discrete-math

[tropic cedar]

The book is wrong for the one-one when saying
x^2+y^2+xy-1=0 has no solutions

It's onto

Take aforementioned limits

[cerulean ore]

It does have solutions?

the book is writing x^3-x but, the question is x^3-3x

[mossy palm]

Kaori, this book is better https://infinitedescent.xyz/

[cerulean ore]

Yeah it's onto, I don't know how limits work but yeah graph says it is

[cerulean ore]

@mossy palm let me try that book

[hasty cargo]

Prove $\frac{1}{n}$ where $n \in \mathbb{Z}$ has a decimal expansion of period at most $n-1$

[vital dew]

Itadakimasu!:

[hasty cargo]

Any help? I tried letting the expansion be $a_1 a_2.. a_m$ but had no idea where to go from therre

[vital dew]

Itadakimasu!:

[reef thistle]

Hmm...

Why is there a periodic expansion in the first place?

(Yeah, I know, just pushing @hasty cargo to the answer.)

[hasty cargo]

Errr, I guess it's rational?..

[reef thistle]

Why should it repeat with a certain period?

(actually there's a preperiod iirc, but let's ignore that first)

[hasty cargo]

Hmm uh, one doesn't the number doesn't divide nicely into 1? As in it's not a factor of 10

I'm not sure urrr

[reef thistle]

(and I think the question wants us to ignore the preperiod)

[hasty cargo]

But yeah I included the preperiod thing earlier as well but it wasn't helping much ^

Oh okay

Wait I meant for the not a factor of 10 as in

[reef thistle]

So, what makes it repeat?

Not a factor of a power of 10, you mean

[hasty cargo]

Yeah^

[reef thistle]

Why should it repeat instead of just going on some non-repeating pattern?

[hasty cargo]

I donkj sdhfjksdhf jl

I don't know :<

[reef thistle]

Think about how you calculate the decimal expansion.

[hasty cargo]

The decimal is an infinite sum?

[reef thistle]

How do you calculate the decimal expansion of 1/n?

[hasty cargo]

You take 1 divide by n

[reef thistle]

...and?

[hasty cargo]

And eventually you get the same remainder

Is that it?

[reef thistle]

So, you do long division, right?

[hasty cargo]

yeah

[reef thistle]

And the remainder eventually becomes the same.

Why?

[hasty cargo]

Because eventually we're going to divide by the same n everytime?

Oh

Wait I think I know

Or I don't actually nevermind jsfh jskhf

I thought it was something like

We divide by the same n

And since n doesn't divide into that

We have to increase that remainder by a factor of 10

And n won't divide into that either

[reef thistle]

So, we have a sequence of remainders.

[hasty cargo]

Yep

[reef thistle]

And you claim that eventually those remainders would be the same

[hasty cargo]

Yes

[reef thistle]

As in, eventually, there would be a remainder that you definitely seen before earlier in the sequence

[hasty cargo]

Yep

I think so you'll keep seeing that remainder at least

[reef thistle]

But not every sequence eventually has a number you seen earlier.

[hasty cargo]

Yeah

[reef thistle]

Like 1, 2, 3, 4, 5, 6... or the prime numbers 2, 3, 5, 7, 11, 13

[hasty cargo]

Oh

[reef thistle]

What is it about the remainders that force them to repeat some number?

[hasty cargo]

The remainders are integers?

[reef thistle]

Well, but the positive numbers are integers, so are the prime numbers.

[hasty cargo]

The dividend doesn't divide into the remainders?

[ripe ferry]

youre really taking the scenic route

[reef thistle]

So what does that tell you about the remainders?

[hasty cargo]

When factored they don't contain the dividend?

Okay wait not factored urrr

They can't be divided unless multiplied by 10

Wait

no hsdlfkjh sdjkfh

[reef thistle]

Hmm, try dividing 3 by 103.

[hasty cargo]

okay so

102 is divisible by 3

but not 103

So we have a remainder of 1

[reef thistle]

Like 3/103, not the other way around.

[hasty cargo]

Oh

3 divide by 103

Urrr

3 is not big enough

So we gotta make it 300

Uhh then we divide 300 by 103 so we get 2 remainder 94

[reef thistle]

What stops the sequence of remainders from just not repeating a number?

[hasty cargo]

94 can't divide into that

We're not getting the same remainders

For now at least

[ripe ferry]

element, gonna be honest
I have no clue where youre going with this

[reef thistle]

What happens if a positive integer sequence does not repeat any number?

[hasty cargo]

It's non rational

[reef thistle]

no, talk about the sequence

[hasty cargo]

Uhh the sequence is random

non periodic

[reef thistle]

Talk about what sort of values you would expect to see in the sequence

[hasty cargo]

Well we really don't know what to expect

It could be 3.1415.. for all we know

[stray reef]

anyone mind telling me what the original q was

[reef thistle]

But you see, 3 1 4 1 5 repeated the 1

[hasty cargo]

Prove $\frac{1}{n}$ where $n \in \mathbb{Z}$ has a decimal expansion of period at most $n-1$

@stray reef

[vital dew]

Itadakimasu!:

[stray reef]

uh huh

[hasty cargo]

@reef thistle Yeah it repeated but 1 that's just a coincidence right or..?

[reef thistle]

But as long as the sequence of remainders repeat a value, it becomes periodic?

[hasty cargo]

No it has to keep repeating the same set of values

14!=15

[naive saffron]

[hasty cargo]

Consecutively

I think

[naive saffron]

I almost read 14 factorial

[reef thistle]

yeah 3 14 15 26 53 58 97 93 23 84 62 64 33 83 27 95 02 88 41 97 repeats 97.

[hasty cargo]

It has to repeat it right next to each other

No numbers in between

yeah

[reef thistle]

Well, let's consider 1/7

0.1428571...

the 1 repeated

[stray reef]

i can't tell what this is even about anymore

[reef thistle]

but it's not next to the other 1

[hasty cargo]

Well we can have numbers larger than 9 that repeat

so we have 0.101010..

Or 0.123123123

[reef thistle]

1/7
0 with remainder 1
10 -> 1 with remainder 3
30 -> 4 with remainder 2
20 -> 2 with remainder 6
60 -> 8 with remainder 4
40 -> 5 with remainder 5
50 -> 7 with remainder 1
10 -> 1 with remainder 3 (remainder repeated here)

I'm talking about the sequence of remainders here

[hasty cargo]

Yeah it goes in a cycle?..

[reef thistle]

Why must there be a cycle?

