discrete-math

[gleaming zephyr]

its really important that word

'not really sure why one value should be able to give two different results'

the definition of a relation is just

a subset of a cartesian product ig

you would have problems with this if ur dealign with af unction

a function

infact a function is just a relation that doesnt allow this not really sure why one value should be able to give two different results

[unreal berry]

well yeah cus it makes no sense as a function

[gleaming zephyr]

yea yea

[unreal berry]

you wouldn't be able to draw it

even a toddler could see that

[gleaming zephyr]

why wouldnt you

what

anywayss

you got it now garllak?

can you write me out an example

of a relation over A

A = {1,2,3}

which is transitive

[unreal berry]

relation over a

[gleaming zephyr]

a relation over A is just from AxA to A

[unreal berry]

I guess R = { (1,1),(1,2),(2,1),(2,2)}

[gleaming zephyr]

what

[unreal berry]

yeah

[gleaming zephyr]

okay

cool

[unreal berry]

so it's ok?

[gleaming zephyr]

yea cool

(1,2) and (2,1) --> (1,1) ig

[unreal berry]

ig?

guess

[gleaming zephyr]

your correct

[unreal berry]

ok

[gleaming zephyr]

anything elwse

what did you not

understand from ann tho

[unreal berry]

not really, I already kind of understand these things, the problem is I can't demonstrate it properly when it comes to the tests

it's only when it's very clear what exactly is happening that i can give you an answer

and I just want to finish this course and move on

[gleaming zephyr]

what course is this

discrete math?

[unreal berry]

yeah

[tawny hollow]

does anyone know finite state automata ?

[unreal berry]

like it's not even knowledge since that would require me to be able to actually apply it

feels more like a poorly formulated IQ test

[tawny hollow]

LOL

[unreal berry]

I have been busy with other courses though

[tawny hollow]

oh ok

[unreal berry]

all exams are like this though

they test you on the one thing you know, or they don't

that's really it

and then it's a question of precision

[tawny hollow]

not sure tbh

[unreal berry]

elaborate

please

It's just it's not really knowledge, it just tests if you can recognize these patterns without actually understanding what they are or used for, memorize an algorithm or just throws you an easy question that you forgot to check and would never be an issue in real practice anyways

[tawny hollow]

oh

[unreal berry]

Neither is worth being entitled 'knowledge'

these are thing good access to data and experience and basic feedback can solve

not true enlightenment

if that is even a thing

[tawny hollow]

[tidal plinth]

@tawny hollow Do you have a question on FA? Just ask

[tawny hollow]

Oh yes let me show you

I'm not sure if this is correct but I'm trying to do (ab)* + (ba)*

@tidal plinth

[tidal plinth]

Is q₀ your start state?

[tawny hollow]

Yes, sorry I forgot to put that initial arrow for it

[tidal plinth]

That's okay. And this is an NFA?

[tawny hollow]

FSA

[tidal plinth]

Deterministic? But you've got multiple arrows with the same label going out of each state

[tawny hollow]

Oh

I keep hearing DFA and nfa

Tho my lecturer only covered FSA

[tidal plinth]

Both are finite state automata (and are actually equivalent in power). They're just different models

[tawny hollow]

Oh ok

[tidal plinth]

In a deterministic finite automaton, each state must have exactly one leaving/outgoing arrow for each symbol in the alphabet (here, one for a, one for b), so that given a state and a symbol, there is a unique next state

[tawny hollow]

Ahh

[tidal plinth]

Okay, here we want (ab)* + (ba)*. It's either one or the other.
So ababab is allowed, and so is bababa, but not abbaab

[tawny hollow]

Yes

[tidal plinth]

Okay, now suppose I have a word in this language, and I tell you that its first letter is a. Then what is the second letter (and should there be one for sure)?

[tawny hollow]

Ys, sigma {a,b}

[tidal plinth]

No I mean, I have a word in this language: a??????. Do you know what the second letter of this word is (given that the first is a)?

Can a be the second letter?

[tawny hollow]

No

[tidal plinth]

So it must be b

And ab is a valid word, so if your machine reads a first (while it's in the start state), then the next thing it reads must be b, and then that string should be accepted

So you should have
(q₀) -a→ (q₁) -b→ ((q₂))
where q₂ is an accept state.
[Obviously this is incomplete, but you must have this much]

But also, the second letter should not be a, so in q₁, if you get an a, then it should go to another state from which you can never get to an accept state, no matter what later input is received

[Such a state is called a trap state. It's simply a state from which all outgoing arrows lead back to itself. So here, we'll have a trap state with two arrows leaving it, labelled a and b, but both loop back to itself. And this won't be an accept state]

So we also have:
(q₁) -a→ (t) ⊃⊃
(where those ⊃s are my way of drawing two self-loops. Label one with a and one with b)

Do you follow so far?

[obtuse lance]

ghosted, rip

[tawny hollow]

Sorry I'm having my lunch atm

@tidal plinth

Brb

[tidal plinth]

Sure, no problem

[tawny hollow]

right im back @tidal plinth

[tidal plinth]

Okay, so read and tell me if you've understood it so far

[tawny hollow]

i dont understand the trap state bit

are you ok going to vc ?

[tidal plinth]

I don't mind listening, but can't talk right now

[tawny hollow]

ah thats fine

[tidal plinth]

Yeah

Hm, the idea of a trap state [which is just a construction] is to have a state that you can never come back from, if you make a fatal mistake

(Which is possible only in some languages)

All as appear before bs, or all bs appear before as
Did I hear you correctly?

Okay, so the string must be either aa⋯abb⋯b or bb⋯baa⋯a
(with some m as and some n bs)

And what's the regular expression for aa⋯a?

And that should be followed by b*

So a*b*

+ b*a*

Yeah, so a*b* ≠ (ab)*

Nope, they're not equal

No, (ab)* means you can repeat ab zero or more times

So that'd give ε, ab, abab, etc.

