discrete-math

[waxen nest]

this whole part?

[snow sleet]

let me try this another way

[waxen nest]

ohh

[snow sleet]

do you see any lines repeated!?!?!?!?

[waxen nest]

i think you mean this part is not neccessary?

right i saw it

[snow sleet]

I'm not talking about this line

the third to last line and the second to last line are the same thing! That's incredibly redundant to state the same thing over again

do you see what I'm talking about @waxen nest ?

[waxen nest]

yea i saw it

i changed my workings

[snow sleet]

awesome

This channel is open for questions

[snow sleet]

@waxen nest wanna see an even quicker way to simplify the expression you were asked to simplify? This way can be done all in one's head

Here was the question you asked:

https://cdn.discordapp.com/attachments/496785905474994186/883156264333496370/unknown.png

[waxen nest]

yeah sure

[snow sleet]

Aight

[vital dew]

logician

[snow sleet]

since R is assumed to be true (because ~P and R was assumed to be true), and R is also the conclusion, this whole statement is a tautology because clearly R->R is always true

make sense?

@waxen nest

[waxen nest]

??

doenst make sense

~P and R wasnt assumed to be true, T was represented as simple statement not true?

[snow sleet]

If V, then U means if V is true, then U is true.

If (~P ^ R) is true, then R is true.

[waxen nest]

why is (~P ^ R) is true in the first place lol

[snow sleet]

read this^ @waxen nest

So IF (~P ^ R) is true, then (~P ^ R) was assumed to be true

[snow sleet]

and then we see this^

[waxen nest]

ohh

[snow sleet]

make sense? @waxen nest

[snow sleet]

So then you see that this^ statement is always true (I'll represent it with a T).

Then you just have to simplify $\neg T\land P$

[vital dew]

logician

[snow sleet]

which is quite easy since that's just $F\land P$

[vital dew]

logician

[snow sleet]

which is just $F$

[vital dew]

logician

[snow sleet]

@waxen nest

[waxen nest]

right got it thanks

[snow sleet]

you're welcome!

This channel is open for questions

[sour arrow]

Not too sure, english isn't a great basis for logic and I think there's a failure here

[last timber]

"For all x in world (if x is not me then x is ugly)"

[sour arrow]

Teacher probably means "Everybody else is ugly" which doesn't rule out that they may also be ugly

Including the "except me" is sus though

[snow sleet]

this only makes sense if he had said "everyone in the world except me is ugly."

yea this is because p imples q doesn't mean ~p implies ~ q

nice lolll

[snow sleet]

lmaoo

so if the sun rises, she became his girlfriend

this can also equivalently be interpreted as "you become my girlfriend or the sun won't rise"

this is correct

∀x(L(x)∧¬O(x)→R(x)) says that all lions that are not old roar, while ∀x(L(x)∧O(x)→¬R(x)) says that all old lions don't roar. So those two TOGETHER explain except in that context

sometimes "except" and "but" are used interchangeably...and sometimes "but" is synonymous with a logical "and"

You're welcome

have a good night!

[waxen nest]

hello there

i need help with these qns

[faint narwhal]

what have u tried?

[sour arrow]

P(U)

[waxen nest]

im not sure how to start tho

not familiar with powersets and idk what am i supposed to do

[sour arrow]

P(U) is the set of all subsets of U

That is:
If A ∈ P(U)
Then A ⊆ U

And vice versa

I wouldn't overthink this though, all the top line is saying is that A,B,C are sets made up from the universe, U

[waxen nest]

oh i see

this is ~B right

or can i write as B'

[sour arrow]

Yes to all

Use the same notation that is on your paper haha

For the assignment anyway

[waxen nest]

what assignment?

[waxen nest]

i mean i tried to copy the symbols from pdf ... but

it didnt work

തሻ൯ܤ∩ܣሺ ∪ ሻܣ∪ܤ൫ሺ ∩ തܤ

when i paste it becomes this ^

[sour arrow]

Use google for notation hunting

"union symbol" gets me ∪

[waxen nest]

think im just going to use B' lol

ok so i need to solve this

[sour arrow]

Gotta know your distributive laws

Pro-gamer tip, can write it like this:
B'((B + A) + (AB'))
To help with distributivity

[waxen nest]

how did you get that

what law is that

[sour arrow]

It isn't one lol. I just relabeled U and ∩

[waxen nest]

oh

my brain isnt really working anymore after trying to do math since 7 hours ago 😩

[sour arrow]

Darn. Sorry we couldn't get through it.

[rustic quest]

Is it just me or is number 3 is not logically equivalent. Idk if I’m missing out on anything

[split drum]

This is kind of a silly question...but this can be written in roster or interval notation, right? I have it this way and a friend has an interval [0, 4]. I think they're equivalent.

[pale epoch]

you should see that ~q suffices to make the LHS true, right?
now consider the case that r is true. if q is true, then q and r is true and thus the LHS. if q is not true, then ~q is true and we have already seen that this suffices

alternatively just make a truth table to convince yourself

[pale epoch]

do your intervals only contain integers?

[last timber]

union is done over x \in [0,1] tho

so presumably it is indeed [0,4]

[split drum]

Ooh, [0, 1] means implies real numbers.

[stray reef]

no, the union of {x, x+3} as x runs over [0,1] would be [0,1] \cup [3,4]

[last timber]

oh yes

i am dumb

[split drum]

Lol, not as dumb as me

Thanks guys

[split drum]

Wait, could you explain this?

[last timber]

well

take any x in [0,1]

say, for example, 1/2

[stray reef]

for each point x in [0,1] your union contains x itself as well as the point 3 units right of x

[last timber]

{x,x+3} is then {1/2, 3+1/2}

[vital dew]

Commander Vimes

[last timber]

@split drum

[split drum]

Oooh, now I see.

[snow sleet]

number 3 is indeed logically equivalent. I'd recommend writing out the appropriate truth table to see this

[simple nova]

https://i.imgur.com/yQ0kkNP.png

Is my thinking here right? I got the cartesian product size which is 6

then 2^6 = 64

[snow sleet]

yes, looks good @simple nova

[young jolt]

im stuck on this question

[young jolt]

<@&286206848099549185>

[fierce osprey]

A logical formula is considered to be in DNF if it is a disjunction of one or more conjunctions of one or more literals.

you can start by ruling out the last two because conjunctions are tying together the literals

and then that odd case of an or not disqualifies option two, I believe

you can have nots within your groups of literals

[gusty gale]

<@&286206848099549185>

[last timber]

@gusty gale what have you tried and where are you stuck

[gusty gale]

I didnt do yet , idk how to do

[last timber]

how is bijective function defined?

[waxen nest]

oh it is one-to-one function

[waxen nest]

this looks similar to my qn

can i get some help too

[meager prairie]

did you end up figurin it out?

[snow sleet]

one-to-one functions aren't always bijective

[snow sleet]

if the function isn't one-to-one or if it isn't onto, then it's not bijective.

[snow sleet]

here's a massive hint...plug in x=1 and record what f(x) is in that case. Then plug in x=-1 and record what f(x) is in that case. Now compare the f(x)'s

[snow sleet]

@gusty gale do you still need help with this?

[limpid topaz]

I have a questions, can I write "The natural number n is prime" as (∀k ∈ N)(∃n ∈ N)(k ≤ n) ([(n/k) ∈ N] => (n=1 v n=k))

[waxen nest]

i think i need to plot the graph out to see?

