help-10

[timid silo]

i get that part

[frosty spoke]

then isn't that just the inductive step

[timid silo]

has 2^n distinct subsets

[frosty spoke]

huh?

[timid silo]

huh?

[frosty spoke]

huh???

what has 2^n distinct subsets

[timid silo]

any set with exactly n-many distinct elements?

[frosty spoke]

yeah

so if we say that a set that has (n+1) elements has twice as many subsets as one with n elements

then aren't we really saying that it has 2 x 2^n elements?

[timid silo]

yes exactly

[frosty spoke]

there's your inductive step

[timid silo]

thats all you need to do to prove it

[frosty spoke]

well you need to actually show that the (n+1) set has twice as many subsets lol

that comes from the argument where you take one element out and consider the subsets that have it or don't

[timid silo]

out of all the subsets, there will be half which dont include this arbitrary element

[frosty spoke]

well you'd need to prove that it's half

[timid silo]

i cant do that without example

[frosty spoke]

yes you can

you can show that the number of subsets which include it and which don't include it are the same

[timid silo]

Like this

[frosty spoke]

hmm no, that's just saying that you can split the subsets into those that don't contain the element and those that do contain it

[timid silo]

Can I use the definition of a power set then

[frosty spoke]

I say "disjoint" there because ultimately we're going to add up the counts of both classes

[frosty spoke]

not sure that would help anything

we're already implicitly talking about the power set, but we don't really have to bring it in

like sure, there's a set of all subsets, and we're talking about the cardinality of the power set, but that's not particularly helpful

[timid silo]

Okay well if a set has 2^n distinct subsets then for this to be true an arbitrary element would have to be in only half the sunsets

[frosty spoke]

that's true, but not proven

would be circular logic

Let's say we have a set A of (n+1) elements

we can partition its subsets into sets C and D, C containing all the subsets that contain the element THE_ELEMENT

and D containing all of the subsets that don't contain it

because C and D don't have elements in common (that's what disjoint means), we can claim that the total number of subsets of A = [number of elements in C] + [number of elements in D]

so note that if we show that [number of elements in C] = [number of elements in D], then we're done here, we've shown that [number of subsets of A] = 2 times [number of elements in D]

and D is just really the subsets of A \ {THE_ELEMENT}, which is a set of n elements

[timid silo]

I think I get this can I try formatting my proof

[frosty spoke]

yeah sure

[timid silo]

im still confused about how to prove number of elements in C] = [number of elements in D]

[frosty spoke]

you can show that the elements in C are just the elements in D with THE_ELEMENT added to them

but this is good progress, you're separating the proof into individual parts that you can then prove

but it's also just kinda obvious lmfao

actually let's not do that subset bullshit

[timid silo]

btw tysm for helping me

[frosty spoke]

construct a bijection between C and D, where to get to C from D, you add THE_ELEMENT to the subset

[frosty spoke]

just trying to keep you from becoming an economics major

and to get from C to D, you remove it

these are f and f^(-1) respectively, and if you show that they're injective

[timid silo]

NO

I AM NOT PROVING IT THIS WAY

[frosty spoke]

LMFAO OH WAIT I FORGOT

LOL

you and your struggles with that

honestly

just say "the elements in C are just the elements in D with THE_ELEMENT added to them"

like it's kinda obvious

you can even say

that f(d) = d U {THE_ELEMENT} is a bijection

and therefore the number of elements in C is the same as the number of elements of D

[timid silo]

stop

i am not using functions

[frosty spoke]

and literally just don't bother proving that it's a bijection

[timid silo]

thats not how it works?

dont u need to prove everything

[frosty spoke]

let's be honest will an overworked TA really bother with a few little proofs by assertion somewhere

[timid silo]

false

not at my school

[frosty spoke]

LOL

[timid silo]

okay im now confused about which way to do it

i rlly not about the functions life

if u get me

[frosty spoke]

hmm are you yourself convinced that the elements in C are just the elements in D with THE_ELEMENT added to them?

[timid silo]

yes

because its trivial

[frosty spoke]

well, you can do this: consider an element in D, called d

show that (d u {THE_ELEMENT}) is in C

and then consider an element in C, called c

show that c \ {THE_ELEMENT} is in D

these are all basically just one-sentence justifications

[timid silo]

so this is not proving bijectivity

[frosty spoke]

no it doesn't involve constructing the bijection

well it's equivalent to proving surjectivity in both directions of that bijection, but you don't have to think about that

[timid silo]

are you ukrainian

[frosty spoke]

no LOL why

[timid silo]

oh okay

that link

[frosty spoke]

oh no that's just the charity I'll donate to for this year LOL

[timid silo]

that is nice of you

are you sure i need to prove the above

it just sounds like basic logic

[frosty spoke]

well yeah it's just a passing remark why these are true

that's basically all proofs, just steps of basic logic

d u {THE_ELEMENT} is in C, because d is a subset of A and so is {THE_ELEMENT}, and therefore d u {THE_ELEMENT} is a subset of A, and since it contains THE_ELEMENT, it's in C

c \ {THE_ELEMENT} is in D, because c is a subset of A, and so subtracting {THE_ELEMENT} still leaves as a subset of A, and since it doesn't contain THE_ELEMENT, it's in D

[brazen belfry]

can you write me an essay Saccharine

[frosty spoke]

no I suck at that

[brazen belfry]

sure

[frosty spoke]

I will leave the caveat that the thing we actually have to show is that every element c can be written as d u {THE_ELEMENT} for some d in D, but this is the same as saying that c \ {THE_ELEMENT} is in D

[timid silo]

i am loosing the will to live

[frosty spoke]

why is that

don't you love cs

lemme explain it without skipping steps

[timid silo]

i go to a school filled with masochists

[frosty spoke]

are you a masochist too

let's define the + operator so when I say $D + k$ I really mean ${d \cup {k}, d \in D}$

[warm shale]

Saccharine

[timid silo]

yes

[frosty spoke]

basically just adding k to every set in D

[frosty spoke]

no wonder why you like cs

and systems

lmk how FPGA programming works out

[timid silo]

i get what youre saying, one set leave arbitrary element, one set take it away, wts they are equal so that

i go to cmu so ill just become a sellout in machine learning alternatively

[frosty spoke]

so what we're really trying to prove is that $D + {k} = C$

[warm shale]

Saccharine

[frosty spoke]

i.e. C is just D but with the element added in

to do this, we need to prove $D + k \subseteq C$ and $C \subseteq D + k$

[timid silo]

isnt this a relation

[warm shale]