[naive saffron]

just consider the group (Z/nZ)^*. Each element has finite order, and the group is finite

I think

[reef thistle]

Let's not drag group theory into this

[naive saffron]

..sure

[hasty cargo]

That's great but

I'm not sure how group's are gonna help

[faint narwhal]

@naive saffron that doesn't matter

[naive saffron]

oh ok

[reef thistle]

Well, one part of that matters

[hasty cargo]

There must be a cycle because uhh there are only so many remainders you can get when dividing by n

Specifically

n-1 remainders

holy shit

[reef thistle]

yeah

[hasty cargo]

Am I crazy

IOR WH sakjhdsjk yhfujosd

[reef thistle]

And then you are done

[hasty cargo]

H IOPFhdsajkl fhsdjklhf

[reef thistle]

congrats!

[hasty cargo]

IH JDSG AJh

OH MYU GOD

IM CRYING

THANK YOU

Not gonna lie I was close to fainting halfway through

[reef thistle]

yeah I really wanted you to try to discover by yourself

[hasty cargo]

Yeah okay I get it thank you so much :>

[olive hamlet]

nice application of pigeonhole principle! alternatively you could use the fact that if somthing has a period of k, then its denom is of the form 10^k - 1, and then use eulers theorem to show this is always possible for some k< n for 1/n

[faint narwhal]

(not necessarily in lowest form)

[olive hamlet]

(except if we have 2s or 5s, but those divide 10^k which can be dealt with)

and yeah that

[cerulean ore]

Guys, are anti/non/asymmetric same?

Our professor has said that they're same even after we showed her explanation on stackoverflow

[stray reef]

they are three different things

non-symmetric: not satsifying the definition of symmetric

asymmetric: if a R b then NOT b R a

antisymmetric: if a R b and b R a then a = b

and your professor really should be fired

[cerulean ore]

asymmetric and non-symmetric looks same

[reef thistle]

So, on the set {1, 2, 3} a non-symmetric relation can be 1R2, 2R1, 1R3.

[pale epoch]

the empty relation is not non-symmetric but it is asymmetric

as it is both symmetric and asymmetric

but it is the only example

[stray reef]

@cerulean ore no

would you say that "not everyone smokes" and "nobody smokes" are the same statement?

[cerulean ore]

Nope.

Let A be any set and R is relation on set A then R is called symmetric Relation iff (x,y) belongs to R => (y,x) belongs to R such that x,y belongs to A.

Is there any example which is non-symmetric but not asymmetric?

[pale epoch]

no

[cerulean ore]

then how did the word came?

[pale epoch]

which word

[cerulean ore]

non-symmetric i.e. not satisfying the conditions

[pale epoch]

it just means not symmetric

to be fair i don't think it is used in literature

i only ever see symmetric and antisymmetric relations, because they are most important

[cerulean ore]

Is an identity relation always antisymmetric?

[pale epoch]

what's an identity relation

[cerulean ore]

Let A be a non-empty set, for every a belongs A we have aRa then that relation is called identity relation on set A i.e. R = {(a,a): a belongs A}

[pale epoch]

if all the elements of the relation are of the form (a,a), then yes

[cerulean ore]

Got it then!

Is types of relations covered in "How to prove it"?

[pale epoch]

yes

[cerulean ore]

Thank you for helping out!

This group is surely going to be the reason behind me passing my exams.

[stray reef]

Is an identity relation always antisymmetric?
you mean equality?

[cerulean ore]

@stray reef Equality relation?

[stray reef]

yes

let's not pretend the relation {(x,x) | x in A} isn't just equality on A

[cerulean ore]

Yep

[weary tiger]

Can a set contain the empty set as an element (not only as a subset)

I can ask that a bit differently: is {{}} equal to {} ?

[olive hamlet]

no. its a set containing an element(namely {}) so it is not the empty set

a set can contain the empty set as an element

[weary tiger]

Oh

Thanks!

Empty sets are wired thought

[ripe ferry]

the set containing no elements is the empty set {}
the set containing the empty set has an element, {{}}
it has empty as an element

yeah its strange at first

[lean leaf]

It's easier to think of sets as boxes. A box can contain an empty box.

[stray reef]

i prefer likening sets to folders

as in, like in a filesystem

[lean leaf]

Hm, that works too.

A directory can contain an empty directory.

[minor radish]

anyone know what the notation 2^∅ means?

is it like the power set of the null set or something

[weary tiger]

the set of all functions ∅ -> {0,1}

but yeah it's also the powerset of the null set

[minor radish]

like the set of all functions that map to 0 and 1?

sorry im kind of confused what that means

isnt the power set of the null set just the null set

[faint narwhal]

The set of all functions with domain the null set and codomain {0,1}

[weary tiger]

no

[faint narwhal]

No,

[weary tiger]

$\mathcal P (\varnothing) = { \varnothing}$

[minor radish]

ah okay right

if a functions domain is the null set how does it map values to the codomain?

since it wouldnt have any values in its domain

[faint narwhal]

There's only one such function

The empty function

[minor radish]

do you have any good readings on that? ive been kind of confused by everything im seeing online

[weary tiger]

well a function can also be considered as a set right

like a function $f: X \to Y$ is a subset of $X \times Y$

[minor radish]

yeah

[weary tiger]

so if you're looking for the functions $f: \varnothing \to {0,1}$, then you are looking for the subsets of $\varnothing \times {0,1} = \varnothing$ which satisfy the property that functions must satisfy

[weary tiger]

there's only one subset of {}, and that is {}
and sure enough, this vacuously satisfies the properties of a function

[minor radish]

doesnt our definition of a function mean that {0, 1} have to be a subset of ∅?

[weary tiger]

why

[minor radish]

since the codomain has to be a subset of the cartesian product of the domain and codomain

unless im interpreting Y is a subset of X * Y wrong

[faint narwhal]

He's saying that a function is a subset of X * Y

[minor radish]

so then Y is the set of ordered pairs for mappings between the domain and codomain?

[faint narwhal]

No Y is your codomain

[minor radish]

oh okay

so f is a subset of X*Y

i see

okay this makes sense now thank you so much

[clear heart]

hi guys id like to discuss tamari lattices

i am part of a widespread campaign to spread ideas surrounding tamari lattices to destroy the PEMDAS vs BODMAS debate

[stray reef]

what's a tamari lattice

also there is no debate those two acronyms are both shitty

[clear heart]

basically conceptualizing sets of non-associative things

and the pemdas vs bodmas debate is like asking whether a cis isomer or trans isomer of any kind is inherently better

[stray reef]

and uh

how's that going to help

and most importantly

what're you gonna accomplish with that

[clear heart]

showing people a cool as heck graph and simultaneously stop arguing in favor of visualizing pemdas AND bodmas as tools to be used whenever appropriate

which in explicit math terms

will not be a lot

[stray reef]

ok wait but like uh

what

[clear heart]

?

