discrete-math

[slate burrow]

Just a quick question, while solving this am i not suppose to solve for x?

[toxic ermine]

Can you help me with this?

[slate burrow]

I can't sadly, but there are others that might help :) read #rules

[toxic ermine]

okay, thanks anyway, ig I'll keep trying :/

[heavy fulcrum]

let's give it a try @toxic ermine shall we?

starting with (a)

take the LHS

then the first bracket says (A-B)

what's the definition of this?

[heavy fulcrum]

@slate burrow the first one, for example, can be interpreted as:

"We get a remainder of 1 when we divide x by 3"

[slate burrow]

right...

so we don't really care what x is?

because when solve it the x is still unknown

[heavy fulcrum]

Yes and No

For example:
We get a remainder 0 when we divide 6 by 3 can be written as
6/3 = 2 + 0 or
6 = (2)(3) + 0

[toxic ermine]

The elements in A but not B

[stray reef]

@weary tiger it is now

[heavy fulcrum]

Exactly!
By definition, in other words (A-B) means: x∈A AND x∉ B

[stray reef]

as far as solutions go i would rather derive the losing positions more rigorously, whence will come both the recurrence listing them all and the win strat.

2 is a losing position, as can be observed directly.

then make a claim: for a losing position L, all positions from L+1 to 2L inclusive are wins while 2L+1 is a loss

and the strat in any position from L+1 to 2L is to cut off a piece of size L

thus making your opponent quite literally take the L

[heavy fulcrum]

@toxic ermine
So on the LHS you have: (A-B)∩(A-C) which translates to:
x∈A and x∉ B AND x∈A and x∉ C
Rearranging gives:
x∈A and x∉ B and x∉ C
Which can be written as:
x∈A and (x∉ B and x∉ C)

[stray reef]

@weary tiger i do hope you appreciate my wordplay somewhat.

[cloud cargo]

the second one, $(A \cup B) - (A \cap B) = A \implies B =\varnothing$ is kinda interesting

[vital dew]

jabra

[heavy fulcrum]

Take out the common thing inside the brackets...
x∈A and x∉(B and C)
now it's clear that x is in A but not in B and C so therefore:
A-(B∩C) is the case, which equals to the RHS

[cloud cargo]

can i ask a question for the second one

[heavy fulcrum]

Yes?

[cloud cargo]

so either A \cap B is disjunct, or it is not. if you suppose that it is, then B={} follows directly since A \cup B = A iff B = {}

now suppose that A \cap B is not disjunct, then there exists an element a in A such that a is not in (A \cup B) - (A \cap B), hence (A \cup B) - (A \cap B) is not equal to A

but this contradicts the hypothesis so A \cap B mustn't must be disjunct right

[heavy fulcrum]

I'm reading it..

That sounds right

Yea.. that's right

[cloud cargo]

oh wait it should be but this contradicts the hypothesis, so A \cap B must be disjunct right

the bottom line

[toxic ermine]

Thanks, I appreciate your help:)

[cloud cargo]

since A \cap B not being disjunct leads to the contradiction in the hypothesis

so basically, $( (A \cup B) - (A \cap B) = A \iff (A \cap B) = \varnothing) \implies B = \varnothing$

[vital dew]

jabra

[cloud cargo]

oh that should be an => instead of <=> in the middle

[stray reef]

are you asking about the expression "take the L"?

or are you asking about how i'm using the letter L here outside of the pun value

cause i say exactly what it means

for a losing position L,

[foggy cloud]

for a_2k+1 I can make the base case but I don't really know how to go after that

like I show that a3 < a1 as the base case but it gets messy after that

[vale cairn]

I just had a go myself and found proving something slightly different is actually a bit easier

Now note we wish to show (a_2k) is increasing, so we need to prove that 1/2 (a_(2k-1) + a_(2k-2)) >= a_(2k-2); that is, a_(2k-1) >= a_(2k-2) for all k, or a_(2k+1) >= a_2k after shifting the index

Try proving that by induction. Then it's fairly straightforward to prove the stuff we need in a)

[stray reef]

wipe it completely

no

sigh let me just write this out

ok nevermind it turns out i dont have the energy to write it out in full

and i only saw your edited version now

i dont feel like fixing the wording

[weary tiger]

hey guys

any idea on what i could do if it was 4^n instead of constant 4

[slow jewel]

Can someone help me with 6(ii)

[simple nova]

Does anyone know loop invariants?

Stuck with no help

[west zenith]

Hint: ||Take the complement||

[slow jewel]

lol

By 3 it must have 35 edges

the complement had 10

so 10 vertices and 10 edges

thats all i know

Cant really proceed from there

[west zenith]

There's another condition on the complement

What does G being 7-regular mean for its complement

[slow jewel]

thats what i was tring to figure out

not sure

maybe n-k-1

not really sure

[west zenith]

You can surely figure out what the degrees have to in the complement

[slow jewel]

no

sorry

[west zenith]

Draw some examples

This is a general phenomon: if you take the complement of a k regular graph on n vertices you get...

[slow jewel]

is it k-regualr

@west zenith

is it k-regualr

[west zenith]

Hum no

Pick any vertex. It has degree k. How many non-edges are there incident to that vertex?

[slow jewel]

n-k

?

[hearty sky]

there exists a professor
there exists a (potentially empty) set of students
that set of students contains the students who like >0 professors
that professor is liked by every member of that set

[marble pivot]

I have to show that E is an equivalence relation

right now I’m trying to intuitively understand the formula

Why did they keep going instead of stopping at (C ∪ C ⁻¹)^2? Doesn’t that satisfy the transitivity requirement already?

[stray reef]

take S = N and C = the relation defined as xCy iff y = x+1

then going up to only (C \cup C^-1)^2 will only get you the relation |x-y| <= 2

[viscid moat]

Given a tabular representation of an operator, is there an easy way to determine if that operator is associative or not?

tabular meaning the value of x op y is found in the row x column y of the table

for commutativity, you simply check if the table is symmetric across the \ diagonal

but is there an easier way that literally going through every combination for associativity?

[patent terrace]

Hi i was given this question

https://cdn.discordapp.com/attachments/825663698911363113/893383602321096705/179505699192176640.png

[vital dew]

ksg

ksg

[pale epoch]

sure, but its unnecessary

oh wait, your M should be natural

yeah, then just round it up

[patent terrace]

so ceiling?

[pale epoch]

sure

you can also floor it and add 1

or just floor actually

but why not add 1 for good measure

[patent terrace]

ok thank you!

[cerulean wind]

just apply the archimedean property

so if e > 0, then there is an N such that 1/N < e. then for any n > N,…

[pale epoch]

(thats a theorem due to eudoxus actually, but it follows directly from the archimedean property)

[last flicker]

do you think it's true or false?

[wind hawk]

I think its true, I cant see an example where its not but Im having trouble proving it

[last flicker]

Well, write down a|2b and a|b in terms of the definition of divides, and see if you can notice anything that could lead you from one to the other

[wind hawk]

Ive done that, so far I got b=((2k+1)(m))/2 k,m being integers but im stuck there

[last flicker]

not quite the right direction. leave it at ma=2b or m(2k+1)=2b and think about what you can draw from the parity of both sides. In particular ||what can you say about the parity of m?|| and after that ||can you use that to rewrite m to get ra=b for some integer r||

I'm sorry for the long response and whatever im struggling to give a hint that isnt just the next step of the proof

i can tell you it exactly if you want but the spoilers are pretty strong hints

If you aren't comfortable with facts like the parities of the sums and products of odds and evens, I'd say a) get on that and b) for now you'd have to expand out m(2k+1)

[cerulean wind]

note that this is an application of euclids lemma

so one could argue based on parity (evenness and oddness of an integer) or based on prime factorizations

[wind hawk]

we havent used the term parity in my class yet so thats throwing me off a little

okay even odd got it

would i have to make 2 cases for this proof?

