discrete-math

[frank zinc]

If it is a circuit it is not a path and visa versa

[last timber]

each circuit is path

but not each path is circuit

[brave lava]

how hard is discrete math out of 10

[sick swallow]

maybe 9 although idk

[unreal stump]

Depends

[lyric pumice]

10 out of 10. It has Millennium Prize Problems among many other long-standing open problems.

http://unsolvedproblems.org/index_files/PerfectCuboid.htm

http://unsolvedproblems.org/index_files/Collatz.htm

http://unsolvedproblems.org/index_files/ChromaticNumber.htm

[frank zinc]

How many possible nxn adjacency matrices are there?

2^(n*(n+1)/2)?

[lyric pumice]

Yes.

[frank zinc]

ty

[lyric pumice]

Certainly.

[leaden moon]

How should I get rid of (P V ~Q)?

[leaden moon]

o

thanks

[weary tiger]

np

[leaden moon]

I was so stupid, I didn't know there was a law to get rid of the T

lol

[weary tiger]

not really a law but when you have a T or F you can simply get rid of it

[leaden moon]

Yeah I didn't know there was a button for that

🤦‍♂️

[weary tiger]

ah ok

[void valve]

Why would this be true? ^* is how they write complement of the set

[stray reef]

do you know the finite version of this

$(A \cup B)^* = A^* \cap B^*$ and vice versa

[vital dew]

Ann:

[void valve]

De morgans?

[stray reef]

yes

[void valve]

Aha! thanks I had missed that each of the subsets on the RHS was complement

[stray reef]

this is just de morgan's for countable unions & intersections

[void valve]

thanks yeah I read it as A_i only on the RHS

thank you 🙂

Wait sorry what do you mean coountable? Like they are countably infinite? Could it be that we have an uncountable number of unions?

[stray reef]

you're unioning and intersecting countably many sets

you could have $\bigcup_{i \in \mathcal{I}} A_i$ for $\mathcal{I}$ a set of any cardinality

[vital dew]

Ann:

[void valve]

Aha so like for example you could have i element or real numbers? What would it mean for the index of the set to be 0.1 for example?

[stray reef]

?

well you'd have an indexed family of sets, one for each real number

for example you could have a family of intervals of the form (0, r) indexed by positive reals

[void valve]

Yeah I dont know I guess I've only encountered integer indices then. So in your example as r could be any real value it wouldnt suffice to only have integer indices, right?

[stray reef]

i mean

you have a continuum-sized family of sets

[void valve]

Yeah so you could have (0,r_1)....(0,r_n) and therefore you would also need real number of indices?

[stray reef]

you're overthinking it

[void valve]

As always hahah 😦

Alright but could you give some simple real-life example where it would be needed?

[buoyant marsh]

Anyone knows about codeword??

[mint bane]

can someone help me translate from english to predicate logic pls, im doing some practice problems and i wanna get em right. the statement is "there are no people who are TA's in this class and didn't get A's", functions that can be used are S(x), meaning that “x has been a student of this class,” A(x), meaning that “x has gotten an ‘A’ in this class,” T(x), meaning that “x is a TA of this clas,” and E(x, y), meaning that “x and y are the same person.”

bottom one is my guess

[quartz wigeon]

Hi, I've been doing this exercise but the -->q keeps confusing me and don't know if I have it correct.
What I'm doing wrong?

[weary tiger]

what is "⇒q" doing inside the table as a column?

[weary tiger]

-->q makes no sense

put a ~q instead

@quartz wigeon

[hollow bloom]

Two graph being bipartite doesn't mean they are necessarily isomorphic right?

[weary tiger]

Certainly not

[naive saffron]

For any n there is a bipartite graph with n vertices: namely the graph with n vertices and no edge

Certainly a bipartite graph with 2 vertices is not isomorphic to a bipartite graph with 3 vertices

[manic tiger]

@quartz wigeon
Are you asking what the --> q means in the truth table, the possible outcomes?

[weary tiger]

"⇒q" should not be in the truth table in the first place!

[manic tiger]

Yeah, should replace that with the negation of q since it's being used down the road

@mint bane if the picture you sent is the translation of there are no people who are ta's in class and didn't get a's then that would be good

some places say it's good practice to not leave negation before quantifiers so you could simplify but that still is right

[cerulean ore]

Is T(n) = 1+2+3+4+....+n equals to Θ(n^2)?

[stray reef]

"equals to"

n(n+1)/2 is Theta(n^2) yes

[cerulean ore]

I am trying to prove that it is Theta(n^2).
So, let c1 * n^2 <= n(n+1)/2 <= c2 * n^2,
If c2 = 1,
n(n+1)/2 <= n^2 for n>=1 .

Now, I have to find a c1 * n^2 such that c1 * n^2 <= n(n+1)/2 .

[unreal stump]

1/2

[stray reef]

^

[cerulean ore]

It doesn't have to be an integer value?

[unreal stump]

No

[stray reef]

it does not

[cerulean ore]

Thank god I asked you guys, thank you! So, for c1 = 1/2, and c2 = 1. The following inequality holds:
c1 * n^2 <= n(n+1)/2 <= c2 * n^2, therefore f(n) is Theta(n^2) .

[stray reef]

yes

[cerulean ore]

Thanks!

[celest geyser]

I have a probability unit in my course, not sure if this is the place to ask, but I posted a question at #help-1 and I was hoping someone can take a look

[celest geyser]

I'm good now thanks

[weary tiger]

I think it's b despite not seeming like it would b

try showing the first expression is equal to 3n C 3

[copper ore]

true

ya they're equal but do you know how to get that top part out of 3n C 3

[weary tiger]

how do you know they are equal if you can't do what you just asked?

[copper ore]

i would never guess the top part is equal to 3n C 3

i just did it

i calculated it

i think you misunderstood me

i said how can you derive the top part lol

top part =

this

i would never guess the top part is equal to 3n C 3

[weary tiger]

like, describe a counting method you mean?

that corresponds with the top expression

[copper ore]

yeah like by reading statement b, I would go straight to doing 3n C 3 not the top part

[weary tiger]

to start, a more suggestive form is probably $$3\cdot {n \choose 3} + 6n \cdot {n \choose 2} + n^2 \cdot {n \choose 1}$$

[vital dew]

Botnuke:

[weary tiger]

try showing the first expression is equal to 3n C 3
also disregard this it sounds like a bad idea now that I think about it

[copper ore]

n^2 should be n^3 btw but yeah

why?

[weary tiger]

uh no?

