discrete-math

[deep rose]

umm

idk if this question goes here

but

how do I define a non-existing metric space?

"Give or define a new non-existing metric or metric space for a specific purpose. This does not restrict it to only a mathematical formulation. Rather, you would imagine or create a new "distance" in your real-life."

[stray reef]

a what?

[deep rose]

this was basically the question, but idk what it even means

[stray reef]

what the actual fuck

[deep rose]

even I have no idea

the professor's lecture didn't even include anything about this topic but this was his report

[bleak void]

I think they just want you to define a metric space.

[deep rose]

so like

i could say, all positive real numbers or some stupid thing like that?

[bleak void]

As long as it's a metric space, sure.

[deep rose]

wait how do I know what is a metric space?

[bleak void]

A metric space is a set S together with a distance function (metric) of type S x S -> R obeying three properties.

Just google it for more details.

[deep rose]

hmm alright

[stray reef]

i mean the "non-existing" part is lie

like*

mega sus

[deep rose]

can there be a metric space to define all existing possible triangles on a 2D plane?

[bleak void]

I'm not sure what you mean.

[deep rose]

like how 3d euclidean space is a metric space

[bleak void]

I suggest not making it too complicated.

[deep rose]

hmm

[bleak void]

The euclidean metric is a well-known metric.

There are versions of it for each dimension n.

[deep rose]

these two were my questions and for the first one I wrote about Euclidean, Leveshtein and Wassestein metric spaces

but for the second question i have nothing to write about cuz i have no idea how to define a metric space

[bleak void]

Pick your favorite set and try to define some notion of distance between its elements.

[dense glade]

if every element in the inside function p is mapped to at at least one element in the domain then q will map every element from from f(p) to one element in the domain of q is this correct or?

[hollow river]

What is f(p)?

Your functions are p and q, so it is important that you stick to those

[simple vale]

Why (7^4)^55 turns into 1^55?

[proven silo]

Because 7^2=9 mod 10

[proven silo]

And 9*9=81 which is 1 mod 10

So 7^4=9 * 9=1 mod 10

[simple vale]

I didn't quite understand. 7^2=9? Why there is 9*9? XD

[light heath]

what do you guys think of my answer to number 44

[coral parcel]

I would say it's because x^4 == 1 (mod 10) whenever x is coprime to 10 -- due to Euler's theorem.

(The totient of 10 is 4).

[neat notch]

would anyone know how to find out the number of ways 12 tasks can be distributed among 6 different people?

[coral parcel]

Do they have to do exactly 2 tasks each?

[neat notch]

for the first question no. But for the second part of the question, they do have to be 2 task each

[coral parcel]

If there's no such restriction, just assign each task independently from all the others and count the total number of all those choices.

With the restriction, I'd start by assigning the tasks as "Alice's first task", "Alice's second task", "Bob's first task", "Bob's second task", "Claire's first task" ... and then afterwards correct for overcounting because nobody cares about which sequence their two tasks were picked in. (If you know multinomial coefficients, that's what this method produces).

[neat notch]

okay thank you

[dense glade]

How many binary operations are there if there is there is an identity element

[coral parcel]

Do you mean something like "for a set of such-and-such size, how many binary operators on the set are there which have an identity element?"

[dense glade]

yes

[coral parcel]

First choose which of the elements in your set will be your identity element. Then choose what your operator will do on each possible input that doesn't have your chosen identity element as one of the two operands. Since a binary operator cannot have more than one (two-sided) identity element, each of the operations you want to count arises in exactly one way by these choices.

[dense glade]

i see

[dense glade]

how do you know

if

a graph has a subgraph isomorphic to k_3,3 or k5

[coral parcel]

Huh, that doesn't seem to have anything to do with counting operators with identity elements.

[dense glade]

no no

LOL

thats a different question

i got your point in the first one

and I was preparing for my final its in 2 hours

so I thought I would ask because I can't tell if

a graph is planar or not

[coral parcel]

Note that "has a subgraph isomorphic to k_3,3 or k5" is kind of a corruption of either Wagner's or Kuratowski's theorem. Neither of the are as simple as looking for a subgraph. One asks wheter one of the forbidden graphs is a minor of the graph, the other whether a subdivision of one of the forbidden graphs appear as a subgraph.

There are algorithms for looking for these configurations, but in an exam setting you shouldn't be aiming to remember and carry out a particular algorithm; graphs will usually be small enough that you should be able to wing it just by looking at the graph semi-systematically.

And if there's 2 hours to the exam, you should now be relaxing your mind instead of attempting any last-minute cramming, anyway.

[dense glade]

Yeaaaah

way too much topics in this class

getting tested on like 17 topics LOL

my brain hurts

[bright moat]

i have this but my teacher said i messed up on the red part

she said i forgot to add the 2k+1 so i did add it to 3k+3

but after that im not sure what to do

[deep rose]

Alright a super interesting question here, I do not know if it belongs in this topic but

For a sequence such as 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, ......

how do I find sum of n terms

I know the nth term can easily be found by round(sqrt(2*n))

but how do I find sum up till nth term?

there has to be a formula to simplify it right

[sturdy walrus]

Hi guys, I'm trying to prove this but I feel like I am unable to provide a more thorough proof? Any tips?

[robust mango]

@sturdy walrus Have u proved that the composition of two surjective functions is surjective? If so, use induction, it should be fairly simple

[fallow zodiac]

Can I get a little help on one of my homework problems? I'm really not sure what I'm doing wrong. I think I'm really close and I just made some small stupid mistake along the way

[stray reef]

,calc -6 * 9 + 9 * 1 * 9

[vital dew]

Result:

27

[stray reef]

@fallow zodiac sounds like a sign error for a_1

[sturdy walrus]

Thank you, this helped a lot. 👍🏽

[fallow zodiac]

I changed the equation to fit -27 so it's 3 instead of 9 now so, -6*(9)^n + 3n(9)^n. but it's still wrong

[stray reef]

oh wait hold on

what was your characteristic equation?

[unreal stump]

(x+9)^2=0 from this

[fallow zodiac]

my characteristic equation was T=9. Cause I did T^2 = -18T -81. Then added and factored and got (T-9)(T-9)

[stray reef]

T^2 + 18T + 81 does not factor to that though

[fallow zodiac]

oh shit I see where I messed up thank you

[zenith oyster]

@coral parcel What do you think of this as a solution?

[coral parcel]

Looks generally right.
The description of what M' does (especially how it manages to write 1 to all the odd cells after it starts that phase in an unknown location on the tape and needs to write in both directions) is slightly handwavy -- I recognize it as text I wrote, which I deliberately made handwavy such that it wouldn't be directly reusable as a homework solution.
In the second-to-last line, return MemoryWatch(M') should probably be something like return MemoryWatch(M',1,1).

[zenith oyster]

Sorry for that. I mean as far as I understand the logic is that if it goes to a former accept state we know that's in an even cell, therefore returning to square 1 would only be to first go one left (then we're in odd) and if that's blank (which it is) it writes 1 and goes in a state that writes 1 and moves one left. If the next square has something in it then it just shifts left... repeat... is that something I should explain in my solution? why would it need to write ones to the right of the accept state?

or wait maybe it can go to the left instead, didn't think about that

[coral parcel]

Yeah. This is not hard to fix up (especially since you don't need to care that the resulting machine is slow), but I will leave it to you to puzzle out the details.

[zenith oyster]

What about this, if it goes into the loop to write into every odd cell. It first make a one jump either left or right then it bounces 2 in the other direction and writes 1 in both, then it will jump back 2 to the right and if there is already a 1 in it it jumps 2 again then repeats this process so it fills both directions, or maybe that's impossible

[coral parcel]

That's not a particularly clear description -- but it could mean something that will work.

