference

[pale epoch]

the first is n choose k or $\binom{n}{k}$

[vital dew]

Lochverstärker

[pale epoch]

i dont know what generalized counting is

[pure estuary]

That second formula looks like you are talking about the multichoose function?

N choose K gives the number of K-subsets of an N-set

Whereas the latter, N multichoose K, gives the number of K-multisets taken from an N-set

Where a multiset is just a set that allows for repeated elements

[wide vine]

maybe one that generalizes to sets with non distinct elements. So permutations or combinations of n total elements with subsets of k and possibly some weird partition of the elements into their own "types" (1 <= #types <=n). Meorsity was telling me this has something to do with generating functions

[obtuse lance]

I wasn't saying that, I was just showing an example of generating functions with binomial coefficients

[obtuse lance]

this looks like what you'd get from doing stars and bars if you've heard of that, since you can think of objects and dividers between them as the things you're arranging

[wide vine]

well nvm

i still need to look at that sometime

[obtuse lance]

generating functions are cool haha

[sturdy helm]

anyone here familiar with fair division?

[vital dew]

Per48edjes

[coral latch]

Whoops, sorry for this y'all

[pastel hollow]

you are cool

[serene perch]

could anyone help me out with proving this proposition? I'm trying to prove it using the well ordering property and I figured out all the steps except when solving for m - 1. By the way, the m is the variable I used as the smallest element in my set

[stray reef]

using the well ordering property???

are you forced to do it that way

[serene perch]

well my prof wants us to do these questions using induction and wop, its practice for my midterm I got very soon

idk its weird why they ask to use both methods instead of just induction

[hidden terrace]

Can anyone explain what is vertex packing for graphs in simple terms?

[stray reef]

the vertex packing problem is the problem of finding, for a graph G, a subset of the vertices of G in which no two vertices are adjacent that has as many vertices in it as possible.

[turbid remnant]

It's the 3rd semester and we have discrete mathematics

It has related topics of CS

[shell stratus]

Hi guys! I'm going to host a debate on set theory. The topic will be 'Is the Axiom of Choice true?' Let's start the discussion! Every time you have to say something, make sure to ping me. Thank you.

[proper marsh]

hello

what is the difference between the universal statement and the universal existential statment

this isn't clear for me

both are referring that all the elements in the set has a certain property

[pale epoch]

the first has two quantifiers

other than that no idea

i never heard those terms, i would wager its not too important

[proper marsh]

which are?

[pale epoch]

$\forall x \in \bR\exists y\in\bR\colon x+y=0$

[vital dew]

Lochverstärker

[pale epoch]

something like that, two quantifiers

[proper marsh]

wow

[pale epoch]

the other is $\forall x\in\bR^+\colon x > 0$

[proper marsh]

and thanks

[vital dew]

Lochverstärker

[pale epoch]

one quantifier

[proper marsh]

is this book in general a good one for a CS student?

i am intending to read and study it

[pale epoch]

i looked at the toc once, it looks decent

it is very long

[proper marsh]

in what terms

[pale epoch]

(and spends half the time on intro proofs instead of anything i would call discrete math)

its just long

[proper marsh]

i love proofs😂

[pale epoch]

you could cover the same content in maybe 30% of the pages

but you have to decide yourself if being wordy is good or bad

[proper marsh]

sure, thank you.

[obtuse lance]

it's probably good for you, go for it have fun

[true venture]

Yo I have the same book, it Def is long. That's good to hear because I learned nothing about proofs in school. I'm only on chapter 2

[hidden terrace]

Can anyone explain what is vertex packing polytope of a graph? (in simple language)

[tall oxide]

how does one answer this?

[neat condor]

Let X be the set of nonempty subsets of R, and let R be the relation on X given by ARB if and only if A ∩ B ≠ ∅. Is R reflexive?

what does this mean

[stray reef]

what does what mean

[neat condor]

X be the set of nonempty subsets of R

this

what is R referring to in here? so confused

[neat condor]

the ans is it is reflexive but i don't get the explanation nor the question

[stray reef]

is it a fancy R? like $\bR$?