Saccharine

[frosty spoke]

uh the subset relation is a partial order, but let's not talk about that LOL

[timid silo]

double containment

[frosty spoke]

right basically if one set contains the other and vice versa

they're equal

[timid silo]

okay hold on

D doesnt have the arbitrary element

thats why we are adding it

to show that it is a subset of C

[frosty spoke]

the subsets of D do not have that element

we're trying to show that C is basically just the subsets included in D, but with that extra element added

[timid silo]

this seems so trival

if we know that D doesnt have the element

obviously unioning it will make it have that

and if c has the element

then it is elqual

since the onlky difference between c and d is that d doesnt have one arbitrary element that c has

but we are unioning it which is basically addition

its logical or but whatever

[frosty spoke]

well, you need to show that if you have an element of C, called c, that it can be obtained by taking something in D and adding the element to it

[timid silo]

something meaning a subset

[frosty spoke]

yes

[timid silo]

but you need to say that c is not in D

[frosty spoke]

well c is definitely not in D

because c contains the fancy element

but all elements of D do not

[timid silo]

yeah thats why you use the union of any subset of D with the arbitrary element

[frosty spoke]

but you need to say that c can be written as a subset without the fancy element + the fancy element

[timid silo]

subset

[frosty spoke]

and really, the way you show this is you show that c = c \ {the fancy element} u {the fancy element}

and note that c \ {the fancy element} is an element of D

[timid silo]

yeah and if D = c \ {the fancy element} then its basically c = D u element

[frosty spoke]

no

D is the set of all little ds

D is the set of all subsets without the fancy element

C is the set of all subsets with the fancy element

you need to show that C = {d U {the fancy element}, d in D}

[timid silo]

yes i understand

[frosty spoke]

the way this is done is you show the double inclusion or whatever it's called

like C is a subset of {d U {the fancy element}, d in D}

and {d U {the fancy element}, d in D} is a subset of C

[timid silo]

double containment yeah

[frosty spoke]

the way you show a subset relation

is by picking an arbitrary element in the first set and showing it's in the second set

so to show the first subset thing, we pick an arbitrary element c of C

and we show that it's in {d U {the fancy element}, d in D}

this is basically saying that we have to show that there exists a d in D such that c = d U {the fancy element}

to show this, we simply exhibit the d itself: d = c \ {the fancy element}

this is in D, because the two properties that define whether something is in D are:

1. it is a subset of A (the whole n+1 set if you remember)
2. it does not contain the fancy element

and therefore we have shown that c is in {d U {the fancy element}, d in D}

that completes the proof of the first inclusion

the proof of the second inclusion is that
if we pick any element of {d U {the fancy element}, d in D}, which we'll call d u {the fancy element}, it's in C, but this is immediately true because d and {the fancy element} are both subsets of A, and therefore their union is a subset of A

and furthermore, d u {the fancy element} contains the fancy element

[timid silo]

let me try

[frosty spoke]

also, there's an obvious nit where you might have to show that D and {d U {the fancy element}, d in D} have the same number of elements LMFAO

but this is logically obvious enough that you probably can get away by just saying they do

but if you wanted to rigorously prove it

unfortunately we're going to have to talk about a bijection LOL

in general, the number of elements in a set is defined by bijections so you're going to have to love them one day

[timid silo]

i think this is mainly confusing because the fancy element is an element and d and c are both subsets and we are just adding or subtracting the element to show it is a subset of whatever

[frosty spoke]

the fancy element is an element of A

d and c are subsets of A, with special properties of course

we're adding the fancy element to d to get something in C

what's confusing in particular?

[timid silo]

Show Base case:\
Our base case is when there is a set with $2^0$ distinct subsets, so this corresponds to the empty set.\\

Show inductive Hypothesis:\
Assume that any set with n elements will have $2^n$ distinct subsets and show that any set n + 1 elements will also have $2^{n + 1}$ distinct subsets. For a set that has a size of $n + 1$, if we pick an arbitrary element to be taken out, half of the subsets will have this arbitrary element, half of them wont. Let the set given to us be set A, which contains n + 1 elements. Now let B and C be other sets. B will contain all the subsets with the arbitrary element, and C will contain all the subsets without the arbitrary element. Because sets B and C are disjoint, the total elements in A consist of the total elements in B + the total elements in C. If it is shown that the number of elements in B = the number of elements C, then the number of subsets of A = 2 times the number of elements in B.\\

Lets call this arbitrary element in A, a. Now pick an arbitrary subset $b \in B$ and an arbitrary subset $c \in C$. Recall that subset b will contain a, however subset c will not contain a.\\

WTS $\exists c \in C$ s.t. $c = b \setminus a$. We know that $b = c \setminus a$ and this is in c because it is a subset of A, and it does not contain the arbitrary element a. Thus have shown that $b \in c \cup a$.\\

Next, any element in $c \cup a$ is in B because d and a are both subsets of A, so their union is also a subset a, thus $c \cup a$ contains a.

latex work

[warm shale]

dopamine

[timid silo]

ayyy

[frosty spoke]

you can use some notation to shorten the number of words if you want

like "the number of elements in B" is written |B| typically

[timid silo]

cardinality

[frosty spoke]

yes

[timid silo]

is the general idea okay

[frosty spoke]

this arbitrary element defines B and C, so you need to pick it before you say B and C

and say something like "let B the set of all subsets of A that contain a" and "let C be the set of all subsets of A that do not contain a"

[timid silo]

OH i see

okay

so i can conclude that i have show that C union A = B

[frosty spoke]

well it's not C union a

it's

[timid silo]

a

[frosty spoke]

{c union {a} for c in C}

[timid silo]

oh

yeah

pokay

this is gross

this shit doesnt come up in industry

no way ill be at google and they will say prove this line of code with double containment looser

[frosty spoke]

LMFAOOO

[timid silo]

but my programming class is actually proof based

[frosty spoke]

well induction does come up sometimes with weird algorithms

[timid silo]

yeah we use induction in my programming class

and next sem when i do functional programming

[frosty spoke]

like you might have to compute something weird with graphs

max-cut problems with graphs are done recursively I think

dynamic programming in general

[frosty spoke]

also you can't choose b and c simultaneously

you can't say that c u {a} = b, rather you say that c u {a} is in B

[timid silo]

wait at the end?

[frosty spoke]

here

you want to show b = c u {a} for some element c in C

[timid silo]

so get rid of the whole recal part?

