discrete-math

[jolly ledge]

fr

why?

ok

what is the element of the set ${{\emptyset}}$

[vital dew]

Kiameimon | Welt Rene (glomed)

[odd garden]

{0}

so f is false because not all elements ?

are the same

[jolly ledge]

not all elements have to be the same.

read carefully

[odd garden]

so f is true

[jolly ledge]

yes

[odd garden]

but g is false because it's missing the equal sign

[jolly ledge]

huh

wdym equal sign

[true pelican]

He probably means $\subseteq$

[vital dew]

_daili

[odd garden]

yes

[jolly ledge]

honestly $\subseteq$ and $\subset$ notations are fucked up because some books mix them up

[vital dew]

Kiameimon | Welt Rene (glomed)

[jolly ledge]

so it depends

if $\subset$ disallows the subset to be itself then yes g is false

[vital dew]

Kiameimon | Welt Rene (glomed)

[native dragon]

Think of {empty set} as an empty box within a big box

The smaller box contains nothing but the large box contains a small box which is still considered something

[odd garden]

how am I supposed to think when solving this exercise

A\B means A-B

[haughty garden]

think about how A and B are related so that each of these statements are true (or at least what sort of properties must occur for these statements to be true)

e.g. for (a), you already know that A is a subset of A U B, but if A U B = A, then what else must be true?

[ember obsidian]

it may help to draw a picture of each statement

[odd garden]

can you please tell me how

[ember obsidian]

try to draw two circles representing sets A,B such that AUB=A

[odd garden]

exercise 20 (a) is this correct?

[iron marsh]

For starters, the piecewise is the same for both cases, so it's a waste to do piecewise.

'onto' is focused on the codomain. So specify a codomain and make sure f doesn't map to all of it

[brave rock]

Just to clarify what dackid is saying here: this function could equally have been written just as F(m) = m+1.

[iron marsh]

In your words @odd garden , what do you think 'onto,' commonly known as surjective, means?

[odd garden]

onto means that for every image has a pre image

correct?

[little prism]

well to be called an image you need to have a pre-image. so no, not quite

[odd garden]

for any y value we can find x

if we don't find then it's not onto

[iron marsh]

Yea, for every output, you can find an input that maps to it

It's really easy to make a map that is not onto, as you just make the whole space larger than the image (what f can map to)

[odd garden]

yes but the questions get harder

[iron marsh]

Start with this one

[odd garden]

so piecewise functions are easier i think

[iron marsh]

No, none of these require piecewise

And your function wasn't really a piecewise function, as it did the same thing in both cases.

[odd garden]

I know that's the point

i could have written it in without piecewise function

[iron marsh]

So why did you?

[odd garden]

Come up with your own function f : N → N that is not one to
one but is onto (Justify)

let's say this exercise

this is done using piecewise function

[odd garden]

makes it safer to cover all possibilites

[iron marsh]

Well you're answer was one to one in that last one

I didn't see the N to N bit, and I can see how you can do this with a piecewise function

Got to be a bit more clever though. F(m)=m+1 ain't gonna cut it

[odd garden]

in this one yes?

[iron marsh]

Yea

[odd garden]

solution was using even/odd but you can make a piecewise function

[iron marsh]

You're piecewise did not work though

[odd garden]

i'm not using my piecwise function ofc lol

[iron marsh]

Then why'd you ask if that was correct?

[iron marsh]

Here

[odd garden]

for this it's correct

how is it not correct ?

[iron marsh]

It is, me missing the N to N bit was causing problems.
But
1: it's not piecewise, so don't write it that way, and
2: Justify why it works.

[odd garden]

Okay thanks

[earnest portal]

I don't like math anymore

[analog token]

do i care?

[coral parcel]

Yeah.

[coral parcel]

Yes.

[coral parcel]

Yes -- if P=NP, then every language in P (other than the empty language and the language of all strings) is NP-complete.

Umm, yes, why wouldn't they?

If P=NP, then it simply isn't true.

Yes, for the third time.

That has nothing to do with the fact that if P=NP then there are plenty of finite NP-complete languages.

[coral parcel]

No, if P=NP, then every problem in P (except for the always-true and always-false ones) is NP-complete.

You definitely should say that you're adding an assumption that P!=NP, because unless that is true, the thing you're being asked to prove is false!

[odd garden]

in the first line I can't understand where we got the 3

3(3an-2)

<@&286206848099549185>

[jolly nova]

its using the condition an=3an-1

[odd garden]

an-1= 3an-2

where did the "3"(3an-2)

[jolly nova]

bruh put an-1=3 an-2 in an=3 an-1

[odd garden]

can someone explain this to me please

[mental pecan]

What are you confused about

[weary tiger]

Is this a good approach to seq probkems?

say you're given numbers a,b,c,d,e,___

and you're supposed to fill in the blank

1. Consider the strict differences in the numbers
2. Consider the strict factors from n to n+1
   If neither the differences or factors follow any distinguishable subpattern

3. Consider the some simple function f(x), such that f(n) = n+1
   where f(x) is some first or second degree polynomial
   (Ex: 2,4,6,10, ___ , f(x) = 2x-2

[obtuse lance]

well step 1 will lead to any polynomial solution, so step 3 is redundant if you're trying polynomials again

[weary tiger]

hmm

rue

i just took a quant oa and i idiotically didnt have a paper out so it proved way too difficult to work it out in my head lol

[obtuse lance]

the strict differences can make any sequence if you have the original sequence too btw, for instance powers of 2

cause these can arbitrarily make any sequence

[weary tiger]

so do you always start with strict differences?

[obtuse lance]

well, the problem is these sequence questions are basically ambiguous

you can make a polynomial that goes through any n points you like to continue the sequence

[ivory badge]

Well that’s just saying it does halt on accepted inputs, and whether it halts on the others doesn’t matter

We have no idea if all A halt on some inputs (not all will)

But the ones in L definitely do and return 1 too

Well, it being RE is because there is an algorithm which will halt on acceptable ones

It’s still RE even if it halts with a negative outcome for all unacceptable ones

You just only need it for the acceptable ones

[ivory badge]

The 1 part is the important part

Since 2 is just saying it doesn’t matter what it does outside L

What might your algorithm be

Well, does this have an issue if A does not halt on an input?

Right, I do wonder if that might be an issue

[ivory badge]

That sounds hard to me idk

[brave rock]

Considering turing machines aren't multithreaded, and definitely can't do infinitely many things at once, you need to be more specific about this.

Hint: the proof I have in mind is similar to the usual one that N^2 is countable.

[iron marsh]

I don't think the 2nd works (it needs to be a bijection)

[sly heart]

Hello guys, does the Hackenbush game have a winning strategy?

[brave rock]

Hint: let's consider a simpler problem. Fix an algorithm A. Can you show that the language of all inputs for which A outputs 1 is recursively enumerable?

[ivory badge]

Well, consider boytjie's smaller problem

Consider a single fixed A, how can you show that the inputs which return 1 through A are RE?

Can you find an algorithm which shows this set is RE?