[vital dew]

Ann

[neat condor]

no it's just a normal one

[stray reef]

show screenshot?

[neat condor]

like i find the statement itself very confusing, R is probably referring to the relation itself

[stray reef]

sounds like you have a conflict of notation

[neat condor]

and X is a non empty subset of R

[neat condor]

ohh

[stray reef]

okay yeah...

they should have used a different letter for the relation

[neat condor]

ikr

but i still dont get the qn

mind explaining it to me

[stray reef]

let's call the relation S

we have the relation "A and B have nonempty intersection"

the question is

is every set related to itself by this

and the answer is obvious: A intersect A = A, and the sets we are considering are nonempty, so yes

[neat condor]

where is this piece of information found in the qn? How do we know that the we are checking the relation of each set with itself

[stray reef]

that's the defn of a reflexive relation, against which we are to check this one.

[neat condor]

ahh okk , but the whole lot phrasing of the qn still confuses me. Are A and B both elements of X?

[stray reef]

yes

[neat condor]

how can X be a subset of a relation which is based on itself

that part confuses me a lot

[stray reef]

.

the question mistakenly uses the letter R to refer to two different things

that is how i interpret it

[neat condor]

oh so the first R is real numbers?

[stray reef]

should be.

it's what i suggested it might be at the start

[neat condor]

ahh ok it make sense now, thanks man

[forest scaffold]

is there any identity for $\sum _{j=1}^NjP\left(N{,}j\right)$ where $P(N,k)$ is the number of k-permutations of N

[vital dew]

Tadashi

[weary tiger]

i don't understand the logic on conditional

[final cliff]

Is there something specific about it you don't understand?

[unique abyss]

Is it the fact that conditional is only false for T -> F

[weary tiger]

not only that

[weary tiger]

I don't understand his logic

[final cliff]

Yes, I'm asking you to try to explain more about what specifically you are having trouble with.

Like, are you getting stuck on why we define it the way we do in general?

[weary tiger]

like how it works

[final cliff]

It's just a definition.

We define "A implies B" in terms of the truth table of the conditional.

This kind of conditional is called the material conditional.

There are other types of conditional.

[weary tiger]

like the inverse ecc..?

[final cliff]

But the material conditional is the one people generally use in math.

[final cliff]

No

[weary tiger]

like what?

[unique abyss]

Well the way I always thought of this was that the conditional was a contract.
“If my it rains, my shirt gets wet.”
The only time this statement is false if it rains and my shirt doesnt get wet.
If it doesn’t rain, then we completely ignore the conditional because im only making claims about scenarios where it rains.

[final cliff]

Stuff like inverse, contrapositive etc are just names for certain formulas involving the material conditional iirc

But there are whole other logical formalisms meant to capture other notions of conditional statements

[weary tiger]

oh , i was confues about that

[final cliff]

Like strict conditionals

[unique abyss]

Don’t confuse him more lol

[final cliff]

Junk like that

[weary tiger]

so in the second case is it true?

[final cliff]

I don't think it will

[unique abyss]

If you mean the case where it rains and my shirt doesnt get wet then that’s false

[final cliff]

I look at the material conditional as kind of a "good enough" approximation of what "A implies B" type statements mean for the purposes of doing stuff like math.

[unique abyss]

If it rains my shirt must get wet. That’s the contract

[weary tiger]

The other one

[final cliff]

For remembering the truth table definition of the material conditional all you really need to know is T->F is the only case where the conditional is false.

[weary tiger]

It is the XOR logic

Right ?

Is it why it is studied for CS ?

[unique abyss]

If it’s a sunny day and my shirt is dry or if it’s a sunny day and my shirt is wet that doesn’t break the contract.
It’s a scenario “outside” of the contract. Remember I’m only making a claim about scenarios where it rains.

