discrete-math

[analog sonnet]

Symmetric difference of A and B?

[pastel temple]

for (B).. is the answer (S/\U)->R or is it R->(S/\U)

[steady axle]

@sour arrow Thanks

[worthy thistle]

Is that statement "for any two odd integers, there exists a third integer the triple of which is equal to the sum of the first two integers" false because 1+1 =2, 2/3 isn't an integer

so proof by counterexample

[copper ore]

how can i prove this?

[sour arrow]

@copper ore
Still need it?

[copper ore]

yeah

[sour arrow]

There's a few obvious simplifications to make right away
¬(p V q) = ¬p ∧ ¬q

As well as
q → p = ¬q V p

So we have on the left,
¬p ∧ ¬q

On the right, (¬q V p) ∧ ¬p

[copper ore]

im trying to solve left only

but yeah

i tried de morgans law first

and couldn't get any further

¬(p V q) = ¬p ∧ ¬q
q → p = ¬q V p
how?

[sour arrow]

First is Demorgan's law

¬ distributes over V and ∧, but will turn them upside down

[copper ore]

yeah i meant the second one sorry

whats the second thing you did

q → p = ¬q V p

[sour arrow]

That's by definition. Definitely worth remembering that one

[copper ore]

what do you mean?

[sour arrow]

Try making the truth tables for both if you don't believe me

[copper ore]

i didn't say i don't believe you lol

i made a truth table already and they have the same true/false values so i know this is solvable

i just want to know what i need to do after de morgans law

idk what you did when you did this: q → p = ¬q V p

what law is that called?

[sour arrow]

I mean a truth table for
q → p = ¬q V p

Try looking up "implication law"

[copper ore]

ohh

ok

i was gonna do that too but it didn't make sense

wait so you use de morgans law first then implication law?

cause wouldn't de morgans law give me ¬q ^ ¬p

can i get some help ?

anyone

[sour arrow]

@copper ore
Hey, back. Still want it?

[copper ore]

hi yes please! 😃

[sour arrow]

On the left
¬(q V p)
= ¬q ∧ ¬p
Using Demorgan's

On the right
(q → p) ∧ ¬p
= (¬q V p) ∧ ¬p
By using implication law

[copper ore]

oh

you can't solve it by working on only one side?

[sour arrow]

You can. But first you'll want to work both sides to see "how", then redo your work in terms of working with only one side

Because it's not obvious how they might relate

If you are going to work with only one side, make it the right side

[copper ore]

what would be the steps if i used the left side, do you know?

[sour arrow]

¬(p V q)
= ¬p ∧ ¬q
= (¬p ∧ ¬q) V (p ∧ ¬p)
= (¬q V p) ∧ ¬p
= (q → p) ∧ ¬p

[copper ore]

how do you figure em out so fast @sour arrow do you use a technique?

like i look at my paper of "Rules of Logic" and didn't see any pattern for getting from step step 2 to 3
= ¬p ∧ ¬q
= (¬p ∧ ¬q) V (p ∧ ¬p)

but like i know it's right cause i did a truth table on it

but how did you solve it?

[sour arrow]

@copper ore
x = x V F

And p ∧ ¬p is false always

I figured it out by simplifying LS and RS until they matched, then I used those steps to go from LS to RS

[copper ore]

i see

@sour arrow can you use distribution law on this

for the LS

[golden cairn]

My head hurts from proofs, does anyone have a guide that is good for understanding proofs better?

if so can you ping me with it

[slow socket]

https://media.discordapp.net/attachments/238956921581862913/537672046751252483/image1.jpg?width=1442&height=382

so i have to show that g maps N onto N
but is not one-to-one

[sour arrow]

g(0) = g(1)
Therefore not one-to-one

[slow socket]

this class is literally giving me depression

wat about show g maps N onto N

[sour arrow]

If you want to map to n, then use n + 1. All ℕ are mappable

[slow socket]

ya but how do i show it

i get that all ℕ are mappable

[sour arrow]

Then you're done

That's what mapping onto ℕ means

Onto = surjective

[slow socket]

ya but im supposed to prove it or some shit

[sour arrow]

g(n + 1) = n for all n ∈ ℕ

[slow socket]

wait where did u get n + 1 from?

the function is n - 1?

[sour arrow]

What is g(n + 1)?

[slow socket]

is that the inverse?

[sour arrow]

No, that's g, when we plug in n + 1

g(n) = max{0, n - 1}

g(n + 1) = max{0, (n + 1) - 1}
g(n + 1) = max{0, n}
g(n + 1) = n

[slow socket]

and why did u plug in n+1, where did u get that

[sour arrow]

g(n) "usually" returns n - 1

How do you get it to return n? Increase n by 1

If you can see why it works, you can see where I got it

Don't get too bogged down in notation here.
If g(n) = n - 1
Then g(n + 1) = n

[slow socket]

how can i do this

[copper ore]

Can i ask a Q or is it one at a time?

[opaque bone]

Isn’t this why we have like... 10 different question channels down below?

[copper ore]

not all 10 channels down below are for discrete math

only this one is named discrete math

[gaunt igloo]

Can you tell me why there is no translation invariant measure on an infinite dimensional Hilbert space

[copper ore]

what

[gaunt igloo]

😰

[copper ore]

i thought your question was something related to my question nvm

[gaunt igloo]

Np

[copper ore]

anyways, how is hilbert space related to discrete math?

[copper ore]

can anyone hepl me

[rare barn]

so what about this formula

[glad oriole]

#❓how-to-get-help is for any sort of question whether its discrete math or not

[cold fjord]

Going into discrete math for the first time, any tips on mindset I could have?

Only because it seems substantially different from traditional calcebreometrig

[paper thicket]

Hello

Can anyone help with this? Idk how to go on about it

Does making truth tables for each and comparing all the components work?

[sour arrow]

Sure, but that's probably more work than you need

You want to find all statements equivalent to A → B

By contrapositive, ¬B → ¬A is equivalent

By implication, ¬A V B is equivalent

And V is commutative, so also B V ¬A

That gives C and E.

[paper thicket]

I see

Thanks for the help, greatly appreciated @sour arrow

[modest zealot]

(p → q) ∨ (p → r) and p → (q ∨ r)

can anyone help me prove this

like it is kinda obvious intuitively, but dunno how tf to prove it

[sour arrow]

LS:
= (¬p V q) V (¬p V r)
= ¬p V q V r

RS:
= ¬p V (q V r)
= ¬p V q V r

[modest zealot]

oOOoohhhhh gotcha

another question, is it logical to negate the other side after negating one side

can you do that to prove stuff?

also, what is the meaning behind implication? how does the truth table work

p q p->q
t t t
t f f
f t t
f f t

like how does having a false p, or false hypothesis make the implication true

[sour arrow]

@modest zealot
Oh boy you're not the first to ask that. It's a very common question

The algebra is better that way. You'll find, in practice, this ends up being a simpler option than if it were instead false.

