discrete-math

[errant bear]

what no, they want it in terms of n

[young wasp]

solve T(n) to get it in terms of n?

i thought that's an easier way to find the relation

*value of Tn in terms of n

@weary tiger i am not sure but something like:

ncr * (n-rc1 (r+1)! + n-rc2(r+2)!...)

[errant bear]

how is it easier lmao, the question is literally asking you to find T(n) in terms of n..

[young wasp]

how would you do it?

[weary tiger]

What?

[young wasp]

ncr ways to choose the set A

minimum no of elements in m would be r+1

no of subsets having r+1 = m elements

would be

n-rc1(you have to chose that extra element that makes up the m subset)

that's my understanding in you example for A= {1,2}

for m = 3 there are 18 permutations 6 each for 123,124,125

and for m = 4 there are 24 each for 1234 1245 and 1235

so for this example : it would be 5c2*(3c13! + 3c24!)

[weary tiger]

So you think we have to account for each case like that?

Why do you multiply them?

[young wasp]

the 5c2? because there are multiple ways of choosing 2 elements from the whole set and for each such selection there would be same number of permutations and this would give you the total number of such permutations

which is what you want to find

[weary tiger]

Alright

[deep tapir]

Thank you!

[idle cave]

can someoen explain russells paradox to me

[vital dew]

Manan.

[gritty crescent]

Also "sets containing themselves" is something not possible(try to think about any set which contains itself)

@idle cave

[worthy shard]

closed formula for this recurrence?

i never saw this kind of question starting at A0

just A1

[pale epoch]

you solve this the same way

[worthy shard]

damn

An = 5 * 2^n + (n/2) * (-2)^n

got that

"determine the largest integer a < ... "

this one i cant even read properly

can someone explain it?

[hallow relic]

: → such that
| → divides
Most likely.

[gritty crescent]

Yes, seems like

[weary tiger]

Let S = {1,2,3,4,5}. A permutation f of S is said to fix i in S if f(i) = i. How many permutations of S fix exactly one element of S?

I tried counting it, but the issue is that there is casework. If we fix f(1) = 1, then f(2) has 3 choices, but then f(3)'s number of choices depends on what f(2) is.

[stray reef]

have you heard about derangements

[weary tiger]

If f(2) = 3, then f(3) has 3 choices. If f(2) = 4, then f(3) = 2 choices.

No, haven't covered that

[stray reef]

damn

that'll be hard to explain from scratch then

[weary tiger]

I can use principle of inclusion-exclusion, combinations, permutations

[stray reef]

but the idea was that you could count the permutations of n-1 points which do not fix any points at all

[weary tiger]

That's what I am trying to do, but there's casework

[stray reef]

okay so like

the number of permutations of 1:n which do not fix any element can be counted using inclusion-exclusion

[weary tiger]

How?

[stray reef]

by applying inclusion-exclusion to the family of sets A_i := {f: S -> S | f(i) = i} and then taking the complement

[weary tiger]

$#\bigcup_{i=1}^5 A_i$?

[vital dew]

n/c

[weary tiger]

But this is going to be huge

It would be like

|A1| + ... + |A5| - |A1 n A2| - |A1 n A2| - ... - |A4 n A5| + |A1 n A2 n A3| + ...

[stray reef]

yeah it is gonna be huge

im trying to figure out how to write it down without a sea of notation

[weary tiger]

I can show you what the book does but it's confusing as well

They also don't explain how the got what they got

[stray reef]

sure

maybe they have a more elegant method for your problem in particular

which btw boils down to n * (number of derangements of n-1 points)

[weary tiger]

They just say that "we see that the number of permutations is..." Which isn't very helpful.

[civic horizon]

5 is pretty small so maybe you could try deriving the recurrence for derangements

[weary tiger]

I did it on paper and I think I get it now

It looks something like this

$\left|\bigcup{i=1}^4 A_i \right| = |A_1| + \dots + |A_4| - |A_1 \cap A_2| - \dots - |A_4 \cap A_5| + |A_1 \cap A_2 \cap A_3| \dots + |A_3 \cap A_4 \cap A_5| - |A_1 \cap A_2 \cap A_3 \cap A_4|$

[vital dew]

n/c

[weary tiger]

Then A_1 means f(1) = 1, so that has 1 choice and the others have 3! choices total

So |A_1| = 1 * 3!

Then A_1 cap A_2 means f(1) and f(2) have 1 choice each, so there are 2 * 1 choices total for the others

But actually

If A_1, A_2, and A_3 are fixed, then doesn't that automatically fix A_4 as well?

So there is some overlap between the case where 3 of them are fixed, and when all 4 are fixed...

[idle cave]

how does

S n (S u T ) = S

[weary tiger]

S U T contains S

So when you take the intersection of something that contains S, with S, you will get only S

[idle cave]

how do u know that S UT contains S

[weary tiger]

What's the definition of the union?

[idle cave]

either one or both

[weary tiger]

What?

[idle cave]

elements that are either in S or T or both

[weary tiger]

Right

So why does S U T contain S?

[idle cave]

because it may contain all the elements of S

i see.

[weary tiger]

Does anyone know how this degenerate conic would look?

If you complete the square you'll probably get a better idea

I did

it equals 0

rather than some constant k

Then it's a point somewhere not a conic

yeah

so how do I graph this

I got 9(x-2)^2 + 25(y-1)^2 = 0

desmos shows nothing

It should show a point

At (2,1)

ok

I'll put that on my graph

@stray reef Any clues about my concern?

[stray reef]

i dont have the energy to write things out in full rn

[hardy remnant]

is there a trick to write out truth tables with more than 3 values? i am getting lost after writing out the values in the table

[stray reef]

are you familiar with binary

[hardy remnant]

yes

im taking logic design rn

[stray reef]

yeah so you can write false as 0 and true as 1

then writing out all the rows in the truth table becomes a matter of counting up from 0 in binary

[hardy remnant]

yeah its just after a couple rows i forget if i already have that written before or not haha

[stray reef]

again, just count in binary

0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
1010
1011
1100
1101
1110
1111

[obtuse lance]

write the columns first instead of rows, since one column is TFTF... the second is now TTFFTTFF... third column is TTTTFFFF...

each time you're just alternating but repeating 1, 2, 4, 8, ... depending on which column you're in

[hardy remnant]

ahh ok, gotcha. alrighty thanks

[coral raven]

from what

you need more information than that

every cyclic group can be said to have 'some integers' in it, in a sense

what's it a subgroup of

yeah i don't understand what you want me to count the elements of

ok so the subgroup generated by (34) will have two elements, (34) and just the identity

since (34) just swaps 3 and 4, so doing it twice swaps them back

the identity is the group element that does nothing

it sends 1234 to 1234

if you do (34) twice, that's the same as doing nothing

i'm still not quite sure what you want to count the elements of

for the subgroup generated by (234)?