[final cliff]

Briscola do you know the definition of the conditional we're talking about?

[unique abyss]

^

[weary tiger]

Yes

[final cliff]

http://3.bp.blogspot.com/_jBZxBm2CpFk/R-blt38veqI/AAAAAAAAAAc/egh0X96UolU/s320/truth+table+conditional.gif

[weary tiger]

Or wait

Yes

The arrow

[final cliff]

Mmm idk if that truth table came out legible lol

[weary tiger]

Yes it is

I mean that

I understood

Thank to @unique abyss

[unique abyss]

Happy to help

It’s something a lot of people get confused

[final cliff]

Okay, yeah. So this table is basically the jist of the defn of the conditional.

[unique abyss]

So you’re not alone by any means

[final cliff]

I usually think of the intuition for it in terms of real life examples.

Like, if I promise you "if you get an A on your test, then I'll buy you pizza."

If you don't get an A and I don't buy you pizza I'm not lying.

If you don't get an A and I still buy you pizza I'm not lying.

If you get an A and I do buy you pizza I'm not lying.

If you get an A and I don't buy you pizza we have a problem though.

I told you I would buy you pizza in that case, but I didn't.

So it seems like I've lied to you in that case.

That's the only false case in the definition of the conditional.

It's the T->F case.

[mint bane]

@rancid pivot @sour onyx i dont wanna clog that channel

[sour onyx]

Post the die again please

Or I can copy it

[mint bane]

if you know that the four is on the right in particular, you know that the third visible face has to be 6

[rancid pivot]

So say you can see 5 and 6. The third side is 3 or 4.

[mint bane]

yeah that i know

[rancid pivot]

No, the third could be 1

[mint bane]

imma just hardcode it probably lol

i meant like orientation

cuz yeah if you just know which ones are visible there are two options

but if you know on which face in particular they are, you can fully determine the third

idk how to do that tho

[rancid pivot]

Unless you know the exact orientation of the die, there's no way to tell which of the 2 numbers it will be

[sour onyx]

Hm say you know 6 and 4 you know opposite six is one and opposite 4 is 3 if we conclude that even and odd sums of adjacent sides must touch (based on this photo) hmmm how do I describe this better

[mint bane]

oh shit i see what you mean

[rancid pivot]

For a balanced d6, you can take any two opposite faces and switch them and the die will remain perfectly balanced.

[mint bane]

say you have the top and the left face

if the sum of those two is odd then the third face must be even

and vice versa

[sour onyx]

Okay so like looking at the 4, to the left is six the sum of those two is 10 which even, on the top we have 5 which summed with 4 is nine which is odd, then the right is 1 and you get a sum of 5 which is odd and on the bottom I believe there is 2 which summed is 6 which is even

[mint bane]

kenshin i love you

[rancid pivot]

Right but there are still two solutions to that

Think of the die in terms on adjacency and opposition. Choose on opposite pair. Each member of that pair is adjacent to the same 4 sides. If you switch the places of these opposite sides, they will still be adjacent to the same 4 sides.

You need a stricter property than balance if you want there to be one answer here

[mint bane]

right but im only worried about non-opposite pairs

[rancid pivot]

OK let me phrase this another way. There are two kinds of balanced d6. You can show this by unfolding the faces. One type is as follows:

6
2 3 4 5
1

1
2 3 4 5
6

If you can see 3 faces like in your picture, where the left face is 4 and the right face is 5, then the top face can be 1 or 6. However, if we have the first type of d6 then it must be 6 and if we have the second type then it must be 1

[rancid pivot]

Unless you choose which type of balanced d6 you're working with, you cannot definitively say which side is there.

[mint bane]

ohhhh you mean like a variety in the actual way the dice are made

yeah i dont care that much about that as long as the results i get consistently agree with only one

im just going by the die i have lol

which seems to be the second type

[mint bane]

i see now what you meant @rancid pivot

took a break and came back and it doesnt work lmao

[rancid pivot]

Lol glad I could help!