[meager prairie]

looks like logician sorta answered

id use more of a definition or uhh

map based approach than graphing

to answer the question

graphing software might make assumptions about domain and such

[waxen nest]

oh okay thanks

[storm osprey]

hi guys i need help on my exercise

Let Q(x) be the statement “x < 2.” What is the truth value of the quantification ∀xQ(x), where the domain consists of all real numbers?

how do i answer thoese question?

[snow sleet]

it's false

because you can find a real number that isn't less than 2

consider x=3

@storm osprey

[snow sleet]

The statement you brought is saying: Every real number is less than 2.

that statement is of course, false

make sense @storm osprey ?

[storm osprey]

if x >2 it still would be false right ?

[snow sleet]

yes, because x=1 is a counterexample for that case

[storm osprey]

haa, i see

[storm osprey]

What is the truth value of ∀xP(x), where P(x) is the statement “x2 < 10” and the domain consists of the positive integers not exceeding 3?
is my answer is correct?

domain 1,2,3

1^2 < 10 true
2^2 < 10 true
3^3 < 10 false

[red nest]

If x2 means x^2, then you should be doing 3^2=9, not 3^3=27

[storm osprey]

if x2 does not means x^2 how the correct answer should be? i because i assuming that x2 is x^2.

[red nest]

I have no clue what x2 would mean if it is not x^2

[storm osprey]

Let P(x) denote the statement “x > 3.” What is the truth value of the quantification ∃xP(x), where the domain consists of all real numbers?

[storm osprey]

that statement consider false ?

[red nest]

The statement is true, because there is a real number greater than 3.

[storm osprey]

if ∃xP(x) change to ∀xP(x) then it would be a false statement?

[red nest]

Yes

[storm osprey]

thank you so much i start to understand a little bit

[snow sleet]

if x2 means x^2 or 2x, then the truth value is true.

[lavish hornet]

hi guys, anyone can help me solve this problem :

1. a semi-competition basketball tournament with 39 teams participating. In this case, one team plays each other once. If each team wins at least twice, show that there are at least 2 teams that have the same number of wins

[stray reef]

how familiar are you with the pigeonhole principle?

[lavish hornet]

hmmm, i know that pigeon is 39 team right ?

[stray reef]

the pigeons in this case would be the teams, yes.

if you wish to put it that way.

[lavish hornet]

we just learned this, and still a little confused

[stray reef]

you can group the teams by how many wins they scored

i.e. the teams who scored 2 wins, the teams who scored 3 wins, etc.

there are 38 such groups as there are 38 possible scores (2 to 39; we are told no team scored only 0 or 1)

[lavish hornet]

So, the pigeon's nest is 38 wins?

[last timber]

yes

@lavish hornet basically restating Ann's words:
For 39 teams there are only 38 possible numbers describing the amount of their wins

[lavish hornet]

ooh okay, thanks guys

[storm osprey]

Hi guys is my answer is correct ?

[gusty gale]

U can refer this

@storm osprey

[pale epoch]

it looks fine

[plush dock]

Heyo, quick question (might be more of a basic algebra problem that I'm struggling with for some reason, rather than discrete, but the context is discrete, I hope it's fine).
In the context of generating functions (combinatorics), I'm going over a solution for a problem and after coming up with the generating function rule, they do this (note: it is missing one ')'): see below.
And while I do understand the right part of the equation, I don't get how they turned it into (1+x^6)...

Thanks a bunch

[vale cairn]

I have a feeling there's a slight typo as you said with the bracket

i.e. i think it should be f(x) = ((1+x+x^2+x^3) + x^6(1+x+x^2+x^3))^4 where you can just factorise out the (1 + ... + x^3)^4 to give the 2nd line

[plush dock]

Ohhh

I get it now. Thank you!

[vale cairn]

np

[mint bane]

it's impossible to use induction over all integers right

[stray reef]

you can use what might be called bidirectional induction

show P(0), and show P(n) implies both P(n+1) and P(n-1)

[mint bane]

thanks ann :) i had a feeling that could be a thing but didnt actually try it, gotta take more of an initiative lol

[weary tiger]

Sn is permutation group

How can I prove this ??

[pale epoch]

maybe one is a subgroup of the other?

[weary tiger]

Hey guys so I solved this (no idea what's it called in english) :

And now I'm trying to solve this one with the same way ( A(x) = Σa(n)*x^n ) but I don't know what to do about the 9^(n-1)

Any ideas ?

[faint narwhal]

it should just be sum 9^(n-1) x^n when you convert into generating functions

[weary tiger]

Yeah but where do I go from there? From what I understand I need to get rid of the variables . Is it something I'm missing?

[faint narwhal]

I'm not really sure what you mean by get rid of the variables

You solve for the a_n

[weary tiger]

the "n"

yeah but isnt the goal to reach a constant number ?

[faint narwhal]

No? You might get something in terms of n

[weary tiger]

Ohhhh guess I misunderstood then , so I just ignore the Σ(9^(n-1)*x^n) all together ?

[faint narwhal]

No? Why would you

[tidal tulip]

i want to see if i am understanding things correctly is it correct to say: 1) empty is a subset of the empty set is true, 2) empty set is an element of the empty set is false, 3) empty set is a subset of the set containing the empty set is true

[pale epoch]

yes

[tidal tulip]

thanks!

[shut jacinth]

can someone help me figure out why these two graphs aren't isomorphic?

i would imagine because all the vertices are deg = 3 you could map all of them to the other

[reef kelp]

Hi everybody! I'm taking a discrete math 1 class and one of our problems is to turn definitions into quantified sentences. The definition I am trying to quantify is "The natural number n is prime." and I wanted to know if my quantified sentence makes sense to you all. I did (∀x)(n/x ∈ ℕ ∧ n > 1 ∧ n/x = k, (k ∈ {1, n}))

[static comet]

Is there an easier way to do this:

[faint narwhal]

It's not necessarily true that the same degrees on the vertices means they're isomorphic

[shut jacinth]

true

[faint narwhal]

The easiest example is a hexagon and two separate triangles

Both on six vertices, all the vertices have degree 1, but these aren't isomorphic

To show that these are non isomorphic, think about cycles

If two graphs are isomorphic and one has a cycle of length n, then the other also must have a cycle of length n

[shut jacinth]

should i look at the maximum cycles in both graphs?

and compare them

[faint narwhal]

Mm I'm not so sure that the maximum cycle length will help here, but that is something that can help in general

[shut jacinth]

oh

so look at the basis of the cycles? (nvm)

[faint narwhal]

I'm not super sure that would be helpful either

Thinking back to your previous idea, instead of looking at maximum length cycles, what about minimum length cycles

[shut jacinth]

trying to wrap my head intuitively why looking at maximum doesn't work and minimum works

[faint narwhal]

I mean, what works is going to depend on what graphs you have. Sometimes maximums will differ and minimums won't

[shut jacinth]

in this case i looked and it does for the minimum (didnt check maximum yet)

left: a-b-j

right: 1-2-3-10

so since there does not exist a cycle in the right graph of length 4 can we say they are therefore not iso?

[faint narwhal]

You have to be a bit more careful than that. You can't just show that there's a cycle of length 4 in the right graph, you need to say why there's no cycle of length 3

But yes, isomorphic graphs must have the same minimum cycle lengths (also called girth)

[shut jacinth]

i see

although im a bit unsure how to show a cycle of length 3 cant be formed in that graph

[faint narwhal]

I don't think there's really any nice way other than brute force

[shut jacinth]

oh i thought you meant to prove it :u

[faint narwhal]

Look at each edge, for example the 1-2 edge, and confirm that 1 and 2 don't share any neighbors. Thus, the 1-2 edge can't be in a 3-cycle

I mean, this is a proof that there's no 3-cycle

[shut jacinth]

i see what you mean now

sorry for the confusion

and thank you for all the help

[faint narwhal]

No problem, the maximum cycle thing doesn't work cause its easy to check that you can cycles of length 10 in both

[shut jacinth]

ah gotcha

[sand cipher]

does that means that?

