precalculus

[charred hull]

i think it's off bots list?

hm

@wraith idol works in #help-9 bot is up

[wraith idol]

Oh

Ok

Good

[charred hull]

ye bot only works in the questions channels

ig it's bc of the !oc and !un commands

[wraith idol]

Oh

Ok

[bold maple]

30x^2+23x+16 / cx+3 = 6x+1+ 13/cx+3
The above equation is true for all values of x ≠ − 3 /c , where c is a constant. What is the value of c?

(A) −18
(B) −5
(C) 5
(D) 18

I dont understand what im supposed to do at all. Can someone please explain this to me?

[charred hull]

oh alright

[bold maple]

<@&286206848099549185>

[proud raven]

[bold maple]

pls help? t.t

[proud raven]

there are ways to do it but what did you try

like

usually on this sort of problems your first thought should just be to try plugging stuff in

[bold maple]

i was thinking about doing long division

[proud raven]

and its multiple choice so

[bold maple]

but i wasent sure if i was supposed to

I dont know where to start or how to do it

[proud raven]

😦

if you just want the answer start plugging in the answers

[finite spire]

Are you sure the above equation is written correctly btw? It looks very wrong

[proud raven]

if you want to understand it get the answer and then work backwards

[finite spire]

$30x^2+23x+\frac{16}{c}x+3 = 6x+1+ \frac{13}{c}x+3$

[obsidian monolith]

Gumiibear:

[finite spire]

Is this the equation?

[bold maple]

that’s the questions

[finite spire]

Ok great. So like jan said, a possibility you can try since it's a multi choice question. Is to simply plug in the given possibilities and find which one is true

Or if you want, we can go through it to find the value of C as if it wasn't a multiple choice question

[bold maple]

can we try to go through it?

[finite spire]

Alright sure

Sounds good

[royal gull]

just take everything into one fraction and solve numerator = 0

[bold maple]

how would i do that?

just move it over and make it negative?

[finite spire]

So starting of, we want to simplify the expression a bit. And mainly to get rid of denominators.

Well the whole problem relies on solving for c. now making one side =0 could be a way. But the important part is to do the simplifications correctly.

So Nicolae. In order to simplify the expression we want to get rid of the denominators. And we do this by multiplying (cx+3) on both sides

Now one important thing to note when multiplying on both sides like that, is to make sure you're not multiplying by 0. Now in this case we're given that x is not equal to -3/c. So we don't have to worry about that

So tell me when you've done that first step

[bold maple]

let me take a picture to show what i did so far

[finite spire]

👍

[bold maple]

i went from right to left

[finite spire]

So what happened to the denominators?

[torn swift]

,rotate

[bold maple]

i took them both out since they were the same

Is that wrong?

[finite spire]

Well was the whole right side the same denominator?

$\frac{30x^2+23x+16}{cx+3} = 6x+1+ \frac{13}{cx+3}$

[obsidian monolith]

Gumiibear:

[bold maple]

no?

[finite spire]

Yeah, so try to not just "take them out"

But rather what we're doing is we're multiplying (cx+3) on both sides

[bold maple]

Would what I did so far be wrong or did I somehow get the right answer

[finite spire]

It would be wrong

[bold maple]

oh

[finite spire]

$(\frac{30x^2+23x+16}{cx+3}) (cx+3)= (6x+1+ \frac{13}{cx+3})(cx+3)$

[obsidian monolith]

Gumiibear:

[finite spire]

This would be what we do. multiply (cx+3) on both sides, in order for the denominators to cancel

But because 6x+1 doesn't have the same denominator

We get:

$30x^2+23x+16= (6x+1)(cx+3)+ 13$

[obsidian monolith]

Gumiibear:

[bold maple]

where did you get 13 from?

or do i use FOIL

oh nvm

i see

[finite spire]

Sorry don't know what FOIL is

[bold maple]

sorry

first, outer, inner, last

I learned it in biology, I think its alg 2

[finite spire]

Ah ok. Well so this first step is the most important. So make sure you understand what I did

So I multiplied both sides of the equality with (cx+3) in order to cancel out the denominators. An important note to take is that you cannot multiply both sides by 0. So we need to be certain that (cx+3) isn't 0. In the exercise it says that x is not defined at -3/c. So that scenario is ruled out. And therefore we get a new equation that is equivalent to the original one

[bold maple]

so i leave cx+3 next to 6x+1 or would i multiply them?

