discrete-math

[signal rover]

Podria servir de ejemplo

[tidal tulip]

k ty

book ask to find gcd of 6,4 and their prime factors and also find the gcd of 30,42 and their prime factors and then ask what i notice between the two problems and ask how i can use it to compute nZ intersect mZ

i have:

gcd(6,4) = 2
prime factors of 6: 2 * 3, so 2 & 3
prime factors of 4: 2^2, so 2

gcd(30,42)= 6
prime factors of 42: 2 * 3 * 7, so 2,3,7
prime factors of 30: 2 * 3 * 5, so 2,3,5

i don’t notice anything and don’t see how to use this to compute nZ intersect mZ

what am i missing in insight here

[wide vine]

Im confused on how 2 and 3 are different if we have the same variable x

they later explained that they could have used and x and y a but it just seems weird as it is currently

like why ever right it like x,x

instead of something like x,y

$\exists {x} \lnot R(x) \land \exists {y} V(y)$

[coral parcel]

The precise choice of variable is not even visible outside the \exists x(...) formula.

[wide vine]

oh

i see

[hard yew]

$\exists$

[vital dew]

Alter

[wide vine]

:V

so do people normally write it like that

[coral parcel]

The only task of the variable name is to help link up the place where you mention the variable to which quantifier tells you how the value come about.

[vital dew]

Brandon7716

[wide vine]

so you probably wouldn't see it ever like this?

as it is redundant?

where x and y are really talking about the same thing (people in this case)

[hard yew]

I almost always see it with only x

[coral parcel]

Sometimes you do. Since it makes no difference for the meaning, it's not common to think all that much about the choice of variable names.

[tidal tulip]

can i get any help with this

[cerulean wind]

try computing 2Z intersect 3Z and 5Z intersect 7Z

are 6 and 4 in the first intersection? are 30 and 42 in the second intersection?

[weary tiger]

Is this right? Th question mark part

[wide vine]

is the triangle the symbol for difference?

[weary tiger]

Yes

I’m confused

[wide vine]

(A-B)

(A-C)

then the finally intersection is what they share

[hard yew]

Is that statement true? The one with difference distributed over intersection?

[wide vine]

oh

[hard yew]

Feels like it should be ||union|| instead

[wide vine]

Alter is saying your original statement is false and the middle intersection should be ||union||

[weary tiger]

Yes that’s what I’m trying to prove

That’s false

But I cannot draw the second statement

[wide vine]

by the drawing?

[weary tiger]

Yes

[wide vine]

well what is your final drawing

without multiple colors overlapping

I would draw out

(A difference B) (still keep c in the picture)

(A difference C) (still keep b in the picture)

[wide vine]

and then see where the 2 pictures have an intersection

[weary tiger]

How is this true

[tidal tulip]

@cerulean wind wouldnt 2z n 3z = {..,0,6,12,18,..} and 30z n 42z = {...,0,210, 420,..} ?

if so i dont see a pattern

[wide vine]

here is a tool that you can use if you want to see if your diagrams are correct. https://www.wolframalpha.com/widgets/view.jsp?id=eb7ef0469ad23a2c5782e8770da04529

just use (union, difference, intersection) and type your sets as (A,B,C,etc)

or in boolean form

symmetric difference also works

[weary tiger]

Do you undersatand why

the C and B ciurcles are shadded here ?

thats all I need to know

[wide vine]

in which picture

[weary tiger]

this doesnt make sense

the common to those 2 is just A

[wide vine]

no clue

[weary tiger]

ok

so I'm right

the book is wrong

I was abou to lose my mind

[hard yew]

What book are you using?

[weary tiger]

ralph grimaldi's

discrete math

[tidal tulip]

can anyone help with this

[weary tiger]

not sure what to do here:

[obtuse lance]

let's focus on part a, try to describe what you think the symbols are saying as best you can

[weary tiger]

well I think it's asking for the union of the ith sets in N? not sure how to compute that though...

so I guess i $\in \mbb{N}$ is ${1,2,3,4,5...}$

[vital dew]

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

[weary tiger]

idk

sorry

so to clarify

I mean

wait how do I latex union lmfao

I only know intersection

$\cap$

[vital dew]

Renegade

[obtuse lance]

\cup haha

[weary tiger]

$A_1 \cup A_2 \cup A_3 \cup . . .$

[vital dew]

Renegade

[obtuse lance]

yeah, try just doing a few of them, just $A_1 \cup A_2$ and see if you can pick up a pattern

[vital dew]

Merosity

[weary tiger]

im just not really sure, $A_1 \cup A_2$... isn't that just gonna be ${0,1} \cup {0,1,2}$?

uhh, one sec

[hard yew]

\ before {

[vital dew]

Renegade

[weary tiger]

man I am confused

[cerulean wind]

start with A_0

[weary tiger]

the thing is

yeah but

[obtuse lance]

there is no A_0

[cerulean wind]

0 is clearly in N here

[weary tiger]

A_0 isn't in N right? since 0 is not an element of N

wha

[obtuse lance]

there is A_1 = {0,1}

[weary tiger]

ah

right

[obtuse lance]

but there's no A_0 by their definition of N here

[cerulean wind]

so then how do you justify the subscript on the union

[obtuse lance]

the subscript is i in N

[weary tiger]

oh

[obtuse lance]

N={1,2,3,...}

so they're fine

[weary tiger]

I think merosity is correct

because

the answer is {0} \cup N

[obtuse lance]

ok back to this

[obtuse lance]

yeah that will be right

[weary tiger]

which must mean that we take into account the definition of A_1

[obtuse lance]

all the A_n have 0 in them

[weary tiger]

yeah so it's kind of like a power set of N I guess???

anyway as for B

[obtuse lance]

no, I think you're confused about what the union means

[obtuse lance]

when you put this, can you simplify this further?

[weary tiger]

the intersection will only yield {0,1}

[obtuse lance]

{0,1} U {0,1,2} is there a simpler way to write this set

[weary tiger]

yeah

0 1 2

[weary tiger]

for B I believe this is the answer

[obtuse lance]

ok good, just making sure, I think we're on the same page now

and yes you're right

[weary tiger]

as with each consecutive term as n tends to infinity we just get additional terms which dont intersect

okok

thank you so much

[obtuse lance]

in this case, A_n has a nice property to it

that $A_i \subset A_j$ when $i<j$

[vital dew]

Merosity

[obtuse lance]

so it kind of might be nice to think in these terms if possible cause unions of sets where one contains the other will always be the larger set, and similarly for intersections the smaller set

[weary tiger]

ah

yeah

that's smart

[obtuse lance]

yeah you're welcome

[weary tiger]

sorry for being stupid, but I've another Q

A_1 U A_2 for example will yield {-2,0,2} U {-4,0,4} so is it like {2n | n in Z} ?

ye

[tidal tulip]

does 3Z include 0

[willow fog]

yes

[tidal tulip]

or nZ in general

k

[willow fog]

if has to

that's the identity element

[tidal tulip]

k that makes sense

[frozen haven]

im so confused in my discrete math class and idk what to do and idk what im even looking at 🙃

[cerulean wind]

the form of k + (k + 1) + ... + n is really close to 1 + 2 + ... + n

can you figure out how they are related

[frozen haven]

that they both end in n?

