discrete-math

[grave sail]

n + n + n + n + n + n + n + n + n + n ... right? each time its O(N) ? or am i thinking wrong

wait

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 /

[little prism]

well 0+1+2+...+9 but same idea

so if you wanna do N pushes, you need 0+1+2+..+(N-1)

do you know a closed form for that?

[grave sail]

is it just a sum?

[little prism]

well it is a sum, sure

but there is a formula for the sum of the first few natural numbers

[grave sail]

n(n+1)/2

okay i see where this is going

thank you

[little prism]

youre welcome

[heavy shore]

Question

If n! is n(n-1)(n-2)….(3)(2)(1)…

Then what is !n

[brave rock]

Not defined.

Some people use that notation to mean the number of derangements of n elements.

[stray reef]

where did you see the notation !n?

[heavy shore]

I heard it’s a thing for some reason

[stray reef]

you heard.

from whom?

in what context?

[frail yew]

he's talking about sub factorials

he most likely derived it from blackpenredpen

[frail yew]

.

[stray reef]

ok but i want to hear him say where he got it from.

and give a link to the video or article or blog post or whatever else

[frail yew]

I'm curious, how did you become a mathematician?

[stray reef]

long story.

[frail yew]

at a seemingly young age also

I'm willing to listen

[stray reef]

i mean first off what do you imagine when you see me put "mathematician" in my bio

bc i might not exactly correspond to that

i am not terry tao or a fields medalist or any kind of public figure like that

[frail yew]

I believe I could figure that out.

[frail yew]

well,

I expect this to be an unacceptable perception of what mathematicians are, however I think that you are given various opportunities to model various situations, occurrences, processes, etc using mathematics

@stray reef

[stray reef]

i mean this is less about acceptability and more "i dont really get you there"

i have a bachelor's degree in math and am currently doing a master's

[frail yew]

ping me

[stray reef]

...

@frail yew

ok there, i pinged you

[frail yew]

hmm

[frail yew]

sorry

your responses didn't appear on my screen

so I thought you had left or something and I asked that you'd ping me so I can return to the conversation

well how did you obtain the job of mathematician if you are still in university?

[stray reef]

what "title"

i do not know of any formal governing body that regulates and distributes the word "mathematician" as a title

[frail yew]

job

[stray reef]

i work as a math teacher.

but i suppose this wouldn't be good enough for your highness.

[frail yew]

I'm sorry

[stray reef]

what a pedestrian, lowly, PEONIC job for somebody to have that claims to be a Mathematician™️, amirite.

[stray reef]

for what?

[frail yew]

implying that you were claiming to be a mathematician

which you aren't

I've learned

[agile magnet]

the "which you aren't" clarification is insane

[brave rock]

Yes

[warm stirrup]

Thank you

[warm stirrup]

Also how to prove something is reflexive, like the mathematical proof. Thank you

[queen mulch]

if anyone can help with the question above!

[warm stirrup]

<@&286206848099549185> pls if anyone can help me solve it

[rapid mural]

start with the definition of reflexive.

[silver jetty]

Is this correct? If not lmk what step I messed up on, TY

[distant bobcat]

For three non disjoint sets $A,B,C$ am I right to assume that $ |A \cap B | + |A \cap C | + |A \cap B \cap C| \leq |A| $ does not hold in general? I am thinking here we are adding $|A \cap B \cap C| $ several times such that its cardinality may exceed that of $A$.

[vital dew]

FredrikPiano

[brave rock]

Even just |A n B| + |A n C| <= |A| doesn't hold in general

[distant bobcat]

How about $ |A \cap B | \backslash {|A \cap B \cap C| } + |A \cap C |\backslash {|A \cap B \cap C| } + |A \cap B \cap C| \leq |A|$?

[vital dew]

Boytjie

[brave rock]

This is not correct notation

But you are trying to write

[distant bobcat]

no, I want with sets

[brave rock]

|A n B| + |A n C| - |A n B n C| <= A

[distant bobcat]

yes

here we dont add duplicates

[brave rock]

And indeed since in general |X u Y| = |X| + |Y| - |X n Y| you can clearly see this works.

This is called the inclusion-exclusion principle.

The only slight leap is seeing that (A n B) u (A n C) is a subset of A.

[distant bobcat]

yes, but if we assume x is an element of (A n B) u (A n C) then it is also an element of A

[brave rock]

'but'?

Yes, indeed as I said it is a subset of A

[distant bobcat]

yes, I understood

another one that I had to use recently was |A n B n C| <= |A n B| and |A n B| <= B

wondering if I can see this as well using the inclusion-exlusion principle. Otherwise the proof is simple using a typical set theory proof

[brave rock]

That is just the fact that a subset cannot exceed a superset in size.

If A is a subset of B then |A| <= |B|.

[distant bobcat]

A superset is a set which contains all the elements of the other set right? Like pacman swallowed the set I guess haha

[weary tiger]

How do you even say these two expressions are equivalent? Can someone pls give me a hint?

[brave rock]

$\sum_{i=1}^n{(2i+1)} = 2\left(\sum_{i=1}^n i\right) + \sum_{i=1}^n(1)$

[vital dew]

Boytjie

[brave rock]

@weary tiger this is what's happening on the top of the fraction, that's all it is.

I've bracketed aggressively, hopefully there's no confusion

[weary tiger]

I see, thank you

[timber barn]

Is f(x) = x^2 -2 surjection and injection?

[brave rock]

What do you think? Recall the definitions of surjectivity and injectivity. Maybe draw a picture of f

Remember, you also need to specify the domain and codomain of f.

[timber barn]

@brave rock I had an exam and I have proven that is injective and surjective but its incorrect so thats why I need help

Even chat gpt says its surjective and injective

[brave rock]

Well it's wrong :) hope that helps

[timber barn]

Oh so its not surjective and injective right?

[brave rock]

If you're not going to engage with my suggestions up here, then we're out of luck, because I'm not interested in just telling you the answer.

[timber barn]

Then just dont reply

I dont get the point of writing a paragraph and not telling the answer because you are not interested

[shadow lion]

what co/domain are we considering?

R?

the domain you consider is gonna affect whether f is injective or not, while the codomain affects whether f is surjective or not

[errant bear]

gives incomplete question
complains when not given the answer

[stray reef]

!noans

[lofty cloud]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[unborn raft]

find the mass of ρ(x,y,z) = xz is a mass density function for T: a solid bound by x^2 + y^2 +z^2 = 4 and z = sqroot(x^2 + y^2). any one can help? I am getting 0

[naive parrot]

does anybody have any resourse for a question bank on mathematical foundations for machine learning

[stray reef]

what is this doing in #discrete-math

when it is clearly #multivariable-calculus

[odd prism]

How can I start to think about this

[elder berry]

# of orange stars + # of black stars = # of stars

[odd prism]

Right we have a square of length n and width n+1

Where each cell has a star

[elder berry]

right, so in total that is n(n+1) stars right?

since each cell has a star, and since there are n(n+1) cells

[elder berry]

do you agree with this?

[strong furnace]

Is there a word for when a function from a set, S to another set Y, has the property that every element of S is mapped by the function

[agile magnet]

Every function must satisfy this

part of the definition of a function f is that for every element x in the domain (the domain here is S), f(x) must be defined

[strong furnace]

ok that makes sense

I wasn't sure if something like sqrt(x): R to R is not allowed

So I guess you would have to say like sqrt(x) positive real to R

[vital dew]

Jacks0n

[half knot]

Ignore above please

[muted moth]

is it better to regard graphs as a triple: the vertex set, the edge set, and a relation/function relating vertices to edges?

[brave rock]

Better than what?

It's a set-theoretic way to construct it, certainly. I'm not sure what the alternative would be.

