discrete-math

[stray reef]

no like

why are you using the same letter on both sides of the sign

are you sure you didn't mean x < y ????????

[honest barn]

Oh

Shit

Right

[rain ridge]

thank you so much for the help, for more context "Give an example of a finite set S, a partial order R on S and an element x of S which is
a minimal element of S with respect to the order R but is not a least element." is the question im trying to solve.

In my attempt to answer question I considered the relation x/y and defined a set S to be {3, 4, 12, 24, 48, 72}

i think this produces a hasse diagram that allows 4 and 3 to be on the same lowest level but numerically 3 is still less than 4

Tbh I barely understand this stuff and have spent weeks teaching myself Discrete Maths

Does what I'm saying make sense?

But again, least really has no meaning when you speak about < because it’s least with respect to a different order
@honest barn maybe I dont understand the topic well enough, and I appreciate the help

[honest barn]

This means the set can't have a least element

a minimal element means that nothing "lies below it"

i.e. x is minimal if whenever y < x, this means y = x

Least means that it is "below every element"

so x is a least element if for all y, x < y

This has to do with incomparability

take the set {3,4,12,24,48,72} under divisbility

we see that 3 is a minimal element since nothing else in there divides 3, except 3 itself

it isn't a least (or minimum) element because 3 does not divide 4

@rain ridge

what I was pointing out earlier is that if a set has a least element, then any minimal element has to be that least element. Let x be a least element, and y a minimal element. Since x is least, we know that x < y, but since y is minimal this means x = y

[rain ridge]

So then have I answered the question as best as it can be answered since the question doesn't allow the relation to have a least element. Thank you for explaining the definitions aswell

[honest barn]

I mean you produced an example

so yeah

Also for division we use the symbol |

so x | y means x divides y

rather than x/y

[rain ridge]

what I was pointing out earlier is that if a set has a least element, then any minimal element has to be that least element. Let x be a least element, and y a minimal element. Since x is least, we know that x < y, but since y is minimal this means x = y
@honest barn Id probably have to include some explanation like this if i was to submit a formal answer right

[honest barn]

no

this is aside the point

Your original question was about a minimal element which isn't the least element

I was pointing out that's impossible since if you had a least element any minimal element is the least element

but that isn't what the problem asked for

it wanted a minimal element which isn't a least element

the first one assumes a least element exists, the latter doesn't

For the latter, you produced an example, and argued why this is an example

i.e. like, 4 is minimal, but isn't least since it doesn't divide 3

that's all you have to do

Although the explanation that 3 < 4 numerically isn't quite right I suppose

I mean this automatically implies that 4 doesn't divide 3 but

¯_(ツ)_/¯

[rain ridge]

Thank you so much for your time and help, really appreciate it

[pliant horizon]

Would this be a propositional function, and not a proposition?

"R^m is isomorphic to the complex numbers"

and would it be a proposition if it was

"R^2 is isomorphic to the complex numbers"

<@&286206848099549185>

[weary tiger]

Can someone explain number 15 to me? I’m kind of confused on the complement.

[stray reef]

what's confusing you about the complement?

[weary tiger]

like does it mean anything outside of the intersection ?

or outside of the union

of a and b

[stray reef]

$\overline{A \cap B}$ means the complement of $A \cap B$, i.e. everything not in $A \cap B$

[vital dew]

Ann:

[stray reef]

it is not $\overline{A} \cap \overline{B}$

[vital dew]

Ann:

[stray reef]

$A \cap B = { 1, 4 }$, so $\overline{A \cap B} = \overline{{1,4}} = {2, 3, 5, 6, 7, 8, 9, 10}$

[vital dew]

Ann:

[weary tiger]

oh so my answer is correct?

[stray reef]

yes

[weary tiger]

oh also 1 more thing

if it were the complement of u

how does that work

[stray reef]

empty set

[weary tiger]

ah ok thx so much!

[pliant horizon]

Would this be a propositional function, and not a proposition?

"R^m is isomorphic to the vector space C over the reals."
And would it be a proposition if it was
"R^2 is isomorphic to the vector space C over the reals."

[weary tiger]

Does any1 know if the books answer to question 40 is wrong cuz I think it is?

i know that an empty set is a subset of any set but not sure bout the reverse

[gritty crescent]

By its very definition, the empty set has no elements.

Thus, no set(other than the empty set) can be its subset, because the empty set doesn't have anything to begin with.

And it's been correctly demonstrated by saying that there's at least one element in A which is not in B

This is a sufficient condition to prove that A is not a subset of B

[honest barn]

Actually

the empty set is a subset of the empty set

This is the only subset of the empty set

[gritty crescent]

Ah yes, let me fix my statement.

[weary tiger]

oh damn

tyty

can u guys help me with another venn diagram question

[errant bear]

i think you misinterpreted your books nswer

[weary tiger]

yea

In a group of students, each student is taking a mathematics course or a computer science course or both. One-fifth of those taking a mathematics course are also taking a computer
science course, and one-eighth of those taking a computer
science course are also taking a mathematics course. Are more
than one-third of the students taking a mathematics course?

ima try to work it out

but im lost

[gritty crescent]

The universal set here is the group of students. Let the number of elements in that set be x(choosing a number such as 100 wouldn't do harm either, but will compromise generality). Proceed with the given information then.

[weary tiger]

ok kool il get back to u

[gritty crescent]

Sure.

[rain ridge]

Suppose that M is a monoid with operation ∗ and identity element e. Suppose also that
a ∗ a = b ∗ b for all a and b in M. Prove that a ∗ b = b ∗ a for all a and b in M.

anybody have a clue on how to start this proof

[honest barn]

are you absolutely certain

that it says a^2 = b^2 for all a,b?

This says that a^2 = e for everything

[weary tiger]

@gritty crescent I’m prolly wrong as hell. Is any of this remotely right

[gritty crescent]

I'm a bit too sleepy to follow at this point. I guess someone else could explain it better at the moment 😅

[honest barn]

Anyway, I guess under that you know

[weary tiger]

lol all good np

[honest barn]

(ab)^2 = abab = e

but then multiply by a on the left and b on the right

you get a^2bab^2 = ab

but then ebae = ab

ba = ab

[rain ridge]

yeah it says a^2 = b^2

Do I suppose that (ab)^2 = abab = e?

[vital wyvern]

hey everyone what does it mean when something is closed under countable intersections?

and unions?