[modest zealot]

is it due to some like convention or something?

to make things easier?

or is there some sort of intuition behind it, because the way I was told was that since the hypothesis is false, any conclusion can be drawn which makes it true regardless of the conclusion

[sour arrow]

In practice, an implication with a false premise is generally ignored anyway. However, for completeness we still have an option for this. This turns out to be the "least ugly" choice, and the algebra works well.

Also note the symbol has no need to match our English interpretation exactly. Think of it as "implies, but a little different"

[modest zealot]

p implies q

hmmmm

i feel like the more i think about it, the more convoluted it becomes in my head

[ivory osprey]

p implies q does not exclude a r implying q

So, when p is true, you know for sure q is going to be true as well, but when p is false, you can't say whether q is false too, because there might be something else going to cause q to be true

That's how I think of it

[modest zealot]

interesting

another question

Rewrite the following statement using symbols (meaning: define some statements p
and q and connect them using quantifiers and a ⇒ symbol): Every continuous function f from R to R is
differentiable at every point.

how would you do this?

p - continous function f from R to R
q - differentiable at every point

p -> q?

my gut however, is telling me is supposed to be the other way

q -> p

[ivory osprey]

p -> q, from the statement, the possibility that a function that's differentiable at every point but is not from R to R exists

But the other way around is not possible

[dapper mirage]

don't forget to quantify!

[dapper mirage]

you can do that with just truth tables

[copper ore]

im not allowed to use truth tables

[dapper mirage]

im not 100% on what you mean "by equivalence", but you could be interested in the truth functional form algorithm

[copper ore]

i have to do it by logic rules

logical equivalence is what i mean

[dapper mirage]

that does sound like you want the truth-functional form algorithm

although, perhaps you mean just by using things like de morgan's laws

[copper ore]

yeah by de morgans and all that etc

[dapper mirage]

have you tried using de morgan's laws on it?

[copper ore]

yeah i di

d

don't know how to solve this though

i just get stuck

[dapper mirage]

ok, what can you see in there that you could apply de morgan's laws to?

[copper ore]

this is what i did

i used implication and then de morgan law

[dapper mirage]

ok i think that is right so far i need to think a bit tho

[copper ore]

okay

i will try to think too lol

[dapper mirage]

😄

[copper ore]

you know what i tried doing but i can't seem to figure out how

so i tried to make all the same letters together so i can give them a T or F

using contradiction

or

law of excluded middle

[dapper mirage]

this is a bit of a tricky one yano xD

[modest zealot]

ok a followup to the previous question

Rewrite the following statement using symbols (meaning: define some statements p
and q and connect them using quantifiers and a ⇒ symbol): Every continuous function f from R to R is
differentiable at every point.
b) Write the negation of the sentence above using symbols, and then write it in english.

how do you write the negation of

p -> q

if p is every continuous function f from R to R and
q is differentiable at every point

do i take the contrapositive?

[copper ore]

yea @dapper mirage it is.

[dapper mirage]

do you have an earlier easier example from your class i can look at to get a better idea what sort of question it is?

[dapper mirage]

ok i got it to disjunctive normal form at last

im going to have to use ^ for and cba with latex

(¬A ^ ¬B) v (¬A ^ C) v (¬B ^ A) v (¬C ^ A)

does that look like something you can work with? @copper ore

[copper ore]

let me see

not really lol

[dapper mirage]

hmm i dont really think it could be simplified much further...

[copper ore]

damn ok

[dapper mirage]

i think what you are supposed to do is look at
(¬A ^ ¬B) v (¬B ^ A) v (¬C ^ A) v (¬A ^ C)
or more explicitly
[(¬A ^ ¬B) v (¬B ^ A)] v [(¬C ^ A) v (¬A ^ C)]
and from there inspect for a specific instance where the statement could be true, and one where it could be false. If it cannot be true then it is a contradiction, if it cannot be false it is a tautology, if it can be both it is a contingency. An informal argument could be:
It could be true simply if the right bracket [] is true, which is true whenever A and C have opposite truth values. So it is at least not a contradiction.
To falsify it, we need both brackets [] to be false. To falsify the right one we need A and C to have the same truth value, either both true or both false. To falsify the left, both (¬A ^ ¬B) and (¬B ^ A) must be false, the only way this can happen is if ¬B is false (law of excluded middle on A) i.e. if B is true. So the sentence can be falsified with A=B=C=T, or B=T and A=C=F. Hence the sentence is not a tautology.
Since it is neither a tautology nor a contradiction it is a contingency.

Again, really this argument really tells us nothing more than a truth table would.

@copper ore i hope this helps, im off to bed now 😃

[tidal minnow]

How can I make a proof by contradiction for "√2 ∉ ℕ ∧ √2 ∉ ℚ "?

[neat siren]

How do I find the time complexity of: $$ T(n) = 2T(n-2) + O(1) $$

[vital dew]

Pastah:

[neat siren]

I know the pattern is i=1 to whatever and $$ 2^iT(n-2^i) + i $$ where T(1) is achieved when $i$ is $log_2(n-1)$

[vital dew]

Pastah:

[neat siren]

oh awit, the reccurence is wrong

[neat siren]

ok well what i wrote was wrong above but would the time complexity be $O(2^n)$?

[vital dew]

Pastah:

[neat siren]

anyhow i guess it is O(2^n)

[dense thicket]

@neat siren do you know the first condtion? T(1) or T(0)

[neat siren]

No, but you could imply it's T(1)=1 or T(0)=1, shouldn't matter since T(n) is a time complexity of a described algorithm

[neat siren]

for the median of medians algorithm, how is it that finding the median for each of the n/5 groups is O(1)? shouldn't it be O(n)?

[azure lichen]

assuming you’re talking about the same algorithm I know, you make groups of size 5

not five equally sized groups

and each of those medians can be found in, I think, 6 comparisons?

for a total of $$6 \left\lceil \frac{n}{5} \right\rceil$$ to find all the medians

[vital dew]

Sascha Baer:

[slow socket]

hey i have this statement
"If you water the concrete, then it grows"
i have to find which is true,
the converse, inverse, or the contrapositive

it will be the inverse right?\

[slow socket]

can someone help?

[weary tiger]

@slow socket The converse and the inverse both have the same truth value. The same with the statement and the contrapositive.

In this case, are we assuming that the statement is true? I don't think watering concrete makes it grow.