[weary tiger]

no

[reef kelp]

No, you don't just distribute the negation unary operator, you negate the logical sentence of "If P, then Q" Here's a truth table

[sand cipher]

so i don't get the hint, should i use that truth table or this

[weary tiger]

well the hint is showing the amount of statements you have

[weary tiger]

and thus the truth and false combinations you'll be using for the arguments

[reef kelp]

Yeah you take each proposition and assign it a letter. Ex. "Gustavo lives in Italy." is a proposition and it could be denoted with the letter p

[weary tiger]

^^

[sand cipher]

i see so it's a hint for the 2nd question

alrite thx guys

[weary tiger]

np

[reef kelp]

Oh yeah I think your problem has a typo too. The second statement should probably be If Gustavo drives a Maserati

[sand cipher]

i'll tell him that xD thx

[stark delta]

Not sure if this is discrete Math..

[weary tiger]

This seems like more #calculus or #multivariable-calculus

[snow sleet]

@sour arrow Here's another instance of that sort of generating function being applied to a similar problem.

[waxen nest]

hey i think i still need some help with this qn

i plotted the graph on desmos but im still trying to figure out whats the domain and range

[pale epoch]

for the domain you have to figure out for which $x$ the term $\frac{5}{\sqrt{2x+3}}$ is defined

[vital dew]

Lochverstärker

[pale epoch]

for the range you can look at the graph and notice that its always above the x-axis (what does this mean?) and consider whats the behavior "all the way to the right" and "all the way to the left"

[waxen nest]

how do i figure that out

im not sure if this is correct

[pale epoch]

well, you divide two numbers right?

when can you (not) do that?

[waxen nest]

divide two numbers? which two numbers

[pale epoch]

5 and sqrt(2x+3) in this case

[waxen nest]

idk

[waxen nest]

isnt range look at y?

domain is x axis right

[pale epoch]

yes, but if the graph is above the x axis that has implications on the possible y values

[pale epoch]

well, is division of all numbers possible?

[waxen nest]

?? what do you mean

[pale epoch]

can you always compute a/b for all a and b?

[waxen nest]

how to get the range and domain ? is the graph even useful to get both range and domain?

or do i need to calculate manually

[waxen nest]

whats a/b ? how did you get a/b?

[pale epoch]

the graph is useful in the sense that it gives you an idea if you know what it means

(and that desmos only shows you the graph for the biggest possible domain)

its just some arbitrary numbers

[waxen nest]

okay so what do i do now

[pale epoch]

tbh it seems like you should go back to the beginning of the chapter and study

[rough crater]

Aₙ is the group of all the strings that are created from the three numbers {-1,0,1} such that from the first number to any point the amount of 1s is greater or equal to the amount of -1s.
Find the generating function that gives us the number of strings that meet these requirements

im not the best with generating functions but i assume this has something to do with the catalan numbers

[stray reef]

group? did you mean set?

[rough crater]

yes sorry

[stray reef]

so... A_n is the set of all strings composed from -1, 0 and 1 for which every prefix has a nonnegative total

did i get that right?

[rough crater]

yep

[stray reef]

hmm

so i suppose we would need some kind of recurrence for |A_n|

[rough crater]

i tried to find one but kept getting stuck

then i realised the whole 1 , -1 is pretty similar to the catalan numbers except that there could be more 1s than -1s

i dont know if that helps at all

[faint narwhal]

seems pretty similar to Catalan's triangle or Lobb numbers

[tawny trout]

Hi, I have a small question and I can't find any resources for this specific case.

Question: if (P and Q), then P
I believe this proposition to be true because if P and Q is true, then so is P. Am I right or wrong?

[vale cairn]

(P and Q) -> P is a tautology yes

although your reasoning for why seems circular tbf

[tawny trout]

yeah, i agree. I wasnt sure if the reasoning was correct. Thank you so much!

[vale cairn]

np

[daring beacon]

can someone explain to me how p and r turned to TRUE here

[waxen nest]

apply demorgan's

[daring beacon]

I don't understand wym cuz I don't see demorgans turning anything to TRUE here

[weary tiger]

No it's called distribution

They applied that law to it

Also, $P\lor{\neg{P}}$ is a tautology which is TRUE

[vital dew]

jswatj

[daring beacon]

Thanks @weary tiger

[daring beacon]

can someone tell me the name of this specific rule

[waxen nest]

yea its called implication

implication law/rule

[daring beacon]

where would you go next from here to prove this is a tautology: (p and r) or ((not p and not q) or (r or q)) ?

[weary tiger]

well

im thinking

[daring beacon]

using substitution

[weary tiger]

you can manipulate like ((p and r) or (not p and not q)) or (r or q)

uhhhh

[daring beacon]

I'm sorry

It's p and not r

in the beginning

[weary tiger]

ok

yeah i can't think rn

[daring beacon]

It's fine thank you for trying

[subtle charm]

Is it possible to simplify the proposition further? ~(a V b V c) V (~c ^ (a V b)) V c

[stray reef]

pretty sure it is

,w simplify not(a or b or c) or (not c and (a or b)) or c

[vital dew]

Results provided by WolframAlpha

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

Requested by Ann#0413

[stray reef]

lmao guess it's a tautology

[grand yoke]

I am having confusion while understanding diff. btw universal statements and existential statements

[stray reef]

"all geese have three feet" vs. "there is a three-footed goose somewhere out there"

[grand yoke]

the first is universal, though it is false statement but the property of having three feet is not there in any goose in a universal set of all geese. am i correct?

[stray reef]

...you are overthinking it but yes the first one is universal

[subtle charm]

How do I go about simplifying it? After I apply de morgan and distributive law, I am stuck with (~a ^ ~b ^ ~c ) V ( ~c ^ a ) V (~c ^ b ) V c

[stray reef]

i'd draw a k-map or something

or a truth table

or do you want to do it symbolically

[subtle charm]

I have to do it symbolically

I have not yet learn what is a k map

[grand yoke]

@stray reef i am kind of getting it

[stray reef]

you can factor out ~c from all but the last term to get (~c & [(~a&~b) v a v b]) v c @subtle charm

[subtle charm]

Ok thanks, I will try it out.

[grand yoke]

universal statements can be converted in conditional statements by rewriting them?

mr. @stray reef ?

[stray reef]

would you call a woman "mr."?

[grand yoke]

oh i didnt check it sorry miss

[stray reef]

anyway yes

[grand yoke]

So if i say that 'all houses are red'. (and lets say in some parallel universe this statement is true) 'if theres a house, then it is red' would mean the same are the first universal statement?

[stray reef]

oh, no. not like that.

"all houses are red" as a conditional would become "if it's a house, it's red"

or "if x is a house then x is red" or whatever

[grand yoke]

hmm

[snow sleet]

"the property of having three feet is not there in any goose in a universal set of all geese" is a confusing statement that can be interpreted in more ways (and perhaps other ways than you meant)...try instead thinking about "there is a three-footed goose" by simply thinking "a three-footed goose exists".

[snow sleet]

this is because ANY house you're given, you are guaranteed that THAT given house is red because "all houses (including the one you're given) are red".

[livid drum]

In spirit they would be the same.
However, in the formalism specifying the languages, their syntax might be different.