[honest barn]

You don't suppose

You know it

since for any element a, a^2 = e^2 = e

hey everyone what does it mean when something is closed under countable intersections?
@vital wyvern This is for a set of sets, suppose that S is a set containing sets. What it means to be closed under countable intersections means that if you have a countable collection of stuff in S, so like A1 in S, A2 in S, A3 in S,... that the intersection of all A_i is still in S

Likewise for unions

If it says closed under finite unions for example, it means for finitely many sets in S, that S still contains the union

and likewise

[vital wyvern]

oh thank you very much

[honest barn]

This is a more general thing

[vital wyvern]

so like what you mean is there will be at least one union in S

[honest barn]

If S is a set with an operation

then for S to be closed under the operation means that for stuff in S, that the output of the operation is still in S

What I mean is

[vital wyvern]

ohh

[honest barn]

If S contains the sets like {0} and {1}

then it has to contains {0,1}

the union of {0} and {1}

An example of a set not closed under some operation is like

Vectors in R^2 under the dot product

if you do the dot product of two vectors, you get a number, not a vector

But if you add two vectors you still get a vector

[vital wyvern]

oh i see

[honest barn]

so R^2 is closed under addition

(This is maybe a bit disingenuous, normally an operation has to map back into the set but let's ignore that)

[vital wyvern]

so if you take the union of two sets

it will still be in S

[honest barn]

which are in S

yup

[vital wyvern]

thanks!!

[honest barn]

but that would imply it for finite ones

but if it's closed under countable unions

[vital wyvern]

gotcha

[honest barn]

you can take the union of a countable collection

etc.

[vital wyvern]

such as 1,2,3,4, etc?

[honest barn]

Uhhh

Countable means like one N's worth

so for each natural number you have a set

[vital wyvern]

can I see an example

sorry words dont really process

[honest barn]

Uhh

Let S_n = {n}

So it's a singleton containing the number n

so the first few look like

{0}, {1}, {2},...

this is countable because you have one N's worth of sets

[vital wyvern]

oh so if S is countable you would have {{1}, {2}, {3},...}?

[honest barn]

Yeah

but I mean if S is closed under countable unions

then if it had {1},{2},{3},...

then it has to have the unions

union*

so it has to contain {1,2,3,...}

versus if you said it's closed under finite unions

it might not have {1,2,3,...}

It'll contain {1,2,3,...,n} for all n

since this can be written as {1} U {2} U... U {n}

so you're doing a union over finitely many sets

[vital wyvern]

okay

because {1} {2} would still be in S

and thats why its closed under countable

[honest barn]

I don't know what {1} {2} is

[vital wyvern]

unions

[honest barn]

that's not a set

{1} U {2} = {1,2}

[vital wyvern]

oh

[honest barn]

I feel as though

you are still misunderstanding

[vital wyvern]

i am quite confused

[honest barn]

Just because {1} in S and {2} in S

and {1,2} in S

doesn't mean S is closed under countable unions

If S is closed under countable unions and {1} and {2} in S, then {1,2} in S

Okay I'm going to write this out formally

[vital wyvern]

sorry 😓

[honest barn]

A collection of sets $S$ is closed under countable unions if the following is true.

[vital dew]

Chmonkey:

[honest barn]

Given a countable collection of sets $A_n \in S$, the union $\bigcup_{n = 0}^\infty A_n \in S$

[vital dew]

Chmonkey:

[vital wyvern]

uh

like the actual union of two sets

are in S?

[honest barn]

not just two

this would imply it for two but

Yes

if you have two sets, their union is a set

you require that that union is also an element of S

[vital wyvern]

oh okay

[honest barn]

Maybe I can illustrate a counterexample

[vital wyvern]

that makes sense

[honest barn]

Let S = {{0},{1}}

[vital wyvern]

oh that would be helpful too

[honest barn]

S is not closed under unions

since {0} U {1} = {0,1} is not in S

It is closed under intersections though

Wait no

it isn't

take {0}\cap {1} = empty set

but the empty set is not in S

[vital wyvern]

Oh

thank you that makes so much more sense

sorry it took so long to explain

[honest barn]

No worries

I feel as though you don't quite still understand what I meant by countable

but that's something you should look into more

This has to do with different sizes of infinity

[vital wyvern]

yeah...when it comes to this kind of math it doesnt quite process the right way

[honest barn]

You can generalize the union operation to unions of infinitely many sets

[vital wyvern]

more of a computational person

[honest barn]

It's just everything that's in at least one set

[vital wyvern]

these things are hard for me to grasp

[honest barn]

Countable collection means you can number the sets

So like your collection looks like {A_i} for i in N

N being the set of natural numbers

[vital wyvern]

so if its infinite

it would not be countable?

[honest barn]

So there's a 0th set, a 1st

no

countable is infinite

it's the smallest infinity

Okay so

For you

If I say like we have a collection of A_i

do you get what that subscript means?

It's a way to like keep track of them, a specific example is A_0, A_1,...

those are all in our collection

For a countable collection just take it to mean that the subscripts are natural numbers

An example of a non-countable collection would be if we let the subscripts be any number

So if we have A_0.32, A_pi, A_1, A_32.6343, etc.

that's not countable

[vital wyvern]

oh i see

and does it have to be sequential to be countable

[honest barn]

What does sequential mean?

[vital wyvern]

like A_1, A_2, A_3, vs A_2, A_1, A_3

[honest barn]

Oh, no it doesn't matter

Since the collection is actually like

the set of all of those

[vital wyvern]

oh right

[honest barn]

formally it's like {A_1,A_2,A_3}

[vital wyvern]

because order doenst matter

right

[honest barn]

so the order doesn't matter

right

[vital wyvern]

makes sense

[honest barn]

so I guess one thing I should mentino is

finite is also countable haha

really it means any collection of size at most as large as the natural numbers

so for your purposes

it's either a finite collection or

if it's infintie we can number them using N

[vital wyvern]

so its not countable if its rational?

[honest barn]

Well...

this is messed up

actually the rationals are countable

there's a bijection between N and Q

But don't worry!

Since this means if you use rational numbers to number out your sets

you could actually renumber them using only natural numbers and still number every set

If you want to learn more about this kind of stuff look into learning some set theory

I think Halmos's set theory is good and is really short + cheap af

[vital wyvern]

So if we have A_0.32, A_pi, A_1, A_32.6343, etc.
so then whats the difference between saying that rational is countable vs this is not countable

[honest barn]

Umm

So what I really wanted to get at is that

If you like set A_x = {x}

where x is any real number

then the collection of all A_x has size |R|

which is not countable

All I mean is that if you want to show something is closed under countable intersections

[vital wyvern]

why is that not countable

[honest barn]

There's no bijection from N to R

[vital wyvern]

whats a bijection

[honest barn]

uhh

one-to-one and onto function

I don't really want to get into this stuff

it's more than I can just explain in a short time

[vital wyvern]

okay thanks for the explanation tho

[carmine vine]

how would I go about approaching this? I know about

i tried drawing a Venn diagram and got 34 but i am unsure about it and need some help 😅

[weary tiger]

so I think what you could do is

find |A∪C| with principle of i/e

then find |A∪B∪C| in the same way (with the formula you posted, to be specific)

number of elements in B which aren't in A∪C is |A∪B∪C| - |A∪C|

then subtract that from |A∪C|?