[copper ore]

oh yeah

nvm

also disregard this it sounds like a bad idea now that I think about it
@weary tiger why?

[weary tiger]

probably algebra hell, when there is a more elegant argument

[copper ore]

true ok

[weary tiger]

hmm not sure where the 3 and 6 are popping out from yet

maybe there is some kind of case splitting going on

I feel like I am just reverse engineering some really inefficient but correct way that someone came up with to count a simple value

[copper ore]

yeah im trying something here too

v inefficient

[weary tiger]

something dumb like subsets that have an element from each different set, subsets that have elements from only 2 different sets, and subsets that have all their elements from one set

and then adding them up

wait that might actually be it

[copper ore]

im basically doing recursion

im turning all of them into this lol

it's taking too long

then when k= 1 i remove the combination symbol

[weary tiger]

I dunno if that's gonna work for an algebraic solution but you can try

[copper ore]

yeah i have no idea

lol

[weary tiger]

something dumb like subsets that have an element from each different set, subsets that have elements from only 2 different sets, and subsets that have all their elements from one set
I think this is it though

for a counting method solution

[copper ore]

i see

sounds shorter than mine

[weary tiger]

yea I regret even suggesting to show it with algebra

[last sigil]

yeah

[copper ore]

do you have a solution with u?

im curious how it looks

[weary tiger]

the last quoted message is pretty much it, but it's very terse

[copper ore]

isee ok

[weary tiger]

$3\cdot {n \choose 3}$ counts the number of subsets that have elements from only 1 set, $6n \cdot {n \choose 2}$ counts the number of subsets that have elements from exactly 2 different sets, and $n^2 \cdot {n \choose 1}$ counts the number of subsets that have elements from 3 different sets

[vital dew]

Botnuke:

[weary tiger]

any subset described in (b) is in one of these categories, so it counts them all

[copper ore]

wow ok

thanks so much

[weary tiger]

someone else should verify this as I could be wrong, but it seems good to me

[brave lava]

https://i.imgur.com/VTTsrDk.png

hey im kinda confused and my teacher's microphone sucks lol

[stray reef]

your second blank should be variable and the third blank should be left

[brave lava]

thanks

https://i.imgur.com/MY5Bdxs.png

for this question

for C

Im confused

I know there is only one empty set

and that its represented by a 0 with a slash through it

[stray reef]

its symbol is not the digit zero

it has its own symbol

$\varnothing$

[vital dew]

Ann:

[brave lava]

yeah that one

for part d

i know all universes have the same empty set

but im not sure what my teacher meant by

"we always denote it"

[stray reef]

i don't know

it looks like there was meant to be a symbol following it

but it got lost

[brave lava]

okay thats fine thanks ill just email him tomorrow

[unreal stump]

What does one empty set for each universe mean?

[brave lava]

we can resubmit anyway

are you talking about the thing in my question?

[unreal stump]

Yes

[brave lava]

like there is an empty { } set for each universe

and an empty set is a unique set, and only one exists

and its unique because it exists for all universes

according to what my lecturer said lol

[unreal stump]

I don't know why we have to deal with "what a universe is" in a math question

D is really weird

[brave lava]

its my first time taking a discrete math course and my book hasnt even arrived so im not even sure either lol

[unreal stump]

Or what denoting it means

[brave lava]

apparently my book is coming from india in 22 days

🤦‍♂️

[weary tiger]

In the 1st part of inequality, would a proof which goes like this suffice: If K(G) is not empty, consider a set of vertices K with |K| = K(G). Such a set exists by definition of K(G). Then this will have some vertices, which will in turn have some edges. and the number of edges incident on the vertices have to be greater than the vertices because of minimality of K. Thus proved

The book has more elaborate proof for the same using different reasoning, so I wondering if there was a simpler one

[stray reef]

what even are kappa, lambda and delta?

[weary tiger]

ah, sorry 😓

kappa is the connectivity

lambda is the edge connectivity

delta is the minimum degree of a vertex

[stray reef]

so kappa(G) is the smallest number of vertices that need to be removed to disconnect G, and lambda(G) is the same but with edges?

[weary tiger]

yup

[stray reef]

and what are you calling K(G)

[weary tiger]

ah, that's kappa

[stray reef]

oh

the "if K(G) is not empty" bit confused me

did you mean if kappa(G) is not 0

[weary tiger]

yes, sorry for my bad english !

[stray reef]

ok i'm not exactly sure what your argument here is

are you just picking out kappa(G) vertices from your graph arbitrarily

[weary tiger]

yup, that's the definition

[stray reef]

so you aren't concerned with whether or not your set disconnects G?

[weary tiger]

we are concerned with that. kappa(G) is the max number of vertices you can remove without concern, and G will still be connected

same with lambda, but for edges

[weary tiger]

ah okay, I think my reasoning is wrong. I assume there will be more edges than vertices because of minimality, but that's not always true

[south marten]

Can anyone check if i've started the proof of demorgan's laws correctly: http://mathb.in/44767

[pale epoch]

let $x \in A' \cap B'$ this means that $x \in X$, and that $x \notin A', B'$ 🤔

[vital dew]

Lochverstärker:

[pale epoch]

also you started with $x \in A' \cap B'$ and concluced $x \in A' \cap B'$

[vital dew]

Lochverstärker:

[south marten]

Oh shoot good point the argument could be circular

@pale epoch is it cicurclar I thought to prove two sets were equal you have to show that one set is entriely contained within the other

[pale epoch]

yes, but you dont do that

[south marten]

Where is the error particularly

[pale epoch]

i told you

also you started with $x \in A' \cap B'$ and concluded $x \in A' \cap B'$

[vital dew]

Lochverstärker:

[south marten]

ahhh okay

[pale epoch]

you have to start with $x \in (A \cup B)'$ and conclude $x \in A' \cap B'$

[vital dew]

Lochverstärker:

[pale epoch]

and then do it the other way around as well

[south marten]

Why can't I start with $x \in A' \cap B'$

[vital dew]

Zophike1:

[pale epoch]

you can, but you have to conclude $x \in (A \cup B)'$

[vital dew]

Lochverstärker:

[pale epoch]

which you did not

[south marten]

ahhh makes sense

[pale epoch]

(in fact, you have to do both directions)

you have only shown $ A' \cap B' \subseteq A' \cap B'$

[south marten]

Yes I know

[vital dew]

Lochverstärker:

[south marten]

Oh 😦

@pale epoch also I was asking if i started the proof the correctly i'm well aware I have to show the reverse inclusion I fixed the issues you mentioned: http://mathb.in/44770

