discrete-math

[snow sleet]

But you haven’t come up with two equivalent claims

U came up with one claim

Your two claims together make up one claim equivalent to this. However, each of your claims on their own aren't equivalent to this.

[tidal tulip]

So how about 2 cases : 1. X is in A (complement) and x is in B (complement) 2. X is in A and x is in B

[snow sleet]

Sure tbh I'm really lost now on where you're going with this

[tidal tulip]

Forget the two claims

This is my claim

And I need cases to prove this claim

Do I need cases at all for this proof? Can I just prove it without any cases? If not how many cases would I need

[snow sleet]

then do this^

[snow sleet]

you would definitely need at least two cases right because you need to show the right hand side is a subset of the left hand side and the other way around

Also note that A intersect B is precisely just A

since A is a subset of B

[tidal tulip]

Ok

It’s gonna be hard doing this in strictly set notation

[snow sleet]

set theory proofs are never fun lol

[tidal tulip]

Idek where to start

[snow sleet]

but it's good practice

[tidal tulip]

😭

[snow sleet]

well suppose X is in B and not in A. for the first case. Then suppose x is in A for the second case

that would be my outline of the proof, the rest is up to you

[tidal tulip]

What’s wrong with the cases I had?

[snow sleet]

first off, case 1 seems really confusing because if x is not in B, then surely x is not in A, so redunant at the very least. Case 2, is also very redundant because if x is in A, then x is in B because we assumed in the first place that A is a subset of B

[tidal tulip]

I see

[snow sleet]

to be more precise I would just stick to doing this

[tidal tulip]

So for these two cases, what would be the “WTS” part ?

[snow sleet]

for the first one, it would be claim 1 here^. For my second assumption it would be claim 2 here^

because either "x is in A and therefore B" or "x is not in A but is in B" those are really the two meaningful cases

I'm gonna be afk for a bit getting lunch and such...but if u need any help in the meantime ping the Helpers

[nocturne grove]

i have to do this exact proof too lol

for this claim, I am getting stuck at $pk = qf, k,f\in\mathbb{Z}$

[vital dew]

42

[nocturne grove]

They are relatively prime so I know i should do something with that from here on

but not sure how

[tidal tulip]

Did you figure out how to solve it?

[nocturne grove]

I havent done it yet

[tidal tulip]

Wait you’re saying I should use these 2 as my cases for the claim?

Didn’t you say it’s not right

[snow sleet]

this is what I was replying too^

[snow sleet]

read this^

After telling you your two claims weren't on their own equivalent to the statement with the equality, I later got the impression that you were wondering how to prove the equality. In which case, your two claims "cases" are a good way to go for a proof.

For example: for sets A and B....."A is a subset of B" isn't equivalent to saying "the set A equals B". However saying "A is a subset of B and B is a subset of A" is equivalent to saying "the sets A and B are equal"

[tidal tulip]

Ok so I’m gonna use these two cases to prove my claim in that case (no pun intended)

[snow sleet]

yes proving those two would prove this^

[tidal tulip]

Ok, now I need a starting point for case 1 in set notation

[snow sleet]

Okay

That should be your first statement for case 1

[tidal tulip]

Can we use small a instead of x

[snow sleet]

that is a small x

$xX$

see the difference^?

[tidal tulip]

I mean a small a

Small “a”

[snow sleet]

no

[tidal tulip]

Ty

[snow sleet]

jk

Let $a\in\overline{(A-B)\cup(B-A)}$.

[vital dew]

logician

[snow sleet]

There's your first statement

[tidal tulip]

Ok

[snow sleet]

So $a\notin(A-B)\cup(B-A)$.

[vital dew]

logician

[tidal tulip]

Right

[snow sleet]

so $a\notin A-B$ and $a\notin B-A$.

[vital dew]

logician

[tidal tulip]

Is there a set notation for “and”

[snow sleet]

intersect

[tidal tulip]

Ok so upside down U

[snow sleet]

$\cap$

[vital dew]

logician

[snow sleet]

yup

[tidal tulip]

Ok

[snow sleet]

let's resume from here

[tidal tulip]

Sure

[snow sleet]

$a\notin A-B$ implies $a\notin A$ since $A\subseteq B$.

[vital dew]

logician

[snow sleet]

recall $A-B=A-(A\cap B)$.

[tidal tulip]

I’m thinking if that makes sense in my head

[vital dew]

logician

[tidal tulip]

Ok yea

[snow sleet]

$A\cap B=A$ since $A\subseteq B$.

[vital dew]

logician

[snow sleet]

now looking at this again

[tidal tulip]

The above is part of the proof?

[snow sleet]

this^?

[tidal tulip]

It’s not letting me go to the message

[tidal tulip]

This one

[snow sleet]

yes I mean this could really be the very first statement above both cases, but yea helpful to put it in your proof

okay so $a\notin B-A$, what does that mean?

[vital dew]

logician

[tidal tulip]

I mean like if i place this step right after the “a is not in A - B implies” line part would it make sense

[snow sleet]

i honestly won't know until I see your final draft of your proof, right now we're going through the proof while I'm giving you pointers on why things are true (not all of these pointers need to be in the proof)

[tidal tulip]

Oh ok

[snow sleet]

milly let's get back to this^

[tidal tulip]

So this means a is not in B - A

[snow sleet]

yes

but be more elaborate

[tidal tulip]

Like if we have an int a, the int a would not be in set B

[snow sleet]

right

[snow sleet]

I would say this implies $a\in\overline B\cup(A\cap B)$.

[vital dew]

logician

[tidal tulip]

That would be true

[snow sleet]

because $a\notin B-A$ implies $a\in\overline B$ ($a$ wasn't in $B$ in the first place) or $a\in A\cap B$ ($a$ was thrown out of the set $B$ when we calculated $B-A$).

[vital dew]

logician

[tidal tulip]

Hmm

[snow sleet]

you following?

[tidal tulip]

I get that a is in the complement of B but I don’t get how a is in A intersection B

[snow sleet]

$B-A$ means take the set $B$ and delete the elements of $B$ that are also in $A$.

[vital dew]

logician

[snow sleet]

Recall $A\cap B=A$

[vital dew]

logician

[tidal tulip]

I think I get it now

[snow sleet]

okay great so that's the end of case 1

we just showed $\overline{(A-B)\cup(B-A)}\subseteq \overline{B}\cup(A\cap B)$

[vital dew]

logician

[tidal tulip]

Correct

Case 2 now is the other way around

[snow sleet]

yes I'll leave that up to you right after I go over one piece we should elaborate more on from case 1

[tidal tulip]

I would say case 2 is similar to case 1

Is it possible that case 2 proof is backwards to what we did in case 1?