well doing (234) twice, (234)(234), will send 2 to 3 and then to 4; 3 will go to 4 to 2; 4 will go to 2 to 3

so (234)(234) is (243)

then doing (234) another time, (234)(243), that's going to be the identity

and then you can stop, because the identity and (234) together is just (234) again

am i going too fast?

no

if you do (12) or (13) or (24) twice it's the same as doing nothing

if you do (123) or (134) or (234) three times it's the same as doing nothing

so (234) means you send 2 to 3, 3 to 4, 4 to 2

(234) again will send 3 to 4, 4 to 2, 2 to 3

so (234)(234) will send 2 to 4, 4 to 3, 3 to 2

ie. (234)(234) = (243)

no

because if you do it again then you get the identity

that's the third one

every group has an identity element, it's part of the definition

in fact an identity element by itself is called the trivial group

just ()? that's just the identity

that's the trivial group

1 element

give me an example of what you want to find, lol

what, a subgroup with two elements?

no groups with 0 elements btw, every group has an identity element

well the only group with 1 element is the one with only the identity

(12) will generate a subgroup with two elements

(123) will generate one with three

(1234) will generate one with four

experience

look

(12) shuffles two elements around in a loop

each time you do (12), they'll move round one

so it only takes two times to get to the identity, and you can stop at the identity because after that you get back to where you started

not a coincidence

for (123), it'll take three applications, because every number needs 3 steps to get back to where it started

what

no, the identity isn't (12)

i don't understand

yes

the identity is the one that does nothing

() does nothing

()(1234) is just (1234)

(1234)() is just (1234), still

no, () is an element

the identity element is an element

the two elements in the subgroup generated by (12) are (12) and (12)(12), which is another way of saying (12) and ()

not replace

yes

[odd epoch]

Any youtube channels that i can learn discrete math ?

[young star]

https://media.discordapp.net/attachments/612696384252018720/847320753916477440/IMG-20210527-WA0000.jpg
does anybody know how to solve this?

[lament zenith]

Is there any way to simplify this solution or a better approach to responding to this?

All I could write down for when looking at this algorithmn

Would be that the lower bound: Will take any input, it will run some constants within the loop represented as c. And for this nested loop, the outer loop will iterate i to n times - until i matches to n. Furthermore the inner loop will also iterate as j to n (being the length of the sequence) and execute constants (mentioned prior as c) such adding the value being inspected from j's inspection of the sequence and its index location, then inspection of thisSum and assigning the maximum sum. This inner loop will iterate until its current index ends at n. Then the outer loop will iterate again and reassign some value being thisSum and once again reiterate this process. This algorithm will at least operate by inspecting some sequence using an outer loop that runs n times and a inner loop that iterates with regards to the out loop making it n^2. Therefore the sum of both the inner loop and outer loop taking at least some input and no less is Omega(n^2).

^ This may be very vague and not well done, since I'm not to sure how to really capture or explain it the best way I want it to be

That being more simple and at times specific when required. So let me know what you think, I'm interested on how I could approach wording these explanations better

[static ivy]

if $f_1(n)=O(g_1(n))$ and $f_2(n)=O(g_2(n))$ then $f_1(n)^{f_2(n)}=O(g_1^{g_2(n)})$
have to prove or disprove this, failing miserably, anyone have any hints?

[vital dew]

MALLOC

[stray reef]

this strikes me as probably not true

n is O(n) and 10 is O(1) but n^10 is not O(n^1)

@static ivy

[static ivy]

ohh dam thanks i see it now!

ty ty @stray reef

[unreal yew]

Can someone help me understand the intuitive proof for this?

There are some explanations about picking from two sets somehow, but I don't really get it at all. The notes I am looking at aren't very good in this part, I think.

[civic horizon]

Imagine you have n hats, of which you choose k

now the left hand side tells you the number of ways to pick your hats

imagine now that you pick one of the hats as your favorite. If you pick your favorite hat in your selection of k hats, then you are basically picking k-1 hats out of n-1 hats. If you dont pick your favorite hat, you will have to pick k hats from the remaining n-1 hats

because you either pick or dont pick your favorite, the sum of these will tell you the number of ways of picking k hats from a total of n hats

[unreal yew]

Oh! That makes sense.

Thank you! 🙂

@civic horizon

[civic horizon]

np

[weary tiger]

I'm trying to do the very last part

How do I even relate a truncated dodecahedron to a planar graph?

Also I think it has f=32, e=90, n=60 (or f=12+k, e=30+3k, n=20+2k where k is the # of vertices truncated)

but I don't see how this makes equality hold?

[weary tiger]

<@&286206848099549185>

[weary tiger]

is anyone good at and willing to explain modular arithmetic, multiplicative inverse mod n, etc in vc?

actually Ill ask in number theory

[tribal gulch]

Hi do you have a concret doubt or is to explain modular arithmetic in general?

[weary tiger]

I think thats more specific

in general

[tribal gulch]

is more group theory

have you done some?

[weary tiger]

no, Im learning modular for my discrete math class

for encryption applications

[tribal gulch]

okay, you can think modularity as a relation between numbers that have the same reste in their eulidean division by a fixed number

[weary tiger]

reste?

[tribal gulch]

for exemple $30=4*7+2$

[vital dew]

carrot137b

[tribal gulch]

sorry, probably I've done a bad translate

let me search it

remandier*

34 also has remander 2 when we divide it by 4

so we say $30\equiv 34\ (mood\ 4)$

[weary tiger]

right

[vital dew]

carrot137b

[tribal gulch]

in general let $a\equiv b\ (mood\ p)$ you have p diferent clases of equivalence

[vital dew]

carrot137b

[tribal gulch]

then you can define the quotient of sets (using that this is relation is an equivalent relation)

I don't know if you were asking for this general explanation or if you have concret doubts

[weary tiger]

mod 4 happens in Z_3 right?

or Z[3]

[tribal gulch]

depends the notation, this is the tipical for groups

[errant bear]

mod n is Z_n

[weary tiger]

ok

[tribal gulch]

yes, thats one notation. I don't know why didn't work the bot, the other one is Z/nZ

[weary tiger]

Z_n = {0,1,...,n-1}

[tribal gulch]

more used in group theory

[tribal gulch]

yes!

[weary tiger]

oki

[tribal gulch]

but you have to be carefull with notation, in this case 1=n+1

normaly I use bars to differentiate concepts $\bar{1}=\bar{n+1}$

[errant bear]

technically Z_n = {[0], [1], ..., [n-1]} but 99% of the time you drop the brackets/overline, which say that they are equivalence classes, because its clear from context what is meant

[weary tiger]

how do you denote {0,1,..,n-1}?

Z^n?