[mint bane]

the problem i ran into is that like even with just the die i have

going in order of top,left,righ - (6,5,3) is one ordering but (5,6,4) is another and the sum of 5 and 6 is obviously odd but it doesnt determine if the last face is 3 or 4

i couldve hardcoded this by now

welp this is no longer a bug, i hereby declare it a feature

[rancid pivot]

Welcome to software engineering lol

[mint bane]

do you know if it's possible then :(

cuz like the kind of die im using is fixed

[mint bane]

just realized king dice is actually a 1 in his face lmao

[cinder gazelle]

Can you help me find the partitions ?

I know the ans but I cant solve it by theoretically

Tag me when you uh solve this

[stray reef]

,rcw

[unique abyss]

I also think part of the issue is that it’s not exactly clear what the “third face” is

The third face has to be adjacent to the two given faces

That leaves two faces as possibilities

Based on the way you’ve described the problem, among these two faces, it’s the one visible to the viewer

[unique abyss]

I think for hard-coding

Let’s see

So every face has four adjacent faces

And there are 6 faces

So 24 total pairs I think

And if 5 is on top and 4 is on the right that’s different from if 4 is on the top and 5 is on the right, if you include the fact that the third phase must be “visible to the viewer”

So the ordering of the pairs matters

[plucky island]

I need help building an understanding of strong induction

[unique abyss]

The difference between strong and weak induction is relatively minor

So if you understand induction in general, you’re 99% there

[plucky island]

I think I understand normal induction and I understand the idea behind strong

But I don’t know what it is

Like I understand that weak induction works when you don’t need to assume it works for all k < that value right?

[unique abyss]

Yeah

And strong induction is when you assume that it holds for values between the base case and the k-th step inclusive

[plucky island]

How do you prove that it works up to the k step?

[unique abyss]

You don’t AFAIK

It’s assumed that it works up to k step in strong induction

[plucky island]

so how is that different then normal

you just assume?

dont you need to do something to get to the assumption?

do you just prove 2 base cases

instead of the first one?

[unique abyss]

The difference lies in the fact that you don’t need to assume that the statement holds for all values from base case to k

Weak means that a statement for n implies n + 1

In strong induction, you are assuming something extra

That the statement is not just true for n

But all values between base case and n

[plucky island]

How do we assume that

By finding multiple base cases?

Is that it?

[pale epoch]

you dont need multiple base cases

the difference is with normal assumption you only assume the previous domino has fallen and show the next will fall and with strong induction you assume all previous dominos have fallen

there is no issue here because strong induction is implied by normal induction

you prove your base case say P(1)

and if you have proved P(n) -> P(n+1) you can use this repeatedly

then you get P(1) -> P(2) and P(2) -> P(3) and so on

actually this explanation isnt best, hmm

(the dominos one is fine)

[unique abyss]

Yeah it’s hard to explain it

But definitely you don’t need multiple base cases

[plucky island]

I understand the dominos part so how can we make that assumption?

[pale epoch]

ok

say you prove something by strong induction

so you prove P(1) and you prove P(1) and P(2) and ... and P(n) -> P(n+1) for all n > 1

so what you did is prove P(1) and P(1) -> P(2) and P(1) and P(2) -> P(3) and P(1) and P(2) and P(3) -> P(4) and so on ...

but each step can also be proven using weak induction just more often

so say you only know P(n) -> P(n+1) for all n instead, call this (*)

and you want to prove P(1) and P(2) and ... and P(n) -> P(n+1) for all n

then what you can do is apply (*) to show P(1) -> P(2), so you now know P(2) is true

[unique abyss]

Yep

Weak implies strong

[plucky island]

Ok I see the difference

[plucky island]

But does weak always work?