[snow sleet]

yes, saying $a$ is in the LHS of the equation, is the same as saying $a$ is not in the union of $A-B$ and $B-A$. Meaning $a\notin B-A$ and $a\notin A-B$.

[vital dew]

logician

[snow sleet]

$a\notin B-A$ is how we got the first case complete

[vital dew]

logician

[snow sleet]

Case 2: let $a\in \overline{B}\cup(A\cap B)$

[vital dew]

logician

[tidal tulip]

Right

[snow sleet]

so $a$ is notin $B$ or $a$ is in $A$

[vital dew]

logician

[snow sleet]

if $a$ is not in $B$, it definitely won't be in $B-A$

[vital dew]

logician

[tidal tulip]

The second step would be that a is not in B

But I don’t get how to write the other half of that

[snow sleet]

if $a\in A$, then $a$ won't be in $A-B$ because $a$ is in $B$ as well

[vital dew]

logician

[snow sleet]

other half of what?

[tidal tulip]

The part after B complement

[snow sleet]

hence $a\in\overline{(A-B)\cup(B-A)}

[tidal tulip]

.

[snow sleet]

what part about this is confusing? where are u getting lost?

do you agree that means $a$ not in B?

[vital dew]

logician

[snow sleet]

or a is in A intersec B which is A

[tidal tulip]

I’m getting a bit confused

Too many things going on lol

For case 2, I have one step written down

And I’m a bit confused about step 2

[snow sleet]

Case 2 means a is not in B or a is in A intersect B

Right?

[tidal tulip]

Are you explaining what the proof will do or is this the title of case 2

[snow sleet]

This is not the title lol

[snow sleet]

It follows from this^

[tidal tulip]

I get that a is not in B because of the complement

[snow sleet]

Alright so a is in B complement or A intersect B, right^^^

[tidal tulip]

Yes

[snow sleet]

Great so a not being in B means a isn’t in B-A

Right?

[tidal tulip]

Sure

[snow sleet]

It also means a notin A-B

Since everything in A must be in B

[tidal tulip]

Yes

[snow sleet]

Great so in this case a is in the set on the LHS of the equation we’re trying to prove

Cuz it’s not in A-B and not in B-A

[tidal tulip]

Right

That’s on the right side

[snow sleet]

No we’re trying to show a is in the set on the LHS of the equation we’re trying to prove

We already proved the other direction….

[tidal tulip]

We’re trying to prove that LHS is a subset of RHS

…no?

[snow sleet]

We already did that

[tidal tulip]

No but the case 2 is switched

[snow sleet]

Read this

[tidal tulip]

Right

But now that case is switched

?

[snow sleet]

What do u mean switched? We need to show the RHS is a subset of the LHS

[tidal tulip]

Yes that’s what I mean lol

I get what u were thinking now

Nvm keep going

[snow sleet]

So we did half of case 2

Now the other half of case 2 is if a was in A intersect B

U following???

[tidal tulip]

Yes

[snow sleet]

K

This means a is in A

[tidal tulip]

Yes

[snow sleet]

So a notin A-B since everything in A is in B and would therefore get thrown out

Following?

[tidal tulip]

Yea

[snow sleet]

Great a in A also means a notin B-A because it would get thrown out of B

So we just showed the RHS is a subset of the LHS

Proof is complete

[tidal tulip]

Ok

I wrote down the cases but they’re in rough copy

And ik I’m missing some steps

[snow sleet]

That’s fine. Clean it up I walked you through the proof so as you’re writing up ur final draft just look back here

[tidal tulip]

For case 1 I think it’s good but case 2 I’m missing steps

And it’s a bit for confusing then case 1

[snow sleet]

Look at our chats lol I don’t wanna say the same thing 3times again when u can just scroll up

[tidal tulip]

I’m not asking for that lol

I can just share the rough is what I’m saying

[snow sleet]

Uh don’t do that

Clean it up

And then it’s complete

While it’s still fresh in ur mind

[tidal tulip]

It’s draining …

[snow sleet]

Yea these proofs are typically boring lol

Set theory proofs…..

[tidal tulip]

There isn’t a set notation for “also” is there?

[snow sleet]

Depends on the context…if a is in A and also in B then use that upside down U

[tidal tulip]

For case 2 I have it as, a is in A which also means a is not in (B - A)

[snow sleet]

There’s set notation for and or and not

That should say if a is in A….

Because the other option was a was in B complement

[tidal tulip]

Ok I’m gonna read case 2 again

Cuz I’m lowkey lost

[snow sleet]

K I’m gonna do some hw and see if anyone else needs some help

[tidal tulip]

I have the good copy (I think) for case 1

[snow sleet]

Lmk when both cases are done

[tidal tulip]

Can we work on them one at a time to prevent confusion

[snow sleet]

I wanna see a complete proof

[tidal tulip]

Besides case 1 is more clear to me atm

[snow sleet]

Sure post a pic of what u got

[tidal tulip]

Ok

Uhh

It’s not letting me upload

… not letting me upload anything …

[snow sleet]

type what you have

[tidal tulip]

That’s too hard

[snow sleet]

ok outline what you got

[tidal tulip]

Shall I try dming it ?

[snow sleet]

sure

you should be able to upload pics here so not sure why that's not happening but yea we can try

[snow sleet]

<@&286206848099549185> Could someone help Milly. I've outlined a proof for Milly, we've been at this for 3+ hours because seems to be having a hard time following along...I don't wanna give the answers outright, hoping someone else here can explain it in a way that may help trigger his mathematical thinking.

[umbral chasm]

Ohmygod

The chat length is so long

[snow sleet]

yeah we've been at this since for 3+ hours

https://cdn.discordapp.com/attachments/496785905474994186/1154922340988223558/Image_2023-09-22_at_7.28_PM.jpg

this is what they were trying to prove^

[ivory badge]

Just say x in the left iff (formula)

and in the right iff wtv

[still isle]

Pls gives me a hint on this. I think we are using Pigeon here but I don't know where to apply it.

[ripe beacon]

A domino tiling problem on 2 x n arrays:
https://stackoverflow.com/questions/37744971/filling-up-2-x-n-array-with-dominos

This gives us the fibonacci sequence.