[frosty spoke]

yeah sure

nobody said that your proof had to be easy to read

just that it has to be valid

[timid silo]

is that sarcasm

[frosty spoke]

no it's the truth LOL

[timid silo]

Use induction to prove for any $n\in\mathbb{N}$, any set with exactly n-many distinct elements has
exactly $2^n$ many distinct subsets.\\

Show Base case:\
Our base case is when there is a set with $2^0$ distinct subsets, so this corresponds to the empty set.\\

Show inductive Hypothesis:\
Assume that any set with n elements will have $2^n$ distinct subsets and show that any set n + 1 elements will also have $2^{n + 1}$ distinct subsets. For a set that has a size of $n + 1$, if we pick an arbitrary element to be taken out, half of the subsets will have this arbitrary element, half of them wont. Let the set given to us be set A, which contains n + 1 elements. Now let B and C be other sets. B will contain all the subsets with the arbitrary element, and C will contain all the subsets without the arbitrary element. Because sets B and C are disjoint, the total elements in A consist of the total elements in B + the total elements in C. If it is shown that the number of elements in B = the number of elements C, then the number of subsets of A = 2 times the number of elements in B.\\

Lets call this arbitrary element in A, a. Let B be the set that contains the subsets that include a, and let C be the set that contains the subsets that do not include a. Now pick an arbitrary subset $b \in B$ and an arbitrary subset $c \in C$.\\

WTS $\exists c \in C$ s.t. $c = b \setminus a$. We know that $b = c \setminus a$ and this is in c because it is a subset of A, and it does not contain the arbitrary element a. Thus have shown that $b \in c \cup a$.\\

Next, any element in $c \cup a$ is in B because d and a are both subsets of A, so their union is also a subset a, thus $c \cup a$ contains a.\\

Thus it has been shown that $c \cup a \forall c \in C = B$.\\

Because B = C, then the sumber of subsets of A = 2 times the number of elements in B. By induction, since the base case and inductive hypothesis are true this implies that any set with exactly n-many distinct elements has
exactly $2^n$ many distinct subsets.

[warm shale]

dopamine

dopamine

[timid silo]

sumber

[frosty spoke]

well okay that's not exactly done

you need to argue that C is precisely the subsets of A \ {a}

this is not a terribly difficult argument

there are a lot of repeated sentences here; can you take them out, so I know what order you're doing things in?

I'd write something like:

1. Let B be the set of all subsets of A that contain a, and let C be the set of all subsets of A that do not contain a.
2. Note that B and C are disjoint, so |P(A)| = |B| + |C|
3. We will now show that B = {c u {a}, c in C}. To do this, we will show [the double containment]
4. Let b an arbitrary element of B. Then we will show that b is in {c u {a}, c in C}, i.e. there exists a c in C such that b = c u {a}. Letting c = b \ {a}, we see that this chosen value is in C, because it is a subset of A and it does not contain a.
5. Forgetting the definitions of b and c in step 4, let c u {a} be an arbitrary element of {c u {a}, c in C}. Then, c u {a} is clearly in B, because it is a union of subsets of A, and it contains a.
6. Therefore, we have shown [the equality in step 3], which means that |B| = |C|.
7. Therefore, |P(A)| = 2 |C|.
8. C is further the set of all subsets of A \ {a}, which is of cardinality 2^n
9. Therefore, |P(A)| = 2 x 2^n = 2^(n+1)
10. Hence, we've proven the inductive hypothesis.

[timid silo]

this is the definition of C though, A without a

[frosty spoke]

no the definition of C is the set of all subsets of A that do not contain element a

you need to at least comment that this is the same as the set of all subsets of A \ {a}

Also, I would highly suggest thinking about this in terms of functions, because it's still not completely rigorous in that you can't talk about cardinality without talking about bijections. The proof also gets a little simpler.

Let B be an arbitrary set with (n+1) elements. By the definition of finite set, there exists a bijection f such that f(B) = A = {1, ..., n+1}. Then we partition the subsets into two disjoint sets C and D, such that C = {n+1 \in c, c \subseteq A}, D = {n+1 \not \in d, d \subseteq A}. C and D are disjoint and C U D = P(A). Further, |C| = |D| because we can construct a functions g: C -> D, where g(c) = c \ {n+1} and h: D -> C, h(d) = d u {n+1}. One can check that these are well-defined and are inverses of each other, and therefore g is a bijection. Then, |P(A)| = 2|D|, and we observe that D is the set of all subsets of {1, ..., n} [proof omitted], so the inductive hypothesis follows.

[timid silo]

I am so bad at functions though

[frosty spoke]

it's not so bad

like you don't need to prove all the little things that they ask

honestly you could probably even get away with B and C are obviously the same cardinality LOL

like there's a certain level of rigour required for each proof, and nitpicking over all the details is annoying because the proofs get long and boring to read

as long as you get the core idea, it's probably fine

[timid silo]

im not going to avoid functions forever

this spring break im just going to be doing work in my dorm

[frosty spoke]

why

you should have fun

[timid silo]

i cant

like its just not possible

[frosty spoke]

why isn't it possible

[timid silo]

because i have a ton of homework

lol even gave us programming hmwk

[frosty spoke]

oh wait spring break doesn't end

the semester or whatever y'all use?

[timid silo]

we dont have quarters like stanford

im a semester child

[frosty spoke]

yknow in my second-to-last year I had a similar thing before spring break

but it was like 3 finals, a paper all going to happen the week before spring break

and I actually didn't know anything in all but two of those classes

I was super stressed out

but then I was like "what's going to happen if I fail everything"

and the answer really was life goes on

so I went and got my Cs and B-s and whatever

[spring trench]

Can you help me please

[frosty spoke]

and it was all okay

was a fun spring break too

[frosty spoke]

yes

[timid silo]

do they really care about that gpa

[frosty spoke]

not employers

[timid silo]

i just want to be a straight b student

[frosty spoke]

if you just want to sell your soul to tech

[timid silo]

yes

sell my soul to quant

[frosty spoke]

better check with your degree program though

[frosty spoke]

idk dopamine are you sure you want to keep taking math

oh wait they hire systems people too

[timid silo]

real

[spring trench]

I need help finding the side in the middle but idk how to

[timid silo]

linux dev for jane street

[spring trench]

I wanna know how to find it

[frosty spoke]

if you become an awesome coder remember u owe me a lot of debugging of my shitty AI code