[pale epoch]

you did not

i mean, now everything you write down is correct except

and hence we have shown that $(A \cup B)' \subset A' \cap B'$

[vital dew]

Lochverstärker:

[south marten]

I'm not understanding how is that not correct because woun't the conclusion follow after showing the reverse inclusion

[pale epoch]

i don't know how to make this more clear

you write

let $x \in A' \cap B'$ this means that $x \in X$, and that $x \notin A, x \notin B$ via definitions.We know that an elements belongs to $A'$ if and only if it is in $X$ and not in $A$ the same observation is applicable to $B'$ This proves that $x \in A', x \in B'$ Finally this means that $x \in A' \cap B'$

[vital dew]

Lochverstärker:

[pale epoch]

or to cut it down a bit

let $x \in A' \cap B'$ [some true statements] Finally this means that $x \in A' \cap B'$

[vital dew]

Lochverstärker:

[pale epoch]

this shows $x \in A' \cap B' \implies x \in A' \cap B'$ or in other words $A' \cap B' \subseteq A' \cap B'$

[vital dew]

Lochverstärker:

[pale epoch]

which is true, but doesn't contribute to the proof you are doing

[south marten]

Okay now I see the mistake i'm sorry : (

[pale epoch]

no need to apologize

[south marten]

How would get the conclusion from what I have so far

[pale epoch]

what does it mean for $x$ to be in $(A \cup B)'$?

[vital dew]

Lochverstärker:

[south marten]

For $x \in (A \cup B)$, $x \notin A$ $x \notin B$ which implies that $x \in A' \cap B'$ also not that $x \in A'$ and $x \in B'$ as well

[stray reef]

are you allergic to the word "and"

[south marten]

nahh I just typing this out on my phone

[vital dew]

Zophike1:

[desert wind]

Hey does anyone have an interesting paper somewhere in the combinatorics/discrete math realm that a beginning graduate student might have to parse through without having to look up >7 definitions? I need just one paper to look at and review but everything I've looked at looks like it'll take 2 hours of reading other papers just to parse.

It's for a Research Analysis, just read and summarize, I don't need to do a lot with it but everything I'm looking at seems waaaay over my head.

[desert wind]

Thanks. Does it still have an impact factor that, y'know, registers? Because part of this is supposed to be a lesson on what papers not to use, what papers to use, etc.

Sorry, being colloquial. XD I meant that as in being prestigious. Yeah, that's pretty low, but right now I might just go with it. Thank you.

Yeah, that gives me something to talk about.

Thank you

[strong wolf]

what the hell am i looking at...

none of this chat makes sense to me

[sick swallow]

ok...

[strong wolf]

lol

[white jetty]

how should I prove this?

[pale epoch]

what have you tried @white jetty ?

[white jetty]

I tried to prove it with one-to-one

since it seems like it's finite

[pale epoch]

what seems finite?

[barren apex]

Alright, quick question with general algorithims

Is the question incorrect with the multiple choices it has?

The unchecked choice shouldn't be there as far as I know

[pale epoch]

seems like it

[barren apex]

¯_(ツ)_/¯

[desert wind]

@white jetty one route might be to consider the binary counting system, and seeing if you can work something with that.

imo it is literally easier to prove your bonus question. XD

[pale epoch]

why

the bonus question has you consider strings thats start with arbitrary many 0's

so you can't just use your suggestion

[desert wind]

Step One: Solve the bonus question
Step Two: Consult answer to Step One. XD

Nah, I getcha

[pale epoch]

that being said the bonus question is not much harder

but the original question is easier and should be solved first

[desert wind]

Anyway, did you get it or do you still need help?

the bonus question has you consider strings thats start with arbitrary many 0's
@pale epoch
Answer to part one has arbitrarily many trailing zeroes. Turns out this doesn't really matter

[pale epoch]

it does though

using your suggestion, you would map 1001 and 01001 and 001001 and ... to the same element

[desert wind]

I'm just saying consider the same general mechanism of add-carry.

Sorry, that was a weird hint.

But yeah, the two parts are really the same, because the set of all that start with one is a proper subset with the same cardinality. To construct the first group, you just need to construct the second, then tack a one on the front.

I unno, I feel like it's a distractor.

Like, imagine the answer to the second question. It contains stuff like the empty string, and every binary string with 0s and 1s included.
Tack a one to the front
Now you have all the answers to part one.

[pale epoch]

this doesn't change the fact that it is easier to come up with the actual isomorphism for the first set...

[desert wind]

Anyway, I'd start with something like this:
Let l be the length of the string.
Consider first the set of single-length strings, {0,1}
Clearly this is a finite set with cardinality 2, and is countable. Namely, we can biject it with {0,1}

Then just work your way from there.

Basically get your 1 length out of the way, then your 2 length, then so on inductively.

If you can partition the number line such that every number ends up in a set of size 2^n, where n is the length of a given subset of all binary strings, then you've basically proved you can do it without having to even specify how.

Ask special permissions for the empty string, and you got part B. Then appeal to "Tack a one on the front" then you got part A.

[desert wind]

@pale epoch thank you for the problem, just decided to write up a solution for my own edification and it was really fun, actually. 😄

Oh wait, you weren't the one who posed it. XD

Thanks anyway, it was fun talking to you!

[fervent briar]

I've got a question about logical equivalence

Question:

Isn't it pretty much just like that?

That's for the second statement

[hollow bloom]

Can anyone find a K 3,3 or K 3,3 modification from this graph?

I have tried stretching vertex a and b but I can't seem to find one

[fervent briar]

I got the answer as 131.916 but it's wrong

I've recounted 3x for this problem

Can anyone help spot the mistake?

[leaden pebble]

@fervent briar it seems you arent taking the correct vectors

you are getting an external angle

[fervent briar]

External?

How would I know if an angle I get is an external or not?

[leaden pebble]

try to draw ab and bc leaving the same point

thats the angle youre getting

[fervent briar]

Like that?

[leaden pebble]

ye

thats not an angle inside the traingle

[fervent briar]

Why?

[leaden pebble]

its the supplementary of b

[fervent briar]

Should it be then BA and BC?

Instead of AB and BC?