[tribal gulch]

as Z_n or Z/nZ

if you don't use the brackets maybe Z/nZ would be more correct, but better use the notation @errant bear has provided

[weary tiger]

Z_n = {{x in Z | x ~ 0}, {x in Z | x ~ 1}, ..., {x in Z | x ~ n - 1}}, where x ~ y is x == y mod n

is this right?

[tribal gulch]

perfect

[weary tiger]

ok

and Z_n = S/~

[tribal gulch]

what is S?

[weary tiger]

quotient set

sorry maybe Z/~

[tribal gulch]

yes, this is more natural for me

[weary tiger]

ok makes perfect sense

now for example

this somehow relates to the euclid algorithm

[tribal gulch]

you can use modular arithmethics in (mod 33) you have the equality 1=3\cdot 33-7\cdot14, in Z_33 the equality is 1=3\cdot 0-7\cdot 14=-7\cdot 14

then the inverse is -7 that is equivalent to 26

[weary tiger]

ok thank you guys @tribal gulch @errant bear

[tribal gulch]

you're welcome. If you want to go deeper into this topic, you can explore the Bézout's identity and theorem. Two examples of this identity appear in this image

[main marlin]

trying to demonstrate that the automorphism groups of a graph with distinct eigenvalues are abelian i have an adjacency matrix A and I know that there exist permutation matrices $P_1,P_2,\cdots$ s.t. $P_1^{-1} AP_1=A$, $P_2^{-1} AP_2=A$, etc, which also demonstrates $P_1A=AP_2$ and $AP_2P_1=P_2P_1A$. I assume from here I've gotta use the properties of the adjacency matrix to show that $P_2$ commutes with $P_1$, but my linear algebra is a bit rusty so i'm having trouble doing so. my professor said to consider the fact that eigenspaces are 1-dimensional but I can't tell how helpful that is here

[vital dew]

panoramatopia

[sudden flax]

what would be the negation of ∃x, x∈A∩B?

from my logic i thought it was Ax, x∉AUB

but I dont really understand negation because sometimes all the operations flip and sometimes only some

[white igloo]

You just flip the existential quantor ∃ in that case. That way you're saying such an x does not exist.

[light radish]

could anyone help me with some predicate logic?

[ember obsidian]

when negating, swap $\exists$ by $\forall$ and $\in$ by $\notin$

[vital dew]

RoκερραJαnpu

[white igloo]

@RoκερραJαnpu is negating ∃ alone, formally correct?

[ember obsidian]

no, each “piece” must be negated

in a relatively simple statement as this you can quickly see it makes sense that the negation of ‘exists x, x in A’ is ‘forall x, x notin A’

[sudden flax]

@ember obsidian so would my my answer be correct? Ax, x∉AUB

[ember obsidian]

no we swap what's said above and no more. that means we don't change the set A∩B

also the forall symbol isn't 'A'

[sudden flax]

oh ok

so i longer question like this ∀x∈R, ∃n∈Z, (xn⩾2)∨(n≠2) would be ∃x∉R, ∀n∉Z, (xn⩾2)∨(n≠2)

for negation @ember obsidian

x^n

[ember obsidian]

if a quantifier uses set membership, eg '∀x∈R', then we don't replace 'in' by 'notin' in the quantifier

[sudden flax]

ok

instead we negate the this : (xn⩾2)∨(n≠2)?

becoming (x^n⩾2) and (n=2)?

[ember obsidian]

think it in 3 pieces. forall x. exists n. x^n>=2 or n!=2

negate each piece then put em back together

exists x (in R). forall n (in Z)

[ember obsidian]

this isn't right

[sudden flax]

∃x∈R, ∀n∈Z, (x^n<2)∧(n=2), i would get this

[ember obsidian]

yes

[sudden flax]

so this is correct?

[ember obsidian]

yes

[sudden flax]

ok. thanks for help man!

[ember obsidian]

no prob

[tame isle]

Someone help

[weary tiger]

Trying to do the second bit

So I can WLOG so that no a_ij=1 (else can get rid of that row and column and use induction)

And I made a bipartite graph with |X|=|Y|=n with edge xy in the graph if a_xy>0 but I'm not sure how to apply Hall's from there

[olive hamlet]

what did you try @weary tiger

or rather i should ask, what do you think. Do you see how if there is a perfect matching in your graph it implies the statement of the problem

[weary tiger]

Yeh so if there is a perfect matching the problem is solved

but what I'm struggling with is showing for all subsets A in X that |N(A)|>=|A|

[olive hamlet]

hmm ill give you a hint

induct on |A|

try out the |A|=1,2 etc cases to get an inutition first or w/e

[weary tiger]

k will do thanks

[slate breach]

hi folks!

Given two sets X and Y, not disjoint $\mathbf{X} \cap \mathbf{Y} \neq \varnothing$, I can define a new set $\mathbf{Z} = \mathbf{Y} \setminus \mathbf{X}$.

If I define the set Z in this way, then $1)~\left| \mathbf{Z} \right| < \left| \mathbf{Y} \right|$ and $2)~ \mathbf{X} \cap \mathbf{Z} = \varnothing$.

[vital dew]

Videlicet

[slate breach]

If I draw out the venn diagrams, I can see it and believe it.

How does one go about proving this?

[slate breach]

lol

[weary tiger]

are X and Y finite?

[slate breach]

yes.

[weary tiger]

Important assumption.

[slate breach]

(i didnt know that, but yes, they are finite)

[weary tiger]

Ok so 1 is obvious right

[slate breach]

it's only obvious cuz i can see the venn diagram.

that doesn't count as a proof though.

[weary tiger]

ok so let |X|=n, |Y|=m and |X and Y| = k

ok wait different approach

actually is fine

|Z|=m-k

so |Z| = m-k< m= |Y|

you there homie?

[slate breach]

got it!

[weary tiger]

2 is even more obvious

[slate breach]

yeah, yes. this makes sense

[slate breach]

but it make sense only if i assume the value of k

[weary tiger]

intersection is nonempty

therefore such k exists

[slate breach]

$k \stackrel{?}{=} \left| X \cap Y \right|$

[vital dew]

Videlicet

[slate breach]

that's what k is?

[weary tiger]

that depends on X and Y

if X={1,2,3}, Y = {2,3} then k is 2

[weary tiger]

k>0

[slate breach]

why do you use the word and?

that means intersection, yeah?

[weary tiger]

yes

[slate breach]

x in X and y in Y?

okay, okay, gotchagotcha

[slate breach]

okay, i'm ready for this one

[weary tiger]

well then do it

its obvious

not even worth the ink

to prove it

[slate breach]

these are all electron and photons flowing through wires and fibers though.