[timid silo]

im rlly good at debugging now

i used to be bad

but now oh only helps if u have multiple failed test cases

[frosty spoke]

you have two angles and a side, so use a triangle calculator or the law of sines on the bottom triangle

okay my issue is that I don't really have test cases for my thing

it computes outputs and gradients

but when you train it, it produces garbage results

[timid silo]

AI is cool but learning how to build a kernal or os or compiler sounds cooler

jk

ai is cool i just get pressured by upperclassmen to not do it tbh

[frosty spoke]

LOL you're a masochist

have you made a heap allocator yet

[frosty spoke]

btw compiler makers have to learn all of this discrete math garbage lmfao

that infix to postfix

[timid silo]

well

i have to choose between my passions and between paying off a lot of college debt

so in the end ill just pick whichever makes the most money, then i can discover my true passions while making bank

[frosty spoke]

damn they charge you money for the torture too

[timid silo]

80k per year

[frosty spoke]

that's obscene

u must have rich parents

hmm idk though, like I sold my soul into cs and yeah it pays a lot of money

but you miss out on the other things in life while you're studying and you won't get it back

and also it's a little bit of an empty life

[timid silo]

no

im an international student

so they force me to pay full tuitiojn

[frosty spoke]

wait what that's weird

also woah where are you from?

I remember the vast majority of intl students at stanford got paid a stipend to go there LOL

[timid silo]

canada

i go to carnegie mellon

their endowment is so small they must make everyone who attends pay full tuition

[frosty spoke]

damn why didn't you go to a uni in canada?

[timid silo]

because why would i not take this opportunity

[frosty spoke]

imagine having a small endowment but being named after one of the richest people to have ever lived

[frosty spoke]

the opportunity to indulge your masochism and make money after graduating?

[timid silo]

sure

something like that

i try to not think about the cost because i dont want an aneurysm

[frosty spoke]

it's okay you can probably pay it off in like 5 years if you work in tech with what they pay now

are you sure you're like mentally okay though

[timid silo]

uh yeah SURE

[frosty spoke]

I know a lot of people who got burnt out really quickly

even I burned out in my last year

[frosty spoke]

this smells like sarcasm

[timid silo]

i dont know how to answer ur question

i guess im doing okay enough

[frosty spoke]

are you taking a decent mix of hard and easy classes

or are all of the classes hard

[timid silo]

its cmu

every class is hard

even the "Easy" ones

im taking 2 maths, programming in c, some BS writing course, and a neuroscience course

[frosty spoke]

is the BS writing course easy enough

[timid silo]

yeah

i suppose

[frosty spoke]

im guessing the math ones are hard?

[timid silo]

every week is the same though

like i have 2 psets due every friday

programming hmwk thursday

programming written hmwk monday

neuro quizzes every class

and they make us write and read material for english

im taking a late day for my programming which sucks so much

[frosty spoke]

why does it suck

maybe you should do the programming assignments with other people in your class

like not cheat on them, but maybe talk them over or something

it seems like you feel like you're overwhelmed

[timid silo]

oh no the programming is not a problem

its not easy, but i can figure it out myself

also really strict AIV unfortunately

you cant discuss anything

im overwhelmed by choice

[frosty spoke]

woah that's kinda dumb LOL

[timid silo]

i chose to do this to myself

[frosty spoke]

well you chose to take the classes, but it doesn't mean that they have to be suffering

[timid silo]

yeah well im take an equivalent of 50 hours per week of work

[frosty spoke]

are you trying to graduate early or something

[timid silo]

no

this is like standard

for everyone

i have to take 45 units to graduate ontime

[frosty spoke]

it's really that much? what is this madness

[timid silo]

i dont have any incoming credits because im canadian and my school didnt have any ap or ib or whatever

[frosty spoke]

oh yeah that's rough

[timid silo]

i can complain about it or i can suck it up

i chose it

[frosty spoke]

but you could probably just take the equivalent classes for APs and breeze them right?

or do you not know calculus and that stuff

[timid silo]

huh naw in my senior year my highschool didnt cover integration

we just did derivatives

i have to do EVERYTHING from scratch

and its infurating when all your friends have done calc 3 and lin alg in highschool

and you can barely do calc 2 and injectivity

[frosty spoke]

INJECTIVITY LMFAO

hmm I wouldn't say you're THAT behind though

high school calc 3 and linear algebra is usually poorly taught

obviously it helps when you're retaking the equivalent in university

but it's ok

[timid silo]

well yeah ofc its ojay

i mean like

this is standard for most people anyways

and i never let anything ever hold me back to begin with

so we are all good

[frosty spoke]

yeah but you seem really dead inside LOL

[timid silo]

considering my highschool background, im doing pretty well i guess

yeah well maybe harvard shoulda taken me or something

i was deciding between cmu and uiuc

so like both were pretty masochistic chocies

[frosty spoke]

interesting, I met a guy from uiuc at a hackathon at berkeley lol

he seemed like he was enjoying life

do you actually like learning math or do you not enjoy it

[timid silo]

dude i would enjoy all my classes a lot more if i had more time

i just dont

[frosty spoke]

hmm maybe you should spend a bit of summer break studying some stuff leisurely

what's your other math class besides this one?

[timid silo]

uh calc 2

:throwup:

[frosty spoke]

are you doing okay in that one

[timid silo]

Calc 2 isn’t hard I just don’t practice enough

[frosty spoke]

practice?

what math classes do you have to take to get a cs degree

[timid silo]

Uh all calc and Lin alg 2 discrete and like probability

Dude mayh isn’t bad it’s just bad when you haven’t done it before and you’re also taking a bunch of other hard courses

I’m not just going to give up on my dream because it’s hard

Like that’s crazy

[frosty spoke]

damn dopamine that's very respectable

I gave up on my master's degree because it was one more quarter of work

[timid silo]

Why did I even work hard in highschool then

[frosty spoke]

and then I realized I was bored LOL

[timid silo]

I worked hard in highschool jusy so i coild be here

[frosty spoke]

yeah that's understandable

[timid silo]

Oh btw thanks for helping me

I appreciate it a lot

[frosty spoke]

no problem

[timid silo]

Do you enjoy this type of math or

[frosty spoke]

hmm well not basic stuff like this because it doesn't have that much insight

but yeah I actually kinda like discrete math, mostly because I'm terrible at it

but then again, all of the stuff involving analysis is neat

[timid silo]