You certainly can't throw away the geometric interpretation though, that's really crucial to the study of graphs. You can write down what that means in terms of the set-theoretic definition, but this is complex.

uh complex as in complicated, not real vs complex lol

[stray reef]

better than what & better in what way

[muted moth]

yeah sorry! I forgot to mention the conventional definition of graphs provided in textbooks

the definition of "a graph being an object containing two sets, the vertex set and the edge set."

are we better off referring to graph as containing two elements, or should we also include the idea of a relation mapping individual vertices to edges or something like that?

[stray reef]

again, better in what way?

[muted moth]

i am just being pedantic

[stray reef]

bruh

[muted moth]

is it more accurate

[stray reef]

[muted moth]

yeah, is it more accurate to think of graphs as a triple?

[brave rock]

More accurate than what

[cerulean radish]

I don't see how that's better in any way

The relation that you are talking about can be derived using the set of vertices and edges anyway, so that's just extra information

[brave rock]

This isn't true, for example multigraphs cannot be described only using a relation between vertices.

[cerulean radish]

Fair

I should stop associating graphs with simple graphs

[stray reef]

if and only if you want to fuel your own pedantry.

[rocky jewel]

is this proof clear enough and sufficient: tex b) uncountably infinite. let s be a string of 0's and 1's, representing a total function f from $\mathbb{N}$ to ${0,1}$ where the ith element of S $s[i]=f(i)$, and let $S$ be the set of all such functions (assuming S is countably infinite). if we put all s in a matrix, we can form a new function $s_n$ by going down the diagonal and setting to the opposite of the element on the diagonal. Since $s_n$ differs from every function in $S$ it is not in S, and so S cannot be countably infinite. the question is to prove:

[vital dew]

Fb_Wdw

[brave rock]

That is sufficient. The explanation is brief but clear enough.

Really it's not a set you want but a bijection, I don't care enough to correct that.

[little prism]

a matrix is usually finite. better to say that you are writing it into a list

[fossil mural]

i'd say the most accurate way to think of it is that it doesn't really matter what representation you use because there are obvious ways to convert between them

the same way that nobody cares if the "real numbers" are dedekind cuts or cauchy sequences or whatever, the word "real number" just refers to the result (it's a complete ordered field), not any particular encoding

if you're like, writing in a proof assistant, or doing model theory or something, then the exact encoding does matter somewhat, but you can just pick one and it probably won't really make a difference which encoding you're using - it's the same kind of choice as "what language are you going to write in", it's a good idea to be consistent but it doesn't really matter to the actual mathematics

if it turns out that it does meaningfully change something, then... just decide on one encoding based on whatever's convenient i guess? or give both encodings names and think of them as related but distinct objects? the important thing is just that, for anything where it does matter which encoding you use, you're consistent about it

for a lot of things you can just, implicitly insert conversions, if you have a theorem that's proven about one encoding and want the same theorem about the other encoding, which is why it doesn't matter

[soft iron]

Hi

Can someone help me

There are some pairs of students assigned with distinct prime numbers that have exactly one student assigned with a composite number between them.

<@&286206848099549185>

[stray reef]

can we see the problem exactly as it appears in the textbook?

cause your logical notation looks strange???

@soft iron

[soft iron]

Ok

Basically can't do prime (x,y) because it's not multivariate as it implies I know x has y

Negation must be processed through

-(P--> Q)

= -(-PVQ) ...... Implication law

Then flips and multiply through

Demorgans , DN

[stray reef]

that is not what i asked for!

!original

[lofty cloud]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[soft iron]

Ok

B)

@stray reef

[stray reef]

hmm

are we allowed <

wait hold on

we don't even have a way to refer to the number assigned to a student

so how can we possibly translate "composite number between [the prime students' numbers]"

i think this problem is screwed up

[soft iron]

It means the same as a number is between two different prime numbers

So, idk about the greater or less than symbols

I'm thinking now on the best alternate

[stray reef]

well then we simply have NO WAY AT ALL of talking about any order relations on students' numbers

like that's a blocking limitation

[soft iron]

Wdym

That's why I put the V 'or' to say the order is unknown

Just Z is between

Wait

I'm still on a

[stray reef]

YOU CAN'T TALK ABOUT BETWEENNESS AT ALL

our variables can only refer to students

we have no way of saying "student x is assigned the number 37"

[soft iron]

The students have different numbers

And the stem tells us that

[stray reef]

do you not understand what i am saying here

[soft iron]

We don't know what the numbers are ONLY that IT the composite is somewhere between

The PRIME

OPTIMUS PRIME

[stray reef]

ok you clearly do not understand me.

[soft iron]

I'm gonna fail this unit

Hehe predicate logic more like esteemed seconds language

[proven ibex]

is the chromatic number of K_n graphs just n?

[cerulean radish]

Should be

[calm chasm]

Say you have a mean of 5.
You're given the n for this. You're told that one of the values x was changed from 10 to 5. How would you compute the new mean exactly?

[stray reef]

well

mean = total/count

when you reduced one of the data points by 5, how much did the total change

[thorny hollow]

Hi

[calm chasm]

True

[thorny hollow]

I am a little confused here for question 2. Does repetition is allowed or not?

"2- Airports are identified with 3-letter codes. For example, the Richmond, Virginia airport has the code RIC, and Portland, Oregon has PDX. How many dif f erent 3-letter codes are possible?"

[calm chasm]

You just reduced it by 5

[stray reef]

yes exactly. the new total is the old - 5, and that's it

[calm chasm]

Wow that was so easy

I was helping someone with this an thought I was forgetting something about summation. But that's all it was.

[stray reef]

yes, repetition is allowed

[thorny hollow]

If repetition is allowed, then thr answer is 26³?

[stray reef]

yes

there are IRL airport codes with repeated letters

Cochabamba, Bolivia is CBB for example

[thorny hollow]

Ic

Thx

[quiet eagle]

Is there a formula for the number of circuits in the complete graph K_n?

[weary tiger]

https://media.discordapp.net/attachments/576511723574525962/1209485456652107796/image.png?ex=65e7181c&is=65d4a31c&hm=db7d10fcc102ee4d78bb6d658f84cf2911f5223cee13a8ba3a377a825c923183&

I attempted to construct such a partition by sth like X-g(Y') and rest of X like constructs where Y' is a subset

Didn't work out

I suspect that we can somehow overkill this using some maximality argument in form of zorn's or sth

Or there might actually be a construction

In any case I am currently stuck and would greatly appreciate some help

[quiet eagle]

I believe it is $\sum_{k=3}^n\binom{n}{k-2}$

[vital dew]

Sciencenjoyer

[grand totem]

Hint: Consider the map $\varphi:PX\to PX$ that sends an $X'$ to $X\setminus g(Y\setminus f(X'))$. Show that this map is monotone (with respect to the inclusion ordering on $PX$) and hence has a fixed point (since $PX$ is a complete lattice). Use that fixed point as your $X_1$ and go from there

[vital dew]

neverbloom6760

[compact owl]

How do we go from the first part of the highlight to the next?

[neon harbor]

x - 3 = -3 * (-x/3 + 1) and move the -3 outside the fraction

[compact owl]

Ah, gotcha. Thank you!

[vernal torrent]

this is a bit of a dumb question but how do you know when to use $n \choose k$ versus $n \times ... \times (n-k - 1) \times (n-k)$

[vital dew]

ImtiSauce

[vernal torrent]

for example, apparently b) is $12 \choose 6$ and c) is $12 \times ... \times 6)

[vital dew]

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

[vernal torrent]

apparently it has something to do with order?

[neon harbor]

(n choose k) is just n * (n-1) * ... * (n - k + 1) / k!, ie. the number of ways to pick k things among n things divided by the number of ways to order k things

so if the order doesn't matter then use (n choose k), for example if picking blueberry donut then chocolate donut is the same as picking chocolate donut then blueberry

[vernal torrent]

That helps a lot, thank you!

