discrete-math

[sleek swallow]

Then search up the construction of the naturals using the axiom of infinity

[pale epoch]

there are multiple ways to do it

[hybrid plume]

thanks a lot guys

that s what i wanted

[sleek swallow]

Also search up the peano axioms if you’re interested in this

Don’t blame you, it is interesting stuff

You’re welcome.

[hybrid plume]

don t forget about sypherz question haha

he posted a question before me

sorry sypherz

[old imp]

chillin man glad you figured your stuff out

[hybrid plume]

i would ve helped you but sadly i haven t had a course regarding ur question

[lyric pumice]

@neon thorn Since

it follows that

So,

Then

[lyric pumice]

You must use the assumption that

and additional information about k to show that

In general, this is arrived at in the form of a sequence of inequalities involving k.

It appears that you have posted for a different problem.

[obtuse lance]

idk hopefully I'm not undermining you help but I did it another way, might as well share it:

$$2^k>k$$ $$2^{k+1}>2k = k+k > k+1$$

[vital dew]

Merosity:

[obtuse lance]

I just multiply the inequality by 2

[lyric pumice]

Alright.

Just keep in mind that k>=1.

[obtuse lance]

good point, I should have put the last inequality as >= instead

[lyric pumice]

You can also prove that 2^(n+1)<(n+1)! for all integers n>2.

[wanton sable]

is there a difference between a = b mod c and a = b (mod c)?

[faint narwhal]

no

[wanton sable]

thanks~

[gusty burrow]

If I have two propositional functions G(x), C(x) that ive already shown to be true for some c, is there something special I must do to show Ǝx(G(x)&C(x))? Or can I just use normal existential generalization

[sleek swallow]

‘Normal existential generalization’?

You’ve find an x = c such that G(c) is true and C(c) is true. Hence, their conjunction must be true for some x = c

[gusty burrow]

By normal existential generalization I mean if you show G(c), you can say ƎxG(x)

Now that ive found those two propositional functions whose conjugation is true for c, I can do that same generalization to their conjugation right?

I couldnt find anywhere in the book that explicity says that but it seems right

Part of my homework is going through existing arguments in english and explaining which Rules of Inference were used in reaching their conclusions

so im just translating all of them into logic and identifying the Rules of inference that way

and the conclusion I wanna arrive at is Ǝx(G(x)&C(x))

[sleek swallow]

Yes that’s correct. You can do that generalization to their conjugation assuming that G(x) and C(x) are both true for the same x

[old imp]

don't apologize for seeking help dude. induction is also a pretty tough concept to grasp at first

[mint horizon]

what does it mean when there's a hat on top of a set?

like... __

[stray reef]

$\overline{A}$?

[vital dew]

Ann:

[stray reef]

like this?

[lyric pumice]

@neon thorn The proof displayed shows that the left-hand side is less than the right-hand side.

2^k+2^k was introduced since it is greater than 2^k+2k+1 and equal to 2^(k+1).

This allows us to establish, or complete, the overall relationship that (k+1)^2<2^(k+1).

[mint horizon]

what does the fancy looking P mean?

[lyric pumice]

Given that k>=5, we have

So

This implies that

Next,

This implies that

Therefore,

This shows that

[sleek swallow]

@iron crescent do not post your question in multiple channels

[hybrid plume]

can someone tell me what is the term used to describe these word formations using a language terms ?

[pale epoch]

this is basically regular expressions

[hybrid plume]

do you have any ressource on the subject

?

[pale epoch]

not really

those are studied in theoretical computer science

see: finite state automata

[lyric pumice]

@neon thorn Yes.

[hybrid plume]

any german speaker here ? i didn t understand the notation

[faint narwhal]

What don't you understand

I don't speak German but I can understand what it's saying

[stray reef]

what does bzgl mean

[hybrid plume]

bzgl in relation to

[stray reef]

and somit?

[pale epoch]

therefore

[stray reef]

and genau?

[analog sonnet]

precisely/exactly

[stray reef]

and you're just asked to compute (X+X^7)(1+X^3+X^4) in that field? lmfao

[bleak pollen]

how can you prove the distributive rule

[analog sonnet]

You mean something like a(b + c) = ab + ac?

[stray reef]

context?

@bleak pollen

[sudden knot]

well presumably to multiply it out, then compute the remainder modulo p(X), the degree 8 polynomial given

[stray reef]

[pale epoch]

tbf he asked about the notation

[stray reef]

hermeskid either didn't see my inquiry or did but is refusing to clarify...

[bleak pollen]

O mb

I was in class

P & (Q or R) -> (P and Q) or (P And R)

I don’t have the and /or notions on my phone

Here is a picture on Wikipedia

I need to write a proof for this rule

[stray reef]

what axioms are you allowed to use

or is it a semantic proof ie by truth tables that you're asked for

[bleak pollen]

No I can’t use truth tables sadly

I have to right it out line by line in new to this type of math btw so bare with my bad terminology

I need to use other rules to prove this rule

I have to start with my hypothesis and go from there

[stray reef]

so then as i said before
what axioms are you allowed

[bleak pollen]

Anything I believe

My professor only said we can’t use the distribution rule to prove its self LoL

I started with de Morgan’s and implications but I’m kinda lost on where to go from that

[weary tiger]

Ok so here is a straightforward question

N identical dice are rolled , number of possible outcomes are?

I just want to know how to interpret the question

Where do I start my journey and make it end at stars and bars

I was thinking in terms of like the total output

So like if I have one die the max value it can hit is 6

So I have 6 outcomes

If two die then the total output's max value is 12

So d1+d2=12

and let mi=6-di

So m1+m2=0

I mean so it hits a wall, is this due to the approach( I highly think it is)

@someone

@stray reef you have a couple of minutes?

[stray reef]

no

[weary tiger]

Rip

<@&286206848099549185>

[analog sonnet]

wdym "possible number of outcomes"

Like, the sum of the pips of the dice that you throw?

[weary tiger]

@analog sonnet pips?

Like a sequence

Where all digits are less than 6

Like for n=1 possible outcomes are 1,2,3,4,5,6

For n=2 it's (1,2),(6,1) etc.

