discrete-math

[azure lichen]

(I assume it probably won’t be, actually)

[unborn mango]

Hey guys, i'm doing sets in my discrete math class. Just started it and it's been a while for me. Does it look to you guys like the problem is wrong?

[opal crescent]

one of the option choices is the answer

doesn't look wrong to me

[unborn mango]

Okay, then i'm wrong, heh

I was looking for the intersect, duh

Thank you

[opal crescent]

🤷

[slow socket]

im working with permutations. just wondering
is this the same

nvm i got it

[ashen magnet]

hi guys
I have -(-inf,0) ∩ Z and i have to list the set
so i did
-(-inf,0) ∩ Z = (0, inf) ∩ Z = Absolute value of Z

Is that wrong? I was told it was wrong without explination

or, would that just be N, for all natural numbers?

i'm a bit confused

[flat tulip]

It would be the natural numbers excluding 0 yes

[sour arrow]

@ashen magnet
What is -(-∞, 0)? That's not a set I know

[ashen magnet]

PJS oooh, i forgot to exclude 0.

Kaynex that is an interval. for example, the set (0,inf) would be all N or all numbers from0 to infinite

[flat tulip]

All real positive numbers

Not necessarily just natural

Natural numbers are a subset tho

[ashen magnet]

oh ok, that makes sense

but going back to my original answer, would the answer (0, Z) be the right notation? or should i say |Z| and 0 ∉ Z

i get the concepts but notation always gets me

or perhaps, (0, N)?

[flat tulip]

Natural numbers excluding 0?

[ashen magnet]

right, but in terms of notation would it be an interval (0, N) or N and 0 ∉ N

[flat tulip]

(0, N) wouldnt make sense, just say n€N, n=/=0

Ive also seen N* being the set without 0 as well

[ashen magnet]

yeah i think it stems from an older definition of the natural numbers but i think the more commonly accepted includes 0. i'd have to check on that.

[ashen magnet]

suppose -((−∞, −5] ∪ (5, ∞) ) ∩ Z
demorgan's (5, inf) ∩ [-5, -inf) ∩ Z

What does one do if the intersection of two intervals is empty??
Would it just be {empty set} ∩ Z = {empty set}
Even if i use associative and ((5, inf) ∩ Z) ∩ [-5, -inf) it still doesn't help :/
Is there a law i am forgetting to apply?

[cold meteor]

if (A U A) is a universal set, what is (A n A)?

[hasty cargo]

A and A are the same set right?

[cold meteor]

yeah

that would just be false right?

[stray reef]

$A \cup A = A$

[vital dew]

Ann:

[stray reef]

$A \cap A = A$

[vital dew]

Ann:

[hasty cargo]

That’s what I was thinking haha

Because the union of two sets includes both sets, however the union of any two set A and A would just be A as the elements in the both sets are the same

The same reasoning for the intersection of both sets, A and A if drawn out on a venn diagram would completely overlap itself and given that A is the universal set, AuA and AnA must also be universal sets

Sometimes I’d just like to call it trivial to avoid the fact I’m horrible at proving it

[acoustic valve]

can i get some help with this? im so confused

I know a) is true because

if R is reflexive, it maps everything to itself, and so R of R would also do that but twice, making it reflexive as well.

but i can't figure out b. (I assume c is the same reasoning as a0

[viral stump]

try an example for (b)

try the easiest antisymmetric relation you can find

see what happens

[stray reef]

unfold the definitions of "R is antisym" and "R . R is antisym"

[viral stump]

extrapolate from there

[acoustic valve]

i think i got it

so

assuming R . R is not antisym

it'd have a relation aRb and bRa where b != a

but both of those relations have to exist in R

which is a contradiction bc R can't have both of those relations at once due to the definiton of R is antisym, right?

:]

[stray reef]

you mean a (R . R) b and b (R . R) a

[acoustic valve]

ye

[stray reef]

it's easier to prove this directly tho

[acoustic valve]

but does it work though 👀

[stray reef]

the argument is kinda shaky. i wouldn't accept that if i were grading homework.

[acoustic valve]

oof

why is it shaky?

[stray reef]

it'd have a relation a(R.R)b and b(R.R)a where b != a
but both of those relations have to exist in R
how does this follow? how does it follow that one must have aRb and bRa?

[acoustic valve]

oh

i confused myself again oof

how would you prove it directly?

[stray reef]

a direct proof would begin like this:
assume R is antisym and suppose a(R.R)b and b(R.R)a for some a, b
from a(R.R)b, deduce the existence of c such that aRc and cRb
...

[acoustic valve]

oooh

then you would do the same for some d b(r.r)d so that brd and dra,

but because r is antisym c must be equal to d and a must equal b

right?

[stray reef]

1. don't mix cases. r and R may very well refer to different things

2. no, it does not follow that c=d just because R is antisym.

[acoustic valve]

oof

[blissful basalt]

Hey guys, having a bit of issues with this question. Wondering if I could have some guidance

(Part A)

It wants to "identify which laws you would use"

I would say:

(A - B) is complement law
A U B is commutative

[slow socket]

@blissful basalt here this might help u

[blissful basalt]

@slow socket Yeah thanks, I think Im on the right track

by saying:

(A - B) is complement law A U B is commutative

[azure lichen]

what is “A - B is complement law” even supposed to mean?

a “law” like this should have another part to the equation

[heady sedge]

can someone help me with this

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

[weary tiger]

i think there's a rule against this kind of posting, but at the same time i think this is more of a probabiltiy question?

even if, i thiiiiiiiink the answer is 4×3×2/(52×51×50) all times 3 choose 52

@heady sedge

[obsidian heath]

In set theory, a null set (denotes by ø) is in every set? As in: ø is an element of {1,2,3} is true?

I guess it depends on the axioms?

I'm just going to say No, it is not an element of it.

{} is not an element of {1,2,3}

[hasty cargo]

I mean I don’t think so tbh

If a set contain a null set

(1,2,3,(),4,6)

[stray reef]

indeed {} is not an element of {1,2,3}

{} is, however, a subset of {1,2,3}, and in fact it is a subset of every set

[obsidian heath]

Ah! That's the one!