[merry crest]

Somebody wanna take a crack at this, I'm stuck

I have tried taking ratios, multiplying terms, taking sums and differences

[haughty garden]

Try convincing yourself that $\frac{1}{x} + \frac{1}{y} = \frac{1}{a}$

[haughty garden]

You can express 1/b and 1/c similarly as well

Convince yourself that 1/x - 1/z = 1/a - 1/c

[merry crest]

Oh, i can see that 1/x + 1/y = (y + x) / xy

[merry crest]

Ok, I've worked it out

i came up with x = (2abc) / (cb + ca - ab)

This means x is Rational, because it is a ratio of integers

Of course a, b, and c can''t be zero

Man, I knew it was something simple I was missing

shame on me

Without fail doing these problems always makes me feel so dumb

@haughty garden I thank you for the nudge in the right direction

[vapid drift]

hello why is the answer d

how did they know its a partition'

[ivory badge]

well, can you think of a single TM which would halt and return a success on the inputs which A returns 1 on?

Also you didn't fix A, you did something like (A, x)

Well, I think we want to narrow it down to only the ones where A returns 1

yeah, true

well if you take 1 as the success, then congrats the set of inputs to A which returns 1 is RE

no?

you just need an algorithm which returns true for things which are in the set

[ivory badge]

well we want the TM to take in A and spit out yea/nay

not a pair (A, x)

[hollow reef]

anyone does turoting in this chat? I am having a HARD time wrapping my mind around basically everything except cipher. pseudocode is also a challenge.

[hollow reef]

*tutoring or a in person math group

[hard stone]

We generally don't do tutoring in this server, but if you have a question, please ask

(Allowing paid tutor service advertisements tends to lead to scams)

[keen topaz]

hey guys. can you check my answer in a problem:

There are 100 points marked on the straight line. How many ways can 10
points be selected from them so that no two selected points are adjacent?
I ve got С(10,91)

[hard stone]

can you explain how you got your answer?

[weary tiger]

other than the endpoints the other 98 points will have only 2 points adjacent to it

I mean at this point you wouldn't even need binomial coefficients to solve it

Oh nevermind I misread the question

Well for those 98 points, you would delete two points adjacent to it every time you select them, so your option for the next point reduces by 2

[keen topaz]

I tried to interpret this problem and decided to position some 'sticks' on 101 places. when I played with this construction I unnderstood that if I had positioned 10 sticks on 91 places I would have got the same result

[weary tiger]

the boundary case is annoying

[keen topaz]

yeah pretty well

[weary tiger]

so you wanna place 1 unit long sticks such that no two sticks share an endpoint?

[weary tiger]

don't you mean 5 sticks? Since each stick selects 2 points?

I might be misinterpreting you since that does select adjacent points

[keen topaz]

not exactly. i would rather say to place 20 stickson 101 place

to separate these 10 points

and this can be eventually simplified to placing 10 sticks on 91 place

if i am not mistaken

e.g. in this case i ve chosen 4 points from 10 and put 8 sticks on 11 places

[weary tiger]

how does this avoid picking adjacent points?

[keen topaz]

we can also simplify this picture and get 1010110 there 1 is two sticks construction and 0 - empty space

and when two '1' are adjacent that means that they separate two points separated by a single dot

[weary tiger]

this might work

[keen topaz]

and if we count these '1' we may get the right answer i hope

[weary tiger]

I would try drawing some counterexamples

[keen topaz]

ok thanks

[weary tiger]

To see if this representation is ok

It seems good

[ivory badge]

You don’t have to do that, you just need to verify that it checks where at least one input spits out 1

Well, the set of inputs which A returns 1 on is RE

Yes, for any A, we have that {x input|A(x)=1} or wtv is RE

Well you can’t just blindly check every possible input

It’s gotta halt and return true for every A with such an input

[ivory badge]

The TM for each A = 1 input set is basically just, run A with x, then check if it is 1

Well this is just the TM for the {x | A(x)=1} set

We don’t know a TM for L

Are there any helpful properties of RE sets you might know

Well, do you know any nice properties of RE sets

Maybe checking your notes back for a handy property or two might help here?

You’ve had no prior exercises or lectures on RE?

[true shore]

true or false and why?

[ivory badge]

This is clearly a question on homework (or worse)

[true shore]

it was graded alredy

[ivory badge]

Can literally see 0/5 top right

[true shore]

yes i got it wrong y im asking captain obvious

[ivory badge]

Well if you get a true/false wrong, that should be a pretty good hint as to the right answer

[true shore]

and a great explanation from captain obvious again

[ivory badge]

Well, what exactly are you asking for then?

[true shore]

how about you read what i wrote this time and not just say something to say it

[ivory badge]

I’m saying if you already have a good idea what the answer should be, since it’s already graded

What exactly is the point of asking if it’s true or false

[true shore]

captain obvious obviously needs to go back to kindergarten to learn how to read

[true shore]

i was just asking why, ok, so if you cant explain it or do not want to then whats your problem?

[ivory badge]

So why do you think it is or is not ambiguous?

[vapid drift]

how to knoe if they are connected

[ivory badge]

well if a R b then a and b are in the same component

[iron marsh]

Hey @ivory badge , I don't know if you remember, but about a week ago I wanted to show that sin(n) is dense in (-1,1).
I wasn't in the right headspace at the time to understand it, but I am now. If you remember how you did it, do you think we could go over it again?

[ivory badge]

Oh yeah sure

Step one: We can turn sin from a function R->[-1, 1]=I into S^1 -> I

Because periodic

[iron marsh]

What is I here?

[ivory badge]

[-1, 1] = I

[iron marsh]

Ah, you're just naming it that.

[ivory badge]

Yeah because lazy

[iron marsh]

Lol, sure. I'm with you. And S^1 is of course the unit circle

[ivory badge]

Ye, specifically mapping based on the [0, 2pi)

I will note further, this is a homomorphism under addition

[iron marsh]

Hold on. what's the action happening in S^1?

[ivory badge]

Circle group

Rotation

[iron marsh]

Ah, so we're sending x to (cos(x),sin(x)). Got it.

[ivory badge]

More or less yeah

Just adding angles is equivalent

[iron marsh]

Well, wouldn't a homomorphism under addition mean cos(x+y)=cos(x)+cos(y)?

[ivory badge]

Nope, because the homomorphism in on R->S1

Next, I showed we can get an integer k such that |k - 2npi| < epsilon for any epsilon, for some n

[iron marsh]

I know if you look at the unit disk in the complex plane instead, x in the reals map to e^{ix}, and that is indeed a homomorphism (under complex multiplication)

[ivory badge]

Yep that’s what we use

Cos(x)+i sin(x)

Or equivalent to it etc etc

[iron marsh]

I'm having a hard time convincing myself of this

You used that dirichlet's approximation thing right?