These two sentences suggest two separate first order languages.
The first quantifies over all houses and has unary predicate R (red).
The second quantifies over all buildings and has two unary predicates: H and R (house and red).

The first sentence symbolically is: "for all x, R(x)";
the second is: "for all x, H(x) implies R(x)".

The "equivalence" between these two sentences is not the usual logical equivalence within the formal system.
These two sentences are stretched over two languages; the "equivalence" is a meta-mathematical one.
Surely, if the first sentence is the only axiom of the first language, and the second the second, any proof of "for all x, A(x)" in the first language can be translated to a proof of "for all x, H(x) implies A(x)" in the second, and vice versa.

[grand yoke]

oh yes 'if theres a house' says abt the existence of house

but there is no need to say that

i am getting it hmm

[subtle charm]

Given P is a class of boys and girls,
Boy(x): x is a boy
Girl(x): x is a girl
Soccer(x): x plays soccer

Can I write the statement; Only boys player soccer

$\forall$x$\in$P (Boys(x) $\Rightarrow$ Soccer(x))

Is this correct? or Do I have to include the girls -> ~Soccer(x))

[waxen nest]

whats P

[vital dew]

Pepega0918

[subtle charm]

P is a class of boys and girls

sorry abt that

[waxen nest]

so x is boy

and x is also girl

wut

so many x haha

[subtle charm]

i mean P is the set of class with boys and girls

[waxen nest]

yea

how do you define the boys and girls respectively?

[subtle charm]

I think the predicate Boy(x) define if x is a boy or girl?

if Boy(x) is true then x is a boy?

[grave sluice]

the statement that only boys play soccer would be the other way around

[vital dew]

Mic

[subtle charm]

hmm why won't Boy(x) -> soccer(x) work?

I am quite confused

[grave sluice]

that would mean that every boy plays soccer

[cerulean wind]

every boy plays soccer is not equivalent to the only soccer players are boys

[subtle charm]

Ah okay. I get it.

[cerulean wind]

every boy on the planet could play soccer, but, for sake of example, your mom could still play soccer with you, and it wouldn't break the statement that every boy plays soccer

[weary tiger]

how to prove this with induction?

[faint narwhal]

If all the vertices had different degrees, what must the degrees of the vertices be? @weary tiger

[weary tiger]

0 to n-1 ryt ?

?

[faint narwhal]

Right

So there are a couple ways to do this, but by induction, you can think about what happens if you remove the vertex with degree n - 1

[weary tiger]

what if there isn't any vertex with degree n-1

[stray reef]

then there's something else you can apply

[weary tiger]

oh okay you assumed every vertices have different degree

[waxen nest]

whats pmi

[weary tiger]

pmi -> principle of mathematical induction

[waxen nest]

oh okay

[weary tiger]

like?

[stray reef]

pigeonhole principle

[weary tiger]

no i dont want to use that

with that it is easy

i just want to do it by induction

[subtle charm]

I know that (-1)^2k = 1 where 2k is an even number. Just wondering is there any theorem I can quote to support it?

[stray reef]

why do you need a theorem

it's obvious immediately from [(-1)^2]^k = 1^k = 1

[subtle charm]

or this will be okay?

Well my school demands it for every step

[stray reef]

can you show the exact thing demanded of you

[subtle charm]

actually, i think this will be more appropriate since -1 raised to any even power is essentially -1 x -1 2k times where 2k is any even number

[subtle charm]

Thanks for the help, but I want to try it on my own first.

Appreciate it.

[stray reef]

are you like

working directly with field axioms rn or something

that would explain why you would be demanded to obsess over the exact reasoning for each step

[subtle charm]

yes haha

[waxen nest]

hi can i check if i can apply distributive on this

in particular this part

[waxen nest]

<@&286206848099549185>

ive tried doing it but im not sure if its correct

can someone help to check?

[iron marsh]

Oh, pretty sure this is just B'

[vital dew]

dackid

dackid

[iron marsh]

Hope this helps @waxen nest

[waxen nest]

what rule in set theory is this?

[vital dew]

dackid

[waxen nest]

A U B = A??

or A U B = B?

[iron marsh]

As long as B is a subset of A

That is important

[iron marsh]

This one, not the second one

[waxen nest]

hmm why does A U B' = B' then

[iron marsh]

Wait, I blundered. The first part is right, but the claim you just made (which I also made) was flawed

[waxen nest]

how do you know B is a subset of A?

[iron marsh]

So A U B' is where it stops

[waxen nest]

so A U B' doesnt simplify to anything?

this is the final ans?

[iron marsh]

No, I made an error here.

[iron marsh]

But this part is right

AUB' is the final answer

[waxen nest]

ok thanks

wait this is weird

[waxen nest]

i tried to plug in this into some online calculator to solve and its correct. it gives AUB'

but if i plug in the original qn, the final ans calculator give is A' U B'

https://www.symbolab.com/solver/set-theory-calculator/\left(\left(A\cap B^{c} \right)\cup A^{c}\right)%5Ccup%5Cleft(%5Cleft(B%5E%7Bc%7D%5Ccap%20A%5Cright)%5Ccup%5Cleft(B%5E%7Bc%7D%5Ccap%20%20B%5Cright)%5Cright)

now im wondering is theres a mistake in my previous working?? or the calculator is wrong

[pale epoch]

you lost the complement on A when you applied the distributive law first

right here

@waxen nest

everything else is fine

[daring beacon]

where would you go next from here to prove this is a tautology: (p and not r) or ((not p and not q) or (r or q)) ?

[pale epoch]

just consider cases

(r or q) is already most of the possible cases, so ...

[daring beacon]

I don't get what you mean

[pale epoch]

there is a finite number of cases to check if this is a tautology

i.e. construct a truth table

if you're smart about it, you can skip most of that by thinking about what cases (r or q) already covers

[daring beacon]

I'm suppose to use substitution only tho

ohterwise I would have already used a table

[outer hinge]

We’re doing things with strings and the Cartesian product and I’m so confused

[weary tiger]

well, what are you confused about

[outer hinge]

I don’t even know what they are and how they relate to Cartesian products or relations

[weary tiger]

Then, you should probably take a look at Cartesian products and relations on their own, there are lots of books on it

[split drum]

Logically, d (the one with a red dot), is true right?

If the premise is false, but the conclusion is true, then the if-then is true?

[pale epoch]

yes

[snow sleet]

vacuously so, so yes it's true

[split drum]

Nice, thanks guys.

[iron marsh]

Gonna be taking this in fall. Any resources you guys recommend to be prepared.

[snow sleet]

We used this text (http://www.ams.org/books/text/041/text041-endmatter.pdf) in our course. I really liked the text, some people didn't since the back of the book rarely gave answers but rather hints which weren't always helpful. There's the link for the free pdf of PART of our textbook... you can get the full text from the AMS bookstore. It's a good text to have if you're into collecting math texts.

Here's the link for book on the ams website https://bookstore.ams.org/text-41/

[iron marsh]

Thanks

[snow sleet]

https://tenor.com/view/michael-scott-youre-welcome-wink-steve-carrell-gif-14492653

[snow sleet]

This channel is open for questions

[oak notch]

Quick question, can someone explain what a pair of arbitrary vertices is? How is it different to an edge when writing it?
Also, what does u,v ∈ V(G) mean in:
Let G be a grapth and let u,v ∈ V(G). u and v are connected if there is a u-v path in G

[faint narwhal]

@oak notch it's just any two vertices in your graph

Saying that uv is an edge means that u and v are connected by an edge, but here u and v don't have to be connected by an edge

[oak notch]

What's the difference in writing uv and u,v?