[leaden pebble]

if you want B, yes

you are getting the angle the line that extends AB does with BC

[fervent briar]

OK, I think I get it somehow

Thank you

[leaden pebble]

extend AB

[errant bear]

think of shifting the vector AB so the origin matches up with BC

[leaden pebble]

and see the angle it forms with BC

[errant bear]

this "new" angle is the one you already found

also this is more multivar calc/LA than discrete math btw

[fervent briar]

also this is more multivar calc/LA than discrete math btw
@errant bear Yes, I got it into the wrong channel since I was doing discrete earlier. Sorry

[leaden pebble]

considering the channels, it may be either LA or geometry in pre

[fervent briar]

It's multi var

[leaden pebble]

I see

[fervent briar]

thanks btw

[leaden pebble]

no problem
always check if the vectors leave the same point

so youre getting the right angle

[mint bane]

whats the difference between proof by contradiction and proving by contrapositive

like im looking at a question that says "prove the statement 'if r is irrational, then r^(1/5) is irrational' by its contrapositive"

my math intuition tells me this isn't the same as if it said prove by contradiction but idk

[pale epoch]

it isn't

proof by contraposition uses the fact that $A \implies B$ is equivalent to $\neg B \implies \neg A$

[vital dew]

Lochverstärker:

[pale epoch]

proof by contradiction, you assume not A and arrive at a contradiction to prove A

[mint bane]

alright i understand that

[pale epoch]

like, in both cases you start by assuming r^(1/5) is rational

but in the first case you deduce that r is rational

in the second case you would use that r is irrational to deduce some contradiction

[mint bane]

so for what i said above, i need to assume that r^1/5 is rational and then deduce from that that r is also rational

and by contraposition that tells us the r being irrational implies that r^1/5 is also irrational

wait did i just describe contradiction by accident? i meant to describe contrapositive lol

[pale epoch]

nah you are good

[mint bane]

ok but then what would the contradiction side of it look like

i wanna make sure i get both rn

[pale epoch]

it's assume r^(1/5) is rational and that r is irrational

and then deduce some contradiction

(in this case probably that r is both rational and irrational)

so yeah, contradiction doesn't really gain you anything in this case

you will have to do the "normal" proof anyway

[mint bane]

so what changes is the statement that gets proven then? the contrapositive proves that not B implies not A but contradiction shows that not B implies both A and not A?

[pale epoch]

in general proof by contradiction you would assume not B and A and arrive at any contradiction

but like, in this case a proof by contradiction is not really possible

or at least i dont see how

[mint bane]

yeah im trying to think in more general terms

[pale epoch]

by that i mean, you could take any "proof by contradiction" and reformulate it to not use contradiction

[mint bane]

do yk an example that might better illustrate it, i just pulled this from a free mit thing lol

[pale epoch]

the thing is, in this case you are already asked to proof a statement of the form "A implies B"

so contraposition is obvious

can't think of an implication that is proved with contradiction

[mint bane]

ohhhhh

[pale epoch]

but like, the standard proof of infinitude of primes is proven via contradiction

you want to prove "there are infinitely many primes", so you assume there are just finitely many

and then arrive at some contradiction

so your assumption must have been wrong

[mint bane]

ok i think im getting it

im gonna do some reading abt it, can i bother you in like 10 min if anything comes up pls?

[pale epoch]

you can try

[mint bane]

also unrelated but what's that books role you have?

[pale epoch]

it's from the testing phase of #book-recommendations

[mint bane]

ah

[pliant horizon]

doing #29 and it feels weird. https://puu.sh/Goc7Y/893585089a.png

I think its this but I'm not sure, and the book doesn't have answers for all problems

$\neg P(1)\lor\neg P(2)\lor\neg P(3)\lor\neg P(4)$

[vital dew]

nix:

[pliant horizon]

these are the answers for some of them https://puu.sh/Gocbi/bad8678a9f.png

[stray reef]

yes, your answer for 29 is correct

[pliant horizon]

okay cool thank you

[keen glade]

I hope Permutations and Combinations doubts are welcome here - I was suggested to move my queries to this channel yesterday.

I'm trying to solve this question. Part A specifically. Here's my approach:
6 distinct toys and 3 distinct boxes such that neither of them is empty. First I focus on filling each box with at least one toy. This can be done in 6C1x5C1x4C1 ways. Three toys remain and they can each go in any one of the three boxes in 3x3x3 ways.

Total number of ways: 6x5x4x3x3x3 ways. The answer given is 540. Could someone please help me realize my mistake here?

[honest barn]

why is the end 3x3x3?

[keen glade]

Because I have 3 toys remaining and since each box is filled with one toy, I have no criteria for putting in these 3 toys - they can go in any one of the three boxes independently

[honest barn]

okay so

you've somehow overcounted by 6

I think this has to do with the initial bit the 6 x 5 x 4

So I think this is what happend

[keen glade]

Okay, please go on

[honest barn]

So what's the logic that it's 6 x 5 x 4?

I imagine for the thing filling the first box

you have 6 options

[keen glade]

Yes

[honest barn]

likewise for the second, likewise for the third

[keen glade]

Yes exactly

[honest barn]

Oof, okay so my original thought isn't right but

I think there's some issue in that like

the way you put in the latter 3 can sort of like, mess up that

Sorry I need to think a bit harder

so I can actually verbalize what I think is going on haha

[keen glade]

I notice that I have changed the way of looking at it - I'm not sure if this helps but I was trying to first see WHAT toys can each box carry. Once I reached 3 toys left, I'm going for WHICH toy can go in which box

[honest barn]

Yeah that's

sort of what I'm getting at

I think like for the first 3 you don't have to fill all the boxes

since in the last 3 you can fill one you missed

but somehow maybe you can quantify this as overcounting by 6

which somehow I think manifests as a 3!

since this is combinatorics lol

[keen glade]

Hahah but that's reverse engineering the solution xD

[honest barn]

I mean

I suppose

haha

So I have an alternative idea

Of how to calculate it just period

[keen glade]

Okay I'm listening

[honest barn]

But I think it's probably important to figure out what's going wrong

[keen glade]

Yes I really want to know what's wrong too

[honest barn]

But we can calculate this as total ways to put 6 items in the 3 boxes

• number of ways in which one box is empty

And I think number of ways in which one box is empty is easy to calculate

[keen glade]

Indeed

[honest barn]

I'm pretty sure it's 3 * number of ways in which one distinguished box is empty

[keen glade]

number of ways in which AT LEAST one box is empty, you mean

[honest barn]

Oh sure

But having two boxes empty

is reallllly easy to count

but nice catch

[keen glade]

Yes

[honest barn]

although I think my method of counting actually includes that

so if the first box is empty

that means all 6 toys are in the other 2

which is just stars and bars on n = 6 and k = 2

and that includes say, first is empty

and second is empty

since that includes counting all 6 toys in the last box

ugh, but then you actually overcount!

lmao

the idea here is, all 6 toys in teh third box

appears both in "ways in which first box is empty" as well as "ways in which second box is empty"

but the overlap is really easy to count

[keen glade]

Yes I see how your approach will indeed get me to the solution honestly - but the fact that I still haven't found the error in my approach is a bit triggering 😅

[honest barn]

Yes I agree

Okay so

so I think there is issue in the initial part

[keen glade]

Hang on, let me just write down SOME possibilities - maybe I'll figure out where the issue is

[honest barn]

You're essentially singling out single elements as like "the one which must go into the 1st, 2nd, 3rd box"

respectively

[keen glade]

Hang on, I think there's another error because I haven't really permuted the toys either?

