discrete-math

[naive gulch]

it says I can use a computational aid but I don't even know where to begin with this, do I just use a site?

[quick folio]

i am getting 184 is that right

and can someone explain the residue class of this

i got that x = -53

but since im in mod 1110

due to gcd(1110,335) = 5

ill have 5 solutions mod 1110

how do i get that 5 solns

[faint narwhal]

Add 1110/5 to your solution

[icy jungle]

Can someone help me with this
Suppose there are 6 questions in an exam paper , find the number of
ways in which a student can attempt one or more questions.

<@&286206848099549185>

[lofty cloud]

From

#❓how-to-get-help

Rule 4

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

[last timber]

anyway

consider ways to select first q

then second

etc

[icy jungle]

you mean like 6!

@last timber

[last timber]

yes

6! is exactly the answer

[icy jungle]

is it not ncr

[last timber]

no, because order matters here

6c6 will give 1

[icy jungle]

its weird because its a 4 mark question

[last timber]

1-2-3-4-5-6 combination is indistinguishable from 2-1-3-4-5-6 but these are two different permutations

[untold crystal]

ok so we say that two sets are equal if they have the same elements via the axiom of extensionality

but what makes it so that two sets aren't equal if they don't have the same elements?

[last timber]

?

wym

[untold crystal]

the axiom of extensionality only says that two sets are equal if they have the same elements, right?

not if and only if

so how do we know that two sets are not equal if they have different elements?

[last timber]

ok so look

sets are equal -> have the same element is just contrapositive

suppose sets are equal and have different elements

but this is contradiction

actually this proof prolly is invalid

oh lol

assume sets have different elements

they cannot be equal since this implies the same elements

qed

[untold crystal]

no

because having the same elements implies that the sets are equal

but nowhere does it say that being equal implies that two sets have the same elements afaik

[quick folio]

how will u describe this relation

answers say transitive

i beg to difer

1R0 and 1R3 has an arrow

but 3R0 has no relation

[faint narwhal]

Uh, that's not how transitivity works

[mint bane]

how can i prove that an integer is always either even or odd without induction

[paper parrot]

Maybe by contradiction? Post the question

[mint bane]

i mean the original is "Prove that there are no integers n such that n is both even and odd"

but i figure that's the same as asking to prove an integer is always even or odd

but professor also said not to cite definition of even or odd???? which ig i kinda get since that's really what the proof seems to be aiming at

[untold crystal]

thx

[last timber]

not exactly

unless u use or in the sens either but not both

[last timber]

how you can show that integer can not be odd and even simultaneously not using definitions of odd/even

[mint bane]

question states "Prove that there are no integers n such that n is both even and odd" and this was professors note

[last timber]

but which definition then u use?

[paper parrot]

Show that this leads to a contradiction:
There is an integer n such that n=2k+1 and n=2k for some integer k

[mint bane]

cant cite those definitions

[paper parrot]

you can

[last timber]

if you have another definition of even/odd from 2k, 2k+1 you can cite it

[mint bane]

i was thinking maybe divisibility

[paper parrot]

"You should work directly with the definition of even and odd"

"You should not cite that every integer can be written as 2k or 2k+1 as that is what you're trying to show"

these are not the same

[mint bane]

oh shit i think ive been reading that wrong then

wow that annoys me

ok i'll try that bc i had an idea for that already merci

thanks @paper parrot

[pastel mica]

i just wanna double check

im allowed to do what i did here right

like chnaging the implication to a ^ and ~

[paper parrot]

when you negate the universal statement, you say that there exists g such that the thing to the right of the comma is false

the thing to the right of the comma is a conditional "if (P and Q and R) then (g=gcd(a,b))"

for the conditional to be false, the antecedent is true but the consequent is false

but you negated the R(g) part as well, which would make the antecedent false

@pastel mica

remember you can convert (a->b) into (~a or b)

just remember to bracket it though because it will be "P and Q and (~a or b)"

[pastel mica]

ohhhh okok

ill try that

this is what u mean right

[paper parrot]

no

the antecedent is the left side of the conditional
a is the antecedent in a->b

[pastel mica]

b is the g=gcd(a, b) and everything else on the left is a?

[paper parrot]

yeah

call that whole thing Z(x)

you start with ∀x, Z(x)

the negation of that is ∃x | ~Z(x)

but I'm not sure if you were trying to negate it or just turn it into an existential statement

If you're trying to convert then its just "~∃x | ~Z(x)"

[pastel mica]

yeah lol i was trying to do a contradiction

but first i thought it would be useful to chnage the inner implication to simplify

i guess it made it even more complicated lol

[paper parrot]

Well I have no idea how you're meant to solve it tbh

[pastel mica]

i kinda just did it throught words cause i didnt how to exaclty do it thought a direct proof

[paper parrot]

but if you're trying to prove the universal, you can't do it by contradiction

[pastel mica]

ohhhhhh

[paper parrot]

you can only prove it false by contradiction

[pastel mica]

is it okay if i prove it true using just words? like i understand that g is the gcd of a and b and therefore a must divide g and b must divide g

and since both of those divide g then the R(g) part must also be true

or is that the incorrect way?

[paper parrot]

sorry idk I haven't done number theory

but you probably need to do a formal proof

[pastel mica]

ahh okok

[paper parrot]

just work out how to express your understanding in propositions

[pastel mica]

thank you so much for the help tho! i really appreciate it

[paper parrot]

no worries

[sand cipher]

hello, i need help with my discrete maths hw, i seem to understand when my professor explains it, but not able to do his work, could someone help me out to understand the question that is given, and so that i could also use this to prepare for the test

[paper parrot]

which question

[sand cipher]

3 & 4

i'm kinda ok with 1 & 2

srry late reply

[stray reef]

this problem gives you the definition of square mates

all it asks for is to show that for any natural x there exist natural y and n such that x+y = n^2

which is a lot easier than it sounds

[fiery vine]

x² - x will work, x>1

[stray reef]

yeah like that

you can also take (x+10)^2 - x

[paper parrot]

2^2 - 2 = 2 but y > x

so you need to show a special case for x=2

I think

[fiery vine]

Yeah but we could use (x+10)² - x too instead

[paper parrot]

sure

[stray reef]

you could also take (x+583)^2 - x

[fiery vine]

(x + a)^(2n) - x

The 4th question tho

I think a = 17

13x + 7 = m•a, m an integer
And 39x + 4 = (13x + 7)•3 - 17 = n•a
And hence 3m•a - 17 = n•a
Or a = 17/(3•m - n)

But a is an integer and 17 is a prime so a = 1 or 17

But a > 1, hence a = 17

@sand cipher here goes the solution

[sand cipher]

thanks you guys, i will review this to understand this chapter more.