[cerulean wind]

thats good to notice

but focus on the beginning of each expression

how can you maybe break up 1 + 2 + ... + n to make one part of it look like k + (k + 1) + ... + n

[frozen haven]

i visually see it but i cant think of anything

im straight up lost in this section of math and idk what its called so i can learn it

[cerulean wind]

1 + 2 + ... + n = 1 + 2 + ... + (k - 1) + k + (k + 1) + ... + n

does that, make things look more clear?

[frozen haven]

uhhhhhhhhhhhhhhhhhh

you lost me

[cerulean wind]

1 + 2 + ... + n = [1 + 2 + ... + (k - 1)] + [k + (k + 1) + ... + n]

[frozen haven]

where are you getting (k-1) from?

[cerulean wind]

you can rewrite k + (k + 1) + ... + n as the difference of two things you know how to calculate

[cerulean wind]

k is some number between 0 and n

if k = 0, then can u see how the problem is already solved?

[frozen haven]

not at all ;-;

[cerulean wind]

um. i dont know how else to break this down then

[frozen haven]

what is this stuff even called

[cerulean wind]

combinatorics

[ember obsidian]

@frozen haven going from 1 to n is the same as going from 1 to k-1 and k to n then adding the results

its written as

1 + 2 + ... + n = [1 + 2 + ... + (k - 1)] + [k + (k + 1) + ... + n]

[wide vine]

How do I read this

It is not true for all x, P(x) is true?

and this one reads

There exist an x such that p(x) is not true?

[coral parcel]

Correct.

[weary tiger]

right so I just wanted to clarify then:

for a) the answer will be ${2n | n \in \mbb{Z}}$

[vital dew]

Renegade

[weary tiger]

and for b), won't it just be ${0}$?

[vital dew]

Renegade

[ember obsidian]

@weary tiger yes

[weary tiger]

thank you!

[weary tiger]

for 6 a) I got $[0] \cup [2,\infty)$

[vital dew]

Renegade

[weary tiger]

is that correcpt?

as for 6b i got [0]

and 7a) I got $\mbb{R} \times \mbb{N}$ but idk

[vital dew]

Renegade

[ember obsidian]

[0] doesnt mean anything as a set

if u mean {0} then 6a,b are wrong. 7a too

[weary tiger]

how can I fix this then?

I'm confused

for 6 a) we have ${0,2} \cup {0,3} \cup {0,4} \cup . . .$

[vital dew]

Renegade

[weary tiger]

right?

[weary tiger]

<@&286206848099549185>

[proven silo]

For 6a) you have $[0,2] \cup [0,3] \cup [0,4] \cup \dots$

[vital dew]

ScapeProf

[proven silo]

{} are used for singletons, [] are used for inclusive intervals

[weary tiger]

ah

so it's [0] U [2,infty)

[proven silo]

Why?

Isn’t 1 in our union?

What about 0.52627277171?

[proven silo]

And again wdym with [0]?

[proven silo]

Can you say what x in [0,2] means for example?

What values of x are in that?

[weary tiger]

ah so it's just all natural numbers U [0] right?

[proven silo]

No, can you answer this?

[proven silo]

And stop typing this

[willow fog]

some of these depend on if 0 is a natural number right

[cunning hinge]

Hey everyone, I just joined the server because of this problem

I know it's simple but I'm just not exactly sure how to write it

[cunning folio]

hi, I hv a small doubt

If I hv a number x, then how to find the highest no of different factors of it such that their pairwise lcm is x

I know answer is the no of distinct primes in its prime factorisation

bt I'm not able to digest it, can someone give a valid point or some proof why this is the only way plz

also I wasnt sure if I should put this doubt here or in #elementary-number-theory so posted it here

[hard yew]

What have you tried?

[cunning hinge]

I wrote P(1,0) ∧ P(1,1) ∧ P(1,2)

[hard yew]

y=0 is not in your given domain

[cunning hinge]

I meant to put 1 2 and 3

With 1 2 and 3 replacing the y in the proposition, would it be correct?

[spring surge]

Given an N-sided convex polygon with more than three sides. Consider all the triangles, whose vertices coincide with vertices of this polygon, but none of these triangles' sides coincides with any side of a given polygon. How many of these triangles exist?

Where can I find more of these types of problems ? Any source which digs deeper into a thinking required to solve it?

[tidal tulip]

im trying to understand notation below with a concrete example

the notation trying to understand is:

[Product i in I] S_i = {(x_i)_{i in I} }

let’s say i have sets

S_1 = {1,2}
S_2 = {3,4}
I = {1,2} (ordered)

then would this be a correct interpretation of the notation below to these sets?

[Product i in I] S_i = S_1xS_2 = { (1,3), (1,4), (2,3), (2,4)}

?

[weary tiger]

can someone help me with this? I don't understand this

[coral parcel]

Not a single word makes sense to you?

(This is more #geometry-and-trigonometry than it is #discrete-math, by the way).

[weary tiger]

nevermind

im stupid

idk why, but this was on my discrete structures hoemwork

[tidal tulip]

does that seem correct to anyone

[weary tiger]

can someone help me on this

[weary tiger]

How can I prove this?

contrapositive

ok

so

thats what I did

but I'm struggling a little bit

@weary tiger

I'm having hard time undrstanding the logic here

it almost feels like it was written by a 12 years old

So

they turned the statement into its contrapositive and proved it

By demorgans law we assume that a|b or a|c

therefore we have two cases

In the solution they proved that in the case where a|b then a|bc

and let the reader figure out the other case a|c

ok

so

I get that they

rewritten

b = ak

but how does ak = b implies (ak)c =...

would u mind explaining it?

because they simply multiplied c onto both sides

if ak=b

then (ak)c=bc

ok

and

thats it?

and from there

they moved the brackets

a(kc)=bc

since kc is an integer

then a|bc

thats it

ah okay

that makes sense thanks a lot

quick thing

if I'm proving something

a bi condiotnal st.

do I have to prove it twice

when it's right and when its wrong?

You have to prove both directions

ok

[weary tiger]

?

nah nah I get it

just suppose 1 side is true

and then derive the proof

then do the oppsite

yep

ok thanks

[weary tiger]

What base gives us this answer

[stray reef]

@weary tiger do you still need help with this?

[weary tiger]

Yes

@stray reef

[stray reef]

ok

do you know how positional notation works?