Or why would they teach strong

[pale epoch]

so you can continue doing this and keep applying (*) to get the strong statement

strong induction is better

you get more assumptions during your inductive step, which is well, stronger

often you have some kind of object that consists of n "pieces" and you want to show by induction that it has some property

so a valid strategy is to break up this object into two smaller objects consisting of n-k and k pieces and use the induction hypothesis on either piece

and then piece together that information somehow

now if you only have weak induction this doesnt quite work, you can only apply the induction hypothesis to smaller objects consisting of n-1 pieces

if you have strong induction you can apply it to any object with less pieces

now you can turn every strong induction proof into a weak induction proof but that hassle isnt worth it

the better question should be "why do they teach weak induction"

[unique abyss]

And note that for proofs where weak induction is used, you can still use strong induction. Strong induction still assumes n-1 like weak does. Just with some more assumptions in between

[unique abyss]

And strong implies weak

It goes both ways

[plucky island]

I get a lot of what you’re saying here but I still don’t understand how to make the assumption that it works how would you structure the proof differently to prove up to k

[pale epoch]

the structure is not different, you just assume more during the inductive step

i agree my explanation isnt great, i kinda lost my train of thought lol

[plucky island]

You don’t need to do anything differently to prove it?

[pale epoch]

maybe the top answer here helps: https://math.stackexchange.com/questions/108297/strong-induction-proofs-done-with-weak-induction

[plucky island]

To prove assumption

[pale epoch]

no

strong induction is kinda just weak induction applied repeatedly

(though a finite number of times)

[pale epoch]

the top answer here shows this kinda

if Q(n) = "P(k) is true for all k=1, 2, ..., n", then weak induction is show P(n) -> P(n+1) and strong induction is Q(n) -> P(n+1)

but if you prove P(n) -> P(n+1), then you can use induction to prove Q(n)

so everything works out in the end

[plucky island]

Hmmmm

Here I have a question from class

Why does it ask to prove a2 a3 and a4

Is that relevant for induction or just a random thing to show you understand how the sequence works

[pale epoch]

,rotate

[unique abyss]

Looks like fibonacci to me

[plucky island]

It was an old midterm question

[unique abyss]

It's just to show you how sequences work

[pale epoch]

it looks like (a) just exists so you convince yourself that the statement you have to prove in (b) is true

or yeah, that

or just easy points

[unique abyss]

@plucky island If you're willing to read alot, you may be interested in this answer https://math.stackexchange.com/questions/1184541/what-exactly-is-the-difference-between-weak-and-strong-induction

top answer

it has a rigorous proof for strong -> weak and weak -> strong

[wicked copper]

Is the set X = {3a + 5b | a, b \in \mathbb{Z}} equal to \mathbb{Z}?

[obtuse lance]

hint: can you make 3a+5b=1?

[wicked copper]

Yes, hence taking some multiple of that will yield any element in Z, thanks 👍

[obtuse lance]

nice

[obtuse lance]

ok now generally what conditions on integers u,v are required so that the set X = {ua + vb | a, b \in \mathbb{Z}} equal to \mathbb{Z}?

[wicked copper]

Hmmmm, u is coprime with v?

[obtuse lance]

yep that ends up being the answer

[wicked copper]

Okay, good to know

Thanks (if it wasn't obvious, this is part of a bigger question, just a spurious thought 😄)

[vital dew]

Amorous aka Lucifer

[stray reef]

oh, that kind of set theory? #foundations may be better suited for it

[rotund timber]

is there a reason why the word "that" is in italic?

[rotund timber]

thanks

[stray reef]

italics are the plaintext equivalent of verbal emphasis

when you pinged me in #precalculus asking where to post set theory questions, i thought you might have been referring to more elementary stuff like set counting with inclusion-exclusion

and the like

[rotund timber]

ohk

[rotund timber]

got it

[rustic zealot]

I'm really lost.. I really don't know where to start. The question is

A rectangular chocolate bar is made up of n squares. The whole bar or any part of it
of the bar can be separated along a vertical line or a horizontal line which separates the ´
squares. If the bar can only be separated one part at a time, determine the division number
that must be done in order to separate the n squares. Use the strong induction to prove
your answer.