[copper ore]

was just on this page lol

[weary tiger]

not sure if this is the most efficient way but sounds like it will work

@carmine vine

[copper ore]

anyone know how a permutation can be regarded as a one to one function

[carmine vine]

Jmm

Hmm

I'll get that botnuke

[weary tiger]

a permutation is by definition a particular type of bijective function

[copper ore]

true

i kinda see it in my head now

[weary tiger]

in some (most?) parts of math, this is the "correct" way to think about a permutation

[carmine vine]

Ok I got 61 but it doesn't feel right

I wish they posted a solution to check

[copper ore]

Let S be a set with n elements and let A be the set of all ordered sequences consisting of exactly k pairwise distinct element

what does set A look like ? An example would be nice

[naive saffron]

Say n=5, k=2, and S={a,b,c,d,e}, then A={ab,ba,ac,ca,ad,da,ae,ea,bc,cb,bd,db,be,eb,cd,dc,ce,ec,de,ed}

[copper ore]

oh i see

thanks

[naive saffron]

The example of k=3 is just too much to write

So I didn't write it out

[copper ore]

lol

yeah

[naive saffron]

But you're welcome

[copper ore]

even 2 is long 👀

[naive saffron]

Yeah ikr

The numbers get pretty big

[copper ore]

i don't get how this is related to the product rule

I noticed that the formula is also the same as counting the number of one to one functions

but just don't understand how this is related to any of that

[copper ore]

wait actually i get it now

took a while haha

[woeful galleon]

knowing that EEPROM can hold 1k bytes

how do i do this.. ? i am pretty confused

[carmine vine]

for the question regardingn parallelograms,

how did they come up wit this solution?

I did 6C1² + 5C1² + 4C1² + 3C1² + 2C1² + 1C1²

which is just 6² + 5² + 4² + 3² + 2² + 1² = 91

[devout vault]

Hey guys I apologize if this isn’t the right place but this problem does come from a discrete mathematics book. Can someone help me understand why this argument can’t be proved valid. A convertible is a fun car to drive. Issacs car is not a convertible. Issacs car is not fun to drive

[stray reef]

just because convertibles are fun to drive doesn't mean they're the only type of car that's fun to drive

[devout vault]

Thanks for the response. That makes sense, is there a way to use the rules of inference to state that or does the answer boil down to we aren’t given enough info about other cars that may be fun to drive therefore no other inferences can be applied?

[stray reef]

well if you want to be formal about it then ig you can say that $\frac{A \to B, \neg A}{\neg B}$ is a non-rule

[vital dew]

Ann:

[devout vault]

Ok awesome thanks so much

Sorry quick follow question to make sure I understand a different problem. There is no rule that because p -> q then q-> p right?

[stray reef]

yes there is no such rule you're right

[devout vault]

Alright thanks for the help!

[mystic bobcat]

(being in Paris) -> (being in France) != (being in France) -> (being in Paris) @devout vault

[gritty crescent]

Is there any particular notation used to denote the contrapositive?

[stray reef]

no

[gritty crescent]

I want to prove that the contrapositive of the contrapositive is logically equivalent to the original statement, but I don't know how to frame this as a proposition in terms of symbols

By its definition, the contrapositive of $P\implies Q$ is $\lnot Q\implies\lnot P$. Applying a contrapositive again simply brings me back, by definition, to the original proposition. What's there to prove?

[vital dew]

Manan:

[stray reef]

you're right, there is nothing to prove

the contrapositive of the contrapositive is not just logically equivalent to the original, it IS the original

[gritty crescent]

Guess the book just wants me to explicitly say that $\lnot(\lnot P)\iff P$

[vital dew]

Manan:

[gritty crescent]

Thanks for helping out :)

the contrapositive of the contrapositive is not just logically equivalent to the original, it IS the original
@stray reef Apart from symbolic representation, there is no fundamental difference between being logically equivalent and being the original proposition, right?

[vital dew]

Zophike1:

[south marten]

Is my proof of De Morgan's law correct ?

[stray reef]

primes are just ' and not ^' in latex btw

[south marten]

Oh thanks yeah I'll have to learn latex for formally but is the proof correct @stray reef

[vast mulch]

how do i solve these?

[stray reef]

these ones in particular?

might help to know that 2^10 ≈ 10^3

[vast mulch]

why start from 2^10?

[stray reef]

it's the power of two most famously approximated by a power of ten, so much so that 2^10 (or 1024) bytes is called a kilobyte

[weary tiger]

Can someone check if i did this right?
In a group of students, each student is taking a mathematics course or a computer science course or both. One-fifth of those taking a mathematics course are also taking a computer
science course, and one-eighth of those taking a computer
science course are also taking a mathematics course. Are more
than one-third of the students taking a mathematics course?

[swift agate]

I need help with this type of problem: Give interpretations to prove that each of the following wffs is not valid. (∃x)A(x)∧(∃x)B(x)→(∃x)[A(x)∧B(x)]

[swift agate]

There exists an x such that a(x) is true and there exists an x such that b(x) is true does not mean that that there exists an x such that A(x) and B(x) is true

[honest barn]

Right, but I think they want an example of such an a(x)

something like

there exists dogs with 4 legs and there exists dogs with 3 legs

but there exists no dog with 4 legs and with 3 legs

[swift agate]

ok thank you for help these type of problems are tricky for me

best way for me to practice is just to see lots of examples i guess

[marble haven]

can anyone help me break up

x is not a member of (A U B)

into it's parts??

i cant figure it out

i'm completely stumped

for example, x e (A U B) = (x e A) or (x e B)

yeah idk

maybe i'm over thinking it. Originally I was just thinking that x not in A and x not in B

ohhhhhh

that makes so muhch more sense

wait so can I say:

(x not in AuB) = ~(x in AuB)
=~((x in A) or (x in B))
= (x not in A) and (x not in B)

[marble haven]

is there some law that states x not in A = x in A'

[quaint star]

Do you mean $x \not\in A \implies x \in A'$ where A' is the complement of A?

[vital dew]

Lunasong:

[quaint star]

Then that is the definition of A'

[marble haven]

oh bet

thanks

[gritty crescent]

What exactly does the set B={x| x is in A and x is not in f(x)} mean?

(This comes up in Wikipedia's proof of Cantor' Theorem)

I can see why the theorem is true( I can create an injection from A to P(A), but it won't be a surjection)

But the notation is confusing me. Can someone help out?

[honest barn]

Recall that

Okay so f is a supposed bijection between a set A and P(A) right?

this means x an element of A, is inside f(x), which is a subset of A since f(x) is in P(A)

@gritty crescent

(Or I guess a supposed bijection, but it matters not)

[gritty crescent]

this means x an element of A, is inside f(x), which is a subset of A since f(x) is in P(A)
I see. How is this fact used to produce a contradiction?

[honest barn]

consider what happens to an element x such that f(x) = B

B is in P(A)

if x is in B then x is in f(x) so x cannot be in B

but if x is not in B then x is not in f(x) so it's in B

[gritty crescent]

Ah, that makes sense. Thanks!