[neon harbor]

by the way, I think you got the answers to b) and c) mixed up

[last dome]

Im a little stuck on the process, from what I understand is you build out a few of the iterations to find a pattern by simplifying but I think im having trouble turning that into an explicit formula

[calm chasm]

When creating cases with absolute values, How do we do that

[rugged dock]

Hi. Recall definition of absolute value func.: $y=x if x\ge 0 or y=-x if x<0$. Now, since we only have two absolute values, we need to prove the following cases:

[vital dew]

javier

[rugged dock]

case 1: $x-1, x+1$, case 2: $-x+1, -x-1$, case 3: $x-1, -x-1$, case 4: $-x+1, x+1$

[vital dew]

javier

[rugged dock]

If you find that one of the proofs for some case is exactly the same for the previous case, just say "similarly, conclusion goes here"

[weary tiger]

#foundations message

[vital dew]

Eso

[paper jolt]

rip i meant to say B \cap Y = Y - A

[grand totem]

That's not true either? If B is empty and Y is inhabited then their intersection is empty, while Y - A isn't

Just realized you corrected the assumption not the conclusion

(So this is indeed true)

[paper jolt]

yeah. how does one show it

[grand totem]

Like one usually shows set equality, show that the left side is a subset of the right side and vice versa

[paper jolt]

hoped there was a more direct way

[compact owl]

The screenshot on the left is the series I got for the generating function on the right. Is my solution accurate?

[wintry cedar]

I dont think you meant to have a + here but other than that it looks good

[lapis mulch]

how do i find the number of spanning trees of G?

[twilit sundial]

casework

particularly on which of the "top" and "diagonal" edges (or both) that you include

[lapis mulch]

how about the minimum weight? it's simply 5 right

[gilded sun]

@agile magnet well yeah,

they got 5^(3) because theres 5 consonants

and 3 vowels

[agile magnet]

No that's not what it means

You chose 3 spots for the vowels to be in correct? That's what the 8 choose 3 is

But you haven't chosen which vowel goes in each spot

How many vowels are there? 5
How many times do we have to pick a vowel? Once per spot so 3 total

Hence 5^3

Makes sense?

[gilded sun]

5 vowels in english language, and each of the three positions for the vowels can be filled with any of the 5 vowels, ok.

[agile magnet]

bingo

So now can you walk through a very similar argument to see how they got 21^5?

There are 21 consonants

[gilded sun]

Okay right, with a google search I can get that, but shouldn't that be included in the question?

[agile magnet]

I guess it assumes that you are familiar with the English alphabet

[gilded sun]

Could we go over some more problems?

[agile magnet]

You can ask them here, if I see them I'll help otherwise someone else will

I'm boarding a flight soon so idk how much longer I'll be able to help

[gilded sun]

So same process. There's 21 consonants in the english language. So for strings with exactly 3 vowels, we choose 3 positions out o 8 for the vowels (8 3) * 5^(3)*(21)^(5)

For strings with 4 vowels, we have ( 8 4) because we choose 4 positions out of 8 fo the vowels, and then we multiply that by 5^(4), and then we multiply that by 21^(4)... So the total number of strings is (8 4) x 5^(4) x 21^(4).

[compact star]

Theorem: Let n be an integer. If n^2 is odd, then n is odd.
Proof: Let n be an integer. Assume n^2 is odd and n is even. Then there is some integer k such that n=2k. Then n^2=(2k)^2 = 4k^2, which is even. Therefore, n must be odd.
Is the proof valid, and which type of proof is used?

[gilded sun]

For 5 vowels, we have ( 8 5) because we choose 5 positions out of 8 for the vowels, and then we multiply that by 5^(5), and then w multiply that by 21^(3) (because there's 3 remaining positions).

Then we add the number of strings from each case together.

[agile magnet]

That's exactly correct

The 3 cases are disjoint so you can simply sum the answer

And you computed each case correctly it seems

[agile magnet]

What do you think?

Does it seem valid? Why or why not?

[compact star]

yeah, 3^2 is 9 and is odd. seems valid.

the proof by contradiction is, 'n must be odd' ?

[agile magnet]

Well this shows that it works when n = 3, do you believe it's valid for all n?

You are correct that this is a proof by contradiction

[gilded sun]

I only now have questions with this. I might be a little lost in my steps since it's late here...

[compact star]

@agile magnet yes, because 3 is odd and all odd numbers exhibit the same behavior

[agile magnet]

3 does not exhibit the same behavior as all odd numbers

For example, 3 is divisible by 3 but 5 is not

So can you be specific about what you mean by same behavior?

[compact star]

@agile magnet yes, because 3 is odd and all odd numbers exhibit the same behavior in relationship to even numbers

[agile magnet]

Only reason I'm being pedantic is because this is important when first learning proofs. I'm sure you know the answer but learning how to phrase that in writing is important

[gilded sun]

What I'm thinking is this: For the 1st student, there's 9 possibilities. 9 -1 = 8. For the 2nd student, there's 8 possibilities, 8-1 = 7. For the 3rd student, there's 7 possibilities, 7-1 = 6. For the 4th student, there's 6 possibiilities, 6-1 = 5. For the 5th student, there's 5 posibilities, 5-1 = 4. For the 6th student, there's 4 possibilities, 4-1 = 3. For the 7th student, there's 3 possibilities, 3-1 = 2. For the 8th student, there's 2 possibilities, 2-1 = 1. And lastly for the 9th student, there's 1 possibility..1-1 = 0.

[agile magnet]

Well really what you need to verify is that each logical deduction in the written proof is correct as written.

[agile magnet]

But by the time you've gotten to student 5, all students have been paired

[compact star]

@agile magnet indeed, I just started discrete math

[agile magnet]

Also after you pair 2 students, then there are only 8 students left

Can you answer the question for if there are 2 students total? Then 4 students total?

[agile magnet]

So the question is asking about the written proof there.

So far you have identified the claim is true and that it's a proof by contradiction

But you're still yet to tell me if you think the proof is correct

So do you think the proof is valid?

[compact star]

Assume n^2 is odd and n is even. 2^2 != odd. Therefore I'm thinking that doesn't look valid.

Should I expect to halt when I identify something that doesn't look valid, or continue with the proof.

If I'm driving down a street, and see a sinkhole. I don't try to drive my Subaru into it.

Exhibit A

[gilded sun]

Could we do this using binomial coefficient computation?

So we can do ((10 2) x (8 2) x (6 2) x (4 2) x (2 2))/(5!)

Simplify that, and find our answer?

[agile magnet]

Yea that's exactly right.

[agile magnet]

You've shown here an example of why the assumption is bad and thus we have a contradiction

This proof says "assume n^2 is odd and n is even" and then shows n^2 is even

Since a number can't be both odd and even, we arrive at a contradiction

So in fact the proof is valid

[compact star]

@agile magnet good point! What I thought I was reading was, assume the result of n^2 is odd and the input of n is even.

Maybe my English comprehension sucks

[agile magnet]

That is exactly what it is assuming

But then it shows that the assumption yields that n^2 is simultaneously odd and even

So the assumption is nonsense

Thus, n must be odd

[compact star]

I went back to the text, but didn't see the area which shows simultaneously odd and even. Can you provide the text

[agile magnet]

Do you see in the assumption where it says "assume n^2 is odd" and then later they show n^2 is even...

It's like 4 sentences man

[gilded sun]

Could I ask questions relating to k-Subsets (or Combinations) of Multisets?

[gritty breach]

for (i), I believe no such graph exists but I can't find a way to explain why concisely. I know that since |V(G)| = 5, a degree of 4 means the vertex must connect to all other vertices. Since there are three of them, then all vertices must have degree >= 3. Is this correct? and what would be better way to phrase this?

[agile magnet]

Yea

[agile magnet]

Nah what you said is perfect

To make it more concrete, you could start with saying "suppose such a graph exists." and end with "Thus, the two degree 2 nodes must have degree ≥ 3, a contradiction"

[gilded sun]

Would we find how many different 5 subsets we can make doing ( 6 1)?

[agile magnet]

How did you get that? Can you explain your work to me?