Why do i need the strong induction ?

[rotund timber]

maybe because weak isn't working

[rotund timber]

what is your approach?

[warped yacht]

Is there a channel for Curve Tracing, asymptotes, radius of curvature

[vital dew]

Amorous aka Lucifer

[rotund timber]

@rustic zealot

this is not my answer btw..just posted this because you were lost

[rotund timber]

and yes you don't need strong induction...it's just because you have been told to do the problem in a certain way by your instructor

[unique abyss]

You are using strong induction above though? Im a bit confused

[wicked copper]

Hmm, is there an identity which states that if a \subset b and b \subset a, then a=b?

[little prism]

that's the definition of a=b

[wicked copper]

Oh, cool

[little prism]

or well an equivalent definition anyway. who knows how your course defined it

[wicked copper]

Just making sure, would this therefore work?

[rotund timber]

i am using but its not necessary

[gentle raven]

Suppose, I have an undirected planar graph of n vertices, what is the maximum number of edges possible?

[little prism]

every edge corresponds to an unordered pair of vertices

how many of those are there?

[gentle raven]

n choose 2 ?

[little prism]

is that a question or an answer

[gentle raven]

i doubt whether it's the answer

[little prism]

why do you doubt it?

[gentle raven]

so it's the answer?

ok

[little prism]

yes it is

[gentle raven]

graph theory is just weird to me

however, i can understand and apply it on the questions

[stray reef]

still, why do/did you doubt it?

[plucky island]

Hey I gotta induction question can you prove something assuming k-1 and then prove k

It’s logically equivalent yeah?

[unique abyss]

Yeah, this is typically what you see in induction

and then k implies k+1

and k+1 implies k+2

thus creating a chain of implications

that's why induction works as a proof technique

[plucky island]

Gotcha so you can either assume k and prove k+1 or k-1 to prove k whatever makes the question easier ?

[unique abyss]

You got it

usually there's no difference between the two choices you presented

[plucky island]

And if you have strong induction with your assumption being 1<i<k can you just prove k from i or do you need to still do k+1 cause logically isn’t it the same as k-1?

[unique abyss]

For strong induction, your assumption is that the statement holds from base-case to k inclusive and then proving k -> k+1

the only difference between weak induction and strong induction is the assumption

[plucky island]

I see

Does that assumption ever let you do i as a k-1 term though

or no

have i and then prove k?

[unique abyss]

can you re-phrase that

If you're saying, assuming base-case to k-1 inclusive and then proving k

sure you can do that

That could be considered strong induction

[plucky island]

So let’s say we have a function that we can satisfy from 1<i<k could you have the function of i be equivalent to a k-1 term then prove k

Does that make more sense

Maybe I can look for a question

Here I found something where I think that’s what they did

[unique abyss]

It's strong induction yeah

Idk why it's assuming the statement for 0 ≤ i ≤ k

and then proving the statement for k

when it's assuming that in the inductive hypothesis

Maybe there's something I'm missing so I'll defer your question to someone else

[plucky island]

Okie well you’ve been good help I’m kinda confused bout it also

[unique abyss]

could just be a typo and that they mean to write 0 ≤ i ≤ k-1

[plucky island]

Yeah I’m not sure

[burnt lynx]

can someone explain how to find equivalence classes for this problem

our professor never went over this and he gave it to us as a hw problem

btw I already showed it's a equivalence relation

[frank cove]

The problem asks you to split up A into a disjoint union of equivalence classes. Recall that for any a∈A, the equivalence class [a] is the set of all b∈A such that aRb. {-3,-1,1,3} and{-2,0,2} are the equivalence classes. (In an equivalence class each element is related to every other element.)