[finite spire]

You want to multiply them out. I just wrote it like that so you saw where the (cx+3) terms went. They cancelled out denominators, but for (6x+1) that doesn't have a denominator to cancel it out, you have to multiply it out.

[bold maple]

that’s what I got when I multiplied them out

,rotate

[finite spire]

Great, so now according to the question. This equality is true for all values of x (except one). So can you find what c has to be for that to be true?

[bold maple]

would i plug it into either side to make it equal the other?

[finite spire]

What do you mean to plug in?

[bold maple]

for c

like 6(5)x^2

[finite spire]

Well you want to find what c is, for this equality to be true

Yeah, so c=5 right?

[bold maple]

yea but would I make it -5 so it cancels out? or would I keep it positive since I will move it over?

[finite spire]

Well you want the equality to be true

So you have $30x^2 + 23x+16 = 6c x^2+(18+c)x+16$

[obsidian monolith]

Gumiibear:

[bold maple]

I have +13

not +16

[finite spire]

+3+13=+16

[bold maple]

ah yes

then yea

[finite spire]

So looking at that you can see two things

$30x^2=6cx^2 and , 23x=(18+c)x$

[obsidian monolith]

Gumiibear:

[bold maple]

Would it be 5 since it would make it equivalent to the other side

[finite spire]

Yup exactly

[bold maple]

ok

thank you for your help

I actually understand now

<#

[finite spire]

Cause that's what we want to find. Which c makes the equality correct 👍

[slim rain]

Need some help. I’m stuck on finding a and c.

[willow bear]

a can be read off almost directly

[slim rain]

The asymptote?

[willow bear]

indeed

[slim rain]

Alright how do I start finding b and c

[willow bear]

there are two nice points your curve goes through

one is (1, -2), as is written explicitly

the other is (0, 5)

[slim rain]

So subbing with f(0) = 5 should give me b?

[willow bear]

yes

[slim rain]

I think I got it, thanks mate.

[grizzled orchid]

petition to change ann's legal name to "mate"

[shrewd urchin]

Can i write set of irrationals as $T={x:x \in \mathbb{R}$ and $ x \notin \mathbb{Q}}$

[willow bear]

no

$\{ x : x \in \bbN \land x \notin \mathbb{Q} \}$ is empty

[obsidian monolith]

Ann:

[willow bear]

because $\bbN \subset \bbQ$

[obsidian monolith]

Ann:

[shrewd urchin]

wait

i mistyped

[willow bear]

if you write $\bbR$ instead of $\bbN$, then yes, that becomes the set of irrational numbers.

[obsidian monolith]

Ann:

[willow bear]

to type braces, you use \{ and \}

if you want the word "and", at least do \text{and} so it doesn't look weird

also

\{

and \}

[shrewd urchin]

OK ann

[willow bear]

\{

and \}

how hard are those two sequences of characters

[obsidian monolith]

Krishna (An average mathy):

[willow bear]

there, FINALLY.

[shrewd urchin]

[willow bear]

i wish i didn't have to repeat myself

[grizzled orchid]

kek

[shrewd urchin]

lol

The code is correct now

[echo plaza]

$\bR\sm\bQ$

[obsidian monolith]

CaptainLightning:

[shrewd urchin]

$T={x:x \in \bR \ \text{and}\ x \notin \bQ }$

[obsidian monolith]

Krishna (An average mathy):

[shrewd urchin]

ann this is correct

[harsh cipher]

why give x,y those ranges? 4000,8000 5, 10 years on y-axis

I can come up with the equation, but I'm supposed to find the intersecting point.