Reading it very closely it seems to be likely that it means something that will work, but you have some work to do making it more explicit. Pseudocode will be useful.

[zenith oyster]

Ok thanks

[worldly knot]

my brain hurts.. how is it true? new lesson

please ping me if u can help 🙂 thanks!

[coral parcel]

How is it not true? Note the "nonneg" in the specification of the domain.

[vital dew]

Anton.

[red nest]

This is bounded above by the 1 + sum of k^{i}/k^n from i=1 to n-1, and below by 1 + the sum of -k^{i}/k^n ,from i=1 to n-1. Both of these have limit 1 as n goes to infinity

[red nest]

Apologies, I think it would be better for my i to range over 1 to n/2 instead of n-1 to get the result properly

[torn escarp]

i'm trying to learn a very fundamental & proof oriented view on solving linear homogeneous recurrence relations with constant coefficients and why the method of the characteristic polynomial actually works. Does anyone have any recommendations on where I could find this type of introduction to recurrence relations?

[quaint notch]

Not sure this is the correct place, but could anyone please explain me why those 2 things are same:

1. The choice axiom.
2. for each f: A->B onto, exist g:B->A one to one such as fg = id(B)

[bleak void]

Well, to define g : B -> A, you need to decide where to send b in B in A. This involves choice.

In the other direction, the g gives you a choice function.

[simple vale]

I have this track and my main issue is the solution of that subtraction (i don't even know where to start) while i think i got the order thing (i think it's 5), can someone explain to me the procedure to solve the subtraction and if the order of [6]_9 is correct?

[pale epoch]

how did you calculate the order?

[simple vale]

I have repeated the binary operation on 6 until i got 0 (that is the identity of +). I got 0 with 6^5

So i think that the order is 5

[pale epoch]

but the binary operation is +

if you did 6^5, you did multiplication

[simple vale]

No, i've done something like 6^5 = 6+6+6+6+6

[pale epoch]

how is that 0

[simple vale]

Ok, so, the binary operation it's a sort of clockwise operation and with so every time we get past 9 we start from 0. So 6+6 should be 12 but in this case it's 2

I've applied the same logic to 6+6+6+6+6

[pale epoch]

oh

you reduced mod 10, not mod 9 then

[simple vale]

I guess so

[pale epoch]

in Z_9 you should probably reduce mod 9

[simple vale]

Mh, how can i do that?

[pale epoch]

you do what you said

except once you get past 8 you start from 0

you should probably review the definition of Z_n

[simple vale]

Yeah, probably

Have you got any resource that you can share me? Like a website that explains it?

[pale epoch]

the keyword you want to search for is "modular arithmetic"

maybe khanacademy but i dont know

[simple vale]

Ok, thanks

Ok, saw that. I don't know how i missed that, it's in the definition of modular arithmetic itself...

This changes the cards on the table, now i've got two orders (3 and 8, i've calculated them with the same method I showed earlier)

[pale epoch]

3 works, 8 does not

3, 6, 9, ... work
the order is defined to be the least of those

[simple vale]

Mh, ok

Thanks a lot

So with this in mind i guess that [2]_9 - [7]_9 is just [5]_9

[pale epoch]

you guess wrong

-5 is not 5 mod 9

[simple vale]

So the "clockwise" thing doesn't matter here

[pale epoch]

it does, but going 5 backwards is different than going 5 forward

also i dont know if this is the best way to think about it

you should just learn how to handle this algebraically

[simple vale]

Mh

[quaint notch]

not sure what you mean

for other direction

[balmy jay]

Say we're throwing $n$ balls into $n$ bins. Let $X$ be the random variable that keeps track of how many balls there are in the first bin. Let $Y$ be the random variable that keeps track of how many balls there are in the second bin. If I were to try to compute $P(X = i, Y = j)$ (the joint distribution of the 2 random variables), would it be valid to compute (total number of ways to distribute $n$ balls in $n$ bins such that there are $i$ balls in the first bin and $j$ balls in the second bin)/(total number of ways to distribute $n$ balls in $n$ bins)? If that is right, how would I go about doing this?

[vital dew]

math man

[balmy jay]

Thanks in advance! 😄

^^ for reference, I got $\frac{{n \choose{i}}{n - i \choose{j}}{n - i - j + n - 2 + 1 \choose{n - i - j}}}{{n + n - 1 \choose{n}}}$ and all the balls are uniformly distributed

[vital dew]

math man

[wide vine]

. "Maximal" means that if any point were added to the region, it would no longer be continuous. In the diagram below, the blue area is not continuous because it is impossible to travel from one part to another without crossing an edge (which is not part of the region). The purple area is not maximal because it can be expanded to include the green area and still remain continuous. I don't get what they mean by maximal really :/

and by "point"

Here is their example

but I don't get the "maximal" part

i understand the continuous part because because you would have to cross through the edge to get to the other blue area.

[wide vine]

ohh nvm

they are saying the coloring region (green) is redundant and that it you can expand the purple region.

[old delta]

i found c = 175, d = 50 to be a solution, but i have no idea what to do now

[austere cedar]

Suppose you have two solutions (c,d) and (c', d'). Then 7(c'-c)-24(d'-d)=0. So you can look for all integer solutions of 7x-24y=0 (with x=c'-c, y=d'-d), and for each such solution (x,y), (c'=x+175, d'=y+50) is a solution for the original question.

@old delta

[simple vale]

The same thing applies when i search for the group generator right?

[main orchid]

Do any of you have any advice to prove these quite complicated combinatorial identities when the identity is given to you

I don't really know how to approach it effectively and I always end up fudging my logic to "explain" why the identity works

especially when the identity is increasingly complex

[deep rose]

does anyone have the formula for sum of roots up to n

[weary tiger]

if ur lookin for something as simple as n(n+1)/2 , then no , if not , then there are nice approximations

there r answers on stackexchange

pick ur fav

[torn escarp]

bumping this ^

[weary tiger]

There's literally a book by Springer called Axiomatic Set Theory : }

[wide vine]

ω(G) ≤ Χ(G) ≤Δ(G) + 1 this would be considered the bounds for vertex graph coloring, correct? ω(G) refers to the largest complete subgraph contained in G and Δ(G) refers to the largest degree vertex of G

Was just looking for what is the lower bound and upper bound (vertex coloring) for any Graph G

undirected graph

want to ensure what I have is correct

[coral parcel]

Those bounds definitely hold.

[wide vine]

Let T be a tree and let u and v be two vertices in T. There is exactly one path between u and v. ```
is this true because trees don't have cycles?

Because based on a bit of drawing, the only way I could think of a tree having an alternative path would there to be a cycle within the tree

and then that would be a contradiction and would not be a tree

[unreal stump]

Yea

If there were multiple paths you have a cycle

[stray reef]

two paths in a tree
up one and down the other
a cycle appears

[balmy jay]

Nice haiku

[stray reef]

ty

[spring surge]

Division operation x/a answers the question how many times i iterate +(a,a) to get x.
log operation give the iteration count on multiplication.

If I define the custom function of my own and ask how many times I self iterate it to reach my number. What branch of maths should I look into?

[pale epoch]

there isnt a field for this

or you would need to be a lot more specific to get a useful answer

why did you define this operation? why do you need an inverse? what properties should it have etc

[spring surge]

Got it. Thanks.

[pale epoch]

if your functions are sufficiently nice, its maybe some part of analysis

like the question whether logarithms exist and whether they are nice is answered in an intro analysis class

[spring surge]

a (x) b = 0.5(a+b); I want to iterate 2 with itself to reach 1.5.

In bruteforce manner I arrive that if I repeat it 2 times I will get 1.5.

Yeah, I feel its mathematical analysis question, maybe some geometric aproach too.

[pale epoch]

hm?

is (x) supposed to be your operation?

[spring surge]

yes

[stray reef]

you're iterating 2 with itself?