[weary tiger]

not worth the e-ink

Ok maybe I was wrong

It's not more obvious than the first one

hmm maybe it is

idk

hard to judge

[civic horizon]

Just note that Z consists of those elements of Y which are not in X

[slate breach]

so for some context, this is the base case for a more general inductive proof, for a lemma i need to prove a result in #linear-algebra

[weary tiger]

lol what

can't see anything posted recently by you

[slate breach]

inductive step

[weary tiger]

inductive step to WHAT

ok lets see

[slate breach]

and this is the problem in axler i need this lemma to do some heavy lifting in: https://cdn.discordapp.com/attachments/540211747613704221/847393956211195934/image0.png

but anyways, im just trying to piece this all together, and this is one of the final pieces to the puzzle.

[weary tiger]

I mean I think you should see it is true in a different way maybe

look at basis of U_i

bases of Ui can coincide

therefore basis of u1 + ... +um is smaller then sum of bases

[slate breach]

lolol. no. don't worry about the axler problem.

dont mind the LA problem.

i just need to know that X intersect Z is disjoint, lol.

[slate breach]

do i just push through definitions?

[weary tiger]

just try

on your own

[civic horizon]

I mean do you need to

Doesnt what i said suffice

[weary tiger]

yeah it does, but this is just unfolding definitions

[civic horizon]

Pretty much

There isnt much to these questions so thats what it comes down to i suppose

[slate breach]

okay, so this is what i've come up with:

$Z = { z \mid z \in Y \wedge z \notin X }$

$a \in Z \cap X \iff a \in Z \wedge a \in X \iff a \in Y \wedge a \notin X \wedge a \in X $.

a in X and a not in X is a contradiction.

[vital dew]

Videlicet

[slate breach]

a \in X and a \notin X ..... only the empty set has this property.

[weary tiger]

Z intersects X is the empty set

this says a is an element of the empty set

but the empty set has no elements

[lament zenith]

Hey guys, from the previous analysis made in (a). I've conclcuded that the lower bound that is appropiate to the running time of the algorithm would Omega(1). Since for any input, if the 1st and 2nd sequence values are a product of the target product. It will return the algorithm output straight away

And for the analysis made in (a). Any input n given, will execute that if condition statement in the inner loop if there exists no sequence that results into the product of the target value. So we can use this understanding to bind the lower bound to Omega(1). Since that the boundary of which the algorithm can run at least. But would that be a appropiate to claim this statement. Or would you say this bound is very loose and not tight?

<@&286206848099549185>

[lament zenith]

I just went back to this question, and figured it is talking about the bounding this algorithm to the "Worst-case time complexity". From that understanding we can say that the asymptotic lower bound that bounds this worst-case time complexity algorithmn would be Omega(n^2). Since the minim requirement this algorithmn needs to run at least would to iterate through the sequence using the index i for the outer loop and the j for the inner loop [Therefore a nested loop]. To find some common value that equals to a target product we're interested in

[dire void]

That’s what I came up with. Using inclusion/exclusion.

Just define the size of X as m* + k and Y as n*+k.

Where n*+k =n = |X|and m*+k = m = |Y|

[civic horizon]

sure

you can also note that $Z = Y \cap X^c$, and so if $z \in Z$, then $z \in X^c$, and so $z \notin X$

[vital dew]

Pappa

[agile citrus]

Hoping someone can help with propositional logic 😓
I have to convert a formula to a CNF (conjuctive normal form) using a CNF algorithm. I get through the first few steps but I'm stuck on the part of the algorithm for distributivity or just not really sure how to arrive at the final step

https://i.imgur.com/LMEkyFK.png

so i have to find the CNF form of (not q and r) implies (p and not q)

through a cnf algorithm but i get stuck from changing it into the last form, I'm unsure where the not q in the last clause goes

[kindred umbra]

I need some help with a question

if anyone can help me please dm

[woeful crag]

Hi

just like how any set is a subset of itself

is any graph a subgraph of itself?

[light ginkgo]

yes ofc

[woeful crag]

ty

[coral bramble]

uh, i need some help with this problem.
Find the number of non-negative interger solutions to this equation: a + 2b + 10c + 20d = 50

[finite cape]

i think this material can help you: http://logic.stanford.edu/intrologic/chapters/chapter_05.html. you should try to use the first distribution rule and then the resolution principle.

[agile citrus]

i ended up solving it like this

i just had to get to that spot where i could apply distributivity on the A

(p^~q)

[agile citrus]

thank you for the link by the way! it's a good read. This topic is kind of rough because it feels like every place teaches it differently so all the resources online are so different and are hard to google 😓

[finite cape]

nice!

[finite cape]

no prob. exactly, i found very hard to discuss it with another people who also learned this topic by themselves due to this difference on the materials online.

[agile citrus]

this link was great exactly for that reason because it's using the exact same way of notation as my university!

[agile citrus]

and my university doesn't mention anything about law of absorption 😓

[jagged junco]

anyone?

[stray reef]

count the edges incident to V1 and the edges incident to V2

those counts must be the same as every edge in G connects a vertex in V1 to a vertex in V2

[jagged junco]

yes, is it with induction? or I can say it according to some

[stray reef]

you do not need induction

you know the definition of a d-regular graph, right?

[jagged junco]

I know there's theorem A bipartite graph contains no odd cycles

if it has to do with that

[stray reef]

you do not need any fancy theorems

keep it simple

[jagged junco]

yes, so every degree of each vertex is d

[stray reef]

yeah.

so there are $\absv{V_1} \cdot d$ edges incident to $V_1$, and there are $\absv{V_2} \cdot d$ edges incident to $V_2$

[vital dew]

Ann

[jagged junco]

I agree

[stray reef]

every edge is incident to both parts and there are no edges incident to only one or neither part

therefore $|V_1| \cdot d = |V_2| \cdot d$

[vital dew]

Ann

[jagged junco]

oh your words made me see it

I understand it now!

thank you 🙂

[light ginkgo]

try contradiction

[stray reef]

you're a little late

[jagged junco]

haha

[light ginkgo]

👍

[jagged junco]

thank you that works too

[light ginkgo]

cmon are you even trying?

if you have a connected graph on n vertices it must have at least n-1 edges

[jagged junco]

Yea I agree, but it said n is the number of connected graph components

if it is n=1 I understand it the way you said

[light ginkgo]

it has k components each are connected. Let n_i be the number of vertices in the i-th component for 1<= i< =k. Now for 1<=i<=k the ith component must have at least n_i-1 edges. What happens when you sum from i=1 to k of (n_i -1)?

[jagged junco]

Oh wow . So we get
sum ( from i=1 to k )(n_i) - k
And then it is |V|-k

but I couldn't think about it if you didn't say it 😭

[light ginkgo]

you have to spend time thinking about the problems

and solving them yourself

it takes time

[jagged junco]

yea you're right I guess.. I feel they speed up with the lessons here, like always every other lesson is completely different subject, makes it hard to keep up

thing is I'm glad someone like you helps me to brighten up those weak points because I don't have any teacher to ask only zoom

thank you!