Can someone explain to me, why we can simply add the probabilities?
So when we tile vertically -> f(n-1) probabilities or options
tile horizontally -> f(n-2) options.

I get that, but why can we simply add these? Is there a proof behind this? How does it work?

[frigid oak]

can anyone help me understand the third and the fourth example?

i want to say that there are no such functions but im sure that im missing something

[safe steppe]

Hi guys, is the only way I can prove a and b using a truth table or is there another way?

[little prism]

well truth table is by far the easiest way

and in the end every proof will be equivalent to the truth table

for a) anyway. for b) you can use a) assuming you already know stuff about 'and'

[safe steppe]

Thanks alot @little prism

[foggy sentinel]

In a workplace is a pool of 8 printers. One of the printers stops working. In how many possible ways can 19 print outs in the que be divided between the 7 functional printers if:

a) at least 1 printjob goes to each printer
b) at most 4 prints go to every printer

I got 177100 on a), is that correct?
I'm not sure how to solve b)

[main cargo]

I have a question regarding proof by induction (the theorem i'm proofing is about the adequacy statement for logic which makes use of trees/semantic tableaux). Basically what it comes down to is, imagine we have tested a property for the base case so it holds k = 0 for example. And then we assume that it holds for k, and I want to prove it for k+1. Can I then say that the property holds for everytihng smaller than k and use that to prove it for k+1 or not?

[safe steppe]

Hi guys, for this question, does my answer look correct, and how do I find the weakest preconditions, is it the top precondition {v<-1 or w>0}

[little prism]

that is called strong induction and is fine, yes

[main cargo]

Okay thank you very much!!!!!

[topaz flame]

how to prove $p∨q, q∨r, p->¬r |- q$ using natural deduction

[vital dew]

thatFirla

[topaz flame]

What I am struggling with is having two disjunctions and I don't know what to do? Because won't I have like 5 outcomes and all need to be true?

[thorny hollow]

Hi

What number is it?

What number would -x be?

Hmm

By substitution

y = 4. There exists only one x which satisfies the given formula

Then x = -4

4 = -(-4)

Is this right?

@pearl fern

@final cliff

sorry for pinging but I am so curious about this

[final cliff]

What are you trying to figure out here?

[thorny hollow]

What number would x be in the formula y = -x if substitute y by 4

And, also,

What is the notation for equinumerous sets…

I will kiss you if you clean my doubts

[final cliff]

You're trying to compare the set of positive "number" and the set of negative "numbers" right? @thorny hollow

[thorny hollow]

No, it is about the print I send above there

The formula

y = -x

It is a bijection
, yeah

Substitute the variable y by 4

4 = -x

What will x be

-4?

[final cliff]

"It can be seen at once that this function maps for instance, the set of all positive numbers onto the set of all negative numbers ..."

This is from your image.

[thorny hollow]

Yeahthen yes I am

[final cliff]

A formula by itself is not a bijection fwiw. Bijections are functions from one set to another that satisfy extra properties.

Here one of your sets is the positive real numbers and the other is the positive negative numbers.

So, you can define a function $f:\mathbb{R}^+\to\mathbb{R}^-$ such that $f(x)=-x$ for any $x\in\mathbb{R}^+$.

[vital dew]

DootDooter

[final cliff]

Pretty sure this is the function they're talking about is all.

It's the same function I mentioned earlier.

For any x in the domain you map x to -x.

Not sure how much my notation makes sense to you.

$\mathbb{R}^+$ is the set of positive real numbers. $\mathbb{R}^-$ is the set of negative real numbers. $f:A\to B$ means f is a function with domain A and codomain B. $x\in A$ means x is an element of the set A.

[vital dew]

DootDooter

[thorny hollow]

This I understand

[final cliff]

If you want to show my f is a bijection it just takes some arithmetic and basic facts about the sets above.

If y is a negative real number, then -x is a positive real number. Notice f(-x)=--x=x.

This shows f is a surjection onto the negative real numbers.

If x and y are positive real numbers and f(x)=f(y), then -x=-y by definition of f, so x=y follows by properties of real numbers.

This shows f is an injective function.

So, f is a bijection

You see what I mean @thorny hollow ?

[thorny hollow]

f(x) = -x

f(4) = --4

so, y = -x in a function notation

would simply be f(x)=-x?

[bold tundra]

I was looking at this question, and wondering why the answer wouldn't be $\forall_x \exists_y (P(x,y)\wedge \neg Q(x,y))$

[vital dew]

bossysine

[tidal bridge]

Hi everyone. I colored the successive neighbours of a tile in an aperiodic hat-tiling and it looks like it produces hexagons. Is there a way to prove that ?

[unborn vapor]

That is correct actually - the only difference is they wrote “not (not P or Q)” and you wrote “P and not Q” but since those are equivalent, the two answers are the same

[bold tundra]

thanks

[rare river]

Hi guys I have a question

proof by contradiction basically means we assume the negation of a statement to be true? And we try to find a single contradiction from one of its premises to disprove that?

basically if we found out that a premise is false then we can conclude that the negation of a statement is actually false from just this singular contradiction

Therefore, if it this statement is false then the negation of the statement should be true instead?

I hope someone could help me stress-test my understanding of the concept, thank you!

[final cliff]

I think you get the main idea. Proof by contradiction is just a technique to show the negation of a particular statement holds.

The mechanics of actually doing that are to assume the statement is true, do some calculations and inferences, derive some contradiction, and then to conclude the negation of your original statement must be true instead based on the contradiction.

You basically said all of this already, just phrased a little differently.

[fallow dune]

man i hate discrete math, especially cuz the teacher mad ass

[tidal prawn]

can someone help me understand what this means?

[snow sleet]

think about it by replacing y with x^2

y must be be the square of x in order for (x,y) to be in R.

that is only (x,x^2) is in R

@tidal prawn

[static bone]

Hi. Can anyone explain where they got the 4 from? is it because its a factor of 8?

[fallow dune]

what does a | b mean?

im thinking its b/a but idk

[faint sphinx]

a divides b

That is, there is an integer k such that b = a*k

[fallow dune]

oh ok thx

[plucky wedge]

I know the answer is b and e, but I was wondering if c and a are the same thing in this statement?

is AxEy == EyAx?

[faint sphinx]

Yeah that's pretty much it. Odd number = can be written as 2k+1, but also means can be written as 4m+1 or 4m+3 (useful to keep in mind for proofs like this!)