[fossil bluff]

how tf did you get that tho

[little prism]

well you could for example just start with one element, e.g. -3 and go through all other elements a in A and check whether -3Ra

then when you are done with that take the next element which isn't already accounted for and repeat

in this case you can speed it up a little by calculating a^2 mod 4 for all a in A and noticing that aRb iff a^2=b^2 mod 4

[quaint notch]

Hey guys, any chance someone knows of a proof or intuition video (well or text) for proving the connections between Zorn's lemma ~ Axiom of choice ~ Well order principle? but with all 6 proofs? 🙂

[final cliff]

Hrbacek/Jech 3rd edition ch 8 covers it if you can't find a video.

[distant iris]

Hey guys. Is anyone here familiar with Sudoku Algorithms? I have a puzzle I'm working on and could use some help. It's in a Sudoku like format and I suck at discrete math. I think some First Order Logic is needed.

Every time I try making expressions they end up stupidly long. Just the first cell ended up taking half my whiteboard.

[limber thorn]

Hello, I am trying to do some math homework and I am stuck on one step. I understand all the way to the associative law, and I know the correct next step is that distributive law, but I don't know why. I can't work out in my head how that was reduced that way using distributive law

this is my distributive law that I am using, I can't see how either one of those are working here

[hardy citrus]

that step actually uses two instances of distributive law

[limber thorn]

ya I figured it was using one on the right and the left

I just cant figure out what they are plugging in for p,q, and r both times

[hardy citrus]

the thing being plugged in for p is the term appearing twice

slash the thing occuring outside the (nested) parentheses in the line before it

[limber thorn]

hmm, are they using the first or second law?

[hardy citrus]

you tell me

one wedge/vee occurs inside the parentheses and one occurs outside them

^and the other

(wedge = /, vee = V)

discord deletes symbols I guess, wedge = upside down V

[limber thorn]

so wedge is the AND symbol?

[hardy citrus]

yes

testing: $$\wedge$$

[vital dew]

AVA

[hardy citrus]

still figuring out discord

[limber thorn]

I don't know, I feel like my brain is melting

I can't find a repeating P in there, it's probably something simple I am missing

the first PˬQ I can't tell how it's turning into what they have it turning into

[hardy citrus]

complex expressions can be brain melting indeed

if you've got a notebook handy, I recommend just writing down the expressions inside the leftmost [] brackets

one on top of each other

and it'll make it easier

and then do the same with the rightmost [] brackets

[limber thorn]

what's messing with me is that hte left most brackets only have two variables

and the distributive law has 3

or are they treating ¬q as an r?

[hardy citrus]

They are

You can always substitute a more complicated expression in for a variable

i.e. we have the commutative law x+y = y+x, and so you can say (2z+xy) + (3y-z) = (3y-z) + (2z+xy)

and you can re-use the same variable too, i.e. "x+x = x+x"

or maybe a better example is going from "(x+y)+z = x+(y+z)" to "(x+x)+y = x+(x+y)"

[limber thorn]

holy crap that just blew my mind wide open

ok so I was overthinking it

[hardy citrus]

prolly

how was that a homework assignment btw? what do you need to turn in?

[limber thorn]

it's this mcgaw hill connect thing, where I place all of those in proper order and go to the next one

they have some unuse dones

I can work out the right answer by elimintation and logic but I wanted to really understand each one fully

[quaint notch]

Can I find it online?

[final cliff]

You should definitely not search for it online on piracy websites. Definitely don't illegally pirate it and read it for free.

[quaint notch]

Ok thank you, I'll go look for it on amazon 🙂

[timber sequoia]

can anyone help me prove

A-(A-B) = A cap B

[hardy citrus]

you can do a standard "let x be in [LHS / RHS]... then x is in [RHS / LHS]"

any luck with either direction?

[silver panther]

I got a different answer to the solution but it seems valid.

Sum of number of correct answers is 720 or more so atleast 1 person who solved 4 or more questions. Say this person solved Q1,2,3,4 (Without loss of generality) then the sum of the number of people who solved 5 and 6 is 240 so at least 1 person in their intersection so we are done.