[stray reef]

visualizing pemdas AND bodmas as tools to be used whenever appropriate
don't get this

does anyone ever even do that

[clear heart]

anyways i don't really have the time to try it out but i wanna see if there are any cool patterns that arise when you compute all possible orders of operations

for some popular formulas and stuff

https://tenor.com/view/look-at-this-graph-nickelback-gif-12683147

[proven shard]

Ok

[minor radish]

how would i go about showing something is irrational?

for example

[faint narwhal]

contradiction

[minor radish]

i understand i have to create a proof by contradiction by setting (2/3)^(1/5) equal to p/q with some relatively prime p, q

so that i can evaluate to 2q^5 = 3p^5 but im not quite sure what to do after

[faint narwhal]

Come up with some contradiction, play around with it

[weary tiger]

$x^5 = 2/3$ doesn't have a rational solution

[vital dew]

Blitzkrieg:

[steel valve]

i have the big dumb and can't do basic logic

shouldn't this truth table evaluate to T, F, F, F

[minor radish]

v is or

[steel valve]

the only condition that evaluates to true is T and T

oh 🤦

[minor radish]

so if p or q is true p V q is true

[steel valve]

i pray for my next quiz.

[minor radish]

easy way to remember is and looks like a capital A

also if you know sets the notations very similar except curved

@weary tiger wouldnt you have to prove that too though

i think its something to do with 2 dividing p^5 cant find out what though

[sour arrow]

Too powerful is the rational roots theorem

Show that 3x^5 - 2 has no rational roots

[faint narwhal]

if 2 divides p^5

what do you know about p being even or odd

[sour arrow]

The only possible rational roots are
±1, ±2, ±1/3, ±2/3
None of them are roots, which you can just show, so the polynomial has no rational roots

[minor radish]

p has to be even since 2 is a prime

[weary tiger]

You can also use that $\sqrt[k]{n}$ is rational if and only if $n$ is a power of $k$

[vital dew]

Blitzkrieg:

[weary tiger]

n, k natural

[faint narwhal]

@minor radish okay, then what do you know about q being even or odd

[weary tiger]

you'd have to multiply by 3 first though

[minor radish]

thats what im stuck on

oh wait nvm

since 2 is raised to the power of 5 q has to be even

[faint narwhal]

and then what can you say

[minor radish]

theres a contradiction since p and q have to be relatively prime

omg thank u so much

[vital dew]

Blitzkrieg:

Blitzkrieg:

[weary tiger]

power of 5

[minor radish]

ohh right

[faint narwhal]

Wait, how do you know that $q^k \mid p^k$

[vital dew]

Zopherus:

[minor radish]

(a mod c) = (b mod c) implies (a^d mod c) = (b^d mod c) right

[weary tiger]

multiply by q^k both sides

[minor radish]

p^k = n q^k

where n is a natural number

[weary tiger]

@minor radish yes, this is why it's called a congruence

or is it the other way

btw. This is a bit redundant, just write $a \equiv b \mod{c}$

[vital dew]

Blitzkrieg:

[weary tiger]

etc.

[minor radish]

I usually do I just don't remember latex commands lol

[cunning dragon]

https://cdn.discordapp.com/attachments/488104216678760469/613834700506529899/378329266171150337.png

How do I prove this equivalence without using Truth table?

[faint narwhal]

use demorgans laws and distributive properties

[cunning dragon]

$\neg(P\land{Q}) \Leftrightarrow (\neg{P}\lor \neg{Q})$

[vital dew]

Melon Husk:

[cunning dragon]

you mean this ?

[faint narwhal]

That is demorgans law yes

[cunning dragon]

Where should I apply it ?

[faint narwhal]

Play around with it

You also have distributive properties to use

[cunning dragon]

AAAAHHH, NOW I SEE IT

$(\neg(P\lor{Q})\lor(P\lor{Q})\land{R}$

[vital dew]

Melon Husk:

[cunning dragon]

so ez

thanks

$T\land{R} \Leftrightarrow R$

[vital dew]

Melon Husk:

[trim oak]

Q: Each user on a computer system has a password, which is six to eight characters long, where
each character is an uppercase letter or a digit. Each password must contain at least one digit.
How many possible passwords are there?

My first stab at it (which is wrong)
$10(26+10)^5 + 10(26+10)^6+10(26+10)^7$

The correct answer is $36^6 - 26^6 + 36^7 - 26^7 + 36^8 - 26^8$
Which makes total sense to me, you filter out any results that don't have a number in them.

I'm just confused because in my head, my way also makes sense (to me) but is obviously wrong. Can anyone point out what I have worked out instead of the correct answer?

[vital dew]

Nokk:

[spring temple]

I think your stab double counts some passwords.

Consider the password 111111

[viral slate]

@fathom knoll @weary tiger

[steel valve]

i don't get proposition logic still

lol

like how the hell do you construct truth tables and use it to solve logic puzzles and circuit gates

i don't see how the solution ties in with propsitional logic

[oak creek]

@steel valve what's the truth table you tried constructing?

also @trim oak did you figure it out?

[steel valve]

i thought a truth table applies here

because it was introduced in the previous section

introduction to proposition logic

[oak creek]

it does

you would form a table for p, q, (p and ~q) ^ (~p and q)

and use that to figure it out

given the addition that q is the statement by person A

[summer dragon]

So how can one find a reccurence relation on the n-digit quatenary {0,1,2,3} sequences with an even amount of 1s?
Any videos or something that can hint me in the right direction

[stray reef]

hmm

let Q(n) be the amount of n-digit quaternary strings with an even number of ones

then 4^n - Q(n) is the amount of n-digit strings with an ODD number of ones

clearly Q(0) = 1 and Q(1) = 3

now

okay ig it'll help to make up some ad-hoc names

if a string has an even number of ones, i'll call it even-unit; otherwise i'll call it odd-unit.

so we can construct an even-unit string of length n from strings of length n-1 like so:

• take an even-unit string of length n-1, and append something other than a 1 to it (3 possibilities)
• take an odd-unit string of length n-1, and append a 1 to it (1 possibility)

so we get

Q(n) = 3Q(n-1) + 1(4^n - Q(n-1))

Q(n) = 2Q(n-1) + 4^n

[summer dragon]

Do you know any videos on the topic?

[stray reef]

no

[summer dragon]

Well thanks :)

[faint narwhal]

@summer dragon You could look at things called dynamical programming

It's essentially an algorithms technique that does this kind of stuff

[reef thistle]

I think you mean dynamic programming

The main idea is memoising the solutions to smaller subproblems

[summer dragon]

Its 4^(n-1) right 🤔

[stray reef]

idk, does it satisfy the recurrence?