[plucky wedge]

I know AxEy ~= ExAy but I'm not sure if this is logically equivalent

oh

I looked it up and AxEy ~= EyAx

but I dont see how they would be different in the question

I think statement c says for all children of an existing parent?
and statement a says there exists a parent that for all children

which says two different things

I'm not sure if I translated that right though, could someone let me know if the quantifiers were translated correctly

[static bone]

Ok thank you

[plucky wedge]

yeah im not sure if c is translated right and if it is then isn't that equivalent to a?

[faint sphinx]

a) is "there exists x, such that for all y, y is the child of x, and y works at a grocery store"

(idk why W(x) became W(y))

[plucky wedge]

I think that might be just my teachers mistake I had a couple of problems on my homework where she would accidently switch variables

[faint sphinx]

c) is "for all y, there exists x such that y is a child of x and y works at a grocery store"

So for a), there would be someone who's a parent of all the children. For c), each child has a parent (but this needs not be the same parent)

[plucky wedge]

oh ok

[faint sphinx]

+whatever about working, but idk that it's right as writen. In either case, this part alone shows they are different

[fallow dune]

Is this an acceptable proof or is there any steps I skipped in the parts after we must show that

[snow sleet]

looks good to me! I'm convinced!

[fallow dune]

👍 thanks for the confirmation

[snow sleet]

I wrote a proof for a problem I assigned myself...was wondering if my proof is convincing to anyone else....I'm convinced but yea wondering if u guys are

pls ping me with whether or not ur convinced....I'm gonna grab some dinner rq and will prolly be helping other ppl in other channels

[fossil mural]

``$|S| = \infty$" is very weird notation

[vital dew]

bee [it/its]

[snow sleet]

i agree lol

[faint sphinx]

I wouldn't call it complete

[fossil mural]

also it kind of feels like you just skipped over most of the proof?

[snow sleet]

I did

LMFAO

no but my thinking is

[fossil mural]

like everything you said is true but you kind of just make one true statement and then jump to "so S is infinite, this is a contradiction"

[faint sphinx]

Whatever your thinking is, it should be part of the proof if relevant, no?

[snow sleet]

since s_i is arbitrary, then for s_l, there's something in S greater than that etc. Since s_j is also in S, then there's something in S less than that etc...

I should prolly right that^

[nocturne grove]

damn thats difficult

[faint sphinx]

That doesn't really fully show it's infinite though

[snow sleet]

I'm using that statement I proved recursively

I proved that u give me any element of S, I can find something in S smaller than it

u give me that "smaller" thing, I can find....

similarly for the greater elements

[faint sphinx]

Like yes that would be correct. But if I was grading it I would give you like 50%

[snow sleet]

Ryx

https://tenor.com/view/byuntear-meme-react-funny-dog-gif-25019867

ok tanks

Yea I mean definitely more detail would be helpful to add in my proof

[nocturne grove]

LOL

[snow sleet]

Ryx also agrees with me that it's correct so I'm taking that win. But always good to hear other ppl's opinions on math work. That's how we get better, right?

[faint sphinx]

I would either prove it inductively, or do something along the lines of that you did to show (and ofc include this part in the proof) that if no max existed, you could then create an infinite, strictly increasing sequence of elements

[nocturne grove]

I agree

I want to be good at this proof stuff even though I am not lol

[snow sleet]

yup i'm implicitly using induction

Aight thanks for the tips Ryx, much appreciated

[faint sphinx]

(and like ofc i know you know; just communicating that in a proof can be hard. I am also a bit of a harsh grader tbf)

[nocturne grove]

Are you guys like profs or something?

[faint sphinx]

4th year undergrad

[nocturne grove]

Oh damn

[snow sleet]

I'm going into a Ph.D. program to be a mathematical logician so slick proofs for things are things I like to come up with. I pretty much leave things in the proof that are sufficient to prove the statement...which I did here too, but making sure the proof flows well and is reader friendly is also good to have other than just correctness.

proofs in textbooks are typically short for a reason

[nocturne grove]

Oh I see, for me I have to add as much detail as possible. Its short but looks complicated lmfao. Also whats a mathematical logician?

Well either way, sounds cool, gl on the phd too

[snow sleet]

thanks. Proofs should be short and sweet. Some like mine are short and to the point, but may lack some flow that would be easier for readers to digest. A very short description of a mathematical logician is someone who uses tools in logic to go about proving things using tricks in logic that aren't typically taught in a intro to proof class. Proof theory is a huge topic as well.

[nocturne grove]

That sounds difficult as shit

But hey seems like you enjoy what you do so I envy you

haha

Seems super cool though

[snow sleet]

for instance like this one I posted^ it doesn't seem very obvious how the logic works, but it's there. This is why chatting with other mathematicians is always helpful...especially if I was teaching an undergraduate course. A professor would def understand my proof, but again if you can't convey your proof to ur readers well, it's not really much good if ur audience is not quite grasping what you're putting down

[jagged grove]

@snow sleet do u happen to know while compiler courses tend to focus on AST and dont use general graphs?

is there no use for general graph theory in those topics

or something along those lines?

i have to take compilers next academic period

[snow sleet]

and that @jagged grove is the only course I got a C in so I should not answer that question

Graph Theory was taught online and I was left teaching myself it so yea, I know when I shouldn't answer questions and send u down the wrong rabbit hole lol

[nocturne grove]

Fair enough, it would also depend on who you are presenting your proof to. I mean I just started proofs so I am not great at them nor am I decent, I am still learning.

[jagged grove]

i got a 7 out of 10 in linear algebra cause i didnt take the final exam

i got a 10 in everything though

[snow sleet]

in graph theory too?

[nocturne grove]

That's why I am supposed to add as much detail as possible ig

[jagged grove]

@snow sleet i got a b+ in the discrete math

[snow sleet]

when you're first starting out yea

[jagged grove]

but we only took trees

so i studied some graph theory by myself

[nocturne grove]

yo im doing that rn

[snow sleet]

ah so u didn't cover graph theory as a course itself

[jagged grove]

no

i might finish all the proofs in my discrete math textbooks

[snow sleet]

One thing graph theory helps us with is some counting problems

[jagged grove]

and look at more advanced stuff

[snow sleet]

i know that for sure

[snow sleet]

good to hear! more exposure the better!