And since the dice are identical

1,2,3,4 and 2,3,4,1 are the same

[analog sonnet]

The dots on the faces of the dice are called pips

themoreyouknow.jpg

[weary tiger]

Oh

The sequence of the pips

All possible sequences of the pips of a given number of dice

@analog sonnet

[hybrid plume]

@stray reef why did u laugh at my question ?

it s not that easy for me

[agile nest]

did you understand the question now? or should I translate it

or which notation were you confused about?

@hybrid plume

[mint horizon]

what does this mean? P({1,2,3}) ?

[agile nest]

power set of {1, 2, 3} ?

[mint horizon]

idk my teacher used it as if it was a set?

[agile nest]

it is

[mint horizon]

ok.

[agile nest]

it's the set of all subsets of {1, 2, 3}

[mint horizon]

ohhh ok. thanks

[agile nest]

so P({1, 2, 3}) = {{}, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}}

[lyric pumice]

@weary tiger The question you have posed involves multisets.

[lyric pumice]

The number of possible outcomes for rolling n dice appears to be the following.

[old imp]

Hoping to get some help on the last steps of my recurrence relation and see whether I’m doing my algebra right, ty in advance

[graceful shard]

Haai〜☆ I have a question about discrete mathematics, how can I enumerate the members of {x ∈ N : x divides 0}? Wouldn’t that mean literally all numbers?

[agile nest]

you seem to have answered your own question

[old imp]

If a|b, ac = b and c = 0 will always work for any a

I’m pretty sure

[graceful shard]

But how do I enumerate it?

Enumerate means to list all the numbers of the set, right?

[agile nest]

yes

it means all numbers in N

so your set is N

[gleaming zephyr]

@graceful shard just writee N

[agile nest]

you need to enumerate N

[gleaming zephyr]

'Wouldn’t that mean literally all numbers?'

no

[old imp]

I’m not sure I don’t have experience with that but wouldn’t it just be x is an element of N

It’s a weird question

[gleaming zephyr]

not all numbers

all numbers in N*

[graceful shard]

Oh oki

Thankz so much for the help

[agile nest]

all strictly positive natural numbers

[graceful shard]

I’ve been stuck on that for hours

[fading widget]

I've tried using the quadratic formula to pick epsilon but im not sure how to proceed

[dawn vine]

@fading widget what is that ^ looking symbol before y = (n+e)^2-n^2?

[sleek swallow]

Logical and

So $(0 \leq \epsilon < 1) \land (y = (n+\epsilon)^2-n^2)$

[vital dew]

Abhijeet Vats:

[dawn vine]

i forgot that's what it was

in that case, it would have been easier to say "there exists epsilon between 0 and 1 such that bla bla bla" rather than saying epsilon greater than 0 WHICH IS between 0 and 1

[lyric pumice]

Let e be epsilon. You can show that the following is true.

[fading widget]

@dawn vine yeah it is

@lyric pumice where do I go from there?

[lyric pumice]

You must demonstrate this.

The problem is tricky.

[fading widget]

How does demonstrating that prove the question?

[lyric pumice]

You will show that some real number epsilon>=0 that fulfills all the requirements must exist.

The epsilon is between 1 and 0 and is the solution to the given equation.

[fading widget]

OH

thank you

[mystic fulcrum]

Hey everyone, I have a quick question on bit strings and predicates that can be used on them. Let's say we have a concatenation predicate ⌢ that concatenates some free strings w and v into wv. We also have a substring predicate that lets us know if some string w is contained within other string v, defined as w⊑v. My question is, how can you build a predicate to satisfy the following:

I guess my question is how can I approach just the problem a in terms of how can we show it if w is instantiated to a string in the set

[feral hearth]

http://prntscr.com/qyk1ri

I'm thinking true but not sure how to explain it.

[craggy gale]

it's pretty clear that strict monotonicity does give injectivity, so you should have a look at surjectivity more closely

[feral hearth]

Would a piecewise function make Bandar incorrect?

Since it could be strictly increasing or decreasing but not all the values in the codomain have an input

[vital dew]

Jack.N:

[lofty iron]

Quick question. I am trying to show that an argument is valid. I reach the point where I have this:

$p \ and \ \lnot p \lor \lnot q$

[vital dew]

Jack.N:

[lofty iron]

Can I use Disjunctive Syllogism to conclude not q despite the fact that Disjunctive Syllogism is defined as:

Given $\lnot p \ and\ p \lor q,\ conclude\ q$

[vital dew]

Jack.N:

[violet tulip]

yes you can. note that by the double negation rule, you can turn $p$ into $\lnot\lnot p$

[vital dew]

patrick:

[violet tulip]

which gives, $\lnot\lnot p$ and $\lnot p \lor \lnot q$

[vital dew]

patrick:

[violet tulip]

you can then just use disjunctive syllogism here as well to conclude not q

[lofty iron]

Awesome Thanks!

[strange zenith]

So this is gonna sound kinda dumb, this is the first question for this chapter, all the other ones I've done without issue, but this one, which looks so simple, is making me want to pull my hair out:

$P \equiv Q$

[vital dew]

Icemasta:

[strange zenith]

Apply commutative rule of ≡ (P ≡ Q ≡ Q ≡ P) where p,q := q, (r ^ s)

[vital dew]

Icemasta:

[strange zenith]

But doing that, I can't seem to make it link back to P ≡ Q

[sleek swallow]

Uh show the exact statement of the problem

Also lmao @lofty iron you don’t need disjunctive syllogism:

$p \lor q \iff \lnot{\lnot{p}} \lor q \iff \lnot{p} \implies q$

By modus ponens:

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

[vital dew]

Abhijeet Vats:

[shut fjord]

Determine whether each of these statements is true or false and provide a justification foreach. a)∅⊂{}

is a) ∅⊂{} false or true?

[soft thorn]

What do you think the answer should be and why?

[shut fjord]

true

[soft thorn]

Why's that?

[shut fjord]

Sorry false

Because its not a proper subset

it would be a subset but not a proper subset

[soft thorn]

Yep

[shut fjord]

that's all I need for a justification?

[soft thorn]

Ehh

Explain why it's not a proper subset

In particular

Are these two sets the same?