[cerulean wind]

if 2b = ak for some integer k, then ak is even and a is odd. k has to be either even or odd. figure out which one works.

[cerulean wind]

nope. you can do it all in one go

[slate burrow]

Question: For what natural numbers n can the following expression "1+2+3+...+n" be a prime number? Example n = 1, it is 1, n=2 it is 3, not a prime.

And for hint it says use arithmetic sums to see how many divisor the results has.

Like does that explanation make any sense?

[pale epoch]

no

a1 is the first n? what?

what is being converted where?

3 is not prime?

[slate burrow]

sorry i mean it's a prime

a1 is basically sum of n=1 which is just 1

i'm not sure if that's the correct equation for arithmetic sums, i have not worked with arithmetic sums before

[pale epoch]

well, it is well known that $$1 + 2 + \dots + n = \frac{n(n+1)}{2}$$

[vital dew]

Lochverstärker

[slate burrow]

is this the formula for sums?

or do you mean that my problem is equal to that?

[pale epoch]

that's a formula for the sum of the first n positive integers

so you have to figure out when n(n+1)/2 is prime

[slate burrow]

right, where the hell did i get that equation from then lool

ok so i believe i know how to prove it

see how many divisor the results has. i think the solution is in this clue right here

[slate burrow]

i'm trying to prove this, and if i use n(n+1)/2 then you can easily show that n is a non prime as long as n > 2 if we use the definition for odd numbers to prove that we can't get a prime number if we take sums of the numbers.

actually does arithmetic series have a formula?

[untold raptor]

hi im trying to prove (P → Q) → (¬Q → ¬P) but only using 3 axioms, modus ponens, hypothetical syllogism and double negation equivalence, i cant seem to find the correct way to start this problem. can someone give me some pointers?

[sly temple]

what would this be?

[pine iris]

recall $P \rightarrow Q$ means $\neg P \vee Q$

[vital dew]

Mns98

[sly stag]

hallow

[tribal zenith]

The solutions said there is only one equivalence class for R6, but I can't see why. Does this relation mean a is related to b if a^k if a multiple of n iff b^k is also a multiple of n?
If we let a to be 6 and b to be 2, aren't they in different equivalence classes? Since 6^1 is a multiple of 6, while 2^1 isn't? 6 is not related to 2?

[untold raptor]

sorry about the late apply but yes we are only allow to use 3 axioms, modus ponens, hypothetical syllogism and double negation equivalence we cant use contrapositive , de morgan law etc

[tribal zenith]

If we have a relation R = {(1, 1), (1, 2), (2, 1), (3, 2)}, is the transitive closure of R {(1, 1), (1, 2), (2, 1), (3, 2), (3, 1)}? Adding arrows to make it transitive?

[stray reef]

no you're missing (2,2)

[storm osprey]

Hi guys, do you guys know a website were i can construct a circuits for this (x + y)’ x

[onyx zephyr]

What is the method or formula for finding all the solutions for a diophantine equation within a range?

Say you had the following question:

i misspelt "i" instead of "in" here but u get the gist

How I do it is put 192y + 165x = 21, euclid's algorithm on (192,165) and then use it backwards to get the "solution" for it which I get

3 = 192(-6) +165(7)

What do I do now to solve it?

And if possible, is there any good resources on the internet for learning/doing these types of questions?

Doing a lot of new number theory/discrete with relations/eqv/diophantine eqv etc but haven't found a good online website for the topic as khan academy doesn't have it which was my go-to.

[pale epoch]

multiply both sides by 7, reduce mod 192

"these types of question" is hard to answer

this is a linear diophantine equation which we understand perfectly, so there isn't really much work you have to put into solving them

there is a theorem that tells you all the solutions

in general diophantine equations are impossible to solve

[onyx zephyr]

Thanks! @pale epoch for answering, sry for late response.

I am not going to lie, I didin't quite understand why Z192 became a variable here. Am I reading the question wrong maybe? What I read is find all solutions to the equation 165x = 21 where x can be any integer between 0-192. Am I reading it right?

[pale epoch]

it asks you to find solutions modulo 192

[onyx zephyr]

165x = 21 (mod192)

?

[pale epoch]

yes

[onyx zephyr]

oh

[pale epoch]

or in other words: when does 192 divide 165x - 21
or yet in other words: when is
$$ 192y = 165x - 21$$
or
$$165x - 192y = 21$$

[vital dew]

Lochverstärker

[pale epoch]

the last thing is called a linear diophantine equation and you find solutions in Z the way you did with euclidean division essentially

[slow jewel]

Can i assume connectedness here

[onyx zephyr]

I see, thanks Loch!

[pale epoch]

molecules are connected, yes

[slow jewel]

thanks

[onyx zephyr]

I was going to ask, looking at the theorems like the following:

the last line, does it say that v = b/gcd(a,b)

[tribal zenith]

isn't (2, 2) reflective?

[stray reef]

this question makes no sense

your relation contains (2,1) and (1,2). if transitivity is to hold, it must also contain (2,2).

[tribal zenith]

oh so we just add (2,2) since R may not be transitive from the start

[stray reef]

R is not transitive, as it happens.

[tribal zenith]

oh okay

how about this

is this transitive?

[stray reef]

the way you presented it is certainly not very easy to parse

but yes it does appear to be transitive

[tribal zenith]

how should we present it? in terms of {(-1 ,-1), (-1, 0), (0,0), (1, 0), (1, 1), (1, -1), (-1, 1)}?

[stray reef]

......maybe try to position the points in a way that makes the arrows less cluttered, idk.

[tribal zenith]

oh ok

[weary tiger]

So if $X$ and $Y$ are monotonic sequences (not necessarily strictly monotonic), then is $X+Y$ a monotonic sequence? And if it isn't, what is a valid counterexample?

[vital dew]

Adam S

[gritty crescent]

@weary tiger Try working out a few examples, in particular, think about how opposite signs on corresponding terms can influence the outcome (combined with the fact that the sequences are not strictly monotone).

A possible solution, look only after you've spent some time trying: ||X=1,1,2,2,3,3,... and Y=-1,-2,-2,-3,-3,... and then X+Y=0,-1,0,-1,0,...||

[weary tiger]

Yeah i figured it out, one possible counterexample would be x_n = n^2 and y_n = -4n

[gritty crescent]

Sounds good

[raw musk]

Hi can someone help me with this question: An automorphism of a simple graph G = (X, E) can be defined as a
bijection f from the set X to itself, such that if the vertices x, y ∈ X
are neighbors in G, then f(x) and f(y) are also neighbors; in other
words, if xy ∈ E then f(x)f(y) ∈ E. Show that for any automorphism
f of a tree T = (X, E), there exists x ∈ X such that f(x) = x or there
exists xy ∈ E such that f(x) = y and f(y) = x (“fixed point property”:
fixed the vertex or fixed edge by isomorphism f).

[sharp basin]

For all real numbersr and c with c ≥ 0, if −c ≤ r ≤ c, then
|r| ≤ c. Can anybody explain the second case that is the case were r< 0 i understand the first case

Its a proof

[vale cairn]

if r < 0, what is |r| ?

[pine iris]

note -c <= r < 0 < -r <= c

[sharp basin]

by definition would it be -r

[vale cairn]

yup

Now we need to prove that -r <= c

Can you see why that's the case?