How do you become good at this

Uh can i ask more questions in this channel too u obviously don’t have to answer

[frosty spoke]

yeah sure

[frosty spoke]

I was much worse than my math major friends freshman year, and I still am

but I think by the end of 3 years, I got as good as my math major friends were freshman year

which was enough to take one or two really hard math classes every quarter

usually these things have a threshold where they just suddenly make much more sense

also, if you've never done proofs before, proof-based mathematics is sometimes really unintuitive and weird, but after you get used to it, it just becomes a mode of thinking

and fwiw, I didn't have the exact wording of the proof immediately from seeing the thing, but you sorta develop an intuition for the key idea and then the rest of it is just formalizing it

[timid silo]

i see

practice and time

[frosty spoke]

not like blind practice, but yeah you'll learn the ways you formalize certain statements

and tbf you're not that far from it, just being able to recognize what needs to be proven and what a proof is is actually a big hurdle for a lot of people

who are just used to mimicking some procedure

[timid silo]

im trying to do Prove that every positive integer is either a Fibonacci number or can be expressed as a sum
of distinct Fibonacci numbers.

[frosty spoke]

hmm

are you supposed to use induction

can't think of a proof for this one off the top of my head

[timid silo]

all of these are induction

im trying to decide if this even is true though

[frosty spoke]

oh yeah if you use strong induction it seems true

cool argument too

you can think of this a little recursively

[timid silo]

anything fibonacci is recursive

what would the base case even be though

[frosty spoke]

base case doesn't matter that much, you'd probably use something like 1 and 2 satisfy this property LOL

the trick here is to write some sort of recursive thing: for N, you can construct the sum as [biggest fibonacci number less than N] + the fibonacci sum of [N - that number]

and remember that strong induction says that if you assume P(1), ..., P(n), and you show that P(n+1), and you show the base case, then it's true for all natural numbers

one thing you have to show is that [N - that number] < that number, but that's not hard to show

[timid silo]

this is so painful

[frosty spoke]

what's painful about it

[timid silo]

uh can we do an example not arbitrary

like idk 10

ah like 2 + 8

[frosty spoke]

yeah basically

let's say you pick a number 155

the first few terms of the fibonacci sequence are here for reference
1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610,

you can construct the sum as 155 = 144 + 8 +3

where you just pick the largest fibonacci number that's less than the thing

and add it on

your job is to formalize this into an induction proof and also prove that it actually works (think about the case where this might fall apart)

[timid silo]

eww why is this like programming

[frosty spoke]

recursion and induction are closely related lmfao

[timid silo]

but how did you generalize picking 8 and 3

[frosty spoke]

okay so it goes something like this

155 = 144 + 11, so now we need to find a sum for 11

so we find the highest less than 11

and that's 8

[timid silo]

i get it

i gotcha

that makes complete sense

155 - 144 = 11, then next biggest num is 8

[frosty spoke]

yes

[timid silo]

and we have 11 - 8

[frosty spoke]

so you know how recursion has the "assume your code will work for everything below this" thing?

same thing here

assuming that you can find a sum for the numbers 1, ..., n, you need to show that it can be found for n + 1

[timid silo]

so the sum from p(1) to p(k) implies p(k + 1)

[frosty spoke]

well more like

the collection of statements p(1), p(2), ... p(k) implies p(k+1)

if you can find a fibonacci sum for each of the numbers from 1 through k, then you can find one for the number k+1

where "fibonacci sum" means sum of distinct fibonacci numbers

[timid silo]

uh taht is weirs

weird

like if you can find the fibonacci sum from numbers 1 to 14 you can also find 15

[frosty spoke]

yes

the construction amounts to taking 15-13 = 2 and then finding the fibonacci sum for 2 and adding 13

[timid silo]

so the algorithm is find the fibonacci number that is less than or equal to the number we have

and subtract that number from the original number to get the next number we are looking for

and then sum this

[frosty spoke]

yes

find a sum for the subtracted number

but you know this exists because you have p(1) through p(k)

[timid silo]

like number - p(k)?

and then like do p(k-1)

[frosty spoke]

let's say you want to prove p(155), assuming p(1), ... p(154)

I'll select 144 as the fibonacci number next to it

then I say

since p(11) is true, I know I can find a sum of unique fibonacci numbers for 11. Let's call that S_11

Then, S_11 + 144 = 11 + 144 = 155 is a sum of unique fibonacci numbers for 155

^ this last statement has a little piece that you need to prove, which is that 144 is not used in S_11 because then otherwise it wouldn't be true

[timid silo]

thank you for your patience

[frosty spoke]

I'm going to go out with some friends, but I wish you luck in your hw

probably will be back in 2 hours or so

[timid silo]

tysm i appreciate your help again

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

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[near helm]

hihi, I just need help with finding the x-intercepts of a parabola

[near helm]

ik its really easy but i cant seem to find any clear explanations on how to do it

and my algebra isnt really that great

.close

[obtuse pebble]

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[near helm]

nvm, sorry

[obtuse pebble]

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[untold quail]

When dealing with estimated frequencies, I am asked to report when the true frequency might be at least 25% and when the true frequency must be at least 25%. What is meant by this? I have an expression for upper and lower bounding the true frequency. Does must refer to a lower bound and might refer to an upper bound?

[drifting wraith]

probably

[untold quail]

Okay, well at least we agree! haha

.close

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[gray meadow]

Homework help. I'm completely lost and would like some guidance please!

[lean acorn]

you find the derivation of that function using implicit differentation

[gray meadow]

I believe it is x/y correct?

[lean acorn]

yes

-x/y

actually

[gray meadow]

Why is that?

isn't it 2x-2yy' = 0

[split pulsar]

No, light's right

It's x/y

[lean acorn]

the derivative equals: 2x + 2ydy/dx =0

you minus the 2x to the other side

then divide 2y

[split pulsar]

It's -y^2

[gray meadow]

the problem is -y^2

[split pulsar]

So the derivative is -2y(y')

[lean acorn]

ohh

i though it was +y^2

mb

[gray meadow]

so y' = x/y

[lean acorn]

yes

x/y

[split pulsar]

And x/y is 8/3

[gray meadow]

so then is it 3 + 8/3(x-8)

[lean acorn]

the formula is: y-3 = m(x-8) to find tan line

obvs replace m with the slop

slope*

[split pulsar]

Yup

Looks like it

[gray meadow]

so it should be 3 + 8/3x - 64/3

[split pulsar]

I think that's good as is

It says an equation, not the equation in a specific form

[gray meadow]

I'll try it

That worked

Thank you guys

[split pulsar]

No problem

[gray meadow]

I'm going crazy with diff

implied diff that is

[split pulsar]

Yeah implicit differentiation took me a while

[gray meadow]

I feel good and then I just bomb it the next time

Thanks!