To figure out how long Jennifer would have to keep car model B to make it cheaper than model A

[blazing parrot]

intersecting point is when Ca(t)=Cb(t)

[willow bear]

well i mean

one would argue that the domain of both of these functions may as well be (0, +∞)

though of course you'll need to pick a window to graph somehow

which i guess can be done by making an educated guess

[harsh cipher]

well now I'm having trouble putting this in the calculator. (put the window ranges as shown in vid)

[shrewd urchin]

Write an inequality

Model A>ModelB
6000$+(800y)$>7500$+(600y)$
Where y is the number of year

$200y>1500 \implies y>7.5 $

[obsidian monolith]

Krishna (An average mathy):

[harsh cipher]

but I'm supposed to find this using the calculator. I'm guessing your method is writing an inequality and then sollving?

[shrewd urchin]

yes

[prisma marten]

14 hour response

[shrewd urchin]

Your arithmetic weak

[prisma marten]

Wait yea I’m a monkey

How did I get 14

[serene heath]

7 times 2

[charred hull]

9+5

[grizzled orchid]

no u

[royal gull]

$\frac{d}{dx} \frac{1}{x}= \frac{\cancel d}{\cancel d x} \frac{1}{x}= \frac{~}{x} \frac{1}{x}= -\frac{1}{x^2}$

[obsidian monolith]

dog:

[serene heath]

cursed

[wraith idol]

Bruh

Too old

[frozen needle]

[serene heath]

[viscid thistle]

Hey

Can anyone help me with this problem

I'm not really aware how to get started with it

Or what to do

[torn swift]

Asking for the derivative evaluated at x=2

You see what f(x) is?

Plug in 2+h and 2 into x

Then find the quotient f(2+h)-f(2)/h

That quotient is called the difference quotient

Then take the specified limit and you’ll have the answer

[pale kettle]

@viscid thistle

[viscid thistle]

@pale kettle

I still don't get it

I got 1

but it isn't part of the answers

;/

[short sorrel]

so you expanded into:

$\lim_{h \to 0} \frac{(2+h)^2 + (2+h) + 1 - (2^2 + 2 + 1)}{h}$

[obsidian monolith]

Namington:

[short sorrel]

right?

[viscid thistle]

ah i see my mistake

thank you

i wonder

if the limit was set at anything higher than 0, how would it affect that equation? @short sorrel

[short sorrel]

then you wouldn't be dividing by 0, so you could just evaluate it directly

no need to pull fancy limit shenanigans

FYI the above limit, (f(x+h)-f(x))/h as h->0, will crop up... quite a bit later if you ever take a calculus class

as in, it's literally the entire focus of differential calculus

[viscid thistle]

hm.. i see

[short sorrel]

though we usually dont think of it as a limit

[viscid thistle]

wait so i just gotta ask

whenever i divided everything by H, i still got 6

which isn't an option

h/h = 1

or am i doing this wrong

lol

[short sorrel]

so we have $$\lim_{h \to 0} \frac{(2+h)^2 + (2+h) + 1 - (2^2 + 2 + 1)}{h}$$

[obsidian monolith]

Namington:

[short sorrel]

lets expand and simplify this

first off, 2^2 + 2 + 1 = 7

$$\lim_{h \to 0} \frac{(2+h)^2 + (2+h) + 1 - 7}{h}$$

[obsidian monolith]

Namington:

[short sorrel]

next is expanding the (2+h)^2

$(2+h)^2 = (2+h)(2+h) = h^2 + h + 4$

[obsidian monolith]

Namington:

[short sorrel]

so we have $$\lim_{h \to 0} \frac{h^2 + h + 4 + (2+h) + 1 - 7}{h}$$

[obsidian monolith]

Namington:

[short sorrel]

does that line up with your work so far?