Okay, we'll still need a trap state, because it's possible in this language to go wrong (irreparably)

I'll explain

What I mean is: Consider the language of all {a, b}-strings that end in a.
Is it possible, when constructing a string in this language, to make a "fatal" mistake?

No, because you can always just add an a at the end and it'll be an accepted string

So in the automaton for this language, you don't need anything like a trap state.

But consider the language of all {a,b}-strings that begin in a. Is it possible to make a fatal mistake here?

If your string begins with a b, then no matter what you do, you can't undo that and make it begin in an a (in the same string)

So if the automaton for this language sees a b when it's in the start state, it should go to a state from which there is no way to later get to an accept state. Makes sense?

Well a trap state is literally a state from which there is no escape — which means all arrows/transitions lead back to itself (immediately)

So you simply make a state in which all leaving arrows loop back to itself

If the alphabet is {a, b}, a trap state, say t, will look like this:
(t)⊃⊃
[Those ⊃s are badly drawn self-loops. One's labelled by a and the other by b]

Meaning, if you're in the state t, and you see a, then it comes back to t. Same for b

Yes, a double-loop here because the alphabet has two symbols

[tidal plinth]

Yes. (Just add the labels)

Okay, now suppose (for a*b* + b*a*), you're reading a string and you see
aba
Is this a valid string (so far)?

Correct, it's not. It's not in the form aa⋯abb⋯b or bb⋯baa⋯a.

And can it be made into a valid string by adding something to it after this point?

Remember, whatever you add, nothing will get deleted

If the string is supposed to be in the first form, that is, aa⋯abb⋯b, then there shouldn't have been an a after the second b (in aba)

And no matter what you add later, that second a cannot be erased, so you can't force the string into the first form.

And it's obviously not in the second form either, because it begins with an a, not a b

So if you see aba, you know the string is not in the language, no matter what comes later

Which means, you immediately send it to a trap state

[Formerly a part of the wall of text]

[ivory badge]

oh boy now that's a wall of text

[tidal plinth]

Guess I best delete that. It's not very useful for anyone else with the other side of the conversation missing

Discord doesn't make that easy… -_-

[pale epoch]

or you just don't care

[tidal plinth]

I wish there was a multiselect delete

[glass badge]

I think I wil need a hint

[sleek swallow]

Huh why did that individual not reply?

[obtuse lance]

voice chat

[sleek swallow]

ah, i see

[tidal plinth]

One-sided voice chat, since I couldn't talk at the time (nighttime here; didn't wanna disturb the wife)

[tawny hollow]

👌

[narrow notch]

Can anyone help me with this 3.6 (b) here?

Aren't SIY and Y two distinctive normal forms?

[analog sonnet]

To me, you'll have to provide some definitions; in particular, the definition of fixed-point combinator, SI and that equality sign with the beta and w subscripts

[faint narwhal]

His exercise is from this book

So all the definitions you need will be in here too

[analog sonnet]

Ah nice thanks

Always wanted to know a little more more about lambda calculus

[stray reef]

uh

your answer is way too large i can guarantee that

why is your denominator 11!*10!?

[faint narwhal]

there are 11! ways to place the A and G beside each other?

how

why is that true?

This is not an explanation at all of why it's 11!

Are you sure there's 11 ways to place the a and g?

Can you list them for me?

that's only one of them

List them for me

There are no a's or g's in the picture

These are the only ways to place a and g next to each other?

No other ways for the letters a and g to be next to each other?

there we go

[wanton sable]

$$A = {n \in \mathbb{N} : 5n + 1}$$, $$B = { 5n + 1: n \in \mathbb{N} }$$

[vital dew]

SushiHammer:

[wanton sable]

are A and B both the same way of writing a set?

or do they mean different things?

[gleaming zephyr]

A does not make sense

B is ok

[stray reef]

[wanton sable]

[analog sonnet]

@wanton sable what's that show in your pfp

[spice lance]

could anyone show me where i went wrong for #21

the problem asks to show that it is logically equivalent (a word cut out in my picture)

[ivory badge]

So is $\leftrightarrow$ an iff

[vital dew]

Darkrifts:

[spice lance]

yes

[ivory badge]

So

What’s the truth value for $\bot \leftrightarrow \top$

[vital dew]

Darkrifts:

[spice lance]

ive never seen that upsides T before

sorry :p

not sure what it means

[ivory badge]

F iff T

Upside down T for false

[spice lance]

oh its F

i should clarify, it evaulates to F

[ivory badge]

It seems like you put true above

[spice lance]

omg, i confused implication

with iff

im a dummy

i was evaluating as if it was ->

not <->

[ivory badge]

you also fucked up the bottom line I think

[spice lance]

wouldnt doubt it

thank you for pointing it out for me

[ivory badge]

Like on the p iff -q one

[spice lance]

ah okay tyvm

[ivory badge]

But yeah, $\iff$ is not $\to$

[vital dew]

Darkrifts:

Darkrifts:

[spice lance]

ah okay good to know

[spice lance]

how can i detemrine if the compound proposition above is satisfiable

without a truth table

nvm watched a video on it

https://www.youtube.com/watch?v=j4hwg5DuQP0

[wanton sable]

i'm confused, shouldn't the absolute value be a positive value since it represents distance? why is it defined as -x for all x <= 0 ?

[sleek swallow]

'positive value since it represents distance' Eh that's a more geometric way of looking at things. You can interpret it as the distance of a point on the real line from the origin.

If you look closely at that portion of the definition, x is negative or equal to 0. If x = 0, then -x = -0 = 0. If x<0, then -x > 0. Hence, -x is, indeed, positive.

[wanton sable]

ah okay thanks! i can see how it works out with your explanation

[stray reef]

if x is negative then -x is positive @wanton sable

[glossy adder]

thanks ann

[modern hamlet]

So I've been working on this for an hour or so

[vital dew]

yohst:
Compile Error! Click the reaction for details. (You may edit your message)

[modern hamlet]

I did this but I don't think it's right

[vital dew]

yohst:
Compile Error! Click the reaction for details. (You may edit your message)

[faint narwhal]

I think this works actually

At least, for this one direction

[modern hamlet]

this is what my TA told me but I guess I just was confused about it

I did x \notin B for the sake of trying to disprove it? but I feel like if I had done x \in B I also could've disproved it in the way I disproved x \notin B.