It's a subset of, not an element of

[versed needle]

can some1 explain this to me pls DM me https://gyazo.com/b0be32847d9c6e2a4bfd1262587c9fd0

how does 5 sub 8 get to be 101

[stray reef]

ok

[unborn jetty]

how is f: r->z f(x)=[x/2] not injective?

this was the question on the assignment and the answer. it seems like its one to one...

[weary tiger]

what is the result when x = 1? x = 2?

[unborn jetty]

1/2, 1

do the brackets do something I'm forgetting?

[weary tiger]

yes

ceiling function

[unborn jetty]

ahhhh

somehow missed that in calc......

[weary tiger]

lol

[unborn jetty]

oh no we went over it in the slides... wow thats a total failure on my part

[weary tiger]

[weary tiger]

The left side of this statement

would it be the green area?

[azure lichen]

well… in what you’ve marked in green there’s parts of C complement

so that can’t be it

oh wait you said left. the intersection of A, B and C is all points which are in all of them

[weary tiger]

So the left side would be

the smallest green part only?

[stray reef]

the small bit in the middle, yeah

[weary tiger]

ohokay gotcha

How about this then

if theres A B and a C

Would that be this then

Since I first subtract what B is sharing with A

then what is standing alone from A

then remove that from A

and am left with that bit

[azure lichen]

I don’t think you’re doing things in the right order

first, calculate the thing in the brackets, the B-A

then, take that away from A, which leaves A unchanged

so you’re left with A

also, why even put a C when there’s only an A and a B involved?

[weary tiger]

I am told there's an A B and a C

[azure lichen]

well, the C is irrelevant

[weary tiger]

Okay I am stuck again

So i have this

And I say that is true

Cause if I assign these values:

A = 1 2 3 4 5 6
B = 5 6 7 8 9 10
C = 9 10 11 12 13

According to my calculations I end up with

A - ( B - C) = 1 2 3 4
(A - B) - C = 1 2 3 4

But I am told it's false?

[azure lichen]

an example does not prove a theorem

for example:
Theorem Adding a number to itself results in the same as multiplying it by itself
Proof 2+2 = 2*2 = 4

@weary tiger

[weary tiger]

Then how would I go about proving that statement?

[azure lichen]

well if it’s false you can’t prove it

a counterexample does disprove a statement

so you can go find one

[cold fjord]

Is a geometric proof valid for showing two sets are equivalent?

As in just representing everything as a venn diagram lol

[azure lichen]

if it’s a rigorous proof and not juts an appeal to common sense

that is, you have to make sure it covers all generality

like: does it also show the case when one set is empty? or included in the other?

a venn-diagram makes some assumptions about the nature of the sets

[cold fjord]

If a graphical depiction physically shows all possible sets then I would say it's accurate

[azure lichen]

yes, I would agree

the main issue is that it’s hard to tell if every possibility has been accounted for

[cold fjord]

For my particular example I'm only working with two sets so it's easier

[azure lichen]

there’s still a priori a lot of possibilities, like, A could be a subset of B. or vice versa. or one of them could be empty. or both. or they could have empty intersection

etc

there’s so many options and you have to make sure the proof is valid for all of them

for some statements that’s no issue

[weary tiger]

@azure lichen let me rephrase myself, how would I go about finding out whether it is true or false

[cold fjord]

I would first do it graphically, Wauner

[azure lichen]

try to find a counterexample. if you can find one, it’s false. if you can’t, try to prove it’s true

or that

[cold fjord]

I mean just for me it's easier to visually see rather than picking and choosing elements to put in each set

[azure lichen]

also, since I recall you having problems doing this stuff correctly, make sure you do the graphical arguments slowly and carefully

with intermediate steps if needed

[weary tiger]

Ah I understand, thank you for your help

[cold fjord]

Then you can get an idea on how to make a proof/counterexample based on what you see

I was considering making a graphical representation to show that $(A-B)\cup(B-A)=(A\cup B)-(A\cap B)$ because I think it's true

[vital dew]

Flip:

[cold fjord]

But otherwise I'm not sure how to prove that

[weary tiger]

@cold fjord

I have tried doing it graphically

I just get kinda confused by it

I may just be looking at it wrong though

[cold fjord]

If you're just toying around with it and trying to determine for yourself whether it's true or not, draw a venn diagram with three circles, one for each set you're working with

And draw it twice, one for each expression

[weary tiger]

Well I usually just draw itl ike this

for A, B, C

[cold fjord]

Yeah exactly

While it may or may not be formal (again this could just give you ideas), you can try shading in the region that represents each rightmost set and then draw the leftmost set around it, if that makes sense lol

[weary tiger]

Not sure I understand

[cold fjord]

So for example (A - B) - C, you would maybe draw an outline around the set C and then shade in the region (A - B) without crossing over into C

Then for A - (B - C) you would draw an outline around (B - C) and then shade in A around that outline

Or just shade in each region independently and supershade the overlaps lol

[weary tiger]

Oh I see

Sorry english is not my native language so I just get confused sometimes

[cold fjord]

It is ok

[weary tiger]

So if I were to draw (A - B) - C

I would first find A - B

which is

everything white

right?

[cold fjord]

Yessir

[weary tiger]

Alright and then I would remove C so I would end up with

[cold fjord]

Ye

[weary tiger]

Ah I get it then

And then A - (B - C) would be

never mind Im confused lol

B-C would be

this?

[cold fjord]

B-C is everything in B that isn't shared with C

[weary tiger]

So this

[cold fjord]

You're right to do what's in the parenthesis first though

[cold fjord]

Ye

[weary tiger]

And then I would remove that from A

so I would get

A - (B-C)

[cold fjord]

A - (B-C) will only have elements in A that aren't in the set of B-C

[weary tiger]

oh right c wouldnt be there either

[cold fjord]

Mhm

[weary tiger]

there

[cold fjord]

That's right yeah

[weary tiger]

Ah okay

I was completely misunderstanding it then

Thanks for your help man

[cold fjord]

No problem, and if you're being asked to prove or disprove, since they're visually different, you know to go for disproving

[weary tiger]

Luckily just being asked if t he statement is true or not

So visually I can see it is false

[cold fjord]

Alrighty lol cool

[cold fjord]

Can a subset of an infinite set (in both directions) be elements from some upper bound counting down tending towards negative infinity?