[shut fjord]

The two sets are the same

this is why we cannot use the 'c'

∅ = {}

NOT ∅⊂{}

[soft thorn]

Fair

I think that's valid enough justification

[shut fjord]

f:R{0}→R

f:R \ {0}→R

^

[soft thorn]

🤔

[sleek swallow]

R{0}?

[shut fjord]

Sorry

my wifi went out, and by the time it came back I already satisfied my question

[austere jolt]

can someone walk me through how to convert a number into an 8 bit quartenary system

[main socket]

I need help translating a sentence to first order logic statement. Sentence is:
There is no set belonging to precisely those sets that do not belong to themselves.
My attempt:
Let F(x,y) mean set x belong to set y. So we get
For all y there doesn't exist x F(x,y) and ~F(y,y)

[sleek swallow]

That seems to be correct at first glance

[main socket]

Thanks for your reply, I am not too confident in my answer

[teal frigate]

for all real numbers x, x/x = 1

is that true?

[soft thorn]

what if x was 0?

[teal frigate]

Oh shit

Thanks

Nice one

[sour arrow]

Yus

Yus

[stray reef]

no

it's 1

dividing 1 by 3 results in 0 threes, 1 left over

[tranquil jewel]

Anyone know if these are the correct choices?, I can post my reasoning for any just not sure if I went about them correctly

[stray reef]

you've got exactly one wrong

7b

[tranquil jewel]

Is my reasoning wrong?

[ivory badge]

What if x = -y or smth

[tranquil jewel]

Ah

So it can be not a=b but still fit the conditions

Alright thanks

[hybrid plume]

is there a turorial for the latex bot in this server ?

most of my exercises are in german

so i need to translate them using the bot and latex

[pale epoch]

it's just latex, so any latex tutorial will do

[hybrid plume]

ok thanks

[hybrid plume]

can someone tell me how to proceed when working on a combinatorics exercise

?

[hybrid plume]

i ve never answered a question correctly

please help me

i know the formulas

i ve done a tone of exercises

still rarely have the right answer

[stray reef]

can someone tell me how to proceed when working on a combinatorics exercise

this question is way too vague to answer as is honestly

[hybrid plume]

ok i will frame it in an other way

i want to know if i m counting right

given a set and conditions

[stray reef]

that's still kind of too vague and there's no good answer to be given here beyond like

idk

double checking your work?

making sure you don't rely on any formulas you don't fully grasp?

[hybrid plume]

do u have some ressouces u studied with for combinatorics ?

[stray reef]

me? no

didn't really ever learn off a book. some things i had originally come up with on my own

and all it took in class was just learning some names

¯_(ツ)_/¯

[hybrid plume]

ok no worries thanks a lot

[lyric pumice]

Some people can become world-renowned researchers in combinatorics without ever having taken a related class.

Sometimes it takes a certain mind.

[woeful galleon]

can anyone give me hint here ?

[gleaming zephyr]

try to unfold definitions

what does it mean for a number to be even

@woeful galleon

then try to arrive at contradictions

if you assume both

[lyric pumice]

Using a modulus is not necessary.

If you understand the context, then "congruence" is almost a hint.

[weary tiger]

x*4=6+6+4 for x=4
if x is another multiple of 4, we get (6+6)(floor(x/3))+4(ceil(x mod 3)) because 3 is the gcd and 12 is the lcm

adding two, we have a multiple of 6

because we can subtract 12 from the 14 because 12 mod 6 = 0

because x is a multiple of 4, x mod 3 is always >0 and <1 so there is always one extra 4

i feel like i overcomplicated things

[vital dew]

work:

work:

[weary tiger]

times

tanks freb

idk anymore

x could be x*4 so yes?

hello

I NEED HELP

but i am confused because my gcd-lcm thing does not work when x is not exponentiezed

#7

what are they asking for?

truth tables? i dont get it

plz @ me if you know what to do

but how

[sleek swallow]

They're asking you to show that operatios between 3 propositions can be expressed as a disjunction of conjunctions

[weary tiger]

9 is simplification

[sleek swallow]

So, you have to come up with a procedure for taking F(A,B,C) and turning it into a disjunction of conjunctions

So whatever you discussed in class, extend it to 3 propositions

[weary tiger]

so (C AND F(A,B,True)) V (NOTC V F(A,B,False)) ???

aha so instead of ceil(x mod 3) it is better to do (4*3)(x mod 3) and because 4*3=12 the whole term can be removed
then only 4^n satisfies 4^n mod 3 = 1 idk why lol but its much simpler and i think you can prove it by induction

[sleek swallow]

@weary tiger what do you think

Does that look like the correct answer?

[weary tiger]

not really

i dont know what im doing

[sleek swallow]

First step is to go back to what you discussed in class for one and two propositions

Literally go back to that first, do the reading, and then come back to the problem

[weary tiger]

ok

im just confused about this form

F(A,B,C)

[sleek swallow]

F(A,B,C) just means that you're connecting three propositions A,B & C using some logical operators

That's all it means

[lyric pumice]

The solution is clear without resorting to fancy moduli.

[weary tiger]

im gonna just guess

thank you all for helping!

yes i am pretty sure i overcomplicated some things

[solemn cairn]

http://puu.sh/F8wzb/650f116d69.png

Can't seem to get the RHS, anyone know where I'm going wrong?

[soft thorn]

associative

hmm

is that so

[stray reef]

$(A \land B) \lor C \neq A \land (B \lor C)$

[vital dew]

Ann:

[stray reef]

@solemn cairn

[solemn cairn]

oh yeah they have to be the same and/or's

thanks

[woeful galleon]

same here, sorry but i went to sleep lol. for even #s we got n=2k and n=2k+1 i am not sure where to go from here tho

[lyric pumice]

@woeful galleon Hi. You can perform a proof by contradiction.

[woeful galleon]

hey! mind giving me a few hints ? it is just so clear to me that i cant prove it using math lol

[sleek swallow]

Use a proof by contradiction. So suppose that x was odd. Then, there exists an r in the integers such that x = 2r+1. What can you do with that?

[hybrid plume]

can someone explain what this set has as elements ?

[lyric pumice]

Assume, to the contrary, that x is even and x is odd.

[hybrid plume]

does this describe a bijective set ?

@hybrid plume

probably

[lyric pumice]

Then x=2a and x=2b+1 for some integers a and b.