[weary tiger]

Heyy i just need help with a discrete maths questions. I don't get how exactly you can write a polynomial function in a set.

The question is C btw

[faint narwhal]

Did you do part a or b? What did you do for those?

[fallow pawn]

can someone quickly explain the difference between these two symbols

[weary tiger]

I've done part A and part B already

[weary tiger]

The first on means "subset "
The second one means "is an element of"

[faint narwhal]

What did you write for parts a or b?

[weary tiger]

I wrote

This was B

[faint narwhal]

Okay, and so for (c), you want that x is a real number

and what does it mean to be a root of the polynomial?

[weary tiger]

Is that when you input a value and the polynomial becomes 0

[faint narwhal]

Yes

It's the values of x such that 1 + 2x + 3x^2 + 4x^3 + 5x^4 = 0

[weary tiger]

Ohhh i see

Would you have to arrange the set in a similar format to summation notation ?

[hidden yacht]

hi, ive drawn an activity-on-node network and a question is how can each activity be reduced by a cost

[faint narwhal]

You don't need to use summation notation

you can just write 1 + 2x + 3x^2 + 4x^3 + 5x^4 = 0

[weary tiger]

Fr?

That's the answer

[faint narwhal]

yes

[quaint star]

Context?

[weary tiger]

guys i need help with

[faint narwhal]

@deft current Given a partition of a set of size n into k nonempty parts, can you see how to get a permutation with k cycles?

I mean, yes, but I don't see how this is relevant for figuring out your inequality

Your inequality is greater than or equal to

Again, you don't need to see that each partition can map to multiple permutations to see that >=

And in some cases, they are equal, like when k = n

Not sure what you mean by that

I'm just saying that the first part is enough to see that >=

you don't need to say anything about partitions potentially mapping to corresponding to multiple permutations

[weary tiger]

I need help for this please IM SO CONFUSED

[weary tiger]

Is the answer to these true, false, false, false?

[coral raven]

is RHS of f even a set

[weary tiger]

if I have a equivalent b mod(m) is a.c equivalent b.c mod(m)?

[obtuse lance]

@weary tiger don't cross post please

[weary tiger]

@obtuse lance sorry I didn't know where to post

there is too many channels

[obtuse lance]

read #❓how-to-get-help

[terse sable]

need help proving that (n+1)*c(n,k)^2/(k+1) is always a positive integer, where 0<=k<=n

[faint narwhal]

yes this is true

[weary tiger]

why tho ? we have proven that if a congruent b mod m then ac congruent bc mod m

[faint narwhal]

wait, what do you mean by . here

sorry I was assuming you mean multiplication

[weary tiger]

yes multiplication

[faint narwhal]

wait, so you've proven it already?

so what exactly are you asking then

[weary tiger]

if we have a congruent b (mod m) then ac congruent bc (mod m )?

[faint narwhal]

yes

you've said you've proven this already

So just look at the proof?

[weary tiger]

ok then why if we have ac congruent bc (mod m) it doesn't mean that a congruent b (mod m)

[faint narwhal]

why do you expect things to go both ways?

it's true that if a = b, then ac = bc in normal numbers

but its not always true that ac = bc means that a = b in normal numbers either

since you can have 1 * 0 = 2 * 0 for example

[weary tiger]

so only for 0 it doesn't work?

does it work for all other numbers?

[faint narwhal]

For real numbers or integers that's true yes

but for modulus it can still be wrong for non-zero

[weary tiger]

what?

you just said yes

[weary tiger]

?

[faint narwhal]

I'm talking about the usual equality of real numbers or integers

where 1 = 1 but 1 is not equal to 5

whereas in mod, you have things like 1 = 5 (mod 4)

[weary tiger]

I don't understand

can you answer this #discrete-math message

[faint narwhal]

For real numbers or integers that's true yes

[weary tiger]

yes

for integers

[faint narwhal]

But not when you're talking about equality mod something

[weary tiger]

what does that mean

[faint narwhal]

For example, you have that 1 * 2 = 3 * 2 (mod 4)

but its not true that 1 = 3 (mod 4)

[weary tiger]

yes

wait

first of all thats equality not congruence?

or is it the same?

[faint narwhal]

they're synonyms

[weary tiger]

2 mod 4= 2

and 6 mod 4= 2

[faint narwhal]

yes so they're equal

[weary tiger]

yes

so 1 mod 4 = 1

and 3 mod 4 =3

[faint narwhal]

which are not equal

[weary tiger]

not equal

so

[weary tiger]

why you say this then

[faint narwhal]

"I'm talking about the usual equality of real numbers or integers"

[weary tiger]

what?

I really dont understand what you mean

[faint narwhal]

I'm not talking about being equal mod something

I'm talking about the usual 1 = 1

[weary tiger]

1=1? so what\

?

I am so confused

[faint narwhal]

For integers a,b,c, it is true that if c is nonzero, then ac = bc implies that a = b

[weary tiger]

ah

but for congruence it is not true

[faint narwhal]

yes

[weary tiger]

wow finally

[weary tiger]

so here you could have said for equalities it is true

[faint narwhal]

that doesn't really make sense

being equal mod n is still an equality

[weary tiger]

actually it is not an equality

it has a different sign

[faint narwhal]

https://en.wikipedia.org/wiki/Equivalence_relation#Equivalence_relations

The sign doesn't matter

[weary tiger]

I mean nvm now I understand it

whatever

thanks

[faint narwhal]

Many people don't write triple equals anyways

[knotty badge]

someone please

[weary tiger]

Ok

start with let x belongs to A

[knotty badge]

ok

[weary tiger]

sorry?

[knotty badge]

?

[weary tiger]

why you said ok?

[knotty badge]

i thought you were explaining it to be, I was following

[weary tiger]

oh lol you just changed username and profile picture?

[knotty badge]

oh haha yes

sorry lmao

[weary tiger]

anyways

let x belong to A

and A subset of B so x belongs to B

and B subset of C so x belongs to C

so x belongs to A and B and C

hence x belongs to A union B union C

[knotty badge]

would it matter if its a proper subset

[weary tiger]

and if x belongs to A union B union C it is sufficient that x belongs to 1 set

[knotty badge]

but could it be any set as opposed to just set C

[weary tiger]

but A and B are subsets of C

[knotty badge]

ohh i think i understand

[weary tiger]

so we proved that x belongs to A union B union C

and that x belongs to C

hence A union B union C =C

[knotty badge]

gracias

[terse sable]

is c(n+1,k)*c(n,k)/(k+1) always a positive integer?