[viscid thistle]

yeah

[short sorrel]

now if we collect like terms

$\lim_{h \to 0} \frac{h^2 + h + 7 - 7}{h} \ \ = \lim_{h \to 0} \frac{h^2 + h}{h}$

[obsidian monolith]

Namington:

[short sorrel]

dividing out h, we should get

oh whoops

i just relaized i dropped an h

lmao

$$\lim_{h \to 0} \frac{h^2 + h + 4 + (2+h) + 1 - 7}{h}$$ should've became $$\lim_{h \to 0} \frac{h^2 + 2h + 7 - 7}{h} \ \ = \lim_{h \to 0} \frac{h^2 + 2h}{h}$$

[obsidian monolith]

Namington:

[short sorrel]

...wow im an idiot

sorry latexing is hard

[viscid thistle]

its ok

take your time

[short sorrel]

$(2+h)^2 = (2+h)(2+h) = h^2 + 4h + 4$

[viscid thistle]

i appreciate it

[obsidian monolith]

Namington:

[short sorrel]

i somehow expanded this wrong lmao

so the h^2 + h + 4 shouldve looked like:

$$\lim_{h \to 0} \frac{h^2 + 4h + 4 + (2+h) + 1 - 7}{h}$$ should've became $$\lim_{h \to 0} \frac{h^2 + 4h + h + 7 - 7}{h} \ \ = \lim_{h \to 0} \frac{h^2 + 5h}{h}$$

[obsidian monolith]

Namington:

[short sorrel]

and then we can divide out the h

$\lim_{h \to 0} (h + 5)$

[obsidian monolith]

Namington:

[short sorrel]

and by direct substitution

we get a limit of 5

sorry, was trying to do math while typing it up in TeX, somehow missed a glaring mistake

but besides that, do you follow?

[viscid thistle]

yeah

man

i relaly appreciate it

guess what

i solved 3 problems and got them right

really REALLY appreciate the help

@short sorrel

so in this case

what would i do

when the limit is at x = -1

[short sorrel]

if f(x) = 10, whats f(-1)?

[viscid thistle]

uhm

still 10?

so A?

[short sorrel]

right

[viscid thistle]

in this case

do i just find it for both f

and g

wait sorry

i mean g and h

and divide g by h

[blazing parrot]

the limit of a product is the product of the limits

[serene heath]

what a dumb options system

[bold maple]

Would this be right since 20/4 is 5

[sharp frigate]

I got B

it's asking for the x value

not the y

@bold maple

[bold maple]

@sharp frigate how did you get 15?

[sharp frigate]

y=20

plug it in

5(x+20)=25

distribute the 5

solve for x

at least i"m pretty sure

idk

[plush trench]

You could also divide both sides by 5 instead of distributing

[sharp frigate]

my math is pretty bad

yeah that could work too

that might be faster actually

[bold maple]

i still cant see how you got 15

[plush trench]

He got -15

If that helps

[sharp frigate]

^

[bold maple]

oh i see

that makes sense

[ivory cargo]

hey

was told i ask algebra 2 questions here is that correct?

[grizzled orchid]

here or #prealg-and-algebra

the line between is fuzzy

[ivory cargo]

gotcha

[tame wedge]

If g(x) = f(x+c) how would I find the inverse of f ?

[willow bear]

the question is really vague, and the only good answer to it is that $f^{-1}$ is the inverse of the function $x \mapsto g(x-c)$, if that exists

[obsidian monolith]

Ann:

[willow bear]

can you post the original problem you were doing, exactly as it is stated? @tame wedge

[willow bear]

k so that isn't what you want then

you want $g^{-1}$ in terms of $f^{-1}$, not what you asked.

[obsidian monolith]

Ann:

[tame wedge]

Right, im still unsure how to answer it nonetheless

[willow bear]

consider that $g = f \circ \varphi$, where $\varphi$ is the function defined by $\varphi(x) = x+c$

[obsidian monolith]

Ann:

[willow bear]

what do you know about inverses and more specifically how they interact with compositions?

[tame wedge]

I can find the inverse of functions if they're specified, I can deal with composite functions but im iffy with some notation and I haven't dealt with inverses of composite functions

[willow bear]

have you not

wait

what

wait so you aren't aware of the fact that for all invertible functions $f$ and $g$, $(f \circ g)^{-1} = g^{-1} \circ f^{-1}$?