[weary tiger]

No

[stray reef]

so you are not even familiar with decimal positional notation?

[weary tiger]

I’m

So like

0123...

[stray reef]

do the words "units place", "tens place", "hundreds place" etc. ring any bells to you?

[weary tiger]

Yes

[stray reef]

are you able to, say, write down the number 623 in terms of its place values?

(decimal 623)

[weary tiger]

Just positional notation term isn’t familiar

[stray reef]

would appreciate an answer to this

[weary tiger]

Ba

Should mean like

Yes I can

[stray reef]

"Ba"?

[weary tiger]

Sorry I’m typing on my iPad

[stray reef]

okay, so you can. please do it now and show me the result

623 = ....

[weary tiger]

6 2a 3a^2

[stray reef]

what's a?

[weary tiger]

Base a

[stray reef]

.

i said the base was ten

are you trying to say that the answer should be 6 * 2a * 3a^2?

[weary tiger]

Plus

[stray reef]

if you mean plus then write plus

[weary tiger]

I m tryinh

[stray reef]

but also, i asked you to do it for the number 623 in base 10

[weary tiger]

I plugged in the keyboard now

Ok so

Ah I see

3 * 10^0 +2*10^1+ 6 * 10^2

[stray reef]

??????

3 * 10^0 + 2 * 10^3 is not 623

please do what i ask you to do.

[weary tiger]

Incorrect ?

[stray reef]

okay now you got it right finally

would be better as a new message and not as an edit, but yes...

[weary tiger]

I knew I’m just struggling with typing on this keyboard it sucks 😦

[stray reef]

okay, so now are you able to write the number 251 (base b) in terms of its place values?

[weary tiger]

Yes

[stray reef]

okay, so do it

[weary tiger]

1b^0 + 5b^1 + 2*b^2

Is this wrong

[stray reef]

this is correct, though it could do with some simplification. but whatever.

are you able to write down the whole equation in terms of b?

[weary tiger]

Ok nice

Yes

[stray reef]

okay so do it

[stray reef]

the very top part of your equation got cut off

please uncrop this

[weary tiger]

Oops

[stray reef]

this is incorrect

double check your right hand side

[weary tiger]

Ok

[stray reef]

in particular also check your handwriting as you managed to make 0b^3 (presumably) look like 0^3 b somehow

[weary tiger]

I think it’s correct now

[stray reef]

now it is correct, yes

are you able to solve this equation for b?

[weary tiger]

I can try

1 sec

Wait am I supposed

To get this

@stray reef ok

I think 8m wrong

Is it just solve for b.

I guess

[stray reef]

double check your algebra

[weary tiger]

Hahaha god I am doing algebra in the dark

1 sec

[stray reef]

also, to answer your question:
yes, if i ask you to solve for b, it means i want you to solve for b.

[weary tiger]

The algb gotta be write this time

No shot

[stray reef]

right*

but yes, now you got the algebra right.

[weary tiger]

Ok

What’s next

[stray reef]

you are not done, obviously

you still have to solve for b

which involves performing the insurmountable, nigh impossible task of solving a quadratic equation

[weary tiger]

Got it

Write

Write? @stray reef

[stray reef]

where did the letter x come from? and how come you keep mixing up "write" and "right"?

write is what you do with a pen
right is the opposite of wrong

[weary tiger]

Yeah I meant b

Thanks a lot

[stray reef]

if you mean b then write b

[weary tiger]

Oh where I come from we spell it write in northern Canada

[stray reef]

i do not believe that

[weary tiger]

Are u northern Canadian

[stray reef]

no

[weary tiger]

Well

[stray reef]

however it does strike me as odd at the very least

[weary tiger]

I appreciate it

Write?

[stray reef]

are you saying that you spell both "put words on a page with a pen" and "the opposite of wrong" as write?

[weary tiger]

Write

[stray reef]

ok so you're being obtuse on purpose

forget it

[weary tiger]

I hope you're not mad at me

[stray reef]

😒

not mad, just annoyed

more so now that i know you were deliberate in your attempts to annoy me

[weary tiger]

I promise iPad corrected it to write like 2-3 times and I didn't even notice until you mentioned it LOL

Anyway I gotta go to bed thanks a again

[devout river]

What if there is no intersection between A ^ B?

Is it null?

and answer will be null symbol

[pale epoch]

if there is no element in $A \cap B$, then $A \cap B = \varnothing$

[vital dew]

Lochverstärker

[devout river]

What method would you guys use to prove this identities?

[stray reef]

$A \cap A = \emptyset$ is not an identity

[devout river]

Im planning on using tables

[vital dew]

Ann

[stray reef]

but any equality between sets can be proved via an element-chasing argument

take an arbitrary element in one set and show that it belongs to the other, and vice versa

[devout river]

Is this correct? ^_^

@stray reef

I dont know what I would with the second one though...

[stray reef]

this is nonsense at best

[devout river]

omg

What do you mean

._.

[stray reef]

i mean exactly what i say

this is nonsense

perhaps this could be read as a truth table that is missing two rows and has multiple redundant copies of the other two

[devout river]

So I should add two more rows?

[stray reef]

you should add two rows and remove the ones that are repeated.

[devout river]

the repeated one is the 4th column right?

[stray reef]

no

please read what i said carefully

[devout river]

Thank you

[sleek loom]

is the relation R=((1,2)) transitive?

and antisymmetric?

[olive wren]

It is (both)

Because the criteria are true vacuously

[sleek loom]

I see, thanks(:

[sleek loom]

This means that the empty set has both these properties as well, right?

[olive wren]

Yeah

[sleek loom]

Niceee

[olive wren]

It’s also symmetric for the same reason

[sleek loom]

Oh nice

I wanted to say it has every property but not every property is true vacuously

I think

Didn’t check

[olive wren]

Yeah it’s not reflexive

[sleek loom]

Oh right

[olive wren]

Unless it’s a relation between $\varnothing\times\varnothing$

[sleek loom]

Isn’t that relation equal to the empty set?

[olive wren]

In which case it is reflexive and has all the basic properties (I’d assume, it probably varies depending on what you mean by “basic”)

[olive wren]

Yeah

[sleek loom]

So how is it reflexive?

[olive wren]

Like the empty set is reflexive if it’s a relation between the empty set and itself

That’s what I meant to say, I just didn’t say it well

[sleek loom]

Oh damn

[olive wren]

Reflexivity requires that given a relation $R\subseteq A\times A$, for every $a\in A$, $(a,a)\in R$.