[lean acorn]

welp you got it

anything else?

[gray meadow]

I might have some in a bit but I'm good for now

.close

[obtuse pebble]

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[twin shale]

Can anyone help me with these two.

[earnest elk]

what are you having trouble with?

[obtuse pebble]

@twin shale Has your question been resolved?

[twin shale]

With the second one.

[twin shale]

I used to do this but I forgot I don’t remember completing a model yk like shading the model

<@&286206848099549185>

.close

[obtuse pebble]

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[burnt blaze]

can someone just sanity check this for me

[burnt blaze]

ill take a photo of my working gimme a min

because you breath in and out

a breath in would be .25L and a breath out would be .25L

so thats the 1/4 infront of sine

i used sine because breathing is a cycle

and i used t*pi/4 because 4pi/4 is pi where the sine wave meets the origin

so one breathing cycle is 180 degrees

is this reasonable

i think the second one is probably more accurate

because its a full cycle

with airflow in and out

[obtuse pebble]

@burnt blaze Has your question been resolved?

[obtuse pebble]

@burnt blaze Has your question been resolved?

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[autumn spade]

this i what i have till now

[glossy basalt]

looks correct

[autumn spade]

yeah but idk how to proceed further

[glossy basalt]

thinking

[autumn spade]

sure take your time, im doing other questions in the meanwhile

[glossy basalt]

numerator: (54)^(n) [(54)^(4m-n)+1]
denominator: (2)^(n-4m) [ (2)^(4m) * (3)^(4m)]

try simplify from that

got sth to do irl ttyl

[autumn spade]

alr i'll try thx!

[glossy basalt]

back

[autumn spade]

okay so two things

[glossy basalt]

yea?

[autumn spade]

1st idk how u got to that

[glossy basalt]

oh...

[autumn spade]

2nd i still cant do it from tthat

[glossy basalt]

hmmm

[glossy basalt]

I'll go through it here then,

so,
numerator

(54)^(4m)+(54)^n

factor out 54^n

then, we have

$54^n \left( \frac{54^{4m}}{54^n} + \frac{54^n}{54^n} \right)$

[warm shale]

OldBiscuit

[glossy basalt]

sorry, slow typer

[autumn spade]

no its okay

[glossy basalt]

then, we have

54^n ( 54^(4m-n)+1)

[autumn spade]

yup makes sense

[glossy basalt]

now for denominator

(2^n)*(3^(12m))

[glossy basalt]

ohhhhh, i got typo on the above

[glossy basalt]

anyways let's continue

(2^n)(2^(4m))(2^(-4m)(3^12m)

(2^(n-4m))(2^(4m))(27^(4m))

(2^(n-4m))(54^(4m)

sorry for the typo

[autumn spade]

what did you do here?

[glossy basalt]

I did times and divided a 2^4m
so basically it's times 1

$\frac{2^{4m}}{2^{4m}}2^n3^{12m}$

[warm shale]

OldBiscuit

[glossy basalt]

which become this

now we have
numerator: 54^n(54^(4m-n)+1)
denominator: (2^(n-4m))(54^(4m))
All good till here?

[autumn spade]

yeah i think so?

im trying on my own what you did

[glossy basalt]

that's a very nice habit

to confirm what someone else did is correct

[autumn spade]

oaky i got it

you can proceed

[autumn spade]

thx :D

[glossy basalt]

let's rearrange all the terms into non fraction way

thats is to add negative to all exponents to the denominator

the whole thing will be
54^n (54^(4m-n)+1) 2^(4m-n) 54^(-4m)

and further simplify to
54^(-(4m-n)) (54^(4m-n)+1) 2^(4m-n)

and i think you get do it from here 🙂

[autumn spade]

yeah i think i got it from here

tysm for the help!

.close

[obtuse pebble]

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[wicked current]

How to do these ?

[glossy basalt]

since the options are pointing to when beta is -2,2

you can first try putting beta =2 or -2 in the system and see what happens

[timid silo]

you can look at it from a matrix perspective and linear Algebre

make the matrix [[1, B-1],[B+1, 3]]

look at this as 2 * 2 matrix

and based on the Cramer( a way for solving this kind of equations) determinant of this matris should not be 0

so B should not be either 2 or -2

for a better explanation you can check this link

https://en.wikipedia.org/wiki/Cramer's_rule

[obtuse pebble]

@wicked current Has your question been resolved?

[timber island]

I would recommend putting the values given in the option. If you don't have them, then there's a separate way

[obtuse pebble]

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[rancid wagon]

can anyone help me understand how I'd calculate this limit if it wasn't given?

https://i.imgur.com/Q5ipJYq.png

[rancid wagon]

it's being used a a tool for a bigger calculation but I wouldn't guess the lim is 0

oh wow and this one that is also given

that is just a curveball to me

https://i.imgur.com/tjdwh6Z.png

[rigid lintel]

perhaps write $\lim_{n \to \infty} \sum_{m = n+1}^{\infty} \frac{n!}{m!} = \lim_{n \to \infty} \sum_{i = 1}^{\infty} \frac{n!}{(n+i)!}$

[warm shale]

heavy

[rigid lintel]

$= \lim_{n \to \infty} \sum_{i = 1}^{\infty} \frac{1}{(n+1)} + \frac{1}{(n+1)(n+2)} \ldots$

[warm shale]

heavy

[rancid wagon]

wait

oh the n doesn't increment?

[rigid lintel]

left of the = sign increments

[rancid wagon]

so if it wasn't part of a limit

what does n start at?

where does it say it starts at 1?

[rigid lintel]

you wouldnt know

unless you define n = 5 or something

[rancid wagon]

so when you did your expansion why is putting 1 there valid?

[rigid lintel]

also n is a constant, it doesnt start anywhere

[rancid wagon]

right, it just is

[rigid lintel]

yes

[rigid lintel]

if m = n+1 and you increment over n, then replacing m by n + i and incrementing over i from i = 1 is the same thing

m would go from m = n+1, n+2, n+3 ...

n+ i does the exact same thing when incrementing from i = 1

[rancid wagon]

okay I get the syntax now

but n goes to infinity

in the lim

so we have inf!/(inf + 1)! summed up

[rigid lintel]

yes

[rancid wagon]

oh because its a factorial

[rigid lintel]

exactly

[rancid wagon]

does it simplify to 1/inf?

aka we're adding a bunch of 0s?