[reef thistle]

Q(n) = 3Q(n-1) + 1(4^(n-1) - Q(n-1)) btw

so Q(n) = 2Q(n-1) + 4^(n-1)

definitely bigger than 4^(n-1)

||Q(0)=1
Q(1)=3
Q(2)=10
Q(3)=36
(4^n)/2+2^(n-1)||

[steel valve]

can r and not p be described alternatively as:

if r then not p

[faint narwhal]

No

[steel valve]

thanks appreciate the quick response

[steel valve]

"For the router to support the new address
space it is necessary that the latest software release
be installed"

q= router supports the new address
r= latest software is installed

i represented this sentence like this:

q then r

but the book says it's supposed to be reversed: r then q

that doesn't make any sense to me because for q to be true in the first place, r has to be true as well.

if r is false then q would not be able to over write the correct intent.

[peak crag]

(r and not p) == not ((not r) or p) == not (if r then p)

[young garden]

Discrete Mathematics and It's Applications by Rosen, good choice

[gray agate]

Hey guys

Find all roots to the equation :
x squared = 71 (mod 77)

Can anyone help out a bro?

[faint narwhal]

@gray agate for this small modulus, there's not really a better option than just trying everything

[gray agate]

Or you just factorise

And then work it out from two smaller moduluses

But then Idk what to do

Can you continue from there

[faint narwhal]

Just try things

If you factorize, your new moduli are just 7 and 11

It's easy to just square everything and see if that gives you a solution

[reef thistle]

Or you can be a mad bro and notice that 77+77+71=225

[gray agate]

That’s not finding all the roots to the equation

If you can actually read the question and solve it like it should be then that’d be great

[faint narwhal]

That does find all the roots to the equation in fact

[gray agate]

I’m looking for all the roots

[faint narwhal]

You just don't really understand what he's trying to say or how this problem works

How many roots do you think there can be

[gray agate]

Idk I’m trying to ask you but I’ve got the solution I just don’t know how to work it out

Can you??

[faint narwhal]

We're just going to guide you through the question

[reef thistle]

If you consider primes, there are only 2 solutions

[faint narwhal]

Not do it for you

[gray agate]

Can you guide me with what I do after I get two moduli factorials

[reef thistle]

What did you get?

[gray agate]

I’ll tell you @reef thistle ,
I got two primes, so: 7 and 11.

Therefore:
x^2 = 1 (mod 7)
And x^2 = 5 (mod 11)

[reef thistle]

yeah, that looks good

[gray agate]

Ok continue

[reef thistle]

can you solve each of them?

separately

[gray agate]

Not sure what to do next bud

I need to find the two roots of the first equation and the two roots of the second equation

How do I do that?

Maybe @faint narwhal can help?

[faint narwhal]

Can you solve the first one?

[gray agate]

No I need help

[faint narwhal]

Think

You can solve it

[reef thistle]

How many possible values are there for x?

[faint narwhal]

You just have to sit down and think

[gray agate]

Element thanks for helping me lol

Ummm 7 possible values and 11 possible values

Is that right

[reef thistle]

You have 7 values to test for the first equation and 11 values to test for the second equation

[gray agate]

yep

[reef thistle]

It's not like x^2+2x+1=0, where you have every possible real number to test.

(for x)

[gray agate]

So I need to find four roots, they say it’s 1, 6, 4, 7. How does that make sense?

1 and 6 for first equation and 4 and 7 for the second

[reef thistle]

Well, now we combine the roots

[gray agate]

But HOW do I get the roots

[reef thistle]

You only have 7 values to test

[gray agate]

So square them and if they equal the equation then it’s valid?

[reef thistle]

yeah, that's all

check if it satisfies, that's all

(of course there are methods where the prime involved is really large, but let's not go there today)

[faint narwhal]

"It's easy to just square everything and see if that gives you a solution"

I did say this ten minutes ago

[gray agate]

Yeah but you decided to prove it to yourself not like element did and actually helped me get to it

[reef thistle]

For further reading where the prime is really large: https://en.wikipedia.org/wiki/Tonelli–Shanks_algorithm

[gray agate]

Thanks element I think this is what I need to deal with however

They are now asking me to obtain 4 solutions from each of the pair

[reef thistle]

Okay there are 4 possibilities, right?

[gray agate]

Right

[reef thistle]

(1 mod 7, 4 mod 11), (6 mod 7, 4 mod 11), (1 mod 7, 7 mod 11), (6 mod 7, 7 mod 11)

Can you find x (mod 77) satisfying each of them?

(separately)

[gray agate]

Well what they’re saying is to then write the gcd (7,11) = 1 as a combination of 7 and 11. Then, by trial and error, we have 1 = 2 * 11 + (-3) * 7

[reef thistle]

Well, you can get that by the extended euclidean algorithm

[gray agate]

How do we get that

Oh

[reef thistle]

basically you do euclidean algorithm, but you keep track of coefficients

[gray agate]

So that’s the EEA result?

The 1 = 2 * 11 + (-3) * 7

[reef thistle]

Mind my abuse of notation here.
gcd(7, 11)=gcd(7, 11-7=4)=gcd(7-4=3=7*2-11, 11-7=4)=gcd(7-4=3=7*2-11, 11-7-(7*2-11)=4-3=1), so 11-7-(7*2-11)=2*11-3*7=1.

So, what you want to do next is notice that to make a number that is 1 mod 7 and 4 mod 11, we just need to add the number which is 1 mod 7, 0 mod 11 with 0 mod 7 4 mod 11.

So, now we have 2 "simpler" problems.
How do we find 1 mod 7, 0 mod 11?
1 = 2 * 11 + (-3) * 7
So what this means is
2 * 11 is 1 mod 7. Note that it is also 0 mod 11.

Likewise, 0 mod 7, 4 mod 11:
1 = 2 * 11 + (-3) * 7
(-3) * 7 is 1 mod 11 and 0 mod 7. Multiplying by 4, we get (-3) * 7 * 4 is 4 mod 11 and (still) 0 mod 7

Understand this?

[gray agate]

Sorry was brb

Umm

[reef thistle]

How about you show me how you understand it by doing another case?

[gray agate]

Yep so you’ve done one case?

[reef thistle]

almost done

just a bit of cleaning up to do

(which if you understand what's going on, you can figure it out)

@gray agate so, did you understand it?

[gray agate]

I’m a bit confused tbh hold on

Sorry

So I’ll tell u where I’m stuck

[reef thistle]

sure.

[gray agate]

So the first 4 roots are the solutions 1,6,4,7. How do we get four pairs out of these roots?

Secondly, after we do the gcd(7,11) = 1 and get the answer 1=2* 11 + (-3)* 7

And then, somehow with one of the pair, we need to use CRT to get the final solution to one out of four cases.

You get where I’m stuck?