[neon pasture]

what

[cold fjord]

Like if $\mathbb{Z}={\ldots, -3, -2, -1, 0, 1, 2, 3, \ldots}$, and $A\subset\mathbb{Z}$, can $A={\ldots, -3, -2, -1, 0, 1, 2, 3, \ldots, a}$

[vital dew]

Flip:

[cold fjord]

I feel like yes?

99% sure yes I'm just double checking because thing

[cold fjord]

Yeah totally ignore that last question

What does it mean for an element to be unique?

[dense thicket]

in what sense?

element is unique when it is the only one that can satisfy a given proprty

[cold fjord]

That makes more sense, I was thinking that a unique element is just an element that doesn't have other duplicates within the set, but that's apparently not even a set if it contains the same element twice according to this book lol

[dense thicket]

well, set {a,a,a,a} = {a}

[cold fjord]

Also makes sense lol

[dense thicket]

yeah sets are cool

[weary tiger]

Can anyone explain to me why

this is transitive

but

this one is not

[slow socket]

so for the 1st pic

u start from the left and look at (x,y), then u look for something that is (y,z) and see if u have (x,z)

so u have (1,3) then u have (3,1) and u have (1,1) so thats correct

and u keep checking

[weary tiger]

My understanding is

its

aRb and bRc -> aRc

[slow socket]

ya

here look at the second picture

[weary tiger]

so aRb would be 1,2 in the second picture

why wouldnt bRc be (2,3)?

oooh right

[slow socket]

wtf my msg deleted

[weary tiger]

if aRb then there has to be a bRa too

[slow socket]

ya so

1,2 then 2,3

then u must have 1,3

[weary tiger]

but I do

thats the second one

[slow socket]

ya u check all of them

is not transitive cuz u have 1,2 and 2,1

but u dont have 1,1

also u have (1,3) and (3,1)
but u dont have (1,1)

(1,3) and (3,1)
(x,y) and (y,z)

then (1,1)
(x,z)

u dont have it

u see it>?

[weary tiger]

I get your point

but

what I dont understand is

so if I check the first

aRb

which would be 1,2

then why cant bRc be 2,3

[slow socket]

ok so aRb = (1,2)

bRc = (2,1)

u understand up to there?

[weary tiger]

no thats what I dont get

why cant bRc be (2,3)

[slow socket]

u have (1,2) u look at what other coord starts with 2

[weary tiger]

so I just take the first one?

cause (2,3) starts with 2 too

[opal crescent]

the thing is you could check it

[slow socket]

aRb = (1,2)
look at wat other coord starts with b
which is bRc = (2,1)

ya u can check that one too

[opal crescent]

but it's not interesting as a counterexample in our case

(1,2) and (2,3) : and (1,3) is in there

[weary tiger]

ah I see

[opal crescent]

that doesn't show the relation isn't transitive

[slow socket]

u check all of them, so it can also be
aRb = (1,2)
bRc = (2,3)
bRc = (1,3)

[weary tiger]

so since I know theyre there it wouldnt make sense to check

so I just check (1,2) (2,1)

[slow socket]

ya check all of them, if 1 of them doesnt work, then is not transitive

[weary tiger]

Okay I see

I could kiss you guys rn lmao

[opal crescent]

[fossil quiver]

you can kiss me

[weary tiger]

Got a test tomorrow and that was the last subject I was struggling with

so hopefully I will do well

[slow socket]

good luck

[fossil quiver]

have fun

[weary tiger]

@fossil quiver sorry my lips are reserved for @slow socket and @opal crescent

[fossil quiver]

:C

unlucky

[weary tiger]

50$ and I might consider a peck on the cheek

[fossil quiver]

that is too much for me

[weary tiger]

how about hmm

5$ and a pack of ramen?

[slow socket]

change to #chill
this is off-topic

[fossil quiver]

OK

[worthy thistle]

For the Summation(j=0 to 2n), (2j+1) = (2n+1)^2; I don't understand what it means

So when j = 0, (2(0) + 1) = 2(0*2) +1) ^2, so 1=1

But when j=1, 2(1+1) = (2(1*2)+1) ^2 so 4 = 25?

[echo lintel]

For every real number x, if x is irrational, then −x is also irrational.

prove w contrapositive

[opal crescent]

what's x + (-x) for any x ? @echo lintel

[echo lintel]

ah i figured it out

it had to do w fractions

[opal crescent]

myeah, x+(-x) = 0 would imply that 0 is irrational if you assume x irrational and -x rational

[weary tiger]

lol

[dusk plaza]

calculus is child's play

[modest zealot]

is the set of an empty set a subset of any set?

{(the 0 with the slash)} a subset of any set?

[modest zealot]

also, wtf does this mean

[glossy adder]

the set of integers n such that there exists an integer k such that n=12k+11

[modest zealot]

sooooo is the set A something like tshi

A = { --- , 11, 23, 34, --- }

to infinity and beyond?

[glossy adder]

ye

[modest zealot]

ahhh okay

yea i was confused

when it said there exists an integer k

i thought that A could be any set with 0 elements, 1 element, 2 elements to infinity

[weary tiger]

also yes the empty set is a subset of every set

[modest zealot]

ah alright

thanks thanks

wait but its the set of an empty set

{{}} if i understood it correctly is a subset of every set

[echo lintel]

https://i.gyazo.com/thumb/1200/d045f66f1dbb638fff5650d9d73036e2-png.jpg

so... i understand how to turn numbers into binary

where does the 66 = 2^6 + 2^1 come from

[weary tiger]

@modest zealot no, {} is

{{}} isn't empty

it contains ∅ as an element

[modest zealot]

oh

hmmmm alright

[twilit wadi]

@everyone can any one explain what is conditional and biconditional statements

[oak creek]

dude. there's 11-12k people in the server. don't try pinging everybody

conditional: if p then q; p implies q

biconditional: p if and only if q; if p then q and q then p

try writing it out on a truth table

[twilit wadi]

@oak creek thanks

@oak creek what is disjunctive normal form and conjunctive normal form

[oak creek]

disjunctive normal form is a form using disjunctions and conjunctive normal form uses conjunctions

google is your friend here

[twilit wadi]

@oak creek thanks

@oak creek feeling little bit difficult to understand these things

[oak creek]

okay so.

you know about truth tables, correct?