[faint narwhal]

how so?

[modern hamlet]

like what if i said like Let x \in B. and then I went since x \in A \setminus B that x \in A and x \notin B but that contradicts our earlier statement that x \in B

so x \notin B is true by contradiction?

or does it not work that way

[faint narwhal]

If you're only assuming that x \in B, how do you know that x \in A \backslash B

[modern hamlet]

oh sorry i guess like w the assume A \ B = B \ A and let x \in A

my bad

[faint narwhal]

but it still doesn't work

if you don't assume that x \notin B, then you can't say that x \in A \ B

[modern hamlet]

why not?

thank u for helping me btw

i really appreciate it

[faint narwhal]

Uh, the definition of A \ B is that x \in A and x \notin B

so unless you assume that x \in B, you can't say that x \in A \ B

[modern hamlet]

but in the case where we assume x \notin B we end up using B \ A where x \in B

does it have something to do with the fact that A\B comes first in that equation A\B = B\A?

[faint narwhal]

no? why would it

A\B = B\A and B\A = A\B contain the exact same information

[modern hamlet]

idk im just grasping at straws

[faint narwhal]

The definition of A \ B is that x \in A and x \notin B

To be able to say that x is in A \ B

you need to know both things

that x \in A and x \notin B

[modern hamlet]

sorry I'm new to proofs so it's just an unfamiliar way of thinking

so say it was B\A = A\B and I use my original answer where I assume x \notin B to make a proof by contradiction. wouldn't I not be able to say that x is in B\A because although x \notin A, x \in B?

mb messed up the end there

[faint narwhal]

how do you know that x \notin A?

[modern hamlet]

bc we assumed that B\A = A\B I guess?

[faint narwhal]

why would that tell you that x \notin A?

[modern hamlet]

uh let me think some time over that

this is hard to process im sorry

that makes more sense but i need to process it

[faint narwhal]

take your time

[modern hamlet]

that makes so much sense

my mind has been blown by this simple fact

thank you so much

[faint narwhal]

no problem

math gets cooler

[modern hamlet]

[faint narwhal]

(and harder)

[modern hamlet]

;-;

[faint narwhal]

(a lot harder)

have fun!

[modern hamlet]

if only i wasn't a math major

[faint narwhal]

you get to experience all the mindblowning coolness at least

[modern hamlet]

lol

[next valley]

And I wish I'm a math major😂

[sleek swallow]

@modern hamlet uh still having trouble with this?

[modern hamlet]

I think I got it

Thanks though!

[vital dew]

presmo:
Compile Error! Click the reaction for details. (You may edit your message)

[sleek swallow]

What do you think?

@spice lance What have you done in order to show that both expressions are equivalent?

[spice lance]

i was wrong it should be

$p\iff \lnot{q} = (p\land{\lnot{q}}) \lor (\lnot{p} \land{q})$

[vital dew]

presmo:

[sleek swallow]

Once again, same question

What have you tried?

If they’re the same, prove it. If they’re not, prove it. You can do it by playing around with the logical function to the right of the equals sign

The way I would do it would be to distribute:

$(p \land \lnot{q}) \lor (\lnot{p} \land q) \iff [(p \land \lnot{q}) \lor \lnot{p}] \land [(p \land \lnot{q}) \lor q]$

[vital dew]

Abhijeet Vats:

[sleek swallow]

So, what can you do with that?

[chilly granite]

I need help with this question

A dec-string d1,…,dn is bad if di=di+1 or di+di+1=9 for at least one i∈{1,…,n−1} and it is good otherwise. What is the number of good dec-strings of length n?

sorry my latex is not very good

[tranquil jewel]

What would b be in this case? (or would this go in proofs-and-logic)

[ivory badge]

x is not necessarily in N

What if you have k as some large number and x= -1

[tranquil jewel]

It would be false then

[ivory badge]

Yes

[tranquil jewel]

So for the case of d, it would not be an assertion though

[ivory badge]

I would say so

[sleek swallow]

Hmm

Okay see, it depends

[sleek swallow]

I can see a situation where d could be an assertion if you were doing a proof by case distinction

[ivory badge]

It’s not an assertion here Vats

[tranquil jewel]

Those are the two examples we were given of not being true/false

[sleek swallow]

But without more context, I suppose d can’t be treated as one here

[tranquil jewel]

alright, also since in e it restricts the x to real numbers it would be false right

[sleek swallow]

sqrt(x^2) = |x|

[tranquil jewel]

ahh so its true then

[sleek swallow]

Is it?

[tranquil jewel]

wait

no

oops

[sleek swallow]

sqrt((-2)^2) = -2, by the assertion. However, sqrt((-2)^2) = |-2| = 2, based on what I’ve written above

[spice lance]

sorry for late reply abhijeet im trying ur method and still stuck

[tranquil jewel]

alr okay

[sleek swallow]

Also, I’d say that for d, you should elaborate on the possibility of it being an assertion in a proof

[tranquil jewel]

How would I do that

[sleek swallow]

@spice lance dafuq, what the fuck are you doing in your first line lol

[spice lance]

demorgans law

[sleek swallow]

You negated the equivalence and equated it to the equivalence

Bruh

Bruh I told ya to distribute

$(p \land \lnot{q}) \lor (\lnot{p} \land q) \iff [(p \land \lnot{q}) \lor \lnot{p}] \land [(p \land \lnot{q}) \lor q]$

[vital dew]

Abhijeet Vats:

[sleek swallow]

Then, distribute again. Notice that you’ll get p or not(p), which is a tautology

So you can just throw it away

Then, all you get in the first bracket is not(q) or not(p), which is basically just p => not(q)

You don’t need to use the De Morgan Laws over here

[spice lance]

ok ill give it a shot

[sleek swallow]

Okay, if you need a proof for it, I can give one to you.