[slow socket]

ya i know that

ya so the contrapositive would be false right?

[weary tiger]

but the statement "if the concrete grows, than you must have watered it" is vacuously true, because concrete doesn't grow.

Yes. The contrapositive is false in real life.

[slow socket]

i was thinking it was the inverse

[weary tiger]

and also the statement "if you don't water the concrete, then it doesn't grow" is also true.

the inverse and the converse have the same truth value

[slow socket]

so it would be both inverse and converse ?

[weary tiger]

yes

[slow socket]

so "If it grows, then you water the concrete"

is true

[weary tiger]

Well, to put it more eloquently, "If the concrete grows, then you must have watered it."

You can figure this out by taking the contrapositive of the inverse.

[slow socket]

cool

can i ask another question?

about using De Morgans Law

i got
"Atlanta is a city and California is a city"
it would be
"Atlanta is not a city or California is not a city" ?

[weary tiger]

well that would be the negation

[slow socket]

ya i have to use De Morgans Law to write the negation

[weary tiger]

If you want to transform the statement into another true statement with De Morgan's Law, it would be

It is not the case that Atlanta is not a city or California is not a city.

but if you're looking for the negation, that's correct

[slow socket]

and it would be true right? cuz one of them is true

[weary tiger]

the negation would be false

oh my gosh

I'm an idiot

it's true

[slow socket]

lol

[weary tiger]

I live in California and it's a state

[slow socket]

xd

dw dude, the problem for the concrete i put earlier

i spent an hour thinking concrete grew if u put water on it

[modest zealot]

guys I don't understand one particular thing

"Every student in this class has taken a course in Java"
U- All People
S(x) = x is a student in this class
J(x) = x has taken a course in Java

How come the solution to this is

(weird looking upside down A)x(S(x) -> J(x))
but not
(weirdlookingupsidedownA)x(S(x) ^ J(x))

[weary tiger]

@modest zealot I assume by ^ you mean AND?

pretty simple

not everyone is a student in the class

[modest zealot]

ye

the statement will never be correct because its for all the people that exist, and then stating the two conditions that they are a student AND taken java right?

wait that doesnt make sense

wat

ooohhhh i get it now

nvm

[dense thicket]

Is cardinality of N^R continuum?

[weary tiger]

Yes

you know that the cardinality of continuum is 2^Aleph_0

and that card IN=Aleph_0

now

Aleph_0<2^Aleph_0

oh wait no

I'm wrong

lul I made a mistake in my thinking

ok wait

ok the cardinality of N^R

is card P(R), the power set of R

because

well as established

Aleph_0<2^Aleph_0

thus

Aleph_0^(2^Aleph_0)<=(2^Aleph_0)^(2^Aleph_0)=2^(Aleph_0 x 2^Aleph_0)

but here's one thing to know about multiplication of infinite cardinals

if k,l are infinite cardinals

then their product k x l = max(k,l)

so Aleph_0 x 2^Aleph_0=2^Aleph_0

therefore Aleph_0^(2^Aleph_0)<=2^(2^Aleph_0)

to show the other way of the inequality

you have

2<Aleph_0

thus

2^(2^Aleph_0)<=Aleph_0^(2^Aleph_0)

[dense thicket]

Wait, so you use that power set of X has the cardinality 2^|X|?

is that always the case?

[weary tiger]

so it seems you didn't study infinite cardinals?

[dense thicket]

Not much, I heard about it but dont remember

theres so much in set theory to learn

[weary tiger]

ok

[dense thicket]

but I do know what you mean later what you wrote

[weary tiger]

oh ic

ok then explanation is simple

you see

if a,b are two infinite cardinals

then a^b is defined to be the cardinal of the set a^b of all functions from b to a

now

given a cardinal k

what can we say about P(k), the power set of k?

recall that 2={0,1}

recall also that there's a nice bijection between P(X) and 2^X

[dense thicket]

ohh so 2 means that?

[weary tiger]

yep

[dense thicket]

k

[weary tiger]

0=empty set

1={0}

2={0,1}

3={0,1,2}

etc

now

P(X) and 2^X are in bijection for any set X

if X is empty we're done

otherwise

here's the bijection

[dense thicket]

is it obvious that this is bijection?

ohh ok

[weary tiger]

yes here it is

$$P(X)\to 2^X$$

[vital dew]

Scientifica:

[weary tiger]

$$S\mapsto \chi_S$$

[vital dew]

Scientifica:

[dense thicket]

this is chararcteristic?

[weary tiger]

$\chi_S$ is the characteristic function of $S$

[dense thicket]

kk

[vital dew]

Scientifica:

[weary tiger]

yes

it's intuitive

a subset of X

[dense thicket]

yeah I see

[weary tiger]

is the same as a function that tells you which elements are in the subset and which are not

so this shows that card P(X)=card 2^X

and by definition

card 2^X=2^(card X)

by definition of cardinal exponentiation

[dense thicket]

wait,. but when is this function 1 and when 0?

[weary tiger]

$\chi_S(x)=1\iff x\in S$

[vital dew]

Scientifica:

[dense thicket]

k

0 in other case

Ok I think I get it

thanks for your help

[weary tiger]

it's a pleasure 😃

[dense thicket]

but R^N would be aleph0, right?

or nah

[weary tiger]

by definition

aleph0=card IN

wait

you talked about N^R first not R^N 😉

[dense thicket]

no if {o,1}^n is cont then R even more tright

oh yeah I even thought like that

[weary tiger]

yay

[dense thicket]

R^N continuum

[weary tiger]

I think that's right

[dense thicket]

N^R aleph? or am I going crazy now?

No its still copnitinum

nvm

if I recall N^N is continuum so its even more continuum

[weary tiger]

(2^aleph0)^aleph0=2^(aleph0 x aleph0)=2^aleph0=c (where c is continuum)

[dense thicket]

ye

tomorrow I have final exam from relations and cardinality things, and the things is I need to have about 25% to pass since I owned the first exam, but I feel like I dont know anything about it. Like finding bijections or some shit, even though its only 25% needed

[weary tiger]

RIP

some cases like the ones we spoke about

it's much easier to do it with cardinal arithmetic than with bijections

also another technique is the Schröder–Bernstein theorem

that's how you show that card P(IN)=card IR

i.e, 2^Aleph0=c

[dense thicket]

k will reserach, thx

[weary tiger]

it's a pleasure 😃

Good luck!

[dense thicket]

thx 😃

[dense thicket]

Why isn't (Z,<=) a Well-founded relation

?

Z being integers

[viral stump]

cuz there's an infinite sequence ... <= 2 <= 1 <= 0

with no infimum

[dense thicket]

Ohh ok

Is there a relation that is kinda well founded but with no supremum?