[sleek swallow]

Don’t give the entire proof away

Let them respond to whatever that has been said first

[hybrid plume]

guys help me out pls

[sleek swallow]

@hybrid plume I’ve not seen that notation before. What’s the context under which it was introduced?

[hybrid plume]

it was in a question let me send it

prove this in english

[woeful galleon]

I am still here, lol . I appreciate all the help

what i am missing to understand is that lets take the first part which is If a number x ∈ Z is even, then it is not odd, i know that an even number is given by x=2k and for odd number is x=2k+1. they are clearly not equal to each other. is that not enough ?

[lyric pumice]

k must be an integer.

x must be equal to x.

x is assumed to be even by some integer and odd by some integer.

[sleek swallow]

@woeful galleon Like Millennial said, x = 2a and x = 2b + 1 for some integers a and b. You need to show that there’s a contradiction. Perhaps try relating a and b in a particular way

[woeful galleon]

Well to show contradiction doesnt a have to be equal b?

If you choose different values for a and b that you can't really prove much

Or I am just stupid.

New material for me lol

[lyric pumice]

No, not necessarily. You would have to demonstrate that a=b.

[woeful galleon]

would setting the equal be the easy way of proving it ?

[sleek swallow]

Try it and see if it works

So now, 2a = 2b + 1

What can ya do with that?

[woeful galleon]

lol, i feel so bad cuz u trying to help me but i honestly have no idea whta i can do with that

if a = b then we get 2a =/ 2a + 1 ?!

[sleek swallow]

Okay, the thing is, we have 2a = 2b + 1. So, what can you say about a-b?

[lyric pumice]

You must manipulate the equation to see if you can arrive at a new equation that is clearly a contradiction.

[woeful galleon]

well, a-b is not 0

[lyric pumice]

What is a-b?

[woeful galleon]

1/2

[sleek swallow]

okay so what's the problem with that?

a & b are integers, so what's the problem with a-b = 1/2?

[woeful galleon]

ohh

1/2 is not an integer

[sleek swallow]

and why's that a problem?

[woeful galleon]

so regardless of what values a and b are.. the answer will not be an integer

we assumed that x was a neg/pos integer

right ?

[sleek swallow]

x is an even integer, by the given hypothesis.

but the problem is that a -b should be an integer

[woeful galleon]

yes (the first hypothesis)

[sleek swallow]

so that's a contradiction

[woeful galleon]

i get what it means now. in terms of work, i would just show that a-b is not an integer which contradicts the "assuming" factor of the hypothesis that x ∈ Z is even, then it is not odd ?

[bleak kite]

How would I prove that 1/2 + 2/2^2 + 3/2^3 + ... + n / 2^n < 2?

Lol idk how to really do induction in inequalities well tho 😦

I don’t know where to start

For the inductive step

[sleek swallow]

There's a related geometric series that you can consider

Look at that first

[little nacelle]

proofs are so aids.

[sleek swallow]

Analyze that and see if you can relate the convergence of that series to the one that you have right now.

[bleak kite]

The 1/2^n that sums up to 1?

I don’t know how to account for the n in the numerator

[sleek swallow]

Consider: $S_n(x) = \sum_{i = 0}^{n} x^i$

[stray reef]

[sleek swallow]

Fml

[stray reef]

that's just (n+1)x^n abhi

[sleek swallow]

Wait fuck my life

[vital dew]

Abhijeet Vats:

[sleek swallow]

Pranked myself @stray reef

@bleak kite But yea use that and then try to get to your sum through the one above

[bleak kite]

What exactly is that tho

I can’t see how I can use that to solve this

@sleek swallow what does that have to do with this problem

[robust mango]

Uh I tried a different approach. Assume it's true for some integer n=k. Then you have k/2^k < 2. What I did was add k/2^k on both sides, so LHS you have (k+1)/2^k, then you multiply/divide LHS by 2. You have 2(k+1)/2^(k+1) on the LHS, you have some stuff on the RHS which you can simplify.

[bleak kite]

How did you get k/2^k < 2?

Oh

It’s just because it’s smaller

[robust mango]

?

[bleak kite]

Idk how to do induction/inequality proofs :(((

[robust mango]

We assume the hypothesis is true for some integer n=k before proving for the next integer, n=k+1

So k/2^k < 2 is true according to our hypothesis, then I just manipulated the inequality to get what we needed, ie k+1 / 2^(k+1)

[bleak kite]

Why would k/2^k be relevant tho

[robust mango]

Well it needs to be true before proving for k+1

[bleak kite]

But wouldn’t it only work for k=1

[robust mango]

Then why are you proving if you think that's true

no it'll work for any integer k >=1

so we prove for k=1(base case), assume it's true for some integer k, then prove for the next one(k+1)

[bleak kite]

I just don’t see how k/2^k < 2 matters

Because they can all be less than 2 but still sum up to more than 2

[robust mango]

If they add up and be greater than 2, then our inequality is wrong. That doesn't happen

My way is not the best way maybe someone else can help you

[sleek swallow]

$S_n(x) = 1 + x + x^2 + x^3 + ... + x^n$

$\frac{dS_n}{dx} = 1 + 2x + 3x^2 + ..... + nx^{n-1}$

[vital dew]

Abhijeet Vats:

[sleek swallow]

@bleak kite

[bleak kite]

I still don’t really know what to do with that besides setting x = 0.5

[robust mango]

I think you should watch some videos on induction

Before proceeding

[stray reef]

S_n is a geometric sum

[sleek swallow]

Look very closely at the sum you have. Look closely at what i've written. Can you form a link?

I can't just give you the answer. You can't just say 'I don't really know what to do with that....'. Think about a relationship between the above and what you have.

[bleak kite]

I know that my sum is equal to the derivative of your sum when x = 0.5

[sleek swallow]

Are you sure about that?

[bleak kite]

It looks like it is

Except I think it’s double it

Wait no

Wait yes

@sleek swallow so uhh what am I supposed to do with the derivative of a sum

[sleek swallow]

Well, there's an explicit form for $S_n(x)$. It's just:

$S_n(x) = 1 + x + x^2 + ... + x^n = \frac{x^{n+1}-1}{x-1}$

So, your sum really is just:

$x \cdot \frac{dS_n}{dx} = x + 2x^2 + 3x^3 + ... + nx^n$

Where $x = \frac{1}{2}$

[vital dew]

Abhijeet Vats:

[sleek swallow]

So what can you do with that?