[ivory badge]

Well ofc it’s true but yeah

There’s infinitely many integers p, q such that |q pi - p| < 1/q right?

| 2 q pi - 2 p| < 2/q

[iron marsh]

I'm not sure why this is

[ivory badge]

This is basically the theorem

[iron marsh]

Let me read the wikipedia article for the approximation thing real quick

[ivory badge]

Multiply by 2

[ivory badge]

You can get p/q with q<=N such that |x - p/q| < 1/N

Make N big

Something like that

[iron marsh]

Okay, now that I believe since the rationals are dense

[ivory badge]

You get the original N statement by pigeonhole I think

[ivory badge]

We get this

p is an integer

And we can make it arbitrarily small, so now we get sequence(s) of naturals where [n] -> [0] in the circle

Because we don’t need the sequence to approximate the sequence 2n pi, we just need each term to get successively closer to a multiple

Next, we approximated rationals by naturals

[iron marsh]

Why do we need to do that?

[ivory badge]

Well idk if it’s strictly necessary, but since we can get n ~ 0, and if we approximate rationals

And R->S^1 is continuous, we should be able to sort of chain these approximations to get n approximating arbitrary reals

To approximate rationals, we need to get |p/q + 2npi - k| < e for some n, for all e

We can get some natural approximations for 2npi less than f, so call that m_f

|p/q + 2npi - k| = |p/q + 2npi - m_f + m_f - k| <= |p/q + m_f - k| + |2npi - m_f|

*+2npi being an equivalence class-y kind of “for some n” thing

We can get k=a/b rational, |bp/q + 2nbpi - a| < 1/b right?

[iron marsh]

what is f here? I don't think it was established yet

[ivory badge]

|bp + 2nbq pi - qa| < q/b

[ivory badge]

Ah right, another epsilon whoops

[ivory badge]

Mix this with our m_f, so we get |bp + m_f - qa| < q/b + bqf

And we can make f low for any b

[iron marsh]

so why does this approximate rationals?

[ivory badge]

I think? Been a while since I remember the convoluted argument I used

[ivory badge]

Because we want a sequence of naturals which get [n] -> [p/q] in the circle

And we know rationals are dense

[iron marsh]

I mean how does this achieve that?

[ivory badge]

Well, we can get arbitrarily close to [0]

And that lets us take very large equivalence classes of p/q

So we can take arbitrarily large rationals a/b approximating it

And then multiply by b which is still fine because it was < 1/b^2

We’d get some q factor but that’s fixed, and f (from our approximation of [0]) can be made arbitrarily small too?

Pretty sure this is close to the kinda silly argument I used

I think I’ve mixed up the order of the approximations

[iron marsh]

Not gonna lie, you kinda lost me here. But I understand what you are trying to do. We can always find n such that [n] is sufficiently close to any [p/q], and that is what gives us that e^in is dense over the unit circle

[ivory badge]

But point is, [x+y] = [x] + [y]

And if our distance between x, y is small enough, then the distance in the circle is pretty similar. It’s not getting any larger I think

[iron marsh]

I understand what's being accomplished here. I'm going to take some time tomorrow and see if I can iron out the details myself. This has been helpful. Thanks sharp 🙂

[ivory badge]

So our approximations are essentially like d( [p/q] , [k] ) < d( [bp], [qa]) + small or smth

Then afterwards Valley came in something about how h(x) = x - floor(x), then h(nx) is dense in [0, 1] or something

So it’s an R/N thing, and you pigeonhole it saying two multiples are gonna be the same in the 1/n size sub intervals like (k/n, (k+1)/n)

And since you can do that for any n, you get nx - floor(nx) < 1/k so that’s your approximation when x = 2pi

And similar from there, which also has some caveats like how the density of h(nx) is uniform across the interval, whereas sin isn’t so much

Which is much cleaner @iron marsh

[iron marsh]

Hmmm, I'll have to take some time to unpack that one.

[ivory badge]

Right but I mean it’s pretty similar it just uses a different coloring

Still approximating things and I think you still need a diagonal sort of approximation? I forget

The natural approximation of things is still a lil tricky to make hit everything?

h(x+2npi) is gonna be dense too in a similar fashion so gg

homomorphism momento

[iron marsh]

I understand the idea for the first one, and I'll need to spend a little time tomorrow digesting the 2nd.
Either way, I understand the general idea of what we did.

[ivory badge]

yee, either way it still relies on getting n ~ 2pi

[iron marsh]

When I say we, I mean you. I contributed nothing here xp

[ivory badge]

Valley pulled out the h(x) one (and a short lil paper on it and its density properties)

Which is way more slick than pulling out the wacky double sequence and getting a cofinal thing in it

[iron marsh]

Definitely. Seems like a pretty clever approach

[ivory badge]

Yeah, whereas mine is uhh

Barely functional

[iron marsh]

I think if you take the time to organize your ideas, I think your approach would work fine.

You're definitely more 'think and type' then 'think, then type.' Hence why there's so many "or somethings" in your argument :p

[sinful rock]

What exactly is the Power Set? I'm getting mixed results when looking it up

[ivory badge]

Right, but it was also like, right when you’re like

[sinful rock]

One video said that given 2 sets A = {1, 2, 5} and B = {3, 4} that the Power Set would be the set of all Natural Numbers

[ivory badge]

“Hey how do I do this” “oh obviously it’s just rational approximation just uhhhhhhh…”

[ivory badge]

It did not, or it was wrong

Very likely, you misheard or misinterpreted what exactly was said

[ivory badge]

But the power set of a set X is the set of all subsets of X

If x in A implies x in X, then A is in P(X)

[sinful rock]

A website said that given {a, b, c}, P = {{a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c}, {∅}}

[ivory badge]

No

{0} is not in there unless one of a, b, c are 0

But the empty set {} is

[sinful rock]

I meant the Null Set

∅

[ivory badge]

yeah, that is very different from {0}

[iron marsh]

Lol, yeah.
Definitely not a complaint btw. Like I said, you've been very helpful and I appreciate you taking the time to delve into this with me again.

[ivory badge]

0 = {} is common though

[sinful rock]

So is the website right

[iron marsh]

NOT a complaint. Shit... 😅

[ivory badge]

I mean a lot of it is just finding the right sequence(s) to show rational approximation is exactly what’s happening

“Yeah bro it’s one of these keys” and it’s an overpacked key ring

[sinful rock]

The Power Set being all unique combinations of the elements in all relevant sets

[ivory badge]

But the general idea is obvious

[iron marsh]

Haha I feel you there 😅

[ivory badge]

The power set is the set of all subsets

When it’s an infinite set, like N, then it’s a lil tricker than calling it combinations imo, since that feels like suggestive language

[sinful rock]

yeah combination wouldn't be right

Because it wouldn't contain {a, b} and {b, a}

[ivory badge]

Those are the same

[sinful rock]

Exactly

[ivory badge]

Order isn’t a thing

[sinful rock]

That's what I was getting at

[ivory badge]

Combinations aren’t ordered either

But consider the powerset of the naturals N

[sinful rock]

Idk if Combinations has a different definition in Set Theory, I just started learning about it

[ivory badge]

It is not just containing all the elements, combinations of 2 elements, of 3 elements, of 4…

Because that contains only finite subsets

And the union of them all is countable, which is a big problem with being a powerset

[sinful rock]

Would't the power set of a continuous set be infinite

[ivory badge]

Btw dackid, consider P(X) countable and if there’s any issues @iron marsh

[ivory badge]

The powerset of any infinite set is infinite

[sinful rock]

I think I get it, thanks for the help

[iron marsh]

Wait, what I'd do to get involved in this? 😛

[ivory badge]

because it’s a silly lil question

Especially consider when with choice

[timber creek]

is injectivity enough for a function to have an inverse?