[sharp basin]

yes is it because c \ge 0

c greter than or equl to 0

[vale cairn]

That doesn't mean -r <= c, since both are positive numbers

Consider rearranging the inequality (or multiplying through -1, being careful, equivalently)

[sharp basin]

r \ge - c

[vale cairn]

yup, which we know to be true

obviously that doesn't constitute a proof, but it allows us to now write up one

Suppose r < 0; r >= -c, so |r| = -r <= c

:)

[sharp basin]

ok that was my confusion

thanks potato

[vale cairn]

np!

what mns said is a condensed version of it but yes

[tranquil flint]

How many cycles of length 2n are in Kn,n

Is it (2n-1)!/2

[toxic ferry]

How can I rewrite m(m-1)...(m-n+1) into a nicer factorial expression?

[obtuse lance]

as a ratio of factorials

think of it as m! with the terms that continue on as starting with (m-n)(m-n-1)...

so you divide those out as m!/(m-n)!

[austere swan]

There's also the pochammer symbol which is the same thing

But it's less common

[mystic elbow]

Hi can anyone please help

<@&286206848099549185>

[pale epoch]

what have you tried?

[mystic elbow]

I’m not able to do it

Like I don’t Understand what it says to do

[pale epoch]

what do you not understand

[fiery hare]

Does any one know any computationally less expensive form of this? $\sum^{\lceil n/2 \rceil}_{k=0}{{n \choose k}k}$

[vital dew]

strexicious

[fiery hare]

(i don't really know where this question fits in. it's related to a programming problem and i may have more in the future, so just give me directions and i will follow in the future)

If I don't have that factor of k then I think I can get away with just doing 2^(n-1)+{n choose n/2}/2

n has an upper bound of 500k

[red nest]

Here is a nice form: If n is even this is $n2^{n-2}$ If n is odd then it is $n2^{n-2}+\frac{n}{2}{n-1 \choose \lceil \frac{n}{2} \rceil}$

[vital dew]

saketh

[weary tiger]

realyl appreciate any help

Assume that the cryptographic key (ie the pair (a, b) in [ax + b] _41) is changed daily. After how many days are you forced to repeat the same crypto? (Alternatively: on how many attempts is the encryption broken?)

[fiery hare]

:o that's a very nice form. and i am sorry to be a bad client but now I realise that k goes up to n, not n/2. don't want to to rob more of your time so can you share how you got this formula so i will try to adapt it to the new constraint

[red nest]

If k goes upto n, then the formula should be $n2^{n-1}$ the way you get that is by saying that $k {n \choose k}=n{n-1 \choose k-1}$ for $k \geq 1$ (the k=0 term is 0 so we don’t have to worry). Now the sum is just $n \sum^{n}_{k=1}{ {n-1 \choose k-1}}}$ which is just $n2^{n-1}$ by adding the binomial coefficients

[vital dew]