[obsidian monolith]

Ann:

[tame wedge]

No I wasn't aware of that

If you want an idea of the level im at: first year uni student - doing a bridging maths course to take first year maths

Just started learning about inverses

[willow bear]

h

this was high school shit for me but ok i suppose

[tame wedge]

Might be why im doing a bridging course

[shrewd urchin]

What does the quesJon mean

[willow bear]

have you ever encountered sigma notation for sums

[shrewd urchin]

Yes

[willow bear]

this is the same but with unions

[shrewd urchin]

I know

[willow bear]

$\bigcup_{r=1}^{20} X_r = X_1 \cup X_2 \cup \cdots \cup X_{20}$

[obsidian monolith]

Ann:

[shrewd urchin]

Okay

Son in total there are 5*20=100 elements in big cup

[willow bear]

no

that'd only be true if all the X_r were disjoint from each other

[shrewd urchin]

🤔

Yes dumb me

[willow bear]

$|A \cup B|$ doesn't always equal $|A| + |B|$ after all

[obsidian monolith]

Ann:

[shrewd urchin]

$|A \cup B|=n(A \cup B)$?

[obsidian monolith]

Krishna (An unexperienced PS):

[shrewd urchin]

@willow bear ?

[echo plaza]

wat

[willow bear]

do you use the notation n() for number of elements

if so then yes

|| is another notation for it, one that i'm more used to

[shrewd urchin]

Yes in my boo it is denoted like thiis

[willow bear]

ok fine

[shrewd urchin]

Book*

How to proceed with this problem

[willow bear]

well, counting the number of elements in S is a good idea, even though it may not be as straightforward as 5 * 20

[shrewd urchin]

I mean this is a example and I can't understand the solution

[willow bear]

that

uh

what the fuck

Since n(X_r) = 5, we get n(S) = 100

what the fuck

the solution is garbage

i am not reading it any further, it is garbage

10 are the number of distinct elements in S
says just FIVE LINES EARLIER that n(S) = 100
just what the FUCK does this book think n() means if not THE NUMBER OF ELEMENTS IN S, WHICH ARE DISTINCT BY MOTHERFUCKING DEFINITION

go trash this piece of shit book you got

[shrewd urchin]

@willow bear are you sure ? This book is used by all students in India. It's written by some top educators of my country.

[willow bear]

yeah well the book is clearly in sore need of a new edition because this solution they wrote doesn't make any goddamn sense

[shrewd urchin]

It the latest one . The first was written in I think 1900s

In late 1900s

Can you give another solution that I can understand .

[willow bear]

ok

one may attempt to count the elements of $S$ by doing $$\sum_{r=1}^{20} n(X_r) = 5 \cdot 20 = 100$$

[obsidian monolith]

Ann:

[willow bear]

but in doing so, each element of S is counted exactly 10 times, as per the problem statement

because each element belongs to exactly ten sets

so $\sum_{r=1}^{20} n(X_r) = 10n(S)$

[obsidian monolith]

Ann:

[willow bear]

and so $n(S) = \frac{100}{10} = 10$

[obsidian monolith]

Ann:

[willow bear]

does that make sense

[shrewd urchin]

Let me read it again

[shrewd urchin]

I think I didn't understand the question

If each element of S belong to exactly 10 of the X_t's and exactly 4 of the Y_r's I don't get this part what does this mean

Explain me this part thanks in advance 😀

[shrewd urchin]

Anyone ?

[willow bear]

@shrewd urchin still here?

[shrewd urchin]

ys

[willow bear]

imagine the elements of S as balls

[shrewd urchin]

ALright

[willow bear]

now, put a sticker on each element of X_1

then put a sticker on each element of X_2

(so that elements of X_1 ∩ X_2 will now each have 2 stickers)

then put a sticker on each element of X_3, and so on

once you do this for all sets from X_1 to X_20, each element will have ten stickers on it

that's what the problem says

[shrewd urchin]

SOOO each set of X_r consisit of 10 values of S

[willow bear]

no!

n(X_r) = 5

okay let me word this another way

[shrewd urchin]

values=elements

[willow bear]

no look

[shrewd urchin]

Yes rephrase the question

[willow bear]

imagine a table

in which the rows are labeled with elements of S

and the columns are labeled with numbers 1 through 20

[shrewd urchin]

alright

[willow bear]

for every $a \in S$ and $r \in {1, 2, ..., 20}$, you put a tick mark in row $a$, column $r$ if and only if $a \in X_r$.