Since there are no elements in $\varnothing$, this is true vacuously

[sleek loom]

Yeah that’s pretty nice, didn’t realize that at first

Didn’t realize there could be different types of empty set

If I’m thinking about it right

[olive wren]

Well it’s the same empty set

[sleek loom]

But in one instance it is reflexive and in the other it isn’t

So there is a distinction

[olive wren]

It’s just that the criteria are dependent on the set that the relation is defined between

[sleek loom]

Damnnnn right

Forgetting the basic things

[olive wren]

Lol it’s not really an issue

[sleek loom]

Could you help me out with an exercise then? I need to prove that if R is transitive, R squared and cubed and so forth are transitive as well

[olive wren]

By R^2 you mean RxR?

[sleek loom]

I understand how to prove it for every R independently but couldn’t generalize it to every power

Yeah

Is that notation normal?

[olive wren]

Yeah

I just wanted to make sure I understood the question correctly

But what do you mean by R^n being transitive? What’s it a relation between?

[sleek loom]

Oh wait I’m not sure, does RxR multiplying relations?

I meant it as R being a relation

Over some set A

Not 2 sets

[olive wren]

Right

But what is like R^3 a relation between?

Because the elements of R^3 are in the form ((a, b), (c, d), (e, f))

[sleek loom]

R and R squared

[sleek loom]

Not sure what you mean

[sleek loom]

Or did you mean what sets does it relate between

[olive wren]

Okay, suppose A={1, 2} and R={(1, 2)} then R^2 = {((1, 2), (1, 2))} and R^3 = {((1, 2), (1, 2), (1, 2))}
So perhaps R^2 is a relation over A^2, but what is R^3 a relation over?

[olive wren]

Yeah

[livid minnow]

I don’t know if this question falls under discrete math but:

I need to find all 5 letter lists such that the letters fall in alphabetical order. Some examples ABCXY would be one such list or BDMNQ would be one, but BAXYZ would not be one

[olive wren]

Can the letters be repeated?

[livid minnow]

nope

[olive wren]

Then think about it like this

Oh wait uh

[livid minnow]

I honestly don’t even know where to start

I’ve thought about getting the permutation of all 5 letter lists then subtracting it from all lists that don’t fall i. Alphabetical order

but then it’s the same problem of not knowing how to find every list that doesn’t fall in alphabetical order

I’ve also thought about it like maybe if I choose B for my first letter I have 26 - B letters left

and so on

but how do I incorporate that process into a list with only 5 letters

[olive wren]

I’ll admit that I don’t know combinatorics

But I found this

https://math.stackexchange.com/questions/2324420/different-possible-arrangements-in-alphabetic-order

[livid minnow]

thanks I’ll look at that

[olive wren]

Oh yeah

Doi

If you can’t repeat any

It’s just the number of ways to choose 5 distinct letters where order doesn’t matter

Because if you choose 5 letters, there’s only one way to put them in alphabetical order

[livid minnow]

OH

so it’s just the combination of 26 and 5

that’s deceptively easy

[sleek loom]

R squared is the empty case in this case, and so is R cubed

The relation R squared is empty because Domain(R)=1

[olive wren]

Sorry I’m a little confused what you’re trying to show here
Is this a question from homework? If so, could you send it perhaps? Thanks!

Like my issue is that R^3 is a tuple with 3 elements in it, so it can’t be between a set and itself, so it can’t be transitive (because to be a transitive relation, it must be a subset of AxA, not AxB)

[tidal tulip]

can someone please check this

[olive wren]

I’ve never seen notation for the Cartesian product of an indexing of sets, but unless it’s defined really unintuitively, this looks right

[tidal tulip]

okay awesome

ty for checking

[olive wren]

Np

[hard yew]

What do you mean by this?

Also, I believe you're talking about this:

A relation to some power is represented through compositions. $\ R^1=R$ and $R^{n+1}=R^n\circ R, \forall n \in \mathbb{N}^+$

[vital dew]

Alter

[hard yew]

Or something else?

[weary tiger]

Would someone help me with how to build intuition to approach these kinds of proofs

the answer says try x=1 and y =1 and x=1 and y=-1
but why? I really can't tell how they come up with these numbers

[pale epoch]

for something like this to hold, a+b and a-b must be related in some way, so you can start about thinking how

then you notice that their sum is 2a and their difference is 2b

and then you write down the definition of gcd, and somehow use this

[sleek loom]

I didn't learn about compositions but that seems about right

I can prove that if a relation R over some set A is transitive then R^2 is transitive by using the property of transitive sets, that R^2 is always a subset of R (not sure how to use the mathematical notation) and the identity that if a relation A is a subset of a relation C and a relation B is a subset of C, then
AB is a subset of C^2

This is proof of the property of transitive relations

This is the proof that if R is transitive then R^2 is transitive

A similar proof could be done for R^3 and so on

So I could prove this for every independent power of R but I can’t figure out how to generalize this to every power. I thought that using induction here, if that is even possible, should work but I couldn’t figure it out

The book I’m studying from has a “solutions” manual in which they just proved these results for R^2 and R^3, when the question asks to prove for every power😐 but maybe I’m missing something there

[weary tiger]

why would 2a and 2b help us at all?

like I can think of why would the relation be

2a and 2b

[pale epoch]

write down the definition of gcd

or just

the gcd will divide a+b and a-b, so also their sum and difference

maybe this is the better approach

[weary tiger]

thats what I did

I said

gcd= C=[ (a-b)x + (a+b)y] and the idea is that for all a and b c=1 or 2

is this correct? thats what i have written down

[pale epoch]

what is [ (a-b)x + (a+b)y]

[weary tiger]

by properties of division

[pale epoch]

no, what kind of object is this

[weary tiger]

if c|a and b

dividend?

[pale epoch]

is it a number?

[weary tiger]

yes

[pale epoch]

what are the square brackets

[weary tiger]

wdym

whats the square brackets

[pale epoch]

[ (a-b)x + (a+b)y]

[weary tiger]

yeah

[pale epoch]

there are brackets []

what do they mean

[weary tiger]

I know

[pale epoch]

i cannot parse this

the gcd is a number

this depends on a, b and also x and y

[weary tiger]

ok

but whats wrong

if it's a number

that depends on a, ba and x and y?

[pale epoch]

there is nothing wrong

[weary tiger]

ok

[pale epoch]

i have no idea what you are saying

[weary tiger]

I have no idea either

I'm confused

why

did u ask me

about the brackets huh

[pale epoch]

i just don't know what kind of object "[ (a-b)x + (a+b)y]" is supposed to be

c = gcd(a+b, a-b) is a number

"[ (a-b)x + (a+b)y]" to me is just a bunch of symbols

it certainly isnt a number

[weary tiger]

ok

if c divides

both

how do you write it down?

and c=1 or 2

right?

we wanna prove that for any a or b

[pale epoch]

$c \mid (a+b)$ and $c \mid (a-b)$

[weary tiger]

c= 1 or 2

[vital dew]

Lochverstärker

[weary tiger]

ok

[pale epoch]

thats why i write

[weary tiger]

combine them

what i wroye up

is literally a combination

of that

[pale epoch]

ok

you can say that if c is the gcd

it will also divide x(a+b) + y(a-b) for all x, y

[weary tiger]

ok

but then I have to find valyues for x and y right?