[weak grail]

Hey people, can some check my work. It is a question on permutations

The question is on permutations:

How many 4 letter words can you make using the word B A N A N A S? repititions/duplicates not allowed.

[stray reef]

can you upload this as a non-HEIC file

perhaps as PNG instead

[weak grail]

@stray reef

[stray reef]

that doesn't look right to me

this assumes all selections of 4 letters will feature the 3 A's (hence division by 3!) and both N's (hence division by 2!) which just is not true

(also you are inconsistent about stroking your 7's; all of them should have a stroke, not just the first one on the page

[weak grail]

Isnt AAAN and NNAA part of the 840 ways?

[weak grail]

Ill take care this time

[stray reef]

yeah but they arent the only ones

so you cant just divide by 12 like you did

[weak grail]

Is it correct till 840?

[stray reef]

well sure but you can't really proceed past that point anyway

also just saying. this is not a seven. this is a one.

[weak grail]

I have 7 letters (n=7)
I have to arrange them in 4 ways (r=4)

[weak grail]

Damn that is bad, ill make sure that wont happen the next time

[stray reef]

i am critiquing your handwriting here.

you should always, always, ALWAYS put a stroke through your sevens.

every time you fail to put a stroke through a handwritten 7, a puppy dies

anyway

one would think that we must go through some casework here

perhaps baesd on whether or not B is included and whether or not S is included

or something

i'm not yet sure what the best case breakdown would be

but i see no way to avoid one

[weak grail]

I dont get it

Ok so you say till 840 it is correct yea?

So i have 840 4 letter words i can make with BANANAS

[stray reef]

no

[weak grail]

But the 840 has repitions in them

[stray reef]

you have 840 four-letter words you can make with $BA_1N_1A_2N_2A_3S$

[vital dew]

Ann

[stray reef]

and no you cannot account for these repetitions with a simple division

[weak grail]

So this isnt a newbie permutations combinations question

[stray reef]

this is not a newbie question, no.

[weak grail]

Where i divide by the duplicates and get the answer

Cuz i just started off with permutations and combinations and i found this question browsing the internet

It dint fit anything i was teaching myself till now

Anyways that also means my intuition isnt correct.

Thanks anyways @stray reef

[limber thorn]

I am trying to figure out why those two are showing green as right here

The "Some" symbol at the beginning has me thinking hte top two are right

like can't there be a rabbit that does hop and one that does not hop and that statement still works?

I am wondering if this thing is just wrong or maybe I don't get something

[ember obsidian]

1st is right

2nd isnt; the statement isnt talking about an animal thats necessarily a rabbit

[limber thorn]

it is though, the 2nd one says the animal has to be a rabbit and it at least once hops

[ember obsidian]

i mean the problem statement

[limber thorn]

seems to fit the SOME in the beginning, again I might be missing something.. But you agree the answers they are giving are wrong right?

Well R(x) makes it a rabbit no?

even if the domain is all animals, the function limits those to rabbits

[ember obsidian]

but thats not all the problem statement says

[limber thorn]

I am getting SOME function of domain is rabbits therefore hops

[ember obsidian]

it says there exists x such that:
if R(x) then H(x)

[limber thorn]

if the animal is a rabbit then it hops

but the backwards E means at least once, not all

or you mean the 2nd statment would need an AND symbool

not an if/then

[ember obsidian]

yes

choice 2 talks about an animal thats a rabbit and hops

[limber thorn]

ok that makes sense

the question feels like its looking for 2 answers but only the first one is correct right?

same with this one, the two red ones are the ones I had as right

[ember obsidian]

yes

[ember obsidian]

reds are right

who wrote these?

[limber thorn]

ok cool

it's some program, that mgraw hill connect thing

I emailed my professor about it

they make you pay for this crap...

[ember obsidian]

hopefully u get points back

[limber thorn]

ya I am sure I will, my professor is pretty cool