[honest barn]

I think you don't have to

because they're distinct

I had that idea earlier too

This is actually how I thought you overcounted by 3! at first

[keen glade]

But being distinct is all the more reason to do so right?

[honest barn]

uhhh, oh

I thought you meant permuted as in

[keen glade]

If they were all identical I'd not need to permute them

[honest barn]

divide by number of permutations haha

since often when things are not distinct you count it as if they were

then divide by overcounting

I think it's not too important to worry about that since there's something else wrong

[keen glade]

yk what, let me draw possibilities. I always get stuck in these kinds of questions, dude. It's so annoying

Get back to you within ten?

[honest barn]

sure

[keen glade]

Okay

[honest barn]

Hey, I figured it out

I can't actually ping you haha

but okay

[keen glade]

Woah what's the issue?

[honest barn]

so first I'll give the solution

[keen glade]

Okay

[honest barn]

so first just pretend that the things aren't distinct

then you know the first part where you fill each box?

[keen glade]

Yes

[honest barn]

this is where 6 x 5 x 4 came from

this actually means nothing

since you can just fill each box with one such element such that they go in that box

if that makes sense?

[keen glade]

Yes yes

[honest barn]

then for the next part

we pretend they're distinct again

then you get 3 x 3 x 3

just as you said

[keen glade]

Okay?

[honest barn]

now let's go and examine

the initial step of putting those 3 into the 3 boxes

this isn't 6 x 5 x 4

it's 6 choose 3

[keen glade]

Yes

[honest barn]

and if you do 6 C 3 * 3^3 you get 540

remember that 6 x 5 x 4 = 6!/3!

while 6 C 3 = 6!/3!3!

So basically I think when you did those 3, you

[keen glade]

No but my issue is - it should be 6P3 not 6C3

As in, you select 3 toys

[honest barn]

But

[keen glade]

Uhh so see

[honest barn]

it doesn't matter

which box the one goes in I think

[keen glade]

No they'll be different cases right?

[honest barn]

now I have managed to somewhat confuse myself again

fdsajkl;

[keen glade]

Hahahah don't worry

PnC is a tough nut

[honest barn]

but

this method does give the right answer

so why

lol

[keen glade]

I was thinking along those same lines rn about 6C3. But you WILL have to multiply it with 3! for permutations...

[honest barn]

but then somehow the 3 x 3 x 3

like accounts for this

somewhere buried in there is accounting for the permuations I think

blech

[keen glade]

Yes, wait hang on I think you're right

T1 - T2 - T3
T4 - T5 - T6

is the same as
T1 - T5 - T3
T4 - T2 - T6

[honest barn]

umm

is that saying like

T1 and T4 are in the first box?

[keen glade]

Yes I'm sorry

[honest barn]

yeah

[keen glade]

It does mean that

[honest barn]

that's kind of what I'm getting at

so like

[keen glade]

And you're absolutely right. The 3x3x3 accounts for the internal permutations

[honest barn]

if you interpret the first line as like the ones which had to go into the thing

[keen glade]

Yes exactly

[honest barn]

it's really really nebulous how though

I think has to do with the shift of "have to" versus "can"

Like if you go by the can for the last 3

to get 3 x 3 x 3

then there's no reason the first 3 have to fill out the boxes

[keen glade]

No no, see you have various ways to select 3 right? So even if you don't have T1-T2-T3 being the same as T2-T1-T3, this will be taken care of when you add in the last toys because that could be taken care of by T4-T5-T3

[honest barn]

oh

right

can you quantify how though?

[keen glade]

I was filling out the first 3 boxes just to ensure none of them remains empty - but then like you said, I can always take all possibilities and then exclude some

Wait let me think

[honest barn]

So I think in the first step you consider

T1-T2-T3 as separate from T4-T5-T6

but both can later become equal

by doing the other one

blech

oh wait but I think I see it!

so the only way two initial things can become equal

okay so umm

[keen glade]

How about 6P3x3x3x3/3!?
The way I see this, you choose three toys and you permute them in all the ways you can to fill in the boxes. Then you fill in the remaining three toys as we decided.

But this includes the cases I mentioned - two other toys being a part of the permutation in the 3x3x3 step accounting for the extra cases, so you divide by the number of ways in which the three initial boxes are permutable - 3! ways

It is essentially 6C3 though so...

I reverse engineered I'm sorry 😅

[honest barn]

Sorry so actually

I don't think

So even if you don't have T1-T2-T3 being the same as T2-T1-T3, this will be taken care of when you add in the last toys because that could be taken care of by T4-T5-T3

This isn't right since

you can't fix those two

like if you chose T1-T2-T3 at the first step versus

T2-T1-T3

no matter what you do

T1 will always be in the first box in the first one, but in the second box in the second case

The only way two of the initial ones can become equal is if you're looking at two disjoint cases

like T1-T2-T3, and T4-T5-T6

and there's exactly 3! like, disjoint ones for any specific initial choice

if that makes sense

[keen glade]

No - what I mean it T1-T2-T3 is one case. Now we have an issue where T2-T1-T3 is another case - assuming T4 - T5- T6 respectively as the next additions

But T2-T1-T3 can be taken care of when we're selecting T4-T5-T3 for example, and in the last step when we're left with T1, T2 and T6 I fill them accordingly

[honest barn]

ugh but this still gets complicated even later on I think

oh I see

[keen glade]

What I mean is, what possibilities we exclude when choosing 6C3 in place of 6P3, we account for all of those in the 3x3x3 step

[honest barn]

oh right

[keen glade]

It's just what you said, it's the internal permutations in the last step that account for those

[honest barn]

because we're busy fuck off

[keen glade]

@weary tiger I'm sorry, please give us 15 minutes?

[honest barn]

No one has to help you

and we're in the middle of discussing something and had been before you came in here

[keen glade]

Chill chill, I think we're almost there?