$2 \otimes 2 = 2$ according to your definition.

[vital dew]

Ann

[stray reef]

one would think you in fact never reach 1.5 no matter how many times you repeat your operation.

[spring surge]

My mistake let me fix it.

I select the zero as starting point. (0+2)0.5 = 1 then (1+2)0.5 = 1.5.

[weary tiger]

There is a sequence of 100 integers. Show that you can choose some (non-zero) number of consecutive terms such that their sum is divisible by 99. [Hint: consider a, a+b, a+b+c, ….]

wait, cant i just choose 99 and be done with it

since that is divisible by 99

[stray reef]

what do you mean by "choose 99"

do you mean that you wish to choose the first 99 integers in your sequence?

[weary tiger]

no

i can choose 1 integer

[stray reef]

the number 99 need not appear anywhere in your sequence.

[weary tiger]

oh right

but a multiple of 99 will

[stray reef]

no, not necessarily that either

[weary tiger]

really?

[stray reef]

yes

all you have is a sequence of 100 integers

[weary tiger]

ok

[stray reef]

a sequence $(a_1, a_2, \dots, a_{100})$ with $a_i \in \bZ$

[vital dew]

Ann

[weary tiger]

and a consecutive sequence where the sum mod 99 is 0

do you think their hint helps? [Hint: consider a, a+b, a+b+c, ….]

[stray reef]

and a consecutive sub-sequence where the sum mod 99 is 0 is the thing whose existence you are to show.

it is a goal and not a given.

[weary tiger]

not really sure how to use it

[coral parcel]

Divide each by 99 and pigeonhole the remainders.

[stray reef]

first you would have to pin down what a, b and c in the hint refer to

and also pin down just what you're taking mod 99

[coral parcel]

Ah, but there has to be something for Crab to do too.

[weary tiger]

99 possible remainders

right

so how many remainders sum to 99?

[weary tiger]

hm

[weary tiger]

they refer to consecutive numbers

[coral parcel]

Perhaps the "consecutive sequence" is what's troubling you? Do we agree that you're given arbitrary integers $a_1, a_2, \ldots a_{100}$ and you're supposed to prove that there exist $n \le m$ such that $99 \mid \sum_{k=n}^m a_k$?

[weary tiger]

yes

[vital dew]

Troposphere

[coral parcel]

So there are how many sums of the form $\sum_{k=1}^m a_k$? And how many possible remainders of these sums?

[vital dew]

Troposphere

[stray reef]

||bar bhtug gb ybbx ng gur frdhrapr bs cnegvny fhzf.||

[coral parcel]

That's some pro-level encryption.

[stray reef]

no it's just rot13

[weary tiger]

rot 13?

[coral parcel]

Rot13 plus spoiler markup.

[weary tiger]

tru

[stray reef]

https://en.wikipedia.org/wiki/ROT13

[muted gate]

Suppose that a relation $R$ is reflexive, show that $R^{*}$ is also reflexive

[vital dew]

texaspb

[coral parcel]

What does R* mean in that context?

[weary tiger]

there are 99 possible sums and 99 possible remainders?

[muted gate]

can somebody help me prove this? I think it just follows by definition, since $aRa \land aRa \implies aR^{2}a$, therefore $aR^{*}a$ but that doesn't seem that correct for me.

[coral parcel]

Why do you only get 99 possible sums?

[vital dew]

texaspb

[muted gate]

it means $R^{*} = \bigcup_{n = 1}^{\infty} R^{n}$

[vital dew]

texaspb

[muted gate]

also sorry to interrupt the ongoing convo

[coral parcel]

So in particular R subset R*.

[muted gate]

yeah

[coral parcel]

Which in itself preserves reflexivity.

[muted gate]

so it just follows by definition? I don't know what else could I be missing here

[muted gate]

in particular because of this

[coral parcel]

It looks like your only miss is that you're overthinking it.

[muted gate]

haha

yeah probably

can I just say that this is true $aRa \implies aR^{}a$ because $R \subseteq R^{}$

[vital dew]

texaspb

[stray reef]

yes you can

and in fact should

[muted gate]

alright!

[stray reef]

anyway,
crab lambda's problem requires a nontrivial insight which i am not sure how to transmit the intuition for

[weary tiger]

,oh maybe not

will each sum have 99 possible remainders

[coral parcel]

Huh? A number has only one remainder, no matter whether you got it by summing something.
(Oops, sorry for the misping, that was a reply to Crab).

[weary tiger]

so a has 99 posslbe remainders

and so does a+b and a+b+c and so on

i think

[coral parcel]

What?

The number a has just one remainder.

How do you imagine it can have 99 of them?

[weary tiger]

but there are many possible number you can put

a

can be any number from the sequence

idk

ok, one possible remainder from 99

[coral parcel]

And how many initial sums are there?

[weary tiger]

100

[coral parcel]

Then think of pigeons.

[weary tiger]

two will have the same remainder mod 99

[coral parcel]

Yes.

[weary tiger]

its not like you can subtract them.

[coral parcel]

Why not?

[weary tiger]

then are they consecutive sums anymore

[coral parcel]

Suppose you have (a1+a2+...+a57)-(a1+a2+...+a19).

[weary tiger]

im confused

hmm i guess you have a consecutive sum

i see it now

thanks for your help

[stray reef]

TIL subtraction is a strictly partial function on Z×Z

[weary tiger]

wah

[coral parcel]

Plus-ultra-finitism.

[weary tiger]

im not that familier with ultrafinitism

[coral parcel]

Good.

[weary tiger]

its something to do with induction being false

or i think thats a part of it

[stray reef]

don't mind me, i'm memeing

[weary tiger]

im interested in set theory tho

i do not, i'm sorry

u can always still get it from places, if u know wht i mean 🙄

[stray reef]

i can explain my joke if you want

[weary tiger]

Please do

[stray reef]

the joke is that your statement of "it's not like you can just subtract them" makes it sound like subtracting two integers may sometimes be impossible (i.e. lead to an undefined result)

hence my comment that subtraction is only a partial, and not a total, function from Z×Z to Z

[coral parcel]

And I continued the joke by alluding to how (some?) ultrafinitists deny that exponentiation is a total function on pairs of positive integers.

[weary tiger]

What a partial and total function?

[stray reef]

do you know what a function is

the formal defn of a function

[weary tiger]

I’m not going to try and guess so no

[stray reef]

damn ok

i was going to say that a partial function is like a function but without the requirement that it be defined at every input

and that when talking about partial functions, those functions which are defined at every input are called total functions to distinguish them

[weary tiger]

i see

[weary tiger]

number of way to distribute 80 balls into 5 distingueable bins such that in any bin there will be no more then 24

[muted gate]

hey guys in the solutions he says that (a) is an equivalence relation (reflexive, symmetric and transitive). I don't get why this is the case! all the other alternatives I understood the reasoning behind them

solution

[weary tiger]

reflexive because xRx for every x, symmetric because only elements are xRx and transitive is also true (the only case with xRy and yRz is when x=y=z)

[muted gate]

ooooh

it's also an equivalence relation because it is of the form (a, b) \in R| a = b

right?

ok!!

got it, thank you

[unreal stump]

I mean equality is a weak form of equality

[muted gate]

I don't understand why (c) is not transitive or symmetric

like I just don't see this reasoning of If $f(x) - g(x) = 1 \implies g(x) - f(x) = -1$???

[vital dew]

texaspb

[muted gate]

how does this make sense

$f(x) - g(x) = 1$ and $g(x) - h(x) = 1$ then, $f(x) - h(x) = 2$ \
I guess in this case he's just doing: \
$(f(x) - g(x)) + (g(x) - h(x)) = 2$ and then somehow extended this to $f(x) - h(x)$??