By the biconditional law, you have:

$(p \iff \lnot{q}) \iff [(p \implies \lnot{q}) \land (\lnot{q} \implies p)]$

[vital dew]

Abhijeet Vats:

[sleek swallow]

Now, you can show, rather easily, that the following is true:

$(A \implies B) \iff (\lnot{A} \lor B)$

Where A & B are propositions

[vital dew]

Abhijeet Vats:
Compile Error! Click the reaction for details. (You may edit your message)

[sleek swallow]

We can apply that to what we have above:

$(p \iff \lnot{q}) \iff [(\lnot{p} \lor \lnot{q}) \land (q \lor p)]$

$(p \iff \lnot{q}) \iff [[(\lnot{p} \lor \lnot{q}) \land q] \lor [(\lnot{p} \lor \lnot{q}) \land p]]$

[vital dew]

Abhijeet Vats:

[sleek swallow]

All I've done is to apply distributivity. Then:

$(p \iff \lnot{q}) \iff [[(\lnot{p} \land q) \lor (q \land \lnot{q})] \lor [(\lnot{p} \land p) \lor (\lnot{q} \land p)]]$

[vital dew]

Abhijeet Vats:

[sleek swallow]

Then, notice that you have two absurdities in there. So, you can get rid of them and you'll arrive at what you wanted to show

@spice lance

[chilly granite]

mainly need help with 9

[sudden knot]

oh baby

another one in 2804

@chilly granite if you're able to do 7 and 8, 9 shouldn't be much more work

like, you can define a dec-string to be strictly-k-bad if for some i, d_i = d_{i+k} or d_i + d_{i+k} = 9

and have a nice formula in terms of k, n for strictly k good dec-strings

then a dec-string is k-good (the normal k-good) if it's strictly N-good for all N from 1 to k

[whole crown]

Are there any kind of YouTubers that teach discrete math or are they a myth

[spice lance]

depends i guess

what u wanna find

ive had shit luck for proofs with equivalence but u might not

[whole crown]

Hmmm

I guess rn I'm doing sets so anything that can help me understand

Safe sets?

[stray reef]

what's a safe set

[vital dew]

presmo:

[spice lance]

and i dont see how

[sleek swallow]

@spice lance Your book's statement is a bit fucked

There exists an n in which set?

Depending on the set and the context, that's true because n = 0 does satisfy 2n = 3n.

[spice lance]

real numbers

and oh i forgot to consider that

ur question is better suited for probability

ask there

i mean i dont disagree, but this could also be rephrased for a probability course

up to you tho

[rigid sentinel]

Can I get help on question A

[pale epoch]

what's giving you trouble?

[rigid sentinel]

why is it saying "x is a odd number" at "A ="

Is it possible that you can link me to a website where I can learn for this level of sets

On youtube I'm just shown the basics

[whole cobalt]

A is a set in thid context

So the problem is saying in the universe 1-10 for all natural numbers, there is a set A that has all the odd numbers

And its then asking you to find the cardinality of that set first

@rigid sentinel make sense?

[rigid sentinel]

Do I have to find out all the numbers for A, B and C before starting the part a) question?

[whole cobalt]

If you're having trouble with the problem sure

[rigid sentinel]

If I'm correct, C is basically 2, 3, 4?

[whole cobalt]

But that shouldnt take long once you get a hold of the problem

Yes

[rigid sentinel]

A is the odd numbers, so 1, 3 , 5 ,7, 9

[whole cobalt]

Yes

[rigid sentinel]

B is 1, 4, 9

[whole cobalt]

Cant help you much with B i have no idea WTF a square number is, im not used to english math

You got that one yourself

[rigid sentinel]

If it means squared numbers between the universe numbers of 1 - 10, it has to be 1, 4, 9. 1 squared is 1, 2 squared is 4, 3 squared is 9.

That's only if the question is asking me that

unless i've misunderstood

[whole cobalt]

Ah right

I've never had to use those ever

You got the rest of the problem figured out?

[rigid sentinel]

So if those values I have are correct for A, B and C, now I can begin to answer part a)

[whole cobalt]

Say if you need more help

[rigid sentinel]

I'm going to skip part a)

because I think I know how to work out part b)

I'll attempt part b) and come back

[whole cobalt]

Well you have already done the hardest part of a

[rigid sentinel]

ah, hardest part of A was just knowing the values of A, B, C?

nearly done on B

[whole cobalt]

Yeah because in problem a its just asking you things like what is the cardinality of A

[rigid sentinel]

B = {1, 4, 9}

P(B) = {∅, {1}, {4}, {9}, {1,4}, {1,9}, {4,9}, {1,4,9}}

That's my attempt for B

and i'm not sure if it needs the cardinality aswell, doesn't say and I'm new to this so I'm not sure if it needs the carditality for those type of questions by default.

[whole cobalt]

How did you get {123}

Out of {149}

Im on phone so CBA with perfect notation

[rigid sentinel]

you're right

sorry

Edited it

[whole cobalt]

You cant tell if its asking you for the cardinality?

[rigid sentinel]

I mentioned cardinality because I'm doing revision using a youtube video and this individual is mentioning cardinality

[whole cobalt]

The notation for cardinality

A is the set

|A| is the cardinality of A

[rigid sentinel]

n = |A|

[whole cobalt]

Yeah n means any number

[rigid sentinel]

2 to power 3 |P(A)| = 8

[whole cobalt]

The cardinality of A id a number

[rigid sentinel]

do i have to mention this stuff?

also can you tell me what the | Means?

[whole cobalt]

But thats not what the problem is asking

[rigid sentinel]

e.g |A|

ok good

[whole cobalt]

I just said it

[rigid sentinel]

I mean

[whole cobalt]

That is notation for cardinality

[rigid sentinel]

just the symbol

|

on its own

ohh

so anything with | | is the symbol notation for cardinality

[whole cobalt]

So if i say what is |A| for your problem what would you answer

Yes

[rigid sentinel]

if you said what is |A| do I say 8?