[honest barn]

I can answer ur question in like 10 seconds tho

if 3 pencils = 96

divide by 3

1 pencil = 32

multiply by 5

[keen glade]

@weary tiger

[honest barn]

I think?

[keen glade]

Yes

[honest barn]

I hope so

at least

[keen glade]

Okay anyway

[honest barn]

Right, so

before we were considering

[keen glade]

Good catch man, that 3x3x3 was a wonderful observation

[honest barn]

T1-T2-T3 and T1-T3-T2 as separate

[keen glade]

Yes

What I mean is, what possibilities we exclude when choosing 6C3 in place of 6P3, we account for all of those in the 3x3x3 step

This

[honest barn]

right

hrm

[keen glade]

I think it satisfies me - what about you?

[honest barn]

I guess I'd ideally want to

[keen glade]

Do you think we're missing out on sth?

[honest barn]

come up with like an... in-words explanation

like the original false one was easy to describe

first we have to fill each box, for the first we have 6 options, then 5, then 4

from there each one has 3 boxes to go into, completely free so we get 3x3x3

[keen glade]

Yes

[honest barn]

I struggle to explain that when we do

6C3 instead of 6P3

[keen glade]

Yes I feel you - it's not as aesthetic, or easily descriptive I guess?

[honest barn]

Yeah, I'm trying to think of this as like

hmm

gah

[keen glade]

Hahaha

[honest barn]

I would be satisfied if like

I took a generic "solution"

like some choices of boxes for each toy

and then can show how given this algorithm so to speak

how we can produce it

but I can't like... pick a generic one

without WLOGing it or something

[keen glade]

Yeah you want a mathematical explanation instead of this verbose one I guess?...

[honest barn]

Idk haha

Gah

Well, if you're satsified with it haha

I would probably go by means of counting the number of ways to fail to fill one box with a toy

[keen glade]

No no, you should be too. We can argue on this further

Yes I can see how your way of solving it is super intuitive

[honest barn]

I tend to like

For combinatorics

[keen glade]

48 - 6h
8 - 1h
32 - 4h
@weary tiger

[honest barn]

I'm not incredibly concerned with understanding every method of counting

[keen glade]

Hmm

[honest barn]

as long as I can find one that works

haha

I guess this just comes from me not wanting to do combinatorics so I'm satisfied with just being able to solve a problem somehow

Wait a second

[keen glade]

I guess, my way to go usually is to use any stupid method but if it is wrong I just want to know WHY that's stupid method is wrong

[honest barn]

Hold up

[keen glade]

Okay?

[honest barn]

I swear to god I came up with a way to count the number of surjective functions

from a set of size n to a set of size k

This is exactly the same

[keen glade]

I have no idea what surjective functions are 😅

[honest barn]

umm

[keen glade]

One-one onto?

[honest barn]

onto

onto functions

= surjective

[keen glade]

Okay yes I remember those

[honest barn]

So like

a choice of which boxes the toys go into

gives you a function

the condition that each box is filled just says they have to be onto

[keen glade]

Yes I see, it's exactly an onto problem!

[honest barn]

okay let me see, I typed up all the combinatorics I did for like

2 weeks lmao

[keen glade]

Man, lovely extension

[honest barn]

Maybe I did that?

I know for sure I did injective functions

or one-to-one

but I don't actually remember if I did onto functions

[keen glade]

So what's your background? Are you into research or..?

[honest barn]

I'm just an undergrad

want to do a PhD

[keen glade]

Ohh okay okay

Niice

So? Any solution which satisfies you yet?

[honest barn]

I am fine with counting the number of ways it can fail

but I looked it up]

and wow

so

[keen glade]

Yeah that's better indeed

[honest barn]

number of surjective functions is complictaed af

https://math.stackexchange.com/questions/264799/calculating-the-total-number-of-surjective-functions

[keen glade]

Oh God I'm gonna pass out seeing that man - not my cup of tea

[honest barn]

So I definitely didn't do that

[keen glade]

Last I studied math at this level it was 4 years ago hahah

[honest barn]

Junior in Uni

[keen glade]

I just graduated this year

Bachelors, I mean

Anyway, let's keep this open for everyone else then. Thank you for your time @honest barn - once again, brilliant catch there :P I'd have never guessed it!

No no, my discussion is over. Feel free to ask @weary tiger :) thanks for being super patient

Ciao

I'm a little busy brushing up my PnC man, just post it here I'm sure there are lots of people who can help!

You could PM me, but I'd answer after I'm done

[honest barn]

No

You want like ∃!

[weary tiger]

To me

[honest barn]

since ∃ means there exists

but nothing about uniqueness

[weary tiger]

how could we write it without !

[honest barn]

but also using your words is prefereable

There is one and only one real solution to the equation x² = 0?"

Is the best way

if you mean in a symbolic form

[weary tiger]

I am trying to learn notation.

Yes ,symbolic form.

[honest barn]

Most of the time we try to avoid logical symbols

if you read math they'll almost always use words

but if you do want to learn the symbolic one, I don't know

Sorry oof

Try like

[weary tiger]

I can't read symbolic lmao.

[honest barn]

#proofs-and-logic

It's more suited

for ther

That says

for all x in R, x is in C

the ∀ means for all

∈ means that an element is in the latter set

OK

I mean

The number 1

is the number 1 + 0i

There's a natural copy of R inside of C

No

for any real number x

x is x + 0i

R sits inside of C

C is the plane

the xy-plane with this funny multiplication

R is the x-axis

and turns out the funny multiplication of C when restricted to the x-axis

operates the same way as it does in R

[weary tiger]

so would you say "for each x" or "for all x"?

[honest barn]

for all x

yeah

yup

Yup

or like

the difference of two perfect squares

if you wanna say it that way

yeah

[weary tiger]

perfect square comes from where?

[honest barn]

Perfect square = square of an integer

by definition

it's just a term

[weary tiger]

thank you i didnt know.

[keen glade]

You might wanna lower the trolling on a serious discord server my dude

[weary tiger]

@weary tiger I think you could do something like this:
$\exists x\in\bR$ such that $x^2=0$, and $\forall x,y\in \bR, x^2=y^2=0 \implies x=y$

[vital dew]

Botnuke:

[weary tiger]

feels filthy typing this

maybe you could replace the and with ⋀ and such that with : or something if you want to be extra filthy and purely symbolic

[vital wyvern]

hello everyone

can someone help me do letter c

not sure how to do with the multi variables

[rain stone]

@vital wyvern if you're struggling on a question like this, I would suggest starting by just writing out some elements

write out a list of examples of elements in A, and elements in B

[vital wyvern]

ok but im confused would A = {(1, 3/2), etc, etc, }?

is that how this is formatted?

and would I have to list every possibility between 1 and 2 but not inclusive of 1 and 2

because isnt that an infinite amount of possibilities?