[vital dew]

texaspb

[meager prairie]

cant you just multiply through by -1

if $f-g=1 \to f>g$ then there's no way that $g \sim f$ by $c$ right

[vital dew]

jan Niku (@pomodoro role)

[muted gate]

oh I guess you can

how did you arrive at $f> g$?

[vital dew]

texaspb

[muted gate]

but yeah if you multiply both sides by (-1) then it holds

now I see it

still don't understand the non transitive part tho

[meager prairie]

yea, i need to mess w it for a sec

[meager prairie]

assume f-g=1 and g-h=1

add these equations directly

you will obtain f-h=2

eg f is not related to 4

to h*

[meager prairie]

if f-g=1 for all integers then f>g seems reasonable hopefully

it gives f-1=g on that domain, so like a literal one translation down gets you g from f

[muted gate]

ok, makes sense

because the relation is for any functions f, g, f - g = 1

obviously f - h = 2 doesn't hold

I appreciate the help

[meager prairie]

[shadow geyser]

Hello there

So if ZAN is this set

I know that the rationals are countable but is this affected by the multiplication of pi as an irrational?

Or does this still remain a countable set despite b(pi)

[civic horizon]

Can you find a bijection from ZAN to Q^2

[shadow geyser]

Q^2?

And shouldn't we be finding a bijection from N to ZAN

@civic horizon

[civic horizon]

Have you seen the proofs that there is a bijection from Q to N and one from Q^2 to Q

[shadow geyser]

No

[proud zinc]

If I have a permutation like (a_1,...,a_n,b_1,...,b_m) how many switches do I need to do to change it to (b_1,...,b_m,a_1,...,a_n)?

Is there an easy formula for this?

[agile magnet]

have u tried to compute any examples?

for example m = 2 and n = 3?

[proud zinc]

Ahhh I got it now

thanks!

[subtle ermine]

I'm struggling with this problem
I know that N must be congruent to 0 mod 3^k, since 2^k must be a factor of N. But I'm unsure how to deal with 3^k as the modulo

I could do N = 0 + (3^k) * M but not sure if this really provides anything meaningful

[elder berry]

Think smaller, don't worry about modulos, but rather about divisibility of N with 3

Since this is from the AHSME, answers are also provided, which can give you a hint

As a further hint, if the value of 3^k seems difficult to find, try to see if ||you can show that the number k can't be big, or rather greater than 1 for this problem||

[subtle ermine]

Ah alr, I saw the solution, but unsure. I was thinking about this in regards to even/odd outcomes but that doesn't help to solve it
I'm unsure how I can show that N Is indeed divisible by say 3, and then not divisible by 3^k where k > 1

Mostly because N is so large so I can't do any arithmetic I think

[elder berry]

Indeed, that's why in a setting for this exam, students naturally think to bound k, or rather to check whether N is divisible by 9 = 3^2, since if not, they would lose a lot of time whereas casework saves much time checking only k=1 and k=0

[subtle ermine]

Ah I see
So I should split into cases
Since k = 0, 1, and then any k > 1 see usually three different cases

And so first I should be trying to find a way to see if this is divisible by 3?

[elder berry]

Yup, see if it is divisible by 3

if not, then k=0 is the answer, if yes, continue to check k=2

[subtle ermine]

Ok
I was reading up Abt divisibity by 3 and read that the sum of all digits must be a multiple of 3

So I'm guessing that will play a role somehow
Or not? It seems like a lot of brute work

[elder berry]

Yup, knowing divisibility of 3 and 9 help a lot here, and indeed at first glance it seems like a big task

that's where problem solving techniques come into play, or a bit* of logical reasoning

either, do the case-work for all the digits (with a nice way of counting)

or just <spoiler for hint> ||sum up all of the numbers from 19 to 92 and check if that number is divisible by 3||

try to problem for a bit, if stuck see the hint and try to convince yourself why it must hold

[subtle ermine]

I think I see what you meant by pattern, aka like we have 10 2s and 10 3s and 10 4s etc all the way to 80

and also 0,1,2,3,4,5,6,7,8,9 repeats

I'm gonna make an expression with this and see if the mod 3 = 0 (the reason I'm thinking of just doing mod 3 is because then I can use smaller congruents of these numbers to make the calculation easier)

[elder berry]

Go for it

[subtle ermine]

It worked!

sDigits mod 3 = 0
sDigits mod 9 = 3

And I think I see why this shows that k can not be greater than 1

Because if K greater than 1

Then N / 3^k = N / (3^2 * 3^(k - 2)) that's allowed I think because we defined K > 1 and so K-2 will be >= 0

And so this would always have N being divided by 9, but since it can't be divided by 9 since that gives a remainder by 3, 3^k can't be a factor!

I was confused at first because I was thinking what if 3^5 can divide N? But that doesn't matter, the point is that 3^2 cannot divide N and for K > 1 you'll always have 3^2 in the denominator

Is that right?

[elder berry]

Brilliant!

Absolutely right

you might as well look at the hint I provided, and also (if you don't know already) you should show why a number divisible by 3 (or 9) must have its sum of digits* also be divisible by 3 (or 9)

[subtle ermine]

Ah I see, gotcha your hint was how I approached it
I'm unsure of how to approach showing the divisibility of 3 rule

Rn what I have is
For some integer x

I'm looking at x mod 10 = a (a is the units, Ik this because I did this in coding lol)
x mod 100 = b , b is the 10s place

Like maybe a + b = 3 m (m is any integer)

And then I use x = a + 10k, x = b + 100c
And maybe try adding or something?
Tbh I'm a bit confused

[elder berry]

So the idea with representing digits as 10^k * d (where d is a digit) is the idea yes

The number then can be represented as 10^n dn + 10^(n-1) d_(n-1) + ... + 10d1 + d0

see if you can simplify this expression modulo 3

hint: ||leave the digits alone, work with 10, 100, ..., 10^n and see what they are equal to modulo 3||

[subtle ermine]

Ok so I was thinking
First trying to make the number simpler
10 mod 3 = 1

10^2 mod 3? well that's just 10 * 10 mod 3 = 1 * 1 mod 3 = 1 mod 3

So I think

All the 10's become 1

(dn + dn-1 + ..... d1 + d0) mod 3

OHHHHHH

So if the original number 10^n dn + 10^(n-1) d_(n-1) + ... + 10d1 + d0 mod 3 = 0

Then (dn + dn-1 + ..... d1 + d0) mod 3 = 0

Because

10^n dn + 10^(n-1) d_(n-1) + ... + 10d1 + d0 is congruent to dn + dn-1 + ..... d1 + d0) mod 3

And with 9

[elder berry]

that is if we assume the number is divisible by 3

[subtle ermine]

10 mod 9 is 1 too

Yep

[elder berry]

In general just 10^n dn + 10^(n-1) d_(n-1) + ... + 10d1 + d0 == dn + dn-1 + ..... d1 + d0 (mod 3)

[subtle ermine]

Alr yeah that makes sense
Very cool!!!!
I only learned modular arithmetic 2 days ago but damn there are so many cool things you can do with it

And just to make sure then, regarding the first problem
Some good take always would be to try splitting large problems into cases (like k = 0,1,2) and try splitting up large problems in general (tbh I just got frightened by the large N so didn't even think to find the sum of diigts at first), and also learn divisibility rules lol

Thank you so much for your help!

[elder berry]

yeah, it really depends on the scope your working with. Timed exams like this one will have tricks like this to simplify your work, and working with the already given answers can further nudge you in the right direction.
N could've been a number so that 3^2022 was its biggest divisor, then more difficult theorems or tricks would be needed in your mathematician toolset.
For these types of problems, the statements seems scary, but the solutions are almost always easy to obtain.
And as a general rule of thumb, you can try out a few cases like k=0,1,2 to get a feel for any problem

Yw

[weary tiger]

in this question

A regular 12-gon ABCDEFGHIJKL is inscribed into a circle of radius 12. [This means that the vertices of this 12-gon lie on a circle of radius 12 and all side-lengths and all angles sizes are the same.] Find the perimeter of the pentagon ACFHK.

what does all side-lengths and all angles sizes are the same. mean exactly

[unreal stump]

I mean if you take any side it will be the same thing

Like a square is a regular 4-gon

An equilateral triangle is a regular 3-gon

[weary tiger]

so this is just a algebra question?