[haughty garden]

You require surjectivity as well

[timber creek]

i see, i'm guessing we specifically want to say that, if f^-1 is an inverse, the codomain becomes the domain?

[haughty garden]

basically yeah

[timber creek]

noted, thanks a lot!

[ivory badge]

It is enough however to get a function g such that g(f(x)) = x, but not the identity on the other direction

[timber creek]

oh ok, ty!

[ivory badge]

I mean, should be fairly apparent how it could be done since injective guarantees uniqueness in the image

Surjective having a g such that f(g(x)) = x uses choice tho

You should immediately see why it implies surjective at least

[grand totem]

And injective => left-invertible requires excluded middle! (And is in fact equivalent to LEM if the set theory has full comprehension)

[ivory badge]

Good ol im f u (B-im f)

That it implies LEM (with comprehension) is incredibly funny to me

[grand totem]

Yeah, though all those proofs involving "(some set theory stuff) implies LEM" are always odd to me since I'm not used to thinking about sets that look like {x | P} (with x not occuring in P), since they're not particularly interesting classically

[ivory badge]

The “we don’t really care about it classically” sets

I’m almost certain there’s a way to make this joke precise and correct

[ivory badge]

There’s probably some relevance to like wacky type-y stuff and related structures

[grand totem]

Maybe that stuff looks less odd in a topos theoretic setting (I believe Diaconescu's original proof of AC => LEM was a topos theoretic one, instead of the purely set theoretic one that's on Wikipedia for example)

[ivory badge]

Right probably

In the same vein as the surjection related one being the category flavored choice

Topos would be like, intuistionistic logic at least, so checks out

Set theory without LEM sounds weird tho

[grand totem]

It sure does

[ivory badge]

Type stuff seems less odd without it imo since there’s not like

Basing the whole thing around \in

[grand totem]

Going back to the injective => left-invertible stuff, I believe this works:
Writing {0 | P} for {x | x = 0 ∧ P}, we set X = {0 | P} ∪ {1} ⊆ {0, 1} and the inclusion i : X → {0, 1} is an injection.
Now suppose there was a g : {0, 1} → X such that g ∘ i = id_X.
If g(0) is in {0 | P} then P holds. If it's in {1} then g(0) = 1 and we claim ¬P holds, since if we assume P then 0 ∈ X and so g(i(0)) = 0, hence g(0) = 0, contradicting g(0) = 1

[ivory badge]

The whole P … contradiction therefore -P one

I’m not about to pull up something on constructive set theory to make sure this follows all the adequate conventions, looks close enough to me

[vapid drift]

why did they divide 3/7 and 9/5 and 3/7 and 9/7

[ivory badge]

They say that 3 does not divide it, and that 9 does not

well, is 5/3 an integer?

Or rather, is there an integer n where 5 = 3n

[vapid drift]

i dont undertand

5/3 is an integeer

[ivory badge]

That’s what the bar means, and they cross it to say that they don’t divide

[ivory badge]

It is not

[vapid drift]

yes it a decimal

[vapid drift]

how did they know ut doesnt divide

[ivory badge]

Because it is not an integer

There’s no n in N such that 5 = 3n

[vapid drift]

how do i know that without guessing

[ivory badge]

because it is obvious to those who are more familiar with it

Also, if you can’t tell

a | b iff b/a is in N right?

If b=na, then b/a = n

[vapid drift]

i see but for this one how is it not the samw

what is s crocidle with aline under it mean

[ivory badge]

That is subset

If you do not know what that means, perhaps you should go back and review earlier sections?

[stray reef]

ok but what the fuck is this problem

[vapid drift]

oh subset is related

[stray reef]

$S \subseteq T$ is literally defined by $(x\in S)\to (x\in T)$ so why the fuck would they make you dance around this with ``indirect proof'' like what is this exercise in pointlessness

[vital dew]

Ann

[stray reef]

oh wait no

[ivory badge]

Contradiction

[stray reef]

you PROVE that S subset T

i misread "assume"

my bad

[ivory badge]

Or rather, contrapositive whoops

Wrong contra

[ivory badge]

Yes?

[vapid drift]

ok but what are they doing

why does g divide x

[ivory badge]

It is 9

Not g

It is in T iff 9 does not divide it

So if it is not in T, then 9 does divide it

Might wanna go back to earlier sections and review a

[fossil mural]

...what's a "continuous set"?

[worthy pivot]

something that does not belong in #discrete-math

[hollow reef]

Write the pseudocode for an algorithm that takes as input a list of n distinct integers and finds the location of the largest even integer or returns 0 if there are no even integers. procedure locate-largest-even ( a1, a2, …, an: integers ). Help me understand this please.

[iron marsh]

Please don't rely on chatGPT to solve math problems...

To check for even, this should be helpful:
Note that $n$ is even if an only if $\floor{\frac{n}{2}}-\frac{n}{2}=0$. You can use this as a condition to check to see if an integer is even.

[vital dew]

dackid

[iron marsh]

In other words, dividing by two gives an integer

[little prism]

any useful language should have n % 2

[iron marsh]

Oh yeah, also mod arithmetic works 🤦‍♂️

[quiet tendon]

are there any tips for these question, if the question is wether it's ture or not

how do I check quickly if it's true

because otherwise I am using half my time proving it while it was false

I know btw first is true second is false but how do isee it fast

[ivory badge]

What is P(A | A)

[quiet tendon]

u mean P(A | AUB)

[ivory badge]

But familiarity is helpful

[ivory badge]

No I mean P(A | A)

[quiet tendon]

oh a question to me?

[ivory badge]

Yes

[quiet tendon]

uhm

1

[quiet tendon]

1>

[ivory badge]

Right, it’s 1

[quiet tendon]

but that is an easy one

[ivory badge]

What’s P(A | A u B)

Considering the definition of P(A | B) it should be apparent?

[quiet tendon]

uhm

[quiet tendon]

well I can apply the definiitonb

but that won't help me I think

[ivory badge]

Try it holmes

[quiet tendon]

holmes?

[ivory badge]

Yes it’s a figure of speech

Just do it

[quiet tendon]

it's 1?

[ivory badge]

Well, try and work it out here

[quiet tendon]

u mean it's correct?

[ivory badge]

I mean to say

How is P(A | B) defined

[quiet tendon]

oh

that is just definition

P(A n B) / P(B)

[ivory badge]

P(A n (A u B)) / P(A u B) here, right?

[quiet tendon]

ye

but that is P(A u B)

[ivory badge]

Can you simplify this

[quiet tendon]

ye ofc

[ivory badge]

Simplify A n (A u B)

[quiet tendon]

P((A n A) u (A n B)

and then

u should see it right

oh wait

[ivory badge]

P(A) / P(A u B)

[quiet tendon]

wait

P(A u (A n B)

right?