I'll keep trying

[light ginkgo]

it is tough but you can do it

[pulsar thunder]

hello! i need help w/ (c). using the formula, the pattern is it's increasing 1 unit from the previous number starting from 1 (so it's 1,2,3,...) but i don't quite know how to explain if it's valid for all natural numbers n ...and letter (f) too :((

[limpid fossil]

any advice on how to prove the property $\phi(mn) = \phi(m) \phi(n)$ for relatively prime integers m and n?

[vital dew]

add

[kindred portal]

any help appreciated for this one

[errant bear]

er, not really a counterexample my bad, but hmm

[kindred portal]

it's in the context of a enumeration assignment

the powerset has 31 + null set of possibilities

[errant bear]

like, 2 subsets of the power set

[tame arrow]

how do I find how many terms are in this sequence?

I came up with the equation An = 5(2/5)^n

but I can't figure out how to solve to get the index of the last value

[kindred portal]

for example in the example they gave

here's the result of all subsets % 31 including empty set

wait proving it's not onto is trivial

because the domain is defined as natural numbers

but %31 only gives 0 - 30

[errant bear]

yes

sorry what is the relevance of your table

[kindred portal]

well there are 31 possible sublists right

of a 5 element list

[errant bear]

uh

[kindred portal]

2^n - 1

[errant bear]

is it

yes

my b

[kindred portal]

sorry that's a shit table i sent

here's a better one

[errant bear]

by sublist you mean element of the power set

[kindred portal]

yeah

pretty much

[errant bear]

and so what is your argument?

[kindred portal]

so the last line here is all the sum(sublist) % 31

and the ones before it are all the sublists listed

there are repeat values

how are we to prove that there will be repeat results

cause although i can say there are 31 subsets, i can't say their %31 results are going to be consecutive

[errant bear]

it might be better to have your output list, not in ascending order

so you can see what sublist corresponds to what sum mod 31

[kindred portal]

ah wait

I know what I can use @errant bear

[errant bear]

what

[kindred portal]

the pigeon hole principle

the powerset consists of 32 sets right

inclusive of the empty

we have 31 possible values

!!

the pigeon principle states that if you're shoving k objects in m holes where k > m

one such hole must have a number of elements of at least the ceiling of k / m

[errant bear]

this would work if you allow the empty set to be in the domain which i mean i guess it would

[kindred portal]

in my course they treat 0 to be an element of natural numbers

¯_(ツ)_/¯

or rather somebody asked a question about whether or not the empty set is a part of it, and the lecturer said yes

so

[errant bear]

im saying like, this works if you let f({})=0 which is kinda sketchy to me, but i guess its okay

oh i see

then yeah thats fine

[kindred portal]

i guess it's just my course's convention ahaha

otherwise i see no other way to prove it easily

[errant bear]

i was trying to php like the 5 elements themselves, but yea its not obv to me

otherwise the answer they want is surely that because they chose mod 31 for that reason

[kindred portal]

yeah

[oblique cove]

is 2 element belongs to this set {{2},{{2}}} ?!

[stray reef]

no

[oblique cove]

no more sorry
{{∅}} ⊂ {{∅}, {∅}} true or false

[gritty crescent]

What do you think?

[oblique cove]

i think it's false @gritty crescent

[gritty crescent]

Why?

[oblique cove]

but it would be true if ⊆ not ⊂

because all elements in set A in set B if we say right side is A and left is B

Distinct B

[gritty crescent]

Interestingly, the set on the right has the same element written twice

So yes, if that's the definition you're using

Then it's false

[oblique cove]

i asked to make sure i'm understand correctly because who suppose to mark my answers say it's true and don't know the difference between ⊆ and ⊂ from rosen book 😦

[gritty crescent]

This is a matter of notation really

[vital dew]

Manan.

[gritty crescent]

So check with the notation your book/class uses and go with it

[vital dew]

Manan.

[dark schooner]

I have a question on set theory

How do u construct an infinite set

[last timber]

@dark schooner there is an axiom for infinity

it can be stated as
"exists I such that {} is in I and if x is in I then x U {x} is in I"

[hot jacinth]

i have a question that may seem totally obvious but i think im missing something right

lets say I have 2 consecutive numbers multiplied, n(n+1). I know you can show its always even because one is even and one is odd so 2k(2k+1) = 4k^2 + 4k which is 2(2k^2 + 2k)

but, to show that n(n+1) is even you cant directly show that it can be written as 2k directly you have to go through the fact all numbers are either even or odd

why is this only possible for divisibility by 2

or rather, lets say im trying to show divisibility by 7

if i cant show directly that its divisible by 7 by writing it as 7k, what am i to do?

theres no "even - odd" version for divisibility by 7

[obtuse lance]

there is, it's just there are more

we call these residue classes, they're described by their remainder when you divide by 7

for instance just like n(n+1) is even you can say n(n+1)(n+2)(n+3)(n+4)(n+5)(n+6) is also divisible by 7

also probably depending on your question, there's fermat's little theorem

if n is not divisible by 7 then n^6 - 1 is divisible by 7

more generally, if n is not divisible by the prime p, then n^(p-1) - 1 is divisible by p

this is part of #elementary-number-theory and modular arithmetic @hot jacinth

[hot jacinth]

Oh interesting

So if you have some expression f(x) you are trying to show is divisible by some number n, then it is not guaranteed that it can be written in the form n*g(x)?

Im in this room cuz this was in my discrete math book 😅

[obtuse lance]

you can sure

I guess maybe it depends what you mean by that form

I'd say n(n+1) = 2*(n(n+1)/2) is of that form

since we can safely say n(n+1)/2 is an integer

in fact we can say it's a binomial coefficient n choose 2

and even better, the binomial coefficients form a basis for all rational polynomials that are integer valued at all integers you plug in

so idk, that's pretty good I guess, you can even represent other things like exponentials this way too

[hot jacinth]

Ah wow ok that makes perfect sense. Thank you very much!

[hot jacinth]

So ive thought about this a lot. How would I take any polynomial and write it as the basis of binomials? @obtuse lance

[dark schooner]

I don’t understand.
Basically I just wanna know if u could create an infinite set via successive addition (n+1) or whether the elements of an infinite set are actualised simultaneously?

[pine spire]

anyone know how to negate the statement "If today is Thursday, then we meet for class"?

[red nest]

Today is thursday and we don’t meet for class

[pine spire]

thank you

also, how to write p XOR q in terms of /\, \/, and ~?

[red nest]

(P or Q)and(not(P and Q))

[pine spire]

thank you

i need help with this problem too

[red nest]

What have you tried

[light ginkgo]

Cmon TRY

[pine spire]

damn you gotta mention me when replying

i already solved it but thanks anyways

[red nest]

Nice

[feral osprey]

What does the uppercase pi means?