[honest barn]

[weary tiger]

@weary tiger I think u got the right answer but not in the right way, although i'm not 100% sure I did it correct either

total students is not M + C + B, it's M + C - B, because M and C each has B counted in their set, so if you add them you add B twice, therefore you have to subtract with b once for the extra b that is added

the general formula for adding together multiple sets that have elements in common is called inclusion-exclusion formula, u can check it out here:
https://en.wikipedia.org/wiki/Inclusion–exclusion_principle

[copper ore]

well i would choose C) but it seems to be 1 off. Can someone else confirm?

[stray reef]

yeah the answer is b not c

consider that positions 58 and 60 must contain 1's if you want your string to be 00-free

[copper ore]

if the length is 77 than the answer should match with fibonacci number 79 though right? Cause i never get it to match

@stray reef

[stray reef]

no, it's smaller

you're not looking for ALL 00-free strings

you're looking ONLY for those which have a zero at the fifty-ninth position

[copper ore]

ohh

snap

okay

[stray reef]

also wait i think i was off by 1

57 symbols, followed by 101, followed by 17 symbols

yeah that means c is the answer

[copper ore]

ok i don't really get how to solve this problem

i was reading it completely wrong earlier

you're looking ONLY for those which have a zero at the fifty-ninth position
@stray reef like what does that exactly mean?

[stray reef]

it means exactly what is said lol

[copper ore]

lol

but like wouldn't it just be 1? I don't get it

[stray reef]

okay so like
here's a bitstring of length 77, broken up into bytes for your convenience but it's still a single bitstring:

11010010 01110111 10110110 10111000 00000011 01101011 00101011 11111100 00001011 01111

can you tell me whether or not the bit at the fifty-ninth position in this bitstring is 0?

[copper ore]

nah its 1

like the bit is 1 ^

[stray reef]

yes

do you still not understand what i meant when i said

you're looking ONLY for those [00-free bitstrings] which have a zero at the fifty-ninth position
even though i meant exactly what i said?

[copper ore]

yeah i get what you mean now

[copper ore]

57 symbols, followed by 101, followed by 17 symbols
@stray reef btw what is technique that you’re using to solve this?

what does symbols mean in this context

[stray reef]

bits

and im not using any techniques

im just using common sense

isn't it obvious that if a 00-free bitstring has a 0 at the fifty-ninth position, then the fifty-eighth and sixtieth positions must necessarily have 1's so as to, yknow, avoid a 00?

[copper ore]

yeah i see that now lol

[vital dew]

Schuams:

[stray reef]

union of two countable sets?

gonna be a bit tricky

[pale epoch]

A union empty set

[signal basin]

A union empty set
@pale epoch ... 😄

[signal basin]

never mind. i did it.

[vital dew]

Manan:

[gritty crescent]

For the special case where both bijections are inverses of each other, the resulting bijection is simply the identity bijection. This is easy to prove.

However, I have difficulty framing a formal proof for the general case. I'd like to get a raw sketch/idea to work out the complete proof.

[unreal stump]

Show the final set is the range of composition

The injection part is simple

[gritty crescent]

Okay, I'll try that. Thanks!

[stray reef]

or you could just give a formula for the inverse lol

[gritty crescent]

Is that sufficient to prove the statement?

[stray reef]

have you already proved bijective iff invertible

[gritty crescent]

Yeah, I can prove bijective iff invertible

Aight, this makes sense

But, this doesn't mean composing two arbitrary bijections is also a bijection, right?

[unreal stump]

You showed composition is invertible

Therefore bijection

[gritty crescent]

I mean, if f:A to B and g:B to A are bijections(A, B are non-empty and equinumerous, but not necessarily identical), then I need to show that fog:B to B and gof:A to A are bijections as well

[unreal stump]

The inverse of first is simply g^{-1}o f^{-1}

[gritty crescent]

But this argument relies on the assumption that fog is bijective to begin with(invertible iff bijective). Since I haven't shown fog is bijective, I can't invoke its inverse.

Or should I explicitly construct its inverse, show that it works, and then assert it is indeed bijective?

[unreal stump]

You construct an inverse explicitly

Yea,latter

[gritty crescent]

Okay, but this seems daunting to be phrased as a formal proof.

[unreal stump]

Sure

[gritty crescent]

I'll give it a try, and send my "proof" here once I'm done. I'd like one of you to check my work.

[unreal stump]

"since fog has an inverse,it is a bijection"

Done

[gritty crescent]

Lmao that seems too dismissive, I'd like a proof which at least shows why fog has an inverse when it hasn't been proven to be bijective. I'm thinking about mapping elements explicitly, then showing how an inverse of fog can be constructed, then assert it is bijective.

[unreal stump]

you constructed it

That's it

[gritty crescent]

Okay, I'll send my proof once I'm done anyway. I don't usually trust my proofs easily.

[unreal stump]

ok

[stray reef]

is xmin supposed to be that marker with the "X 0.037, Y 2.839" window next to it?

[vast mulch]

yea

[stray reef]

then no you have not marked it correctly

[vital wyvern]

can i ask

why is a sigma field a sigma field

whats the point of having one

[pale epoch]

whats a sigma field?

sigma algebra? its an ingredient to define a measure on a set

[sleek island]

Could anyone explain why it's twice here?

[analog sonnet]

Each cycle that contains v in particular contains two incident edges to v; in other words, the moment you register a cycle containing an incident edge to v, you will register the very same cycle containing another incident edge to v

That's why you count each cycle through v twice

[sleek island]

Thanks!

[analog sonnet]

💊

[copper ore]

would i use induction to solve this?

[weary tiger]

can't u just try n=0 and n=1 for instance and see if they match with the given definition

[copper ore]

but that doesn't show that it's true for all integers n>= 0

[weary tiger]

ah u have to motivate, i thought they just wanted you to choose one of the answers and that was it

ye i guess u can try induction for each one of them and see

it's just that i did it the lazy way by just finding if they were not equal, which if u try for n=0 and n=1 if you will see that some of them can't be equal to the original formula

[copper ore]

yeah

it'd be cool to know the induction way tho

[sour arrow]

Indeed, I would try a few to see if they're not equal, and attempt induction if I think they may be

[weary tiger]

ye since it's just a multiple choice question, it's always easier to first try find contradictions

[copper ore]

yeahh but

it'd be cool to know the induction way tho

[sour arrow]

Aight, so let's do one of these.
I'll use P(n) to mean "the proposition is true for the number n".

For part a)
P(n) only if f(n) = 6(4^n) - 2^n

Base case, check P(0)

[copper ore]

f(0) = 5

[sour arrow]

Oops, misread the question

See the edit

For the recursive definition
f(0) = 6

For the function they want us to check
f(0) = 5

So it fails the base case

[copper ore]

ok

[weary tiger]

ye in this case all of them will fail the base case

oh except for b)

and c) lol nvm

[sour arrow]

So let's go onto b)
The proposition is that the recursive definition is the same as
f(n) = 7(4^n) - 2^n

Again, trying f(0) in both definitions shows that they both evaluate to f(0) = 6, so P(0) is true

[copper ore]

ok

[sour arrow]

And that's the base case done.