saketh

If k goes upto n, then the formula should be $n2^{n-1}$ the way you get that is by saying that $k {n \choose k}=n{n-1 \choose k-1}$ for $k \geq 1$ (the k=0 term is 0 so we don’t have to worry). Now the sum is just $n \sum^{n}_{k=1}{ {n-1 \choose k-1}}}$ which is just $n2^{n-1}$ by adding the binomial coefficients
```Compilation error:```! Extra }, or forgotten $.
l.55 ...ust $n \sum^{n}_{k=1}{ {n-1 \choose k-1}}}
                                                  $ which is just $n2^{n-1}$...
I've deleted a group-closing symbol because it seems to be
spurious, as in `$x}$'. But perhaps the } is legitimate and
you forgot something else, as in `\hbox{$x}'. In such cases
the way to recover is to insert both the forgotten and the
deleted material, e.g., by typing `I$}'.

Preview: Tightpage -1310720 -1310720 1310720 1310720
[1{/usr/local/texlive/2020/texmf-var/fonts/map/pdftex/updmap/pdftex.map}]```

[fiery hare]

Oh shit you are right. I had this but I was staring at the (red boxed) factor and I didn't realise it was essentially (n-k) = (n-1) - (k-1)

[red nest]

Yeah, it’s a little tricky. Btw you don’t have to worry about wasting my time with this question, I basically had to figure out the second version of the question to get the answer for the first version

[leaden dust]

I'm struggling here with the bi conditional

Also not sure if I should try to simplify the bigger one or expand the smaller one

[weary tiger]

quick question

do the outside symbols mean round down?

ok yeah it does mean round down

[pale epoch]

yeah and whoever wrote that can't tex properly

[leaden dust]

Anyone?

Dying over here lol

[tranquil flint]

Anyone know how to count how many cycles of length 2n in Kn,n 🥺

me stuck

[harsh geode]

yes it is the "floor" symbol

[copper oasis]

Is x^k log n O(x^k)?

No, right? because x^k logx > x^k

forall x > some value c_0

[toxic ferry]

Can anyone give me an idea where to start by proving that f is surjective? I know the definition, but the recursion is really throwing me off.

[red nest]

here is a hint: If $f^{n}=Id$ then $f(f^{n-1})=f^{n-1}(f)=Id$ so these functions are inverses of each other

[vital dew]

saketh

[toxic ferry]

Where did you get $f(f^{n-1})=f^{n-1}(f)$ from?

[vital dew]

bunkermush

[toxic ferry]

It should be $f(f^{n-1})=f(f(f^{n-2}))$

[vital dew]

bunkermush

[red nest]

associativity of composition

In any case you don't need the second equality for surjectivity, I just put it in there to say that f and f^{n-1} are inverses of each other

[weary tiger]

Can someone explain why if P implies Q is a contradiction, P must be a tautology?

[toxic ferry]

do you mean commutative?

f(g(x))=g(f(x))

[red nest]

no

f(gh)=(fg)h, associative

[toxic ferry]

that requires three functions. I only see two: f^n-1 & f

Sorry this is frustrating I always struggled with functions

[red nest]

ok so the point is that we have f(f^{n-1})=f(f(f^{n-2}))=(ff)f^{n-2} and we basically kkep repeating this until we get f^{n-1}f. But don't worry too much about this point, it is entirely irrelevant to the proof that f is surjective

[short steeple]

i have a combinatorics question, so i'm not sure whether to ask it here or in #combinatorial-structures; i'll ask it here first and if it belongs in the other chat please let me know :)

i have to come up with a combinatorial proof for this recurrence relation for Stirling numbers of the second kind: for positive integers $n$ and $k$ with $n\geq k$,
$$S(n,k)=\sum_{i=0}^{n-1}\binom{n-1}{i}S(n-1-i,k-1)$$

[vital dew]

Snodlop

[short steeple]

my current solution starts with placing n-1 items in k-1 boxes, and observing that the last item has (n-1 choose 0)=1 place it can be placed

but i don't think that argument can be used once i move on to placing n-2 items into k-1 boxes

any help will be appreciated, thank you :)

[short steeple]

what i mean by this is -> from placing in the n-2 items in k-1 spots to satisfy S(n-1-i,k-1) for i=1, i'm not sure how i'm supposed to count anything relating to (n-1 choose 1)=n-1 from distributing the remaining two items

[short steeple]

it has been around half an hour and unfortunately i am still stumped, so i'm going to have to drop an <@&286206848099549185> on this one, sorry

[north dust]

This combinatorics class I'm taking is killing me

[short steeple]

same

apparently it only gets worse from here for my class lol

[north dust]

The double summations are wack

Ahhh, good luck though!

[short steeple]

thank you! you as well

[golden thunder]

does turing machine need to finish all input and land in an accepting state to accept an input?

or

only need to land in an accepting state to be accepted

[scarlet current]

i got a question regarding sets. I need to show that the cardinality of a set is less than the cardinality of its power set. If i show that a function f from S to P(S) is injective this shows that the cardinality of a set is less than OR equal to the cardinality of its power set, right? Do i need to show that the function is not surjective to confirm that the cardinality of a set is less than AND not equal to the cardinality of its power set?

[golden thunder]

formally yeah

but

power set contains atleast the amount of items in the original set

and then +1 for empty set (then + more)

so thats enough

so for example, {a,b}. powerset = {({},{a},{b}) ,{a,b}}

[gritty crescent]

Yes, you need to demonstrate that it cannot be surjective.

Sorry, I don't know how to approach this problem.

[weary tiger]

Hello. I have a general question about the binary representation of integers. Given three consecutive integers, n, n+1 and n+2, and given their respective decomposition into sums of distinct powers of 2, how can I show that every power of 2 does not appear an even number of times in the sum n + (n+1) + (n+2)?

[slate burrow]

I want to solve for a and b, i'm wondering is there a way of turning the plus sign to multiplication?

I tried different ways, the only thing that makes sense. Is that addition can be written as multiplication, for instance if we have 4+4 we can write it as 2x2x2 but i'm not sure how i can implement that here

[tribal meadow]

wouldnt u use logarithms to solve for a

[slate burrow]

that was my first thought, but i don't think it works, since we'll get ln(1) which is 0

the only solution i can come up with is b,a=3

[vale cairn]

Are a and b integers

I mean I assume they have to be

[slate burrow]

natural numbers

[vale cairn]

ye

[slate burrow]

yeah natural numbers

[vale cairn]

Anyway, hint: 2^a = b^2 - 1

you can now consider the factors of both sides

[slate burrow]

i see now

[vale cairn]

epic

[slate burrow]

thx

[vale cairn]

np

[slate burrow]

how did i not see it xD

[vale cairn]

aha dw

[slate burrow]

so when we convert it to b^2-1 it's no longer addition, right?

after we factorize it*

[vale cairn]

i'm not sure what you mean by 'no longer addition' here

[slate burrow]

one sec

[tribal meadow]

use difference of two squares

that will do that

[vale cairn]

oh lol yeah fair that must be what it means ig

so yes difference of two squares

[slate burrow]

right

i see now

[vale cairn]

epic

[slate burrow]

thanks a lot everyone

[vale cairn]

np

[slate burrow]

ohhh this problem is very similar to catalan's conjecture, no?

The only solutions are if a = 3 and b=3

[onyx zephyr]

When doing permutations and combinations, I use the nRp and nRc formula. But what do I do when I get questions like this that add new rules?

"no two or three may be married to each other", with this exception the formula doesn't work. What's the workaround?

Answer became 10x8x6 incase someone curious

[vagrant tiger]

how do i read this statement

[pale epoch]

there exist $x_1, x_2, ..., x_8$ such that $\alpha(x_1, \dots, x_8)$

[vital dew]

Lochverstärker

[vagrant tiger]

so there exists x1-8 such that alpha x1-8 is true?

i believe the statement tells us which tuples are true trying to figure out how to parse it

[pale epoch]

well, yes

\alpha(...) is just some relation

not further specified

the tuples themselves aren't true

[vagrant tiger]

hmm the tuples are proven t/f by the statement no?

[pale epoch]

no, the tuples are just tuples

ordered lists of elements

\alpha is like a function that assigns a truth value to them

[vagrant tiger]

yes

[pale epoch]

it could be a statement like $x_1 \leq x_2 \leq \dots \leq x_8$ or $x_1 = x_2 = \dots = x_8$ or some wild stuff like $x_1 > x_2 \land x_2 \neq x_3 \lor \dots$