[mint bane]

what is the loop invariant here

is it just that $j \leq n$

[vital dew]

nitezba

[mint bane]

oh wow that's weird lol

[proven shard]

Loop invariant

[sand cipher]

need help with those 3, for the last one there is a mistake, t is also an integer

[obtuse lance]

hint for 5: factor n^2-9

then because it's a prime number that tells you something about what these factors can be

[sleek turtle]

can i get some help

[sand cipher]

sorry for late reply, if i do factor it would be (n-3)(n+3)
but how would i prove that p=7 from that since it would be
p=(n-3)(n+3)

[obtuse lance]

@sand cipher

tell me what the factors of a prime number are

[sand cipher]

well since it's prime wont it be 1 and themselves?

[obtuse lance]

aha good

so those 2 factors can only be 1 or p

[sand cipher]

yep

[sand cipher]

but after know what the factors are what to do?

[errant bear]

you know that either (n-3) or (n+3) = p, and the other equals 1

[sand cipher]

alrite

got it

thx @obtuse lance @errant bear

[minor lake]

Ok so,

I must create a secret code that has 6 letters and 7 digits and I'm being asked to find how many secret codes I can dial if everything is mixed together

I already answered two questions

1. If letters and digits are together, how many codes. I found $2\cdot(26^6\cdot 10^7)$

[vital dew]

Chad

[minor lake]

2. If digits only should be together, how many codes. I found $7\cdot(26^6\cdot10^7)$

[vital dew]

Chad

[minor lake]

But I can't find the last one where letters and digits can be mixed together

I guess that this situation would include the two previous cases but I can't really think about the other cases, I tried but it didn't work

[last timber]

then u just do codes over 36-letter alph

[minor lake]

aleph being (26^6 * 10^7) ?

[last timber]

i mean if you allow letter and digits to be mixed you not even care whether it is digit or letter

[minor lake]

hmm shouldn't I since there are only 6 letters and 7 digits

it*

hold on I need to think

yeah I'm not sure to understand what you mean by 36-letter aleph

[last timber]

oh wait lol nvm i misread

ok so then i guess it would be just some summation

[minor lake]

lol

so I just read one message that my teacher sent to everyone

so apparently the 8c is "hard" and we need to use combinatoins

ill start from there kek

[weary tiger]

So I managed to do the second inequality, but I have no clue on how you get the first 🤔

[last timber]

i prolly would do induction

[weary tiger]

How do you induct both n and k at the same time ?

[last timber]

fix one variable and do induction on another

[weary tiger]

So first show by fixing k, then show by fixing n ?

[last timber]

nah

i mean fix e.g k

and do induction on n

since k is arbitrary you basically will have what you want

[vapid light]

You can just directly show this

The way you got the second inequality is the same way you get the first inequality

[weary tiger]

Indeed

[weary tiger]

Came here to say FUCK DISCRETE thank you.

[remote mulch]

lmao

continues working on that discrete maths stuff

[hybrid tendon]

Is there any way to find the possible number of topological orderings of a tree. (that is a tree like looking undirected graph

don't tell me I need to go to #combinatorial-structures

[weary tiger]

n! > 5^n for every integer n>= ?

What would be the base case(first integer that makes this true?

[obtuse lance]

I say just start at n=5 and keep trying numbers until you find it

I pick 5 because 5! is clearly smaller than 5^5

and if that's not clear to you, you should try to think about it and see if you can figure out why it should be obvious to you

[weary tiger]

@obtuse lance are you able to tell me if it is negative ?

I know it’s nonnegative lol

Nonpositive I meant

[obtuse lance]

'it'?

[weary tiger]

I don’t see an integer where this is true

[obtuse lance]

just use a calculator and keep trying

[weary tiger]

I did and the greater n went up, the less true the statement got

And 0-5 is false

[obtuse lance]

yeah, explain why

without computing it how do you know

[weary tiger]

I’m not trying to prove that the statement is false though, I’m trying to find an integer that makes the statement true

[obtuse lance]

if you don't understand the simple case you have no hope

you won't see why it is eventually true that n! > 5^n

5! = 1*2*3*4*5 while 5^5 = 5*5*5*5*5

5^n is only multiplying by 5s each time you increase n

5! is multiplying by a larger number each time you increase n

so while it starts out smaller, it starts to multiply larger and larger numbers

the 100th term of 100! is 100 while the 100th term of 5^100 is just 5

you see what I mean?

[weary tiger]

Yes I see thank you

I got n=12 for the lowest possible integer to make that statement true

[weary tiger]

if a congruent b (mod m) then a^c congruent b^c (mod m)?

[errant bear]

yes

[weary tiger]

ok thanks

ok in reverse if a^c congruent b^c (mod m) does that mean that a congruent b mod(m)?

@errant bear

[quick folio]

does anyone know how to do b without venn diagrams

[balmy hornet]

Does this proof by contradiction make sense? If you can’t read my hand writing please tell me 😂

[mint bane]

can someone explain the difference between a set and proper subset colloquially.

[mint bane]

the negation of p implies q is p and not q

[balmy hornet]

can you reiterate that?

[mint bane]

$\lnot (p \implies q) \equiv \lnot p \land r$

[vital dew]

nitezba

[last timber]

[sly fractal]

why does (¬q ∨ p) = p

[paper parrot]

Because if p is a proposition, then it must either be true or false

consider the truth table for (¬q ∨ p)

oh sorry I misread

[sly fractal]

oh i didnt update this fully for full context

why would ((!q or p) or (q implies r)) equal to (p or (q implies r)) but im confused how it turned into p for '(p or (q implies r))'

[paper parrot]

alright give me a sec to think about it

you can change (q implies r) into (~q or r)

then you just have a long statement of ors

~q or p or ~q or r

get rid of the first ~q because it appears twice

p or (~q or r)

p or (q implies r)

@sly fractal that's it

[sly fractal]

Do you know the specific laws or rules used for this so I can search it up and try it myself

[paper parrot]

the equivalence between a conditional (a -> b) and (~a or b) isn't a law, but you can see it by a truth table

but after that, you have
(~q or p) or (~q or r)
Then you use the associative laws to rearrange this into something like
p or ((~q or ~q) or r)

[sly fractal]

oh ok thank you!

[paper parrot]

Then idempotent law to get rid of one of the ~q's
p or (~q or r)

then change the (~q or r) back into a conditional by logical equivalence

and that's it

you got it

[short steeple]

i'm stuck on this hw problem:

Is it true that, for each positive integer $n$, there is a simple connected $4$-regular planar graph that has at least $n$ vertices? Either construct a sequence of such graphs where the number of vertices increases without bound, or prove that there is an upper bound on the number of vertices in such graphs.