[pale epoch]

i mean i think this is already overkill

[weary tiger]

well I have to

prove that it's =1 or 2

in generality yes

u are right

but I think 1 is by def since they are relatively prime

right?

then I just need to prove that for some a or b it will be =2

[pale epoch]

how is that supposed to work

if you think its 1 by definition, it can never be 2

[weary tiger]

yes

you're right

[pale epoch]

but that is not correct, both cases occur

[weary tiger]

hmm because we are not talking about a and b

(a-b) and (a+b)

[pale epoch]

yeah, but as i said

you collect some facts

so if c divides a+b and a-b

it also divides their sum and difference

which is 2a and 2b

[weary tiger]

ok and gcd of (2b,2a) = 2 or 1

[pale epoch]

its not both

[weary tiger]

can I put the 2 out

[pale epoch]

but the only numbers dividing 2a and 2b are 1 and 2 (assuming a and b are coprime)

[weary tiger]

yea because I was gonna say the same

gcd(a,b)= 1 by def becuse it tells us they are coprime

so (2a,2b)= 2 * gcd(a,b)=2?

is this proper?

[pale epoch]

gcd(2a, 2b) = 2, yes

the argument also works

but tbh you should take at least one step back

and review and work with the definitions a lot more

i suggest proving that if a divides b and c, then a also divides b+c and b-c

since this is used here

and review some information you have about the gcd

you clearly arent comfortable working with them

[weary tiger]

hmm well

I find these division proofs kind of tricky

idk how to get better at them

like in set theory equality proves has a theme

but these ones are all over the place

(I often forget what I'm trying to prove)

[pale epoch]

in general you use the definition a lot

gcd divides both numbers

and it is the greatest of all that do

and then you have a toolbox of certain facts

how divisors behave

i.e. if a number d divides a and b it also divides all linear combinations of a and b

[weary tiger]

and the goal to show

if the gcd divides a and b then it also divides the linear combination?

[pale epoch]

the goal in general in math is to reduce the complexity (subjective) of the problem using your toolbox

[weary tiger]

but thats given

[pale epoch]

and the gcd dividing 2a and 2b is "less complex" in the sense that only one unknown appears

tbh i am not sure if this explanation is good

[weary tiger]

the only thing I' missing

from the entire proof

tbh is

[pale epoch]

i am sure that you should review the definitions and lemmas you know related to the gcd

and at some point you just see such things

[weary tiger]

I get that we are trying to show that if c divides a and b then it divides ax+/-by and so c in this case c will always be 1 or 2

now the part I'm trying to understand

why by the difference or addition we got that c=1 or 2

but I will get it

thanks a lot

[pale epoch]

at that point you still havent used the fact that a and b are coprime

so you need to find some way to do that

[weary tiger]

yeah if they are coprime

they can't have 1 or 2

it must be 1

and thats what pushed you to think

then maybe the + or / of both terms could give us that c=2

[wide vine]

in statement like this,

the $\lnot \exists$

[vital dew]

Brandon7716

[wide vine]

is only for the x?

it is saying there is no such x where the proposition is true?

predicate*

How would I read this statement in english?

There exist no x but some y such that (!p(x) and q(y)) is true?

[shell lagoon]

In a standard deck of 52 cards, there are 4 suits (Spades, Hearts, Clubs, Diamonds) and 13 cards in each suit (no Jokers). How many 7-card hands contain at least one card of each suit?

intuitively speaking.. how should I begin thinking about this?

I really have no intuition when it comes to these counting questions

at least one tells me that we can take the set where we have 2 of each + 3 of each maybe? and then that would be.. approximately right but yeah

like at least 1 is the same as adding 2 + 3 + 4.. 7

[wide vine]

How is this different from $\exists {x} L(x) \land \forall {y} (((x\neq y) \longrightarrow \lnot L(y))$

[vital dew]

Brandon7716

[wide vine]

I guess I am just asking why do we have to have the $\exists {x}$ all on the outside

[vital dew]

Brandon7716

[wide vine]

or is this so we know that this x variable acts on the whole predicate?

I just am wondering we if are need the parethesis like they are

$\exists {x} L(x) \land \forall {y} ((x\neq y) \longrightarrow \lnot L(y))$

[vital dew]

Brandon7716

[wide vine]

I guess my question is... is my predicate the same as the one pictured or does my predicate not make sense

for refernce to clear it up, here was the definition of L(x)

:keke

[hard yew]

Yes, you need the

[wide vine]

is that because it tells the reader where the x comes in

like it is still part of the exists part way at the start

[hard yew]

Otherwise the scope of the exists quantifier only applies to the first predicate

[wide vine]

okay

[hard yew]

In the version that you have your second x has no relation to the first x

[wide vine]

Yeah, that is what I was wondering about

[hard yew]

But you need them to be the same in order to define what it means "exactly 1"

Also, not sure if you weren't supposed to use it but there is a predefined quantifier for "there exists only one unique x" or "there exists exactly one x"

[wide vine]

well it has not shown up

I feel like that might be something similar to an xor?

[vital dew]

Alter

Alter

[wide vine]

hmm

is the first one more common than the second

[hard yew]

yeah

[hard yew]

It is defined through "for all" and "exists" in this way

[wide vine]

seems useful

[hard yew]

I haven't seen it used often but it's good to know

[tidal tulip]

does this proof work:
i want to prove

let U be a universal set and let F={S_i : i in I} be a set of subset of U, where I is some set. let J be a subset of I.

show

[intersect i in I] S_i subset [intersect i in J] S_i

proof: suppose J subset I

let x in [intersect i in I] S_i

then by definition of intersection x in S_i for all i in I

since x in S_i for all i in I

then x in S_i for all i in J since J subset I

therefore by the definition id intersection x in [intersect i in J] S_i

[wide vine]

can some explain how (4) is not false?

I think im not sure the difference between

$(x \neq y) \land \lnot P(x,y))$

vs

[vital dew]

Brandon7716

[wide vine]

$(x \neq y) \longrightarrow \lnot P(x,y))$

[vital dew]

Brandon7716

[wide vine]

oh nvm

I miss read my rows and columns

[tidal tulip]

can anyone check this

[sleek loom]

Kinda hard to read

[olive wren]

Looks good, but you should learn latex so these things are more understandable
Also, you wrote “suppose J subset I” on the first line of your proof which is redundant because it’s a given

[stray reef]

alright so there's this problem that came up in my math class in reference to a different problem (which has since been solved)

this seemingly simple problem has kind of eluded me

An odd number of dominoes are arranged horizontally on a chessboard (8x8) without overlap. Prove that there exists at least one column in which the number of cells covered by dominoes is odd.