One is an edge and the other is an arbitrary pair of vertices?

[faint narwhal]

I mean, these things don't really mean anything by themselves

It's only when you say uv is an edge, or u,v are vertices in G, that these things mean something

[oak notch]

Okay then. What if I write: "Let u,v ∈ E(G)." What does that mean? Is the syntax correct?

[faint narwhal]

Mmm

I think people would know what you meant if you wrote this

But people just tend to write uv or maybe (u,v)

It kind of boils down to how you've defined E(G) to begin with

[oak notch]

I see. How would you write it? "uv", "(u,v)", or "u,v"? Do the formats mean the same thing?

Like "Let (u,v) ∈ E(G)." --- "Let uv ∈ E(G)." --- "Let u,v ∈ E(G)."

All these 3 mean that it is an element in the edge set of G?

[faint narwhal]

I guess your question comes down to how we define a graph and how we define V(G) and E(G)

[oak notch]

Yeah

[faint narwhal]

So Wikipedia defines it in this way

For them, edges are just sets of two vertices, and so they would write {u,v} in E(G)

For others, especially if you care about directed graphs, want edges to be ordered pairs, since sets don't care about ordering so you can't say an edge points a certain way

In that case, if your edges are ordered pairs, you would write (u,v) in E(G)

However, people get lazy and just start writing uv in E(G) when it's understood that they really mean {u,v} or (u,v)

[oak notch]

This clears it up

[faint narwhal]

No one writes u,v in E(G), not that I've seen

[oak notch]

Does u,v in V(G) usually mean that it's a pair of arbitrary vertices in V(G)?

Or does that depend on the context

[faint narwhal]

Yes

Also technically, writing u,v in V(G) means that u and v can be equal unless stated otherwise

[oak notch]

I see. Thanks

I'm going on to a different topic here, but what about this one?
"Hey guys, this is about the characterization theorem in Graph theory. Can anyone please explain to my why {x,y} should not be equal to {u,v}? I've been watching the lecture my teacher provided but I still can't understand it. He also mentioned about intersection being empty or 1, but can't be 2. Can someone also explain the intersection part, please?"

Case 1: {x,y} =/= {u,v}
  > u =/= x and u =/= y

I asked that question yesterday, but no one was able to answer it

[waxen nest]

okay thanks for pointing that out

can someone check is this correct?

[faint narwhal]

So does your professor never address the case where u = x and v = y?

[oak notch]

Well, he just described that u cannot be x and y (or v cannot be x and y)

[faint narwhal]

Yeah I'm confused as to why your professor said that

[oak notch]

Then after that, we considered G - u.
"Since x,y ∈ V(G - u), then x and y are connected"

[faint narwhal]

yeah weird

you definitely need to consider the case where x = u and y = v

[oak notch]

Oh

After that, he considered that case

What I'm asking is, can you explain case 1?

I think if I get case 1, I can get case 2

[faint narwhal]

What don't you understand about case 1?

[oak notch]

Why can't u = x or u = y

Or like why {x,y} should not be equal to {u,v}

[faint narwhal]

what

It can

The point is that you're splitting it into two cases

And these two cases cover all the possibilities

Maybe as an example, if you wanted to prove something for all integers. You could split it into two cases. Case 1, the integer is odd, case 2, the integer is even

If you prove both of these cases, then you've proven it for all integers

[oak notch]

Hmmm. I guess what I'm trying to say is that I don't understand how to prove case 1 in the characterization theorem

[faint narwhal]

Oh

Yeah, so like they say x,y in V(G - u)

since V(G - u) is connected, there's some path between x and y

but this gives you a path between x and y in G as well

[oak notch]

So in case 1, u should not be equal to x and y and v should not be equal to x and y because we will be deleting u/v?

[faint narwhal]

yeah, this argument doesn't work unless x,y is in V(G - u) or V(G - v)

[oak notch]

Ooh

Ah, it makes sense now. Thanks for answering everything Zoph

Oh wait, one last, can you also explain the second case?

[faint narwhal]

What's the second case

[oak notch]

Case 2: {x,y} = {u,v} (x = u, y = v)

Then P1 and P2 were concatenated because it had a common vertex which was w.

[faint narwhal]

Right

What are you confused about?

[oak notch]

Why are we supposed to concatenate G -u and G- v though?

[faint narwhal]

Huh? We're concatenating two different paths to give us a path from u to v

[oak notch]

WHy didn't we do the same for case 1?

[faint narwhal]

Because there was no need to

[oak notch]

Ah cause u and v aren't x and y

Is case 2 something like transitivity in equivalence relations?

[faint narwhal]

I mean, being connected is an equivalence relation on the set of vertices yes

So yeah, you can think of it that wag

[oak notch]

I see

Okay then. That made things a bit clearer

Thanks for answering everything Zoph

[waxen nest]

hi anyone can tell me the ans for these qns?

[last timber]

no we do not just give out answers

also @waxen nest for what is this

[waxen nest]

what is this?

this is math

[last timber]

yes i mean where is question from

[waxen nest]

my school past year paper

im practicing for my exam TOMORROW!

My answer is:
a) private key is (35, 65)
b) encrypted ciphertext for 15 is 20.
c) decrypted original message for ciphertext 5 is 60.

i would appreciate if anyone could help to verify these answers 🙏

my workings are abit too long so i shall not post it here to flood the chat

[sharp ledge]

Let f : (−2, 3) → R defined by f(x) = x^2
Determine the image of f. Is f injective?
Is f surjective? Is f bijective?

[sharp ledge]

Will the image be a range ? since the x is from -2 to 3

[weary tiger]

How different is discrete maths from normal maths

?

I just started learning it

And give me any tips or strategy for it

Pls :)

Thank u

[sour arrow]

Why isn't discrete normal? What did it do to become a misfit?

[weary tiger]

Ok discrete maths vs calculas etc etc

[pale epoch]

it might serve as an introduction to proof-based mathematics

my tip is to do the homework

[waxen nest]

hi can help me?

[weary tiger]

What is logical equivalence?

(I've watched a few vids and read few sites i can't understand)

[weary tiger]

...

[weary tiger]

...

[weary tiger]

😂😂😂😂😂😂😂😂

[weary tiger]

What does it mean p biconditional operator q is a tautology

[desert edge]

Could I get some help with understanding solving logical propositions exercises

we are using some notations im not familiar with and my professor basically skimmed through it

I understand up to this point

And I know this is the solution.
I also see the rules applied but I dont quite get how

[snow sleet]

Then image is the range, yes

[snow sleet]

Plug in x=-1.5, then plug in x=1.5. What do you notice about f(x) for each of these values of x? This is for the part that determines whether or not f is injective. For surjectivity, find a real number that f does hit...in other words find a real number that never equals x^2 for any x in (-2,3).

[snow sleet]

It means that p happens exactly when q happens.

[weary tiger]

What does it mean here by

Taking p as common

And why put 1 there

[weary tiger]

?

[sour arrow]

@weary tiger
Two statements are logically equivalent when their truth table matches.

[weary tiger]

Yea thanks

[sour arrow]

Wait, that proof is really messed up. That from the book?

Oh they're proving the absorption law and failing lol

[snow sleet]

[p v (p ^ q)] is equivalent to [(p^1) v (p ^ q)], which is equivalent to [p ^ (1 v q)]. (1 stands for 'true' there btw)

[p ^ (1 v q)] is then equivalent to [p ^ 1], which is equivalent to p.

(1 v q) is equivalent to 1 btw since (1 v q) will always be true.