[obtuse lance]

means product

the p|n means what we take the product over

we only look at the primes p that divide n

and multiply 1-1/p together

for instance phi(12) = 12 * (1-1/2) * (1-1/3) since the only primes that divide 12 are 2 and 3

[feral osprey]

Thanks

[obtuse lance]

you're welcome

[unreal stump]

Try showing (n k)<=(n k+1) for k<n/2-1

Consider (n k+1)/(n k) and do some manipulations to show it's >=1

[tranquil flint]

Hey guys for the pigeon hole principle, the statement is if there are n>m pigeons being fit into m holes than at least 1 hole as more than 1 pigeon

right

but everytime i think about it, im like why cant we fit all the pigeons into 1 hole

i mean unless the statement requires each hole be filled

i dont get the intuitive nature of the theorem

Like if there are more people in london than max hairs on the human head

[coral raven]

if you fit all the pigeons into one hole

then that hole has more than one pigeon

[tranquil flint]

then 2 ppl have same amount of hair

Oh

LMFAO jesus christ ok i am so dumb

...

[coral raven]

don't sweat it

[tranquil flint]

havent slept in 24 😭

[unreal stump]

Why is (n/2+1)/k > n/2+1?

You could just do if n/2>=k
Then ((n/2)+1)/k=(n/2)/k +1/k>1

[smoky kiln]

How would I write this sum out as a double sum in ( i ) and ( j )? ( \sum\limits_{ \substack{ i , j \geq 0 \ i + j \leq n } } a_{ij} )

[vital dew]

g_ijoe

[pale epoch]

$\sum_{i=0}^n\sum_{j=0}^{n-i}a_{ij}$ should work?

[vital dew]

Lochverstärker

[smoky kiln]

ok it makes sense, thanks

[weary tiger]

how can I show a tree with even number of edges has a vertex of even degree?

if I assume all deg(v) are odd then I can conclude there are even number of vertices but idk what next

probably that forces a cycle but idk

[obtuse lance]

doesn't every tree have a vertex of odd degree, not just ones with an even number of edges?

[weary tiger]

sorry homie I meant even

[obtuse lance]

two formulas come to mind

for a tree |V| = |E| + 1

and generally speaking,

$\sum_{v \in V} \deg(v) = 2 |E|$

[vital dew]

Merosity

[weary tiger]

ye I tried using this one

didnt rly know about the one for trees but I see

[obtuse lance]

so |E| being even means |V| is odd

now assume deg(v) is odd for all v

so you're summing an odd number of odd numbers and getting an even number

that's the contradiction

[weary tiger]

wait why

[obtuse lance]

this one

[weary tiger]

oh from the first one

[obtuse lance]

yup

[weary tiger]

how come I didnt know about this formula hmm

[pale epoch]

how did you define tree?

[obtuse lance]

graph with no cycles

or you asking ledog

[pale epoch]

yes

[weary tiger]

• you can go from any vertex to any other

[pale epoch]

you can use that formula to define tree (in the finite case)

and depending on what you used, its more or less easy to prove

(so maybe its part of the exercise)

[weary tiger]

maybe ye ill try to prove it

[hot jacinth]

In class today this proof for the inverse of f o g being f' o g' assuming f and g are bijections was shown today

This particular one i sourced from stackexchange but it was basically the same

[weary tiger]

ok ez induction

[hot jacinth]

But i think im a bit slow today because I have no idea how this shows that the (f o g)' is f' o g'... lol

could someone walk me through it? thank you in advance 🙂

[pale epoch]

i don't think there is more one can say about this other than the first line

[obtuse lance]

looks like you're not even reading it because the inverse has f and g inverse reversed

[hot jacinth]

doesnt this imply the order doesnt matter?

[obtuse lance]

no

think of g as putting on socks and f as putting on shoes

f o g means you put on socks then shoes since it starts right to left

then you take off your shoes then take off your socks in reverse order, not take off your socks before your shoes

[hot jacinth]

Ok, with you so far

So in that first line when they move the brackets

thats not changing the order, just the grouping

?

[obtuse lance]

the brackets aren't super meaningful because function composition is associative

[hot jacinth]

Wait wait

ok at the risk of embarassing myself

doesnt associative mean that the order doesnt matter?

[weary tiger]

associative means it doesnt matter where you put the parentheses, so like x(yz)=(xy)z

[hot jacinth]

Oh im thinking of commutative then

[weary tiger]

yes

[hot jacinth]

OK then back to the inverse of the composite is the composite of the inverses proof

Maybe Im just confused because i dont see someting like (f o g)' = ... = ... = I and then f' o g' = ... = .. = I

instead there is g o f o f' o g'

unless

OOOOHHH nevermind

i see

sorry 😅

[weary tiger]

it says "write down the variance" which kinda implies its trivial

but it's not trivial right unless im missing something?

[light ginkgo]

Yea finding the variance isnt trivial

I dont really think theyre trying to say its trivial thogh

[pale epoch]

what's n and m?

[light ginkgo]

What

[pale epoch]

someone redacted a question

[weary tiger]

ah thanks

[obtuse lance]

just came to mind, a more general perspective on the formula for trees earlier is to use euler's formula |F|-|E|+|V|=2 since a tree can be thought of as having just 1 face, so |F|=1 recovers the formula

[light ginkgo]

number 5

[lilac delta]

can somebody explain why this is considered Anti-symmetric?

[stray reef]

antisymmetry isnt a matter of consideration, its a matter of being

why don't you tell us what's making you cast doubt on this relation's antisymmetry?

[feral osprey]

Can an Order of an object withing a group exist only when the objects be desribed by another object to the power?

[light ginkgo]

What?

An object can always be described by another object to some power

[feral osprey]

oh, thanks @light ginkgo

We studied order using the order of an object via his power to become the identity object

Just wanted to understand it accurately

[light ginkgo]

Im not really sure what your asking can you restate it?

[feral osprey]

can an object of a group have it's order described not via it's power to the function the group is assigned?

(G,*,e) g belongs to G

can o(g)=n be described not by g^n=e?

[coral raven]

how can something not be described by a definition of the thing

[light ginkgo]

Im sure there are other ways to describe it but it isnt really useful to do so.

[feral osprey]

@coral raven So by definition it's true

I see, thanks

[lilac delta]

is it antisymmetric because it doesn't include {1,3}?

[stray reef]

()

(1,3)

[lilac delta]

yeah, thanks.

[pure harbor]

If i have a nth order linear recurrence eq, im thinking it always have n linindep solutions? But how would you prove this

nvm i think i got it

FTA and then do some rearranging

[unreal stump]

You can convert a homogenous linear reccurence into a ODE

So,Your problem gets reduced to solving an ODE

[pure harbor]

thanks, ill look up that

[unreal stump]

See exponential generating functions

[weary tiger]

Hi, I am a little confused on this question. I tried drawing a graph representation of the relations but, I'm a little lost on the information given and how they relate.