[jagged grove]

ive looked at a thing called theory of ordered sets

[snow sleet]

Are you talking about the WOP?

the Well-Ordering Principle

[jagged grove]

that i do know

is every subset of the naturals has a minimum

[snow sleet]

nonempty subset

[jagged grove]

minimum is a n such that n < a for all a

[snow sleet]

but yea

[jagged grove]

yeah forgot that

but yeah

sec

[snow sleet]

<= actually

[jagged grove]

true

[snow sleet]

because n<n certainly isn't true

[jagged grove]

yeah

[snow sleet]

since n is in the nonempty subset

[jagged grove]

and were including n as an a

[snow sleet]

aight yea so u know the WOP then

[jagged grove]

in that for all quantifier

[snow sleet]

yup

[jagged grove]

sec

[faint sphinx]

posets are fun

[nocturne grove]

this some next hieroglyphs

what are these words bruh

[faint sphinx]

I am mentoring a directed reading program on them

[snow sleet]

noice noice

partially ordered sets

[nocturne grove]

whats that?

[jagged grove]

i mean the stuff on that book

[nocturne grove]

liek certain elements are sorted or something?

partially idk

[snow sleet]

pretty sure partially ordered sets are called posets right @faint sphinx ?

[faint sphinx]

correct

[jagged grove]

so ordered sets would be a subset of posets?

[faint sphinx]

It's a set with an "ordering" on it, which satisfies just 3 axioms/assumptions

[nocturne grove]

i should finish this last problem, no more distraction (i don't know waht im doing)

[snow sleet]

relations and stuff pretty much

but they're not equivalence relations

[jagged grove]

order i would assume

[faint sphinx]

ordered = poset (or a special type of poset) here. More generally, there are preorders, where you don't assume antisymmetry, but those are less important

Just different naming conventions

[jagged grove]

seems like the kind of thing that would be important for scheduling systems

[snow sleet]

me and judson have bad notation in common LMFAOO

[faint sphinx]

I do not know or care for applications 🧍‍♂️

[jagged grove]

i gotta make a living

[snow sleet]

how are u with counting problems?

[jagged grove]

describe a counting problem quickly

[snow sleet]

combinatorialists are always in demand

[jagged grove]

is it like

permutations

and combinations?

[snow sleet]

yes

[jagged grove]

i know the basics from discrete

but i have not gone in depth

[snow sleet]

For instance, If S is a set with n elements, how many functions f: S^2 -> S are there?

[jagged grove]

is s^2 a cartesian product?

[snow sleet]

yes

[jagged grove]

n

or at least n

i would pressume

[snow sleet]

SxS=S^2 just like R^2=RxR

[snow sleet]

way more S^2 is a set of ordered pairs of elements of S

[jagged grove]

so n^2?

[snow sleet]

and they are sent to any one of S places right?

[snow sleet]

nope more

[jagged grove]

mm i havent counted possible functions

sec let me think

[snow sleet]

S^2 has n^2 elements

[jagged grove]

ohh

and its permutations right?

[snow sleet]

and each of those elements can go to any one of n places

[jagged grove]

the order doesnt mather

yes

so nxnxn

[snow sleet]

answer would be .........

?

[jagged grove]

n^3

[snow sleet]

oh god

[jagged grove]

im bad?

[snow sleet]

https://tenor.com/view/byuntear-meme-react-funny-dog-gif-25019867

[jagged grove]

i know

[snow sleet]

bro missed the final exam in discrete too

nah jk

[jagged grove]

dont tell me were dealing with factorials

wait

we are

[snow sleet]

the answer is nxnxnxn....xn n^2 times right?

so $n^{n^2}$

[vital dew]

logician

[snow sleet]

nope

[jagged grove]

so i can choose a n

n to the 2 times

is what u mean

[snow sleet]

each element of S^2 ( and there are n^2 of these) goes to any one of n places so n times n times n.... (n^2 times)

(a,b) can go to n places

one of n places

(x,y) can go to one of n places as well

so can all of the ordered pairs

so n^(n^2) is the answer

[faint sphinx]

Now define this for infinite sets 🗿

[snow sleet]

https://tenor.com/view/dog-meme-gif-25326537

LOL

[jagged grove]

i dont even know where to start

[snow sleet]

I would start with

crying

LMAO

[jagged grove]

nahh more like

a book on combinatorics

[snow sleet]

and then I'd take a break for like years

it always does the trick

LOL

[faint sphinx]

tbf i don't know like cardinal arithmetic really. But the set of functions A --> B is really just the set B, producted with itself |A| times (lmaoo choice moment for infinite sets)

[snow sleet]

$\infty^\infty$

[vital dew]

logician

[snow sleet]

LOL

[jagged grove]

so |A||B| ?

[snow sleet]

no |B|^|A|

each element of A can go to one of |B| places

[jagged grove]

ahh

its cause their functions

[snow sleet]

yes

[jagged grove]

so a can only go to one

[snow sleet]

yes

[jagged grove]

for each possible function

[faint sphinx]

infinite sets have cardinals, not size "infinity"

[snow sleet]

uh

[snow sleet]

bijections the answer fr

[faint sphinx]

That's the size of A x B

[jagged grove]

yes\

but we dont want that

cause thats not a function

a function is a subset of AxB

[snow sleet]

I think that's true? Functions are relations so

[jagged grove]

yeah but it has more to do with the fact that if (a,b) but if a is in b we cant have (a,a)

[snow sleet]

I'm asking how many functions we can have....they don't have to be injective

two elements fromt the domain can go to the same element in the co-domain that's totally fine

[jagged grove]

i edited

[snow sleet]

yes

[jagged grove]

cause thats not what i was trying to say

[snow sleet]

https://tenor.com/view/byuntear-meme-react-funny-dog-gif-25019867

ok

[jagged grove]

one element in a cant have to images in b

thats the important thing here

[snow sleet]

yes

wait are u still trying to count that function question I asked or are u thinking of something else?

[jagged grove]

im trying to figure out where the n^(n^2) reasoning came from

[snow sleet]

it came from the fact that the domain has size n^2 and the codomain has size n.

Each element of the domain can be sent to one of n places in the codomain

so u multiply n...n^2 times

[jagged grove]

question

[snow sleet]

yes

[jagged grove]

does this counting behave as a function itself?

[snow sleet]

what do u mean "this counting"

I mean sure given different n, n^(n^2) should be different right?