[inland imp]

proof by contradiction? if all the columns have an even number of dominoes then the total would be even

[stray reef]

the total of what exactly

sure, if every column has an even number of covered cells, then the total number of covered cells is even - but that's even no matter what

[inland imp]

wait so each domino covers 2 tiles?

i think i misunderstood the question

i have an idea of how the proof would go

since the dominos don't overlap you can visualize it as something ilke this

each ring represents a domino and counts towards the columns on both sides

you can start with the first column and since there's only 1 stick that you can put rings on to affect it the number of rings on stick #1 must be even

and then induct from there until you reach the last one where you must have an odd number of rings left

and so you'll have the last one be odd no matter what

and then contradiction

[sleek loom]

does "arranged horizontally" have anything to do with the question?

[stray reef]

yes

in case it wasn't clear: it means each domino is placed horizontally

[inland imp]

like this

excuse my terrible ms paint skills

[sleek loom]

oh i see

[austere cedar]

You can also prove it in the following way: Sum up the number of dominoes in every odd numbered column. So in total, you count every domino exactly once, as each domino covers one tile in an odd numbered column and one in an even numbered column. So the sum equals the total number of dominos and is therefore odd. This means that at least one of the summands has to be odd, i.e. there is a column with an odd number of dominos.

[stray reef]

oh yes that's right

[tidal tulip]

thanks for checking! just tex’d it up, how is it looking now

Let $U$ be a (universal) set and let $F= { S_i : i \in I }$ be a set of subsets of $U$, where $I$ is some set. Let $J$ be a subset of $I$.

Proof:
Let $x \in \bigcap_{i \in I} S_i$, by the definition of intersection $x \in S_i \forall i \in I$. Then $x \in S_i \forall i \in J$ since $J \subseteq I$. Therefore by the definition of intersection $x \in $\bigcap_{i \in J} S_i$. Thus $\bigcap_{i \in I} S_i \subseteq \bigcap_{i \in J} S_i$

[vital dew]

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

[olive wren]

Looks good!
(Just two notes: you have an extra $ which is giving you an error. It’s the $ in x\in$\bigcap…
And I find it more readable when you put the \forall before the statement, (like \forall i\in J: x_i\in S_i), but that’s up to you ig)

[tidal tulip]

awesome, thanks!

i need to prove the other direction as well, but i don’t have time atm. i’ll try it later today and tex up my attempt. thanks for the tex tip

[olive wren]

You’re trying to prove that
[ \bigcap_{i\in I} A_i \subseteq\bigcap_{j\in J} A_j\implies J\subseteq I ]
?

But that’s not true. Take $A_1=\varnothing, I=\varnothing, J={1}$

[tidal tulip]

let me write it out

$\bigcup_{i \in I} S_i \supseteq \bigcap_{i \in J} S_i$

not sure why that’s not displaying

[olive wren]

You have a space before the $

[vital dew]

strings

[tidal tulip]

what does that mean rhen

[olive wren]

That’s not the other direction though? It’s a different question altogether

[tidal tulip]

ok i was wrong in phrasing it that way

that was my error not the way the problem was given

[olive wren]

Oh okay no problem

[tidal tulip]

and by proving what i just did and this part what does these two statements mean taken together? i’m trying to get an understanding

[olive wren]

You can’t really say anything about $I$ and $J$.

Like if you take my example from before, you see that this holds, and $I\subseteq J$

But if you take $I\supset J$ with a non-empty $S_i$, your expression will still hold.

So you can’t determine whether $I\subseteq J$ or $J\subseteq I$

[tidal tulip]

ok i will think this over when i have some time. have to go i’ll bbl

[olive wren]

Okay bye!

[tender scarab]

Ans of 3(c) should be 140?

5x4x7

[sleek loom]

bruh

[crystal bone]

help

No, its first day of school and they let us answer that so sad

[sleek loom]

so sad

[weary tiger]

how should i go about this?

[olive wren]

Uh wait

Is that a test?

[weary tiger]

no

[olive wren]

Oh

Why does it say (15 points)?

[weary tiger]

hw

[olive wren]

Okay

Did you see my answer or do you want me to resend it?

[weary tiger]

i didn't see it

[olive wren]

Okay

There may be a better way, but suppose $(Q, \Sigma, \delta, q_0, F)$ is an automaton that accepts $L$. Then you can create a new automaton $(Q', \Sigma, \delta', q_{00}, F')$ where:
[ Q' = {q_0\mid q\in Q}\cup{q_1\mid q\in Q} ]
[ \delta'(q_i, \sigma) = \delta(q, \sigma)_{i+1 \bmod 2} ]
[ F' = {q_0 \mid q\in F} ]

The idea is that each state is a copy of a state in the original automaton and keeps track of whether the current length is even or odd.

[stray reef]

is a fully formal construction of the DFA required

[olive wren]

I just couldnt find the right words to explain my idea tbh

[ancient heath]

hey, im peer reviewing some proofs for class and im trying to figure out how they came up with $1000! + n = n(1000!) / n + 1$. Does anyone have an idea?

[vital dew]

Lunarros

[stray reef]

@olive wren here is how i would phrase it:
take an automaton that accepts L; split each of its states into two copies, of which one gets labeled even and the other odd
each state transition in turn becomes two state transitions, one from even to odd and the other from odd to even
the new start state is the even version of the old start state (since we start at 0 symbols read, an even number)
and the accepting states are the even versions (and only the even versions) of the old accepting states

[coral parcel]

That's not what they came up with. The rewriting was $1000! + n = n(1000! / n + 1)$. Note different bracketing on the RHS.

[vital dew]

Troposphere

[coral parcel]

(The proof is missing an argument that 1000!/n is an integer, though, especially since n can be larger than 1000).

[ancient heath]

I'm mostly just confused as to how they rewrote the LHS to the RHS. lmao I'm just trying to figure out if its just me being dumb or if maybe its something they should review

[coral parcel]

Um, put the n outside a parenthesis.

Do you disagree that the rewriting is valid, or are you just questioning where they got the idea to do that particular rewriting?

[ancient heath]

Mostly just questioning where they got the idea to rewrite it like this

[coral parcel]

They want to show that the LHS is divisible by something, so a natural way to do that is to rewrite it as a product.

[ancient heath]

ohh ty so much lmao im so dumb

[weary tiger]

@cosmic mesa

[cosmic mesa]

yo was up @weary tiger

[weary tiger]

loving discrete math rn

[cosmic mesa]

when there is moose there is goose

[weary tiger]

true?

disjunctive sylogism

[cosmic mesa]

https://tenor.com/view/calculating-puzzled-math-confused-confused-look-gif-14677181

[weary tiger]

set proofs ---> logic

[olive wren]

mmm that's well very well put, thank you!