[weary tiger]

(IN,<=)

[dense thicket]

instead of minimum

[weary tiger]

it's well founded

but there's no supremum in IN

[dense thicket]

I mean if thats a thing

well founded relation but instead of lack of infimum we have lack of supremum in definition

[weary tiger]

oh wait

wtf wait

I'm confusing well founded relation with well-ordered lul

mb

[dense thicket]

no thats right, N is well founded

well ordered is well founded and all elemnts are comparable, right>?

[weary tiger]

oh yes well-ordered is a particular case of well founded

well guess well ordered is a well founded linear order?

[dense thicket]

yup

[weary tiger]

the order may not be linear

so that's a well founded order

[dense thicket]

guys do you like kuratowski-zorn lemma?

[weary tiger]

The Axiom of Choice is obviously true, the well-ordering principle obviously false, and who can tell about Zorn's lemma?

that's what Jerry Bona said xD

[dense thicket]

what does that mean

[weary tiger]

you sure know axiom of choice?

[dense thicket]

zorns lemma is equivalent to axiom of choice right?

[weary tiger]

yes

the thing is

if you think about it

axiom of choice is so intuitive

well-ordering is so counter-intuitive

and we get absolutely no intuition about Zorn's lemma lul

that's what the quote means

[dense thicket]

ok TRUE

[weary tiger]

xD

well despite Zorn's lemma being difficult to grasp (at least for me), I still like it

it has wonderful applications

it allows to do maths

like

[dense thicket]

and why do people call it zorns lemma and not kuratowski lemma or kuratowski zorn lemma?

[weary tiger]

in algebraic geometry, it's important because then we can speak of maximal ideals anytime we wish

ah dunno

that's more of a history question

[dense thicket]

kuratowski got cucked

[weary tiger]

according to Wiki: "Proved by Kuratowski in 1922 and independently by Zorn in 1935"

yeah it's weird

life is unfair lul

maybe there are other reasons

[dense thicket]

maybe zorn's proof was cooler

or no one undedrstood kuratowski's proof lul

[weary tiger]

xD dunno

or maybe Zorn was more popular or known than Kuratoswki

[slow socket]

This is class is difficult for me

I got my first test in one week

We started doing proofs and is cray

Crazy*

[dense thicket]

yea, it fucks with your head the first time you see it

but that's the case with every new subject I take

also, if we have set N^N then it means these are all sequences that can take naturals as values, right?

also meanign every function N -> N ?

[weary tiger]

yes that's right

if A and B are sets

A^B is the set of all functions from B to A

[azure lichen]

Zorn has a much cooler name

[weary tiger]

MAX

[azure lichen]

it means “anger” in german

[weary tiger]

lul

[azure lichen]

and the guy was in fact german

[weary tiger]

luuuuuuuuuuuuul

[azure lichen]

or wrath, rather

[weary tiger]

awch

[dense thicket]

guys I have a question

[azure lichen]

question: what is discrete math actually? from what I can tell we kinda had its contents distributed over various subjects

[dense thicket]

what does * mean here and fin under power set

[azure lichen]

{0,1}* is the set of all finite strings with 0s and 1s

dunno why it’s labled such

[dense thicket]

OK!

[azure lichen]

as in what the * itself really means

[weary tiger]

t!wiki discrete

[uncut grove]

📖 | ** https://en.wikipedia.org/wiki/Discrete **

[azure lichen]

but that’s what I’ve seen that set called

[dense thicket]

So I guess a finite set

[azure lichen]

no, not finite!

[weary tiger]

t!wiki discrete mathematics

[uncut grove]

📖 | ** https://en.wikipedia.org/wiki/Discrete_mathematics **

[azure lichen]

it’s a union of infinitely many finite sets

it’s the unions of all “strings” of length 1, 2, 3, …

…it’s nbasically the binary numbers

[dense thicket]

no but * has to be of cardinality less than naturals

right?

[weary tiger]

I think why it's called denoted A^*

it's kind of like

you know

a finite string of length n

[azure lichen]

• is not a set

[weary tiger]

is an element of A^n

then A^* is like

plug in any integer in *

[azure lichen]

yea that makes sense

[dense thicket]

ohh ok but each natural can be represented as set

[weary tiger]

$$A^*=\bigcup_{n\in\mathbb N}A^n$$

[vital dew]

Scientifica:

[azure lichen]

beat me to it

[weary tiger]

here $\mathbb N={0,1,2,\dots}$

[vital dew]

Scientifica:

[azure lichen]

heresy!

[dense thicket]

Ok and P_fin(N) ?

[azure lichen]

finite subsets of ℕ

[dense thicket]

k

[azure lichen]

P(N) is the set of all subsets, and the fin indicates finite

[dense thicket]

also, thats true that |P(N)|=|R|=continuum, right?

[weary tiger]

y

[dense thicket]

then P(P(N)) is bigger than continuum?

[weary tiger]

y

Cantor's theorem

card P(X)>card X

[azure lichen]

yea, powerset(X) > X

[weary tiger]

even in infinte case

[azure lichen]

for all sets X

where > is comparison of cardinalities ofc

[weary tiger]

y

[dense thicket]

also if aleph 0 is |N| then aleph1 is continuum?

[azure lichen]

undecidable

aleph1 is something else

aleph1 is just whatever the second smallest infinity is

but we can’t know whether continuum is the second smallest

for reasons beyond my understanding

[dense thicket]

and thats an open problem whether there is infinity between |N| and |R|?

[azure lichen]

no

it’s not open

it’s undecidable

[dense thicket]

k

[azure lichen]

do we know of any set with cardinality aleph1?

continuum hypothesis

[dense thicket]

k

have you ever seen the proof of that being undecidable?

like how does that work

[azure lichen]

as said, beyond my understanding

it’s just basically… math works out fine either way

[weary tiger]

advanced set theory

you'll need to pick up a Jech for this one

dunno if he does the proof

here's what I mean by Jech

https://www.springer.com/fr/book/9783540440857

one day I'll read it

that's my dream 😍

[azure lichen]

one of my life goals is staying the heck away from set theory

[dense thicket]

I wish I understood set theory

[weary tiger]

no one does

[dense thicket]

but everyone does better than me

[azure lichen]

friend of mine decided to do a reading course on axiomatic set theory

and he said I should stay away from it

pain lies that way

[weary tiger]

beauty lies behind pain :3

[dense thicket]

btw what do you guys major?