@weary tiger see above^

[snow sleet]

they're not failing lmaoo. Their proof is solid

it's just a little hard to see the jump from [p v (p ^ q)] to [p ^ (1 v q)] without noticing [p v (p ^ q)] is equivalent to [(p^1) v (p ^ q)]

[weary tiger]

Hmmmmmmmm thanks

[snow sleet]

You're welcome!

[vital dew]

A Kid Named Galois

\begin{align} 
\neg{(\exists{x}\forall{y}[p(y)&\implies{\forall{zq(z)}])}}
\\ \forall{x}\exists{y}\neg{[p(y)&\implies{\forall{zq(z)}]}}
\\ \forall{x}\exists{y}\neg{[\neg{p(y)}&\lor{\forall{zq(z)}]}}
\\ \forall{x}\exists{y}[p(y)&\land{\exists{z\neg{q(z)}}]}
\end{align}
```Compilation error:```! Missing } inserted.
<inserted text> 
                }
l.60 \end{align}
                
I've put in what seems to be necessary to fix
the current column of the current alignment.
Try to go on, since this might almost work.```

[tender hearth]

I'm uncertain about negating nested quantifiers. Are my last two steps incorrect? I'm trying to negate it. pls help me

\begin{align}
\neg{(\exists{x}\forall{y}[p(y)&\implies{\forall{zq(z)}])}}
\ \forall{x}\exists{y}\neg{[p(y)&\implies{\forall{zq(z)}]}}
\ \forall{x}\exists{y}\neg{[\neg{p(y)}&\lor{\forall{zq(z)}]}}
\ \forall{x}\exists{y}[p(y)&\land{\exists{z\neg{q(z)}}]}
\end{align}

[vital dew]

A Kid Named Galois
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[tender hearth]

<@&286206848099549185> pls help me ^

[pale epoch]

looks good

[tender hearth]

Thank you!!

[daring beacon]

can someone explain to me why only T implies F == F and not also F implies T:
for instance, if I use the example: if police does not see me speeding (F) and I get a speeding ticket(T) should be false right ?
unless the policeman made a mistake ?

[snow sleet]

(True -> False) is by definition False since (True -> False) is equivalent to [~True v False], which is equivalent to (False v False).

(False -> True) is true (vacuously)

use the fact that (p -> q) is equivalent to (~p v q) to see this

[kind hollow]

is this class really hard? Im taking it with Calculus 1 at the same time

[snow sleet]

it's hard for some, but quite natural for others. I thought the course was pretty easy.

[vivid locust]

Use p->q ≈ ~p^q and expand each

[waxen nest]

okay

[waxen nest]

isnt it or instead of and

[snow sleet]

(p->q) is equivalent to (~p v q)

[snow sleet]

yes

[kind hollow]

I have no choice, just needed someone’s two cents. Thanks

[snow sleet]

You're welcome lmao

[vivid locust]

Oops my bad

[snow sleet]

all good it's a simple error any of us could've made

[faint narwhal]

Prove that k = 3 first

[vital dew]

Abacus Paradox

Abacus Paradox

Abacus Paradox

[faint narwhal]

Yep

[lament copper]

hello im kinda having trouble understand what is being asked here I did the truth table though

i really dont understand what im being asked

like prove why the last colomb answers are true or false or?

yea

like this ?

or using laws?

sorry for the spam

this would be idempotent laws

would that be it?

So pretty much step by step and come to the solution of it being biconditional?

this is all the hw

i dont know either

[devout root]

Is there an identity like $\binom{n}{2} + n = \binom{n+1}{2}$ or maybe someone can offer an idea at a proof.

[vital dew]

jm

[faint narwhal]

Just write out both sides and do some algebra @devout root

[devout root]

I know.. but I'm lazy :).. thanks ❤️

[dire wren]

Just finished a proof for the start of my Combinatorics class (Pigeonhole Principle). It's a bit late, would appreciate a second pair of eyes to check for any errors. Thanks. Attaching below

[faint narwhal]

Yeah seems good

I think I'd justify a little more why

The thing with zeroes at the end being divisible by 1969 implies that the thing without zeroes at the end is also divisible by 1969

[dire wren]

Yeah that makes sense

Thanks ^^

[waxen nest]

any idea how i can proof this

[devout root]

Ok.. that wasn't terrible.

[snow sleet]

this can be done in a slightly simpler fashion since $\frac{n(n-1)+2n}{2}=\frac{n(n-1+2)}{2}=\frac{n(n+1)}{2}$

[vital dew]

logician

[waxen nest]

hey i need help with this

[weary tiger]

@snow sleet do u happen to know asymptotic functions at all? like big O or omega?

[snow sleet]

uhhh I meannn I know aymptotic functions from calc...I'm not sure what you mean by "big O or omega"

[weary tiger]

big O notation

more like theory of computation

algorithm analysis

[snow sleet]

oh

yea no

sorry

[faint narwhal]

What do you need help with?

@weary tiger

[vital dew]

Pepega0918

[weary tiger]

its this question

[subtle charm]

Sorry about that

go ahead

[faint narwhal]

Yeah, there's not much more to this than just writing out the definitions

[weary tiger]

yeah it seems to be that way, but my proof doesn't really make note of some variable that they want me to

pepega can go

[subtle charm]

Hi @weary tiger if you are okay, can I post my question?

[weary tiger]

ill ask about it later

[subtle charm]

okay

Let P be a set of athletes in the Olympics
Soccer(x): x is a soccer player
Swimmer(x): x is a swimmer
talk(x,y): x talked to y
How do I go about writing the statement that every soccer player talk to a swimmer during the Olympic?
My solution is this:

$\exists$x$\in$P(Swimmer(x) $\land$ [$\forall$y$\in$p(Soccer(x) $\Rightarrow$ talks(x,y)])

[vital dew]

Pepega0918

[subtle charm]

Is my understanding right?

[faint narwhal]

No this isn't correct. In words, what you've written down is that there exists a swimmer that every soccer talked to

Do you see why?

[subtle charm]

Initially, I wrote it in a different manner. Let me typed it out.

[cerulean wind]

thought they wrote "there exists a swimmer in P and if that swimmer is also a soccer player, then they have talked to every person in P"

[faint narwhal]

Ah you're right

[cerulean wind]

@subtle charm you should start your solution by writing for all x in P, if swimmer(x), then ...
(roughly, i dont know what you are and aren't allow to quantify over)

[subtle charm]

$\forall$x$\in$P(Soccer(x) $\Rightarrow$ [$\exists$y$\in$P(Swimmer(y) $\land$ talks(x,y))]

[vital dew]

Pepega0918

[subtle charm]

not sure if this is right?

but i think this is wrong

since if the hypothesis is false then the statement is vacuously true?

[cerulean wind]

yea that looks right now. for every person x in P, if x plays soccer, then there is a person y in P that is a swimmer who talked to x

[subtle charm]

This is my original solution but I kept thinking that if Soccer(x) is false then won't the statement be true by default?

[cerulean wind]

if there is no witness x to soccer(x), then the statement is vacuously true.
like, if for some reason, soccer got cancelled and there were no soccer players, then it would be vacuously true... sort of

[subtle charm]

so if x is not a soccer player then does it mean that x also talks to a swimmer?

$\forall$x$\in$P([$\exists$y$\in$P(Swimmer(y) $\land$ talks(x,y)] $\Rightarrow$ Soccer(x) )

[vital dew]

Pepega0918

[subtle charm]

does this even makes sense?

[cerulean wind]

no. lets say x1 is Usain Bolt and x2 is Simone Biles, and Usain Bolt talks to Katie Ledecky but Simone hates swimmers and doesnt talk to any of them.

soccer(x1) and soccer(x2) are both not satisfied, but x1 talked to a swimmer and x2 didnt talk to any swimmers.

it could go either way (unless you know more information)

[cerulean wind]

this says if a person talked to a swimmer, then they play soccer

[subtle charm]

Yeah, I think I get the idea.

[minor pumice]

yo is this proof by induction?

if so then what is the induction hypothesis?

assume true for some prime k > 3 in positive integers

stuck after that

[civic horizon]

A direct proof is the way to go

[minor pumice]