[vital dew]

Snodlop

[short steeple]

from what i can tell, it should be possible to construct a graph with an arbitrarily high n vertices, but a) i wouldn't know how to conceptually draw such a graph, and b) i feel like there's a cleaner way i can answer this question than just a brute-force example

the "at least n vertices" is tripping me up i think, it makes me want to skip straight to a sufficiently large n

if anyone can help please @ me i am about to fall asleep but i will read it in the morning if i don't figure it out right now

[plush nymph]

,help

[vital dew]

A brief description and guide on how to use me was sent to your DMs!
Please use ,list to see a list of all my commands, and ,help cmd to get detailed help on a command!

[icy ridge]

Ok

[quick folio]

can someone tell me why this is wrong

1R{1,3,4}

isnt connected

i cant see

[weary tiger]

Is that from university

[mystic moss]

Hm the construction here is pretty nice actually, I wouldn’t call it “brute force”

It’s kind of like how you would construct a 3-regular graph with n vertices for any n (without the planar restriction)

And then once you try and change the 3-regularity to 4-regularity you immediately solve the planar issue (although it won’t work for all n anymore, it will work for arbitrarily high values)

[quaint star]

Is that just the union?

The union just equals S, so yes the cardinality is the cardinality of S

[stray reef]

$S \cup S \cup S \cup \dots \cup S = S$

[vital dew]

Ann

[stray reef]

so yes

[quaint star]

^

[balmy hornet]

Would it be due to the idea that since it is being duplicated that it would be just counted as one?

[quaint star]

The union is the set of all elements that belong to any of the sets that you are taking the union of, but there is only one set, viz S. So it's just the set of elements of S

[balmy hornet]

Thank you Lunasong and Ann!

[weary tiger]

when the importer is sus

[balmy hornet]

Since cos is offset by 1/2 would it be |S| = 2m + 1/2?

[last timber]

@balmy hornet is m assumed to be integer?

well if it is integer, then cos is zero exactly at odd multiples of 1/2

that is at pi/2 and 3pi/2

in each full circle you have 2 such multiples

@balmy hornet now look, say i have cos(2x)

it will be zero at pi/4, 3pi/4, 5pi/4 and 7pi/4

because all of these are in [0, 2pi] and multiplied by two give exactly zeroes of cox

now, can you consider zeroes of cos3x

and think what i am trying to point out?

no

no no no

how you can have noninteger amount of roots

i mean if you want consider how m affects the range of mx if x varies from 0 to 2pi

yes

for x you have two zeroes

for 2x you have four zeroes

[balmy hornet]

3x then 6 zeros

[last timber]

yes

look, cos (mx) will have zeroes whenever mx = is odd multiple of pi/2

actually better way to explay

[last timber]

in each complete circle from 0 to 2pi cos x has two zeroes

multiplying x by m makes mx in cos traverse m full circles

thus you have m*2 roots

[balmy hornet]

|S| = 2m ( with no addition or anything?)

[last timber]

yes

[balmy hornet]

Ah ok ok i get it now. thank you!

[last timber]

yw

[simple nova]

Is my assumption correct?

I assume its false since it is the empty set

But I mean it might be true since what else could you possibly get

[weary tiger]

can someone tell me how the intuition for the first line equivlency using laws of algebraic proposition?

[short steeple]

$\emptyset \setminus {\emptyset} = {a\in\emptyset | a\notin{\emptyset}}=\emptyset$

[vital dew]

Snodlop

[short steeple]

since there is no a that is in the empty set

[simple nova]

So it is True that that it gives back just the empty set

[short steeple]

i would believe so

[short steeple]

you lost me, sorry

[simple nova]

@short steeple https://i.imgur.com/dOhsDR7.png

Is my answer correct? I believe I did the work but this one is confusing for me

[short steeple]

it looks correct yeah, i trust your work

[tepid fulcrum]

What are the GCD and LCM of this pair of integers?
2 * 3 * 5 * 7 * 11 * 13 and 2^11 * 3^9 * 11 * 17^14
how would i do this problem?

[errant bear]

since you have the prime factorization you can just take the product of the common factors

[short steeple]

wait i just double checked, i don't think it is correct

you can use the distribution rule to simplify it, ((p or q)and (p or r)) is congruent to (p or (q and r))

[lucid ingot]

"i trust your work" 👀

[simple nova]

@short steeple Yeah I think it might be B right?

[tepid fulcrum]

so will it be 2* 3 *11?

[short steeple]

that's what i got

[lucid ingot]

gcd*

[mystic moss]

Ah my bad ;-; what constructions have you tried so far for the problem?

[quaint star]

The GCD takes the smallest exponent for each prime number (if you take a smaller exponent it won't be the greatest divisor, and if you take larger exponents it's not a divisor ), and the LCM takes the largest exponent of each prime factor (smaller exponents means it's not a multiple, larger exponents means it's not the lowest multiple)

[short steeple]

i solved the inequality for the minimum number of vertices, so i know it's at least 6 vertices for 4-regular planar graphs

[simple nova]

@short steeple Yeah I just re-did the work and got htat

Thanks for help!

[mystic moss]

Inequality?

[short steeple]

i can construct 3-regular graphs semi easily, they can even maintain planarity, i just wouldn't know how to add the extra vertices and edges in to make it 4-regular

[tepid fulcrum]

so would i do 11 * 17?

[mystic moss]

Okay so what’s the construction

[short steeple]

yeah, planar graphs have 3n-6 edges, counting vertices gives 2n edges

*at most 3n-6

[quaint star]

@tepid fulcrum no, but this channel is very busy now. Try a questions channel.

[tepid fulcrum]

i just posted it in #help-6

[short steeple]

this pattern can go indefinitely, you make it 3-regular by constructing a cycle that touches every leaf

[mystic moss]

Woah that’s very cool actually

[short steeple]

please excuse my awesome photoshop skills

[mystic moss]

I was thinking of a different construction, so that’s why my suggestions didn’t help 💀

[short steeple]

it is kinda cool but unfortunately i don't think it's very helpful

ah damn

[mystic moss]

Yeah so here the thing grows by adding leaves right

[short steeple]

yeah

[mystic moss]

But say you start with just an even length cycle

How can you change it from 2 regular to 3 regular

[short steeple]

so for 2n cycles

you can add an edge from i to i+n

[mystic moss]

Mhm exactly

So picture this for like a hexagon

Now if you wanted to change the 3 regularity to a 4 regularity for that hexagon

How would you do it

[short steeple]

i guess if you went by the same methodology, for 3n cycles, you could add edges between i and i+n, and i and i+2n

oh and then you can add vertices in the middle to fix planarity

for the intersections

[mystic moss]

Maybe that works? I don’t know if each intersection in the middle will have degree 4

Maybe you can make it that way

[short steeple]

you can probably shimmy the edges around

i'll need to draw it fancy in latex

[mystic moss]

Yeah I guess

But that’s too much work

[short steeple]

i'm guessing you know an easier way to go around the intersections?