[sage island]

how do you feel about pings tho

[oak creek]

issa lot kek

[twilit wadi]

@oak creek I know truth table

[oak creek]

you don't have to ping me each message, just send each message please

okay so: can you pull up a truth table for 3 variables p, q, and r?

[twilit wadi]

yes

it contains 2 to the power of 3 rows

[oak creek]

okay so

look at this truth table

DNF is basically

a manner to get the "formula" of the end value

without knowing it

through the truths

so for the first truth value

it can be summarized as: p and q and r

correct?

[twilit wadi]

it is p or q and r I think

[oak creek]

that's. not the way you would structure DNF

basically: you would look at all the truth values

and strictly structure them using ands inbetween each

so the first truth would be p and q and r

second would be p and not-q and r

third thruth would be not p and q and r

then you would space each out with "or"

$(p \land q \land r) \lor (p \land \neg q \land r) \lor (\neg p \land q \land r)$

oof

[vital dew]

Colorodo Brown Stain:

[oak creek]

so that would be disjunctive normal form

basically, it's a disjunction of conjunctions

[twilit wadi]

yes it is disjunction of conjunctions

[oak creek]

do you understand DNF now

[twilit wadi]

I will go through the material and get back to you

thanks

[oak creek]

yeah

[obsidian rampart]

list the elements of the state space for flipping three fair coins at once. what is the state space?

im looking through my notes but i cant find it.

[sour arrow]

I'm guessing that's the event space, it's the set of all possible outcomes

[obsidian rampart]

like the percentages?

so .5 x .5 x .5?

[sour arrow]

Where's 0.5 come from?

[obsidian rampart]

50% chance

heads or tails

[sour arrow]

You have a 100% chance of landing heads or tails

[obsidian rampart]

what

the chances of a single coin are .5 so wouldnt you just multiply the probabilities?

[sour arrow]

I'm jk of course. No, the set of all possible outcomes is the list of the outcomes

[obsidian rampart]

ok

[sour arrow]

Heads, tails, tails is in the event space

[obsidian rampart]

so like this

[sour arrow]

Or state space? I've never heard it said like that

[obsidian rampart]

0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 1 ```

[sour arrow]

Yeah, that works!

[obsidian rampart]

0 being tails 1 being head

ok

that makes sense

[sour arrow]

Wow you typed that fast

[obsidian rampart]

it was either that

or the .5

professor mentioned it briefly in class, but didnt go any further than that.

this actually makes sense now that i think about it lol

[steady axle]

Does this server not cover Theory of Computation?

[pastel temple]

https://cdn.discordapp.com/attachments/538229935999811585/560103068746645504/unknown.png

for problem d) , what exactly is a | b ?

[hasty cargo]

A divides B

Meaning B/A is some internet

internet

integer

[pastel temple]

ok ty for your response but how come (1,0) is a possible pair?

[hasty cargo]

hmm what the heck

Wait

[pastel temple]

the answer in the textbook is
{(1,0),(1,1),(1,2),(1,3),(2,0),(2,2),(3,0),(3,3),(4,0)}

[hasty cargo]

0/1

is 0

Which is I dunno subjective but I don’t include 0

But I guess right

since it doesn’t give any remainder

[pastel temple]

but since its (a,b) E R, that means (1,0) is a=1 b=0

which is 1/0 in this case

[hasty cargo]

no but a divides b

So we have b/a

[pastel temple]

ah i see so its somewhat inverted

[hasty cargo]

Yeah

[pastel temple]

a | b, a is the denominator

[hasty cargo]

Yup

[pastel temple]

okay how come (2,1) isnt a pair?

[hasty cargo]

1/2

[pastel temple]

1/2 is fine no?

[hasty cargo]

Is not a internet

[pastel temple]

ouuuuu

[hasty cargo]

Integer

i actually can’t spell

[pastel temple]

1/3 is?

[hasty cargo]

remember

b/a

not a/b

[pastel temple]

oh its 3/1 technically

yes yes

ty bonappetit!

[hasty cargo]

Bon appetiteeeè

[stray reef]

@hasty cargo what do you mean you don't include 0

0 is an integer

[hasty cargo]

Really

I thought it wa subjective

[stray reef]

no

[hasty cargo]

Okay cool

Shit

[stray reef]

whether or not 0 is considered a NATURAL NUMBER is up to convention

but it is an integer and everyone agrees that it is

[hasty cargo]

OH YEAH THAT ONE

Yes yes

Shit my bad sorry

[stray reef]

0 is as much an integer as 4, -17, or 2209

[pastel temple]

Answer says Transitive and Reflexive but why isn't it also Symmetric?

[neon pasture]

it is

[pastel temple]

How would you go about solving a problem like this?

What I did was do R = {(a,a),(a,b),(b,b),(b,a)}

is that correct?

And from there check whether it complies with the conditions of a reflexive, transitive, symmetric and/or anti-symmetric relation

[stray reef]

it's not symmetric tho

just because everyone who visited a also visited b doesn't mean everyone who visited b also visited a

is that correct?

no

what you did is not just incorrect, it's nonsensical

[neon pasture]

oh shit I misread

my bad

[weary tiger]

a

[twilit wadi]

@oak creek conditional statement if then

is it similar to if else statement that we use in programming

[civic dagger]

This is a postcondition to a piece of code. If r = FALSE must the other side of the biconditional also be false or does it not matter since this statement only refers to r = TRUE?