[bleak kite]

Integrate?

Shhh I’m tired and trying to do math

@sleek swallow plez explain I’m brain dead

Hmm

[sleek swallow]

Do you understand what has been written above?

So:

$x + 2x^2 + 3x^3 + ... + nx^n = x \cdot \frac{dS_n}{dx} = x \cdot [\frac{(n+1)x^n}{x-1} - \frac{x^{n+1}-1}{(x-1)^2}]$

At $x = \frac{1}{2}$, we have:

$\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3} + .... \frac{n}{2^n} = \frac{1}{2} \cdot [\frac{-2(n+1)}{2^n} - \frac{4}{2^{n+1}}+4] = -\frac{n+2}{2^n}+2 < 2$

[vital dew]

Abhijeet Vats:

[sleek swallow]

@bleak kite

[hybrid plume]

this doesn't make any sense. the power set should be bigger

[stray reef]

should it?

why?

[hybrid plume]

it s written by the professor

he said

this

because each subset can be identified with its characteristic function

[stray reef]

you have $|\bN^\bN|$ on the right, not $|\bN|$.

[vital dew]

Ann:

[stray reef]

it's true that $|2^A| > |A|$ for any set $A$, but this doesn't apply here.

[vital dew]

Ann:

[craggy gale]

what your Prof said does imply this
https://cdn.discordapp.com/attachments/496785905474994186/675993096273526794/Capture.PNG

[hybrid plume]

and in this context what does the n powe4r n mean

i know it s a bijective function

from n to n

[stray reef]

no

it's not "a" bijective function

it's the SET of ALL functions from N to N

[hybrid plume]

oh

[stray reef]

$X^Y$ is the set of all functions from $Y$ to $X$.

[vital dew]

Ann:

[hybrid plume]

so it s bigger then the power set

makes sense

thanks a lot

[tacit willow]

Could anyone give me a negation example of If it is sunny and windy today, then today we go sailing or kite flying.

[stray reef]

wdym by "negation example"

[tacit willow]

How would you negate the sentence_

[stray reef]

how would you negate "if A then B"?

[tacit willow]

if not A, then not B

[stray reef]

nope

the negation of "if A, then B" is "A, yet not B"

[tacit willow]

aha, alright. Thanks !

another question d. ¬∃𝑥(¬𝑄(𝑥) ∧ (𝑃(𝑥) ∧ 𝑅(𝑥))) No athletes who does not use performance enhancing drugs wins a medal

Does this look right?=

P(x) is "x is an olympic athlete", Q(x) is "x uses performance enhancing drugs" and R(x) is "x wins a medal". Universal set is set of all humans.

[sick fiber]

https://math.stackexchange.com/questions/178605/number-of-bit-strings-with-3-consecutive-zeros-or-4-consecutive-1s

Can someone help me.walk through this recursive solution ?

[weary tiger]

Hi everyone

How can I solve this

2x+3 congruent to 2 (mod 5)

or 2x+3 = 2 (mod 5)

[lyric pumice]

Hi.

Reduce the equation algebraically until you obtain an equality with a particular congruence class that is easy to evaluate.

@weary tiger

[weary tiger]

so is it just resolving 2x = -1 (mod 5) É

?

@lyric pumice

[lyric pumice]

Perhaps.

[copper ore]

~(S --> G) == (~S --> ~G) ?

are those two equivalent ^?

or does the arrow change when you start to put the negation inside

[weary tiger]

Perhaps? 😆

[copper ore]

lol

[weary tiger]

@copper ore your thing is pretty easily found with a truth table tbh

[copper ore]

oh yeah true

forgot about that

lol

thx

woah it's different

my mind is blown

[lyric pumice]

@weary tiger There are several possibilities for obtaining a useful congruence class.

You should end up evaluating for x outside of that class.

[weary tiger]

... I know that

I want to know what to do with the +3

Because to remove the 2 I have to multiply the inverse modulo 5

So maybe to subtract I have to do some weird thing also

And google only has simple examples like 3x = 2 (mod 5) and such

[lyric pumice]

I obtained the solution without multiplying.

[weary tiger]

That's awesome

Is this 1995? Am I playing Myst the puzzle game ?

[lyric pumice]

Keep reorganizing the equation.

[lyric pumice]

@teal latch Rewrite.

Type the problem.

[lyric pumice]

Start performing the Boolean algebra.

Continue.

Get those zs.

[lyric pumice]

@teal latch I'm sorry.

Use the addition rule.

[tacit willow]

is this correct?

[lyric pumice]

@tacit willow What do p and q mean?

[tacit willow]

aha, sry

p means 'he is rich, q means 'he is happy'

[lyric pumice]

What does "not q" mean?

@iron crescent Which problem is hard?

[lyric pumice]

That problem is nice.

There are no fancy tricks required.

You can solve it by using sets.

[charred cargo]

10 points) Use the method of direct proof to prove the following statement:
∀m, n, l ∈ Z,(m is even ∧ n is odd) → (ml + n is odd)

would someone be so kind as to walk me through this?

[narrow escarp]

dont know what direct proof is but m is even so write it as 2p for some p. n is odd so write it as 2q+1 for some q. then ur ml+n will look sth like 2s+1

[charred cargo]

<@&286206848099549185>

dont know if i should have done that

but

u guys help right?

[tough locust]

What's wrong with nnut's hint

@vernal violet you can prove it several ways, one way is induction

[charred cargo]

ok so would it look something like this ((2i) & (2i+1)) → (2i+1)

[tough locust]

What does the & and arrow mean

[charred cargo]

& means and

this

∧

[tough locust]

What does it mean that two numbers are and

5 and 3 implies 9

?

[charred cargo]

?

it means i add them?

[tough locust]

You wrote 2i and 2i+1 imply 2i+1

[charred cargo]

i thought that it meant both sides need to be true

[tough locust]

How can a number be true

Or false

[charred cargo]

i dunno man

thats where im stuck

[tough locust]

@vernal violet look up "triangular numbers" on wikipedia

Kozz you say

Let m=2k for some k and n=2r+1 for some r

[charred cargo]

ok

[tough locust]

Then you show 2k(ell)+2r+1 =2s+1 for some s

And that proves the claim

[charred cargo]

2k(ell)

?