[mystic moss]

Yeah so draw this idea on the hexagon

For like all vertices

[short steeple]

yeah i got it

[mystic moss]

What does it look like

[short steeple]

i curved the edges slightly to avoid the 6-way intersection but without that it'd look like a hexagon where each opposite vertex has an edge connecting them

through the middle

[mystic moss]

Hm wait is this for the 3-regularity construction or the one you just mentioned above

[short steeple]

oh wait i was wrong yeah

[mystic moss]

Awesome

Now make it planar

[short steeple]

wait i can do that

but like

what about n=9

[mystic moss]

Right so we won’t do exactly this 3n construction

It’ll still be on even cycles actually

[short steeple]

so 6n?

[mystic moss]

Uh 2n should work I think

[short steeple]

oh i get it

oh oh oh oh oh

that's so sick

oh wow

it just clicked

thank you

[mystic moss]

Np have fun

[lucid ingot]

ignore my prev replies

it is alright

you just negate the first term

[balmy hornet]

just before ∃?

[lucid ingot]

for all f, [there exists p. (not) loves(p,f)]

[lucid ingot]

ignore this; im kinda braindead rn. my prev msg is its negation

[balmy hornet]

wait

[lucid ingot]

there exists p*

(missed an f, but yeah :p)

[balmy hornet]

ooo wait that f ok ok i know what you mean

[balmy hornet]

Thank you!

[mint bane]

how do i approach this

[candid pier]

I see you drew a venn diagram. Did that help shine light on part a.) ?

Consider two sets that do not have an intersection. E.g., consider the sets B = {1, 2}, C = {3, 4} and A = {2, 3}.

[mint bane]

ok ik a is false

bc if i have a set that's in a and b then it'll have some elements that are in a and some in b so it wont be a subset of a or b

[candid pier]

Yep -- so long as you have a counterexample that's enough to constitute a proof.

[mint bane]

ok then b i feel is true i think

but proving is different than disproving lol

[candid pier]

Proving the contrapositive seems straightforward to me.

[mint bane]

[candid pier]

Assume A U B ≠ B

[mint bane]

contrapositive is A U B ≠ B implies A is not a subset of B

is that really more straightforward?

[candid pier]

Maybe -- it is when I wrote it down lol.

In words, if A U B ≠ B, then there is some element of the union that is not in B. That means the element must exist in A. If it exists in A, then A cannot be a subset of B.

I'm sure you can write a similar proof using the conditional.

[mint bane]

so in other words when we assume the contrapositive we're assuming A is bigger than B

but from that alone shouldnt it follow that A cannot be a subset of B

[candid pier]

Not necessarily bigger than B -- A could very well have no elements in common with B.

[mint bane]

ohhh gotcha, it could just be {3} U {4}

[candid pier]

Yes.

How did you arrive at 2(a + b + c) + 1?

Oh ok -- did you mean 2(a + b + c + 1) + 1 ?

[balmy hornet]

Whoops yes lol i didnt realize i was missing a 1

[candid pier]

Hm.. looks correct. What is the question?

Did you want to represent it in some other way?

[balmy hornet]

how can you represent the sum of the above three odd numbers?

can i just imput any odd number? I dont know if thats what the question is asking

[candid pier]

I think you already answered the question haha

2(a + b + c + 1) + 1 is precisely how you would represent the sum of 3 odd numbers. The odd numbers being: 2a + 1, 2b + 1 and 2c + 1

[balmy hornet]

OHHHH ok ok. wow i didnt realize i answered both doing it like that. So if i wanted to simply prove it to show it would come out odd i would just input odd numbers?

[candid pier]

if you wanted to prove that the sum of 3 odd numbers is also odd?

Yes, you did that. You showed that the sum of 3 odd numbers is 2(a + b + c + 1) + 1, which is of the form 2m + 1.

[balmy hornet]

ah ok ok thank you!

[candid pier]

Sure thing

[mint bane]

the word "some" would be an existential quantifier right?

like "some statements are logical"

[candid pier]

yeah

[mint bane]

i know this is rules of inference but i never know how to start these

[candid pier]

If you take the first and third premises, what can you conclude about q?

Then use information with premise 2 to conclude the result.

i gotta head out for now.

[mint bane]

aw rip thanks anyways

youve been a tremendous help :)

[tawny wigeon]

hello

so I have a Digital Control systems course

which I guess applies as applied math

any good book recommendations?

[quaint star]

@mint bane why is b F?

[mint bane]

i thought it might be some jankiness trying to trip me up with the difference of something being a subset and something being an element of a set

[haughty garden]

{2, 3} being a subset of A would also imply that {2} is a subset of A which would imply that 2 is an element of A

[mint bane]

ah

so i was overthinking it

cool lol

[haughty garden]

If it was something like {2} is an element of A, then you’d be correct

[mint bane]

also w the statement "no student can both pass and fail the same test" how can i write that in propositional logic

im gonna make p(x,y): x fails test y

would it just be $\lnot \exists x (P(x,y))$

[vital dew]

nitezba

[mint bane]

what's tripping me up is the "same test" part

[candid pier]

can you just add another proposition? namely, NOT P ?

[opaque pilot]

Sorry what is your question ?

[untold cargo]

Can someone help me out with bijections? I am getting the concept sort of but am having trouble executing it. The sample problem im working on is a function with the interval (2,5) to (4,9). I think I drew the function correctly and showed 1:1 but not to sure how to start onto

[candid pier]

you have to demonstrate that for every element in (4, 9), there exists some element in (2, 5) that maps to it. do you have an explicit function you are working with?

[untold cargo]

yes, my function is 5/3x + 2/3

[candid pier]

consider the inverse of that function

[untold cargo]

ok, I think this is where im stuck. I have my inverse y = (3x-2)/5, but how do I plug the numbers back in to show onto

[candid pier]

if you keep the variables consistent you'll see that you've already solved the problem.
y = 5/3x + 2/3
x = 1/5(3y - 2)

this second statement maps elements from (4, 9) [that is, y] to (2, 4)

the fact that the function is invertible is enough to tell you that it is onto and one to one. but to make it clear, if you choose any y, you can find an x (using the inverse) such that f(x) = y

you know this function is "onto" because for every element, y, of (4, 9), you can find an x, of (2, 5) that maps to it. that's what the inverse does.

[untold cargo]

Thanks so much for your help, I think I solved it correctly, I really just didnt understand that an element y of (4,9) is just anything in the interval, that was the reallllly big picture part that I was missing

[candid pier]

oh yeah sorry

[untold cargo]

I ended up saying that 4 < y < 9 and then plugging all of those into the inverse, but you said I would be right if I just plugged in any of those?

all of those meaning 4, y, and 9

that gave me 2 < x < 5, which lines up great

Just need to find the inverse, put one element from (4,9) and show that it gives an element in (2,5)

[candid pier]

yes that's pretty much it

your function takes care of the end points so i just focused on the concept of onto

[tacit wharf]

can someone help me understand what this means

the left side means in a group of n+1 we pick k right?