[stray reef]

yes, the other side of the biconditional is false

[azure lichen]

think of it that way: if the right side was true, then the left must be too
hence if the left is false, right must be false too

[twilit wadi]

@civic dagger thanks

can you people suggest me any books or online material to read to learn about conditional statements

[slow socket]

im doing a probability problem

i have "a fair coin is tossed four times, consider the events"
A = {'exactly one of the first two tosses is heads"}
B={'exactly two of the four tosses are heads"}

i found P(B) = 3/8

but idk wat to do

is asking if A and B are independent

is the P(A) = 1/4?

[stray reef]

no, P(A) is not 1/4

[slow socket]

how can i do this

[steady axle]

You flip a coin 2 times, and you have two possibilities. Yes?

It is either heads or tails for each flip

What is the chance one of the first two flips lands on heads?

Does that wording help you better? @slow socket

[slow socket]

Wat i did was 1/2 * 1/2

Idk this is so confusing

[slow socket]

P(A) = 1/2?

[copper ore]

is this a good place to ask about showing that a function is big O?

[slow socket]

@copper ore yes

[copper ore]

how do i do this?

[neon pasture]

use the technique demonstrated in class

depends on how much calculus you know

[golden depot]

assuming this is the right spot for a set theory question,

let A be a set s.t |A| = n, n in omega, and s in A, how do I prove that |A{s}| = n - 1?

|A{s}|=n-1 *

|A \ {s}|

[stuck bough]

Anyone here know anything about Hamming Code? I am confused as to what they are and how to actually find the codes

[glad oriole]

@golden depot what’s omega?

[rare barn]

the set of natural numbers I suppose

you want to prove that (A \ {s}) and {s} are disjoint,

and that |{s}| = 1

and then use the additivity of the cardinality

(which you might also need to prove)

@golden depot

[ashen magnet]

I have a quick question when counting min/max cardinality of the intersection 2 sets

|A| = 10 |B|=15
My answer:
Largest value |A ∩ B|
Consider A ={1,2,3,4,5,6,7,8,9,10}
And B = {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}
at most 10 elements can be in the intersection
So Largest value |A ∩ B| = 10

Smallest value |A ∩ B|
If A and B are disjoint then |A ∩ B| = 0
Otherwise the smallest possible is 1.

Is my reasoning okay?

[sour arrow]

Yes

You don't need the line considering the possibility that A and B are not disjoint, since that's not the smallest value

[ashen magnet]

oh i see. thanks for the feedback

[void valve]

$\begin{matrix}\mathbb{R}_+: 1&2&3&4&5&...&n\S: 0.1&0.2&0.3&0.4&0.5&...&0.n\end{matrix}$

[vital dew]

kjsahd:

[void valve]

This is probably stupid but is this a valid bijective function to show that R_+ has the same cardinality as the interval 0<x<1 ?

[ashen magnet]

quick question, if i have 5 slots and 3 can be 3 letter "a"s the amount of words with 3 letter "a"s is 15*26^2, right?

[ashen magnet]

or rather, 15(25^2) because a is only allowed to appear 3 times in that 5 char string

[weary tiger]

why 15? I think it should be 10, because 5c3 = 10

@ashen magnet

[ashen magnet]

oooh you right

so 10 * 25^2

[weary tiger]

pretty much

[ashen magnet]

This is my last question tonight. I really have trouble when it comes to applying it to functions.

I have f: X -> Y
X{1,2,3,4} Y {a,b,c,d,e}
find the number of functions based on the restriction
a) no restriction: 5^4 = 625
b)injective: P(5,3) = 120
c) f(1) = a: C(1,1) * 5^3 = 125

I'm really not sure about that answer c i gave.
I was also thinking C(1,1)*C(5,3) but that gives me 10 which doesn't seem right...

My thinking is one slot is taken as f(1) = a so the total number would just be c(1,1) * 5^3 as it doesn't matter for order and it doesn't have to be injective

[weary tiger]

well now you only have to consider {2,3,4} since 1 is already assigned as you said

each of the inputs on {2,3,4} have 5 possible outputs, so it's just 5³

[ashen magnet]

mmm yeah that is what i found to make the most sense. it's just my text book had me confused with some examples having results like c(n,r) * c(n,r) or c(n,r) + c(n,r)

[weary tiger]

sad

[ashen magnet]

what's wrong?

[slow socket]

ur book is sad

[ashen magnet]

Stop guys, I paid good money for it D:

I get better understandings of concepts on here though.
My professor likes giving strange practice problems I can't find anywhere else. Even in the book. so it's nice getting things cleared up on here

[modest zealot]

any1 got any clue how to prove something super obvious

[weary tiger]

use the definitions

[modest zealot]

the definitions?

[weary tiger]

definition of complement

[modest zealot]

x is either in A or not in A

?

[weary tiger]

you skipped a step but yeah, that is what the union will be

[modest zealot]

but how does saying x is eitheri n A or not in A prove that its in the universe

i feel like im missing something here...

[weary tiger]

because those are the only 2 possibilites that you can have

either it is in A or it is not

[modest zealot]

do i have to show both inclusions for something like this?

prove its a subset in one way and the other way

[weary tiger]

$A \cup A^c = { x | x \in A \lor (x \in U \land x \notin A) }$

[weary tiger]

then use distributive laws

[modest zealot]

hmmmmm ok...

[modest zealot]

i donz get this

[ivory badge]

It's basically saying you can get arbitrarily close to the edge

[craggy gale]

it means for any x of S, there exists an open interval centered at x that's entirely contained in S

[modest zealot]

mmmm u guys mind showing an example?

im still a bit confuzzled

[craggy gale]

]-1, 1[ is open 🤔

[-1, 1] isn't

[modest zealot]

wait WHAT

reverse brackets!!??!?!