[tough locust]

$2k\ell$

[vital dew]

gfauxpas:

[tough locust]

I didn't want to type "l" because it looks like "I"

So i said ell

[charred cargo]

so i can turn the arrow into an equals sign

[tough locust]

@vernal violet https://proofwiki.org/wiki/Closed_Form_for_Triangular_Numbers here's like 6 proofs

You may not understanding all of them and that's okay

Just try to understand the direct proof for starters

@charred cargo I wouldnt use "and" and "implies" here,I would use English

[charred cargo]

?

that is english

[tough locust]

Well that's just 1+1+1+...+1 n times

[charred cargo]

elaborate please

[tough locust]

You're not using & and -> properly and there's no reason to use that notation for this proof anyway

If you add 1+1+...+1 n times you're counting to n 1 number at a time so you get n

You said 2i & 2i+1 -> 2i+1, if i =3 then you're saying "6 and 7 implies 7"

It doesn't make sense

[charred cargo]

indeed

[tough locust]

Yeah the triangular number is the trickier one

[charred cargo]

so implies essentially equates to an equals slign?

sign

[tough locust]

No, implies means that if statement before the arrow is true then the statement after the arrow is true

[charred cargo]

you said that doesnt make sense in this context

[tough locust]

Like (my dog is red) & (my cat is red) implies (i have a red pet)

Because you were dealing with numbers, which can't be true or false

[charred cargo]

so then in this context

its an equal sign?

[tough locust]

What is?

[charred cargo]

implies

[tough locust]

5 & 6 = 6 doesn't make sense either

[charred cargo]

ok

[tough locust]

"5 and 6 is the same number as 6"

You're using notation from logic for numbers

"2 and not 7 if and only if -11"

Doesn't make sense

[charred cargo]

ok

2k(ell)+2r+1 =2s+1 for some s

how is the l in there

m= (2s) for some s
n = (2i + 1)
ml + n = (2i+1)

right

[tough locust]

They told you to consider $m\ell+n$

[vital dew]

gfauxpas:

[tough locust]

If n=(2i+1) and ml+n=(2i+1) then n=ml+n

That's not good

[charred cargo]

help me understand how you got that equation

step by step if you will

[tough locust]

You said n=(2i+1), I'm quoting you

You also said ml+n=(2i+1), im also quoting you

[charred cargo]

i see now how that doesnt make sense

thanks

[tough locust]

[charred cargo]

but help me understand how you got that equation

[tough locust]

If a=b and c=b then a=c, that's how = works

= means the left side and the right side are the same object

Different names for the same numbers

[charred cargo]

i dont get how that helps me derive this formula

could you help me with something else?

im not understanding this

(15 points) Use the method of direct proof to show that for any positive
5-digit integer n, if n is divisible by 9, then some of its digits is divisible by
9 too.

[soft thorn]

🤔

[charred cargo]

n = (some positive number) + 10000

right

[soft thorn]

Well, non negative

But sure

[charred cargo]

but hten i dont understand how to turn the second half into a logical statement

some of its digits is divisible by
9 too

how

like if i had the number 43255

[soft thorn]

If you're allowed modular arithmetic

This is pretty straight forward

[charred cargo]

umm

halp

I have no clue

[soft thorn]

Are you allowed modular arithmetic?

[charred cargo]

i dont even know what that means

but

[soft thorn]

Actually, it wouldn't be too bad without modular arithmetic either

[charred cargo]

i guess

[soft thorn]

Consider writing your number digit wise

Say the digits were a, b, c, d, e

How could you write n in terms of these?

And e

[charred cargo]

wouldnt it an,bn,cn,dn,en

[soft thorn]

No

and also in that specific order

Not that the ordering is important

For example

Any three digit number

I can write a * 10^2 + b * 10 + c

Right?

Then a will be the digit in the hundreds place

[charred cargo]

i guess

[soft thorn]

b the digit in the tens place

c the digit in the ones place

So how could you write n

[charred cargo]

i honestly dont know

[soft thorn]

Think about it

[charred cargo]

still cant figure it out

@soft thorn

you still there

[soft thorn]

Think about it very carefully

How can you write n in terms of a, b, c, d and e

[proven girder]

Would someone mind correct me if I am using De Morgan's Laws incorrectly here, please:

I believe it's true, but after doing a bit of searching I'm not sure if it's true as the second statement is in parentheses as oppose to an example:

[soft thorn]

the example is correct

the solution looks less correct

[proven girder]

Yeah, which is what led me to believe mine isn't too correct as if we apply the negation the "or" should change to "and" and the negations would be applied to j and k if I'm not wrong

But this also doesn't seem correct, so now I'm sandwiched on what is correct

[soft thorn]

that looks better

to give a sanity check

you can always think

what if j was true and k was false?

or other similar combinations

this won't necessarily tell you if something is correct

but it can very well tell you if something went wrong

[proven girder]

So in other words I could verify my answer with a truth table

[soft thorn]

a full truth table

would verify if it was logically equivalent or not

yes

a quick check wouldn't involve doing the whole thing

[proven girder]

Yeah if I wasn't able to solve it by just looking at them then that would've been my second option, but that would've took a bit longer

Thank you

[lyric pumice]

First, make three overlapping sets.

Put 6 objects into the region where all the sets overlap.

Put 8-6=2 objects into each region where two sets overlap.

Put 15-(6+(8-6)*2)=5 objects into each region where a set does not overlap with another set.

Then put 30-(6+3(15-(6+2(8-6))+8-6))=3 objects outside of the sets.

The probability is 3(15-(6+2(8-6)))/30=1/2.

[proven girder]

Can someone approve of my negation (b)

[analog sonnet]

I think I'm missing some context about this problem

[tacit willow]

Does this look correct?

[sleek swallow]

Lmao

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

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

[vital dew]

Abhijeet Vats:

[sleek swallow]

Also, your truth table isn't correct because there are supposed to be 8 combinations of truth values for p,r & s. So, you missed out FTT and FTF

@somber ember

But anyways, none of that matters cos you can 'demonstrate' it using the equivalence above. I don't know what 'laws of equivalence' they're talking about. They're probably literally referring to the laws of boolean algebra, in which case that's some weird terminology.