[rigid lintel]

yup

well 1/inf + 1/(inf+1) ...

but you get it

[rancid wagon]

well I get the first one

second one still iffy 😄

[rigid lintel]

second limit is a similar sttory

[rancid wagon]

maybe its a syntax issue

does the 2nd one say

take the sum and then multiply it by n

or is the n somehow part of the sum?

[rigid lintel]

yes

n is a constant so you can bring it inside

[rancid wagon]

then isn't it indeterminate as n -> inf?

[rigid lintel]

ehh not really

just bring it inside the sum

[rancid wagon]

😭 that's what I was asking about if its legal or not

how would you bring it inside the sum anyway?

[rigid lintel]

$\forall n \in \mathbb{N}, n \sum_{m=n+1}^{\infty} \frac{n !}{m !} = \sum_{m=n+1}^{\infty} n\frac{n !}{m !}$

[warm shale]

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

[rancid wagon]

(n * n!)/m! doesn't seem right

[warm shale]

heavy

[rancid wagon]

is that... true?

[rigid lintel]

well yes

n is just a constant here

[rancid wagon]

oh wait is that jus the same thing as n (a/m + b/m + c/m) = na/m + nb/m + nc/m

[rigid lintel]

yeah

sort of

[rancid wagon]

I meant like the concept of distributing inside of the sum operator

[rigid lintel]

yes

but we have the sum to infinity here so you have to be slightly more careful

[rigid lintel]

this is rigorous enough dont worry

[rancid wagon]

ok it's a little better now like I kind of see it but I'm still not 100% on the 2nd limit goes to 1

the very first element of the sum is just

(inf * inf!)/(inf + 1)! right?

[rigid lintel]

yes

[rancid wagon]

just that element by itself what does it equal?

trying to picture infinity factorial is a bit of a mindfuck

[rigid lintel]

how about this

$\lim_{n \to \infty} \frac{n \cdot n!}{(n+1)!}$

[warm shale]

heavy

[rigid lintel]

this should be easier to think about

[rancid wagon]

that's what I was thinking about

that's what I meant when I wrote what I wrote lol

just that alone is already tripping me up

it's like (inf * inf * inf - 1 * inf -2 * inf...)/(inf + 1 * inf * inf - 1 * inf - 2...)

[rigid lintel]

recall that $(n+1)! = (n+1) \cdot n!$

[warm shale]

heavy

[rancid wagon]

is what I wrote correct?

the expansion

[rigid lintel]

yes thats correct

[rancid wagon]

if that is correct then it all cancels out to 1/inf + 1

which is squarely 0 right?

[rigid lintel]

well not quite

but doing arithmetic with infinity is kind of bad

[rigid lintel]

do it like this

[rancid wagon]

sry it wouldn't actually cancel to 1 / inf +1

it would cancel to inf / inf + 1

which I would then say = 1

[rigid lintel]

yes

very well

[rancid wagon]

but in the original limit

we're taking the sum of all of those 1s

as n goes to inf

[rigid lintel]

not all would be equal to 1

[rancid wagon]

so wouldn't we be adding 1 inf times?

oh wait do the rest equal 0

[rigid lintel]

yes

[rancid wagon]

2nd elm is (inf * inf * inf - 1 * inf -2 * inf...)/(inf + 2 * inf + 1 * inf * inf - 1 * inf - 2...)

then it cancels to inf/inf + 2 * inf + 1

which is 0

ok I see it

[rigid lintel]

please dont do arithmetic with infinity though

[rancid wagon]

how else

would I get to the answr 😭

I only do canceling

or for purposes of canceling

[rigid lintel]

the whole reason we have limits is to argue with infinity in a good way

[rigid lintel]

like here

[rancid wagon]

when you say don't do arithmetic

I know inf +/- n = inf

then I basically just cancel out the infinities and see what happens

that is incorrect?

[rigid lintel]

just dont do arithmetic with infinity lol

[rancid wagon]

wait is inf +/- {constant} not just inf?

inf absorbs it right?

[rigid lintel]

well yes

[rancid wagon]

that's what I'm saying

[rigid lintel]

but its bad fashion

[rancid wagon]

the + 1 + 2 etc I was writing was for illustration

lol

[rigid lintel]

inf/inf can be anything

do you see

[rancid wagon]

uh

[rigid lintel]

we have limits for this

take for example $\lim_{n \to \infty} \frac{2n}{n} = \frac{\infty}{\infty}$

[warm shale]

heavy

[rigid lintel]

but that limit = 2

not 1

thats why its bad to do arithmetic with infinity

[rancid wagon]

yeah I see that limit = 2

pretty clearly lol

I think ik what you mean

alr, thanks for the help m8

[rigid lintel]

np

[rancid wagon]

.close

[obtuse pebble]

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[foggy dawn]

C(x)=a+b(x) for cost, fixed costs, than unit cost * units made.. However, how can I use it to find the weekly gross pay of an employee

[frosty spoke]

lol you'd better give much more information than that

I could have high total or fixed costs, and still pay employees nothing

[foggy dawn]

Hourly rate, and hours worked.. So I suppose the fixed cost is 15 and what changes is the hours worked.. That's really it, I know you can just multiply 15*80 and get the gross pay but I'm curious to know if advance maths can also be used to calculate that

[stable barn]

can i get a hint to determine the value of y

[hoary mirage]

y=2x

[stable barn]

im confused

is it y=60

[tranquil quiver]

this isn't even your help channel bud

[stable barn]

then what do i do

sorry

[tranquil quiver]

#❓how-to-get-help

[frosty spoke]

is gross pay not literally just the hourly rate times the hours work

[foggy dawn]

Yea

I was just curious to see if there were other ways of calculating the same results

So looking online I found that formula

[frosty spoke]

you need to say what information you have

if you have someone's yearly pay, and you know that they're paid the same every week

then that's another way?

[foggy dawn]

Actually sounds easier to implement into my code, but yes it is another way.. Thank you I know what I'm asking is sort of an unconventional way of doing things

I'm probably just over thinking everything any way

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[obtuse pebble]

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[obtuse pebble]

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[violet mulch]

hello

[violet mulch]

i have an assignment im having some trouble with

is to find the maclaurin series of a piecewise function

(ln(1+x^2))/x^2 when x is not 0

1 when x is zero

now I have found the maclaurin series of ln(1+x^2)

how do I proceed to 'add' the /x^2 to the series?