[faint sphinx]

Better solution: Consider the set of functions from A --> B, where |A| = N and |B| = M. rather than working around the square in there

[jagged grove]

well M^A

[snow sleet]

no

[faint sphinx]

And justify why the | {set of functions A --> B} | = M^N

[jagged grove]

it would be m^(n^2) if it were a cartesian product

this is not

[jagged grove]

now this

[snow sleet]

in Ryx's example the answer should be M^N.

[jagged grove]

yes

cause codomain is not a cartesian product

[snow sleet]

Think about how many elements are in our domain

and how many elements are in our codomain

and use the reasoning that each element of our domain can go to only one element of our codomain

those three things are the only things needed for this problem

[jagged grove]

the power set is not needed?

[snow sleet]

you already noted S^2=n^2, and that's correct

[faint sphinx]

dandida is this what you are thinking of

[snow sleet]

power set is the set of all subsets....yea no we're not doing that here

How about I write S like this instead

[jagged grove]

sec im going through the justification for m^n

[snow sleet]

S={s1,s2,...,sn}

Given $S={s_1,s_2,s_3,\dots,s_n}$, count how many functions $f:S^2\rightarrow S$ exist. Recall $S^2={(s_i,s_j):s_i,s_j\in S}$.

[faint sphinx]

functions S --> {0,1} frfr

[vital dew]

logician

[snow sleet]

yeaaaaa buddddddy Ronnie Coleman

[jagged grove]

do u happend to whatch dota streams or something?

[snow sleet]

dota streams?

[jagged grove]

yeah ronnie coleman is a meme within the admiral bulldog stream

also the culturist

[snow sleet]

nah I just watched Ronnie deadlift 800lbs and called it "lightweight baby"" ain't nothin to it but to do it

lmao true

[snow sleet]

I typed this up to help u think about where each element from S^2 goes when it gets sent to S

[jagged grove]

there is a choosing for each element in the codomain of n

and since we have s^2 elements in the codomain

s can get map s^2 times

[snow sleet]

nope

[jagged grove]

in the domain

[snow sleet]

for each element of the domain [which is of the form (s_i,s_j)] it can be sent to any one of n places. Say it gets sent to s_1. Then for the next ordered pair, it can be sent to one of n places, say s_1 again...and so on

so all the ordered pairs each have n options for where they can go

[jagged grove]

wait

im gonna change what i typed earlier

[snow sleet]

k

[jagged grove]

done

[snow sleet]

uh okay, as long as it makes sense in ur head....what would the answer then be

[jagged grove]

for what rix mentioned?

[snow sleet]

no for my problem lol that u were doing

[snow sleet]

this one^

if u answer it for mine, you'll def be able to answer Ryx's much simpler problem

[faint sphinx]

no

bad

backwards

[jagged grove]

s^(s*s)

[faint sphinx]

Do mine, and conclude logicians as a special case lol

[snow sleet]

the reasoning goes both ways. If u generalize ur reasoning

[jagged grove]

for urs

[snow sleet]

looks good s=n I'm assuming

[jagged grove]

yes

[snow sleet]

perfect

[jagged grove]

if it were lets say

SxSxSxS...xS -> S

it would be S^(s*s...*s)

[snow sleet]

so that's S^ what power?

u gotta define that power of S

cartesian product*

[jagged grove]

im mixing a indexx

in the notation

[snow sleet]

uh

How bout this:

For $|S|=n$, count the possible functions $f:S^k\rightarrow S$

[vital dew]

logician

[snow sleet]

k and n are natural numbers

[jagged grove]

S^(S^k)

[snow sleet]

uh

S is a set

S^S doesn't make sense

[jagged grove]

n^(n^k)

[snow sleet]

yes

that's the answer

[jagged grove]

now for riz

[faint sphinx]

rizz

[jagged grove]

the only way of justifying that is by using induction

ryx

sorry

[snow sleet]

that's how I read it at first lmao

rizzard of oz

[jagged grove]

so is induction a valid approach

or not?

[snow sleet]

why would u even....

sure yes that would work

[jagged grove]

can u put the statement back in

so i can put it on paper

and actually write a proof

[snow sleet]

you could fix one of the sets and induct on the number of elements in the other set

[snow sleet]

Ryx's or

[jagged grove]

yes

ryx's

[snow sleet]

Yea I'll type it

[jagged grove]

thats my idea

thx

[snow sleet]

Given nonempty finite sets $A$ and $B$ with $|A|=N$ and $|B|=M$, count all the possible functions $f:A\rightarrow B$.

[jagged grove]

ur missing the second part

[vital dew]

logician

[snow sleet]

u mean the answer?

[jagged grove]

no

justify

[snow sleet]

justify? what am I missing lol?

I'm so lost

[jagged grove]

this

i meant to prove that by induction

[snow sleet]

Given nonempty finite sets $A$ and $B$ with $|A|=N$ and $|B|=M$, count all the possible functions $f:A\rightarrow B$. Justify why this answer is $M^N$.

[vital dew]

logician

[faint sphinx]

ooh no not induction

[jagged grove]

why?

[snow sleet]

shhhh

[faint sphinx]

That doesn't really fit with this one

[snow sleet]

jk

[jagged grove]

him

m and n are naturals

[snow sleet]

it does if u fix the correct set. My prof did this once

[faint sphinx]

Yeah but you don't need it anyways, and I don't think it helps here

[snow sleet]

She's tryna geek out on proofs

let her do induction

https://tenor.com/view/dog-meme-gif-25326537

[jagged grove]

u do know

im a he

[snow sleet]

oh

woops sorry

[jagged grove]

its ok

[snow sleet]

let him do induciton

great I can't spell now

induction*

[jagged grove]

writing give me a sec

[snow sleet]

Think about which set ur going to fix the cardinality of

[jagged grove]

writing it

sec

almost finished with n = 1

will share

[snow sleet]

Ping me when u have a proof I’ll be grabbing some food in the meantime but yea I wanna see this. I recommend induction on |the cardinality of A.|

Shoot was trying to make that one of those black things that u can open up to see

[jagged grove]

for n = 1

ill try for n+1

or i was thinking of using n-1 instead

@snow sleet

[snow sleet]

This looks fine for n=1, but the ordered pairs look backwards and the last line “which means…” isn’t convincing me that the number of functions when n is 1 is precisely m

[jagged grove]

thats what was bugging me

how can i express the counting of the possible functions

like i have shown that for a function there are m elements sure

its what i figured

[snow sleet]

If A={a}, (a, b_1) etc

[jagged grove]

ahh

ok

[snow sleet]

U do it how u want, some people do with words or just use ordered pairs like I did

(a,b_1) represents one function

[jagged grove]

one thing though is this still discrete math?