[obsidian monolith]

Ann:

[willow bear]

that way, each column, corresponding to one of your sets, will have 5 ticks in it, in total.

[shrewd urchin]

You mean tick the

elements which are in S

[willow bear]

no

you don't tick the elements

you tick the cells in the table

[shrewd urchin]

yes that cell in which the element is

[willow bear]

no, the elements are the row labels!

sigh

[shrewd urchin]

leave that will understand it later

[viscid thistle]

how can i prove that the nth prime lies between n^a+1 and n^b where a = max(ln(p(n)-1)/ln(n)) and b= min(ln(p(n))/lnn)

can't do tex

[sharp marsh]

In triangle ABC, b = 12, c = 8, and <B = 40*. Find the missing side and angle measures

So would I just do Sin40/12 = SinC/8

Like how do I do that part when I multiply both side by 12

Or is it better to just flip the sin to the bottom part to start

Cause I'm getting like C = 26.66666667 and that doesn't sound right

Or do I get Sin40 = (SinC * 12)/8

[uncut mulch]

rearrange the equation to isolate C

[sharp marsh]

uh

How

[uncut mulch]

similar to how you isolated sin40 but do it for C instead

[sharp marsh]

Multiply by 8

[tidal sequoia]

I just learned about factorials and combinations and permutations on this online class can somebody help clarify something

[pale kettle]

[tidal sequoia]

so for permutations

n!/(n-r)! is the formula. is the (n-r)! to account for the each used up slot in an order?

I'm not sure I fully understand the formula

like r should be the number of things you want in an order it sounds like, so you do (n-r)! in order to cancel out the rest of n! after you got what you need?

[pale kettle]

kind of?

The way I think about that formula is

you're picking r things out of n objects

so for the first object, you have n choices, then you have n-1 choices

etc. all the way down to n - r + 1

[tidal sequoia]

like I'm thinking of this reasoning, because if you have n=10 and r=3, you're only selecting 3 objects, so 10!/(10-3)! = 10!/7!

which is effectively 10x9x8x7!/7! = 10x9x8 (because at this point multiply after the 8 wouldn't make sense because 10x9x8 accounts for the 3 different objects (r)

[pale kettle]

Right, that's kind of the same reasoning as what I said

[rugged ice]

Can someone give me a tl;dr on reading limits? I'm pressed for time and this is the last day of the class.

[grizzled orchid]

wot

wdym "reading limits"

[serene heath]

speed limits?

just read the number lol

[grizzled orchid]

[rugged ice]

[grizzled orchid]

no literally we don't know what you mean

[serene heath]

@rugged ice give example?

[obsidian monolith]

JY1853:

[hybrid charm]

$\lim_{n\to \infty}(\frac{1}{n+1}+\frac{1}{n+2}+\dots+\frac{1}{2n})$

[grizzled orchid]

[obsidian monolith]

Nick (Far From Home):

[grizzled orchid]

acceptable

[echo plaza]

those smol brackets

[limber bone]

isnt that integral 1/2n

XD

[echo plaza]

$\lim_{n\to\infty}\br{\frac1{n+1}+\frac1{n+2}+\cdots+\frac1{2n}}$

[obsidian monolith]

CaptainLightning:

[echo plaza]

$\lim_{n\to\infty}\sum_{k=1}^n\frac1{n+k}\=\lim_{n\to\infty}\frac1n\sum_{k=1}^n\frac1{1+\frac{k}{n}}$

[obsidian monolith]

CaptainLightning:

[echo plaza]

in case anyone actually wanted help

[hybrid charm]

oic \br

$\br{\frac{x}{y}}$

:

[short sorrel]