[weary tiger]

(-1, 1)

the interval does not contain the endpoints

[modest zealot]

Oh wait thats it?

an open set is just a set that doesnt contain its endpoints

[ivory badge]

If you want a concrete example of what it means, if you have (0, 2) then if you have 1.9999, there exists an r (0.000000001 for example) such that (1.9999-0.000000001, 1.9999+0.000000001) is a subset of (0,2)

[craggy gale]

reverse brackets have more charm tbh

[ivory badge]

tbh

[weary tiger]

french notation tbh

[ivory badge]

I typically use parenthesis because habit but reverse brackets look cooler

[craggy gale]

(a, b) is ambiguous

[weary tiger]

lmao ordered pairs

[ivory badge]

true, but depending on c o n t e x t

[modest zealot]

oh so its one of those limit things

[craggy gale]

it could be pair of things

[modest zealot]

it like approaches that value, but not quite that value

[weary tiger]

imagine using (a,b) when you're discussing open sets in R^2

[ivory badge]

It's not a limit per se, it's saying you can always get closer to the endpoint

[modest zealot]

but its NEVER the endpoint

incredibly close, but never

[ivory badge]

Yeah

Closed sets you can take the point at the end and there will be no open set which contains it and is inside the set

[modest zealot]

(a) Let a, b ∈ R. Show that the interval (a, b) is open.

For example, how would u show that?

[ivory badge]

Basically it's open if any object within can be covered by an open subset of the original one

So you show it is open either by 1) it's open because they use notation that says "this is open" or 2) it's open because there's always a open subset which covers any number within

[modest zealot]

lets say we use method 2 (i dont think method 1 is gud enugh)

[ivory badge]

It's not exactly good but I highly doubt i'm getting all the context of what you have

[modest zealot]

o

[weary tiger]

use the definition of openness

[craggy gale]

take x in (a, b)
(x-a)/2 > 0
(b-x)/2 > 0
the minimum of those two is > 0
it works as the r of the def

[ivory badge]

that works

What's this for anyway, a set theory course?

[modest zealot]

why does that work?

o for some mathematical reasoning class

[ivory badge]

ah

It works because x+r is not >b and x-r>a

[modest zealot]

like the thing that keeps coming back is limits limits limits

[ivory badge]

and it forms an open subset ergo (a,b) is open

[modest zealot]

FK THIS IS SO CONFUZZLIZNLZING

[ivory badge]

It's not too bad once you "get it"

[modest zealot]

can u explain it in a dumber way

[ivory badge]

Definitions of concepts tend to throw people for a loop till they realize what it means

Closed means you can put a definite maximum (and minimum) in this context

open is not closed

You can't say the "maximal value of x < 2"

but you can for x<=2

[modest zealot]

ok i get that, sets n shit

[ivory badge]

Closed means "this has a definite endpoint"

[modest zealot]

but i cant connect that with the definition

[ivory badge]

The definition says there is no endpoint definitely because you can always get closer (there is always an r such that x+r < Y)

You can't always get closer to 2 in [0,2] because it definitely ends

[modest zealot]

there will always be some r value, that you can put between a value (x) and a value (b) for example

[ivory badge]

yep

[modest zealot]

Oh

ok

we gettomg somewhere

gimme a sec now

[ivory badge]

Take the unit interval (0,1). there's always a lesser value because there is no minimum because 0 is not included

Intervals like this is basically "exploit the properties of real numbers"

[modest zealot]

ok yup i got part a

its a bit similar to the answer some1 gave above

but we choose r to be (b-x)/2

and r to be (x-a)/2

such that

x+r < b and x-r > a

x is an arbitrary element

and i think that proves openness right?

im basically saying that between x and b, there will always be a value in between no matter what for any x, b

[ivory badge]

Ye

[modest zealot]

awesome awesome

[ivory badge]

That's literally what it means: you can always get closer

[modest zealot]

gotcha

tyty

[ivory badge]

Again, no pressure, definitions of "new" concepts tend to throw people for a loop, esp when it's formalizations of things they thought they knew

[modest zealot]

ye ye

[ivory badge]

This server tends to be helpful if you have questions, but if you have questions on some specific thing (like problem (a)) they tend to like using #question-whatever

[modest zealot]

o

naw its k

one more question

how do u negate the statement

i kinda have an idea

but im not sure

ik for sure its something along the lines of

there exists an x in S that for all r less than 0

is that correct for now?

or is it for all r greater than 0

[weary tiger]

also

closed does not mean 'not open'

it is just an unfortunate choice of word

[modest zealot]

o

[ivory badge]

yeah it's a bit more in depth as a being the complement of an open* set but uh

[weary tiger]

there are sets which are both open and klosed

[modest zealot]

wdf

bro wtf is mathematics

[weary tiger]

it is mathematics

[ivory badge]

ikr

Anyhow, negations aren't that bad

[modest zealot]

i always ALWAYS have problems doing negations

because i dont know whether to switch things or not

[ivory badge]

Basically: there exists some x such that for all r, (x-r,x+r) is not a subset of S if I remember right

[modest zealot]

for all r greater than 0 or less than 0

[ivory badge]

Correct me if I'm wrong of course, but I believe it should stay greater than

since ya know

(2, 1) isn't the best thing

My logic is a bit rusty and rough so don't take what I say at face value, actually look at it and make sure

[modest zealot]

yups will do

[weary tiger]

memes

what can you say about the openness of the empty set? its closedness?

[modest zealot]

yes

wait

wat

ok now im confused

[weary tiger]

remind me of the definition of open

[modest zealot]

that u can ALWAYS choose an r value

such that x+r will always be less than b and greater than a

[weary tiger]

well yes, that's the second part of it

what's the first part

[modest zealot]

uhhh

for any x value?

or for all x values

[weary tiger]

right

for all x in the set

but there are no elements in {} in the first place

so it's vacuously true

[modest zealot]

uh huh...

[weary tiger]

kek

[modest zealot]

wait fk it lemme send u the whole problem, im confused as shit

how tf do u do b and c

c i kinda have an idea, but its prob terrible af

im a very confused potato atm

[ivory badge]

So B isn't that bad, it's just logic stuff so i'll leave that alone for now

So consider (a,b] and the definition of open

Is there any point x such that there exists no r?