[snow sleet]

(a, b_2) another etc

[snow sleet]

Sure is

Discrete math is the overarching area combinatorics is the more precise area

[jagged grove]

ur right my pairs are backwards

[snow sleet]

Im glad u fixed the right set tho ahaha

[jagged grove]

i was just sort of thinking how i would expand the terms

lol

let me keep going

[snow sleet]

K

[jagged grove]

that is corrected

now going for n+1

@snow sleet

[plucky wedge]

did she mean m + 2 = 2?

[snow sleet]

looks good to me

[jagged grove]

wait im going through branshi stuff

that inequality look sus

to me

[snow sleet]

nope this proof is right

[plucky wedge]

oh n is also 0 I see

[jagged grove]

i was thinking more like n+2m+1>= n+2*1+1

[snow sleet]

n being a perfect square means its nonnegative

[jagged grove]

since m>=1

[snow sleet]

technically ur right

that should be a >= instead of > there

[jagged grove]

u go spreading that formalism on everyone jo

jeez

lol

let me go back to the other proof

[snow sleet]

lmao

[plucky wedge]

why did she replace m with 1 in n + 2 * 1 + 1

[snow sleet]

to show it's at least that number

on her way to showing it's greater than n+2

because m is at least 1

[plucky wedge]

oh ok I see

[snow sleet]

again it shouldn't be a strict inequality between n+2m+1 and n+2*1+1, but nevertheless ur teacher prolly already knows about that typo

since m could be 1

[plucky wedge]

hopefully this gets easier with practice

[snow sleet]

it will

different ppl prove things differently

everyone has their own flavor

sometimes it's hard reading other proofs

[jagged grove]

thats why reading different books makes things easier

[snow sleet]

I'm typing up a proof by induction for our problem rn

[jagged grove]

im doing the same thing

i think my idea has a hole

ill show u once im done

[snow sleet]

let's see who finishes first

I started after u so

[faint sphinx]

(this formula still makes sense when A or B is empty btw)

[snow sleet]

done

[jagged grove]

damn

give me a sec to show u how i was approaching it

ill finish the step where im at

[jagged grove]

i was thinking the exact same thing but was expressing it in a cumbersome way

let me show u

[snow sleet]

aight

[jagged grove]

idk if u get the idea

cause trying to write the next terms would be a pain

[plucky wedge]

why are we supposing n^3 + 5 is odd?

since its contradiction aren't we supposed to say n3+ 5 is even and n is odd

[jagged grove]

cause 5 is odd

ohh

[snow sleet]

U could define T more concisely as $T={(a_i,b_j):i\in{1,2,3,\dots,n},\ j\in{1,2,3,\dots,m}}$

[vital dew]

logician

[snow sleet]

nope proof by contradiction only assumes the negation of the consequent not the negation of the atecedent

[jagged grove]

it has to do with the way implication works

if a then b

[snow sleet]

we can assume n is an integer and that n^3+5 is odd...prove by contradiction means we also add the assumption n is not even (i.e., n is odd in this case)

[jagged grove]

but you proof that a then not b leads to a contradiction

then the case true then false disappears

[plucky wedge]

oh ok I'll write that down

[jagged grove]

so ur left with truth

[jagged grove]

this is the thing though, sometimes i have the right idea but idk how to express it in a way in which im minimizing the amount of effort it takes to write it

[snow sleet]

let me give u an example of a proof by contradiction

[snow sleet]

just takes practice

[jagged grove]

do the inexistence of the square root of 2 in the rationals

[snow sleet]

done

[jagged grove]

are u a phd student by any chance?

[snow sleet]

almost

[jagged grove]

at masters?

[snow sleet]

@plucky wedge

[snow sleet]

nah I'm a senior undergrad rn

But let me write a proof that more fits the structure of the statement you showed.

[jagged grove]

doing it with no suprema and axiom of completeness

that is new

nvm

ur proving that is irrational

not that it exists

[snow sleet]

Given $x\in\mathbb R$, if $0\leq x<\epsilon$ for each $\epsilon>0$, then $x=0$

[vital dew]

logician

[snow sleet]

@plucky wedge I'll write a proof by contradiction for this problem

[jagged grove]

they dont see neighboorhoods in discrete how u going to show that?

[snow sleet]

\begin{proof}Suppose $x\neq0$. Then $x>0$, and so $0<x<x$, a contradiction. \end{proof}

[jagged grove]

nvm

[vital dew]

logician

[plucky wedge]

I understood until for such k, q^2 = p^2..

also whats | in 2|p

[jagged grove]

divides

[plucky wedge]

oh ok

[snow sleet]

q^2=p^2/2 from the equation 2q^2=p^2

[plucky wedge]

oh

[snow sleet]

Are u looking at the first or second proof for the sqrt of 2 being irrational?

they're both by contradiction

but yea which one are u looking at

[plucky wedge]

first one

[snow sleet]

k

[jagged grove]

there are 2 cases here

x and -x

[plucky wedge]

the math makes sense now but what is gcd(p, q)

[jagged grove]

greatest common divisor

[snow sleet]

greatest common factor

[snow sleet]

nope we're given x>=0

[snow sleet]

ur not wrong to say greatest common divisor

GCD literally stands for that

it means p and q share no other positive divisor other than 1

gcd(p,q)=1

but I showed they share 2 as well

a contradiction

[jagged grove]

in analysis sherbert uses what u showed alot for reaching contradictions

things like the limit of a convergent sequence is unique

and stuff of that nature

but he shows it using neighborhoods

idk why

like epsilon neighboorhoods

[snow sleet]

the definition of a sequence converging literally has epsilon neighborhoods in the definition of convergence

the definition says

[plucky wedge]

so it shows 2 is a common factor of p and q but why do we say that p and q share no other positive divisor other than 1?

is it implied in the proof somewhere

[jagged grove]

primes

[plucky wedge]

or a definition of something

[jagged grove]

or relative primes

something along those lines

[snow sleet]

let $(a_n)_{n=1}^\infty$ be a sequence in $\mathbb R$. Then we say the sequence converges to the point $L\in\mathbb R$ if for every $\epsilon>0$, there is a natural number $N$ for which $|a_n-L|<\epsilon$ for all $n\geq N$. In this case we say $a_n\rightarrow L$