Now, for the inductive step we assume that P(n) is true, and use that to show P(n+1)

What does P(n) say?

[copper ore]

the proposition is true for n

[sour arrow]

Actually let me rephrase I'm not being clear haha

[copper ore]

npnp

[sour arrow]

P(n) is assumed true, which means
f(n) = 7(4^n) - 2^n
For our specific choice of n. f(n) being the recursive definition of course

P(n+1) means that:
f(n+1) = 7(4^(n+1)) + 2^(n+1)
Which we would like to show

.
f(n+1) = 4f(n) + 2^(n)
By our recursive definition

[copper ore]

ok

I actually got stuck on the part after this lol

i couldn't make it show that they were equal

[sour arrow]

Indeed, I don't think we can

[weary tiger]

its kinda lika

and then u have to show that the f(n) = (7*4^n)

which is like another induction inception

i wonder if u can just apply strong induction and show for 2 base cases that they are not equal? the formula and the original formula

[copper ore]

.
f(n+1) = 4f(n) + 2^(n)
By our recursive definition
@sour arrow wouldn't it be f(n+1) = 4f(n) + 2^(n+1)?

[sour arrow]

Yes haha

Sorry about the slow pace and mistakes, I'm a little bit busy

[copper ore]

no problemo

[weary tiger]

Not sure but maybe if u then again apply induction on f(n) = 7*4^n, and try for the base case when n=1 because f(n) is only defined for n>=1, then you will see that f(1) = 7 *4^1 => 4 * f(0) + 2^1 = 26 and not 7 * 4=28

[copper ore]

are you talking about option b)

[weary tiger]

From what Kaynex said earlier:

`f(n+1) = 7(4^(n+1)) + 2^(n+1)
and
f(n+1) = 4f(n) + 2^(n+1)
By our recursive definition

We can write f(n+1) = 7(4^(n+1)) + 2^(n+1) = 4 * (7 * 4^n) + 2 * 2^n

And we can also write our recursive defintion as f(n+1) = 4f(n) + 2^(n+1) = 4 * f(n) + 2 * 2^n

So we want to show:
4 * (7 * 4^n) + 2 * 2^n (this is the option b) = 4 * f(n) + 2 * 2^n (this is our recursive definition) => (7*4^n) = f(n)
`

ye option b

and then as i said earlier

if u then again apply induction on f(n) = 7*4^n, and try for the base case when n=1 because f(n) is only defined for n>=1, then you will see that f(1) = 7 *4^1 => 4 * f(0) + 2^1 = 26 and not 7 * 4=28

it's weird though since we are applying induction twice here, I've never done that so idk if it's correct. Would be cool to here what kaynex has to say about that

[copper ore]

We can write f(n+1) = 7(4^(n+1)) + 2^(n+1) = 4 * (7 * 4^n) + 2 * 2^n

should be

$4\left(7\cdot :4^n-2^n\right)+2^{\left(n+1\right)}:=:7\cdot :4^{\left(n+1\right)}-2^{\left(n+1\right)}$

[vital dew]

The-Elite:

[weary tiger]

ye i might have misread option b lol

[copper ore]

np

but anyways..

the answer is b) (i checked the answer key)

the problem i have is showing that LHS = RHS

but... yes i never did induction twice either

[weary tiger]

okay i think i got it now @copper ore :

(option b):
f(n+1) = 7*4^n+1 - 2^n+1
= 4 * (7 *4^n) - 2^n+1

(recursive definition):
f(n+1) = 4 * f(n) + 2^n+1

And we assumed n to be true meaning that:
f(n) = 4 * f(n-1) + 2^n (recursive defition)
= 7 * 4^n - 2^n (option b)

Using our assumption we can get something like this by just moving -2^n from the RHS to LHS:
f(n) + 2^n = 
4 * f(n-1) + 2^n + 2^n = 4 * f(n-1) + 2^n+1 = 
7 * 4^n

So we substitute 4 * f(n-1) + 2^n+1 = 7 * 4^n into the equation of option b for when n+1:
4 * (7 * 4^n) - 2^n+1 =
4 * (4*f(n-1) + 2^n+1) - 2^n+1 =
4 * 4 * f(n-1) + 4 * 2^n+1 - 2^n+1 =
4 * 4 * f(n-1) + 3 * 2^n+1 = 
4 * 4 * f(n-1) + 3 * 2 * 2^n = 
4 * 4 * f(n-1) + 6 * 2^n = 
4 * 4 * f(n-1) + (4+2) * 2^n =
4 * 4 * f(n-1) + 4 * 2^n + 2*2^n =
4 * (4 * f(n-1) + 2^n) + 2^n+1 = 
4 * f(n) + 2^n+1 (this is the recursive definition for when n+1) = 
f(n+1) 

So we have shown that starting from option b when n+1 we arrived at the recursive definition for when n+1. 
First we tested for the base case which showed to be true, then we assumed n was true and using that assumption we showed that it was also true for n+1, thus completing the induction.

[sick nebula]

hihi, can anyone explain the intuition behind the binomial theorem? what exactly does the number of k element subsets in an n element set have to do w the coefficients of each term in an expanded binomial?

[unreal stump]

$(1+x)^n$ can be written as $(1+x)(1+x)....(1+x)$ n times.
Coefficient of $x^k$ is number of ways you can choose k elements from the total of n elements

[vital dew]

DrunkenDrake:

[copper ore]

@sick nebula it’s the number of ways each term is made using FOIL

you see there are 2 ways for x and y using FOIL

1 way for x and x

1 way for y and y

that’s where the coefficient comes from

[brave lava]

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

im confused for the one i got wrong?

what is this symbol called ^

nvm

its called a wedge

i got it

[sick nebula]

@copper ore thank you!

[marble pivot]

I thought a field had to include a multiplicative inverse for every element

Why didn’t they include 0 in the second table

[pale epoch]

what's the multiplicative inverse of 0 in the real numbers?

[marble pivot]

There isn’t one

[pale epoch]

indeed

the additive identity can't have a multiplicative inverse

[marble pivot]

Ah

So then a field only has multiplicative inverses for every element except the additive identity?

[pale epoch]

yes

[marble pivot]

lit

[gritty crescent]

the additive identity can't have a multiplicative inverse
@pale epoch Does this hold true for any arbitrary field/whatever it's called?

[weary tiger]

Yes it does hold for arbitary fields, does not hold in arbitrary ring (there is a zero ring in which 1=0)

[gritty crescent]

Interesting

[leaden moon]

I’m stuck, can anyone give me a hint?

[gritty crescent]

Do you want to prove why $$(A\lor B)\land(A\lor B')\iff A?$$(Also, $B'$ is the negation of $B$, I assume?)

[vital dew]

TedNotKaczynski:

[leaden moon]

Yeah

[gritty crescent]

If it's any help, $$(A\lor B)\land(A\lor B')\iff A\lor(B\land B')$$by the distribution law

[vital dew]

TedNotKaczynski:

[gritty crescent]

Can you continue from here?