[modest zealot]

when x = b?

that was my original thought

when x = b

u cant do shit

since r > 0

[ivory badge]

Yep

[modest zealot]

wait thats it

wut

wut

wut

[ivory badge]

therefore, there exists some x such that r does not exist

therefore it's not open

[modest zealot]

broooooo i cant believe i got it with my whack idea

sick

[ivory badge]

yeah it's a lot more straighforward than it appears under formal symbols

Personally, I prefer the order of Seeing the formal way -> intuitive explanation -> connect to the formal definition so you can properly understand

[modest zealot]

i go more like
understand the shit, wut it means, and then start making ideas

[ivory badge]

Fair enough

So you got c, go back to b

[modest zealot]

aight

[whole sinew]

What is discrete math

[ivory badge]

math but without continuous

[whole sinew]

waow

[weary tiger]

lmao

[hexed rune]

Not true

Every function is continuous in discrete math

[ivory badge]

h m m

[weary tiger]

[modest zealot]

anyone here want to explain relations of a set

and how there are 2^(|A|^2) number of relations of a set A

where |A| represents number of elements

[weary tiger]

So you're asking how to derive the general formula for the number of relations of a set of a given size?

[modest zealot]

yes

basically

LOL

[weary tiger]

i mean

consider the definition of relation

Doesn't it depend on whether the set is reflexive etc

AxA has |A|² elements

and a relation is a subset of AxA

so 2^(|A|²) possible subsets

kek

[modest zealot]

a relation is ALWAYS a subset of AxA?

wut

[weary tiger]

obviously

[modest zealot]

i think im missing something

is it okay if you offer an example?

[weary tiger]

example of relation?

[modest zealot]

that proves there r that many elements

[weary tiger]

wat
do you want me to prove there are |A| * |B| elements in A x B

[modest zealot]

like for example set A = {1, 2, 3}

wait

nvm

so in a nutshell, a relationship is basically how the function maps to itself

or how the function maps to another function

[weary tiger]

no

[modest zealot]

awww fuk

[weary tiger]

it is a subset of the cartesian product

[dense thicket]

relation is basically grouping elements that are not distinguishabkle from some point of view

[modest zealot]

why tf is that useful?

[dense thicket]

relations have some useful properties

[modest zealot]

hmmm

so a subset of a cartesian product has properties like reflexive, symmetric, n stuff like that

[dense thicket]

it can

depends on how you define a relation and what subset it is

[modest zealot]

some relations can be written as a function

[dense thicket]

yes

[modest zealot]

oh god lemme c if this clicks in my brian

this is some whack shit

connecting subsets of carteisna products with functions and relations

[dense thicket]

I mean there are many relations , you can consider two natural numbers being in relation lets say a and b if a is smaller than b

[modest zealot]

but how tf is that connected to sets?!?!?1

im not understanding the connection between sets and relations

oh wait...

[dense thicket]

well function is defined by taking element from one set and spitting out element from the other

[modest zealot]

r sets in relations, usually defined as like real numbers, natural numbers, and stuff like that?

like there r relations in sets like the real number set

[sour arrow]

Let A = {1, 2, 3}
Let B = {a, b, c}

Some relation could be {(1, c)}
This is a subset of A×B
It essentially says "1 is related to c"

[modest zealot]

OH

[sour arrow]

A relation with no added structure is a very loose concept, where a function is much more concrete and we typically work with them instead. A function is a special type of relation.

[modest zealot]

and for functions, we kinda assume the universe we r working with is usually real numbers

or the set

[sour arrow]

If you're doing calculus, yes

[modest zealot]

ahhhh

that makes a ton of sense

[sour arrow]

If you're doing a lot of other math, almost never

[modest zealot]

w0t

o ic wut u mean

any1 mind explaning how the fuck this everything but reflexive

this shit fucking with my brain hard

[sour arrow]

Let A = {a, b, c, d}
You can see that's the set above.

Let's think about the relation from A to itself.
This is by definition, a subset of A×A.

According to the picture, this relation is just {(a, a)}
That is, a is related to a.

[modest zealot]

yes

[sour arrow]

"Reflexive" means every element is related to itself.

This isn't the case, since b isn't related to b.

[modest zealot]

i understand the reflexive part, for somethign to be reflexive, every single element of a set must map to itself

[sour arrow]

Yus yus

[modest zealot]

i dont udnerstand the last 3

lmao

how tf those r all YESSSES

[sour arrow]

Symmetric
If aRb, then bRa

[modest zealot]

clearly not tho

[sour arrow]

Where not?

[modest zealot]

no symmetry whatsoever

WAIT

ITS TRUE

BECAUSE aRb is false

this is some whack logic yo

[sour arrow]

Every relation aRb has a symmetric relation bRa

This is true, simply because there doesn't exist any aRb relation lel

[modest zealot]

the picture above, there isnt an element that maps to anything else, which implies symmetry

yes

yes yes

understood

one question

[sour arrow]

All three follow that logic

[modest zealot]

for antisymmetry, is the definition like

if a maps b and b maps a, then that implies a = b

[azure lichen]

Every relation aRb has a symmetric relation bRa
aRb isn’t a relation, it’s a truth value or sth; R is the relation

[modest zealot]

so like symmetry, antisymmetric, and transitive all have the implication logic

and since the hypothesis or initial statement is false, the whole statement is true

[sour arrow]

Very nice catch, that is true.

[azure lichen]

yea, empty truth or whatever it’s called

vacuous truth?

also reminder that a relation is just a subset of the cartesian product

and aRb is really just a shorthand for (a,b) ∈ R

[modest zealot]

one more question

how come this isnt transitive

clearly

a -> c, and c -> b

which results in a -> b

[azure lichen]

well, the explanation is wrong, so

[modest zealot]

o

wat

[stray reef]

[modest zealot]

👀

[azure lichen]

maybe they meant for the top arrow to go the other way

[modest zealot]

o

LOL

[azure lichen]

or sth

[modest zealot]

but for the sake of the diagram only, its transitive right?

[stray reef]

the relation as drawn is transitive.