[gilded sun]

There’s 6 unique letters. Since we want exactly 5 letters, a letter is excluded from each subset. So we have 2-1 = 1, (2 being the choices to include or not include in the subset)

[agile magnet]

I don't get the relevance of the 2 - 1 = 1

Choice to include what in the subset?

And where does the 6 choose 1 come from you never explained that to me

[gilded sun]

I’m confused, does the entire expression not make sense, or just 1?

[agile magnet]

You say "we have 2 - 1 = 1" and then say 2 is the choice to include or not include in the subset

Include or not includewhat? You're talking about some object but I can tell what you're talking about

[agile magnet]

And then here you said the answer is 6 choose 1 and you never brought up that expression in your final answer

I'm not saying you're wrong but you need to be able to justify your work. I'm not going to sit here and just say yes or no your answer is right. This is also how you check your own work for yourself, which is the end goal

[gilded sun]

I don’t think you think my answer was much of a justification then.

[agile magnet]

You haven't told me what the 2 - 1 = 1 means

You said it's including or not including something

Idk what that something is

[gritty breach]

How can I improve on this answer? It feels too wordy or informal or something... I attached the question for reference.
Consider a 9 vertex graph G with 4 vertices of degree 6 and 5 vertices of degree 5. Then the total edges of G |E(G)| can be computed as 1/2 E deg(Vi) = ... = 49/2. However, it is impossible to have half an edge, so we must either deduct or add a degree to a vertex. Since all vertices must be of degree 5 or 6, we have only two options: deduct a degree from one of the 6 degree vertices, resulting in six 5 degree vertices and three 6 degree vertices; or add a degree to one of the 5 degree vertices, resulting in four 5 degree vertices and five 6 degree vertices. In either case, there are at least 5 vertices of degree 6 or at least 6
vertices of degree 5.

[coral raven]

i'd just say: the only way the conditions cannot be satisfied is if there are exactly 4 vertices of degree 6 and 5 vertices of degree 5

but then the total degree is odd, but total degree is always even, contradiction, so we are done

[gritty breach]

wow very concise, nice! I've always been dog shit at worded answers

[spark gazelle]

does anyone know of a good cheat sheet website/website that would help me in making a cheat sheet on induction and homogeneous and non-homogeneous recurrence relations?

[forest breach]

Having trouble starting with the following proof, I already tried contrapositive.

If 3n^2+n+2 is even, then n is odd

[stray reef]

well can you write down the contrapositive of this stmt? @forest breach

[forest breach]

If n is even, then 3n^2+n+2 is odd

Then I did

Let n be an even integer
n = 2k
v = 3(2k)^2+(2k)+2, where v is some integer
v = 12k^2+2k+2, v = 2(6k^2+k+2)

The issue is this proves that 3n^2+n+2 is even, not odd

[stray reef]

ok well alright

was your question "Prove this"

or was it "Prove or disprove this"

[forest breach]

It was "Write a proof for the following statement: If 3n2 + n + 2 is even, then n is odd."

[stray reef]

lmao

they're making you prove false statements??

[brave rock]

It is often hard to prove false statements

[stray reef]

you sure nobody made a typo at any point down the line?

errata in the book etc?

[forest breach]

https://i.gyazo.com/673051eedeabadba4f4f5606811f6893.png

[stray reef]

[brave rock]

You should email whoever set this worksheet with a counterexample highlighting the mistake.

[agile magnet]

[stray reef]

yeah is emailing possible

if not you are just fucked

but HOPEFULLY it's incompetence and not outright malice

[agile magnet]

Did they mean the converse?

[fossil mural]

i mean if you can't email them you could just submit a counterexample instead of a proof

[agile magnet]

if n is odd then 3n^2 + n + 2 is even

Because that's true

[fossil mural]

in fact 3n^2 + n + 2 is even for all n

[agile magnet]

What's written is not true lol

[forest breach]

[agile magnet]

It isn't tho

[brave rock]

Just do this

[forest breach]

I know, that's why I was so confused, because there is no way to write a proof for

"If 3n^2 + n + 2 is even, then n is odd."

[brave rock]

No point getting worked up about the fact whoever set this task made a mistake, just try to resolve it

[forest breach]

Thanks for the help all

[compact owl]

e^x is the exponential generating function that produces the sequence on the right. How can I rightward shift the sequence by 2 0's?

Having a hard time understanding the integration aspect of rightward shifting...

[brave rock]

Well, the sequence corresponding to an e.g.f named f(x) is f(0), f'(0), f''(0), .. etc.

So let's say F is the integral of f with F(0) = a for some constant a. Remember when we integrate a function we can add a constant to let F(0) be whatever we like.

So what's the sequence corresponding to F(x)? The first value is a, and the rest is exactly the same as f(0) just shifted by one, since differentiation undoes integration.

So OK, with that in mind, how do we get the EGF of the sequence 0, 1, 1, 1, ...?

Well we have f(x) = e^x, and we want to find the integral F(x) with F(0) = 0.

Now the integral of e^x is e^x + c, so we choose c = -1.

So the sequence corresponding to the EGF e^x - 1 is 0, 1, 1, 1, ...

Now it's your turn! Use this to find the EGF of 0, 0, 1, 1, 1, ...

[compact owl]

I would assume that the EGF would be e^x - 2 in that case

[brave rock]

I do not like guesses.

If you don't have good reasoning, I don't consider it an answer. Use the explanation I gave you to work it out.

[compact owl]

First off, thank you for the thorough explanation! Highly appreciated. Based off the logic you have provided, I would say that to rightward shift by another 0, we could use the integral of F(0) with C being 0 as well. So the integral of e^x - 1 is e^x - 1 + 0

[brave rock]

This is not correct, try again.

I don't think you believe that the integral of e^x - 1 is e^x - 1.

[compact owl]

Am I on the right track though? As in attempting to find the integral of (e^x -1) + C

[brave rock]

I put out steps above

Maybe you should think about why I chose c = -1 in that specific example.

[compact owl]

Is it because you are comparing the difference between f(0) = 1 and F(0) = 0

So c becomes -1

[brave rock]

You should reread this from the top, I think

I don't like guesses!

[compact owl]

Ok so, as you've mentioned the integral of f(x) = e^x is e^x + c. This becomes F(x)
So we are trying to find F(0) = 0 + F(1) = 0

[regal pelican]

Using mathematical induction, show that (6^n) -1 is always divisible by 5, given n is a natural number.

[fossil mural]

well that's false so it's going to be rather difficult to prove

[regal pelican]

oh wait mb

[regal pelican]

not anymore

[fossil mural]

ok yeah that's true

[white bear]

How would I go about evaluating this?

I dont see a way to do partial fraction decomp...

[quasi jewel]

I have a question on Gale Shapley-- i know that it always returns the matching where the proposing side gets their best possible option in their preference list

In the sense that best(m) gives the best option s.t. a stable matching exists where m is paired with that option

But I dont understand intuitively why best is an injective function

And I would like a hint on how to prove this using that intuition because I have none rn

[quasi jewel]

Nvm figured it out 💀💀💀

[stray reef]

$\sum_{k=n+1}^{2n} \frac{1}{k^2-1}$ and THEN pfd?

[vital dew]

Ann

[stray reef]

reindex, basically.

[mighty ingot]

anyone knows how to do this

[stray reef]

a nonzero number of people probably do!

[mighty ingot]

im stuck at proving w^6+x^6+y^6 +z^6 mod 8 is even

[stray reef]

are you trying to prove it's even for all possible combinations of w, x, y and z?

[mighty ingot]

yeah

[stray reef]

(w,x,y,z) = (0,0,0,1) should blow that claim to smithereens then, shouldn't it...

so the question is, why are you trying to prove a false statement?

well, that or you didn't answer my question truthfully.

[mighty ingot]

yes!

[stray reef]

that was a "why" question, not a yes-no question.

why are you trying to prove a false statement?