[open narwhal]

What does it mean when they are asking for an inverse of a statement? For example: "If Samantha lives in Paris, then she lives in France."

[sleek swallow]

Let $p \implies q$ be a given implication. Then, the inverse of the implication is just $\lnot{p} \implies \lnot{q}$

[vital dew]

Abhijeet Vats:

[sleek swallow]

I'm not sure what you mean by 'inverse of a statement'. Inverses of implications are defined but i've not seen the inverses of other kinds of logical expressions being defined.

[open narwhal]

We are learning in class about the inverse, converse, contrapositive, ect of statements and i keep getting them all confused on how to write them

[sleek swallow]

It's very simple. Let $p \implies q$ be a given implication. Okay?

[vital dew]

Abhijeet Vats:

[sleek swallow]

These are just things you have to get used to, so don't sweat it too much

I'll go through them with ya, yea?

[open narwhal]

Id appreciate that

[sleek swallow]

So, the above statement is any given implication. p and q are just propositions. To keep things simple, we'll say that they are primitive propositions, though there really isn't anything stopping them from being logical functions of two or more primitive propositions.

Then, the converse of $p \implies q$ is just $q \implies p$.

[vital dew]

Abhijeet Vats:

[sleek swallow]

The inverse of $p \implies q$ is just $\lnot{p} \implies \lnot{q}$

[vital dew]

Abhijeet Vats:

[sleek swallow]

The contrapositive of $p \implies q$ is just $\lnot{q} \implies \lnot{p}$

[vital dew]

Abhijeet Vats:

[sleek swallow]

The point is that this is just new language that you have to pick up and it's not gonna happen in a day. So, keep using it and refer back to the definitions when you're using them or when you encounter them in questions. You'll get used to them slowly.

I should say that I do NOT condone any kind of willful memorization of this or anything else. Instead, work it into your vocabulary in a natural way by using it in the correct context.

[open narwhal]

honestly your explination helped a lot to understand the thought process behind it

thank you

[sleek swallow]

You're welcome. Don't memorize any of this, use it and make it a natural part of your language.

[tacit sandal]

Hi,
I need help with the following question:

I have the following values for S: S = {(1,3), (1, 4), (1, 5), (2, 5), (1, 1), (2, 2), (3, 3), (4, 4), (5, 5)}
Show a Hasse Diagram for S```
I've done it like this, is it correct?

[weary tiger]

Hi everyone

Why is (~A + A * ~B) == (~B + ~C) * A

[sleek swallow]

[weary tiger]

I can rewrite it maybe it'll help

(A * ~B) + (A * B * ~C) == (A * ~B + A * ~C + B * ~A)

[sleek swallow]

No i mean

What the fuck are you writing lol

[weary tiger]

Boolean expression?

~ is for NOT

[faint narwhal]

• and * are pretty standard symbols for 'or' and 'and' respectively

[sleek swallow]

Are they? Idk I've not really seen them used in logic

[faint narwhal]

mostly in CS classes

Not really in math classes

[sleek swallow]

Oh okay okay

Person, post the exact statement of the problem

[weary tiger]

I want to simplify

(A and ~B) or (A and B and ~C)

the answer is

(A and ~B or A and ~C or B and ~A)

but how? I'm stuck at this step

I have tried "A and (~B or B and ~C)"

but then I'm still stuck

[sleek swallow]

Okay so:

$[A \land \lnot{B}] \lor [A \land B \land \lnot{C}] = A \land [ \lnot{B} \lor (B \land \lnot{C})] = A \land [ (\lnot{B} \lor B) \land (\lnot{B} \lor \lnot{C})] = A \land [\lnot{B} \lor \lnot{C}]$

[vital dew]

Abhijeet Vats:

[weary tiger]

Thanks boss 👍 ☕

[sleek swallow]

This is for CS, yea? Then you'll have to convert it into the notation that's used most commonly in your CS class

[weary tiger]

Yea no worries

I was stuck at this step, converting it to the right notation is no prob

Thanks again

[sleek swallow]

You're welcome.

CS has some weird notation for this, not gonna lie

[faint narwhal]

Makes a bit more sense in terms of bits if you let 0 be false and 1 be true

[hollow bloom]

If we say p only if q, is that the same as p implies q?

[sleek swallow]

Yea

[weary tiger]

@sleek swallow

How did you distribute

[hollow bloom]

Thank you!

[weary tiger]

um

[sleek swallow]

Distribute?

@faint narwhal hmm alright lel that's interesting

[weary tiger]

(~b or (b and ~c)) === ((~b or b) and (~b or ~c))

[sleek swallow]

Yea what about it

I distributed as per normal using the distributive law

[weary tiger]

but there are three variables there

~b, b, and ~c

none are common to both sides of the +

of the or *

[sly pewter]

Isn't distributive law like for three variables

[weary tiger]

omg

😆

fail

[sleek swallow]

$p \land (q \lor r) = [p \land q] \lor [p \land r]$

[vital dew]

Abhijeet Vats:

[sleek swallow]

Also, no

Distributivity can work for two variables as well

It's just that there's a simplification you can do

And i really shouldn't be calling them variables. They're propositions or, at the very least, open sentences

[sly pewter]

I mean I think they meant p, q, r as variables

[weary tiger]

thanks everyone

[sly pewter]

Not really variables ye

[hollow bloom]

If r is a sufficient condition for s does that just mean r implies s?

Just another fancy way of saying if r then s?

[sleek swallow]

Indeed

@sly pewter yo mamma wasn't a variable

Prank

Relax

Memes

[hollow bloom]

oooo

thaats kinda memean

[sleek swallow]

But yea, those are all equivalent ways of saying r implies s

[hollow bloom]

gotcha

[sleek swallow]

But usually, we stick to the symbolic convention, since it's easier to do algebraic manipulations on the symbols

[hollow bloom]

thanks bb

what do you mean

by symbolic convention?

like the arrow

?

[sleek swallow]

Well, we usually say $p \implies q$

FUCK

[hollow bloom]

F

[vital dew]

Abhijeet Vats:

[sly pewter]

Do you mean your mamma was variable?