[reef thistle]

Well, what I did was effect Chinese Remainder Theorem

but being a bit more clear on what's going on

[gray agate]

Yeah CRT = Chinese remainder theorem

[reef thistle]

Earlier I mentioned that the 4 roots are (1 mod 7, 4 mod 11), (6 mod 7, 4 mod 11), (1 mod 7, 7 mod 11), (6 mod 7, 7 mod 11)

That you have?

[gray agate]

How’d you get that?

Sorry if I’m missing out on something obvious

They just seem to appear

Outta nowhere for me

The only 4 roots are 1,6,4,7, how do you get 4 pairs out of them

I may have to DM you this later , I added you.

Loving the help so far though 👌🏽👌🏽

[faint narwhal]

1 and 6 are roots to the first equation

x^2 = 1 mod (7)

and 4 and 7 are roots to the second equation

[reef thistle]

So, that's just all possibilities

[stoic kite]

Can someone help me with this

Let p and q be (not necessarily different) prime numbers. Find all solutions to diophantine equation px + qy = pq.

[weary tiger]

gcd(p, q)|pq

[stoic kite]

so this is enough

just to write this

if there is no numbers specified

just primes

[weary tiger]

https://tenor.com/view/umm-confused-wtf-blinking-okay-gif-7513882

[stoic kite]

i mean is this formal solution

[weary tiger]

ax+by = c, we proceed by checking if gcd(a, b)|c

if p = q
x+y = p

then solutions are given by
x = t
y = p-t

[stoic kite]

yes but i am interested in solution whitout inserting numbers

[weary tiger]

if p != q

[stoic kite]

ok

[weary tiger]

ok

we see that x = 2q, y = -p is a solution

[stoic kite]

hm i will think about it

[weary tiger]

https://en.wikipedia.org/wiki/Diophantine_equation#One_equation

[stoic kite]

thanks

[weary tiger]

u = p/gcd(p, q)

v = q/gcd(p, q)

if those are different primes then that's just p and q

[stoic kite]

ok

i get this

[weary tiger]

general solution is ((2+k)q, -(k+1)p)

or you can shift k a bit

((1+k)q, -kp)

looks nicer

[stoic kite]

so this is the same thing

((2+k)q, -(k+1)p) = ((1+k)q, -kp)

[weary tiger]

for any k integer ((1+k)q, -kp) is a solution, and those are only solutions

[stoic kite]

ok thanks

[weary tiger]

no it's not the same

[stoic kite]

oh

so the difference is the shift

[weary tiger]

but sets are the same

[stoic kite]

ok

[weary tiger]

set of ((1+k)q, -kp) for k integers

and

set of other ones for k integers

[stoic kite]

ok

is the whole explaination connected

if i add up what you have told me line by line

[weary tiger]

maybe

[stoic kite]

on which level is this

is it for first year in college

[weary tiger]

hmm

[stoic kite]

the explaination details

[weary tiger]

I guess

[stoic kite]

ok

[weary tiger]

level is something like high school or higher

[stoic kite]

ok

thanks

[weary tiger]

it wouldn't be hard to teach a high schooler about linear Diophantine equations, but does anybody do that?

[stoic kite]

in my case no

[weary tiger]

here you go

[stoic kite]

at least not much in depth

thanks anyways

[young garden]

Are ¬∀x (C(x) ^ B(x)) and ∃x ¬(C(x) ^ B(x)) equal?

[west hedge]

yup

[naive saffron]

Damn how do you type those

[young garden]

I have a keyboard fitted especially for propositional logic and quantifiers, utf8 compliant

Search it up

[obsidian tendon]

just l a t e x

$\neg ~ \forall x (C(x) ~ \land ~ B(x)) \text{ and }\exists x ~ \neg(C(x)~\land ~B(x))$

[vital dew]

hegel:

[naive saffron]

no u sloth

Actually

$\neg @obsidian tendon$

[vital dew]

Whoever:

[obsidian tendon]

unacceptable

[dapper rose]

no double u

[cerulean ore]

There is no domain/range given for the function

so, should I directly find the inverse?

[stray reef]

you should assume the domain to be the largest subset of the number line possible

that is to say, in this case, it'd be (-∞, 8/3) ∪ (8/3, +∞)

however irrelevant that might be

[cerulean ore]

Noted.

For proving it to be reflexive:
Let x ∈ N
Since, R = {(x,y): x ∈ N and x-y is divisible by 5}
=> (x,x) ∈ R because (x-x) = 0 which is divisible by 5
Therefore, R is a reflexive relation on N.

[stray reef]

wording could use some work

but the idea is correct overall

[cerulean ore]

What else do you think would work?

[cerulean ore]

Let A denotes divisibility by 5 and B denotes divisibility by 9 then,
Integers neither divisible by 5 nor by 9 = 1000 - n(A) - n(B) + n(A intersection B)

[naive saffron]

Why not use R\{8/3}

[dapper rose]

R doesn't have the line not good enuff

[cerulean ore]

n(A) = 1000/5 = 200
n(B) = 1000/9 = 111
n(A intersection B) = > not divisible by 9 and not by 5 => not divisible by 45
=> n(A intersection B) = 1000/45 = 22

Is this way correct?

[faint narwhal]

A intersect B should be both divisible by 9 and 5

[cerulean ore]

oh yes, that's a mistake.

n(A) = n(A-B) + n(A intersection B)
=> n(A-B) = n(A) - n(A intersection B)

that's it?

n(A intersection B) is actually divisible by 9 and 5 both, so, 1000-n(A intersection B) would be not divisible by both

= 1000 - 200 - 111 + 22 Should work for first part

[cerulean ore]

No domain or range is not given

How do I solve this?

[stray reef]

if your function is given by a formula, assume the domain to be the largest subset of R on which the formula is defined, and the codomain to be R, unless otherwise stated

the domain as it happens is R itself for both of these

[cerulean ore]

x^3 - 3x = x(x+1)(x+1), since the function is 0 at 3 points therefore, the function is not one-one. Right?

[stray reef]

x^3 - 3x is not equal to x(x+1)(x+1).

[cerulean ore]

Sorry!

Going to take break after solving the last question

but, since it has 3 roots therefore, it is not one-one, right?

[stray reef]

well, DOES it have three roots?

[cerulean ore]

If R is a partial-order relation on set S
=> R is anti-symmetric
Since, R is anti-symmetric.
=> for every x,y ∈ S, (x,y) ∈ R and (y,x) ∈ S such that x = y
Since, (x,y) ∈ R therefore, (y,x) ∈ R^-1
Since, (y,x) ∈ R therefore, (x,y) ∈ R^-1
=> (y,x) and (x,y) ∈ R^-1 such that x=y
Therefore, R is anti-symmetric.

x^3 - 3x = x (x^2 - 3) = x(x+sqrt(3))(x-sqrt(3)), yes it does?