[weary tiger]

among the few math books that I read from cover to cover: Elements of Set Theory by Herbert B. Enderton

LOVELY

❤

[azure lichen]

third semester bsc math

[weary tiger]

gave me orgasm

[dense thicket]

first here

[weary tiger]

2nd year masters, majoring in model theory, with applications to dynamics and combinatorics (but I'll try to shift to applications to group theory)

[azure lichen]

probably gonna drag the bsc out to 8 semesters instead of the usual 6, too many fun things to choose

and I wnana get a broader overview before I start my masters

do more of the “general” classes you know?

there’s too many things

[weary tiger]

AWSOME

tbh that's what I really wanted too T_T

[azure lichen]

I’m also very interested (but extremely bad) at theoretical physics

so I’m definitely taking classes that help with a deeper understanding of that

[weary tiger]

like I hate how much general stuff I still don't know

T_T

oh just don't speak about theoretical physics to me lul

gonna be like Kreygasm

[azure lichen]

diffgeo, functional analysis, I also wanna do sth with group theory (there’s a course for physics masters on group theory and its applications in physics that I’m eyeing)

[dense thicket]

yeah physics is cool once you understnad it, actually took history of physical sceinces next semester, we'll see if its cool

[azure lichen]

I had to take some mandatory physics classes

[weary tiger]

yay group representations are VITAL for physics

[azure lichen]

which were not really my thing

[weary tiger]

diff geo and functional analysis are a must I guess

depends on the field maybe?

[dense thicket]

why is functional analysis a must?

[weary tiger]

quantum mechanics

[azure lichen]

cause hilbert spaces are what quantum mechanics breathes, basically

but afaik they also show up in other places

[dense thicket]

Ive heard its very hard

[weary tiger]

everything you didn't study yet is hard lul

did you study multivariate analysis?

[azure lichen]

I saw a poster for a course “mathematical themes in general relativity”, which I’d love to have taken but ofc I’m a few years behind ^^ and they had as prereqs diffgeo I&II and functional analysis I&II

[weary tiger]

RIP

[azure lichen]

and idk how you can possibly do those both without doing an extra year

[weary tiger]

can't

[azure lichen]

cause each of those courses is 10 credits, and a year is 60

[weary tiger]

we did those in 1st master year

[azure lichen]

here they’re both third year bsc courses

[weary tiger]

I did only one course on functional analysis though

[azure lichen]

so usually you’d have to pick one or the other

cause you have other things you also have to do

[weary tiger]

lul F

[azure lichen]

you have to also do at least seven credits in an applied math core subject

and write a thesis

and go to a seminar

I think that’s about it

[weary tiger]

tbh doing Riemannian geometry and functional analysis (like

[azure lichen]

so I think technically you could pull it off but you won’t have time for anything else

[weary tiger]

(like C* stuff, weak topology, distributions...)

on 3rd year after HS

[azure lichen]

better imo to just stick around longer and do a lot of fun stuff

[weary tiger]

is impressive

[azure lichen]

I mean I’m amazed at how far I’ve come already

[weary tiger]

fun stuff

number theory

[azure lichen]

analysis I&II was basically purgatory

that exam man

[weary tiger]

oww

[azure lichen]

I think I have light shell shock

[weary tiger]

oof

[azure lichen]

4h exam on a one year course, almost all or nothing

you could in theory get sth like a 20% bonus thorughout the year

which I more or less had

but everything else came from the exam

[weary tiger]

RIP

[azure lichen]

oh I mean I did alright. my worst grade so far but still a respectably high one

I just absolutely sucked at the proofs section

got like 5/20 points there

on a 60 point exam

[weary tiger]

F

proofs are hard the first time we learn to write them

[azure lichen]

(one third computations, one third “small proofs” and one third theory from the lecture)

I aced the last bit, got like 2/3s on the computations and the small proofs went horribly

yea and my school’s approach to proofs is a two week intro to logic, proofs and naive set theory and then we just start with the interesting material glhf

(linalg and analysis prof collaborated to get that intro out of the way quickly before starting their actual classes)

oh and let me tell you one thing: cantor-schröder-bernstein is not something you wanna see in week two

[weary tiger]

NOICE

lul same intro for us too -_-

[azure lichen]

I mean I found it alright

I mean the first weeks in analysis and linalg aren’t really very difficult stuff so you have plenty of time to warm up

oh I hear this year’s freshmen got the AoC ⇒ Zorn’s Lemma proof in week two

[dense thicket]

cantor bernstein is probably the hardest proof ive seen in myu calsses

[azure lichen]

I think there’s sth of a tradition to put one proof freshmen can’t follow in week two

[dense thicket]

since I havent seen kuratowski zorn lemma proof

[azure lichen]

it was kinda funny, in Algebra I (3rd semester) we looked at AoC, Zorn and Well-Ordering and we pretty much just ignored AoC⇒Zorn, and proved Zorn⇒Well-Ordering as homework (Well-Ordering⇒AoC is trivial). then I hear from my friend in first year that they did AoC⇒Zorn ^^

[dense thicket]

lol

like a week ago we did zorn=> AoC

[weary tiger]

back

oh that's noice

we didn't have any of this stuff lul

I was in a faculty "oriented to applications"

[azure lichen]

zorn⇒wellorder is kinda nice tbh, so I’d just go do that and then AoC falls out pretty much immediately since you can wellorder and then pick the smallest element in every set

and that’s a choice function

[weary tiger]

so we didn't even study group theory, ring theory, ...

T_T

[azure lichen]

aw that is sad

groups are fun

I wanna do more group theory

[weary tiger]

hell yah!

ME TOO ❤

I think that's what I'm doing for PhD

[azure lichen]

I don’t care for most of algebra but groups and in particular group actions seem really nice

also lie groups

[weary tiger]

Model Theory + Group Theory= ❤

[azure lichen]

I wanna learn about lie groups

[weary tiger]

lie group=differential manifold + group

[azure lichen]

I know that much

we did a tiiny intro in classical mechanics

just enough to confuse everyone

I’m trying to take methods of mathematical physics II next semester (part I was mandatory, part II isn’t), but idk if I can handle the extra work load

since I’ll be tutoring too

but that’s mostly on lie groups

[weary tiger]

NOICE

NOICE NOICE NOICE

t! NOICE

t!yt noice

[uncut grove]

📽 | ** https://www.youtube.com/watch?v=Akwm2UZJ34o **

[azure lichen]

何

[weary tiger]

noice

[azure lichen]

but yea I’ll see

if I can handle it great, else I can just drop it

idk how much extra workload tutoring will be

[weary tiger]

just, HAVE FUN!!

😉

[azure lichen]

it’s two hours classes + correcting homework

if I end up with a small class, it’ll be fine. if I end up being the most popular tutor somehow I’ll have a problem

(the students can freely change between different TAs)

[dense thicket]

wait is N^R bigger than continuum?