[shrewd obsidian]

it's not

[quaint summit]

how are you getting from the original question you posted to

im stuck at proving w^6+x^6+y^6 +z^6 mod 8 is even

[regal pelican]

can someone help with this please

[thorny hollow]

Hi

[coral raven]

what have you tried

[thorny hollow]

Would the answer for 4 be 4(13!/8!)?

[vestal bronze]

it would

[thorny hollow]

Is the answer for 5 2(26!/21!)?

[vestal bronze]

yeah

[thorny hollow]

why

Nvm

[thorny hollow]

Is the answer for 6 (52!/47!) - (48!/43!)?

[vestal bronze]

no that's a different thing

[thorny hollow]

How

Which is the answer for the question 6?

My reasoning was to subtract the number of possibilites that may contain queens with the number of possibilities that doesn't contain queens, so I arrived at the solution: (52!/47!)-(48!/43!)

[vestal bronze]

so you counted hands that have queens, which is a different problem

it says exactly, not "at least'

[thorny hollow]

Oh

[stray reef]

yeah you need to read the problem in its entirety and not miss those words

also for the love of god please don't let the problem number and your potential answer run together

for 6, is the answer 52!/47! - 48!/43! ?
is the answer for 6 equal to 52!/47! - 48!/43! ?
will the answer for 6 be 52!/47! - 48!/43! ?

three possible rephrasings

[thorny hollow]

Thx

[regal pelican]

I've proven the base case but for the k+1th step I'm not sure what I should put on the right hand side

[uneven violet]

To get started, try this:
$$6^{k+1} - 1 = 6\cdot 6^k - 1 = (5 + 1)\cdot 6^k - 1$$

[vital dew]

OneTrackPony

[quiet eagle]

Do you know if the order of the indices in the three for-loops of the Floyd-Warshall algorithm affects the relaxation process?

[uneven violet]

It doesn't affect the relaxation process, but for practical implementations it can affect performance.

(since multi-dimensional arrays are usually stored in memory in row-major order, and memory caches like good spatial locality, so iterating through consecutive elements in memory is preferred rather than jumping around)

[calm chasm]

I'm going to use a trivial proof technique here. I just need to show that q is true. Now, is x being a real number enough to say that x can be rewritten as a fraction

[little prism]

no, not all real numbers are rationals

[calm chasm]

True

Darn

[rugged dock]

What does the hint say

[calm chasm]

Basically hinted at the proof technique being trivial

[rugged dock]

Proof. Clearly this isn’t true. QED

[calm chasm]

You're such a funny guy

Little rascal

[little prism]

when is p=>q true

[calm chasm]

Easier to say when it's false. It's false when p is true and q is false.

For every other case it's true.

[rugged dock]

Are you reading velleman

[little prism]

ok and what is the case here?

[calm chasm]

Wait

Maybe this is a vacuous proof.

[calm chasm]

I'm thinking it's p = t and q = f

[calm chasm]

Na it's a book my professor wrote.

[little prism]

"2 is irrational" is true?

[calm chasm]

Oh wait

If I write 2 / 1 is that still considered rational

[little prism]

yes

all integers are rational

[calm chasm]

I see

Noyed

X is a real number here

[little prism]

you can also write it as 4/2. or 2000/1000

[calm chasm]

Yeah

How do I say if q is true or false

[little prism]

do you need to say?

[calm chasm]

No

I'm thinking about the question more

[little prism]

you cant say. you dont know enough about x

[calm chasm]

Hmm

[calm chasm]

Well what about x being a real number. Not ever real number could be rational right

So it's not default true

[little prism]

yes

but you dont care

[calm chasm]

Noted

[little prism]

p is false

thats all you care about

[calm chasm]

How about I negate p and show that not p is true

Can i just start like that?

Sometimes when I write these proofs I'm not sure how to start it.

Could I just start by saying p is false. Not p is true and show that 2 can be written as a fraction

[little prism]

yes

[calm chasm]

Bet

[hallow flicker]

Explain this part please? I don't understand how this relation is reflexive.

[stray reef]

!nogpt, generally.

[lofty cloud]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[stray reef]

but in this specific case, it happens not to have messed up. [because you were lucky.]

can you tell me which part of chatgpt's output confuses you?

here maybe i'll break it down into bite sized pieces for you:

[1] For any integer a, we have a - a = 0.
[2] 0 is divisible by 3.
[3] Therefore, a - a is divisible by 3 for all a.
[4] Therefore, (a,a) belongs to R for all a.
[5] Therefore, R is reflexive.

@hallow flicker can you read through this and tell me the earliest point here that you don't understand?

[quiet eagle]

ah, interesting. Thankss

[static briar]

can someone explain how thsi was found, he said it was becuase 15 = 1 mod(7) but I dont know how that fact makes 15x thing go down to the x part with 4mod(7)

[brave rock]

if a = b (mod m) and x = y (mod m) then ax = by (mod m), I hope you are aware?

So if 15 = 1 mod 7, then what does this tell us about 15x.

[static briar]

x = 25mod(7) which means x leaves remaidner of 25 when divided by 7 meaning x = 4mod(7)

[calm chasm]

Question. The hint says I don't need all these inequalities. But I'm not sure how to take that hint. Can I just break up the inequalities and use them?

[rapid mural]

by def we have floor(x) leq x

can you think of something similar with ceil(x)?

[calm chasm]

I wrote the def for both!

[rapid mural]

well you have

1. floor(x) leq x
2. x leq ceil(x)

can you see how to proceed

[calm chasm]

Not exactly. I want to combine these somehow. But the hint stumped me

[rapid mural]

well both inequalities are true

so what can you deduce

[calm chasm]

I don't know...

Are you trying to say they are logically equivalent

[rapid mural]

if a leq b and b leq c

what is the relationship between a and c

[calm chasm]

What is leq?

[rapid mural]

<=

[calm chasm]

Oh

[calm chasm]

Wait so I can split up the inequality from the definition.

Hold on. I'm going to think about this again.

[calm chasm]

In the book I'm using, I can use the fact that x is a real number, we can combine 1 and 2 and get
Floor(x) <= x < ceiling(x).

[rapid mural]

does this solve your problem

[calm chasm]

Not quite

I need to get the floor(x) where x is.

[calm chasm]

Did I make the wrong move here

I can try and do it on my own from here. I think I just needed to start it

[calm chasm]

Okay I think that was actually enough to solve my problem.

Unless the floor(x) fails to be <= the ceiling(x)

[stray reef]

http://dontasktoask.com/

[woeful copper]

For this question I induct over n by proving

$\left(\left(\bigcup_{i=1}^{n}A_i\right) \cup A_{n+1}\right)^c = \left(\bigcap_{i=1}^{n}(A_i)^c\right) \cap (A_{n+1})^c$

right?

[vital dew]

ℝage-E | Lv35+x-Exp:5770+11000

[woeful copper]

I am 99% sure that is the case, I just have trouble setting up the proof. How would I set it up so that it's coherent?

[prime grotto]

wait rage

(give me a second so i can look up the name)

you know de morgan's law?

[woeful copper]

Yup that was exercise 1.2.3, did that.

[prime grotto]

alright so

you also know

that the union of 2 sets is also a set, right?

[woeful copper]

Yup

[prime grotto]

oh wait you wrote something similar above

didn't see

[woeful copper]

Oh right, I could have just used A_n and A_n+1 instead of the mess above

[prime grotto]

yeah that works too

[woeful copper]

So I know I have to prove this and I have proved it, I just have trouble in setting up the induction proof (idk how)

I know how it works

[prime grotto]

so first you prove base case (de morgan's law) which you already did

[woeful copper]

Alright

[prime grotto]

now let's say you have a sequence of sets A_n

now use induction hypothesis

it works for up to first n sets

now you want to prove it works for first n+1 sets

then you use the union = set and de morgan's law to prove it also works

that's a sketch though

[woeful copper]

Ah, so basically
Base Case (n=1,2)
-prove de morgan's laws
Induction Hypothesis
-assume it works for n sets
-then prove for n+1
By induction over n, proved for all neN

[prime grotto]

yep!