[unreal stump]

Well it's a geometry probelm

[weary tiger]

yh

smth like this

[unreal stump]

Yea

[stray reef]

yeah, this is a geometry problem

[unreal stump]

Yea

[weary tiger]

hm

how can i find the perimiter

lol

[stray reef]

must stress that this is a geometry problem and hence a poor fit for this channel, but

think about angles

maybe it might be helpful to give a name to the center of the dodecagon

[weary tiger]

hm ok

[weary tiger]

how will they help

[stray reef]

they will help you find some side lengths

[weary tiger]

how

can we do one sidelength together?

[stray reef]

sure, which one?

[unreal stump]

Don't you need to know sine rule for this

[weary tiger]

AC?

[stray reef]

not in this case no

[stray reef]

ok sure

do you have a name for the center of the picture? if yes, i'll use your name. if not, i'll give my own name and we will continue anyway

[weary tiger]

O

$O$

[vital dew]

crab λ

[stray reef]

ok, you called it O, great.

consider triangle OAC

tell me what you know about it

[weary tiger]

OA=12

[stray reef]

true

[weary tiger]

thats all i know

[stray reef]

no it's not

you know some more things

[weary tiger]

OB = 12

[stray reef]

B is not a point that's of relevance to us rn

go one further

[weary tiger]

OC=?

[stray reef]

yes, what is OC?

[weary tiger]

idk

hence, the question mark

[stray reef]

O is the center of the circle

[weary tiger]

oh wait

its 12

[stray reef]

OC is 12, yes

[weary tiger]

geez

how did i not see

[stray reef]

there's something else you know about triangle OAC

[weary tiger]

uhh

hm

$n \choose k$

[vital dew]

Jester

[weary tiger]

hoh

[stray reef]

@weary tiger #latex-testing

[weary tiger]

O i C mb

does it have to do with O?

[stray reef]

yes

[weary tiger]

its not a right triangle...

i dont see it

[stray reef]

what is angle AOC?

[weary tiger]

idk

tbh

[stray reef]

it's 2/12 of a full turn

to put it in a way bordering on literal

[weary tiger]

oh yh

[stray reef]

in other words it's 60°

you have an isosceles triangle in which the angle at the apex is 60°

[weary tiger]

2/12 * what

[stray reef]

how many degrees make a full turn?

please tell me you know this.

or admit you don't, to my inevitable disappointment.

@weary tiger ?

[weary tiger]

360

ofc i know that

ah i see

[stray reef]

and yet it took you all of five minutes to recall

[weary tiger]

its 60

got it

[weary tiger]

no i didnt

i left for 5 minutes

[stray reef]

60 degrees yes

[weary tiger]

so we combine these three facts to find AC

AO=12
CO=12
Angle AOC=60

[stray reef]

well, sure, but unless you have access to the law of cosines (which i suspect you do not!) you will not be able to do any number-crunching.

[stray reef]

i was expecting you to conclude from here that the other two angles in triangle AOC, being equal, also measure 60° each, and hence triangle AOC is equilateral.

[weary tiger]

ah i see

[weary tiger]

@stray reef stuck on finding CF

We know that CO is 12, and FO is 12, and angle COF is 4/12*360=120.

this is not an equilateral anymore

[stray reef]

angle COF is not 120°

it is not 4/12 of a full turn, but only 3/12.

[weary tiger]

oh

ok so then CF is the hypotenuse

[stray reef]

COF is a right triangle, yes.

[weary tiger]

huh its not a nice number

[stray reef]

you could have and should have just left it as $12\sqrt{2}$

[vital dew]

Ann

[weary tiger]

ok

is the final perimiter 36+24sqrt2

[stray reef]

parentheses, but yes.

[weary tiger]

ok

thanks

[shadow geyser]

Hi there

I have a question regarding this

So I know that the set of rationals is countable as we were taught about how to go about that in class

But Since pi is an irrational and the multiplication of a rational and irrational results in an irrational and the addition of a rational and irrational results in an irrational

ZAN would technically be a set of irrationals

Would this result in ZAN[x] being uncountable?

Or should I go about it thinking that pi is a constant and since a and b are the only changing coefficients

We can still consider these coefficients as countable such that ZAN[x] is countable?

I'm not sure which way to go about this exactly

[placid mason]

Consider the set {sqrt(2)}. This is a set that only contains irrationals, but it's countable (it's size 1).

What's in the set doesn't affect the cardinality of the set, I think you're right to say ZAN is countable

[shadow geyser]

hm

we assume that this is all irrationals though

like the set of all irrationals is uncountable

[placid mason]

This can't possibly have all irrational numbers. ZAN doesn't even have sqrt(2): if it did then this implies sqrt(2)=a+bpi which implies (a+b*pi)^2-2=0, which implies pi is algebraic. But Pi is transcendental

[shadow geyser]

Ah

I see now

But even so

ZAN is still a set of infinite irrationals though

Or does that not matter considering pi is transcendental

As you said

[placid mason]

Consider the set {n*sqrt(2)}. This is an infinite set of rationals, but it is countable.

[shadow geyser]

Infinite set of irrationals you mean

[placid mason]

Yes sorry about that

[shadow geyser]

Nw nw

So okay

How can we say that it is countable though?

If you don't mind me asking

[placid mason]

That's not necessarily easy, it depends what was taught in your class

[shadow geyser]

All we learned is that if we can show there exists a bijection

Then the set is countable

[placid mason]

Hmm seriously? Then I can see this question being very difficult

First you probably want to prove or find a proof of this general theorem "a union of countably many countable sets is countable"

[shadow geyser]

I think I figured it out

It's sort of scuffed as in not exactly a bijection but

I am going to say that

If we say that a and b are within the rationals

Let's take pi to be a constant

We know that the set of rationals is countable

So based on the variables a and b we would have ZAN to be countable

So ZAN[x] being a set of polynomials with countable coefficients would make the set countable

[placid mason]

Everything you said is true

But I'm concerned there is some big leaps

[shadow geyser]

It is a little bit scuffed indeed

I'm not sure how else to describe ZAN to be countable though

At least we haven't learned anything major regarding set theory

Discrete math goes into the bare minimum

[placid mason]

Fair enough, in that case yeah unless you're super clever this is a hard problem

The steps you want to fill in your proof are: why does Q countable imply ZAN countable? Why does ZAN countable imply ZAN[x] countable?

[shadow geyser]

🤣

I do not know how to do that

But you are right

[placid mason]

They're actually due to the same reason, so really you just need to figure out the first question

[shadow geyser]

Those two aspects is what I'm missing

I know why the set of Q is countable

But I do not know how it implies that another set would be countable

In this case ZAN

[placid mason]

Is it possible to prove that Q^2 (which is the set of rational coordinates) is countable? If you can do that, it will help

[shadow geyser]

Yeah I don't know either

We haven't really delved deep into this material

[placid mason]

It's hard without any more theory, you might just want to Google and learn some more theory

[shadow geyser]

Oh you're saying

Countable union of countable sets

So the composition will result in a bijection

I see now

Perhaps I will search some stuff online, but I appreciate all your help noneheless @placid mason

[placid mason]

No problem, good luck

[floral field]

If we have a connected planar graph G and each face of G has either 3 or 5 boundary edges, why must the number of faces be even

[prisma canyon]

can anyone explain to me how to solve this.
If i toss 3 dice, whatis the probability that the total is 7

there has to be some easier way then just writing out the chance of everythung happeneing and counting right?