[weary tiger]

yes

we proved it earlier

that card N^R=card P(R)

but card P(R)>card R (thx Cantor)

[azure lichen]

btw why does this channel (along with #foundations) not have a description?

[weary tiger]

that's what we call discrimination against minorities

[azure lichen]

yes, I openly discriminate against mods who don’t write channel descriptions :^)

[weary tiger]

@sudden knot

[sudden knot]

hi

[weary tiger]

hi 😃

[azure lichen]

or admins I guess if only admins can do that?

[sudden knot]

u can openly discriminant against me

[weary tiger]

as he said: "btw why does this channel (along with #foundations) not have a description?"

[azure lichen]

hi what’s your discriminant

[weary tiger]

😢

[sudden knot]

oop

1

[weary tiger]

😮

[sudden knot]

I dunno what to put in it

[weary tiger]

t!wiki discrete math

[uncut grove]

📖 | ** https://en.wikipedia.org/wiki/Discrete_mathematics **

[weary tiger]

t!wiki formal logic

[dense thicket]

Any help with this one?

I mean if we take chain fron minus infinity to something wouldnt that mean there isnt smallest element?

[pearl basin]

Can someone offer some guidance in solving the recurrence relation

T(n) = 10T(n/2) - 16T(n/4) + n
Initial conditions T(1)=1, T(2)=10/3. The textbook gives the hint: For finding a single solution, multiplying your initial guess by c*log_2 (n) may help. (this means nothing to me)

My strategy on easier problems is to make a diagram of the tree of the recursion for a few steps, find the total value on each level of recursion, then find a summation expression for the total value of all the levels of recursion. Usually the top of the sum symbol is log of something. Then if I bring the sum to a closed-form expression and plug it back into the recurrence it works! Then I can plug the initial condition in and solve for the constant.

[sour arrow]

Let 2ˣ = n. Then,
T(2ˣ) = 10T(2ˣ¯¹) - 16T(2ˣ¯²) + 2ˣ

This is now a linear recurrence. In order to better show it off, let g(x) = T(2ˣ):
g(x) = 10g(x - 1) - 16g(x - 2) + 2ˣ

Can you do that one?

@pearl basin

[pearl basin]

Yes, I can! That’s brilliant. Though the section of the textbook is about mergesort-like recurrences and doesn’t cover solving them with its characteristic root so I think that’s not what I’m supposed to do.

[sour arrow]

Hrm. I'm not sure how to go about it then. I'm not up on that.

[shadow dragon]

p: Sam had pizza last night.
q: Chris finished her homework.
r: Pat watched the news this morning.
Express, in words, the
statements represented by the following formulas
(p ∧ q) ∨ r

how would you do that one

[weary tiger]

It's a bit hard to be unambiguous about it.

One or both of the following is true:

1. Sam had pizza last night, and Chris finished her homework.
2. Pat watched the news this morning.

[azure lichen]

“Either sam had pizza last night and chris finished her homework, or pat watched the news this morning, or both”

either…or helps with making the scope unambiguous imo

but you have to add an …or both to ensure you don’t mean XOR

[slow socket]

would the truth table for
p XOR p
be like this
0 1
1 0
?

[weary tiger]

no, 0 ^ 0 = 0

[slow socket]

?

0^0 = 1

[weary tiger]

no, 0 xor 0 is 0

[slow socket]

why u saying is 0 tho

[weary tiger]

because it is...

like that's how xor is defined

[azure lichen]

0 XOR 0 is indeed 0
XOR is defined to be 1 if exactly one of its inputs is 1

if you’re looking for

p q | p?q
0 0 | 1
0 1 | 1
1 0 | 1
1 1 | 0

that would be NAND, also written with ↑

NAND as in “not and”, because it’s equivalent to ¬(p∧q)

[slow socket]

https://cdn.discordapp.com/attachments/238956921581862913/540229564505391105/image0.jpg

im asked to verify Modus tollens using truth tables which is [(p -> q) ^ not-q] => not-p
did i do something wrong, cuz is not the same

[sour arrow]

What you're thinking of is
(p → q) ∧ ¬q ⇔ ¬p
And you're right, you can't prove that because it's not true

Instead
((p → q) ∧ ¬q) → ¬p
Is true if, wherever ((p → q) ∧ ¬q) has a 1, ¬p also has a 1

[slow socket]

so i can say, is not a tautology?

[sour arrow]

((p → q) ∧ ¬q) → ¬p
Is a tautology yes

[slow socket]

wat

so i did it wrong?

[sour arrow]

No

((p → q) ∧ ¬q) → ¬p
Being a tautology doesn't mean that
((p → q) ∧ ¬q) and ¬p
have the same truth table

[slow socket]

it means is always true?

[sour arrow]

It DOES mean that, whenever ((p → q) ∧ ¬q) is true, then ¬p also is true

[slow socket]

but is not in my table

[sour arrow]

Your table isn't done?

[slow socket]

wait do i have to do the whole thing
((p → q) ∧ ¬q) → ¬p

?

[sour arrow]

You could, and that would give you [1,1,1,1]

Note I didn't say that ((p → q) ∧ ¬q) and ¬p have the same truth table

I said that, whenever ((p → q) ∧ ¬q) is true, then ¬p also is true

¬p can be true when ((p → q) ∧ ¬q) is false

[sour arrow]

Perfect, looks good to me

[slow socket]

i was thinking that they have to have same values in both

so it doesnt matter if ((p → q) ∧ ¬q) is false and ¬p or
((p → q) ∧ ¬q) is false and ¬p is false

[sour arrow]

A → B is true, given that when A is true, B is also true

[slow socket]

oh i see

so if u compare in the table

((p → q) ∧ ¬q) → ¬p

is all gon be true

[sour arrow]

Yus, if you did make it, and you can choose to if you really want to show it off, that column will be true

[slow socket]

cool, thanks

[slow socket]

is there any other way to proof this absorption law without using truth tables
[p v (p ^ q)] <=> p

[near pilot]

can anyone help with this?

[slow socket]

too small

[near pilot]

one sec my bad

is that alright?

[slow socket]

ya

[near pilot]

any ideas? im new to graph theory and my teacher is pretty terrible

i can do part a by showing which edges should be removed on a diagram but idk if there is an algebraic way to do it without using the formula given in part b

[tight marsh]

@near pilot You can use fact that tree always have |V| - 1 edges

[azure lichen]

and of course that any graph with the same vertex set V is contained in the complete graph, so you can actually get that tree with just removing edges

[golden cairn]

I ¬2k

Don't shoot me, its totally funny.