[vital dew]

logician

[snow sleet]

you can assume that all common factors of p and q have been taken out aside from 1

in other words, u can assume the fraction is a simplified as possible

[jagged grove]

wouldnt u have to add something like

[snow sleet]

that's an assumption u can make without losing cases in ur proof

[jagged grove]

withouth loss of generality

[snow sleet]

nope

not needed

[jagged grove]

ok

[snow sleet]

u could and u wouldn't be wrong

but it's not needed. If I give u a fraction u can assume it's in it's most simplified form, otherwise simplify it until u get gcd(p,q)=1

[plucky wedge]

oh ok so we are assuming that p/q is in it's most simplified form but since we can take out 2 it's a contradiction

[snow sleet]

yes

[jagged grove]

yes

[snow sleet]

I assumed p and q were relatively prime (meaning they share no other positive factor other 1), and yet I showed p and q are both even and hence divisible by 2>1

yup that's it

[snow sleet]

@jagged grove do u see why epsilon is needed now lol

[plucky wedge]

oh ok I see

[snow sleet]

my second proof for that makes use of the fact that I can assume p is minimal

given a nonempty set of positive integers, the well-ordering principle allows me to make such an assumption

[jagged grove]

well cause we need any epsilon greater than 0

by def of convergence

we should probably tell ryx that we showed the problem suggested

[snow sleet]

yes and $|a_n-L|<\epsilon$ is equivalent to saying $L-\epsilon<a_n<L+\epsilon$

[vital dew]

logician

[jagged grove]

yeah

theres a theorem that states that

[plucky wedge]

is proofs usually the first unit in discrete math

[jagged grove]

that can be shown using the properties of the absolute value

[snow sleet]

yes

[jagged grove]

depends on whos giving it

[snow sleet]

and the fact that epsilon was positive

[snow sleet]

not usually. usually it's set theory

and then logic

and then proofs potentially

[jagged grove]

and some colleges dont even emphasize proofs

like they mostly give u induction

and call it a day

[snow sleet]

in their defense

[jagged grove]

at least thats what they did in my college for SE majors

[snow sleet]

u can't bring someone from intro level to wizard level in even a semester

[faint sphinx]

[jagged grove]

but i took intro to mathematical thinking for math majors

so

[snow sleet]

proof theory takes time to learn

[jagged grove]

true

[snow sleet]

lol

[jagged grove]

lol

is the math sorcerer in this discord?

guy is litt

[snow sleet]

and even profs aren't quite at the finish line of wizardly level. There merely much further along the math proof experience level than students are

[snow sleet]

no clue

[jagged grove]

at UCR where i did math (i can actually go back whenever i want i just dont have the money to. If i get a permanent position that might be different though).

every teacher needs a phd

to give a undergrad level course

its mandatory

idk how common that is

[snow sleet]

well yea because profs are representing the college and ppl pay a lot of money to go to universities/colleges. They want their profs to be experts in their fields

[jagged grove]

in UCR its barely paying i could take an entire undergrad semester for about 300 dollars

so

this is american mind u

and i had no scholarship

[snow sleet]

gotcha

UCR university of ?

[jagged grove]

let me send u a link

[snow sleet]

k

[jagged grove]

http://www.posgradomatematica.ucr.ac.cr/index.php

so this is the page for their grad curriculum

thats a little bit more expensive

[snow sleet]

costa rica

[jagged grove]

but it has all teachers that are working in there

yes

[snow sleet]

looks nice

I wanna live in costa rica lol

it's like the bahamas

[jagged grove]

http://www.posgradomatematica.ucr.ac.cr/index.php/administracion/personal-docente

those are the teachers

[snow sleet]

I mean think about it this way: If you're a prospective student looking to apply for colleges and one college has professors all with Ph.D.'s while another has only some with Ph.D.'s, that could be a deciding factor....so not to surprising to see colleges wanting profs with Ph.D's

[jagged grove]

yeah but look at the colleges though

[snow sleet]

of course

[jagged grove]

the ones from latin america are top latin american unis

and most of them are from france and the us

[snow sleet]

but yea aside from campus and such, I'd wanna be taught by an expert especially since I'm paying a lot for education

[jagged grove]

where are u at the moment?

[snow sleet]

I got a proof for u to try @jagged grove

[jagged grove]

sure

[snow sleet]

west coast united states

Show that there are infinitely many even numbers greater than two that can be written as the sum of two primes

prove this directly

[jagged grove]

isnt this a variation of the fundamental theorem of arithmetic?

[snow sleet]

not really. I made up this problem

[jagged grove]

ij

ok

[snow sleet]

that has to do with a unique product of primes.

Ryx let him think

[faint sphinx]

two cannot even be written as the sum of two primes unneeded condition

[snow sleet]

I wrote it like that for a reason.....

Goldbach's conjecture.....

[snow sleet]

it was also my way of stating positive evens

[faint sphinx]

fair

[snow sleet]

I'm curious to see what dandida comes up with

U can assume there are infinitely many primes

[jagged grove]

so this is different from goldbachs cause of the infinitely many?

[snow sleet]

more or less yes

[faint sphinx]

goldbach is that every even > 2 can be writen like that

[jagged grove]

yeah

for all

[faint sphinx]

this just just infinitely many

[jagged grove]

basically

[snow sleet]

My hints for u @jagged grove are u can prove this directly and u may assume there are infinitely many primes

[jagged grove]

so would showing that there exists more than 1 be enough for this?

[snow sleet]

no

u need to show infintely many exist

[jagged grove]

this is what i dont get though

i dont recall being introduced to the infinitely many

concept

[snow sleet]

okay,

so

if I asked u to prove there are infinitely many odd numbers

u could state, let z be an integer then 2z+1. Moreover, 2z+1<2(z+1)+1.

There are infintely many integers

and I showed there are infintely many odd numbers I can construct

does that make sense?

[jagged grove]

yes

so basically showing that for any i can have another one

[snow sleet]

okay use a similar method of "construction" for when u show that there are infinitely many even numbers greater than two that can be written as the sum of two primes

[snow sleet]

yes there are infintely many distinct integers, and I showed that that corresponds to distinct 2z+1. In other words if u gave me distinct z and y, then 2z+1 would not equal 2y+1.

So we haven't lost the infinitism so to speak

[jagged grove]

ok i got it sec

@snow sleet