[balmy jay]

There's a theorem that says the if you add up, over all faces, all the number of edges surrounding a face, the resulting number is 2 times the number of edges. Since each face has an odd number of edges and the sum has to be even, we have that there must be an even number of faces since adding up an even number of odd things results in an even thing. If we had an odd number of faces, you'd have an odd number of odd things which would mean there'd be the contradiction that odd = even

^^ also the reason for that theorem is cause every edge is double counted by the 2 faces that are on opposite sides of it

[wide vine]

are faces the same as regions?

@balmy jay or is this something differnt

[balmy jay]

Depends on what you mean by regions

But yeah it's probably the same thing

[wide vine]

well because I just went through I think the same thing you are talking about in my book

[barren bluff]

handshaking lemma but on faces

yes

[wide vine]

although it went on to discuss number of edges in a planar graph

[barren bluff]

yes that result is because each faces have edges that form a cycle and each cycle must have at least 3 edges hence that inequality

in a connected planar graph

[wide vine]

do these rules apply for only simple graphs?

[barren bluff]

yes

[wide vine]

okay

[wide vine]

I guess you could know you have (6)^n total possible outcomes and then you just need to divide how many ways you can get a sum of 7 from 3 dice from that?

and then I guess once you have your know tuples that would sum up to 7 you would compute all their permutations

Showing that an expression is unsatisfiable by considering all possible settings for the variables takes time proportional to 2^n for an expression with n variables. There may be some shortcuts possible, but no one knows a provably efficient method to solve SAT, despite years of research effort. In fact, many computer scientists believe that no such efficient method exists. Has there not been any proof on the matter? It would seem like once you have reduced your boolean expression you will have to check each case to ensure it is unsatisfiable

I do get you can reduce your space of 2^n

if you happen to realize one of the variables is redundant but once it is simplified then you have to check the 2^n for the variables you have left?

This seems similar to figuring of if an expression is a tautology and after simplifiying it, I would assume you would have to check all the possible combinations.

[wide vine]

are there more advanced models for scheduling or is this as good as it gets?

[weary tiger]

How can one assign digits to letters so that TWO + THREE + SEVEN = TWELVE? Find one solution. [Each letter corresponds to a different digit and each digit is represented by exactly one letter.]

this feels like a bunch of trial and error

how do we do it

[stray reef]

well you could expand everything in terms of place values and collect like terms and treat it as a diophantine equation in however many variables w/ the constraint that they must be all different and between 0 and 9 inclusive and that T and S aren't 0

the value of T can be reduced by noting that the left hand side will be too small to equal the right hand side if T is too large

[weary tiger]

wah

[stray reef]

i said some words.

[weary tiger]

yes

[stray reef]

those words may sound kind of salad-y this time

for this i apologize

i meant that you can expand TWO = 100 * T + 10 * W + 1 * O and likewise for the other numbers

[woven cape]

I would start by examining the case T=1

[weary tiger]

no i understood that

i dont understand how to treat it as a diophantine equation

i mean, i used a diophantine equation before

[stray reef]

diophantine equation = equation where the unknowns are integers

[weary tiger]

dont know how to apply it here

[stray reef]

also,
1 is the ONLY possible thing T could be. the LHS cannot be bigger than 99999+99999+999 and T=2 would force a stricter upper bound on the LHS that would force it below 200,000

[weary tiger]

"the LHS cannot be bigger than 99999+99999+999" i understand, "T=2 would force a stricter upper bound on the LHS that would force it below 200,000" i dont

[stray reef]

T=2 implies LHS <= 99999+29999+299

which is less than 200k

[weary tiger]

oh right

[woven cape]

You also only need to find a solution so you dont need to work through all the possibilities

[stray reef]

by the way, there are ten letters

[weary tiger]

so every letter gets a digit

[stray reef]

wrong way around

every digit gets a letter

[weary tiger]

right

[weary tiger]

but why does that matter

[stray reef]

the value of T is determined uniquely

T cannot be anything bigger than 1, and it also cannot be 0

oh actually hold on

something else can be deduced by looking at the addition problem in column form and thinking about carries

[weary tiger]

hm

so

THREE
SEVEN
TWO
+

[weary tiger]

so its not true?

[stray reef]

what's not true?

\begin{tabular}{ccccccc}
&&&&T&W&O \
&&T&H&R&E&E \
+&&S&E&V&E&N \
\hline
&T&W&E&L&V&E
\end{tabular}

[vital dew]

Ann

[stray reef]

this is what i have written down on paper, or rather a reproduction thereof

[weary tiger]

i bet H is 0 or smth

actually, H is 0

[stray reef]

couldn't it also be 9 with a carry from the last column over

[weary tiger]

oh

well, from the first column

we know O+N=10

[stray reef]

plus, we have T=1

[weary tiger]

you never fully explained

[weary tiger]

^^^^

[stray reef]

the value of T is determined uniquely
T cannot be anything bigger than 1, and it also cannot be 0

[weary tiger]

what does that mean

[stray reef]

it means exactly what is said

T cannot be anything bigger than 1

[weary tiger]

but why

[stray reef]

and T also cannot be 0

the LHS cannot be bigger than 99999+99999+999

[weary tiger]

right

[stray reef]

the LHS cannot be bigger than 200,997 therefore T is at most 2

and T=2 would force the LHS to be at most 130,797, which is less than 200,000

which again is impossible

[weary tiger]

and 29999+99999+299 is not bigger than 99999+99999+999

[stray reef]

have i now explained to you why T = 1?

[weary tiger]

so T=2 works

[stray reef]

.

[weary tiger]

isnt it fine if its less?

19999+99999+199 is less

[stray reef]

the ENTIRE left-hand side is less than 200,000 and you want it to have a 2 in its hundred-thousands digit?

do you hear yourself?

[weary tiger]

oh right

[stray reef]

T=1, and the ten-thousands column has to have a carry, therefore S has to be at least 9 without a carry from the thousands column, or at least 8 with.

if there is no carry from the thousands column (i.e. H = 0) then we have S = 9 and thus W = 0, but we can't have two different letters both standing for 0.

therefore the thousands column has to have a carry, therefore H = 9 and the hundreds column also has a carry.

do you understand this or not

[weary tiger]

i say W is 0

and S=9

[stray reef]

W is what?

you realize that also forces H=0?

[pastel hollow]

My book shows that
$$\sum_{k = 1}^n k\binom n k = n2^{n-1}$$
by differentiating both sides of the special case of the binomial theorem
$$(1+y)^n = \sum_{k = 0}^n \binom n k y^k$$
and then plugging in $y = 1$. Then they say you can use similar reasoning to show that
$$\sum_{k = 1}^n k^2\binom n k = n(n+1)2^{n-2}$$
But I can't get it to work. how do you prove this?

[vital dew]

EdgarAlnGrow

[weary tiger]

oh yeah

[stray reef]

@pastel hollow probably some linear combination of the first and second derivatives of (1+y)^n

[weary tiger]

then 1+S=11

right

to carry over the one

[stray reef]

i believe that we, or at least i, have arrived at this so far:

\begin{tabular}{ccccccc}
&&&&1&0&O \
&&1&9&R&E&E \
+&&8&E&V&E&N \
\hline
&1&0&E&L&V&E
\end{tabular}

[vital dew]

Ann

[weary tiger]

s=8?

[stray reef]

yes

[weary tiger]

then that would be 9

oh carry over

[stray reef]

no, we have a carry from the thousands column.

yes.