[vital dew]

Lochverstärker

[vagrant tiger]

how do the quantifiers apply to the relation

[pale epoch]

what do you mean?

this is just another statement, that tells you that there are 8 elements that are related in some way specified by \alpha

[vagrant tiger]

there exist $x_5,

[uneven belfry]

hi mathematicians

me need help

Derive $(k+1)^{3} - k^{3} = 3k^{2}+3k+1 ,from ,the ,value ,of \sum_{k=1}^{n}k^{2}.$

does someone know where to begin with

[vital dew]

Emma762

[elder berry]

Do you perhaps need to find the sum of k^3?

Since you don't need the value of k^2 to show that equality, just expand the left hand side

[uneven belfry]

alrighty lemme do it rn

[elder berry]

Or maybe it is asking you to determine a formula for sum of k^2 using that equation

[uneven belfry]

3k^2+k +1

[obtuse lance]

yeah, question sounds backwards

if you take that polynomial equation and sum over it the left side becomes a telescoping series and simplifies nicely

[uneven belfry]

is derive and lead out the same here?

lead out ... from the value of

[elder berry]

Lead out is really strange wording

[uneven belfry]

we havent seen telescoping series as of rn

if i expand lhs i get 3k^2+k+1 so we get 3k^2+k+1=3k^2+3k+1

if k>0 then rhs should be bigger

i think

[void vale]

Ive been stuck on this problem all day 😦

[quartz musk]

@void vale for the initial conditions, since n must be greater than or equal to 3, n would be 3

initially

n-1 would be a2, n-2 would be a1. Substitute in the values and you have your answer

[uneven belfry]

it says find the infimum right?

since n element of natural numbers the domain would be IR (no complex numbers) so how do u find the highest lower boundary or it doesn't have an inf?

[cerulean wind]

what? the domain is the natural numbers

and yes it says find the infimum of the set of all rationals of the form 1/(n^2 + 1) with n in N

[uneven belfry]

if domain is natural numbers

then it would be [0,+inf[ this means 0 is inside

wouldn't inf be 0 zero

since its the biggest lower boundary

for us N means 0 is part and N^+ 0 isn't

[last timber]

@uneven belfry it does have infimum. find it.

[exotic jacinth]

Hey there, I really need help with my discrete math homework... Am I allowed to hire someone? There is 10 questions in total.

[cerulean wind]

no, you can’t hire people from this server but you can ask homework questions here and people can offer help

[safe hawk]

i am stuck 😦

[stray reef]

what are you stuck on?

the rather odd directions they give you?

@safe hawk

there are at least two different ways you could be stuck.

either "i have no idea what to even do" or "i'm confused at what the 'using acid order linear blah blah blah' stuff is about"

or it may be something else entirely.

would you be so kind as to enlighten us what it is?

oh, who am i kidding. i'm a fool to ever expect such levels of cooperation from anyone.

[onyx zephyr]

Kinda rude not to answer you I agree @stray reef

[stray reef]

i've accepted that people generally don't want to answer a rabid lunatic with female hysteria such as myself.

[onyx zephyr]

:/

[slate burrow]

Wtf

I don’t think its gender related. Or i hope not

Some people just ask and don’t answer afterwards which is like wtf

[pale epoch]

it happens to me too, mostly when i ask people what they have tried or tell them to try anything 👻

[onyx zephyr]

I have a question. I am doing combinations and permutations and got to a word problem I got stuck on.

"Let M be the set of all words that can be made with the letters RAGGARGRANEN"

Q1: Amount of words?
A1: 12!/3!3!3!2!

Q2: How many words in M contains the word GRENAR?

How am I supposed to think for the second question?

To save some time, RAGGARGRANEN = 3R 3A 3G 2N E

GRENAR = 2R A G N E

[radiant thorn]

How do I prove B has the winning strategy if G has a perfect matching ?

[radiant thorn]

<@&286206848099549185>

[gritty crescent]

Let GRENAR be X, and consider the permutations of X with the remaining letters, accounting for repetitions

[stray reef]

@radiant thorn write down B's winning strategy in terms of said matching.

[subtle charm]

Can someone explain to me why A is false? I don't get it.

[pale epoch]

set P(x) = "x is even" and Q(x) = "x is odd" over the domain of integers

[subtle charm]

okay, but if I pick an x say 1, then won't the LHS be true and the right hand side be true too?

[pale epoch]

you don't get to pick an x

there are quantifiers

LHS is "every number is even or odd" which is true

RHS is "every number is even or every number is odd"

which is not true

[subtle charm]

Oh i get it now. Thanks a lot!

[slate burrow]

How can i prove that this equation has only one solution and that is only if a,b= 3?

$2^{a}+1=b^{2}}$

[vital dew]

Steppaz
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[slate burrow]

My reasoning is that 2^3 is the only value for a that you can have if you want to get b. Because as a increases in value the difference between the values of a and b increases as well

This is like catalan’s conjecture. They just changed the variables

[slate burrow]

<@&286206848099549185> sorry for the tag but i am still confused on how i can answer this problem ^^

[safe marlin]

hi need some help please:
how can i know what is the inverse function of this function:/
f(x) = 2*[x] - x

[coral raven]

consider that 2^a = b^2 - 1, does that help?

[slate burrow]

Oh, yes I've already done all that :) but I am stuck at the part where i have to explain why a,b=3 is the only solution :D

[coral raven]

ok so what did you do with 2^a = b^2 - 1

[slate burrow]

This problem is very similar to Catalan's conjecture, but i don't understand the reasoning behind the conjecture

so

[coral raven]

yeah

[coral raven]

so the conjecture is unproven, and this is a specific case of it, so don't worry about catalan's

[slate burrow]

I reasoned that 3 is the only solution, since as a and bgrow larger and the difference becomes more noticeable between them

[slate burrow]

welp

[coral raven]

that's nothing to do with it

[safe marlin]

@coral raven i beg you need ur help haha man

[coral raven]

one sec sorry

[safe marlin]

ur beast xd

[slate burrow]

the inverse is y = f(x)

[civic horizon]

catalan is proven

[coral raven]

it's not

[safe marlin]

here's my question:
hi need some help please:
how can i know what is the inverse function of this function:/
f(x) = 2*[x] - x

[coral raven]

not right now sorry

[slate burrow]

so it has nothing to do with the difference

[coral raven]

nothing to do with the difference between a and b, no

[austere swan]

Catalan's conjecture was proven in 2002

[coral raven]

as a grows, b must grow much faster to catch up, but there's nothing wrong with that

[slate burrow]

Does it have something to do with odd and even numbers in (b-1)(b+1)

[coral raven]

really? damn

i could have sworn

[civic horizon]

i mean for (b-1)(b+1) to be a power of two both b-1 and b+1 have to be powers of two

[coral raven]

oh well it's really advanced mathematics right

anyway

if (b-1)(b+1) = 2^a

[slate burrow]

yes

[coral raven]

b-1 and b+1 must be factors of 2^a

[civic horizon]

there are only two powers of two which are 2 apart

[coral raven]

but what do you know about the factors of 2^a

they must be powers of 2 themselves

[slate burrow]

right

wdym powers of 2?

[coral raven]

like

[slate burrow]

and which are those?

[coral raven]

b-1 = 2^c, b+1 = 2^d

[coral raven]

bruh

[slate burrow]

xd

like

im losing my mind

right

right right

sorry

lol

[coral raven]

ok, if you've got that then you can take it from there right

[slate burrow]

yea perhaps, let me see what i can do with that information

I totally knew that "different between a and b" had to be wrong but it lowkey felt right

omg

i see it now

i now understand what "it has to be power of 2"

so its 3 since (3-1)=2 which is 2^1, and (3+1)=4 which is 2^2

[coral raven]

basically

[slate burrow]

nice

so this only happens for 3 then right?

[coral raven]

let x = [x] + r, so [x] is the integer part and r is the decimal

then y = 2[x] -x = 2[x] - ([x]+r) = [x] - r
then y + 2r = [x] + r = x, does that help?

[safe marlin]

wait so whats the answer

im confuse xd

[coral raven]

so i gave you x = y + 2r, is that not enough

[safe marlin]

well the final answer is:
= -2[-x] -x

[coral raven]

ok

[safe marlin]

but idk how

need to understand why xd

[coral raven]

so x = y + 2r