[near pilot]

okay im assuming thats for part a, how about part b?

thanks btw

[azure lichen]

nah, that works for both, if you solve (b) you get (a) for free

[near pilot]

alright thank you

[slow socket]

a | b is "NAND" https://cdn.discordapp.com/attachments/238956921581862913/540292656836771871/299175087389802496.png

is there a name for this law?

[sour arrow]

NAND is "not and"

[slow socket]

it is?

why do u say is not?

[slow socket]

p and (p or q) = p
how does this equal p

[weary tiger]

equal is sorta like biconditional

p and (p or q) => p

and if we know p, then p and (p or q)

since we've showed both directions of the biconditional, they're equal

[slow socket]

im confused

[weary tiger]

Two propositions are equivalent if we can derive the second from the first and the first from the second.

[slow socket]

but how is it prooved

i dont get ur wording

[weary tiger]

If you can prove the second proposition from the first, and you can prove the first proposition from the second, then they are equivalent.

[slow socket]

i got
p or (p and q) <=> p

wat i did was use the distributive laws and got
p and (p or q)

and now idk

[azure lichen]

p and (anything) ⇒ p

for the obvious reason that if you want p and stuff to be true, p must be true

[slow socket]

ya but i have to prove it

[azure lichen]

truth table is a proof

[slow socket]

i cant

i have to use logic laws

[azure lichen]

what

[slow socket]

https://cdn.discordapp.com/attachments/440214120361492500/540428872903032832/image0.jpg

[azure lichen]

the heck are “logic laws”, did they just take the logic out of logic

[slow socket]

lol

[azure lichen]

man I am glad I never had to go through bullshit like this

a truth table is a valid proof and would take half a minute to write out

[slow socket]

i know

but i have to use this shit

[azure lichen]

so you want to reduce what p and (p or q) ­↔ p to 1?

[slow socket]

so the question is

[azure lichen]

(I see you make a distinction between ⇔ and \↔, this is a foreign world to me)

[slow socket]

well the whole question rly is, "Verify the following absorption laws"
p or (p and q) <=> p
p and (p or q) <=> p

<=> means "implies"

[azure lichen]

no it doesn’t, it means “is equivalent to”

implies is either ⇒ or ­­­\→, I don’t understand this “there are two arrows” distinction

I’ve only ever used the double arrows

[slow socket]

ya so is saying to prove that is logically equivalent

idk wat to do

[azure lichen]

your amazing table has awfully few things that could be used to reduce the complexity of a statement

[slow socket]

😦

[west crater]

Is this false?

Because it doesn't imply that it was ExP(x) only that it was one of those

[azure lichen]

this is false, yea. it would have to be and to drop that, not or

1. can be true without P(x) ever being true

[slow socket]

so Sascha u dont know how to do my problem?

[azure lichen]

if I did I’d have helped you

I could probably figure it out with enough time but I don’t see the value in trying myself sorry

[slow socket]

i understand

thanks

its late too

[azure lichen]

no, it’s early

I just woke up

[slow socket]

oh

xd

is 3 am here

[azure lichen]

well, early, it’s 9am

six hours eh? that would put you on the atlantic coast of the americas somewhere?

[slow socket]

east coast

got my discrete math class in 8 hours

[azure lichen]

it’s funny how technically you added exactly no new information to my statement and yet you did

[slow socket]

lol

[azure lichen]

since “east coast” and “atlantic coast” should be the same thing but the former refers to the east coast of the US

[ashen magnet]

hey guys, i've been working on a problem that asks to show ¬((¬p ⇒ ¬q) ∧ (¬q ⇒ p)) is equivalent to ¬p
i can't figure it out, Instead of the equivalence being ¬p i end up with p ∨ T here's what i did. I must be using the wrong approach, overlooking or making a mistake somewhere

[weary tiger]

I think you have your implications messed up

$p \implies q = \neg p \vee q$

[vital dew]

TendentiousTorturousTopics:

[ashen magnet]

could one say p -> q = p && !q

that's the logic i was using

[sharp raptor]

i think thats wrong

p -> (q = p && q)

thats it

[weary tiger]

no that's wrong

[sharp raptor]

X & true = x

[weary tiger]

p->q is not equivalent to p and q

p->q is equivalent to not (p and not q)

which is equivalent to (not p or q)

[sharp raptor]

!p | q == p => q

[weary tiger]

The standard substitution is $\neg p \vee q$ for $p \implies q$

[vital dew]

TendentiousTorturousTopics:

[ashen magnet]

is there any equivalent statement that uses the and operator?? or should i how i approach this problem

change*

[sharp raptor]

you can use demorgan to turn the or into an and

[weary tiger]

(not p or not q) and (q or not p)

same thing

[sharp raptor]

for that particular problem

[light path]

the truth table for implication is true in 3 possibilities

[sharp raptor]

I'd say try replacing the =>'s into ors

and just look at what you get

[light path]

so you cant just write a single and statement

[weary tiger]

so therefore p

[sharp raptor]

keep in mind that
(A | B) & (A | !B) == A

[weary tiger]

If you rewrite the implications, you get $(\neg p \vee \neg q) \wedge (q \vee \neg p)$

[vital dew]

TendentiousTorturousTopics:

[ashen magnet]

i see, my implications were wrong then. i'm going to write it out and see what i come up with

[sharp raptor]

a=>b
!a | b
!(a & !b)

[weary tiger]

Your whole expression becomes $\neg((\neg p \vee \neg q) \wedge (q \vee \neg p)) = \neg(p)$ by Nick12's comment

[vital dew]

TendentiousTorturousTopics:

[sharp raptor]

that's how you can turn the or into an and btw

using demorgan's law

[weary tiger]

yeah but it's useless to turn the or into an and if it's stuck inside a negation

[sharp raptor]

but idk why youd want that

ye

[light path]

hey nick you should try to write this in latex

$\land \lor \Rightarrow$

[vital dew]

ixsetf:

[weary tiger]

hmm are \land and \lor the same as \vee and \wedge?

[sharp raptor]

lets see
$\vee \land \wedge \lor$

[light path]

pretty sure yes

[vital dew]

Nick12:

[light path]

looks like minor formatting differences

theres also $\lnot$

[vital dew]

ixsetf:

[light path]

writing things that way will be easier than trying to use ! => etc

[ashen magnet]

i reduced the problem to $\ (\neg p \wedge q) \vee (p \wedge \neg q)

i just don't see where to go. The or is messing with me

[sharp raptor]

$ (\neg p \wedge q) \vee (p \wedge \neg q)$

[vital dew]

Nick12:

[sharp raptor]

are you sure that's right

I think you might be missing a not on the second p

[ashen magnet]

i see where i messed up

you're right, p has a not

i accidentally gave the second implication's p a not

[sharp raptor]

have you tried distribution?