[weary tiger]

from previous column

[stray reef]

O+N=10 as you noted

the remaining unassigned values are 2, 3, 4, 5, 6, 7

[weary tiger]

let me just write this down

[stray reef]

sure

i'll just continue copying my findings from paper to here

from the tens column, V is either 2E+1 or 2E-9 depending on whether or not we have a carry
the possible pairs of values for (E,V) are: (2,5), (3,7), (4,9), (5,1), (6,3), (7,5)

[weary tiger]

why do we know W=0 again

[stray reef]

1 + 8 + carried 1

[weary tiger]

oh right

[pastel hollow]

Is there some kind of linear system I can solve to find the linear combination?

[stray reef]

write out the first and second derivatives themselves then it should become clearish

[pastel hollow]

I did

[weary tiger]

@stray reef mind giving a hint on what to do next

[stray reef]

do some casework

i can tell you that at least one of the four cases will rule itself out

@pastel hollow you might want to go to a #help-_ channel for the time being assuming crab and i don't get their problem sorted out

[pastel hollow]

whatever, just forget it

[stray reef]

no need to be so passive-aggressive

[pastel hollow]

it wasn't meant to be

[weary tiger]

i feel like its bad to start with E

i think I should start with V

hmm i still dont see what to try out

@stray reef ideas?

[stray reef]

.

the possible pairs of values for (E,V) are ...

[weary tiger]

ok

there are all the pairs i got (2, 4), (3, 6), (4, 8

[stray reef]

you yourself said O+N=10

[weary tiger]

actually (2, 4), (3, 6), (4, 8), (6, 2), (7, 4), (8, 6)

[weary tiger]

oh

[stray reef]

surely the ones column ought to have carried a 1 into the tens

yet you forgor that

[weary tiger]

so (2,5), (3, 7), (6, 3), (7, 5), (8, 7)

[stray reef]

no

8 is already used

so you cannot have (8,7)

[weary tiger]

oh right

now we just try?

values

[stray reef]

yes, and see what can be said about the other four letters

[weary tiger]

i found none to work

for some reason

[stray reef]

did you arrive at a contradiction in all four cases?

[weary tiger]

well i know for certain there is a contradiction in (6, 3)

i think i run into a contradiction with 3,7

though i cant exactly describe it

[stray reef]

what is your contradiction for (6,3)?

[weary tiger]

R will have to be 8 or 9, which are taken

[stray reef]

why will R have to be 8 or 9?

your hundreds column will be 1+R+3+1, which can give a carry with R as low as 5

[weary tiger]

where did you get 5

V is 3

[stray reef]

edited

[weary tiger]

why would V carry over

[stray reef]

well you yourself said you're considering E=6, V=3

look at the tens column: 0+6+6+(carried 1) = 13...

i mean, i arrived at a contradiction in (6,3) too, but for an entirely different reason

[weary tiger]

yes

[weary tiger]

oh

why>

[stray reef]

why what?

[weary tiger]

i gtg to soon and i dont feel like working on this anymore

[stray reef]

are you asking me why i, being unawa-

ok fine then let's forget it entirely

the only case that doesnt rule itself out is (3,7) and it gives you two solutions

[weary tiger]

i have to explain the contradiction tho?

[stray reef]

||104 + 19722 + 82526 = 102352
106 + 19722 + 82524 = 102352||

[weary tiger]

right?

[stray reef]

of course you do

the contradiction i arrived at in (6,3) is that the unassigned digits left are 2, 4, 5 and 7, no two of which sum to 10 and hence there's no way to assign values to N and O

[weary tiger]

i keep getting stuck here

TWO = 10O
THREE = 19633
SEVEN = 8373N
TWELVE = 103573

I cant find a N+O=10 that works

i have a 2 and 4 left

only

i need to leave a 4 and a 6

aha

i think i have it

hm

[stray reef]

available 234567
possibilities for (E,V) = (2,5) (3,7) (6,3)* (7,5)*

      1 0 O
  1 9 R E E
  8 E V E N
-----------
1 0 E L V E

========================

(2,5) --> available values 3467

      1 0 O
  1 9 R 2 2
  8 2 5 2 N
-----------
1 0 2 L 5 2

6+R = L+10 => R=L+4 => R=7, L=3

      1 0 O
  1 9 7 2 2
  8 2 5 2 N
-----------
1 0 2 3 5 2

{O,N} = {4,6} in either order => two solutions!

104 + 19722 + 82526 = 102352
106 + 19722 + 82524 = 102352

========================

(3,7) --> available values 2456

      1 0 O
  1 9 R 3 3
  8 3 7 3 N
-----------
1 0 3 L 7 3

{O,N} = {4,6} (only pair of digits summing to 10 from available)
=> {R,L} = {2,5}, but this gives either 1+2+7=5 or 1+5+7=2, neither of which is true

========================

(6,3) --> available values 2457, no pair summing to 10 for O and N!

========================

(7,5) --> available values 2346, carry into hundreds column

      1 0 O
  1 9 R 7 7
  8 E 5 7 N
-----------
1 0 E L 5 7

{O,N} = {4,6} (only pair of digits summing to 10 from available)
=> {R,L} = {2,3}, but this gives either 1+2+5+1=3 or 1+3+5+1=2, neither of which is true

[weary tiger]

oh so when (2,5) works

not (3, 7)

[stray reef]

yeah my bad sorry

i had "only one non-contradictory case" in mind and copied the wrong one

[weary tiger]

@stray reef can you help me out here

TWO = 10O
THREE = 19R22
SEVEN = 8252N
TWELVE = 102L52

[stray reef]

i already said all i had to say

you declared unwillingness to take the problem to its conclusion so i just dumped all my reasoning out

[weary tiger]

oh wait i found it

\begin{tabular}{ccccccc}
&&&&1&0&4
&&1&9&7&2&2
+&&8&2&5&2&6
\hline
&1&0&2&3&5&2
\end{tabular}

[vital dew]

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

[weary tiger]

wtf

\begin{tabular}{ccccccc}
&&&&1&0&O
&&1&9&R&E&E
+&&8&E&V&E&N
\hline
&1&0&E&L&V&E
\end{tabular}

[vital dew]

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

[weary tiger]

@stray reef how did yours work

nvm let me ask in #latex-help

[stray reef]

\\

[weary tiger]

?

\begin{tabular}{ccccccc}
&&&&1&0&O \
&&1&9&R&E&E \
+&&8&E&V&E&N \
\hline
&1&0&E&L&V&E
\end{tabular}

[vital dew]

crab λ

[stray reef]

\\ makes a newline

[weary tiger]

there we go

[stray reef]

or a new row in array environments

[weary tiger]

\begin{tabular}{ccccccc}
&&&&1&0&4 \
&&1&9&7&2&2 \
+&&8&2&5&2&6 \
\hline
&1&0&2&3&5&2
\end{tabular}

[vital dew]

crab λ

[weary tiger]

how did you do you background color

hmmm

i copied yours exactly

\begin{tabular}{ccccccc}
&&&&T&W&O \
&&T&H&R&E&E \
+&&S&E&V&E&N \
\hline
&T&W&E&L&V&E
\end{tabular}

\begin{tabular}{ccccccc}
&&&&1&0&O \
&&1&9&R&E&E \
+&&8&E&V&E&N \
\hline
&1&0&E&L&V&E
\end{tabular}

[vital dew]

crab λ

crab λ

[silent rivet]

𝐵 ={𝑛 ∈ 𝑁 | 𝑛 𝑖𝑠 𝑎 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑒 𝑜𝑓 5:2 ≤ 𝑛 ≤ 19}

Would B = {2,7,12,17}

[muted tendon]

Does this work? Felt a bit too easy lol

[coral parcel]

Neither 2 nor 7 nor 12 nor 17 is a multiple of 5.

[coral parcel]

Yup, that will work.

With infinite sets we can even make A=B -- for example let A=B=Z and set f(n)=2n and g(n)=floor(n/2).