[sleek swallow]

Oh shit rekt

[hollow bloom]

Can't call something that don't exist a variable

[sly pewter]

Sorry for shitposting here btw

[sleek swallow]

But yea, we stick to that

[hollow bloom]

Mkay

[sleek swallow]

Cos it's easier to say that:

$p \implies q = \lnot{p} \lor q$

[vital dew]

Abhijeet Vats:

[hollow bloom]

Oh ya

[sleek swallow]

In symbols than it is to write out the words for it lol

[hollow bloom]

I learned that just few hours ago

Helped me with my homework

LOL

is texit

in latex

[sleek swallow]

Latexit

[hollow bloom]

$\sim p$

[vital dew]

Tamaarine:

[hollow bloom]

oo

[sleek swallow]

Gosh, I'm wishing to have homework at this point

Jesus

[hollow bloom]

lol

very funny

😦

[sleek swallow]

It's true lol, I'm dying to get my first midterm and first homework

But 7 months to go before that happens

[hollow bloom]

wot

why aren't you in your semester?

[sleek swallow]

Cos i'm dying in the army lol

[hollow bloom]

ooo

[sleek swallow]

So 7 months to go before i fuck off to uni

[hollow bloom]

Salute

Can't find a emoji that does that

o7

There

[sleek swallow]

Yo mamma does that

Prank

[hollow bloom]

how could this happen to me

i made my mistake

your mum is a contradiction \s

[sleek swallow]

Oh shit rekt

[hollow bloom]

good roast

very well done

i know

wait @sleek swallow do you mind taking a look at this one

A sufficient condition for compoundXto boil is that its temperature be at least 150◦C

If p is the compound x to boil, q is temperature to be at least 150

I said it was the same as the conditional

[sleek swallow]

So a sufficient condition for p is q

[hollow bloom]

Mhmm

[sleek swallow]

So what's the conditional you can write?

[hollow bloom]

I wrote p implies q

For the conditional

Then for that sufficient condition

I also wrote p implies q

My bad should have give context

[sleek swallow]

should have give context

Prank

[hollow bloom]

react to myself

feels bad

[sleek swallow]

Uh okay, so let p = compound X is boiling and q = temperature must be at least 150

Then, the statement is saying that p => q. Hence, p is sufficient for q and q is necessary for p.

Option a is just the converse and that doesn't have to be true just because the conditional is true.

Option b is the contrapositive of the implication. That means it has to be true.

Option c is just rephrasing the implication so it's the same as the implication.

Option d is the inverse of the implication and that doesn't have to be true.

Option e is just saying that q is a necessary condition for p, which is exactly our implication.

Option f is saying that q is a sufficient condition for p, which doesn't have to be true.

Sorry took a while, I'm out in the field now

[hollow bloom]

No but jesus

i didn't expect you to verify all for me

wtf

i had all of the mdone

just f

wanna make sure

but i give my sincere thanks

now i can check my answer

pog

[sleek swallow]

Check mine cos i'm in a bit of a rush so i wrote them down quickly

Make sure i didn't make any mistakes lel

I'll check it later myself

[hollow bloom]

okay a-d

is exactly the same

but e and f

i flipped

I was basing it on this shiet

[sleek swallow]

$\lnot{r} \implies \lnot{s}$ is the inverse of $r \implies s$

[vital dew]

Abhijeet Vats:

[hollow bloom]

Mhmm agree

Which is false

[sleek swallow]

And the thing is, it's the contrapositive of $s \implies r$

[vital dew]

Abhijeet Vats:

[sleek swallow]

So r would be a necessary condition for s and s would be a sufficient condition for r if s implies r were true

[hollow bloom]

im having a mind fuck right now

can you elaboate

a little more

you are like my professor

im dumb

[sleek swallow]

I am flattered that you would compare me to a professor lol

Especially since I'm only a dumb army boi

[hollow bloom]

bruh

i barely

understand shit you just said

you at least know

what you are saying

[sleek swallow]

I can go through the basics with you

[hollow bloom]

if you have the time

please

[sleek swallow]

But ultimately, this is just new language and you'll only get used to it through repeated use done over a period of time.

[hollow bloom]

2 weeks in man

and professor gave out homework in latex already

never touched latex before

[sleek swallow]

Okay, so $p \implies q$ is a proposition iff p is a proposition and q is a proposition.

[vital dew]

Abhijeet Vats:

[sleek swallow]

So far so good?

[hollow bloom]

Proposition is just true or false

Right?

[sleek swallow]

Yes, a proposition is just a sentence that's either true or false and nothing more.

[hollow bloom]

Gotcha

[sleek swallow]

Now, one can ask what a 'sentence' is but I don't think that'll concern you unless you take linguistics or pure logic.

In any case, there are several different ways to talk about an implication or a conditional.

So, $p \implies q$ is the same as:

1. If p, then q.

2. p if q

3. p is a sufficient condition for q

4. q is a necessary condition for p

FUCK

[hollow bloom]

Nope I can read it

Don't worry it

[vital dew]

Abhijeet Vats:

[hollow bloom]

👌

So q is necessary c ondition for p

is the same as p implies q

gotcha

[sleek swallow]

So, there are other types of implications that have specific names

You'll pick these up over time, as you solve more problems

So, suppose that $p \implies q$ is our given conditional. Then, we have the following special conditionals:

1. $q \implies p$ is the converse of $p \implies q$

2. $\lnot{q} \implies \lnot{p}$ is the contrapositive of $p \implies q$

3. $\lnot{p} \implies \lnot{q}$ ist he inverse of $p \implies q$

[vital dew]

Abhijeet Vats:

[hollow bloom]

converse flip

[sleek swallow]

Once again, don't memorize these. Learn them over time and make them a part of your language

[hollow bloom]

inverse just negate

awwww

i was memorizing them

[sleek swallow]

Nope

Include what I've said above as a part of your notes

[hollow bloom]

I'm going to have to write them down

[sleek swallow]

But don't memorize them willfully. Weave them into your language

[hollow bloom]

Okay I tried

Really

In class when professor said it first

About the converse and contrapositive

I tried my best to understand it

But the thing is

It really gets confusing sometimes

[sleek swallow]

Of course it does

It's a new language