[whole cobalt]

No

😄

[rigid sentinel]

2n is |A|

[whole cobalt]

Its ok

[rigid sentinel]

n = |A|

not sure

[whole cobalt]

Tell me what is the cardinality

[rigid sentinel]

The cardinality is 8

[whole cobalt]

I mean

What does cardinality mean

[rigid sentinel]

I think it means the number of values, there's 8 in my answer

no no

the number of values

in B

so 3

[whole cobalt]

In B its 3 yes

What about A

[rigid sentinel]

ah, not attempted A

[whole cobalt]

Attempted A? Whats that lol

[rigid sentinel]

oh

ohh

ok

In A it's

5

[whole cobalt]

Hell yeah

|A| = 5 is the first part of problem a

[rigid sentinel]

so |A| is basically asking me, what's the cardinality of A?

[whole cobalt]

Yes

So what is the next part asking you?

[rigid sentinel]

so I know that the ∩ symbol means what they both have in common

[whole cobalt]

Yes

[rigid sentinel]

so depending on how many numbers in common, the cardinality is the number of values in common?

[whole cobalt]

Thats it

[rigid sentinel]

and same goes for C

now my issue is, how am I supposed to get my working out onto text

I'll attempt it now and come back

[whole cobalt]

Do you need to do that?

We didnt lols

[rigid sentinel]

really?

[whole cobalt]

But good luck

[rigid sentinel]

yep working out is needed

ok i'ma attempt it now

Part a)

B = {1, 4, 9}
C = {2, 3, 4}

|A| = 5
|A|∩|B| = 2
|C| = 3```

I'm trying to think about how I can put the working out into text

[whole cobalt]

|A| = amount of elements in set A = 5

Idk that would be more than enough for me

[rigid sentinel]

oh

i just found out in a past paper

Tbh you're right, it looks fine to me aswell having it as how I put it

[whole cobalt]

Looks good

[whole cobalt]

Might have more luck in the #probability-statistics channel with this question

[rigid sentinel]

' means the values that aren't in the U (universe numbers)

[wheat crypt]

Ooh thanks, my bad @whole cobalt

[rigid sentinel]

C = {2,3,4} and U = {1,2,3...,10} which means C' = {1,5,6,7,8,9,10}

[whole cobalt]

Probably, we used different notation in our course

[rigid sentinel]

I think you used "C"?

as the notation?

[whole cobalt]

But no idea what C\B means

[rigid sentinel]

you used this?

[whole cobalt]

No

[rigid sentinel]

Ok scratch that, anyways the B x C

I did some research

[whole cobalt]

Yes?

[rigid sentinel]

B = {1, 4, 9}
C = {2, 3, 4}

so my attempt was this

B x C = { (1,2), (1,3), (1,4), (4,2), (4,3), (4,4), (9,2), (9,3), (9,4)}

does it look right

[whole cobalt]

Yep

[rigid sentinel]

the brackets are correct?

or do they need to be {}

[whole cobalt]

That looks right to me

Also you can just multiply the cardinality of them to get the cardinality of A x B

So 3 times 3

[rigid sentinel]

ah

Also

[whole cobalt]

= 9 in cardinality of theor cross product

[rigid sentinel]

A = {1, 3, 5, 7, 9}
C = {2, 3, 4}

A ∩ C = {3}

[whole cobalt]

Yes

[rigid sentinel]

Going to do research on the \ question

[whole cobalt]

No idea what it is

Never saw it in my course

I suppose it could be an XOR but who knows

[sleek swallow]

That's your set difference

[sleek swallow]

Yes, that's the same thing as the set difference.

[rigid sentinel]

Cheers boss man

C = {2, 3, 4}
B = {1, 4, 9}

C\B = {2,3}

[sleek swallow]

Indeed.

[rigid sentinel]

If it was B\C would it be different?

B\C = {1,9} ?

[sleek swallow]

Yes. In that situation, you'd want all the elements that belong to B and do not belong to C.

[rigid sentinel]

Ah

X = {0,1,2,3...,15}

So F can only be the prime numbers in that list

F = {2, 3, 5, 7, 11, 13}

X = {1,2,3...,20}

So G can only be the multiples of 3 in that list

G = {3,6,9,12,15,18}

X = {1, 2, 3, 4, 6, 9, 12, 18, 36}

H = {1, 2, 3, 4, 6, 9, 12, 18, 36}

[whole cobalt]

looks good to me

[rigid sentinel]

Sweet

I'm going to have a break for today, I have a headache, thank you for all your help mate

[whole cobalt]

yeah np

[errant bear]

why do you list x as a set?

[rigid sentinel]

I don't write X in the working out when I'm submitting it.

that was just for the people monitoring my work in this channel

I've worked out what the values of F, G ,H are

However in the next part(B) of the same question it still uses A, B and C

Do I replace A with F, B with G and C with H?

[pale epoch]

seems like b) has nothing to do with a)

and it just wants you to draw venn diagrams with labels A, B, C

[rigid sentinel]

That's really confused me

so for b), I'd just draw a Venn diagram for (i), how would I illustrate it? Just plot on the A, B and C?

I've done this so far

[errant bear]

you would illustrate it by shading in the area, or "elements" of the set being described, its just asking for the visual representation

[pale epoch]

yeah, that's what i think is being asked

like, in general a venn diagram for 3 sets would look like

maybe with a big circle around for the universe

then you can label each circle differently with A, B, C

and shade the appropriate area

[rigid sentinel]

Ah, so for part a), it's that Venn Diagram that Lochverstärker has just showed me and for part b) it's to shade in which is what fei has told me

[errant bear]

yea, which section(s) would that be

[rigid sentinel]

Are you asking me which sections I'd need to shade in?

Also, would I need to shade in (AnB) in a different colour to nC'

[pale epoch]

you have to read the whole expression

[rigid sentinel]

1 colour?