I only know how to do induction and counter example

can u give the first step or something

do I just do let p = 5 and q = 2p + 1

[civic horizon]

Go case by case

[minor pumice]

p is prime and q = 11 is prime

[civic horizon]

What if p = 0 mod 3

Yeah so the statement is true

That is not a counterexample

Thats just an example

[minor pumice]

?

[minor pumice]

are you trying to get me to do proof by counterexample?

[civic horizon]

No

[minor pumice]

wdym

[civic horizon]

I want you to consider cases

So we show that it is impossible for p to be 0 or 1 mod 3

Which means it has to be 2 mod 3

[minor pumice]

oh I get it

[subtle charm]

Hi, can I check if the -> used in propositional logic translate to subset in set theory?

if so, is the statement below true/logically equivalent?

A $\cup$ B $\rightarrow$ A $\equiv$ A $\cup$ B $\subseteq$ A

[vital dew]

Pepega0918

[subtle charm]

I'll check back again in the morning so thanks in advance to whoever answers this!

[slate tree]

is x^3 *(log(x)) the same as log(x^4)?

I know 2x^2 + x^3 logx is O(x^4) but I'm trying to understand it better

[weary tiger]

no

log(x^4) = 4logx

[slate tree]

hmm okay

is it because with an exponent outside the log it makes the log grow quicker than it would so the n goes up? I think I remember something about that.

[weary tiger]

no clue what you mean tbh

[slate tree]

idk originally I thought that problem would be O(x^3)

[weary tiger]

its O(x^3 logx)

[slate tree]

the book says x^4

[weary tiger]

It is also O(x^4)

It is also O(x^12312731273)

[slate tree]

oh bc the problem is asking for the least integer n for (x^n)

[lofty night]

So I’m new to discrete math and I was wondering if someone could help me with truth tables

And converses

[simple nova]

https://i.imgur.com/1YRh6mc.png

Getting stuck on this one since it has a biconditonal

[tawny walrus]

Watcha need help with specifically? Basically the rundown is, for every given letter in a truth (typically, p, q, r, etc), you assign it a truth value. So basically if you have p and q, you will have your truth table for those two letters be TT, TF, FT, FF. Then solve for the and column, so it would be true when p and q = TT respectively, and false for all other combinations. and so forth. I know it's a bit of a clunky answer, but if you have any specific questions I can help

[lofty night]

So I used that and i thank you for your help but I need to find the converse of the statement “if the Eagles win the Super Bowl that I am the president”

How would you do that?

[tawny walrus]

So lets convert your current form to p --> q. So now a converse if basically flipping the p and the q, so your new sentence would be "If I am the president, then the eagles win the super bowl", hence your new form would be q --> p

[lofty night]

So the converse is the opposite?

That makes sense

Thank you

So I now know how to make a truth table but how do you prove something using a truth table? The question is p → q ≡ q∨ ∼ p

Would it be done the same way?

[tawny walrus]

Yes. So create a truth table for p → q, then either in a separate truth table or the same one do q ∨ ~p. Then in the case of a truth table, if the T an F values are equal (going up and down, which I believe are columns, my English isn't too great even though I'm a native speaker) then they are logically equivalent

Heres a truth table I've solved out before, you can see where I drew the two arrows at the bottom and put the logically equivalent sign, that is an example of if a statement is logically equivalent

[lofty night]

So just make two truth tables and see if they match?

[tawny walrus]

Yea, or do it in the same one like I did. Saves you from having to do the p q and r columns again

But also, the whole truth table doesn't have to match, just simply the columns with p --> q, and q or ~p

[lofty night]

Okay so I’d look like this

Right?

[tawny walrus]

Yes, so then the two statements are logically equivalent

Btw, how do you type if --> symbol, and the and and or symbols? I always have to look them up then copy and paste them😆

[north harness]

"If a quadrilateral is a square, then it is equilateral. A rectangle is not equilateral"

Ik this sounds dumb

but i put

"Therefore a rectangle is not a square"

Which is techically true ig

[tawny walrus]

Lemme think this out

So

p --> q

p being a quadrilateral is a square

q being, p is equilateral

Where does the rectangle come in here

[north harness]

Uh

Its a detatchment law

So

Actually

Um

thats just the question

Lol

[lethal frost]

hello

is this correct?

as an interval notation

[near tide]

Hey there,
I'm wondering if anyone can help me with this.
Rough translation of the text:
The propositional functions F and G are given so that
.....
Find the largest domain M that includes real number pairs (x,y) and fulfils the following
.....

[tawny walrus]

Yep, looks right to me.

[rustic quest]

Having trouble with a formal proof for this one, can someone help me?

[cerulean wind]

hint: ||(p v s) --> r <==> (p --> r) ^ (s --> r) and ~r --> ~s <==> s --> r||

[rustic quest]

The exact part im stuck at is when (p --> r) ^ (s --> r)

ooh

i see now

what

do you call that when (p v s) --> r <==> (p --> r) ^ (s --> r)

im trying to do it with rules of inference

proofs by cases?

thanks

[desert edge]

how exactly do I know when to use the law of excluded middle?

[pale epoch]

what do you mean? you use it every time you do a proof by contradiction

[desert edge]

sorry i just recently got introduced to logic

with discrete math and its troubling me more than I expected it to

i pretty much understand every rule here till classical rules:

we are proving logical propositions with some weird tree notation

[pale epoch]

🤔

[desert edge]

example

Its proved from bottom to top

i didnt understand anything from my teachers explanation

[pale epoch]

i'm afraid i cannot parse this weird notation

[desert edge]

i dont blame you

im having a very hard time finding help cause theres barely anything on it

its not even in my book

[hazy sand]

so I know you can multiply a matrix by a scalar, but is there a similarly understood operation for sets?

[pale epoch]

in general, no

[hazy sand]

if someone were to want to, would it be more appropriate to say that it's not permitted, or is it just not specifically named as an operation on sets?

[pale epoch]

if the elements of your set can be multiplied by scalars, then its easy to extend that notion to sets

but in general, there is nothing you can do that makes sense

[hazy sand]

okay

[pale epoch]

(the word scalar already implies you are talking about vector spaces, so in that case there is a "natural" way)

[hazy sand]

yeah i'm not particularly sure what to refer to a single number as

[pale epoch]

well, how would you multiply the set {green, blue, banana} by 3?

[hazy sand]

ah yeah ok

for the sake of discussion, suppose it's a set of numbers

[pale epoch]

well, then there is a notion of multiplication

and you would define $x\cdot A$ as the set ${x\cdot a \mid a \in A}$

[vital dew]

Lochverstärker

[hazy sand]

ah okay

[pale epoch]

for some set of numbers A

in most cases anyway, thats what people would assume unless you mention something different

[hazy sand]

for some set whose elements are real numbers, if that set were multiplied by zero, the result would just be the single-element set {0}, right?

[pale epoch]

ye

[hazy sand]

mk

and generally there's no mathematical operation that should increase a set's size

[pale epoch]

why not?

the powerset does this

[vale cairn]

what you're currently talking about has very little to do with set size (number of elements)

[pale epoch]