[ivory badge]

Ignore my idiocy rq

[quiet tendon]

P((A u A) n (A u B))

[ivory badge]

A n B is a subset of A

[quiet tendon]

both shold hold

so this is p(A)

[ivory badge]

So A u (A n B) = A

[quiet tendon]

y

P(A) / P(A uB)

is the answer

so that is always bigger

ye

[ivory badge]

Yep, that’s P(A | A u B)

[quiet tendon]

I see it now

but how do u see it fast

[ivory badge]

ye, or equal

[quiet tendon]

and also for second one

[quiet tendon]

I know second one is false

but if I am gonna prove it

it wastes time

[ivory badge]

Because it’s clear if you think about how the P(|) is defined

[quiet tendon]

it means given

[ivory badge]

A makes up a bigger portion of A u B than the whole set, heuristically

1 - P(A^c)

what is this

[quiet tendon]

wait isn't A u B bigger

[ivory badge]

It is bigger than A

But smaller than X, so to speak

[quiet tendon]

P(A)

[ivory badge]

Right

[quiet tendon]

so how do I see it?

[ivory badge]

What’s P(A n B | A)

[quiet tendon]

with definition

uhm

let me think quickly

P(A n B)

[ivory badge]

Right

Since it’s already a subset of A

[quiet tendon]

/ PA)

btw

[quiet tendon]

yes

but u divide by p(A)

so then it's not that easy to see

P(A n B) >= P(A)^2 is also still hard to see right

[ivory badge]

But it’s not immediately clear it’d be bigger, so you can try and show it false

[quiet tendon]

oh so if u can't see a good way to prove it

u should check if it's false?

[ivory badge]

Well, if you don’t think it should be true

Then showing it false makes sense

How would you show this one false

[quiet tendon]

well that is easy

[quiet tendon]

but what is the usual way to choose

because I can just try some numbers

and adjust the number the way it will help my proof

is that good way?

[ivory badge]

If you think it shouldn’t be true then try showing it false

[quiet tendon]

but that is hard to see as u just saw with second example idk if it's false/true

[ivory badge]

Well, do you think it should be true?

[quiet tendon]

well I alr know the answer

[ivory badge]

If P(A n B) >= P(A)^2 always

Then what if B empty?

[quiet tendon]

ye I guess that makes sense

[ivory badge]

There’s no easy way to tell in general

Whether something is true or not

Just gotta think if it should be or not and proceed to try that direction

[vapid drift]

why is x or z colum like that

i dont get why its like that. i thought its x or z first x and z is both true. so why is it true

[final cliff]

Is there a specific row you are confused by? When I go thru the columns for x and z and compare them to the column for x v z everything looks fine.

To be clear, x v z is true if either or both of x and z are true. So, rows in the columns of x and z that have a T under x, or a T under z or a T under both x and z will have T in the x v z column.

[final cliff]

xor stands for exclusive or. A xor B means A is true or B is true but not both.

[final cliff]

https://i.ytimg.com/vi/JQ0JOfxUAgY/maxresdefault.jpg

Here's some truth tables. The or table has 1 when A=B=1. The xor table has 0 when A=B=1.

[vapid drift]

Is this possible shape

,rotate

[vapid drift]

did i draw the picture right?

[final cliff]

Pretty sure you did it right, unless I'm miscounting somewhere.

You can just read through your adjacency matrix, for an undirected graph entry i,j of A has a 1 in it if there is an edge between whatever node corresponds to i and whatever node corresponds to j.

[vapid drift]

how did they know that it was the naswer

[ivory badge]

You can make an adjacency matrix in this case

But those sets in P are the sets where a, b are in X iff (a, b) is in R

[final cliff]

You can also compute these out by hand. 1 only relates to itself so one of your classes is just 1. 2 relates to 3 and no other things relate to 2 or 3 besides themselves so another class is the set with 2,3 and so on.

For a bigger relation you might have lots of tuples to check though which can become unwieldy.

Because you might have some goofy thing like 1 relates to 2 relates to ... 1000000.

[vapid drift]

wait so i can make matrix out of this and see which ones connect?

[final cliff]

You can also draw out the graph of the relation I think. Each equivalence class will contain all the nodes of a connected component of your graph.

[final cliff]

See what I mean?

[vapid drift]

how to i read this so the answer is b

[final cliff]

You mean option d?

[vapid drift]

yeah

[final cliff]

Look at the bits that are all connected together

[vapid drift]

is the answer just a simplified version of it

[final cliff]

List off the set of nodes in each one. Each of those sets is one equivalence class.

So for example one of the graphs has just the node 1, so one of your equivalence classes is the set {1}

Another one of the graphs contains the nodes 2 and 3, so one of your equivalence classes is {2,3}

The last graph contains the nodes 4 and 5, so your last equivalence class is the set {4,5}.

[vapid drift]

i see i understand thank you for the detailes olution

]may i ask one more quetsion?

[final cliff]

You see how I constructed these graphs from the given relation right?

[vapid drift]

yes that makes sense

[final cliff]

Kk

[final cliff]

Yeah, I may not be able to answer it though lol.

[vapid drift]

the answer only is theroy and i dont understand how he got them

[final cliff]

What is your question about this?

[vapid drift]

sorry just finding the solution

will post solution

[ivory badge]

Well forget the solution

Try and answer it now

[vapid drift]

i dont understadn what the guy is writing

[final cliff]

Maybe I'm dumb but I don't see how A is correct lol

[ivory badge]

It isn’t

[final cliff]

The remainder of f under division by three is fixed

f can't map to certain integers

[ivory badge]

However ||g is surjective||

[ivory badge]

This is wrong on A, however G and H are correct

Is this your work?-

[final cliff]

My hint for understanding this problem is try to work out A,B,D and E first. The rest is just definitions.

f can't be surjective by the reason just mentioned.

(Can you find a concrete example of this?)

[vapid drift]

i only know these definitoons with truth tables

[final cliff]

f can't be injective because the floor isn't injective

[vapid drift]

surjective when the last column is all true right?

[ivory badge]

What?

[vapid drift]

but how to put in truth table

[final cliff]

No, a function is surjective when every element of the codomain gets mapped to.

[ivory badge]

No???

As it says on the image, when the range = codomain

The image is wrong though

f is not surjective

[final cliff]

For example consider floor(1) vs floor(1.5)

[ivory badge]

And to show it is not surjective, what value of x satisfies f(x)=1

So, is g injective?

Why or why not

[vapid drift]

i dont undersatnd

[final cliff]

Do you know what injective means?

[vapid drift]

injective when truth table is all false

[final cliff]

No

A function is injective when no two distinct elements of the domain map to the same thing.

Probably you need to review definitions first before you can make progress on this problem.

[final cliff]

Is this the definition they gave you or just some diagrams to show you examples?

[vapid drift]

definition from notes

[final cliff]

This doesn't seem very good for a definition

Okay, well usually people define a function $f:A\to B$ as injective if and only for all $x,y\in A$, $f(x)=f(y)\implies x=y$. This is equivalent to saying for all $x,y \in A$, $x\neq y\implies f(x)\neq f(y)$.