Could any one help walk me through it?

[tardy remnant]

what does it mean for a system to be consistent?

p^q to be a tautology?

[forest lodge]

Is {1, 2, 3} a subset of {2}

I know
{2} is a subset of {1,2,3}

[pale epoch]

{1, 2, 3} is not a subset of {2}, since for example 1 is an element of the former but not of the latter

[forest lodge]

Ty!

[hasty glade]

Hi

Is this statement true?

[feral osprey]

How do I prove?

And is the reverse true?

[obtuse lance]

are you familiar with euler's theorem?

also I assume gcd(a,n)=1

[feral osprey]

yes

It's a theorem that for each prime that is the product of n you calculate n(1-1/p) for each prime?

[obtuse lance]

sort of

that's more like describing how to compute phi(n)

which is called euler's totient function

but euler's theorem is if gcd(a,n)=1 then $a^{\varphi(n)}=1 \mod n$

[vital dew]

Merosity

[obtuse lance]

it's a kind of generalization of fermat's little theorem

[feral osprey]

huh

can it be also true for if gcd(a,n)=2 then $a^{\varphi(n)}=2\mod n$

[vital dew]

Nightchanger

[obtuse lance]

sure, it's more like if gcd(a,n) is not 1 then you end up with a divisor of n

like take n=4 and a=2

1=3 mod phi(4) but 2^1 != 2^3 mod 4

[feral osprey]

I don't get the example

oh. I see

[feral osprey]

so how does it relate to my op?

[obtuse lance]

knowing this is the key to answering it

you can think of 1 as being a^0, so you can think of stuff living in the exponents as mod phi(n) arithmetic

[feral osprey]

I see thanks

[obtuse lance]

you're welcome

[weary tiger]

Hi, in the context of binary relations; when we examine them to see if they are symmetric, reflexive, or transitive, these are traits that we look for after the relation is established correct? Not something that we are trying to create when relating two sets?

For example if I am given some relation, I would have to logically determine if they have any of those traits?

[lament zenith]

I believe so, if you have a relationship between two sets and have drawn this out. You look at if they're symmetric, reflective or transitive. If not they're the opposite or neither. For example a graph can be either me symmetric, anti-symmetric or neither.

[lament zenith]

Yes

[weary tiger]

Right, I was just pondering that and couldn't fully grasp it until now.

[lament zenith]

Do you have the example of the relation

Because I'm not to sure if I've mixed something up with binary relations

ok dw

[weary tiger]

I am referring to binary relations

[lament zenith]

Yeah, just wanted to double check if I'm not talking about some other topic about it

[weary tiger]

but nah, no example just thinking about it broadly is all, now I am understanding previous questions/my final coming up

nahh I think we are thinking of the exact same thing lol

thanks for taking the time to chat about it

[loud sonnet]

Let $n,k$ be positive integers with $k<n$, then consider the process where a person flips a coin and stops if they get exactly $k$ consecutive heads, or if they get to flip it $n$ times (what happens first). Then, how many ways can this happen?

[lament zenith]

Glad i could help!*

[vital dew]

MisterSystem

[loud sonnet]

I was trying to solve this using generating functions somehow

and studying the case where k=2

But nothing so far

*whatever happens first,

[dense garnet]

hi, does anyone know about binary tree here? ive got a few questions to ask

[stray reef]

ask your questions

[dense garnet]

so im tryna figure out why im not getting this array

the question asks you to pre order the binary first

i so pre ordered it

from this

(9,5,8,13,20,1,12,22,15)

to this

(9,5,1,8,13,12,20,15,22)

then the question asks you to read the binary in reverse post order

so im expected to get this

(9,12,15,22,5,13,1,20,8)

however, idk what step im missing here

i keep getting different answers regardless of using all the methods available

i dont how it goes from 22 to 5 and back to 13

[stray reef]

wait

can you show the tree you're working with?

@dense garnet

[dense garnet]

idk if i drew it correctly

but hopefully it is

[stray reef]

eh?

wait

the problem gives you a binary tree and asks you to do things with it, yes?

[dense garnet]

no its sorta like a challenge question

it asks you draw a binary tree using pre order

[stray reef]

okay can you show the question in full, exactly as it is stated

i want to see the exact wording of the question

i have a strong feeling there's some details you're not mentioning that may be important

[dense garnet]

aite hang on a sec

its about decrypting a message

im quite new to the topic but im assuming each alphabet can be represented in the order of numbering?

like A equals to 1

B equals to 2

[stray reef]

each letter

and uh

i don't see why you would need to replace them with numbers, really...

[dense garnet]

haha i guess thats just my way of understanding

[stray reef]

let me try

[dense garnet]

but im quite stuck on inverting them

been researching and still found no clues so now im finally questioning that i may have done sth wrong

[stray reef]

i mean ok first off

there is no requirement anywhere that the nodes of a binary tree should always be numbers

they can hold any kind of data you like

[dense garnet]

oh i see

[stray reef]

anyway, i'm going to try and reconstruct the tree rn

[dense garnet]

okie

omg thank you so much for helping

;-;

[stray reef]

yeah this is what the tree ends up as

[dense garnet]

oh thats entirely different...

[stray reef]

the first nodes i placed are the i (root), h (leftmost) and o (rightmost)

then i worked from there

there are now two ways to get the reverse post-order reading:

either read it in post-order and then write what you get backwards,

or read it in pre-order except taking right before left

[dense garnet]

aite lemme pre order it first

is it i, e, h, m, t, a, l, v, o

for the pre order part

[stray reef]

that is an l, not a c

but other than that, yes, of course the preorder is iehtamlvo. the problem gives you that!

and i made use of it

[dense garnet]

i got this

from post ordering it

h, t, a, m, e, i, v, o, l

[stray reef]

this is incorrect

in a post-order the root should come at the very end

you got the e subtree right though.

[dense garnet]

oh

it starts from h tho right?

left, right, root

[stray reef]

as i said,

you got the e subtree right

[dense garnet]

okie

so the m and l subtrees are incorrect?

[stray reef]

there is no c

okay

look

[dense garnet]

oh sorry

i meant l

[stray reef]

you got the first five characters right in your postorder

[dense garnet]

ohhhh

[stray reef]

the postorder is htamevoli

[dense garnet]

i is the root

root

[stray reef]

yes

i is the root

so it should go at the end

as i told you before

[dense garnet]

omg ;-;

why am i so dumb

oh wow now that makes sense why ive been getting wrong answers

i constructed a wrong tree in the first place

[stray reef]

yes, your tree was wrong

[dense garnet]

lemme try with the other one

but omg again you're a life saver

thank you so much for helping ;-;

[tardy remnant]

Can I get some help for the injection and surjection part of f(x) pls

[weary tiger]

@north stone please stop dming me to do your exams

[north stone]

ok troll

[weary tiger]

your name is durag phil thats cultural appropriation man

[north stone]

<@&268886789983436800> can somebody do something about this blabbering troll?!?!?