or just the operation "add an element that was previously not in the set" increases a sets size (as long as it was finite)

[hazy sand]

i suppose i meant traditional mathematics problems, such as addition and multiplication

i suppose i should just ask outright the problem I'm addressing:

I noticed some classmates having a discussion involving division by zero, and (with a bit of my own refinements, I must admit; some concepts they expressed were wrong enough to disregard the discussion outright,) they seem to have come to the conclusion that ℝ * 0 = {0} ⇒ {0}/0 = ℝ

what'd be an efficient way to disprove this, assuming they're only just now taking Calculus II

[pale epoch]

division by 0 in a field (like R) is undefined

this situation does not improve when you try to divide whole sets of real numbers

[hazy sand]

yeah the problem is they've come to the conclusion that UNDEF = ℝ by combining {n : n ∈ ℝ}/0 = UNDEF with {0}/0 = ℝ

[pale epoch]

{0}/0 is just a collection of symbols

it has no mathematical meaning

[hazy sand]

well they used 0/0 more specifically, I had changed it to make multiplying a set by a number result in a set

[pale epoch]

0/0 also has no mathematical meaning

[hazy sand]

that'd be what i'm trying to convince them of

[pale epoch]

well

thats a bit more technical

generally we define division by multiplication of inverses

the (multiplicative) inverse $a^{-1}$ of a number $a$ is defined as the (unique) number such that $a \cdot a^{-1} = 1$

[vital dew]

Lochverstärker

Lochverstärker

[pale epoch]

if now $0$ has a multiplicative inverse $0^{-1}$, then $0 \cdot 0^{-1} = 1$, but then also $a = a\cdot 1 = a \cdot (0 \cdot 0^{-1}) = (a \cdot 0^{-1}) \cdot 0$ by associativity and commutativity and finally $(a \cdot 0^{-1}) \cdot 0 = 0$ and thus $a = 0$

[vital dew]

Lochverstärker

[pale epoch]

which is kinda bad

actually there are probably easier contradictions if you allow division by 0

[hazy sand]

yeah like the 0=1 'proof'

[pale epoch]

this

yeah, so

if you allow division by 0 you will have to give up many other rules of arithmetic

[gloomy beacon]

How do I apply that coefficient formula on here?

[finite sapphire]

Is that discrete math?

[gloomy beacon]

my bad

[pale epoch]

this is discrete, its essentially a counting problem

what have you tried? whats the coefficient formula?

[gloomy beacon]

https://www.gstatic.com/education/formulas2/355397047/en/binomial_theorem.svg

It would be this one right?

[pale epoch]

that would be my approach, yes

its called the binomial theorem/formula though

[gloomy beacon]

ah right so the problem for me is the x on the outside of the brackets

[pale epoch]

well the "x" in the image is your 5x^{-2} and the "y" in the image is your -7x^3

[gloomy beacon]

Ohh so would I just exclude the x that isn't part of the brackets?

[pale epoch]

oh

sorry i misread

so if you expand the (a-b)^11 thing

you get a sum

and multiply this whole thing by x

now what does that do to the exponent of x of the summands?

[gloomy beacon]

+1?

[pale epoch]

ye

[gloomy beacon]

the power would go up right?

ah

[pale epoch]

yes, by 1

[gloomy beacon]

So I use the formula with the powers just increased by 1

?

[pale epoch]

you can ignore the x in front of the brackets and find the coefficient of x^13 instead

since multiplying by x will turn the x^13 into x^14

[gloomy beacon]

oh right

damn thanks that's perfect

cheers for the help

[velvet gate]

Hey can someone help me with this question

I want to know if im doing it right

or wrong

so far

[zealous crane]

Is there any chance anyone with some decent discrete knowledge would be able to hop in a vc for a little bit and help me out with something, I'm somewhat struggling

[agile magnet]

just send your question here

[zealous crane]

It's these two problems that I'm struggling with

For #3 I'm just confused on even how to start because I'm still unsure of how to get r when it's not in either of the given bits of information

And for #4 I think it's false, but I'm still just a touch confused as to how to get there, here is my work so far though

[hidden totem]

so obviously this is a simplified version of the faulty proof with a=b, it's using the same concept of factoring difference of squares

however, if we're gonna simplify it to the point where we're just using regular numbers, then this is very roundabout way to prove that dividing by zero breaks arithmetic

0 = 0
0(1) = 0(2)

divide zero:
1 = 2

[floral lily]

r = a student read 25 books or less this year

when it says "read 30 books this year" how can i show that its more than 25, to me the ¬r would imply the student didnt read 25 books or more than 25 books, how can i properly display this?

[dry oar]

hey, would this be a false quantifer?

[snow sleet]

uh

I'm very confused about the notation being used there

looks like someone tried to use set builder notation

while trying to incorporate predicate logic

[dry oar]

i'm confused as well 😓

[rich siren]

is this transitive because technically if a can be both x/y while also being z, then

x(a),y(b) - y(b),z(a) - x(a),z(a)
x(b),y(a) - y(a),z(a) - x(b),z(a)

Is there an easy way to disprove/prove transitive properties?

or do you just have to basically map everything and try to find one that doesn't fit
(x,y), (y,z), (x,z)

[faint narwhal]

u basically have to try everything

Remember that a relation being transitive means that every x,y,z makes it work, not that there exists x,y,z that makes it work

[rich siren]

So every time both a x,y and y,z relation exist, a x,z must exist

I understand it I guess, just figured there'd be a less annoying way to check

[faint narwhal]

heres a dumb way

u can represent a relation as a matrix of 0's and 1's

if u square this matrix and look at the non-zero entries of this squared matrix

all u need to check is that the non-zero entries of this squared matrix only appear in spots where the original matrix was also non-zero

[rich siren]

that's too big brain for my small brain

[waxen nest]

if A = {1} R = {1,1} . is it reflexive, symmetric or transitive?

[rich siren]

so here I could just say

x(a),y(a) and y(a),z(b) but not x(a),(b)

@waxen nest I believe it's all 3
reflexive because all elements relate to each other
symmetric because all elements x(1),y(1) and y(1),x(1)
and transitive because there's only one ordered pair so it can't even break the rule

but im a set nub too

[waxen nest]

ahh right i think it should be all 3 too

[dry oar]

can anyone help me with this question?

[waxen nest]

ohhh @dry oar our profile pic looks similar

[snow sleet]

if p is true, then r is true, and so s is true, and so u is true. If q is true, then t is true, and so u is true. In either case, u is true.

for p v q to be true, at least one of p,q must be true btw

[tall lodge]

Wtf

[snow sleet]

so if p is true, you need modus ponens, and hypothetical syllogism

if q is true, you need modus ponens, and hypothetical syllogism as well

there's a law that says (s v t) -> u is equivalent to [(s-> u) ^ (t-> u)]...I'm forgetting what that law is called but you'll need that law also

all that really matters with equivalences is that you know them...not necessarily that you can name them. But for your case, it sounds like you're at the stage where you must name them....a few years later on, I doubt you'll need to know the names of the laws.

[polar hound]

Hi. Let A be a set. What is A union A and what is A intersection A? Is it just A?

[tough root]

I think it is A

[stray reef]

yes it is just A @polar hound

[polar hound]

Alright thanks

[tough root]

What is a cardinality for set of all quaternions?
Is same as cardinality od all complex numbers and real number?

[faint narwhal]

Yes