try working it out from there

[safe marlin]

i mean is it a good idea to a new parameter (r) ?

[coral raven]

yes

[safe marlin]

im confuse xd

how im suppose to get it to = -2*[-x] -x

[coral raven]

so i switched x and y initially because we're finding the inverse, so when you switch it back it'll look similar but different

and after that you can sub r out again by using the fact that x = [x] + r

so it's perfectly possible

[safe marlin]

i dont know

it dont give me the same answer xd

[coral raven]

so what do you get after doing that

[safe marlin]

i get it to :
= 3x -2[x]

the final answer is:
-2[-x] -x

@coral raven any ideas pleasE?..

[weary tiger]

How many different reversable affin functioner E(x) = [ax + b]_1000 are there?
1000 = 2^3 * 5^3
abs(U_1000) = fi(2^3 * 5^3) = fi(2^3) * fi(5^3) = 1 * 2^2 * 4 * 5^2 = 400
U_n is the subset of all elements from Z_n which are orthogonal with n, such that x in U_n => gcd(x,n) = 1
That they are reversable means that a decryption function D(y) exists, right?
-> It is equivalent to answering: In how many ways can a and b be choosen?

[coral raven]

the difference is because r is the non-integer part of x, not of y, so just switching x and y directly doesn't quite work

[safe marlin]

idk what to do to be honest..

[coral raven]

let y = [y] + s, and try and relate s and r?

[safe marlin]

lemme try it

@coral raven it will just become more comlicated man

complicated

[coral raven]

might be a red herring, hold up

ok so let's start over for clarity

x = [x] + r

and y = [x] - r

you basically want to get r and add it to y twice

so r is just how much less than [x] y is

and [y] will always be equal to [x] - 1

because y = [[x] - r] = [[x-1] + 1-r] = [x-1] + 1-r

so [y] + 1 = [x]

[coral raven]

and given this, r = [x] - y = [y] + 1 - y

so as x = y + 2r:
x = y + 2([y] + 1 - y) = 2[y] + 2 - y

and this works

[safe marlin]

but it dont give the same answer

in the end.

[coral raven]

gimme a sec to explain why

[safe marlin]

the answer is:
= -2[-x] -x

[coral raven]

so now switching this round

we get y = 2[x] + 2 - x

your answer is -2[-x] -x

[safe marlin]

without the 2

[coral raven]

done

[safe marlin]

idk need to read again what u explained im confused to be honest

[coral raven]

so basically we just want to show that

2[x] + 2 = -2[-x]

ie. [x] + 1 = -[-x]

wait

so f(2.4) = 1.6
then -2[-1.6] - 1.6 = -2(-1) -1.6 = 0.4

so this doesn't actually work

by [x] do you mean the floor function

[safe marlin]

weird

[coral raven]

or the truncate function

[safe marlin]

ye

[coral raven]

which one

[safe marlin]

floor

[coral raven]

oh well if it's floor then it works

[safe marlin]

for example:
[2.4] = 2

i didnt understand well what u said haha

[coral raven]

if it's floor, then -2[-1.6] - 1.6 = -2(-2) - 1.6 = 2.4 and it all works out

[coral raven]

look

essentially this function just flips the bit after the decimal point

it takes 2.4 = 2 + 0.4

and it gives out 2 - 0.4 = 1.6

right?

so to get back to 2.4, we need to add the bit after the decimal point twice

[safe marlin]

wut u mean

[coral raven]

1.6 + 2(0.4) = 2.4

[safe marlin]

?

[coral raven]

which bit don't you understand

[safe marlin]

idk man

the function is:
y = 2[x] - x

[coral raven]

yes

[safe marlin]

idk how you do it idk

[coral raven]

which part of my shortened explanation below here did you not understand

[safe marlin]

its all g man im just giving up of this question ty

[coral raven]

ok

[safe marlin]

what you mean here?

..

[coral raven]

so x = [x] + r

yes?

[safe marlin]

ok?

[coral raven]

do you understand

[safe marlin]

ye..?

[coral raven]

ok

so y = 2[x] - x = 2[x] - ([x] + r) = [x] - r
so y = [x] - r

got that?

[safe marlin]

ye

[coral raven]

ok

so if x = [x] + r
and y = [x] - r

then x = y + 2r

[safe marlin]

why?

oh

nvm

okay?

[coral raven]

ok

[safe marlin]

i got this

[coral raven]

so now it gets a little more complicated, but basically we want to find r starting from y

because then we can put that r into x = y + 2r, and get x from just y

[safe marlin]

so i need to put in this x = y + 2r this r = [x] -y ?

[coral raven]

something like that

[safe marlin]

k?

[coral raven]

so to find r starting from y

basically y = [y] + 1-r

[safe marlin]

how u got this?

[coral raven]

ok so

y = [x] - r = [x-1] + 1-r

and so [y] = [x-1]

[safe marlin]

how u got that [y] = [x-1]

[coral raven]

so if y = [x-1] + 1-r

basically [x-1] is a whole number right

it's an integer

[safe marlin]

idk how you got this still [y] = [x-1]

[coral raven]

ok but you agree that [x-1] is an integer

?

[safe marlin]

ofc everything under []

is int

[coral raven]

yes

now, 1-r =< 1

so if y = [x-1] + 1-r

then [x-1] is the largest integer below y

[safe marlin]

okay but how u know 1-r < 1

[coral raven]

but the largest integer below y is the same as the floor, so [y] = [x-1]

[coral raven]

r is always positive

[safe marlin]

why lol

it can be 0

[coral raven]

oh, right

yeah i shoulda been using leq

ok i edited it

[safe marlin]

so its 1-r <= 1?

[coral raven]

yes

[safe marlin]

ok?

so [y] = [x-1]

[coral raven]

wait no

[y] = [x-1] except when y = x

[safe marlin]

thats when r = 0

[coral raven]

yes

but the formula works in that case also

so we don't need to worry about it

so assuming that r > 0 and [y] = [x-1]

[safe marlin]

ok?

[coral raven]

so anyway now we can say:
y = [y] + 1-r

so r = [y] + 1 - y, right?

so now we have r in terms of just y

[safe marlin]

must be a more simple way maybe?

[coral raven]

eh

[safe marlin]

this is very complicated way tbh

[coral raven]

i was breaking it down into very small steps and going very slowly so you would be sure to understand

it's not so bad

[safe marlin]

but wait

[coral raven]

anyway now we have r in terms of just y

[coral raven]

?

[safe marlin]

if i do x = 0 and x = 2

it will give me y = 0

in both of em

[coral raven]

x = 0 and x = 2?

[safe marlin]

so how this func have a inverse func

[coral raven]

what?

[safe marlin]

ye

[coral raven]

no

f(2) = 2

right?

[safe marlin]

yeye

[coral raven]

ok

so we have r

then x = y + 2r

so x = y + 2([y] + 1 - y) = 2[y] + 2 - y and we're done

[safe marlin]

2[y] + 2 = -2[-y]?

[coral raven]

yes

[safe marlin]

hmmm

[coral raven]

because [y] + 1 = -[-y]

[safe marlin]

thats true

[coral raven]

so the full proof in one big chunk without words looks like:

x = [x] + r
y = [x] - r = [x-1] + 1-r = [y] + 1-r
r = [y] + 1 - y
x = y + 2r = y + 2([y] + 1 - y) = 2[y] + 2 - y = -2[-y] - y

so not too long

there probably is a slightly faster way but i don't see it

[safe marlin]

@coral raven i almost understand everything man ur beast im just gonna ask about what u said that 2[y]

that 2[y] +2 = -2[-y]

i mean if we take y = 1 its not gonna work?

since 4 is not equal to 2

[coral raven]

oh, right

[safe marlin]

so

[coral raven]

well this is for [y] != y

[safe marlin]

r is should not be equal to 0

[coral raven]

if [y] = y, then that whole argument doesn't quite work

[safe marlin]

this is true

[coral raven]

but the formula works anyway

it's simple to show

[safe marlin]

if r is between 0 and 1

[night jackal]

https://gyazo.com/aea78e12d2bfad8863f3bed372e53db9

what proof method is used here?

[pale epoch]

they tell you what they use next to the lines?

[night jackal]

i meant like is it a direct proof?

[elder berry]

What is bugging you? Is it the summation notation?

[coral raven]

do you know what induction is

[coral raven]

well there's your problem

google around until you find a video you like or something

[oak notch]

Hello. How do you measure a cycle? Do you count the number of vertices or the number of edges?

[obtuse lance]

what's the difference?

[oak notch]

As far as I can tell, there isn't any. Which one do you count though? (as preference)

[obtuse lance]

try to prove the number of vertices is equal to the number of edges for a cycle

they are equal

[oak notch]

Alright then. Thanks!

[oak notch]

Does anyone know what the name/term of this equation is?

It's for getting the max number of edges with n vertices

[obtuse lance]

you could write it as n(n-1)/2 and call it the nth triangular number I guess

but number of edges on a complete graph of n vertices is fine too

[oak notch]

Is there a technical term for it?

On a different note, I like your username. I haven't encountered that word before

[obtuse lance]

idk what you're looking for