[pale epoch]

shade in $(A \cap B) \cap C'$

[vital dew]

Lochverstärker:

[rigid sentinel]

Sweet

[pale epoch]

the statement $\cap C'$ makes no sense

[vital dew]

Lochverstärker:

[pale epoch]

(it is not a set)

[rigid sentinel]

I'll attempt the shade and then come back

[errant bear]

ok

[rigid sentinel]

that's what I shaded in for this question

[rigid sentinel]

You want me to stick with this one?

[sour arrow]

Yes, that's perfect

[rigid sentinel]

Is it correct?

[sour arrow]

Yes. (A ∩ B) is the space A and B shares, then ∩ C' means to avoid any space that has C

[rigid sentinel]

I've done this so far, I'm not sure how to do the \A'

[pale epoch]

what you shaded for now is not $B \cup C$ though

[vital dew]

Lochverstärker:

[rigid sentinel]

oh

[rigid sentinel]

I'll come back to that later

F = {2, 3, 5, 7, 11, 13}

G = {3,6,9,12,15,18}

H = {1, 2, 3, 4, 6, 9, 12, 18, 36}

is this right

venn diagram for the values above

[dusk sandal]

2nd one

[rigid sentinel]

Cheers, but have I answered the question correctly?

[dusk sandal]

i didnt check if your answer was correct but i'd assume so?

[vernal musk]

I’m having a hard time and I only have 2 attempts left

[sour arrow]

@rigid sentinel
Yes that looks good

[rigid sentinel]

Sweet

I'm going to have a long break, thanks for the help.

[vernal musk]

Plz I need help ^^

[sour arrow]

What's "div" here?

Like "n1 div 2 = 0"
What does that say about n1?

[zinc pewter]

Don't understand what this question is asking

p means "I learned to play a musical instrument" r means "I own a guitar"

so would my answer be something like "I learned to play a musical instrument" and "I do not own a guitar"

since negation

I'm not sure if I'm understanding the question correctly

[sleek swallow]

Yea that’s correct

[wanton sable]

can someone help me check my answers to this please (my answers are to the right). tbh i didn't really understand this question and it seemed too simple which probably indicates i did something wrong here

not sure if i'm writing the intervals correctly and my answer to 4. is probably wrong too

[naive saffron]

Not an answer to your question, but $|x|=\begin{cases}x&\text{if }x\geq0\-x&\text{if }x\leq0\end{cases}$ is sacrilegious as when $x=0$ both choices are valid

[vital dew]

Whoever:

[ivory badge]

but then |x|=0 under both choices

[naive saffron]

still

[wanton sable]

wat

[naive saffron]

It's against rules

#rules

[real brook]

Solve the RH

[small shard]

@naive saffron mb ill delete

@weary tiger can i join u

[real brook]

correct

[small shard]

im a second year in university

[naive saffron]

same

[small shard]

and im stuggling on my last 2 assignments for finite math

[naive saffron]

[real brook]

yeah

[small shard]

my grade is currently a 90 and im afraid to drop from a 90

[real brook]

just ask and someone will help for free

(maybe)

depends on if they can answer

[small shard]

kk lol ill ask in the morning im go bed rn

[real brook]

Ok, ttyl

[small shard]

but ty for the info ill be back tmr

[real brook]

cool

sounds good

[wanton sable]

if i have a value like 0 <= x <= -4 this indicates x is the empty set right

since u cant find an x that satisfies that equality

[real brook]

I’d say so, yes

[naive saffron]

0 <= -4 is a contradiction so yes

empty set

[real brook]

Whoever

what’re you learning?

[naive saffron]

#advanced-analysis

[real brook]

Nice

[analog night]

how to check if these degrees of verticies make a graph?

what is the method?

[pale epoch]

havel hakimi algorithm

at least in general, sometimes you can get it quicker

e.g. using handshaking lemma

[analog night]

thank you

one other question

is there a Hamiltonian cycle in this graph?

[sour arrow]

Yes, go around the outside of the triangle

[analog night]

what do you mean?

[weary tiger]

Can someone help me with this

7^82 mod 100

I got 45

But its not correct can someone explain why?

[last sigil]

Why don't you explain what you did?

[dusk sandal]

I need to prove that $(2^a-1) \textrf{mod} (2^b-1) = 2^{a \textrf{mod} b}-1$

[vital dew]

sockratees:

[dusk sandal]

so i wrote that $2^a-1 = k_1 (2^b-1)+2^{a mod b}-1

$2^a-1 = k_1 (2^b-1)+2^{a mod b}-1$

[vital dew]

sockratees:

[dusk sandal]

which is equivalent to

$2^a-1 = k_1 (2^b-1)+2^{a-k_2 b}-1

$2^a-1 = k_1 (2^b-1)+2^{a-k_2 b}-1$

[vital dew]

sockratees:

[dusk sandal]

but i really dont know where to go from here

[reef thistle]

okay this is equivalent to what you had

if a<b you are done, so try the other case

@dusk sandal

[dusk sandal]

wait this is done?

like i get the other cases

but i dont get how this is a completed?

[reef thistle]

what happens if a<b?

@dusk sandal

what's a mod b?

[dusk sandal]

a mod b - a

😗

= *

[reef thistle]

a mod b = a

[dusk sandal]

but like i get the other cases

that's not what's confusing

[reef thistle]

how do you do the other case?

[dusk sandal]

a mod b = 0

but like again

[reef thistle]

not that, the other case is a>=b

[dusk sandal]

it doesnt have to be does it

you can separate into cases

a < b

a mod b = 0

[reef thistle]

you can split that into more cases

[dusk sandal]

and b > a

[reef thistle]

but really not that much point

no, b>a is the same as a<b

[dusk sandal]

or i mean

a > b

where a mod b \not = 0

[reef thistle]

okay...

so which cases are you done with?

[dusk sandal]

i understand the a mod b = 0 case

as well as the a < b

im confused with the last case

[reef thistle]

what have you done so far for the last case?