[sweet ingot]

Witht he function we need to prove injective
f : Z → Z where f (x) = x + 1
This is the answer we are given

If f (x) = f (y) then x + 1 = y + 1.
Hence x = y. So f is 1-to-1

But it feels like I can just do this for stuff that isnt injective as well

f : Z → Z where f (x) = x^2

If f (x) = f (y) then x^2 = y^2.
Hence x = y. So f is 1-to-1

but the second example is wrong right?

[coral parcel]

How do you get from x²=y² to x=y?

[sweet ingot]

because -2 and 2 both go to 4

well i guess im wrong

I square root both sides

[coral parcel]

But the square root of a² is not necessarily the same as a.

They differ when a is negative.

[sweet ingot]

yeh I get that

[coral parcel]

In the original example you know x+1 = y+1.
Subtract 1 from both sides of that equation.
You now have x+1-1 = y+1-1.
However, it holds for all x that x+1-1 = x.

And similarly that y+1-1 = y.

Whereas it's not always the case that sqrt(x^2)=x.

[sweet ingot]

ah ok i see now

sqrt(x^2)= x or -x

so its not injective

so is it correct to say the square root of 4 is +-2

[coral parcel]

The square root of 4 is unambiguously 2.

That also makes the compound statement "sqrt(4)=2 or sqrt(4)=-2" true.

because only one of the things you connect with "or" has to be true.

[pastel hollow]

The two - signs cancel, so you can just write it as "the square root of 4" 👍

[sweet ingot]

oh its specifcally in this example because x is not squared yet sqrt(x^2)=x

[coral parcel]

sqrt(x²)=x is not true in general but it happens to be true when x=2.

[sweet ingot]

right because when x is -2

sqrt(-2^2) = -2 is false

[coral parcel]

Right. Except you should write it as (-2)^2 because -2^2 means -(2^2).

[sweet ingot]

oh ok

negation is higher in the order of precedence?

[coral parcel]

No, lower.

I suppose from a strictly formal point of view it is correct to say sqrt(4)=±2. But it is true in the same way that "2 plus 2 is 4 or 317" is true.

[sweet ingot]

oh right so yeh it does 2^2 first then negates it

[silent rivet]

Thank you. Day 5 of discrete maths 5, 10, 15

[coral parcel]

Yeah.

[stable vine]

Can someone help me with these 2 questions

[pale epoch]

what do you think?

[stable vine]

Upper on should be false

While down is also false

My question is when is a set a subset of the power set?

Like if i had {{2}} in the first question then would the answer be true?

@pale epoch

[pale epoch]

yes

everything you said is correct

i think you get it 👍

[stable vine]

Thx

Last question

Would the answer be true here?

[pale epoch]

yes

[lost sequoia]

does anyone knows how to find and explain the answers for me?

my head is gonna explode soon because of this question and from another question

[weary tiger]

no , counter-example: we're gonna use a smaller set {1;2;5;4;7;3}
f(1)=0 and g(1)=3
f(2)=38 and g(2)=0
f(5)=0 and g(5)=238
f(4)=33737 and g(4)=0
f(7)=0 and g(7)=23438
f(434)=0 and g(3)=0
it is clear that (for all x in {1;2;5;4;7;3}) (f(x)=0 or g(x)=0) is true
but (for all x in {1;2;5;4;7;3} f(x)=0) is false and (for all x in {1;2;5;4;7;3} g(x)=0) is also false
conclusion: the two statements arent equivalent

this is just an example to help you see the difference, and not a formal answer.

[lost sequoia]

I understand the last 3 lines but, is there any reason or explanation for the f(2)=38 or g(7)=23438?

[weary tiger]

oh sorry, it is just a choice

[lost sequoia]

its fine its fine

[weary tiger]

you can define a function however you want

[lost sequoia]

that's just what I dont understand

[weary tiger]

for example

we choose f(1)=2 and f(q)= maths
f : {1,q} -> {2,maths}
x -> f(x)
f is still a function

takes inputs

and gives outputs

and a set can contain anything

[weary tiger]

im sorry what is

the function thing?

[lost sequoia]

just the part where you "created" the functions for the value

like f(5) = 0

[weary tiger]

okay

so

[lost sequoia]

oh and why is the first pic false and the second pic is true? its a or operation no?

[unreal stump]

First one is you pick some x ,then either f(x)=0 or g(x)=0

Second one is either f or g is the zero function

[weary tiger]

we have f(2)=38 which is not 0 , so we can't say that for whatever x in {1;2;5;4;7;3} , f(x)=0

[unreal stump]

An example of first is f(x)=1 if x=2 and 0 otherwise and g(x)=0 if x=2 and 1 otherwise

Not in second

[weary tiger]

Drake did give u another explanation, to help u see things in a different way that might suit you

[lost sequoia]

Im so confused @-@
now I understand why discrete math is hard

cause when my lecturer teaches I can understand all

but not the work

[unreal stump]

Ok,So you know circuits?

[weary tiger]

i think a real life example would work best

im trying to come up with one

[lost sequoia]

you mean or, and, not gates?

[unreal stump]

Yea

[lost sequoia]

yup I know

learnt it quite a lot this semester

[unreal stump]

So you have 2 circuits A and B.
And you send a 0 signal and a 1 signal

If A doesn't light up for 0 but B does and B doesn't light up for 1 but A does that would be case 1

[lost sequoia]

ok, so far so good

[unreal stump]

Because given whatever signal,atleast one of them doesn't light up

Case 2 is no matter what signal you give one of A or B or both won't light up

[lost sequoia]

why?

hmm, is there a way to draw it in a diagram?

[unreal stump]

I mean you know truth tables right?

[lost sequoia]

this is case 1 right?

so in this B is a not gate no?

[unreal stump]

Actually yea

Didn't realise that

A is the nothing gate

[lost sequoia]

then we have a or gate at the intercept at the end

[unreal stump]

Thing is at no point you want both of them to light up

So the net gate will be NOT(A and B)

Like the whole expression corresponds to this

A corresponds to id
B corresponds to not

[lost sequoia]

id?

[unreal stump]

I guess you call it buffer gate?

The gate that does nothing

[lost sequoia]

ohh

so the entire circuit will become just a not gate?

[unreal stump]

.

For all inputs,either A or B doesn't light up

That's how you read the expression

Given an input you can find one that doesn't light up

[lost sequoia]

https://tenor.com/view/anime-confused-loading-paper-thinking-gif-23758618

but A and B are the gates no?

[unreal stump]

Yea,But then there is this expression

\forall x ( f(x)=0 or g(x)=0 )

This whole thing corresponds to the circuit which can be represented as NOT ( A AND B )

Where A corresponds to f and B corresponds to g

[lost sequoia]

my brain is exploding

lemme reread a bit

[lost sequoia]

wont it be come 1 1 1 0 at the output?

[unreal stump]

Yea

[lost sequoia]

that's what we are looking for?

but how is it = \forall x ( f(x)=0 or g(x)=0 )?

[unreal stump]

Well Take f(x)=1 if x=1 and 0 everywhere else
And g(x)=0 if x=1 and 1 everywhere else

Same effect as our circuit construction

Think everywhere else as "0 input"

[lost sequoia]

||Im definitely the idiot in my class ||

[unreal stump]

You know it's funny that my construction is inherently circuit based

I didn't intend that to be at all

x=1 -> like high input->1 level
x!=1 -> like low input->0 level

[lost sequoia]

[lost sequoia]

ok then?

[unreal stump]

Well what did we discuss?
A is buffer gate and B is Not gate

So here it corresponds to
f(x)=1 at x=1
f(x)=0 at everywhere else

B corresponds to
g(x)=0 at x=1
g(x)=1 at everywhere else

This will satisfy first case

[lost sequoia]

is there a way to like understand how does that become this? QwQ