[faint narwhal]

It's the number of ways to pick k things from a group of n+1

so the left hand side is just a number

[tacit wharf]

oh i think i see

normally in probability the + means or and the * means and

what would the divide on the right mean?

does it have any special meaning?

wait logically wouldnt it mean something is repeated?

[faint narwhal]

in what context

[tacit wharf]

like

normally when you want to find the probability of thing 1 happening and thing 2 happening you multiply the probability of the two right?

and vice versa for + and or

how does the divide play into effect?

like im trying to wrap my head around what the equation really means

am i thinking about it wrong?

[tight lotus]

so this one makes a lot more sense if you already have an intuitive grasp of the explicit formula for n choose k

n!  /  (n - k)!  /  k!

division is like removing over-counting,

if for everything you should have counted once, you counted b times, you should divide by b

[candid pier]

in this case, the left hand side is counting how many ways you can choose k objects from n+1. the right hand side counts the same thing in a different way. pick the first object. you have n+1 ways of doing that. then determine how many ways you can pick the remaining objects. that's n choose k-1. this method overcounts by k because for every fixed object (the one you picked first), you can swap it with one of the k-1 objects.

[tacit wharf]

thats actually really clear

[candid pier]

took me a while to reason it out. i drew some pictures

i think some people call them 'story proofs', where you simply explain what is going on in words. sometimes it is clearer that way. sometimes.. not so much

[tacit wharf]

it helps me with simpler equations like this one lol

tysm

[last timber]

igues induction would help

and pigeonhole

[naive gulch]

How are they getting from (a^4 ) ^3 to the rest of the problem?

[prime parcel]

If that's confusing u then you can just do it like a⁴= 1(mod5)
So a⁸=a⁴(mod5)=1(mod5)
So a¹²=a⁴×a⁸=a⁴(mod5)=1(mod5)

[naive gulch]

I'm a little lost on the second line

a^8 is congruent to a^4 mod 5?

[crisp ridge]

hi guys, how do I show that ¬p→(q→r) and q→(p∨r) are logically equivalent without using truth table?

[stray reef]

are you allowed to use $(A \to B) \equiv (\neg A \lor B)$?

[crisp ridge]

no

wait honestly theres no instruction about that, it just says proved that its logically equivalent

[vital dew]

Ann

[stray reef]

wrong symbol fuck

anyway uh

if theres no instruction against it then i'd just apply that over and over

[crisp ridge]

yah I guess so, but how do I do that?

[stray reef]

$\neg p \to (q \to r) \ \equiv \neg\neg p \lor (q \to r) \ \equiv p \lor \neg q \lor r \ \equiv \neg q \lor (p \lor r) \ \equiv q \to (p \lor r)$

[quaint star]

Last line should be an implication

[crisp ridge]

O_O srsly this is it? ._.

[stray reef]

fuck

[vital dew]

Ann

[stray reef]

there we go

but yeah this is it

[crisp ridge]

can u pls explain like the process on getting it?

[stray reef]

dunno

i just expanded the implications using that formula i posted above

[quaint star]

You don't really have a lot of rules for manipulating inequalities, so it's best to rewrite inequalities in that form, do the manipulation, and then go back to an inequality if you need to

[crisp ridge]

oh yep i guess its simpler this way thanks

[weary tiger]

Did you do (a) and (b)

[grave grotto]

I tried to solve a)

[weary tiger]

Well for (a) you basically are trying to show that at most 5 variables of 6 will be set to 1

well what's the best case for setting 5 variables

[grave grotto]

Well, if we set any 5 variables we get a value > than 40.

[weary tiger]

yeah so there you go

same goes for b

[grave grotto]

So for a) I need to prove that they are primal infeasible?

[weary tiger]

what is primal infeasible

I think you just need to make an argument that picking all 6 variables will never work, so clearly you need to pick at most 5

[grave grotto]

Wouldn't the given constraint be smaller rather than smaller or equal than 5 if that was the case, if we pick any five then the constraint is wrong after all.

[weary tiger]

that's true but you're proving a weaker condition

like that's what (b) asks basically

[grave grotto]

Well b) starts from j=2, so X1 is 0. Which might be a tad different

[grave grotto]

@weary tiger I did some contradictions to show that if the given constraint was false then there would be a contradiction

What about question c) , any ideas on how to proceed there?

[vernal linden]

Hey guys,
I have an exercise to prove that
|P(A)| = |P(A n B)| * |P(A\B)|
A and B are some finite sets.

Can anyone point me to the right direction?

I was thinking at first maybe to prove it by contradiction.

But I'm not sure how to proceed

[limpid fern]

@vernal linden bro

are you still there

does P mean power set

so you're trying to prove the cardinality of the power set of A is equal to the cardinality of the power set of A intersected with B multiplies with the cardinality of the power set of A difference B

right?

so a set theory proof

well remember what a power set it

a power set is the set of the subsets of another set

the cardinality just spits out the number of elements in that set

so I think you could just do a direct proof

I think that would be quicker

anyways, hopefully that helped

[vernal linden]

Yes

[limpid fern]

oh good

I guess you could try to experiment with 2 sets and see what happens

[vernal linden]

Yeah I created my own 2 sets just with made up values to see how it's working, I do understand it but I'm not sure how can I prove it

[limpid fern]

hmm

[vernal linden]

Direct proof also sounds better

[limpid fern]

I'll try to think of a way

give me a moment

[vernal linden]

Okay, just don't give me the solution, just a hint :)

[limpid fern]

ok

OOOOOH

I thought of something I remembered

the cardinality of a power set of always gonna be $2^A$

[vital dew]

Modular Code

[limpid fern]

2 raised to the power of the number of elements in the normal set

so $2^{|A|}$

right?

[vital dew]

Modular Code

[limpid fern]

2 raised to the power of the cardinality of the original set

I just found this random picture from flammable maths

but it explains it

card means cardinality

the bars mean cardinality as well

hopefully that helps

[vernal linden]

Hm... It give me an idea 💡

Thank you! I'll try to do it

[limpid fern]

👍

[paper parrot]

looks good

[weary tiger]

@paper parrot does the second one also look good because theres a cross through the proper subset symbol?