[leaden moon]

Oh what the..

Thanks

[gritty crescent]

Np :)

[leaden moon]

so much lol

[vital dew]

same:

same:

[sick swallow]

woah it is same

[weary tiger]

w o a h

[true comet]

Is there a way to solve "Show that (p → q) ∧ (p → r) and p → (q ∧ r) are logically equivalent" using logical equivalences? I could pretty easily do it using a truth table but while the other exercises specify "using a truth table" this one does not.

[pale epoch]

consider that $(p\implies q) \iff (\neg p \lor q)$

[vital dew]

Lochverstärker:

[pale epoch]

@true comet

[true comet]

Okay, I can work with this, thanks!

[pale epoch]

although truth table is fine

[strong wolf]

I spy notability

[vital wyvern]

hello everyone

can someone help me on my hw

oh mb

[last timber]

,rotate

[last timber]

what is limit of this

[vital wyvern]

Thanks 😅

[last timber]

as k to inf?

nvm

[vital wyvern]

I think so yeah

[last timber]

i am dumb lol

[vital wyvern]

Because it implies that k = positive integers

Lol

[last timber]

ah no ok

A_k indeed is subset A_{k+1}

so union would be largest set of this obviously

[vital wyvern]

so am I like adding the probabilities since

its a union

[last timber]

what is limit 1/k as k to inf?

[vital wyvern]

0

?

[last timber]

so you would have (0, 3) as union

[vital wyvern]

yes

[last timber]

that is

[vital wyvern]

the limit?

[last timber]

limit is (0,3)

[vital wyvern]

0.3?

why not anything else

oh

OH

[last timber]

and igtg sorry

[vital wyvern]

it was that easy

okie thanks tho!!

[last timber]

yw

[vital wyvern]

so for question 3 part b

would that just be 0

for the pic I sent

[marble haven]

do questions about partitions fall under discrete mathematics?

[weary tiger]

not especially? depend though.

[weary tiger]

what does the upside down T mean

[delicate ridge]

it basically means contradiction

[weary tiger]

okay tyvm

[stray reef]

it means False

[nocturne carbon]

Calculation:
    7 · 8
  =⟨ Fact `8 = 7 + 1` ⟩
    7 · (7 + 1)
  =⟨ Fact `7 = 10 - 3` ⟩
    (10 - 3) · (7 + 1)
  =⟨ “Distributivity of · over +” ⟩
    (10 - 3) · 7 + (10 - 3) · 1
  =⟨ “Distributivity of · over -” ⟩
    ((10 · 7 - 3 · 7) + 10) - 3
  =⟨ “Identity of ·” — twice ⟩
    10 · 7 - (3 · 7) +   10   -   3
  =⟨ Fact `3 · 7 = 21` ⟩
    (10 · 7) -   21  +   10   -   3
  =⟨ Fact `10 · 7 = 70` ⟩
    (70)     -   21  +   (10   -   3)
  =⟨ Fact `10 - 3 = 7` ⟩
    (70     -   21)  +        7
  =⟨ Fact `70 + 21 = 49` ⟩
    49     +        7
  =⟨ Fact `70 - 28 = 42` ⟩
    56

Can someone please explain what ive done wrong, im new to discrete math

[serene basalt]

7 · 8 = 7 * 2 * 2 * 2 = 14 * 2 * 2 = 28 * 2 = 56

[nocturne carbon]

yea ik there is a alternative way to do itt, bt my professor gave us this and told us to fix the error

[short hamlet]

idek how to start this

i think base case: n = 1. start at the only even number

<@&286206848099549185>

[olive sun]

I know it's in Spanish but I'll translate it. What I have to do is prove that the first proposition is logically equivalent to the second one by using succession of statements. Now, I know how to turn everything into the other side with the exception of the "If, then" into the "And" from the second proposition. I'm genuinely lost, is there like a law that allows to do that, cause I'm losing my mind over this.

[tawny oracle]

Hello! Just wondering if anyone can help me understand why the last question is true. I would like to know if the apostrophe next to the“ O’ “ is what makes it true.

[weary tiger]

probably means just the complement to O denoted as O'. And u can see that O has odd numbers. So the opposite of that would be even numbers

https://en.wikipedia.org/wiki/Complement_(set_theory)

and the larger set which is implicitly defined in this case is U

so O is a subset of U, containing the odd numbers. U contains both odd and even numbers. When you take the complement of O, then you get all the other elements that are not in O within the larger set U. You can think of it as removing all the elements of O within U, so what is remaining will be the even numbers. @tawny oracle

[weary tiger]

if you try to visualise it it would look like this, where U is the larger set containing both of them

[errant bear]

is U the universal set here?

[weary tiger]

ye i guess so if you mean by the larger set that they are contained within

it seems that they implicitly defined it as U = {1,2,...,10} as the universal set for this assignment

[formal tulip]

What is the symbol between x and A and also what is it saying?

[weary tiger]

Probably “element of” in this context

[humble hornet]

Alittle help with imduction

Need guidance

[haughty garden]

You might want to use strong induction

More specifically, assume that f(n) and f(n - 1) both hold true. Then use those two statements to show f(n + 1) = blah

[humble hornet]

Hmm

[vital dew]

opengangs:

opengangs:

[haughty garden]

Should be a few lines of working

[humble hornet]

Idk about strong induction yet

[haughty garden]

Oh that’s sort of what the hint is gauging at

I just flipped the indexing around a bit

You just assume the statement holds for two cases instead of the regular one

Strong induction assumes that every case preceding n is true which is where your assumption of two cases come in

[humble hornet]

Oh i see

[hollow saddle]

I've been so stuck on question 12. I was able to do 11, but I don't think I'm understanding the question itself. Am I finding what I did previously just without the not negation?

[pale epoch]

you want to express $\neg P$ only using NAND and parentheses

[vital dew]

Lochverstärker:

[hollow saddle]

so P nand P would not make sense? This is all so new to me haha

[pale epoch]

why do you think it would not make sense?

[hollow saddle]

No parenthesis

I figured that PnandP would not be an acceptable answer

[pale epoch]

that would surprise me

also, you could just add parentheses

((P) NAND (P)) 🤷

[hollow saddle]

ahhh, so if I use (not) PnandQ for the first one on 11, then if I were to rewrite it without the negation not, It would be (PnandQ)nand(PnandQ)?

no...

wait yes

lol

[pale epoch]

ye

this has applications in circuit building

because you can build wtv you want if you have only NAND and no NOT gates

[hollow saddle]

Ohhh, okie! Thank you so much (:

[weary tiger]

So uh

I described a specific issue I had in #precalculus and I dont think I need to write it all out again

[gusty crag]

@weary tiger it was something about pi in binary?

[weary tiger]

Yeye

Sequential calculation that can be expressed using relatively simple mathematical functioms

[pliant horizon]

Quick question: What is the codomain of the exponential function? I thought exp was R->R with the range being R^+ but my discrete teacher said differently.