[vital dew]

dootdooter

[final cliff]

The latter implication says informally that if x and y are different elements of the domain of f, then they map to different things when you plug them into f.

Mmm, do you know the bold arrow notation I am using?

It means implies.

A function $f:A\to B$ is defined to be surjective if for any $y\in B$, there exists an $x\in A$ such that $f(x)=y$.

[vital dew]

dootdooter

[final cliff]

All this means informally is that if I give you any y in the codomain (which is B) of the function f, then you can find me an x, where plugging x into f makes f spit out y (i.e. it makes f(x)=y true).

[vapid drift]

the arrow is implies

?

[final cliff]

This one $\implies$ means implies.

[vital dew]

dootdooter

[vapid drift]

and implies means : if this then that?

[final cliff]

This notation $f:A\to B$ means f is a function with domain $A$ and codomain $B$

[vital dew]

dootdooter

[final cliff]

In other words f maps the set A to the set B.

So, $\to$ and $\implies$ are different arrows with different meanings.

[vital dew]

dootdooter

[final cliff]

Yes that is correct

[vapid drift]

:0

[ivory badge]

Ok English doesn’t seem like a first language here, but I’m curious what truth table you’re referring to tbh

[vapid drift]

i usually make a truth table but im not sure how to in this question

[final cliff]

Yeah, for this kind of problem you won't always be able to build a truth table to solve things. Because you may have infinitely many elements in your domain, and you care about whether or not the propositions in the definitions for things like injectivity and surjectivity hold for all of them.

[final cliff]

For example $f:\mathbb{R}\to\mathbb{R}$ defined by $f(x)=3x$ is injective, but to show it is injective you need to show this holds.

[vital dew]

dootdooter

[final cliff]

The problem with truth tables here is that you can't make a truth table with one row for every choice of x, there are too many real numbers for that.

[vapid drift]

i see so to see if its injective would i plus in the same number for both equations?

[ivory badge]

No

[final cliff]

These aren't good examples for showing something is injective because neither f nor g is injective here.

[ivory badge]

You would say f(a) = f(b) and try to derive that a = b (which is false here, try a=1, b=1.1)

[vapid drift]

what is a and what is b

[final cliff]

Are you familiar with proving "for all" statements?

[vapid drift]

not by the name

[final cliff]

Mmm, okay well if I want to prove this

[final cliff]

I need to prove this holds for my f(x)=3x

I can try particular cases like f(x)=f(y) implies x=y where x=1 and y=2 and junk, but that doesn't tell me enough information

Because the actual definitions for injectivity starts with "for all x,y" not for x=1 and y=2.

So you have to deal with x and y as general elements of R to show my f(x)=3x function is injective.

[ivory badge]

They’re arbitrary variables. If it works for any choice of a, b then it works for all of them at once

[final cliff]

You assume x,y are real numbers, then you assume f(x)=f(y).

By definition of f this tells you 3x=3y, by algebra this tells you x=y.

So, then f(x)=f(y) must imply x=y.

[vapid drift]

ok so i do plus in a random number and if ther equal they are injective ?

[final cliff]

Since this implication holds for any choice of x,y we know my f(x)=3x function is injective.

[final cliff]

No that won't work.

The trick is to do with not actually specifying a particular value of x and y

Like, imagine I want to prove to you that all your friends like to read comics. And I do it by picking some friend of yours named jim who likes reading comics, but I ignore all your other friends.

Have I shown every single one of your friends likes reading comics?

[vapid drift]

no

[final cliff]

Yep, what if you have a friend named sally who hates comics?

So, we have to pick elements of the domain of f in general here.

Then you use known properties of f and of the elements of the domain of f to prove f(x)=f(y).

[vapid drift]

are you trying to sleep?

[final cliff]

I probably will get off here in a sec.

[vapid drift]

I understand iwill review what you posted and review the notes/fundemntals

thank you for the help

[final cliff]

Try these: prove f from the reals to the reals defined by f(x)=3x+1 is injective. Prove f from the reals to the reals defined by f(x)=x^2 is not.

[ivory badge]

Can you show whether or not g in your exercise is injective, too!

[final cliff]

The first one will be similar to my f(x)=3x proof. The second one you will need to find a counterexample to the definition of injectivity.

[final cliff]

This is also a good example to try.

[fossil mural]

what about if L' isn't in P or in NP-complete?

also was this proof generated by chatgpt

we assumed that at least one intermediate language is finite (namely L), which is not the same as all of them being finite

if we assumed that all intermediate languages are finite and derived a contradiction, the result of that would just be that at least one intermediate language is infinite, not necessarily that all of them are

chatgpt doesn't know what maths is so when it generates text that looks like a proof it's generally nonsense that just resembles a proof, instead of an actual proof

yes
if you assume that an intermediate finite language exists, and get a contradiction, then that means no intermediate finite language can exist

yes

...the problem is that the reasoning it used to get a contradiction isn't actually correct

i also just noticed another problem, which is that for an arbitrary language L in NP, L' might not be in NP
it's definitely in co-NP, ...but it's actually unknown whether that's equal to NP

[vapid drift]

i have another question

[fossil mural]

ok well yes, if L is finite then L' is definitely in NP

but that's because all finite languages are in P, and the complement of a language in P is in P

[vapid drift]

what does the 6 represent

[fossil mural]

{2, 3, 4, 5, 6, 7} has 6 elements

[vapid drift]

what does the 9 represnt

[fossil mural]

and chatgpt's reasoning in case 2 is definitely wrong

Case 2: L' is in NP-complete
This would also imply that L is in NP-complete because we can simply negate the output of the verifier for L'. This contradicts our initial assumption that L is not in NP-complete.
"we can simply negate the output of the verifier for L'" nope, if you could do that that would mean NP = co-NP

[fossil mural]

there are 9 places that you could put something in a string of length 9

L is finite so it is poly searchable
this is the correct proof of that statement btw

[vapid drift]

how did he know to do 7!

[fossil mural]

L is finite, therefore it's in P, therefore it can't be intermediate

[fossil mural]

because after placing 2 characters of a 9-character string there are 7 others, and we want to know how many ways those 7 characters can be arranged

[vapid drift]

are those two characters a and e?

[fossil mural]

...yeah that's the thing
it knows enough about what proofs "look like" to generate things that look like proofs, but not enough to generate actually correct proofs

[fossil mural]

yes

[vapid drift]

how do we know a is the first letter it can accept and e the last letter

[fossil mural]

"the first occurrence of letters (i.e. the leftmost letter) must be A, and the last occurrence of letters (i.e. the rightmost letter) must be E"

[vapid drift]

how did they get those ones

the 25 and 36 you already explained the rest i know

15

[fossil mural]

$\binom{6}{4}$ is the number of ways to choose $4$ things from the set ${2,3,4,5,6,7}$ which has $6$ things

[vital dew]

bee [it/its]

[fossil mural]

to get that that's equal to 15 you would use the formula for n choose k, or just a calculator

,w 6 choose 4

[vital dew]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by bee2660#0

[inner schooner]

@vapid drift شوف الخاص

[vapid drift]

where did the 2 come from

[fossil mural]

6-4

the general formula is $\binom{n}{k} = \frac{n!}{k!(n-k)!}$

[vital dew]

bee [it/its]

[vapid drift]

ah and 6-4 is 2

[fossil mural]

yep

[vapid drift]

thank you i will do the question by myself now i thanking you

[quiet eagle]

Any set B and its power set B' have no element in common right?