[paper parrot]

I've never seen that symbol before, but I'm assuming its crossing out the 'equals' part of the subset symbol so that it means proper subset

so it must be false

[weary tiger]

okay thanks

[scenic folio]

can someone help me with this? "Prove that 31|(6a+27b+7c)"

[errant bear]

it doesnt as is

[naive saffron]

a=-1, b=0, c=1, then 6a+27b+7c=1 but 31 does not divide 1

[obtuse lance]

a=b=c=1 you get 40 which isn't divisible by 31 either

[errant bear]

it does if u try hard enough 🥺🥺

[obtuse lance]

9/31 is an integer when I believe 😌

[last timber]

just define integers to be rationals you know

[naive saffron]

just work over a field and everything is fine

you see fields are nice since they are euclidean domains and thus they are unique factorization domains

[last timber]

just work in degen ring

[crisp ridge]

does anyone know how to show (pVq) V [(¬p) ^ (¬q)] is a contradiction using truth table (truth matrix)?

this is what ive got so far
P Q
T T
T T
T F
T F
F T

[paper parrot]

@crisp ridge You don't need repeat rows for the inputs, you have two TT rows and two TF rows, and no FF row. The number of rows should be 2^n, where n is the number of statement variables.
Also, as you wrote it, that isn't a contradiction

p q (p∨q) (~p∧~q) (p∨q)∨(~p∧~q)
T T T F T
T F T F T
F T T F T
F F F T T

It's a tautology. If you meant to put an 'and' between the first and second half, then the last column will indeed be false for every row, making it a contradiction.

[crisp ridge]

oh my bad I took the wrong truth table 😂 yah theres no truth table provided it just say show it is a contradiction

thanks so much

[paper parrot]

nw

[errant bear]

did u just manually type/format that..

[paper parrot]

yes

[glacial veldt]

hello

[glacial veldt]

i was wondering if someone can help me with this?

thanks

how would one determine if its true or false

[candid pier]

a true antecedent cannot imply a false consequent. everything else is fair game and considered 'true'

[glacial veldt]

ok i see thank you so much!

@candid pier

I was wondering if D would be false

[candid pier]

since monkeys cant fly (not yet anyway), then the implication is true

[glacial veldt]

ahhhh i see

ok cool got it!

[cloud terrace]

Would anyone be able to explain what either a minimal element is or what it means to cover a set? I feel like c,d should be minimal and cover intersection K which is just c

[opaque chasm]

Provide an example of a probability space and events where the events are pairwise independent, but not mutually independent. Try to make your sample space as small as possible. Show that your example is correct.

anyone got any ideas for me?

[glacial veldt]

hey guys 🙂

need help with this one how would i know how many rows

[minor lake]

consider a)

you'd need one row for p, another for not p and a last one for p implies not p

so 3

[glacial veldt]

i saw online that there is a formula

[glacial veldt]

i got this one 🙂 thx anyways

tbh i am struggling with this one

[terse ginkgo]

In a hand of 5 cards (out of an ordinary deck of 52 cards), find the probability that there are 2 spades, 2 clubs, and 1 heart.
In a hand of 5 cards (out of an ordinary deck of 52 cards), find the probability that there is a suit with 2 cards, another suit with 2 cards, and yet another suit with 1 card.
i know i have to break it up
but im kinda lost

discrete probability

so idk if i post here

[terse ginkgo]

did it

[deft current]

@glacial veldt start with columns for the atomic propositions and then work from there e.g. for a) p is atomic so start with that, then not p, then the and. So you work inward to outward if that makes some sense

[paper parrot]

the number of rows is determined by the number of statement variables. You need a row for each possible combination of statement variables, so more statement variables means more rows. The number of rows is 2^n where n is the number of statement variables.

[paper parrot]

The trick is to break them down into simpler units, like the left and right side of a conditional. Assign truth values to those, and then just use those to work out the truth value for the whole conditional

If it helps, you can even put in a column for the negations of the statement variables (~,p,~q...) so its easier to work out the other stuff

[autumn pebble]

If I am not mistaken this relation on the set is just reflexive right?

Oh is it also vacuously transitive?

Vacuously reflexive*

And not transitive at all

Vacuously symmetric* lmao

[zinc moss]

how would one prove this is a tautology

[wicked frost]

p -> q is equivalent to ~p v q

@zinc moss

[minor lake]

I'm being asked to find how many hands contain exactly one six and 3 diamonds

I split the problems into two cases, one where the 6 that I pick is a diamond and the other where it's not a diamond

6 is a diamond: 1 (the picked diamond) * 12C2 (the two other diamonds left) * 39C2 (the two cards left to complete an hand)
6 is not a diamond: 3 (types of card except diamond) * 13C3 (the 3 diamonds to pick) * 38C1 (the one card left to complete the hand)

i just noticed i forgot to remove some 6s

[wicked frost]

how many cards are in a hand?

sorry if this is standard but idk

[minor lake]

how many fingers do you have on your hand

im trying something rn

[glacial veldt]

@deft current @paper parrot thanks guys appreciate it

[wicked frost]

ayo fr?

no wonder its called hand lmao

[minor lake]

yea lol

[wicked frost]

ok youre right yeah

!!THUMBS UP!!

[minor lake]

stil not working breh

30030 + 48906

[wicked frost]

what

is that not the answer

[minor lake]

REEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEE

the sum of the two cases

i keep trying new shits and every time i do i realize im dumb because i previously forgot something to take in account

see i just noticed that it should be 35C2 wtf i did

36*

[stray reef]

can you please not

[minor lake]

?

am screaming because im done since i tried 2836494 stuff

[stray reef]

can you please not scream like that

its the autistic screeching meme

[minor lake]

huh?

waddafoq i didn't even know that to the point that I would sometime use it in real o_O

well thanks for enlightening me. decidedly i learn stuff everyday

(welp am a roblox player since 2012 probably explains why)

[crisp ridge]

hellow guys, does anybody knows how to determine the following if it is tataulogy? (Using truth table)

1. (P -> Q) ^ (Q -> R)
2. ~P ^ (P -> Q)
3. ~[(P ^ Q) -> R] <-> ~[~(P ^ Q) V R]
   heres the given truth table https://gyazo.com/73505f191689c0e93d9d5141495c8881

[pale epoch]

add at least 3 more columns for the statements you want to consider

fill in

see if its always true

[crisp ridge]

true for all the columns or rows?

[paper parrot]

Break the statements up into simpler parts. So break the first one into a column for (p->q) and a column for (q->r). Work out the truth values in those columns for each row, and then add another row for the full statement, which is just the conjunction (the "anding") of those two columns. Work out the truth values for that last column. If it is true in every row, then its a tautology. If its false in every row, then its a contradiction.

[paper parrot]

rows

For number 3 you'll probably want to break in into even smaller pieces

[crisp ridge]

ok thanks

[bronze bronze]

hey, is this sentence preposition ? "elephants are smart animals"

[unreal stump]

No,smart is ambiguous

[bronze bronze]

thanks bro

[shell robin]

Y'all got a good source to learn Graph Theory?

[gritty crescent]

You could ask for book recommendations in #book-recommendations .