[ashen magnet]

that's what i'm trying now, i see i can try it now that my p has the not it was missing

yup so i get not p and true which i believe is not p, right?

here's what i got then

[light path]

a good way to check your work is to use a truth table on your first and last steps

if the truth table changes

you changed the meaning of the statement, so you made a mistake somewhere

[ashen magnet]

I'm not sure i get what you mean. i think the goal was to prove the statement was equivalent to not p without using the table. the final step shows that the statement is indeed equivalent to not p. As such, i'm not sure how there is an error. each step checks out to me after i look it over again

[light path]

right, I said the truth table is a way to check that your steps are valid

the problem saying not to use the truth table means that you shouldnt use a truth table as a replacement for these steps

using it to check your answer wont matter

[ashen magnet]

I see. sorry for the confusion

[light path]

np, gl with your work

[ashen magnet]

so I have a statement and i want to find the negation. ∀y ∈ Z ∃x ∈ R (x^2 >= y)
I’m thinking ∀y ∈ Z ∃x ∈ R (x^2 < y) but I'm also thinking ∀y ∈ Z ∄x ∈ R (x^2 >= y)

I think it has to be the first one i said because i follows ~(p->q) = p and ~q
Negating statements like these always confuses me. is there any tips or methods on handling these?

[sharp raptor]

isnt it
∃y ∈ Z ∀x ∈ R (x^2 < y)

iirc
NOT(∀xP(x))↔∃xNOT(P(x))
NOT(∃xP(x))↔∀xNOT(P(x))

[wary salmon]

how to prove an irrational number raised to an irrational power can give you a rational number using sqrt(2)^sqrt(2) in a case analysis

[viral stump]

If its rational you win

Otherwise take power to sqrt2 again

And you win

[tidal minnow]

Does anyone have a guideline to becoming a pro in discrete maths?
Didn't pass the exam, so I have 6 months time to get an A

[verbal furnace]

Share your exercises.

[modest zealot]

whats the difference betwene subsets and proper subsets

they seem like the exact same thing...

[ripe cave]

Proper subsets of X are subsets of X without the actual set X

[modest zealot]

could u provide an example? most of the eaxmples ive seen were confusing af

[ripe cave]

So taka ${0, 1}$

[modest zealot]

yes

[vital dew]

Gonzo17:

[modest zealot]

Mhm

[ripe cave]

the subsets of this set are $\O, {0}, {1}, {0, 1}$

[vital dew]

Gonzo17:

[modest zealot]

Yes

Cuz u can fit those sets inside B

I mean A

Matching elementa

[ripe cave]

The proper subsets are all subsets excluding the actual set ${0, 1}$

[vital dew]

Gonzo17:

[ripe cave]

so $\O, {0},{1}$

[vital dew]

Gonzo17:

[modest zealot]

Oooohhhhhhh

Y is that important???

[ripe cave]

Sometimes useful to consider, tbh i haven't seen it a lot

Mostly just making notation shorter

[modest zealot]

Icic

Tysm for clarifying it, appreciate it

❤

[ripe cave]

np

[modest zealot]

when you have cartesian product

AxBxC

which one goes first?

or precedence?

[azure lichen]

neither. A×B×C creates 3-tuples (a,b,c)

while e.g. (A×B)×C would be a tuple of the form ((a,b), c)

(as in, the elements of those sets)

(have that form)

and since there’s a trivial bijection since all of these we simply don’t care

[modest zealot]

oh so in short

it doesnt really matter

[azure lichen]

pretty much

well, it probably ends up mattering with infinite products because there everything goes weird and you suddenly have to mumble “axiom of choice” every time you do something with it

[modest zealot]

ah okay gotcha

[wanton sable]

aren't these just the same thing?

[viral stump]

time and space?

no

for example sorting algorithms usually require O(n) space and O(n logn) time

or the easier ones O(n) space and O(n^2) time

where you just swap things around

[wanton sable]

oh shit its u

👀

thanks jacob tho

i have another question

this is a weird one

V[j][k] gives the number of the last game of the most productive k game stretch

V holds the number of wins christine has had per game in her soccer carreer

basically theyre saying, her latest (call it j) game has had the highest 4 wins in comparison to her j/2 game's highest 4 wins

so, she's at the peak of her career

however they're asking me if she should retire?

it seems pretty obvious that she shouldn't, since her latest 4 games have been her highest scoring games of her career

i don't understand why they're asking such a obvious question i must be understanding this wrong

wait maybe this is a soccer question

when do soccer players retire

[wary salmon]

can someone guide me on question c

if L(a) = L(b) then a must equal b even without the union expression

[sudden knot]

you need the union

the union part allows you to make the deductions that a ≤ b and b ≤ a

or replace ≤ with R

[dusk rain]

can someone help me with this proof/give me a starting point?

i think this is supposed to be a proof by contrapositive

[sour arrow]

Do you know what the contrapositive statement is?

[dusk rain]

if n is not prime then 2^n -1 is not prime right?

[sour arrow]

Yus!

Is that any easier to prove?

[dusk rain]

i think so

but i'm not sure where to start

[sour arrow]

I looked it up. https://math.stackexchange.com/questions/319963/if-2n-1-is-prime-from-some-integer-n-prove-that-n-must-also-be-prime

[dusk rain]

oh thanks

[violet heath]

Lmfao

[slow socket]

lol

[modest zealot]

Find A^3 if

A = {a}

wtf does this mean

power set? cartesian product???

[craggy gale]

I would say cartesian product tbh

[modest zealot]

ah ok

tyty

[modest zealot]

does anyone have time to explain generalized intersections and unions?

[wild flame]

Anyone here can help me with these quantified statements?

https://i.imgur.com/AhGsf0M.png

is that okay?

[earnest oxide]

Display the graph for the following relations:
(a) ρ = { (X, Y ) ∈ A × A | X ∩ Y = ∅ }, where A = P ({1, 2}) could someone explain how i graph a relayion like this

[sharp raptor]

@wild flame do you still need help?

[wild flame]

Yes! @sharp raptor

Mainly to make sure I presented the information properly.

Ws that screenshot the proper way to write my quantified statement? if I' mwrong, don't tell me what the answer is, just tell me why I'm wrong.

[sharp raptor]

I'm not sure if that's wrong, but afaik a lot of the time quantified vars are presented in an $\exists x \in S$ if S is your set

[vital dew]

Nick12:

[wild flame]

file:///C:/Users/mathu/Downloads/Discrete_Math_Long_Assignment_1.pdf