[fossil mural]

consider B = {empty set}

[quiet eagle]

okay but for every nonempty set B that is true, correct?

[fossil mural]

B is not empty

it contains a set that's empty, but it's not empty

by "{empty set}" i mean the set containing the empty set

[quiet eagle]

haha

[fossil mural]

if B is the empty set then that's not actually a counterexample because the empty set doesn't have elements in common with anything (because it doesn't have elements)

[quiet eagle]

yea

But in general!

for normal sets like {1,2,4,e}

[fossil mural]

are you saying that sets that contain sets aren't "normal"...?

[quiet eagle]

don't cancel me for that pls

So, there is a set B, which has no element in common with its powerset

[fossil mural]

...yes, there is at least one set with that property

[quiet eagle]

I went back by a lot

[fossil mural]

...by the way are you actually sure that 1, 2, 4 and e definitely aren't sets?

[quiet eagle]

Why would you ask that? Is {1} equal to {{1}}??

[fossil mural]

no, because then 1 would be a set that contains itself and that can't happen (assuming that this is ZFC)

but 1 might be equal to {2}

or, probably more likely, it might be a set containing some other object that isn't a real number, like maybe 1 is the set containing the empty set

[quiet eagle]

ah phew

[fossil mural]

although if this is ZFC then real numbers are definitely sets because everything is a set

[quiet eagle]

Is that so!?

[fossil mural]

the thing is

you have to get the real numbers from somewhere

[meager prawn]

If you take an intro to set theory, you'll see it's the standard way of defining them

You have the empty set and everything is built from it

[quiet eagle]

that sounds elegant. I'm hyped for set theory.

I've got the real numbers defined by using sequences and Cauchy-Stuff I think

[fossil mural]

alright so if 1 is some sequence... is a sequence a set? how did you define sequences? :)

[quiet eagle]

I forgor, it's been some time :(

[fossil mural]

the reason that the real numbers don't "feel" like sets is that generally we don't care what the elements of 1 are
but they could still exist

[quiet eagle]

"they could exist"? What could exist?

[fossil mural]

elements of 1

although in general if you have to worry about whether real numbers are actually sets you're generally doing something wrong (unless you're constructing the real numbers but in that case you know what the elements of a real number are)

"are B and the power set of B disjoint" is just a weird question to ask if B is a set of real numbers, the answer would not be meaningful or useful

[quiet eagle]

I see

I am in fact a bit puzzled by an exercise

Let $A$, $B$ be sets and $f : A \rightarrow B$ be a mapping. The power sets of $A$ and $B$ are $\mathcal{A}$ and $\mathcal{B}$, respectively. We consider the mapping $g : \mathcal{B} \rightarrow \mathcal{A}$, $B' \mapsto f^{-1}(B')$. Show: \
(a) $f$ is injective if and only if $g$ is surjective.

[vital dew]

Sciencenjoyer

[fossil mural]

...alright
what specifically are you confused about?

[quiet eagle]

So I suppose B' is element of power set B?

[fossil mural]

yep, B' is an element of the power set of B

equivalently, it's a subset of B

[quiet eagle]

Then why should an a in A exist such that f(a) = B'

Regarding the definition of g

[fossil mural]

i think f^-1 here is supposed to be preimage, not inverse

[quiet eagle]

Heh I've already been through that. Was told not to assume an inverse exists in this server

[fossil mural]

$f^{-1}(B')$ is the set of all $a \in A$ such that $f(a) \in B'$

[vital dew]

bee [it/its]

[quiet eagle]

And we have clarified there exists at least one set B, which is disjoint with its power set

[fossil mural]

that's true but not relevant

[quiet eagle]

No way

[fossil mural]

it doesn't matter whether B' is in B or not because f^-1(B') would still mean preimage and not inverse, that's just what that notation means in this case

[quiet eagle]

ye

It may be the case that B and B' have no element in common. So if b is element of B and b' element of B' and f(a) = b, then f(a) !=b' for every a

[fossil mural]

B and B' definitely do have elements in common unless B' is empty

B' is a subset of B

[quiet eagle]

But every element of B' is a set

[fossil mural]

...oh wait hang on the notation here is confusing

[fossil mural]

here we're using B' to represent an element of the power set of B

and the power set of B is $\mathcal{B}$

[vital dew]

bee [it/its]

[quiet eagle]

oh yeah sorry

I ment the fancy B

[fossil mural]

ok so yeah it might be true that elements of the power set of B are also not elements of B, ...but also it wouldn't be relevant

[quiet eagle]

In that case g would map nothing though

[fossil mural]

what...?

why?

[quiet eagle]

because no f(a) would be equal to B'

[fossil mural]

that doesn't matter

because f^-1(B') doesn't mean inverse, it means preimage

we're asking whether f(a) is an element of B'

[quiet eagle]

B' is an element already

[fossil mural]

...i don't even know what that means

everything is an element of something, why would that matter

[quiet eagle]

lol I ment B' is an element of the fancy B (power set of B)

[fossil mural]

why would the power set of B be relevant here

we have f, which is a function from A to B, and B', which is a subset of B

[quiet eagle]

here

[fossil mural]

and we write f^-1(B') to mean the set of a in A such that f(a) is in B'

why would it matter whether f(a) is equal to B'?

[quiet eagle]

brr I am steaming

So I was looking at the definition of g

In order for its codomain to be nonempty there should be at least one f(a) = B', correct?

[fossil mural]

no

[quiet eagle]

I dont mean the codomain but the value set

[fossil mural]

(range)

still no

[quiet eagle]

Is that the term for that?

thanks

[fossil mural]

i'm assuming you mean "the set of values that can be outputs of g", and if you do then yes that's the range

[quiet eagle]

yeees, thanks. Good to know

[quiet eagle]

How could there be a preimage f-1(B') if no f(a) =B' exists?

[fossil mural]

because a preimage isn't an inverse

f^-1(B') means the set of a such that f(a) is AN ELEMENT OF B'

being an element of B' and being equal to B' are not the same at all

[winter tangle]

5x =40x ^-½

Pls help

[quiet eagle]

lol

[quiet eagle]

As I already mentioned I am steaming a bit. Let me think a little ._.