[wooden acorn]

How are combinatorial proofs supposed to be formatted? Like what type of proof is wanted from those? Not sure what the difference is compared to a regular proof.

[coral raven]

in what context

[wooden acorn]

For proving an identity

[coral raven]

idek

[weary tiger]

@wooden acorn probably give an example

It is just a normal proof though

[gritty crescent]

Induction, generally?

[wooden acorn]

It might help me if Ik what RHS and LHS are, but I will send the problem rn

[unreal stump]

Counting

[wooden acorn]

This is the identity

[unreal stump]

Yea, That's induction

[wooden acorn]

Oh ok

[unreal stump]

You could also use generating functions,but you probably don't know them

[wooden acorn]

Yea generating functions are next week

What does it mean by: hint: count the number of ways to choose n people from a group having n women and n men?

Do I have to usean analogy?

Cool I didn’t realize they had LaTex built into discord

[unreal stump]

The sum could be rewritten as ${{n}\choose{0}}{{n} \choose{n}} +{{n}\choose{1}}{{n}\choose{n-1}}...+{{n} \choose{n}}{{n}\choose{0}}$

Now count the number of ways of choosing n people from n males and n females

You could choose 0 male and n female,or 1 male and n-1 female... n male and 0 female

[vital dew]

DrunkenDrake

[unreal stump]

But, That's the same as choosing n people out of 2n people

[wooden acorn]

I think my only question for the rest is what LHS and RHS mean? I think I know how I’d do everything else.

What do they stand for

[unreal stump]

You mean a combinatorial interpretation?

Left hand side is counting by taking cases

0 men n women
1 men (n-1) women
2 men (n-2) women...

Right hand side is just counting n people out of 2n people as a whole

[wooden acorn]

Oh ok I get it now

Thank you

[unreal stump]

Also,I will give you an exercise to think about regarding generating functions

Let's say you have 2 polynomials (1+2x+3x^2) and (3+2x+1x^2). Try to predict the coefficient of each power of x in their product without doing the complete thing

For example you don't need to do the whole thing to say the constant term is 3

(ordinary)Generating functions are sorta built on this idea

[vernal linden]

Can anyone give me a hint on how to prove this?

$if\ A \Delta B \subseteq A \Delta C\ then\ A \cap C \subseteq B\subseteq A \cup C$

[vital dew]

therealcain

[vernal linden]

A, B, C are subsets of the universal set

I'm pretty sure i need to prove that $A \cap C \subseteq B$ and then prove that $B \subseteq A \cup C$

[vital dew]

therealcain

[vernal linden]

But i'm getting lost

[unreal stump]

Let x be in B and not in A U C, then (A∆B) contains x,while (A∆C) doesn't

Do a similar thing for the other inclusion

[vernal linden]

Okay, thank you!

I tried to draw a venn diagram but it made me even more confused

[unreal stump]

Venn diagrams are usually not useful

[vernal linden]

Yeah i'm starting realize that they're only useful for showing definitions

[distant bobcat]

If you have a directed cycle for the PageRank algorithm will then this diverge as you are stuck in a loop following a simple path?

[thin fossil]

Not sure if this would be discrete math or proofs. How is this 4 conditionals?

Or generally if I were to say A = B iff C = D iff E = F

[deft current]

@thin fossil A = B <=> C = D
there's the => case and the <= case

C = D <=> E = F
again, there's the => case and the <= case

[thin fossil]

Thought it was this. Got it. Thank you so much!

[naive gulch]

I know the prove route via induction but for find a formula what exactly is it asking?

[pale epoch]

it wants you to compute it for small n and guess

[naive gulch]

I feel so dumb rn isn't 1/n(n+1) the formula for each next number

[deft current]

well yes but think of the sum as a function of n

[naive gulch]

Ah so for a small n the function is the sum for 1/(1*2) onwards to the small n?

[deft current]

yes till the last term which is 1/n(n+1)

e.g. f(3) = 1/1x2 + 1/2x3 + 1/3x4

if f(n) is the value of the sum above,
try calculating f(n) for n =2,3,4,5 (small n) and see if there's a pattern in the values

[weary tiger]

Does anyone have a good resource for learning how to prove whether a function is bijective, surjective, or injective? I'm kind of struggling with the book that I have currently haha

[deft current]

is there a specific question or example you're struggling with?

[weary tiger]

I'm just finding proving problems involving injectivity/surjectivity/bijection to be very challenging to think about, especially in relation to the problems that we are doing for homework in my class haha

let me see if I can find a problem that I find tough

I suppose this would be a good start, since I'd rather work on the basics and build up from there

[deft current]

how would you start it off

[weary tiger]

Well, to prove that it's bijective, first I should prove it's injective, then prove it's surjective

From what I could find, injection basically entails

Assuming a function f: A->A then for all a,b in A, if f(a)=f(b) then a = b

[deft current]

yes

[weary tiger]

Although, I'm not sure what that would entail for R-{2} -> R-{5}

[deft current]

So to prove f is injective, you want to show the above holds

Let a,b in R - {2} and suppose that f(a) = f(b)

Prove that a = b

[weary tiger]

(one second)

[deft current]

Had a go?

[weary tiger]

Sorry, I had to help my siblings with something haha, but

let's give this a shot

5a+1/(a-2) = 5b+1/b-2

(5a+1)(b-2) = (5b+1)(a-2)

simplified

-10a+b = -10b+a

-11a = -11b

a = b

[deft current]

yes good

[weary tiger]

How would I go about surjectivity?

[deft current]

something similar

recall the definition and go from there

[weary tiger]

Alright! I've gotta finish up some homework on Cardinalities of Sets so I'll get back to you after that if you wouldn't mind

I'll probably just text something in here

may likely show up here often for Final Exam prep haha

Wish me luck aye

[worn vale]

Hey, what kind of language would the following expression accept?
(0+λ )(1*10)*1*

[last timber]

@worn vale wym

this is regex

regex

[worn vale]

@last timber That question is taken straight from the book that my math course is using. I guess that they want me to draw the graph(?) that the expression generates. Im not entirely sure.

[last timber]

do you know what notation they use mean?

[worn vale]

English is my second language so I might be bad at explaining stuff like this. Here is one example from the book.
picture (a) => 0*1*
picture (b) => (1+0(0+1))(0+1)*

"Formulate regular expressions that are recognizable by the following machines"

[last timber]

ye

correct

so what do you want?

[last timber]

in here?

[worn vale]

Ye Im struggling with that expression.

[last timber]

well

empty string is in your language

0 is also

[worn vale]

Does that mean that the first state of the machine accepts either a 0 or nothing?

[last timber]

initial state of a machine is also final

acceptable

[worn vale]

Ah yeah.

[last timber]

since empty string is in language

otherwise you parse 0