[rain stone]

yep, there are an infinite amount @vital wyvern

but

you can describe them with the interval notation

examples just help you get a feel

and yes, you're exactly right; A is ordered pairs of real numbers, where the first component is between 0 and 2, and the y-component is between 1 and 2

[wild flame]

The booklet does not appear to mention what to do in a set-notion when referring to multiple variables.
What does 'ab' mean?
https://i.imgur.com/gd1sYRm.png

I'm not used to set-notation builders having two variables

[weary tiger]

'ab' means a times b I think

in more plain english it says: all products you could make with 1 element from A and 1 element from B

you can find these exhaustively by going through
1⋅2, 1⋅3, 1⋅4, and
2⋅2, 2⋅3, 2⋅4

[wild flame]

Ah. So I would have to first find the element(s) that is in both Set A AND Set B

Then multiple them, correct?

[rain stone]

@wild flame no. You just need a to be in A, and b to be in B. You don't necessarily need a to be in B, or b to be in A

[wild flame]

Ah.

so the Set AB = {2, 3, 4, 4, 6, 8}

Since there are repeating elements. It becomes:

AB = {2, 3, 4, 6, 8}

Right? @rain stone

[vital wyvern]

@rain stone Ohh wait so I think I sort of get it thanks for talking it out with me! ^^

[weary tiger]

I think we did a problem like that in class but not really sure

Yes, but it has to come from "Mathisfun"

Yes

Yeah, I think I got it. I'll try it out thanks

[brave lava]

https://i.imgur.com/jv93yaG.png

im not sure my last one is correct

for the third fill in the blank

[stray reef]

it is correct

[brave lava]

thanks Ann

[brave lava]

https://i.imgur.com/eqWirQW.png

i think top 2 are true and e

idk about d an c

[unreal stump]

Isn't that personal?

[brave lava]

what

[unreal stump]

Preference of an "L shaped approach" vs a side-side derivation

[brave lava]

he talked about it in the lecture and he said L shaped approaches are better

[unreal stump]

Plus,You can make any proof wordy

[stray reef]

what the fuck is an "L-shaped approach"?

[brave lava]

https://i.imgur.com/hvGE9lv.png

on the right side

is the L approach

[unreal stump]

What?

[brave lava]

the left side is the traditional approach i used from precalc and other classes

the right side is the "L" approach

[unreal stump]

Pretty sure it's completely personal

[brave lava]

https://i.imgur.com/ovBZSW1.png

not according to him lol

https://i.imgur.com/eqWirQW.png

any help with this

[unreal stump]

That's impossible

Sorry

[brave lava]

what

[unreal stump]

There are semantic problems like "what counts as a definition"(If I define a particular function,Is that a definition,by your sense?) ,"how are fundamental properties different from definitions"?"What is a L shaped approach"?

Don't think anyone can help you with those

[brave lava]

which once do you think are not right

[unreal stump]

AE ,will not bother with B,C or D

C,D depends on the definition of a definition

[brave lava]

damn

[unreal stump]

B is personal

[brave lava]

why is A not true

i thought if we prove something

then we can use that proof in future proofs

[unreal stump]

Sorry AE are true

[brave lava]

lmao

i got it right anyway

4/4

LOL

that was aids

https://i.imgur.com/Pf2jPjN.png

i think im just going to read my book and watch the trevor guy on youtube for discrete math

[unreal stump]

Try zach star. He might be an engineer, but he does good math videos

[brave lava]

my teacher is the type where you dont learn anything after 30 mins lol

he rambles a lot too and never sticks to a point

do you have his playlist

for zach star

and discrete math

https://i.imgur.com/dqUQ00Y.png

for this one

i think its false b/c you cant assume one side is true

you gotta show both sides are equal

thats the whole point of a proof

i got it right

[burnt herald]

can someone explain why the answer is B?

does this qn belong here or proofs-andlogic?

[honest barn]

Proofs and logic I think

[burnt herald]

Proofs and logic I think
@honest barn Thanks

[weary tiger]

Does this look right

[weary tiger]

And what about this

[open narwhal]

I'm having some issues determing if a function is 1-1 and onto. Would anyone be willing to help me out in a questions channel or voice channel?

[pallid lintel]

whats the function

[weary tiger]

@open narwhal

and a bit larger for the people in the back

[gritty crescent]

Wouldn't this proof work if I just chose k!+1, k!+2, ..., k!+k instead?

[stray reef]

what do you think k!+1 could factor into

[gritty crescent]

Ummm right

Should've thought harder; thanks for pointing out!

Is k!+1 a prime for any k>3?

[unreal stump]

11

[gritty crescent]

Ah, I see. Thanks!

Is there a way of determining this?

Or you just knew 11 works?

[unreal stump]

It is an unsolved problem

[gritty crescent]

Oh, I see.

[misty merlin]

Hello anyone available?

I need help with an assignment question

[stray reef]

[misty merlin]

Don’t give me an answer but any guidance is appreciated

I don’t know where to start. I understand what adders are. How do I approach this question?

[stray reef]

hint: 3x = 2x + x

2x can be calculated with no adders or half-adders

[misty merlin]

If I adjust the question so change 3x to 2x and 0 =< x < 8 to 0 =< x < 4

Can u build a circuit for that?

Idk how my circuit to answer the actual question should be

How do u build a circuit for 2x?

Do you just draw x1 and x2 lines across and to output f(2x)

At the beginning I thought of doing x + x + x idk if I was on the right track

[stray reef]

2x is just x bitshifted left by one

[south marten]

Can some check if my proof of $(i)$ in the pigeonhole principle is correct: http://mathb.in/44843

[vital dew]

Zophike1:

[weary tiger]

Huh?

[south marten]

@weary tiger that is the negation of Injective yes ? I'm prettty sure since I looked up the defintion online

[weary tiger]

First, what is the word surjective doing there?

[south marten]

oh my bad sorry typo

[unreal stump]

Typo

[south marten]

I'm sorry :c http://mathb.in/44852

[weary tiger]

Am I losing it or have you written a definition for injective, rather than not injective?