[woeful copper]

De morgan?

[prime grotto]

i mean the proof is a sketch, you can extend it a little if you want

[woeful copper]

Oh my bad, my brain for some reason inserted guy in the middle

I read it as sketch "guy" though

[prime grotto]

brain moment

[woeful copper]

lul, thanks a lot hsf!

[prime grotto]

no problem!!!!

[dire stag]

How do I fnd the sequence for 3,7,17,...?

The terms increase by 4, 10, 16, 22. And each of those increases by 6

But how is this related to the above questions?

[uneven violet]

If the sequence 4, 10, 16, 22, … is called c_n, then b_n = b_(n-1) + c_(n-1) for n > 0, and b_0 = 3. Thus

b_n = c_(n-1) + … + c_0 + 3.

So the n’th term of b is related to the previous sum.

The c sequence is arithmetic, and so you can find its sun.

[ocean socket]

what exactly is "a set of order 6"

[icy swift]

i might be wrong but it could mean that the set has 6 elements

https://gyazo.com/4f6dc83843cc93560e2bf513e0420937

https://gyazo.com/68875eb8ad1585dd645a03f206d0917e

i dont understand the logic of the solution,

the part where it says "view all the numbers in the set as different elements"

then you'd be overcounting a set like {1,1,1,1,2,2} 24 times

actually nvm

i think i get it

[ocean socket]

i must say i doubt it. its mentioned in relation to a set which can never be a vector space

[icy swift]

hmm

[ocean socket]

although i see that would be the standard

[icy swift]

could you maybe post the question in its full

or the sentence that its mentioned in

[ocean socket]

Q: "what is a set that can never be a vector space" ans: "a set of order 6" if it is what you say it is, just saying not order 1 or order infinity would be eneough but idk.

ty

[uneven violet]

The order of any finite field F is the power of a prime, say q=p^k. If a vector space V over F has finite dimension n, then the size of V is q^n.

Since 6 is a product of two different primes, it's not possible as the dimension of a vector space.

[ocean socket]

yeah i just got it explained in linalg channel ty

[uneven violet]

Ok, np.

[ocean socket]

saving the answer tho

[brave rock]

Cross-posting and its consequences have been a disaster for the human race

[ocean socket]

well i didnt know it would escalate 😅

[uneven violet]

Just put "Cross-posted in LinAlg" or something, if you for some good reason decide to cross-post.

[compact owl]

My solution (in red) is the sequence for the ordinary generating function at the top. I used to partial fraction decomposition to solve it. Can one of you please verify if my solution is accurate?

[copper tulip]

is this diff eq or discrete structures or neither

[fossil mural]

out of those probably "discrete structures"? (but in the sense where like, it's about things that are discrete, it's not necessarily related to anything in particular that someone decided to call "discrete structures" because sometimes people give things names that don't make much sense)

[woeful copper]

You are correct, discrete structures include combi, grapj theory and proofs and logic

Basically the purpose of this channel :p

Unless it's another topic I'm unaware of

[weary tiger]

I love

I will have an exam like this in two weeks

Anyone have any recommendation for other exams I could take to practice?

oh wait tahts homework

https://file.groupme.com/v1/98615836/files/2e085f6b-bad2-4ac2-9a69-1126e25274b1?access_token=tcKql0M5MipDezxTuyvBVIAExRcS7jxNfeWVfQUf&omit-content-disposition=true like this

[calm chasm]

Did I set this up correctly

actually

would it make sense to work on cases?

[weary tiger]

𝐴 ⊆ 𝐵 = ∀𝑥 ∈ 𝐴(𝑥 ∈ 𝐴 ⟶ 𝑥 ∈ 𝐵)
Is this equivalence correct?

Is there another definition for subset eq

[dense basin]

this is the one afaik

[weary tiger]

Am i missing a step?

[fossil mural]

well that's true but it's a somewhat weird way to phrase it

you're essentially saying "any element of A that is an element of A is an element of B"

when you could just say "any element of A is an element of B"

i.e. $\forall x (x \in A \to x \in B)$, or equivalently $\forall x \in A (x \in B)$

[vital dew]

bee [it/its]

[crimson zenith]

I can't find a formal definition of a constant function anywhere. Is it a simple as, "For any values x1 and x2 in the domain, f(x1)=f(x2)? Or is it more complicated?

For example, given the definition I just gave, that would mean a domain with a cardinality of one would make any function f constant regardless of what f(x) is. Is this intended?

[sick grail]

yes

the definition you give is fine

[rapid mural]

a more obvious definition could be

exists y in Y such that for all x in X fx=y

[crimson zenith]

interesting

[stray reef]

Is it a simple as, "For any values x1 and x2 in the domain, f(x1)=f(x2)?
what other complications do you think should be there

[weary tiger]

So I'm working on a (high school) math assesment which I chose to do on decompression models (you know, the ones used in scuba diving), and the aim of my IA is to use an existing discrete model to develop a continuous one. Though, the model is recursive, and I'm at a point where I'm just stuck staring at my screen trying to figure out how to continue. I might've chosen a topic that's too advanced for me, but I want to go through it, I just need a nudge in the right direction.

if anyone is interested in helping me (for some reason) or giving me ideas, please dm me 🥲

[zenith marsh]

u guys have any tips or roadmap for discrete math? i have been in this course for like 2 days / 6 periods

😁

[lone verge]

how do i find a set of positive integers X and Y and such that there exists integers A and B so that AX + BY = 2, and the GCD of X and Y is not equal to 2?

[uneven violet]

I'm answering what I think you are asking: You want to find integers X and Y with GCD(X,Y) different from 2, such that there exists integers A and B with AX + BY = 2.

You can take X = 1, Y = 2. Then GCD(X,Y) = 1. Now take A = 2 and B = 0. Then AX + BY = 2.

[lone verge]

Yes thats what i was asking

it looks so easy now in hindsight

😅

thank u

[uneven violet]

np

[astral hatch]

Does a universal quantifier make a predicate a proposition?

ie, is a predicate like (E(x) -> W(x)) changed into a proposition when we have ∀x(E(x) -> W(x))

[brave rock]

Yes

[astral hatch]

Is that true for a existential quantifier too

[brave rock]

Yes

[astral hatch]

How does that work

Doesnt the truth value of the inside predicate still depend on the values of E and W

[brave rock]

Uh, yes?

[neat oasis]

I asked this in help channels but realised it may be a bit too advanced for there, could someone help me here?

[stray reef]

$S \geq f(x)$ is cursed notation.

[vital dew]

Ann

[stray reef]

actually a lot of this is kind of cursed. did you get this from your textbook?

[neat oasis]

No, this is not direct translation, let me try to get it a bit more thorough

[stray reef]

!original

[lofty cloud]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[neat oasis]

this is better right?

[stray reef]

show ALL of it.

yes even if it isnt in english.

[neat oasis]

it's in another language

ok then

[stray reef]

If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[neat oasis]

Here there's also the inferior sum, but that part of the exercise I didn't need any help for

[stray reef]

ok, right.

[neat oasis]

although it could have some relation ig, so the result is 2^|A intersection x|

[stray reef]

so you missed a subscript.

and also $A \in P = 2^{[n]}$ is not at all the same thing as $A \in 2^{[n]} = p$. the way you wrote it, it was confusing as shit.

[vital dew]

Ann

[stray reef]

just had to point that out.

[neat oasis]

I didn't wrote it out like that sorry, I asked in help channels and they translated to latex as best as they could

[stray reef]

this image is already clearly latexed

ok so f(x) is the function that returns 1 if x ⊆ A and 0 otherwise. ok.

and we want to find the sum of $f(y)$ for $y \geq x$, where $x$ is fixed, and that'll give us $S_{\geq}f (x)$.