[quiet eagle]

Oh yeah that would be the case if B' was a set. Which it is

okay fine

Would the scalar number 2 be a subset of {{1};{2};{1,2}}? @fossil mural

[fossil mural]

well it would be if 2 = {{1}}, otherwise no

..."is 2 a subset of [...]" is a pretty weird question though

[quiet eagle]

haha I realized it

I have an appointment now. I'll come back with a fresh mind sometime later

Thanks for now

[warm fog]

Hello,

I am a newbie to the field of inferring preconditions and am trying to wrap my head around it. Currently, I am trying to do the questions given over here - https://www.cs.williams.edu/~freund/cs326/ReasoningPart1.html
In the following question -

    x = x - 2;
    z = x + 1;
    { z != 0 }

Why is this happening -

    { x != -1 } =====> Shouldn't this be x!=1
    x = x - 2; 
    { x != -1 }
    z = x + 1; 
    { z != 0 }

[unreal stump]

What is initial x

[warm fog]

It can be anything ... imo

[unreal stump]

Ah mb

Yeah it should be x != 1

That's a typo

[warm fog]

Can someone please help me come up with pre conditions for these equations -

x = 7*k+5
b = 2*a-6
y = 3*b+2
{(x + y) > mu} 
z = x + y
{z > mu}

[vague scaffold]

alright guys. i've asked a number of mathematicians what I thought was a basic proposition, but it turns out there's significant difficulty in my question. Is this the proper place to ask questions about projective geometry/algebra?

[vapid drift]

how did he know to add 21 and 4

[fossil mural]

tbh just ask the question
if this turns out to not be the right channel then we can tell you where to put it once we know what the question is actually about because we've seen it

[vapid drift]

I mean alot of numbers add of to 21

[ivory badge]

He wants 21 because the goal is multiples of 21, the 4 is what’s left

[vapid drift]

i see thabk you

how is this true

i understadn that when i plus in 1 for the firsdt one but the second one?

[ivory badge]

It is the inductive hypothesis isn’t it

[vapid drift]

yes

[ivory badge]

What don’t you get

[vapid drift]

21 times 5^1 is amultiple of 21?

yes it is im stupid

[ivory badge]

[vapid drift]

does this mean if both x and y are true or one of them is true the answer is true

[ivory badge]

If either of X or Y is true, then X v Y is true

When both are true, that satisfied it

[vapid drift]

then whats the differece is it to this

[ivory badge]

That is an entirely different thing

That’s always true when Y is false

It’s only false when Y is true and Z is false

[vapid drift]

would the correct wording would be

[vapid drift]

like this

how would you say it like that

[ivory badge]

If Y implies Z?

[vapid drift]

I have another questiob

i have truth table now what does it prove

[ivory badge]

Does it prove the final statement?

[vapid drift]

how would i know

[vague scaffold]

Here is my question.

The projective harmonic conjugate uses this rule. I wrote a script to produce all of the whole/natural number possibilities on a number line. That is to say, if you make A=0, and solve for all whole number possibilities (I used python), the next picture is the coordinate locations on a number line that are valid for the remaining points. There are two valid solutions for every final value, because it works forward and backward.

I have found that every valid harmonic conjugate produces a length exactly half that of a pythagorean triple's perimeter. The first values, 0-2-3 produce 6 for the final value, which is 1/2(3+4+5) the first pythagorean primitive.

This process creates a value that is exactly half the perimeter of all of the pythagorean triples, and only creates these values.

--- Why??---

here's the python script if you care to follow up

This seems like a hint at a serious truth about numbers to me

anybody willing to say i'm at least not insane to wonder this?

[ivory badge]

Well do you have any notion of what it means to be a valid argument

[vapid drift]

no

[ivory badge]

Then go back and review

[ivory badge]

Do you understand why

If not, go back and review so you can see why the rows where they’re all 3 true are the important ones

[vapid drift]

oh that photo was from the notes

[vapid drift]

i still cant fin dthe patterm

[urban berry]

@wise quartz

its done

i think he ment 2sqrt(k+1) in the comment

[vapid drift]

can i see the original question

[ivory badge]

The pattern is all three are true, and those are the only rows where it happens

And it compares it to the truth of r, which is what they’re trying to derive

[urban berry]

its d

[vapid drift]

for this table?

[vapid drift]

what does this mean

[urban berry]

never delt with matrix before

[urban berry]

best bet might be posting it in #1021175428326633542

[vapid drift]

butt this is discrete math]

[ivory badge]

What does the degree sequence mean

[vapid drift]

i dont know

[ivory badge]

What do you think the critical rows are for it

[vapid drift]

i also dont know

what are the rows im comparing

[ivory badge]

Well try and think what they should be, checking your examples

[ivory badge]

What’s significant about the columns highlighted for the critical rows

[vapid drift]

they all are true

[ivory badge]

No

The columns

That’s is correct that the critical rows have all these three columns true

But that’s not what I’m asking, why is it those 3 in particular

[vapid drift]

r is true in those rows

right?

[ivory badge]

That’s not why they’re important volume

They’re important because those are your P1, P2, P3 and r is you C for an argument

And you show that having all of them true means that C is true

[vapid drift]

c is only true for these coloms

[ivory badge]

Columns are vertical chief

Rows are the horizontal ones here

[vapid drift]

wait how is their r my c

[ivory badge]

It is not

It is a C for a different P1 P2 P3

[vapid drift]

so i have to look at the coloums of c and compare it to z

and if z and c are true

[ivory badge]

Why Z?

What

[vapid drift]

becuase they are comparing to r

[ivory badge]

What’s the connection there

The order of things on the table doesn’t matter

You want P1, P2, P3 all true to imply C true right?

Do review old notes some, though

[vapid drift]

So these two

,rotate

[vapid drift]

sorry i know im dumb

[ivory badge]

Why did you just copy down the table

[vapid drift]

i circled the ones that are true

[ivory badge]

There are only two rows where P1, P2, P3 are all true though

[vapid drift]

yeah i circled them

[ivory badge]

So is C true on those rows

[vapid drift]

only one of them

[ivory badge]

So it is not

So it is not a valid argument

gg

[vapid drift]

for this one do i find the sum of all degrees and divide by 2

and if its odd it doesnt exist?

[ivory badge]

Well the sum of all degrees divided by 2 isn’t odd here, it’s 11/2?

[vapid drift]

is it supossed to be odd or integer the rule

[ivory badge]

Idk, check your notes

[vapid drift]

for this one

[ivory badge]

I meant check your notes on this part in particular

[vapid drift]

i figrued it out

[ivory badge]

But good

[vapid drift]

i couldnt find anything im just hoping it wont appear in 10 hours

[ivory badge]

Should’ve kept better notes fr

[vapid drift]

fr

[ivory badge]

Are K and L isomorphic

And are G1 and G2 isomorphic

[vapid drift]

what makes something isomorphic?

[charred gale]

I believe K and L have a different number of connected components, and that G_{1} has no vertex of degree 4

[ivory badge]

If I can map each point on the left to a unique point on the right, and send edges to unique edges between them

Essentially, if you can move the points around into the same shape, including the edges

[ivory badge]

Trying not to just give him answers and all, so as to help figure out why they’re the answers

[vapid drift]

why cant the answer be 8 and 7 why is it 5 and 7

[ivory badge]

Well, does 3 divide 8 or 7?

And does 6 divide 8 or 7?

[ivory badge]

There can be multiple pairs of elements

[vapid drift]

How to simplify

,rotate

[vapid drift]

i dont know why i just asked that

[vapid drift]

good morning

can someones double check thsi one

can someone tell me why the last column

what is that