[dusk sandal]

the latex above

[reef thistle]

how about try subtracting 2^(a mod b)-1 from both sides?

then all you are doing is proving that something is divisible by 2^b-1

[dusk sandal]

hmmmm

i see

i'll try that

thank you

[swift bone]

hey guys

what book do you suggest to learn discrete math

[gleaming zephyr]

rosen is the classic

[rigid sentinel]

is this correct

for only (BUC)

[gleaming zephyr]

complete it

[rigid sentinel]

ok

these are the exact same thing?

Just different symbols

[pale epoch]

yeah, just different notation

[rigid sentinel]

Ok I've made a full attempt

I've shaded in all objects that belong to B and C, so even if it's the smalls gabs that belong to A, it won't matter as they still belong to B and C

[gleaming zephyr]

ur missing stuff

[rigid sentinel]

then \ means to shade BANDC not A

[gleaming zephyr]

x is in B or C and x is not in A complement

[sleek swallow]

(BUC)\A' is the set of all objects that belong to BUC but do not belong to A'.

[gleaming zephyr]

x is in B or C and x is in A

shade all

[rigid sentinel]

the outside?

outside the circles

[gleaming zephyr]

no whay

[rigid sentinel]

I'm confused, do I shade in A aswell?

[sleek swallow]

No. First of all, where is B U C in the diagram?

[rigid sentinel]

left and right circles

all of that

[sleek swallow]

Okay. Where's A'?

[rigid sentinel]

I did research on the symbols, it says not A

and it says not A twice

/ and `

[sleek swallow]

:/

[rigid sentinel]

Am I wrong

\ means objects that belong to B or C not A

and ` means objects that dont belong to A

according to this

https://www.rapidtables.com/math/symbols/Set_Symbols.html

[pale epoch]

well, you shaded $B \cup C$ now

[vital dew]

Lochverstärker:

[pale epoch]

the problem is, that you now need a universe to even make sense of A'

[rigid sentinel]

ohh

i think i understand

[pale epoch]

i guess you could solve it without

[sleek swallow]

you can't

A' can only be defined with reference to a particular universal set

Let A, B & C be subsets of some given set U

[pale epoch]

i mean the overall question can still be solved, as A, B and C are implied to be subsets of the universe

but it would be more instructive if you give yourself a universe

maybe a circle around what you already drawn

then i would ask you to shade in A', maybe with a different color

[rigid sentinel]

a square?

[pale epoch]

so the whole square is your universe?

[rigid sentinel]

and i shade the inside of the square

or does it need to be a circle?

[pale epoch]

no, the shape is irrelevant

square is fine

can you shade A'?

[rigid sentinel]

that's what i've done

everything apart from A

\A means not A, A' means not A

[pale epoch]

what about the stuff that is in B or C, but not in A

[rigid sentinel]

So I did research

BUC means everything that belongs in B or C

so they shaded in parts of A aswell

But does the /A and A' counter this?

and what if it was just \A on it's own and not A'

and what if it was just A' and not \A

[pale epoch]

honestly what

[rigid sentinel]

What a headache

Is my answer wrong?

If it is, I'll have another attempt

[pale epoch]

what question is it the answer to?

[pale epoch]

then it's wrong

as i said, i would advise you to first shade in A'

(in a different color than $B \cup C$)

[rigid sentinel]

You mean do A' first?

[vital dew]

Lochverstärker:

[pale epoch]

yeah, you did $B \cup C$ correctly

[vital dew]

Lochverstärker:

[pale epoch]

i wanna see if you can do just A'

[rigid sentinel]

But in the final solution, it would be one venn diagram

correct?

[pale epoch]

yeah

[rigid sentinel]

this is A'

[pale epoch]

no, this is $B \cup C$

[vital dew]

Lochverstärker:

[rigid sentinel]

then it has to be this

that's A'

[pale epoch]

depends on your universe

if the whole square is your universe, then no

[rigid sentinel]

this is all it is to the question

[pale epoch]

if only the circles are your universe, then it is correct

[rigid sentinel]

My teacher hasn't said anything about how the venn diagram should look

I'm just sticking to 3 normal circles

I'll just ask him, cheers though, we've agreed on what is BUC and \A'

[pale epoch]

well ok

the black part is $A'$ and the green part is $B \cup C$

[vital dew]

Lochverstärker:

[rigid sentinel]

In your past, has your teacher told you that you must do 2 different colours?

[pale epoch]

no, i am just doing this to help you understand

[rigid sentinel]

would I lose marks for not stating which part is BUC and which part is A`

[pale epoch]

i assume no

if you final answer is correct

[rigid sentinel]

or can I just display it as this

Ok

I'll double check with my teacher anyway

[pale epoch]

well, this is not the final answer

i was trying to help you get to the final answer by using colors, but if you dont care about that, then nevermind

[rigid sentinel]

Yes I understand what you mean

You was trying to help me understand for which part is which, as the final answer in this just to me looks like it's BUC eventhough I need to understand why it is the final answer.

If that makes sense, I understand why it's the final answer.

[gleaming zephyr]

BUC/A'

shade all

[errant bear]

that isnt the final answer...

[stone nymph]

not sure if this is the right place to post this
I just started with proving correctness of algorithms with induction and loop invariants. But I am not sure how i should prove a recursive one with induction. This one for example
double expRecursive(double x, int n) {
if (n <= 4) {
return expIterativ(x, n);
}

return expRecursive(x, n/2) *
       expRecursive(x, (n + 1)/2);

}
It is given that integer divison rounds of to zero and that expIterativ is correct. Prove the correctness of expRecursive

[pale epoch]

you need to find an invariant, prove it holds (lets say at the beginning of each function call) and prove that on termination that gives the desired result

[rigid sentinel]

@errant bear is it wrong?

I just scrolled up, according to fei, this isn't the final answer for (BUC)\A`

[errant bear]

you want elements that are in either B or C, or both, and not in A'

[rigid sentinel]

yes we went back and forth with this, I also tried this and this was incorrect

[peak gull]

It is wrong because everything you shaded IS in A' because it is not in A.