[vital dew]

Ann

[neat oasis]

yes

[stray reef]

this i believe boils down to the number of sets y such that x ⊆ y ⊆ A.

which would be 2^|A \ x| actually, not 2^|A ∩ x|.

[neat oasis]

Yeah I saw that too, but I was not able to give a value

[neat oasis]

no, no, that's for the inferior sum

inferior sum being for <=

but 2^|A\X|? that can't be, there's a bunch of ways you can get it to give 0

like

in [3], S>={1,2}({2,3})=0

[stray reef]

oh, hold on.

it's 2^|A \ x| if x is a subset of A

otherwise it is 0

that should check out

[neat oasis]

oh

that has

[stray reef]

the existence of at least one such y implies x ⊆ A

[neat oasis]

a lot of sense,

like I saw 2^|A\X| before but discarded it because of the 0 cases

ok ok

thanks

[astral hatch]

but doesnt that imply that its then a predicate?

[brave rock]

No

The truth value of a sentence always depends on its content. I feel like you're not saying what you really mean, or perhaps I am simply not understanding.

[golden nacelle]

One of my first proofs I wrote in my disc math class, can someone verify it?

[stray reef]

logic holds up, but style is not great.

but for a first(ish) proof it's fine.

[golden nacelle]

What can I improve?

[stray reef]

less symbol soup, i'd say.

and less assignment of letters to represent statements.

i could rewrite this proof to make it better on style, but it's midnight here.

if you're ok waiting like 7-8 hours for me to wake up tomorrow, i can do it in the morning.

[golden nacelle]

Fair enough, I was just tryna represent them all as propositions and proceed

[golden nacelle]

yep that works

[stray reef]

i mean tbh like

this is fine as scratch work

[calm chasm]

if you have 3^(n+1) > 2^(n )* 3 why can we change 3 to 2 ?

My understanding is that it's just because the inequality is still true but being able to just change the numbe feels like magic.

Like if I have 10^(n+1) > n^(2) * 10 and I'm doing proof by induction and need 10^(n+1) > (n+1)^2 can I just do that?

I see now

as long as the inequality still holds when doing proof by induction, I can do this

Example:

[zenith quail]

relatable tbh

[calm chasm]

Does anyone know of a video that explains universal quantifier proofs? I realized I don't understand how to prove them.

[dawn gulch]

Hello everyone

Anybody here know discrete math

I guess nobody is alive

Hm

[dawn gulch]

Anyone?

[lone verge]

what does (x,y) having domain A x A mean?

[final cliff]

Pretty sure they just mean the set of values that pairs (x,y) are chosen from

[lone verge]

So that means (x,y) can be (1,2), (2,3) (4,3) etc.?

[final cliff]

Yep

[lone verge]

can it be (1,1), (2,2)?

[final cliff]

Yes, it can be any element from the cartesian product of A={1,2,3,4} with itself.

[lone verge]

Thank u

so if it said domain A and not AxA, i couldnt take (1,1) right

[final cliff]

Yeah, (1,1) is not an element of A={1,2,3,4}.

Sometimes people will talk about a "universe" or a "domain" or a "domain of discourse" in the sense of "the set of elements we take values for our variables from"

So, it would be good to double check your definition for domain here to be safe.

[lone verge]

Noted

[calm chasm]

I tried getting the lhs a few ways but none of them worked.

Could I get a hint?

[rugged dock]

Did you complete the square

[stray reef]

??

[weary tiger]

My first thought is to find for what values 4x is greater than x^2 then show that adding 5 to x^2-4x makes the value greater than or equal to zero

I suck at discrete math tho

Also $f'(x) = 2x -4$ might be helpful

[vital dew]

vote.gov Register to vote.gov

[lone verge]

anyone know how to show that LHS = RHS?

[stray reef]

looks like some annoying but not terribly hard algebra

[lone verge]

i tried making their denominators the same but it just results in so much stuff that im lost

[stray reef]

say, where'd this come from?

[lone verge]

induction proof question

[stray reef]

this?

[lone verge]

yes

thats the question

[stray reef]

i have a feeling that there is a better way to do this

[lone verge]

what do u propose?

[stray reef]

why not decompose 30/((2r+1)(2r+5)) into partial fractions

[lone verge]

what do u mean by that

[stray reef]

https://en.wikipedia.org/wiki/Partial_fraction_decomposition

[lone verge]

sorry idk how to do this 😭

[stray reef]

hm

$\frac{30}{(2r+5)(2r+1)} = \frac{\tfrac{15}{2} \cdot 4}{(2r+5)(2r+1)} = \frac{\tfrac{15}{2} [(2r+5) - (2r+1)]}{(2r+5)(2r+1)}$

[vital dew]

Ann

[stray reef]

so you can do this hack, i guess

you'll get $\frac{15}{2} (-1)^{r+1} \paren{\frac{1}{2r+1} - \frac{1}{2r+5}}$ as your summand

[vital dew]

Ann

[lone verge]

it looks more complicated to solve now 😓

sorry im slow

[stray reef]

hm

it's hairy, yeah

[lone verge]

nvm

i figured out my mistake

[dawn gulch]

Hello

[stray reef]

hello, do you have a math question to ask?

[dawn gulch]

Yes I do

Because im preparing for a quiz

[stray reef]

!da2a

[lofty cloud]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[dawn gulch]

And it’s hard to get something so simplified

Sorry

[stray reef]

you should send the question upfront

[dawn gulch]

I am just flustered

Give me moment

[stray reef]

http://nohello.net/ etc.

[stray reef]

ok sure. ping me when you've sent the question.

[dawn gulch]

I am trying to simplify truth tables because logical equations are hard to understand as symbols not as words and sentences.

I understand the 3 symbols but what I want to do is to solve a truth table by using analogies or sentences to prove or disprove something based on the table itself.

@stray reef

[stray reef]

...

again,

are you or are you not looking at a specific problem

[dawn gulch]

Not

[stray reef]

ok then come back when you do?

right now it's almost impossible to understand what you want

[dawn gulch]

Then do you want me to clarify

[stray reef]

i want a concrete problem that you tried to solve and ran into issues.

you have already confirmed that you are unable to supply such a problem.

[lone verge]

I understand that as n gets bigger, the mod of the nth term gets smaller, and that every 2nd term in the sequence is negative, but I cant connect it to the in between 0 and 1, and the in between 1 and 2

[dawn gulch]

I am saying that I have a hard time breaking down logical equations to complete truth tables

[calm chasm]

I didn't but I'll try. Right now it's not obvious to me why that would work.

If my goal is this form,
(x+a)^2+b
I'd still have an issue.

[agile magnet]

Can you give an example problem?

It'll be easier for us to help you if you give an example problem, your attempted solution, and where you got stuck

[short ridge]

the number of rows you will need is 2^k where k is the number of variables. and then setup all combinations of true false and keep plugging into the equation and evaluating the truth

[rugged dock]

I said that because it’d put it in an easier form to digest .and work with

Probably would lead to some clear path

[calm chasm]

Okie. I'll give it a try

[somber fox]

I don’t understand the setup here. Any square be an be divided into n squares? Where n >= 5? So it’s asking to show that any square can be divided into at least 5 squares? What’s the connection of this with k + 3?

[brave rock]

Well firstly, the problem says n > 5, not n >= 5.

The problem says this: choose some n with n > 5. Then a square can be subdivided into exactly n squares.

So it's not at least 5, it's exactly n where n > 5 is of anyone's choosing.

I won't discuss the k+3 bit, that's the hint.

[somber fox]

Thanks for clarifying. I think this helps.

[somber fox]

I wonder the requirement for 5. Is it necessary? Or it’s just how they constructed the problem? Because of n = 4. You can subdivide a square into 4 squares.

[amber wolf]

The symbol in part b. has put me in quite the pickle.

What is it? I can’t find a mention of it anywhere in our coursenotes. I originally took it as a symbol for a subset of but apparently that is not the case

Please ping in reply! Thanks.