[errant bear]

[north stone]

😋

[gritty crescent]

Can you share message details through ModMail?

[weary tiger]

what is mod mail

how

[ember obsidian]

@graceful condor

just dm this

[north stone]

no

im over it

[weary tiger]

what details do i send

wait nvm

[north stone]

crime report subject #5000B-4

[weary tiger]

thank you for the money

ive never been paid to do an exam before

[ember obsidian]

did you seriously accept money to do someone else's exam?

[weary tiger]

thats not what i said

i said i have never been paid to do an exam before

it still stands true

[gritty crescent]

Can you cut if off regardless?

[weary tiger]

i am going to report them to their dean

[gritty crescent]

[gusty ermine]

hello im just wondering can someone explain why AxB is only these 4 elements?

[ember obsidian]

ask asked above, please send details of your exchange with phil to @graceful condor

[weary tiger]

what details needed

there is private information that i dont feel comfortable sharing

[gusty ermine]

why only (0,2) (1,9) (3,2) and (0,9). what happened to (1,2) (9,2) (1,6) and the others?

[gritty crescent]

A screenshot of concerned messages with their username would suffice

[weary tiger]

nvm i think theyre just giving you a general example

[gusty ermine]

then wheres (0,6)?

[weary tiger]

they sent me payment message in their dm

what do i do

[gritty crescent]

If you're not going to substantiate your claims, I don't think there's anything we can do.

[weary tiger]

they go to nyu

i have their test

i think its better to just email the school dean

[gritty crescent]

Do that if you deem necessary, but refusing to let us know about the details yet continuing to complain about it is making us suspicious of you as well.

I strongly recommend sending a screenshot of concerned messages to ModMail. You can block out parts of the text you find private/confidential other than usernames.

[gusty ermine]

^i do agree

[north stone]

this guy is a fraud and a cheat

hes been stalking me for ages

[weary tiger]

i told you if you send me your test i will report you

[gritty crescent]

A function is a special subset of A×B, where each element of the first set(domain, here A) is mapped to a unique element in the second set(codomain, here B). Since (0,2) is already in f, you can't have (0,anything else) as well.

[north stone]

@ing me in every server

[weary tiger]

we only have 1 mutual server and its mathematics??

you dm'd me first asking abt your stupid calc 3 test

[gritty crescent]

Alright, since neither of you is willing to cooperate, I find no choice other than muting you both.

[gusty ermine]

so is there a particular reason 0,2 is first? Like when creating the subset do u need to go from left to right?

so say like its A = {1,2,3} and B = {4,5}
would u map something like
F = { (1,4), (2,5)}?

cause what i dont understand is 0 appears twice in the f = (0,2) and (0,9)

and then 2 appears twice as the 2nd value as (0,2) and (3,2)

[gritty crescent]

No, that was just an example. You could have (0,6) too, the important thing is:

1. Everything in A must be mapped to something
2. The element being mapped to should be unique

[gritty crescent]

This is acceptable for a function

In fact, all the elements can map to the same thing in the codomain

For your sets A and B, {(1,4),(2,4)} is not a function since it doesn't include any pair with 3. {(1,4),(2,4),(2,5),(3,5)} is not a function since 2 is being mapped to two elements.

[gusty ermine]

oh so in order to get a function u need the same amount of elements in the first set and the second?

[gritty crescent]

That would be incorrect, since some elements could get mapped twice at the cost of another element not getting mapped at all

As I said

Every element in the domain should be mapped to something

And that "something" should be unique

Let me draw a picture to clarify this visually

[gusty ermine]

ok so, everything in x needs to map to a y, and the y needs to be unique?

[gritty crescent]

Yep!

Do notice that two different x can map to the same y

But the same x can't map to two different y

And every x must be mapped to some y

[gusty ermine]

oh ok

so like does that mean the f in the question isnt what A->B is equal to

like A->B isnt a operation/function (multiple divide etc)

[gritty crescent]

It is an operation in a sense, but a "unary" operation(an operation that works on a single input value), as opposed to "binary" operations like addition, multiplication, etc.

[gusty ermine]

its just given as the question to be those 4 different sets?

hm but like

couldnt i make my f something like (0,9) (1,2) (9,6) (3,2)

like theres a bunch of combinations that i could map all x to y and make sense the same x isnt mapped to 2 different y

[gritty crescent]

Sure

[gusty ermine]

hm i see

[gritty crescent]

All of them are functions from A to B

[gusty ermine]

man i was so confused i thought it was like matrix multiplication or something

and then couldnt find any pattern in f's elements

[gritty crescent]

Here's a pictorial way to think about it: everything in A has an arrow going out of it, but no more than one either. Two different values in A can have arrows pointing to the same thing in B.

[gusty ermine]

right yea that clears stuff up

but u could have 2 mapped to 6

[gritty crescent]

Sure

[gusty ermine]

it would still be a function

[gritty crescent]

Yeppp

[gusty ermine]

and everything could be mapped to 5

and still funciton

but u cant have A = 1 and A = 2 mapped to 1 B value

[gritty crescent]

(A function where every x maps to a different y is called an injective/one-one function)

[gusty ermine]

right right

thank you very much manan

[gritty crescent]

That's...what my image clarifies? XD f(1)=f(2)=5 there

[gritty crescent]

No worries; goodluck!

[candid musk]

Hello

Can someone help me with this ? 😫

<@&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.

[light ginkgo]

what is S geometrically

[candid musk]

Its a set

[light ginkgo]

geometrically

[candid musk]

Not really mentioned

[light ginkgo]

yes you have to think about it lol

[candid musk]

Yeah I guess lol

[light ginkgo]

why dont you draw the interval and fix some values for a and b and see what the set S looks like and compute sup S and inf S for that specific case of S. Then try to see if you can generalize it

[candid musk]

I don't have much idea cuz we are taught this new

and I wasn't able to get this kind of Problems

[light ginkgo]

ok for a=0 and b = 1 what is S?

[candid musk]

its a point

Origin (0,0)

[light ginkgo]

how? the point 1/2 would be in S

ok S is basically the interval (0,1)

[candid musk]

Ahh yeah my

bad

[light ginkgo]

ok now what is sup of (0,1) and inf of (0,1)

[candid musk]

Sup is 1

and inf is 0

[light ginkgo]

ok now you just need to prove it for arbitrary a and b

the idea is the same

[candid musk]

Like how, Can u show some working regarding?

[light ginkgo]

you know b is an upper bound so show that if c < b then c is not an upper bound

a is a lower bound show if a<c then c is not a lowerbound

[candid musk]

Hmm