[rigid sentinel]

I don't know, I'm clueless now

A' means everything that is not in A

so that's what I shaded

[shy anchor]

Question: how many ways can you scatter 12 Identical balls to 8 different cells, so that the first 2 cells have at least 10 balls combined

[peak gull]

@rigid sentinel it's like a double negative... You want everything in BUC but not in A', so really you want everything in BUC that IS in A.

[rigid sentinel]

Yes which is what I did here

and some people are saying that this isn't the final answer

[peak gull]

But you still have some stuff that is not in A, specifically all that stuff that is in BUC but not in A.

You need to shade ONLY the stuff that is not in A'.

[rigid sentinel]

that's the final answer?

Not in A means B and C so I shaded in B and C

[errant bear]

no it doesnt

thats literally the most important rule in basic logic

If I don't have any money, does that mean you must have money?

no

[peak gull]

@errant bear you talking to me or him???

[errant bear]

him

[peak gull]

Oh, gotcha...

[errant bear]

yea

[peak gull]

It's been a hot minute since discrete math so just making sure lol.

[errant bear]

yeah, urs is correct

[rigid sentinel]

I was looking at this, it says all objects that do not belong to set A, so I initially shaded everything that didn't belong to A

[native dagger]

Does anyone know any good resources for binary operation questions? I tried the Rosen textbook but the topic was not there.

[errant bear]

yes

but \ means you are removing those

[zinc pewter]

Need help with these problems

<--> means if and only if I'm pretty sure in grammer

so I think the answer for b is I learned to play a musical instrument if and only if I can read music

[teal canyon]

yes

[zinc pewter]

Oh sweet

look at me go

and for c)

Yeah I got no idea

Hmm

[teal canyon]

-> is if ..., then ...

[zinc pewter]

"I can read music and I own a guitar then I learned to play a musical instrument"

[teal canyon]

it's q /\ r there

[zinc pewter]

oh

"if I can read music and I own a guitar then I learned to play a musical instrument"

[teal canyon]

yes

[zinc pewter]

sweet

ty

[zinc pewter]

my propositions are "if there are maple trees"

and "it is spring"

so if p is "if there is maple trees"

and q is "it is spring"

and r is "we can make maple syrup"

would this be correct?

[sleek swallow]

Indeed

$p :\iff \text{if there are maple trees}$

[vital dew]

Abhijeet Vats:

[sleek swallow]

$q :\iff \text{if it is spring}$

[vital dew]

Abhijeet Vats:

[sleek swallow]

$r :\iff \text{we can make maple syrup}$

[vital dew]

Abhijeet Vats:

[sleek swallow]

That's how you should define propositions

[zinc pewter]

Oh thanks for the tip that is much better

p <--> you are never bored

q <--> if you have a cellphone

r <--> a good book

correct?

[sleek swallow]

you mean (q or r) => p?

[zinc pewter]

yeah

logical or symbol

[sleek swallow]

$(q \lor r) \implies p$

[vital dew]

Abhijeet Vats:

[zinc pewter]

yeah

V is "or" in language

[sleek swallow]

Indeed

[zinc pewter]

so it evaluates to "if you have a cellphone or a good book then you are never bored"

I actually forgot negation

so it would be negation p

[sleek swallow]

No. You defined p as you are never bored

[zinc pewter]

yeah that's true

is it okay to have a negation element tied to a proposition

probably is

[sleek swallow]

negation element?

[zinc pewter]

not is negation

[sleek swallow]

You mean $\lnot{p}$

[vital dew]

Abhijeet Vats:

[zinc pewter]

yeah

[sleek swallow]

is it okay to write that? Indeed, it is

[zinc pewter]

https://cdn.discordapp.com/attachments/496785905474994186/668253853157818385/340191992636243968.png

is this right

[sleek swallow]

Correct

[zinc pewter]

My prof wants it as a atomic proposition I think

so It would be

p <--> you are bored
q <--> if you have a cellphone
r <--> a good book

and my final answer would be

not sure if that still works

[sleek swallow]

Uh more like this:

$p :\iff \text{you are bored}$

$q :\iff \text{you have a cellphone}$

$r :\iff \text{you have a good book}$

So:

$(q \lor r) \implies \lnot{p}$

[vital dew]

Abhijeet Vats:

[sleek swallow]

Uh your phrasing was just a bit off but that probably works

though I should say that the negation of p is, more appropriately, 'you are not bored', not really 'you are never bored'. The former is a statement about a specific point in time while the latter needs to be quantified

[zinc pewter]

That is true

probably better going with the first answer then cause that's a really good point

[sleek swallow]

mmh

[sour arrow]

That says "I have air miles or I haven't flown in a plane"

You want to say "I have air miles AND I haven't flown in a plane"

@zinc pewter

[zinc pewter]

yeah you're right

[dim magnet]

I'

I'm going to assume this is the best place to ask a question about set theory?

Specifically about arbitrary unions

ok

So I'm a bit confused in regards to

What it means by

"empty collection"

is it just a set containing the empty set?

{{}}

[sleek swallow]

Nope. If it's an empty collection, it means that it contains nothing.

It is literally empty.

[dim magnet]

Then how is it a collection?

[sleek swallow]

We tend to make a distinction between sets of objects and sets of sets. The latter is usually given the name 'family of sets'.

So, the empty collection is the family that contains no set.

Well, that's the same thing as asking 'Why is the empty set a set?'. The answer is that if you look at the axiom of specification and have a set A, then you can define the empty set by means of a contradiction. Since the axioms allow it, it is a set that exists.

[dim magnet]

Why not call the empty collection an empty set then?

[sleek swallow]

Some authors do that. If I'm remembering this right, Halmos doesn't make the distinction in his book. He refers to sets of sets as just sets, instead of families. I've watched lectures where the lecturer doesn't make that distinction.

Does that clear up your doubt?

[dim magnet]

Yep

Out of curiosity

Since I haven't done much outside of elementary set theory