[stray reef]

bad wording

in each line, either the "since" or the "therefore" will have to go

also idk why you put commas after each "since"

this entire proof sounds like an attempt to conform to some sort of rigid format

[timid elk]

In Prim's algorithm, if the edge with the least weight creates a cycle, do I choose the immediate next one, or do I go back to the previous node and choose the immediate next there? I'm kind of confused.

[reef thistle]

prim's algorithm we grow the minimum spanning tree from a source

[timid elk]

I know

[reef thistle]

I'm trying to remember what I learnt

[timid elk]

oh

[reef thistle]

so, we keep a priority queue of points where we store the minimum edge needed to reach that point from our visited points

when the edge with the least weight creates a cycle, just move to the next edge in the priority queue

[timid elk]

Can I use an example to make sure I understand?

But maybe move over to questions

[reef thistle]

Yeah sure

[cerulean ore]

@stray reef actually, I was trying to be concise.

[stray reef]

wasn't a very good attempt honestly

[cerulean ore]

I will get better at it, I am trying my best.

How you would have written it?

[cerulean ore]

1st question, part a

Uploading ;-;

What do you guys think? Did I solve it correctly?

[stray reef]

the $\cap$ should be $\land$

[vital dew]

Ann:

[stray reef]

and no, you should've written $(A \land \neg B) \rightarrow \neg (C \land D)$

[vital dew]

Ann:

[stray reef]

it may be the case that the butler did not do it nor did he return to his hotel room

(i.e. not C and not D)

while your thing fails that

[cerulean ore]

Ohhh

I have noted that capped mistake.

So, (A and ~B) should be the or of all three, right?

He did it and didn't go or he didn't do it and go or he didn't go and he didn't do it

[stray reef]

you can do that but it's way simpler to just say NOT(he did it AND went back)

[cerulean ore]

Wow

You're genius!

[weary tiger]

lol

[cerulean ore]

I really didn't see it until she wrote that

[nocturne cypress]

I don't understand how to solve this

[glossy adder]

first thing: do you think it's true or not

[reef thistle]

true? Give a proof.
false? Give a counterexample

[nocturne cypress]

I don't think so? My understanding of the first condition is that if two sets in the collection share elements the two sets referred to are actually the same set, and the second condition is that if two sets share elements, such as S_1 and S_2, then S_1 is equal to S_2

[reef thistle]

Okay, give a counter example

[nocturne cypress]

So the collection of sets {S_i, i in I} where I = {1,2}, S_1 = {1}, S_2 = {1} can't exist with the first condition but can with the second?

I don't know if that even makes sense

[analog sonnet]

It does

That’s a valid counterexample

[nocturne cypress]

ok cool

ty

[cerulean ore]

So, I had an exam of DM today, I have scored full.

[reef thistle]

congrats

[cerulean ore]

Thank you very much! You guys should come to India and teach us.

[faint narwhal]

Do most Indians speak English?

Your english is pretty good

[cerulean ore]

Not everyone is good at speaking English overhere but, most of them understand it.

In reputed colleges you can expect them to know English.

[faint narwhal]

Huh interesting

[cerulean ore]

We really need good teachers, unfortunately teaching has become a source of employment only. One who doesn't gets into a company joins teaching.

You guys have a passion as no one joins a study group that too on discord.

[reef thistle]

A good idea to discuss math here then

[steel valve]

\

i don't get how those two F's dissappear somehow

can you group the F's together using associative laws because they both share "or" to negate each other?

that's what it seems like to me

[sour arrow]

@steel valve
That's not negation law? Basically, A OR F = A

[steel valve]

🤦 alright thanks

[timid elk]

So there's this variation of the Tower of Hanoi problem. And the solution given by the writers is the following.

Which makes sense, but I've written down a different one and I was wondering if it is correct. In the handwritten solution I have, it says that you need 2n-2 moves for the discs (all of them, expect the base), 2 times. Plus the 2 base discs. So it expresses it in this weird notation:
T[2n]=2T[2n-2]+2
The brackets are indexes. Which also makes, sense, I guess? Not sure, I need your opinion.

[faint narwhal]

Your's is basically the same

It's just a bit awkward because your relation T is only defined at the even numbers

[timid elk]

I've also taken a note that T0 is 2. Would that be correct? It doesn't make sense to me.

[faint narwhal]

I mean

Hard to say, the n = 0 case doesn't really make sense, so just plugging it into a formula might give you some weird number

[timid elk]

Well the book says T0=0 for the normal Hanoi problem, so I wouldn't change that. It's just that I use it later to calculate the closed-form expression.

[faint narwhal]

I have no clue what you're saying

[timid elk]

Is it my English or the math? 😛

[faint narwhal]

the first

[timid elk]

Long story short is that I need T0 to finish this exercise and I'm not sure about its value.

[faint narwhal]

It seems dumb if they specifically ask for that

The n = 0 case does not make sense in the context of the problem

[timid elk]

You know what a closed-form expression is, right?

I'm asking because terminology might be different in English.

[faint narwhal]

yes

[timid elk]

Well, in order to calculate it, I need to specify T0.

With the method I'm using, at least.

[faint narwhal]

I'm not sure how that could be the case

Why you couldn't start at T1, or maybe T2 in your case

[timid elk]

Well, I do start at T1, but the recurrence relation has n-1 😛

So I need a value that makes sense for T0.

[faint narwhal]

Then start at T2

[timid elk]

But it's still a recurrence 😆

Is there something I don't get?

[faint narwhal]

Just use T1 as your base case

You don't need to use T0

[timid elk]

Actually, T2. Because I work with T[2n]. So, for n=1 I'd need to find the moves for 2 discs. So, 2 moves, I guess? T2=2.

Yeah, it's weird. I might be going with the writer's solution.

[faint narwhal]

The writer and your solutions are the exactly same

just substitute 2n for n everywhere

[timid elk]

My solution was T[2n]=2T[2n-2]+2. If I do that to the writer's solution I'd get A[2n]=2A[2n-1]+2, no?

[faint narwhal]

Yeah sorry, just divide everything by 2

[timid elk]

Yeap, that's it. Thanks for your time and help!

[crystal bough]

A map from \mathbb{N} \to \mathbb{Q}N→Q can be described simply by a list of rational numbers. If this list contains each rational number at least once, we can remove repeats to obtain a bijection \mathbb{N} \to \mathbb{Q}N→Q.

For a rational number \frac abba​ (in lowest terms), call |a| + |b|∣a∣+∣b∣ its height. There are finitely many rational numbers of each height. Hence, if we list all the rationals of height 1, then the rationals of height 2, then the rationals of height 3, etc., we will obtain the desired list of rationals.