[unborn vapor]

the symbol likely does mean subset, but the set you listed is an element of A, not a subset of A

[amber wolf]

Would a subset of A be {2,7}?

Maybe I'm not clear on the definition of a subset

Because in our course notes, the subset symbol is that symbol with a line underneath

[unborn vapor]

Yes! If B is a subset of A, that means that all of B’s elements are also elements of A. You could also think about it as A with potentially some elements removed. Your answer is not a subset because 4 is not an element of A

[amber wolf]

Ah, the elements of B in this case would have to separately be elements of A then?

[unborn vapor]

Yes

Another subset of A is {2, 3 {2, 3, 4}}, or even {{2, 3, 4}} though that last example is harder to parse and might require a bit more thought

[amber wolf]

Very nice, you explained it well thank you!

[unborn vapor]

No problem!

[elfin furnace]

How to find G+? And is G^2 correct

[agile magnet]

your adjacency matrix looks fine, I'm not checking your matrix multiplication you can do that yourself

what is the definition of G^+?

[elfin furnace]

Transitive closure

[agile magnet]

ok, so can you draw that graph out for me?

[elfin furnace]

Idk how

I just wanna know how I can find the transitive Claire

Closure

The textbook barely explains it

It just says it’s the union of G^2, G^3, and G^n

How do you compute that value or matrix

[calm chasm]

What I've learned from math text books is its probably in the text book. You probably just missed something or didn't fully appreciate what you read.

Happens to me often

[agile magnet]

Ah ok so I think there's a better way to view it

like that definition is not wrong

but it doesn't really explain the "transitive" part

So another "combinatorial" way to view G^2 is that it is a modification of the definition of the adjacency matrix G

rather than G_ij = 1 if i -> j by one edge

we have G^2_ij = 1 if i can reach j in two edges

G^3_ij = 1 if i can reach j in three edges etc etc

so that's where the textbook definition comes from

but what I think is a nicer combinatorial definition that is that G_ij = 1 if i can reach j in a finite number of edges

so j is reachable from i

this is where the transitive part of the name transitive closure comes from

because reachability is transitive

if i can reach j and j can reach k

then i can reach k

does that help you come up with the matrix for G^+?

[calm chasm]

I don't have the time to help much. But I'm pretty sure this is just a permutation with restriction. Have you tried drawing it out? Or how about just making the numbers smaller so you can start with an easier problem.

[rose creek]

im not sure if im supposed to permute the men and women, multply that by the # places a women can be seated

[haughty garden]

this looks like tedious inclusion/exclusion

[glad panther]

do you guys have good resources to practice mathematical, strong, and structural induction?

I don't think induction is intuitive at all, at least for me. I don't really understand it

[silver panther]

Mathematical Induction: https://www.youtube.com/playlist?list=PLWItv3oXcB2XWzvJIAMujaMoHwUQx32Sw

@glad panther these videos are good, make sure to try lots of problems after, there are also lecture notes with problems for these videos...

The creator also uses induction to prove irrationality of pi

[viral crown]

andreescu's mathematical induction: a powerful and elegant method of proof (or something)

it's primarily intended for competition math audiences but i find it serves perfectly well as a resource for a lot of induction problems

[neon harbor]

You can pair each woman with a man, so you end up with 9 pairs plus 5 single men, and the solution should be straightforward from there

[edgy lintel]

im having trouble displaying a collection of pairs on a given interval in set builder notation

[stray reef]

!xy

[lofty cloud]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[stray reef]

@edgy lintel

[stray reef]

hold on

[stray reef]

can you explain in more detail what pairs your set consists of?

or show a graph with them all marked?

[edgy lintel]

Multiples of 1/2

[stray reef]

...

[edgy lintel]

It comes from this problem

[stray reef]

oh my god why didn't you post the problem in the first place.

wait so ok. you want to approximate this integral via the midpt rule with 6 subintervals, yes?

[edgy lintel]

yes

[stray reef]

ok now tell me

what's the length of the full interval of integration?

don't answer any more or any less than i ask you!

[edgy lintel]

12

[stray reef]

good. your interval does have length 12.

you split it into 6 equal length subintervals. what is the width of each subinterval?

[edgy lintel]

2

[stray reef]

uh huh

and can you write out in a list the intervals themselves and their midpoints?

i think you will find that it will differ from what you sent earlier.

[edgy lintel]

ill be back ,my boss is making me put on a fur suit

[stray reef]

what kind of boss is that? are they a furry themselves?

[edgy lintel]

no no i work for a party business

well

i think my boss might be a furry 🤔

woaw 1 /2 does not equal 2 🤦‍♂️

my bad

[stray reef]

indeed, they are quite different.

[stray reef]

@gritty garnet it's true that ∅ is a SUBSET of any set, but it is NOT true that ∅ is an ELEMENT of any set.

[wet flame]

is (b+c)* equivalent to (c+b)* (regex)

[haughty garden]

yes

[wet flame]

hm ok

https://bakkot.github.io/dfa-lib/regeq.html this website said otherwise which is weird

[haughty garden]

oh they define it to be concat

[wet flame]

ah i think its using + as repetition not or

ye

[haughty garden]

yeah if it’s used as concatenation, then they are not equivalent

[wet flame]

how would i describe "all words where num of a’s + num of b’s is odd" in regex over the alphabet {a,b,c},
i guess it would start with something like c*(a+b) but idk after that

[agile magnet]

split it into cases

based on the parity of the number of a's and the parity of the number of b's

create a regex for each case, combine appropriately

[wet flame]

so like 3 a's, 3 b's, 2 a's 1 b, 2b's 1 a ?

[agile magnet]

well that's not the only possibility

do you know what I mean by parity?

[wet flame]

just odd/even?

ah

there can of course be more than just 3 a's/b's ye

[agile magnet]

yea

also I'd ignore c for now

figure out the regex ignoring c's

and then add then later as necessary

[wet flame]

how would you suggest combining the different regex for each case

[agile magnet]

what do you think?

I mean what ways are there to combine regular expressions to form a new one?

[wet flame]

my initial thought would be to just union them together

[agile magnet]

lets start there

[agile magnet]

why?

[wet flame]

so you can pick any of the cases

[agile magnet]

sounds good to me

[wet flame]

im a little stuck, if we just consider we only have a's and b's, we just need even a's odd b's, or vice versa right? Im not sure on how to express even or odd in regex

[weary tiger]

Can someone give me a simplification of what they are saying here? First they refer to congruence class. Then they say m pairwise disjoint equivalence classes modulo m. Whats the difference between "equivalence classes" and "congruence class?"

[agile magnet]

No difference

They should have kept consistent wording

Equivalence class is a more general term I guess

When talking about modular arithmetic, we say "x is congruent to y modulo m"

Hence congruence class

But modular congruence is an equivalence relation (easy exercise)

So we may also say equivalence class

[weary tiger]

Thank you

@agile magnet when you say modular congruence is an equivalence relation, is it because of the proof being this theorem ...

Behind*

[agile magnet]

Well you want to use that "Consequently" sentence to prove it's an equivalence relation

[dire stag]

How do I find a0?

[stray reef]

you find the n'th term as instructed then plug in n=0...?

but the problem also doesn't say to find a_0.

[dire stag]

I think that I need a0 in order to do the polynomial fitting

In the previous examples that I did on my own, I think that I always started at a0, not a1.

This is also the last problem on the homework, the previous problems also didn't mention anything about the first term being a1 and not a0

[stray reef]

why? what method are you using?

[near stream]

how would i use membership tables for this?

[compact owl]

What sequence does the generating function f(x) = (3-2x) * 5 generate?

I'm slightly confused as there's no division here.

[brave rock]

Remember that the ordinary generating function of the sequence $f_0,\ f_1,\ \dots$ is the infinite sum $f_0 + f_1x + f_2x^2 + f_3x^3 + \cdots$

[vital dew]

Boytjie