She also said a function must be onto to have an inverse, but then that seems to contradict the existence of the log function

[naive saffron]

Well like you said already, you defined exp as a map from R to R, so the codomain is R by definition

The image of the function, however, is R^+

and also yes a function must be onto to have an inverse

But remember the definition of an inverse function

$g:B\to A$ is the inverse of $f:A\to B$ if $g(f(a))=a$ for all $a\in A$ and $f(g(b))=b$ for all $b\in B$

[vital dew]

Whoever:

[naive saffron]

and $\log$ does not satisfy this definition because $e^{\log x}$ is not equal to $x$ for all $x$, in particular this does not hold for any negative $x$ since it doesn't even make sense

[vital dew]

Whoever:

[naive saffron]

However it is nevertheless a left inverse of $e^x$, since it is the case that $\log(e^x)=x$ for all $x\in\bR$

[vital dew]

Whoever:

[naive saffron]

@pliant horizon

[weary tiger]

i dont understand this example

moreover i dont understand reflexive

[honest barn]

wdym?

the example just says you're looking at things being equal mod 7

and for any number x, x = x mod 7

[merry token]

hey in a proof what does hexibiting mean

the context is

"Disprove the following conjectures by hexibiting a counterexample:"

I just negated the statements and proved that the negation is true

[weary tiger]

maybe a typo for exhibiting?

[merry token]

ohhhhh

that would make sense

in that case negating it and proving that the negation is true should be sufficient right?

[weary tiger]

yeah, just prove that the conjecture doesnt hold

[honest barn]

Lmfaooooo

that's insanely funny

hexibiting

I'd have been so confused too

[merry token]

yeah threw me for a loop lmao

[pliant horizon]

I think this is a relatively stupid question but I was wondering if someone could explain why exactly this proof works.
It's to prove that for a real number x, if x^3 is irrational then x is irrational. It's pretty easy to prove by contrapositive (if x is rational then x^3 is rational) but I don't see exactly how that proves the original proposition. I also get that the contrapositive is equivalent but how this particular contrapositive is equivalent is not clicking for me. Maybe I messed up the contrapositive though

also i meant to respond to you earlier @naive saffron but I got distracted. What you said makes sense, and I had already thought that the log was a pseudo inverse of some kind since not all the conditions were met. I'm glad that my intuition was mostly correct. thank you!

[naive saffron]

You're welcome

And you can think of your contrapositive proof in this way

Suppose for contradiction x^3 is irrational and x is rational, then x^3 is rational since x is rational, which is a contradiction

So x can't possibly be rational

@pliant horizon

[pliant horizon]

oh

yeah that makes perfect sense

[naive saffron]

sorry for some reason made so many stupid typos

very tired

[pliant horizon]

oh np i didnt even notice

[naive saffron]

You can always think of contrapositive proofs this way

[pliant horizon]

yeah thats perfect thank you so much

❤️

[glossy pollen]

Hello #discrete-math,

I am seeking some help understanding the function:

The question at hand is to rewrite this functions so that it has less error when being estimated by a computer

The book gives this explanation:

I have to replicate this process for class and do not understand how it can be justified that the new function is better than the old function beyond just testing cases

[stray reef]

the reason you're getting loss of precision is because you're subtracting two large numbers that are close by

@glossy pollen

[glossy pollen]

so subtraction is specifically the issue?

what if the expression was as follows:

where y is equal to sqrt(x) but could be negative or positive.

could we reliably reduce the error not knowing if y is being added or subtracted?

[stray reef]

where y is equal to sqrt(x) but could be negative or positive.
what

[glossy pollen]

i know i know

[stray reef]

make up your mind, is y a separate input or do you have like a two-valued function

what do you even mean by "not knowing if y is being added or subtracted"

[glossy pollen]

if y is negative then -(-y) = +y

[stray reef]

...

[glossy pollen]

i might not really have the language to ask what im thinking

[stray reef]

are you saying you have a function of the form $f(x) = x(\sqrt{x+1} \pm \sqrt{x})$ but it's given to us as a blackbox or what

[vital dew]

Ann:

[stray reef]

so like all we know is that either there's a plus there or a minus but we don't know which

or do we know even less

[glossy pollen]

yea i suppose so

[stray reef]

then evaluate f(1) or something and check if it gives you sqrt(2)-1 or sqrt(2)+1

and act accordingly

[glossy pollen]

haha yea i guess im aksing something absurd

nevermind

[stray reef]

you are

[glossy pollen]

thanks for the answer

[pale epoch]

because implication is defined that way

$(p \implies q) \equiv (\neg p \lor q)$

[vital dew]

Lochverstärker:

[pale epoch]

and when p is false, not p is true, so you get that

it even has a fancy name

'principle of explosion'

[weary tiger]

where all discrete math is used in computer science?(excluding graphs)

[pulsar timber]

trees

[weary tiger]

anything else?

[pale epoch]

what do you consider discrete math?

finite state machines are discrete in nature

and a very big part of computer science

so are pushdown automata

arithmetic in finite fields is used in cryptography

e.g. diffie hellmann is based on the theory of quadratic residues (number theory)

other crytographic protocols are based on finding discrete logs in finite fields

which is also number theory, but as the name suggests it deals with discrete structures

[weary tiger]

thanks for the answer

[glossy pollen]

It depends more on what you consider to be computer science imo

If you think computer science is making a website with prebuilt libraries that hold your hand the whole way then discrete math has not much to do with that.

If your doing basically anything else, then the finiteness of the bits matter, and using those finite bits to produce more and more elegant information is the business of computer science.

[neat holly]

Quick question: Can someone guide me through what this looks like and why?

[unreal stump]

A consists of all pairs (x,x),such that x is a rational number lying betweem 0 and 1(both included)

B consists of all pairs (x,1-x) such that x is a irrational number in the same interval

[neat holly]

To my understanding, they form an X (that is, y = x, y = 1 - x) and both A and B never intersect. Would I be correct in that assessment?

[unreal stump]

Yes

Because a rational number is never an irrational number

[neat holly]

Oh, I see! The format of these problems sometimes still gets me.

B could have something like (pi, 1 - sqrt(2)/2), I assume.

[weary tiger]

can anyone give me a hint ?

[pale epoch]

consider that division by 0 is undefined

[weary tiger]

hmm okay thank you

so the last statement, (let a and be equal 1) cannot be true because otherwise we'd be dividing by 0 in the previous step (divide by a - b)?

[errant bear]

yes

[weary tiger]

thank you 🙂

[errant bear]

but do you know why you cant divide by 0

[weary tiger]

because it is undefined behaviour?