[haughty coyote]

just decrease each power of x by 2 in the series

[violet mulch]

lmao that makes sense

thanks

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[grizzled robin]

can someone help me answer this?

[obtuse pebble]

@grizzled robin Has your question been resolved?

[grizzled robin]

<@&286206848099549185>

[obtuse pebble]

@grizzled robin Has your question been resolved?

[grizzled robin]

<@&286206848099549185>

[grizzled robin]

.close

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[finite pollen]

If x = 9 is the chord of contact of the hyperbola x^2 - y^2 = 9, then the equation of the corresponding pair of tangents is?
I am on laptop rn, so can't like capture my attempts, but here's what i did: I know that combined equation of pair of tangents from a point say h,k which should be outside hyperbola is SS1 = T^2
i just can't find that point h, k , please give me hint, not solution on how can i find the point from where tangents are drawn

[grizzled shore]

at what points does x=9 and x^2 - y^2 = 9 intersect

[finite pollen]

9, 6root2

• or - 6root2 that is

as its a vertical line

[grizzled shore]

what are the tangents to those 2 points

[finite pollen]

didn't understand? can you elaborate

[grizzled shore]

what equation is tangent to the 2 intersections

isn't that what the question is asking

[finite pollen]

it is asking combined equation of pair of tangents from an external point

ok never mind, just got a similar example in my textbook...thanks for your time sir, i appreciate it.

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[shadow lava]

is the reciprocal also considered the opposite? x and 1/x

[fading quest]

the inverse?

[latent walrus]

the reciprocal is the multiplicative inverse if thats what you mean

[shadow lava]

why do we say reciprocal?

shouldn't we say inverse?

[fading quest]

remember that inverse is not the same

you can call it the multiplicative inverse if you want

but its shorter to say reciprocal

[shadow lava]

asking because of this: https://youtu.be/GpWCFoCznGI?t=10

"implicit is the opposite of explicit", and it shows "1/explicit"

[fading quest]

ehh, opposite is probably not the wording i would use

[shadow lava]

the "reciprocal" of explicit would be better wording?

or the "inverse" of explicit would be even better wording?

i thought inverse and opposite are sort of truths, in a way. + is the opposite of -. they cancel each other out.

or they "reverse" each other

[fading quest]

if you were to find the inverse function of f(x)=x then what would that be?

it would also be x, its just what the meaning inverse means in the context

is x the opposite of x?

[shadow lava]

you swap x and y to find the inverse

it would be a one-to-one function I think?

[fading quest]

but implicit and explicit arent even inverses or reciprocals or anything like that, not sure why they would write it that way

[shadow lava]

passing vertical and horizontal line test

[shadow lava]

oh OK

maybe I should watch 1brown3blue on implicit differentiation?

tends to go a bit overkill, sometimes I get lost watching his videos

[royal basin]

you got his name backwards.

[fading quest]

3b1b is always my go to for understanding subjects but generally the vids are better once uve worked with them a bit

[shadow lava]

it's commutative lol

• 3blue1brown, rather

does anybody know what it means?

[royal basin]

does anybody know what what means?

[shadow lava]

3blue1brown

[severe reef]

I think colors of the markers he uses

[royal basin]

you mean the origin of the name?

[shadow lava]

yes

[severe reef]

Nvm that was the other guy

[royal basin]

iirc it's because he has heterochromia, i.e. two different colors in his irises

i remember him talking about it in a q&a somewhere but i cbf to find the link to it

[shadow lava]

Ohhh I see

[fading quest]

huh, thats a cool name origin

[shadow lava]

that makes sense why the logo looks like an eye

[royal basin]

yeah, that's one of the possible coloration patterns

[shadow lava]

whenever I see an eyeball I think of Cosmos for some reason

[royal basin]

as it happens, i have some form of it too.

[shadow lava]

never noticed, the Cosmos title "eye" is kinda 3blue1brown

great TV show btw, I would recommend watching the original series first (PBS) from 1980

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[teal smelt]

Is it possible to have a point on a graph where the first derivative is 0 but is not a local maximum or minimum?

[fading quest]

,w x^3

,w graph x^3

[warm shale]

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[teal smelt]

what is the point called

where it remains at 0

[fading quest]

i think those are called saddle points, i know they are in 3d but not 100% about 2d graphs

[teal smelt]

ty

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[untold oasis]

Can someone please confirm I did this correctly?

I am trying to find d/dx of 3xy = 4 with respect to y

[obtuse pebble]

@untold oasis Has your question been resolved?

[autumn gale]

3xy^2 you mean? you've done derivative with respect to x btw

[obtuse pebble]

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[soft wind]

A man can paint a room by himself in 8 hours while his son can can paint the same room in
12 hours. The man works by himself for 3 hours and is then joined by his son. How many
hours will it take them to finish painting?

[soft wind]

i don't get this one

[haughty coyote]

look at the percentage of work each one does in 1 hour and figure it out from that

[soft wind]

i did this
3/8 + x/12 = 1

then i got 7.5 and added 3 so i got 10.5 hours

[haughty coyote]

they work together

so it isn't x/12

the dad doesn't stop working

[soft wind]

is this correct?

[haughty coyote]

did to take this remark into account ?

[haughty coyote]

father and son working together can't possibly be longer than father alone. Think for a minute

[soft wind]

(1/8 + 1/12) x-3 = 1?

[haughty coyote]

where did the first 3 hours go

assuming you forgot the parentheses

[soft wind]

is the final answer 6?

[haughty coyote]

yes

[soft wind]

i still don't understand how to get it tho

[haughty coyote]

+3/8

[soft wind]

(1/8 + 1/12) x + 3/8 = 1?

[haughty coyote]

(x-3) still

3/8 is the work done by the father in the first 3 hours
then they work x-3 hours at a rate 1/8 + 1/12

[soft wind]

wait ok so (1/8 + 1/12) (x - 3) + 3/8 = 1 right?

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[boreal sleet]

Is this right?

[supple granite]

absolutely not

[boreal sleet]

How do we divide then

[high lily]

what's the original question, what exactly are you being asked to do

because as is, there isn't much to do with the original fraction
$$\frac{x+1}{18x+8}$$

[warm shale]

ℝamonov

[high lily]

where's your fraction coming from

[boreal sleet]

u=9x^2+18x