[azure lichen]

yea

what ann said

[modest zealot]

ok if i understand this correctly

A = {a, b, c, d}

for something to be transitive, ALL ELEMENTS have to map to itself

[azure lichen]

that would be reflexive

[modest zealot]

i mena reflexive

mbmb

for something to be symmetric

u only need one set of symmetry? like (a maps to b) and (b maps to a)

u dont need symmetry to happen on all the thingies

?

[azure lichen]

no, you need every set of symmetry

if a→b, then also b→a

for any a,b

[modest zealot]

ok so
A = {a, b, c, d}

is there a difference between
a->b and b->a

or

a->b and b->a AND a->c and c->a

like does the thing only NEED one symmetry for it to be identified as symmetric?

[azure lichen]

well, they’re two different relations

[modest zealot]

or does it need symmetry on ALL the elements

[azure lichen]

both are symmetric if that’s all there is to them

but a→b, b→a, a→c is not

because a→c has no counterpart

[modest zealot]

so as long as there is something like a->b, there needs to be a counterpart

OH

THAT MAKES SO MUCH SENSE YO

if you have it, then u need the reverse, if u dont, then its all good

BRO

THIS MAKES SENSE

[azure lichen]

in particular, if you don’t have it, then you also can’t have the reverse

[modest zealot]

yes

[azure lichen]

(because the reverse would also be an “it”)

[modest zealot]

so u either have the a->b and b->a or u dont have any at all

[azure lichen]

aye

[modest zealot]

cuz if the relation ONLY has a->b, then its antisymmetric

[stray reef]

a -> b and nothing else?

yes

[modest zealot]

what if it has a->b and like b->c

[azure lichen]

antisymmetric just means if you hvae a→b then you can’t have b→a

[modest zealot]

is that antisymmetric?

[azure lichen]

that wuold be antisymmetric, again assuming nothing else in there

[modest zealot]

ok ok ok ok

so its one of those things

oh my

this makes so much sense

its the implication condition shit all over again

DEFINITIONS SO FUCKING IMPORTATNTSONGSRG

[azure lichen]

someone’s learned an important thing about math :P

[modest zealot]

BROOOOOOOOOOOOOO

I feel like a god now

bow to me u peasants

ok im sry

im out

[azure lichen]

aight now find relations on the set of natural numbers which are:
-symmetric but not transitive or reflexive
-transitive, but not symmetric or reflexive
-reflexive and transitive but not symmetric

[modest zealot]

hmmmmm can we assume set A = {a, b, c, d}

or is there another set i shud consider that'll make this easier

[azure lichen]

on the set of natural numbers

was specified

[modest zealot]

oh

fuck

[stray reef]

why don't you try finding relations for all 8 combos of presence/lack of refl, sym and trans?

[azure lichen]

honestly yea, why not

[modest zealot]

what the fuck is going on now

im so confused

[azure lichen]

some you can even find well-known ones

like, ones that you know from primary school

[stray reef]

we're giving you exercises

[modest zealot]

oh god

[stray reef]

well, sascha gave you one, and i extended it

[modest zealot]

i was NOT prepared for this

[azure lichen]

yea I just copied an exercise I got when we first learned about equivalence relations

(equivalence relation fulfills all three of reflexive, symmetric and transitive)

[modest zealot]

👀

[azure lichen]

(they’re very useful)

(like, very)

(because they generalize the notion of =)

[modest zealot]

-symmetric but not transitive or reflexive
a+b = 5

-transitive, but not symmetric or reflexive
a+b > 5??

-reflexive and transitive but not symmetric
idk

best i can do for now

[stray reef]

Find eight relations $R_0, R_1, \cdots, R_7$ on the set of natural numbers $\mathbb{N}$ such that the following properties are satisfied: \\
\begin{tabular}{c|ccc}
$i$ & $R_i$ refl & $R_i$ sym & $R_i$ trans \\
\hline
0 & No & No & No \\
1 & No & No & Yes \\
2 & No & Yes & No \\
3 & No & Yes & Yes \\
4 & Yes & No & No \\
5 & Yes & No & Yes \\
6 & Yes & Yes & No \\
7 & Yes & Yes & Yes \\
\end{tabular}

[azure lichen]

that second one is false

consider:
a = 1
b = 5
c = 1

then a+b > 5
b+c > 5

[vital dew]

Ann:

[azure lichen]

but a+c = 2

so that’s not transitive

[modest zealot]

oof

hmm

holy shit

what the fuck are those 8 exercises

bro my brain

is gonnna explode

[stray reef]

do them one at a time 😛

[azure lichen]

I gave you exercises 2, 1 and 5

on ann’s table

[modest zealot]

ok lemme figure dis shit out

but b4 that

lemme finish my goddamn hmwrk

[azure lichen]

but yea, do them one by one in whatever order

[modest zealot]

GAAA

[stray reef]

should you choose to do them, make sure to specify which one you're going for 😛

[modest zealot]

btw r there any tricks ot finding relations like that?

or just brute fore

force

[azure lichen]

not really, but check some well-known ones

[stray reef]

^

[azure lichen]

like, what relations on ℕ do you know? that are like, common

[stray reef]

(don't overthink that)

[modest zealot]

like functions?

[azure lichen]

no, relations

[modest zealot]

lmao no lcue

[azure lichen]

I mean, functions can be seen as relations too

specifically the graph of a function is a relation

(the graph being the set of tuples (x, f(x)) for all x)

but let’s not go too complicated

a = b would be a relation

as an example

[stray reef]

it's often referred to by its symbol alone if it has a symbol

[azure lichen]

perhaps the example

[modest zealot]

uh huh....

wait owuld a relation be like =, >, <?

operations?

[azure lichen]

operations are functions

but those three are relations

[modest zealot]

so there is something simpler

that are just relations

[azure lichen]

=, > and < are all relations

+, -, * … are not

they’re just functions, and binary ones at that so that would make for lousy relations :P

[stray reef]

the "output" of a relation is a statement

in some sense

[modest zealot]

bro wtf math am i studying...