binomial coefficient? combination? n choose 2?

[oak notch]

Okay, how would you call the left expression? The (n 2)

How would you say it?

[obtuse lance]

n choose 2

[split drum]

Does this seem valid?

[upper wolf]

If x is a positive real number and > 0, then x-1 is not necessarily >= 0. For instance, x = 0.5 is in R+, but x-1 < 0.

[split drum]

Frick

[split drum]

Could you provide a hint as to where to go from here? I've been looking at this for a while.

[upper wolf]

Well, to salvage what you currently have, I don't see that you're actually using the fact that x-1>=0 anywhere in your argument. Removing that portion does not invalidate that (x-1)^2>=0 because you already state that any real number squared is >= 0.

[split drum]

Oooh, OK. So I can just say that since x is real number, then x-1 is a real number and then (x-1)^2 >= 0?

[upper wolf]

Yep, no problem there. The real numbers are closed under subtraction.

[split drum]

Awesome, thank you!!

[split drum]

<@&286206848099549185>

Does this proof make sense logically? I thought it was good, but then a friend said I need more cases than this. 😕

If anyone answers, please ping me. I'll be back to check in 5.5 hours after I get some sleep.

[steel granite]

how can i solve recurrence relations
(that is CS math

[onyx zephyr]

Anyone know a place I can pay a 1 on 1 tutor for really basic discrete math learning? Got some old exams I am practicing on could use some guidance.

[pale epoch]

try a messaging board from your uni

i think a local person is the best bet

[onyx zephyr]

I don't mind doing it online through voice even without a camera really should not be an issue.

If anyone is interested please let me know.

[tribal zenith]

Let A = {1, 2, 3}. Consider C = {y ∈ A : x R y} : x ∈ A for relation R.
If I define the relation R to be > , would C be {{2, 3}, {3}, {}}? Or {{}, {1}, {1, 2}}?

[gritty crescent]

The relation is defined on elements of the set, why are you considering subsets?

Oh mb

I think I misread slightly

I find the notation slightly unusual, just to be clear, $C={y\in A : xRy\text{ for some }x\in A}$?

[vital dew]

Manan.

[pale epoch]

randoms are just more likely to scam you

hence i suggested locals

[split drum]

Can someone tell me if this is a correct statement: $n^4 &\equiv n^2 \pmod{4} \ = 4|(n^4 - n^2)$

[vital dew]

Umiguess4000
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[split drum]

Which can then be rewritten as $4k = (n^4 - n^2)$?

[vital dew]

Umiguess4000

[onyx zephyr]

.

@pale epoch when i did this problem u told me to "reduce by modulo 192" in order to find all the solutions after doing euclid backwards (extended). What do you mean by that mathematically?

How would I go about that etc

[plain ridge]

How would I go about this?

[obtuse plaza]

Is there any site or app that can give a answer of questions?

Of discrete maths?

[coarse cave]

working with factorials, how would it be set up?

[vale cairn]

there are only 4 things to calculate so you can just do it manually xd

i.e. by exhaustion

[last granite]

maybe figuring out some upper and lower bounds for the recursion might work,even guessing. like sqrt(n)<sqrt(n+100) and sqrt(n+100)<n/2 (asymptomatically)

btw check out cormens algorithms book for more on these

also you can keep in mind subtracting a lower order term from a solution does not change the solution

maybe the lower order term can help you deal with the algebra

[quiet sapphire]

Anyone can help me with this?

<@&286206848099549185>

[deft dock]

i believe for the first you write the combinations of the gates that make the function true (a value of 1)

for part a

so for example, (P and Q and R) make the function true

and so does (P and Q and not R)

so write out the combinations that make the function true, and then combine them with an or, as any one of those will make the function true

and i think for B you use logical equivalences to simply part A even further, and then ig draw it using logic gates

no clue about 4

[slate burrow]

I'm pretty sure my base case is not correct but i'll fix it later

i don't know how i can get rid of a in the last step :/

[true zinc]

anyone here good at logical proofs

[hidden crystal]

<@&286206848099549185> can someone explain this and some other problems to me

@distant glade

@rugged forge

@iron marsh

[cerulean wind]

shtaup

[hidden crystal]

can you help? @cerulean wind

[cerulean wind]

no stop tagging people. your question will be answered soon enough. why is it so urgent?

[hidden crystal]

Ok. @solid trench

@mighty cape

[coral raven]

<@&268886789983436800>

[ember obsidian]

@hidden crystal stop pinging random users

[hidden crystal]

why?

is that against the rules?

[hazy pine]

yes.

[ember obsidian]

yes

it's also common sense, it's very obnoxious

[hazy pine]

not to mention against basic decency

helpers are volunteers, dont expect them to respond to your beck and call.

also wait 15 minutes before pinging helpers

[hidden crystal]

i'm looking for the rule preventing me from doing that and I don't see it

[lofty cloud]

From

#❓how-to-get-help

Rule 4

If your question has not been answered for a minimum of 15 minutes, you may use the Helpers tag once. Please do not try to bump your question using this ping unnecessarily. Do not abuse this ping. Do not individually ping users with the Helpers tag without their express permission.

[hazy pine]

there you go.

[hidden crystal]

oh ok i didn't see that

i didn't know asking for help too much is against basic decency in a discord server for helping people with math

[ember obsidian]

we're not talking about asking for help

[coral raven]

don't ping specific ppl, if they feel like helping they'll help anyway

[ember obsidian]

it's about obnoxiously pinging every online user on the list

we have a designated helper ping for this reason that you have already (prematurely) used

[hidden crystal]

i only ping helpers

[coral raven]

you already pinged helpers

why ping ppl more than once

[hidden crystal]

because i didnt know if the helper ping worked

[coral raven]

??

why tf would it not work

[hidden crystal]

usually a little bot says its pinging helpers

i didn't see that

[ember obsidian]

no bot here confirms that

[coral raven]

yeah, what

[hidden crystal]

ok, i didnt know that

[ember obsidian]

it works if you see the green highlight

[hidden crystal]

i'm guessing dm'ing helpers isn't acceptable either

these decency laws really prevent getting help in a server for math help

[coral raven]

pinging so many ppl doesn't actually make it more likely for you to get help

[hidden crystal]

asking more people for help repeatedly doesn't make it more likely for you to get math help?

ok.

[coral raven]

yes

it's rude

[hidden crystal]

then don't volunteer to be a helper?

if you don't like people asking for help

[coral raven]

you can want to help ppl and also not want to be pinged multiple times immediately

[hidden crystal]

some questions are more urgent than others

[ember obsidian]

jman you're on thin ice. keep complaining about the rules while not acknowledging the fact you abused the server & you're out

[coral raven]

you probs shoulda asked earlier then

anyway

wrt. your actual q

what have you tried

[hidden crystal]

whatever.

[coral raven]

bruh

[cerulean wind]

that method of increasing your chances at getting help seems to be working really well

[hidden crystal]

oops

[cerulean wind]

hint: factor out a 3 and then note that 18 = 9 * 2

[hidden crystal]

ok

[smoky horizon]

you could also treat it as an algebra problem if you can't see the pattern
6m(3m+24)=9x, you're trying to solve for x

x=(1/9)*6*m*(3*m+24)
x=(2/3)(3m^2+24m)
x=2m^2+16

[hidden crystal]

<@&286206848099549185>

I don't get it and google isn't helping.

i'm sure its easy

[weary tiger]

If you divide that by 4a, do you get an integer

[hidden crystal]

i can't divide 2a by 4a

it won't fit @weary tiger

[weary tiger]

“That” is “2a*34b”

[hidden crystal]

i can't divide 2a by 4a

[weary tiger]

Simplify 2a*34b

$2a\cdot 34b=?$

[vital dew]

Tesseract

[hidden crystal]

2(a * 17b) ?

how do i simplify.

[weary tiger]

$2a\cdot 34b=(2\cdot 34)ab=?$

[vital dew]

Tesseract

[hidden crystal]

ok, i never simplify like that

but now that?

[weary tiger]

what is 2*34

[hidden crystal]

38

68

(68)ab

[weary tiger]

Now, divide by 4a

$\frac{68ab}{4a}=?$

[vital dew]

Tesseract

[hidden crystal]

17b