[errant bear]

what do you mean

[weary tiger]

not quite sure, other than that i was told that its undefined

[errant bear]

sadge

[gusty crag]

@weary tiger are a and b supposed to be rational numbers, real numbers, or something else

[weary tiger]

doesnt say

thats the entire problem

but im satisfied with my conclusion

[pale epoch]

its fine, it doesnt matter what numbers a and b are

if you want to know, why division by 0 is undefined, its because there is no number x, such that 0*x = 1

as 0*x = 0 for all x and 1 is different from 0

[gusty crag]

it does if a and b come from a structure which has zero divisors

[pale epoch]

no

[gusty crag]

you're right

I was thinking about something else

[pale epoch]

division by 0 is undefined in any ring

even if there are zero divisors

unless maybe you consider the zero ring, but that is madness

[gusty crag]

I thrive on madness

[midnight plover]

My teacher got the opp sign

but my distributive is diff than his so what did i do wrong

[haughty garden]

Both ways won’t work because they mean different things.

Distributive law is: $a \land (b \lor c) \equiv (a \land b) \lor (a \land c)$

[vital dew]

opengangs:

[haughty garden]

It’s basically saying “a and EITHER b or c must true” which is the same as saying EITHER (a and b) OR (a and c) must be true

[spring mica]

Why does line 1 have ^

and line 4 and 5 have ->

i thought line 1 would also use ->

<@&286206848099549185>

[viral patio]

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.

[woeful jewel]

@spring mica you should learn your defintions again.

[gritty crescent]

@spring mica The premise is the statement: There exists a student x who is in the class AND has not read the book.

[spring mica]

Then why does line 4 have a -> @gritty crescent

shouldn't that also have an AND

which ones @woeful jewel

have any good resources for me

[brave lava]

https://images-ext-2.discordapp.net/external/rFeDd5pQgYqLD9qoqJ3c7OTNCpaUckoa-VY0rDsrEVE/https/i.imgur.com/A7aIGAU.png

yo can anyone help with these

[brave lava]

<@&286206848099549185>

[near root]

True or false:
A = {1,2,3,4} B={1,2,3,4,5,6} number of functions f:B->A that fulfill |f[A]| = |A| is 4^3 * 3!
Could someone please explain to me this?

[stray reef]

what is f[A] here

[near root]

Pretty sure it's the img?

[pale epoch]

why just pretty sure?

[near root]

Last time I meddled with Ann I was called out

I got called out*

[pale epoch]

they just want to know what f[A] is

[stray reef]

^

[near root]

Isn't |f[A]| just A^2?

[pale epoch]

[near root]

Since that's all the possible functions so...

[pale epoch]

what

[near root]

What do you mean what

[pale epoch]

i want you to explain more

|f[A]| is probably some number

A^2 is not

[near root]

|A^2| is 256

[stray reef]

[pale epoch]

lets start at the beginning

[gritty crescent]

As far as I can tell, |f[A]|=|A| means all such functions f:B->A such that f is surjective/injective?

[pale epoch]

what is f[A]?

[near root]

Img of A

[pale epoch]

ok

what is |A|?

[near root]

Number of elements inside of A

[pale epoch]

which is?

[near root]

4

[pale epoch]

so you want that the image of f has cardinality 4

as f is a function B -> A, that means that f must hit every element in A

or in other words f is surjective

so the task is to count surjective functions B -> A

did you cover stirling numbers?

[near root]

Uhh lemme check

Yep

Glad I know how it's called in English now

[pale epoch]

then my hint is to use that

i.e. figure out how stirling numbers relate to surjections

[stray reef]

A = {1,2,3,4} B={1,2,3,4,5,6} number of functions f:B->A that fulfill |f[A]| = |A|

A is a subset of B so technically you just need f restricted to A to be a bijection

taken literally

[pale epoch]

f(5)=1, f(6)=2, f(1)=3, f(2)=4, f(3)=f(4)=4 satisfies the condition, but f restricted to A is not a bijection?

[stray reef]

does it?

wait hold up

[pale epoch]

image of f is all of A

[stray reef]

f[{1,2,3,4}] = {3,4}

it says f[A] not f[B]

[pale epoch]

oh

i cannot read

disregard everything i said lmao

ann is right

[near root]

Uhh

What

F can't be bijection though since |B| > |A| or am I too pepega to understand

[pale epoch]

hence

f restricted to A

[near root]

So can I just calculated the total amount of functions and substract the bijections and multiply them by the remaining numbers that remain in B in the power of the times they appear?

[pale epoch]

what

the total amount of functions between what

subtract the bijections between what

[near root]

so the task is to count surjective functions B -> A
@pale epoch
For the record is Stirling's numbers the n, c, p or d?

[pale epoch]

i told u to disregard what i said, i misread the question

A is a subset of B so technically you just need f restricted to A to be a bijection
this is the real hint

[near root]

How do I restrict it tho?

[pale epoch]

https://en.wikipedia.org/wiki/Restriction_(mathematics)#Formal_definition

[near root]

Right so I made the domain smaller to be equivalent to A, now I get a bijection function, the possibilities for that is 4! am I still missing something?

[pale epoch]

yes

what to map the other elements of B to

[near root]

So b has remaining 2 numbers each that can go to the remaining numbers

Therefore it's 4 in the power of 2 I think times the 4!?

[pale epoch]

ye

[near root]

Eyyy

[pale epoch]

that's your given solution as well

[near root]

You take paypal?

[pale epoch]

what?

[near root]

For helping me

[pale epoch]

money? nah, im good

[brave lava]

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

can anybody please help me

ive been stuck on this

[pale epoch]

this should just be reading your script

[brave lava]

i watched his lecture and it didnt help

other wise i wouldnt be asking

[pale epoch]

what symbols from arithmetic are used in boolean algebra?

[brave lava]

addition

multiplication

subtraction

[pale epoch]

ok so +, * and -

anything else?

[brave lava]

nope

[pale epoch]

no numbers?

also what is "dot" that you filled into first blank?

[brave lava]

the symbol

of multiplication

he kept repeating it in his lecture so it was a guess lol

[pale epoch]

it's not an element of the set though

[brave lava]

nope

[unreal stump]

Subtraction?

[pale epoch]

you are looking for a symbol x, such that a + x = a

for every a that is in the set

[brave lava]

https://i.imgur.com/z6uHJbY.png

@pale epoch

[pale epoch]

the identity of multiplication is not dot

[brave lava]

lol

i dont know what else to put

[pale epoch]

whats the identity of multiplication?

for what x is a*x = a for all a

[brave lava]

1 . a = a

[pale epoch]

ye

[brave lava]

so third box is 1.a = a

lol?

wtf

[pale epoch]

no

it's 1

thats the multplicative identity

in boolean algebra, we use 1 to represent "true", 0 for "false"

multiplication is "and" and addition is "or"

[brave lava]

yeah

[pale epoch]

then you get stuff like A or FALSE = A as A+0=A

and well, we reuse 4 symbols

multiplication dot, addition +, 0 and 1

the first 2 blanks are about the "additive symbols", the last 2 blanks about the "multiplicative symbols"

[brave lava]

i have this right now

[pale epoch]

should be fine 🤷

[brave lava]

last one was wrong

maybe i have to put