Thus, we conclude \mathbb{Q}Q is countable.

but height 1 will be infinite and so will height 2 and so on

[weary tiger]

why will the set of rationals of height 1 be infinite

there's only 1 such rational

[stray reef]

bad paste

[crystal bough]

ah right
my bad

[stray reef]

also, yeah, there's only one rational with height 1

0/1

aka 0

[crystal bough]

I forgot that we are taking absolute values

[stray reef]

it

literally says it right there...

|a| + |b|...

[weary tiger]

lol

[weary tiger]

can anyone help me with equations

[faint narwhal]

But also read #❓how-to-get-help

[dull tinsel]

What I want to know is, how I do prove ¬p ≡ something with (pΙq)

we good bois, i figured it out

[weary tiger]

you just sub in logical values for p, q

[acoustic valve]

How do I prove that the number of elements in the set {ε,a,b,c,....z}^n is equal to (26^(n+1)−1)/25? Where epsilon is the empty string.

[faint narwhal]

What does that even mean

[acoustic valve]

in terms of string concatenation rather than the cartesian product

e.g. epsilon + a = a, because epsilon is an empty string

[faint narwhal]

I'm uh, not sure that's true? Like the final counting result

For 1, it's not an integer so

[humble bridge]

How do I prove that two expressions are NOT logically equivalent without a truth table?

Like, with equivalent proofs I just rearrange the expressions on each side until they equal one another, but how do I know where to stop with two expressions that are inequivalent?

[reef thistle]

well, for truth table

you can stop when you have T for one expression but F for the other expression

in fact, you can just state the assignment where they are not the same

[stray reef]

@faint narwhal are you sure $\frac{26^{n+1} - 1}{25}$ is ever not an integer for $n \in \bbN$?

[vital dew]

Ann:

[faint narwhal]

Yeah no I'm dumb

[cerulean ore]

Thank you guys for helping! I have scored full in first test.

This is the part that I've to cover now

Which book should I follow now?

[azure lichen]

idk but send your university a book about kerning

[stray reef]

it's spacing that is the problem here

not keming

[azure lichen]

you made me do this https://www.reddit.com/r/unicodecirclejerk/comments/cyeikt/m_latin_small_ligature_rn/

[stray reef]

cursed

[cerulean ore]

Lol

Would guys please recommend now?

As I can't study from the shitty reference book that is being used in our University.

[cerulean ore]

That answers it^

[loud ingot]

Do you guys agree with what I have for work?

Not sure if I should write 1/x such that x is of rational numbers set (Q)

or keep what I have: x is of an integer set

[stray reef]

${ \tfrac1x : x \in \bbQ, x \geq 1}$ would not describe the same set as ${1, \tfrac12, \tfrac13, ...}$.

[vital dew]

Ann:

[loud ingot]

Do you think it's because of the first value, 1?

[stray reef]

what are you talking about

it's impossible for a number to be an integer and yet not be rational

all integers are rational

no, i'm talking about the fact that ${ \tfrac1x : x \in \bbQ, x \geq 1}$ contains things you don't want it to contain

[vital dew]

Ann:

[stray reef]

such as 4/5

or 3/8

or 2/11

[loud ingot]

I see, so it's better to keep what I have originally

right?

[stray reef]

it's not that it's BETTER

it's that what you wrote is CORRECT

[loud ingot]

I see

[stray reef]

while replacing the Z with Q would make it WRONG

[cerulean ore]

@faint narwhal

Do you know any good book that I can follow for this? (Unit 2)

For the first unit I followed "How to prove by Velleman" it was amazing.

[faint narwhal]

Think that was my recommendation

The book I used for this stuff was Applied Combinatorics by Trotter

It definitely has the first part of what you're looking for

[cerulean ore]

I loved your recommendation, I am thinking about to buy the hard copy of it but, it is too expensive

[faint narwhal]

For the second part, you'll want a book on abstract algebra. Maybe Pinter's book

Yeah that's the pain. I'd buy all the books if they weren't so expensive

[cerulean ore]

It got me into proofs that even if the questions are not from set, relations I still try to solve them

[faint narwhal]

It's a very nice book

It helped me switch from cs to math

[cerulean ore]

Seriously, Computer Science to Maths?

With time I've also developed so much interest in Mathematics.

[faint narwhal]

Yeah I did a ton of competition math in middle school/early high school

Then got super interested in cs for late high school and applied to college with cs as my major

But that book along with some early math classes really convinced me to do math

[cerulean ore]

So, you're doing graduation in Math?

[faint narwhal]

Yep

[hallow blaze]

hey guyz

How does one say in logic (symbolic )form

I bought a lottery ticket and may have won a million dollars

[hallow blaze]

Let p and q be the propositions p: I bought a lottery ticket this week. q: I won the million dollar jackpot.

Also how do u express these proposition as biconditional

[loud ingot]

I think a is infinite, but finite in some cases, b. is infinite, and c is finite rather than infinite

Does anyone agree with my answers or think otherwise? This question has me confused

Any help appreciated

[tranquil cargo]

Q^c for example

[stray reef]

@loud ingot you are only correct on b

[weary tiger]

A^c must be finite, for it is only the empty set, no? @stray reef

[stray reef]

no

jbc A is infinite doesn't mean A=U

like let U = N and let A be the set of all even numbers

you just said odd numbers don't exist @weary tiger

[weary tiger]

LOL

😦

[loud ingot]

@stray reef so a and c should both be infinite then?

[stray reef]

ok wait

you said "infinite, but finite in some cases" for (a)

that's confusing but it is technically correct

(c) can be both finite and infinite

[loud ingot]

Could you explain why for c? Just so I’m clear @stray reef

[manic surge]

AUB can also be finite , right?

when A = B = U

[stray reef]

no it can't, U is infinite and AUB has A as a subset

@loud ingot do you want an example of A \cap B finite or A \cap B infinite

[loud ingot]

let’s try infinite @stray reef

[stray reef]

A=B

since A \cap A = A

[manic surge]

ah, sorry I meant A cap B 😅

wait, that's infinite either

nvm

[loud ingot]

@stray reef while we’re at it, what would finite look like?

[stray reef]

A = the set of even integers, B = the set of odd integers

i invite you to determine what the intersection of these might be

[loud ingot]

so it would be an empty set

for finite

[stray reef]

no, just because a set is finite doesn't mean it's empty

you can get finite sets of any cardinality this way

but if you're talking about my example specifically, then yes, my example is the empty set.

[blazing light]

I'm taking intro to discrete maths and came across this problem in my textbook. how do you solve it? i tried expanding algebraically

[faint narwhal]

You can do it by expanding algebraically

[blazing light]

How?