[weary tiger]

Is this an integer

[hidden crystal]

yes

[weary tiger]

So, does 4a divide 2a*34b

[hidden crystal]

yes

[weary tiger]

There you go!

[hidden crystal]

thanks @weary tiger

[neon fulcrum]

Write the negation of each of the following logical expressions so that all negations immediately precede predicates. In some cases, it may be necessary to apply one or more laws of propositional logic.

know how to make the entire expression negative
I know some of the laws but I don't know how to distribute the negative through the expression?
This is what I got so far

If someone can tell me whats next I'll know the rest

[neon fulcrum]

<@&286206848099549185>

please help ME!

if you want ofc

[weary tiger]

Change the quantifiers

then do the implication next

[neon fulcrum]

Is this what you mean?

@weary tiger

[haughty garden]

The first negation shouldn’t be there

You would have pushed the negation symbol inside when you swapped the quantifiers

[neon fulcrum]

What is that law called?

I have no idea how to distribute the negative

[haughty garden]

It comes directly from the negation of quantifiers: [ \lnot \forall x. P(x) \equiv \exists x. \lnot P(x) ]

[vital dew]

opengangs

[neon fulcrum]

question ^

[haughty garden]

Yeah

[neon fulcrum]

ohhhhh

did not know that

What is there were two quantifiers?

forgive the terible ms paint

[haughty garden]

Note that what you really have is [ \lnot \left(\forall x. \exists y. P(x, y)\right) \equiv \exists x. \lnot\left(\exists y. P(x, y)\right) \equiv \exists x. \forall y. \lnot P(x, y)]

[vital dew]

opengangs

[haughty garden]

So you can think about it as "pushing the negation symbol inside"

but in doing so, you flip the quantifiers

[neon fulcrum]

so confusing

[haughty garden]

It comes from your negation laws on quantifiers

[neon fulcrum]

so basically to distribute the negative you inverse the quantifiers and negate the last thingy that isn't a quantifier?

[haughty garden]

[ \lnot \left(\forall x. P(x)\right) \equiv \exists x. \lnot P(x)] and [ \lnot \left(\exists x. P(x)\right) \equiv \forall x. \lnot P(x)]

[vital dew]

opengangs

[haughty garden]

it might help to think about what your statements are saying and why the negation is what it is

[neon fulcrum]

like the sentence version?

[haughty garden]

Your statement is saying there is an x so that no matter what value of y I pick, P(x, y) -> Q(x, y)

So if I want this to be false, it must be the case that "for any x I pick, I can always choose a y such that P(x, y) does not imply Q(x, y)"

[neon fulcrum]

Ok. I will work on this a little bit

let me try and see if I can do it

[haughty garden]

I think once you understand how to convert the sentence version into its negated form, the symbolic counterpart will make a lot more sense

[neon fulcrum]

how is this?

[haughty garden]

looks good

[neon fulcrum]

do you know what that step is called bytw??

distribution of negative?

[haughty garden]

I don't think we've ever called it by a name, I think we've referred it to as "pushing the negation in" or something

[neon fulcrum]

ok, thank you very much for your help @haughty garden you rock!

[mystic elbow]

Hi

Can anyone help

[elder berry]

The hint itself explains what you should do really well, what have you tried?

[elder berry]

A lot of things are wrong here

1) You shouldn't argue the case when P(k+1) is 2^(k+1) x 3^(k+1) 
since P(k+1) is actually 2(k+1) x 3(k+1)
2) This should be true for any board, not just 2k x 3k, but also 2k x 3m, or 2k x 3(k+100) which you haven't considered
3) There are other errors like `split the board in half which would yield P(k) case` which is untrue, consider splitting a 2x3 board in half, it simply doesn't make sense because 3 is not even

Have you tried using the hint they provided and follow that?

[mystic elbow]

Hmm I see I correct them btw could u check some of the other questions pls

[jagged junco]

Hi.
Let n be native number. How many subsets of {1,2,...2n} the sum of their elements is even? (Don't give in your answer sigma, not allowed)

---

I understand that we need to divide into cases by the amount of odd numbers we have in the subset but I don't understand how to start.. please help?

[ornate cliff]

S = Ƿ({a, b, c, d, e, f, g, h, i}) [ Ƿ: Power set], (A,B) belongs to R if and only if |A| = |B|.

I’m confused about the condition “(A,B) belongs to R if and only if |A| = |B|”

Does this mean you can only have a subset with the same cardinality/number of elements e.g.
R = {{a}, {b}, {c}, {d} … }
or
R = {{a, a}, {a,b}, {b, b}, {b, a} … }

[worthy mist]

it means if you have two sets A and B and their cardinality is equal then R contains the tuple (A, B) so among other things R contains ({1,2,3,4}, {a, b, c, d})

[ornate cliff]

Thank you!

[paper adder]

-(Ax.P(x))

[valid loom]

Let A = {1, 2, 3, 4, 5, 6, 7, 8}.
Show the steps to obtain (3,5,7,8)∘(1,3,2).

guys anybody knows how to answer this question

I kinda confuse with the ∘ symbols

[gritty crescent]

@valid loom Are these cyclic permutations?

And if so, what's your convention for interpreting them (compose them left to right, or the other way)?

[valid loom]

@gritty crescent yes I think so since it has ∘ symbols but I didn,t really understand how to do it

[gritty crescent]

It still depends on whether your convention is to read permutation compositions right to left (more common and likely here, given the composition circle) or conversely.

I'll assume the former.

So the first parentheses is (132). What this means is that 1 is mapped to 3, 3 is mapped to 2, and 2 is mapped to 1. Everything else is mapped to itself.

Now in the next cycle (3578), 3 is sent to 5, 5 is sent to 7, 7 to 8, 8 to 3. There's one thing to notice though: the first cycle sent 1 to 3, so it essentially 1 that got mapped to 5. Can you now describe what the final permutation cycle would be?

I usually find it helpful to draw diagrams and chase where each element is going with arrows

[valid loom]

@buoyant escarp I dont really get about the final permutation cycle

@gritty crescent btw thanks that a lot of help tho

[crude nymph]

Q : Write a proof of the following argument: Every pet in the pet shop is a cat or a dog. Every cat in the pet shop plays with yarn. There is a cat in the pet shop that naps a lot. Therefore, there is a cat in the pet shop that play with yarn and naps a lot.

I just need help in which I should use a conjunction or a conditional for "Every cat in the pet shop plays with yarn" and "There is a cat in the pet shop that naps a lot"
So are both of them conjunction or conditional cause Im getting mixed up between it

I think that they are both conditional statements am i wrong or right?

[haughty garden]

You would probably use De Morgan’s law to push the negations and then apply distributivity

[crude nymph]

Yo guys

https://cdn.discordapp.com/attachments/895686154442145862/895687363211198514/4521512.PNG

[meager prairie]

yo

[crude nymph]

this is for rules of inference if I had smth like c(a) V d(a) changing it to c(a) would still work cause of the addition method or nah?

[haughty garden]

c(a) V d(a) doesn’t imply that c(a) needs to be true

Addition says that if a proposition is true, then the disjunction of that proposition with another is true

So if P is true, then P V Q is also true

But the converse doesn’t need to hold

~(p ^ q) is a proposition

So you can apply distributive law twice

R is just a placeholder

You can treat ~(p ^ q) as r if you like

Actually you would only need to apply distributive on the second expression

~(p V q) V (p ^ q)

By De Morgan’s, you have (~p ^ ~q) V (p ^ q)

Then applying distributive, you have
(~p V (p ^ q)) ^ (~q V (p ^ q))

The red line would be correct

It would apply to the entire expression if the negation symbol was outside

Yup looks right

[crude nymph]

So nah it wont work is what your saying

Cause i am trying to prove c(a) is true

Maybe I did something wrong in my steps

[haughty garden]

c(a) V d(a) is not enough to suggest that c(a) is true

You need c(a) V d(a) and ~d(a)

[crude nymph]

Ok

∀x(c(x) V d(x)) im trying to break this down using the rules of inference

[haughty garden]

Once you apply De Morgan's law, you would get (~p V ~q) V (p V q) which is just (~p V p) V (q V ~q) which is a tautology (here, since everything is with respect to V, you can switch the orders around)

yeah you could jump from the line to the result

[wild peak]

guys how do i do this