[south marten]

Yeah I think I put down the wrong defintion

[weary tiger]

It should be something like “there exists unique a and a’ with f(a)=f(a’)

[south marten]

@weary tiger I sent an image of the correct defintion

[weary tiger]

Yes that one looks better

[south marten]

Yeah that's what I applied to say $f$ is not injective i'll have to change the link

[vital dew]

Zophike1:

[south marten]

@weary tiger aside from that is the proof correct ?

[weary tiger]

I couldn’t follow it so probably not

But maybe I am just not understanding what you’re trying to do

[misty merlin]

@stray reef Thanks btw

[south marten]

What i did was a set up the set $X$ with different number of elements than $Y$ then used the definition of not one-to-one to show that $f$ is not one-to-one

[vital dew]

Zophike1:

[weary tiger]

I get that, but you never mentioned pigeonhole principle or anything like that

So I’m not sure what the reasoning was

For concluding that f is not one to one

[south marten]

To conclude that $f$ is not one to one it comes down to seeing that

[vital dew]

Zophike1:

[south marten]

$f(x_{1}) = f(x_{n+1)}) \implies x_{1} \ne x_{n+1} , \forall n \geq 1$

[vital dew]

Zophike1:

[weary tiger]

Actually I think I may have realized what you did now

And it’s not right if that is the case

Lemme reread

Ok so...

This is a faulty assumption

[south marten]

Oh how to do I fix it 😦

[weary tiger]

f(x1) must equal f(k) for SOME k in X, it does not need to equal f(k) for ever single other k in X

Err wait

Btw

f(x1) needs to be general

It’s not like every element needs to break the injective property

Just 1

Well, 2 really

[south marten]

oh that's fair

[weary tiger]

As they would come in a pair

[south marten]

My assumption was faulty since I said that f(x1) = f(k) for all x but that's not the case

[weary tiger]

All k, but yes exactly

[unreal stump]

What are you trying to show, exactly?

[south marten]

$f(x_{1}) = f(x_{n+1)}) \implies x_{1} \ne x_{n+1} , \forall n \geq 1f(x_{1}) = f(x_{n+1)}) \implies x_{1} \ne x_{n+1} , \forall n \geq 1 $

[vital dew]

Zophike1:

[south marten]

^ This was what I was trying to get at

[weary tiger]

Use pigeonhole principle to show that f from X to Y with |X|>|Y| cant be one to one, drake

[unreal stump]

You could simply write "there doesn't exist n"

Such that the condition is satisfied

[south marten]

Oh makes a sense damn I can't belive my proof was destroyed 😦

[weary tiger]

Happens to everyone

[south marten]

@unreal stump I could write there dosen't exist an $n$ such that $f(x_{1}) = f(x_{n+1)}) \implies x_{1} \ne x_{n+1} , \forall n \geq 1 $

[vital dew]

Zophike1:

[south marten]

@weary tiger i'm trying to prove PHP you can't use what you want to prove

[weary tiger]

“For all” is not exactly your friend here

I was responding to a question from drake

[unreal stump]

$\nexists$

[south marten]

ahhh okay okay

[unreal stump]

Are you trying to restate injectivity condition?

[weary tiger]

Are one line proofs with pigeonhole principle accepted? Or do you have to go through function constructions first?

[south marten]

http://mathb.in/44854

[weary tiger]

I gtg for a bit, but if you think about, this question is pigeonhole principle stated pretty much exactly, but in the language of functions

[south marten]

@unreal stump I think I got the proof for $(i)$

[vital dew]

Zophike1:

[unreal stump]

Not exactly,how do you know f(x1)=f(x(n+1))?

[weary tiger]

And you were asked to use php

[unreal stump]

It could be f(x2)=f(x(n+1)

[south marten]

Oof true

[unreal stump]

For some i,f(xi)=f(x(n+1)

[south marten]

ahhh that makes sense

@unreal stump http://mathb.in/44855

[unreal stump]

Or better,Let f be injective,then f should have a range of (n+1) elements,but range has only n elements

So can't be . Therefore f is non injective

(because f(x1),f(x2)... Should all be distinct)

[south marten]

That makes a lot more sense I see where my error was I made the faulty assumption that f(xn)=f(x(n+1)) for all given $n$ which is not the case

[vital dew]

Zophike1:

[south marten]

@unreal stump is the proof okay now

I think i may have put the wrong definition for injective again

[unreal stump]

Could be that f(x2)=f(x3)

And f(xn) not equal to anything else in the set

[south marten]

yeah hense why you said for some $i$ that f(xn) = (x(n+1))

[vital dew]

Zophike1:

[unreal stump]

It should be $f(x_i)=f(x_j) \exists i,j$

[vital dew]

DrunkenDrake:

[unreal stump]

Like not necessarily n+1,and i

[south marten]

but would that be fine ?

[unreal stump]

$f:{1,2,3,4} \mapsto {1,2,3}$
$f(1)=1,f(2)=2,f(3)=2,f(4)=4$

[south marten]

so i'm pressuming yes to say for some given $i$ $f(x_n) = x_n+1$ because you did say that eariler I belive @unreal stump

[vital dew]

Zophike1:

[unreal stump]

Well,it could be any 2 elements

But those two elements will map to same thing

[south marten]

yeah I figured thanks for helping me you have any advice with counting arguments i'm pretty weak at them 😦

[unreal stump]

Contradictions probably

[south marten]

it seems easier to say something is "not" in the context of a counting argument then it is to try to prove it directly

also how close was my orginal proof to a correct solution ?

[unreal stump]

I don't know. I can only say whether a proof is right or wrong

[south marten]

it's fair

[rain ridge]

concerning the field of discrete mathamatics, could anyone provide an example of an partial order relation that has a minimal element in regards to the order relation that is different to the least element of the set the relation acts on.

[honest barn]

What

To have a least element implies you already have some sort of order

The only way you could make this work is if you’re considering some “natural” order in regards to

least element of the set the relation acts on
And then a different order, but this is really pointless since you’re considering two orders

Take like N ordered by divisibility and then 1 is a least element with respect to this, and then if I just arbitrarily consider the partial order < which states
x < y if and only if x = y then every element is minimal, so take 2, 2 is minimal with respect to < but not the “least” element

But again, least really has no meaning when you speak about < because it’s least with respect to a different order

[stray reef]

x < x if and only if x = x

[honest barn]

In fact, for any set, and any element of the set you can make a partial order which has that as the least element.

What’s wrong with that Ann?

It’s a partial order, no?

[stray reef]

i mean

x < x???

what

[honest barn]

I defined an order

Usually you define an order using <

Sometimes when it’s a partial order you use <=, but it’s perfectly acceptable to use <