[stray reef]

so to de-clutter that, you want to show that $\sqrt{x^2 + y^2 + \frac{x^2y^2}{(x+y)^2}}$ is always rational for $x, y \in \bQ$ with $x + y \neq 0$?

[vital dew]

Ann

[stray reef]

may i suggest combining the terms under the square root into one fraction?

[stray reef]

and what did you get?

[stray reef]

wonderful, so you had symbolab do it instead of doing the algebra yourself?

not to mention that symbolab left it in a form where it's difficult to understand anything

[tidal tulip]

can i get a proof check:

Let $U$ be a (universal) set and let $F= { S_i : i \in I }$ be a set of subsets of $U$, where $I$ is some set. Let $J$ be a subset of $I$.

Show that:

$\bigcup_{i \in I} S_i \supseteq \bigcap_{i \in J} S_i$

Proof: Let $x \in \bigcup_{i \in J} S_i $, by the definition of intersection $x \in S_i \forall i \in J$. Since $J \subseteq I$ then $x \in S_i$ for some $i \in I$. Therefore $x \in \bigcup_{i \in I} S_i$. Thus $\bigcup_{i \in I} S_i \supseteq \bigcap_{i \in J} S_i$.

[vital dew]

strings

[torn tendon]

i think you used the wrong notation on 1st line of proof for intersection
other than that looks good

[tidal tulip]

oh snap

i used union when i meant intersection

ty

all together for this problem i proved (where J is a subset of I)

$\bigcap_{i \in I} S_i \subseteq \bigcap_{i \in J} S_i$

$\bigcup_{i \in I} S_i \supseteq \bigcap_{i \in J} S_i$

[vital dew]

strings

[tidal tulip]

what does that mean intuitively

[craggy juniper]

this just means they’re equal

[rigid oriole]

check it

[tidal tulip]

yeah what does it mean lol

[craggy juniper]

still don’t see anything wrong with what i said

[rigid oriole]

what are equal

[craggy juniper]

oof i read it wrong

read the union as intersection

[tidal tulip]

yeah what does that mean tho

i haven’t seen anything like it before so not entirely sure

even tho i proved it both ways

[weary tiger]

Is there a quick way to determine whether these are true or false?
https://gyazo.com/24f86f45acf3a6974e794af9ed2d0302

[severe swan]

If I have the set A = {Ø} then would the cardinality be a) 0 because the set contains no elements or would it be b) 0 because the empty set does not count as an element? Or would it be something else?

[gritty crescent]

A contains an element.

A box containing an empty box is not empty itself.

[gilded finch]

Yes, juste use properties of O, Theta, Omega and their "little" variations

[weary tiger]

Is B a valid way to prove a function is one to one?

[ember obsidian]

actually no theyre not logically equivalent

[weary tiger]

I was thinking no because p->q is not the same as ~p->~q

yea

Thank you!

[wide vine]

is this something I would have to remember by their name

or just understand them and then apply them?

a few of the names make sense but some of them I have no clue how they got that name

[weary tiger]

Is this statement true? I have a feeling it’s true

simply because there is an infinite amount of integers less than 5

So you can always find a y less than your X

[gritty crescent]

Yes

In particular you always have x-1<x<5

[stray reef]

@wide vine unless you're planning to do a lot of work in logic specifically, no, you do not need to remember these by name.

[weary tiger]

Thanks!

[whole hamlet]

How many ways can you choose a task-force of 8 members from the committee where 6 are employees and 2 are administrators? can someone help me with this problem?

[inland imp]

what does the committee consist

i'm just abusing words at this point

[latent grove]

seems like a perms/combs problem!

8C6? im not sure..but u can check out intro permutations and combinations to get to being able to solve ur problem

[severe swan]

How would I put this into set builder notation?

{0, 3, 6, 9, 12}

Would this be correct?
{3n | n = 0, 1, 2, 3, 4}

[hollow timber]

Yes

[weary tiger]

Can someone help with this? I don’t quite understand it

[stray reef]

well, one would think the question is:

When you write out all 4-subsets of {1, 2, 3, ..., 10} in lexicographic order, in what position does the set {2, 3, 9} appear?

however, as written, the answer is nowhere, because {2, 3, 9} is not a 4-subset of [10].

@weary tiger was this your point of confusion?

[gritty crescent]

How does lexicographic order on a single set work anyway?

I've only seen it when we have some product of sets and then we consider order componentwise, moving left to right

[stray reef]

i think you would compare the smallest elements, then the second-smallest, then the third-smallest etc until you found a pair that differs

[severe swan]

If I have to identify the sets where 2 is an element, why does this count? {0, 1, 4, 9, ...} Can someone explain it to me please

[stray reef]

can you show the whole problem?

@severe swan

as written, {0, 1, 4, 9, ...} does not sound to me like it contains 2 as an element.

[wide vine]

Im confused

so you have to write out explicitly that c is an arbitrary element

or else it is assumed to be a particular?

[wide vine]

because how the question is stated it isn't clear at least to me that c is defined to be a particular or arbitrary

is it because it says "Hypothesis"

ahh

I think I realized my mistake

the parts in blue are outside the scope of the actual rule but only used to help the reader know

[hard yew]

Does this say that you choosing "incorrect" is correct or that "correct" is the answer they were looking for?

[wide vine]

me choosing incorrect was correct

[hard yew]

I see

[wide vine]

but I initially picked correct because I didn't know that c is a particular

but I guess if you state it as a "hypothesis" then it is

[hard yew]

So you're given that c is an element, and that P(c) is true. So you know that there's some element for which P(x) is true (namely the element c). However you can't conclude that P(x) is true for all x just because it is true for c

[hard yew]

It depends on how it's stated

[wide vine]

Okay I see, so when we use c it is a placeholder for some particular element in the domain?

[hard yew]

In this case "c is an element" refers to an individual element. If it had been "c is any element", then forall P(x) would be true

[wide vine]

okay okay.. Yeah that was my issue

[hard yew]

What is your question?

[severe swan]

If I have to identify the sets where 2 is an element, why does this count? {0, 1, 4, 9, ...} I don't understand why {0, 1, 4, 9,...} counts if there is no 2 in the set.

[hard yew]

You were asked to select all the sets that contain 2 as an element. {0, 1, 4, 9, ...} does not contain 2. You did not select that set. Your answer is correct.

That is why your answer is correct

[weary tiger]

Hey. Forgot to mention there is a typo. It should read 3-subset. I’m confused as to how to approach it

[stray reef]

it sounds like you are misreading the feedback given to you by the homework/assignment system

you left {0,1,4,9,...} unchecked, indicating that 2 is not an element of this set. the homework system highlighted this answer option in green, meaning that you got it right and 2 is, indeed, not an element of {0,1,4,9,...}.