@faint narwhal

[faint narwhal]

you should be able to write a_12 in terms of a_1 and a_0

You know a_12 and a_0

just solve for a_1

[weary tiger]

@faint narwhal so

that seems like a lot of algebraic rearrangement to do by hand

is there a numerical method to kinda bruteforce this

[blazing light]

He’s my tutor we’re both stuck ☹️

[faint narwhal]

I'm saying it's theoretically possible to do, I'm not saying it's the nicest way

[weary tiger]

sure, sure

but is there a subject i can study up on in particular to be able to solve it more elegantly

[blazing light]

Do you know a nicer way

[faint narwhal]

Okay, it's actually not too bad to brute force

However, you should go the other way

Just let a_1 = x

and brute force your way up

so a_2 = 4x + 4

[weary tiger]

yeah, i reexpressed as a_1 = expression

[faint narwhal]

Then a_3 = 36x + 36 + 6x = 42x + 36

[weary tiger]

i got up to having it expressed in terms of a_5

[faint narwhal]

Every single term is just going to be linear in x, shouldn't be too bad

[weary tiger]

and decided that maybe what i was doing was a little unorthodox

[faint narwhal]

This is probably the nicest way

[weary tiger]

alright. it seems like something i can generalize into a python function without recursion

[faint narwhal]

It's doable by hand in like 5 minutes

[weary tiger]

or with tail calls at the worst

yeah, but the prof will want it done in python

thanks for insight

@faint narwhal ah i figured out a method to describe it algorithmically due to your noting that it stays linear to x

thank you

[weary tiger]

or rather linear to a_1

[fiery hearth]

hey

could someone help me out

?

[stray reef]

not if you don't post your question

[fiery hearth]

okayy

so

i cant seem to figure out

how to solve the Recursive Staircase Problem

i can figure out how many ways there is to solving it

but not how each way solves it

yeah

it can only go 1 or 2 steps at a time

but what i need now is each step

so for 4 it would be

1,1,1,1 - 1,2,1 - 2,2- 2,1,1 -1,1,2-

but i dont know how to get that

:/

[fiery hearth]

yeah i got that

but

im trying to program it

so it gives me each path

[weary tiger]

@weary tiger hi

[wintry sparrow]

Hi

can someone help me

with these

for 8, i did n!/(n-r)! and n^r but im not exactly sure

[weary tiger]

heyo

[reef thistle]

looks okay

[loud ingot]

Does anyone know what would be a good example for this

A - B =/= B - A

I found this one trickier than I though

[west hedge]

what are A and B?

sets?

[loud ingot]

Let me see

Here it is 4. (B)

[west hedge]

ok

did you try something?

[loud ingot]

Yeah. I tried seeing how different set values of A and B might affect it, but they’ll be the same no matter what I realized. I thought of saying ‘let A be the set of all even integers’ and B would be for all odd integers, but then they would equal out to an empty set. That’s as far as I managed to get

[west hedge]

well for one thing, it looks like the problem wants you to choose A and B to be subsets of {1, 2, 3, ... , 9}

maybe try an example where at least one of the two sides isn't empty?

and another thing, if A is the set of even integers and B is the set of odd integers, what is A - B?

[loud ingot]

A - B would then be an empty set?

[west hedge]

can you tell me what it means for x to be in A - B?

[loud ingot]

What would x be? Not following srry

[west hedge]

I mean the definition

A - B is a set, to define it you need to say what it mean for an element x to belong to it

[loud ingot]

So for x to be in A - B, it would be saying: the compliment of B, where B is the set of all odd integers. Therefore, x would need to be odd integers

Is that right?

[west hedge]

let me save us some time
x in A - B means x is in A and x is not in B

if B is odd integers, then x in A - B means x is even and x is not odd

compare that with what you just said

[loud ingot]

Yeah probably just read it wrong then

[west hedge]

can you think of any even numbers that are not odd?

[loud ingot]

2, 4, 6, 8 and so on?

[west hedge]

right

so is A - B empty in this case?

[loud ingot]

Yes

[west hedge]

we agreed

x in A - B means x is even and x is not odd

you just gave me a whole list of even numbers that are not odd

therefor....

[loud ingot]

I must be missing something here

Seems simple

[west hedge]

you basically just said A - B contains 2, 4, 6, 8 but you also claim that A - B is empty

[loud ingot]

Alright yeah I see. So B - A would contain 1, 3, 5, 9

[west hedge]

yes

so is A - B = B - A in this case?

[loud ingot]

Nope, due to odds and even integers

[west hedge]

right

of course you can use the subsets of even and odd integers of {1, 2, 3, ... , 9} in the actual problem, since it wants subsets of U

[loud ingot]

I see. So when I write this, and just to make sure I’m correct

A - B

I’m saying that this is the compliment of B, and that I’m looking for values of subset A that share with subset B

Right?

[west hedge]

you can refer to it as the complement of B (in A) yes

but A - B is not what you said in the second part of that sentence

the elements shared by A and B make up the intersection not the complement

A - B consists of the elements of A that are not shared by B

[loud ingot]

I see now

[west hedge]

👍

[timid elk]

can (2a)!/a! be simplified?

[faint narwhal]

kind of, not really

[timid elk]

Would it be more readable if I left it like this?

[faint narwhal]

Yes

[timid elk]

Okay then, I won't dwell on it. Thanks.

[timid elk]

1*2*2*3*3*3*4*4*4*4...
is the product of n^n, right?

Snap

Okay I fixed it

[faint narwhal]

What?

[timid elk]

Is it not?

[faint narwhal]

I don't even know what you're trying to say

[faint narwhal]

Yeah sure that's right

[timid elk]

Does this have a known closed form expression? Google or Wolfram alpha aren't helping me.

[faint narwhal]

Are

you serious?

[timid elk]

Yes I am, why? Is it an absurd or a ridiculous question?

[faint narwhal]

it is, think about the question for a second

[timid elk]

Oh you probably mean it's a stupid one.

[faint narwhal]

Did you realize why

[timid elk]

I'm not quite sure. Maybe there is no such thing as convergent products?

[faint narwhal]

Even if there were, would this converge

[loud ingot]

Does anyone know how absolute value may affect 2. (a)? First time seeing this

[loud ingot]

I see, so how would this affect my answer?

New to that topic @weary tiger

@weary tiger so I know how to find A x B for that specific question, but I’m not sure how cardinality, or power of a set would alter the question

Ok, so is this question, A x B, any different from this one, | A x B |

Please keep in mind I’m very new to discrete math

Ok, so how would find 2. (a) in the pic above

Yep

Hm so cardinality is basically the number of elements in a sec

Set*

So am I finding A x B, then finding the number of elements in the result @craggy sail ?

Alright, but would my way of approaching this problem be correct^^?

[faint narwhal]

Yes

[hollow osprey]

hi could anyone help me with this?

what i know so far is that i have to plug in 100 for the n terms in both of the algorithms but i am not sure what 10a asks for

does it mean to take the ratio of (10(100)log_2(100))/((1/2)(100)^2) ?