[weary tiger]

Think I understand my lex questions. I do have another one that I don’t understand. Hope someone can help out a bit

[severe swan]

Thank you Ann

[wide vine]

P(x) and Q(x) have to share the same domain, right?

right?

might be better to ask #math-discussion and see where they might take you

[tidal tulip]

what’s an intuitive understanding of

$\bigcap_{i \in I} S_i \subseteq \bigcap_{i \in J} S_i$

$\bigcup_{i \in I} S_i \supseteq \bigcap_{i \in J} S_i$

[vital dew]

strings

[pale epoch]

is there more context?

specifically how are I and J related

[tidal tulip]

yes J is a subset of I

[pale epoch]

S_i are just any sets?

[tidal tulip]

Let $U$ be a (universal) set and let $F= { S_i : i \in I }$ be a set of subsets of $U$, where $I$ is some set. Let $J$ be a subset of $I$.

[vital dew]

strings

[pale epoch]

intuitive reason for the first is "intersecting with more sets can make the result only smaller"

is the second correct as stated?

[tidal tulip]

yes

[pale epoch]

because intersection are always smaller than unions

unioning with anything always makes the result bigger (or leaves it unchanged) and intersecting with anything can only make the result smaller

[tidal tulip]

okay what you said makes sense

i am trying to relate it to those two statements

[pale epoch]

i asked bcs the first is \cap on both sides

and the second is \cap vs \cup

[tidal tulip]

yeah that’s how it’s stated

i had to prove both

Let $U$ be a (universal) set and let $F= { S_i : i \in I }$ be a set of subsets of $U$, where $I$ is some set. Let $J$ be a subset of $I$.

Show that:

$\bigcup_{i \in I} S_i \supseteq \bigcap_{i \in J} S_i$

Proof: Let $x \in \bigcap_{i \in J} S_i $, by the definition of intersection $x \in S_i \forall i \in J$. Since $J \subseteq I$ then $x \in S_i$ for some $i \in I$. Therefore $x \in \bigcup_{i \in I} S_i$. Thus $\bigcup_{i \in I} S_i \supseteq \bigcap_{i \in J} S_i$.

[vital dew]

strings

[pale epoch]

yes that works

[tidal tulip]

k sweet

ty for helping me understand what it’s saying

[weary tiger]

Sorry to bother, but does anyone know how to approach this?

[olive wren]

If I had to guess, it’s the derivative of (1+x)^n at x=4 (I’m too tired to check)
But you should do what they said and expand (1+x)^n and take its derivative

[ancient heath]

is R transitive in this statement?

[heavy tinsel]

How do I compute this?

[coral parcel]

Do you know how the digital sum relates to mod 9?

[heavy tinsel]

Like 2139138 = 2000000 + 100000 + 30000 + 9000 + 100 + 30 + 8?

[coral parcel]

That's some of the way. Now notice that, say 30000 = 3·9999 + 3 and 9999 disappears mod 9.

[heavy tinsel]

ohh ok, I see. I think I was trying to do that

tyty

[sly geode]

How do i do this

[coral parcel]

Do what? What are you trying to achieve with that sentence (fragment)?

[sly geode]

ur using that to do a solution expression

[coral parcel]

Hmm, is this a database exercise and you're supposed to construct a relational-algebra expression for the English description of a query?

[sly geode]

i think so because the sentence thats given we have to use that to form an expression

[coral parcel]

First find an expression for the set of pages containing the given words.

Meanwhile, also find a way to construct an expression giving you people who have endorsed a page in a given set.

[sly geode]

ah ok i think i get it now

[heavy tinsel]

Is there another way to calculate 10! mod 17 besides first computing 10! then doing the same trick?

[coral parcel]

Do you know that a·b mod 17 = (a mod 17)·(b mod 17) mod 17?

[heavy tinsel]

This is whats listed in the textbook, nothing else

[coral parcel]

This is just property b, rephrased with a binary mod operation, then.

What it all comes out to is that if you're evaluating some complex expression made of additions and multiplication and you're only interested in the residue modulo 17 at the end, you can choose to replace an intermediate result with its residue mod 17, and that won't affect the final result.

So you can start multiplying together 1·2·3·4·5·6·7·8·9·10, but each time you get an intermediate result that is larger than 17, do a reduction step to keep the numbers you need to multiply nice and small.

[heavy tinsel]

I will try that and see what I get, ty

[obtuse lance]

||I was gonna suggest -1/6! with wilson's theorem and counting backwards for fewer terms might be another way but you're not ready for that yet :P||

[coral parcel]

Indeed, Merosity.

[obtuse lance]

plus inversion might be a pain anyways

[distant bobcat]

I read somewhere that a 3-connected graph has 2 internal faces (excluding the infinite face). I'm not sure where they are getting this from or if its right? We need to remove three vertices for the graph to be disconnected at minimum and obviously the vertex connectivity is bound by n-1 vertices as a max. I tried using Euler' s formula, but I did not get anywhere.

[coral parcel]

What does "face" mean here? Is it tacitly assumed that the graph is embedded in the plane?

[maiden drift]

I have the following question: I have a closed set K in the plane which has the following property. If I take any point x in the plane, then there exists a unique point p_x in K such that |x-p_x| < |x-k| for every element in K where k \neq p_x. Any ideas to show why K then has to be convex?

I tried supposing a contradiciton that K is not convex, so there exists k_1,k_2 in K which has a line segment between them not completely in K. I was thinking maybe there were a point in the line segment not on K which would contradict the uniquessness property above. However, I don't think this is a good idea since it doesnt seem to use the fact that K is closed at all

[coral parcel]

I don't think x can necessarily be a point on the line segment between k1 and k2.

One thing you can use K being closed for is to argue that WLOG you can choose k1 and k2 such that they are in K, but the segment between them is entirely outside K. (I don't know if that actually helps, but it seems to make the situation more concrete in my mind, at least).

Another thing the closedness might possibly help with is prove that p_x as a function of x must be continuous.

[maiden drift]

How does closeness help in proving continuinty?

[coral parcel]

I'm not entirely sure; it is more of a gut feeling at this point.

[obtuse lance]

I think this does it. If it's not convex, we can pick two points a,b on the boundary such that there is a point c between them that isn't in K. We know we can find an interval of points on that line segment because there's an open ball around it not contained in the closed set K. Now we can pick points within this interval that are closer to a than to b and closer to b than to a. But this violates the uniqueness of p_x. So that means our assumption it's not convex is false.

[maiden drift]

by that you mean the point c lies in the segment between a,b?

[obtuse lance]

yeah

[maiden drift]

I don't see the contradiciton

[obtuse lance]

I think I was vague in my wording too, by "open ball around it" I mean around c

"exists a unique point p_x in K"