...yes, thats what that example is proving

[shrewd urchin]

oK

[ivory cargo]

is there a mathbot

how do i use it

nvm

[short sorrel]

$\Gamma \varnothing \circ \ell \mathbb{I} 5 \emph{H} m 0 \bR + \alpha \emph{L}$

[obsidian monolith]

Namington:

[willow bear]

what

[grizzled orchid]

no u

[shrewd urchin]

[shrewd urchin]

A={2,6} , B={2,4,6},C={2,4,6,8...},D={6} so can i write that $D \subset A \subset B \subset C$

[obsidian monolith]

Krishna (An unexperienced PS):

[shrewd urchin]

is it ok to write chain of subsets ?

[proud sparrow]

Looks okay

[echo plaza]

\{

\}

[valid flint]

yes thats a perfectly fine notation since the subset symbol is associative

[proud sparrow]

and transitive

[echo plaza]

calling the subset symbol associative is a hot take

[valid flint]

transitive is what i meant. i keep mixing this shit up lol my bad

[willow bear]

yes, that's exactly how you write inclusion chains anyway

$A = \{ 2, 6 \} \\
B = \{ 2, 4, 6 \} \\
C = \{ 2, 4, 6, 8 \} \\
D = \{ 6 \} \\
D \subset A \subset B \subset C$

[obsidian monolith]

Ann:

[shrewd urchin]

Let A, B, and C be the sets such that A ∪ B = A ∪ C and A ∩ B = A∩ C. Show
that B = C.

[short sorrel]

consider what it means for sets to be equal

[shrewd urchin]

Help

[limber bone]

A subset B means all elements of A are in B

this is probably proper subset which means they are not equal

since all elements of A are in B A-B means all elements in A but not in B

but all elemenets in A are in B

so A-B is the empty set

i think the excercise wants you to like prove A --> B ig

[willow bear]

@shrewd urchin you're going to need to prove a total of four implications

one possible pattern would be $\mathrm{(i) \implies (ii) \implies (iii) \implies (iv) \implies (i)}$

[obsidian monolith]

Ann:

[bold maple]

5x+2y=−16
y−3x=3

would i multiply the bottom equation to be able to do subtraction/addition by -2

[uncut mulch]

have you tried/did it work?

[bold maple]

I have been trying to work it out, but I forgot that if I multiply one side do I have to multiply the other? @uncut mulch

[uncut mulch]

multiply both sides to keep the equation consistent

[willow bear]

if you do something to one side of an equation

then you have to do the same thing to the other side

otherwise, the equation gets messed up!

[bold maple]

Do I need to multiply by the same number for both equations?

[uncut mulch]

do you mean sides?

(because multiplying the other equation by that same number doesn't help you in any way)

[tawdry current]

You only multiply the other equation - as long as you multiply both sides of one equation you are fine.

[steady plaza]

how do i expand this?

[pale kettle]

use some log rules

[steady plaza]

yeah nvm it was so ez #backtoschoolszn lol

[distant hearth]

Why is precalc a fucking thing

[acoustic laurel]

exactly dude

[shrewd urchin]

I have two statements

$A \subset B$

[obsidian monolith]

Krishna (An unexperienced PS):

[proud sparrow]

and?

[shrewd urchin]

$A-B=\varnothing$

[obsidian monolith]

Krishna (An unexperienced PS):

[proud sparrow]

Are they equivalent

[shrewd urchin]

I have to show that both are equivalent

I wrote a proof

[proud sparrow]

Okay

[shrewd urchin]

please check if the reasoning is valid

[proud sparrow]

Sure

[shrewd urchin]

let me rite it down

here

[proud sparrow]

Let's see your rites

[shrewd urchin]

Given: $A \subset B$ To show: $A-B-\varnothing$ \ Proof:Lets assume that $A-B \not= \varnothing$ this implies that there exist x such that and $x \in A and x\notin B$ \ But as $A \subset B \implies A-B=\varnothing$

[obsidian monolith]

Krishna (An unexperienced PS):

[proud sparrow]

Where did you get $A \subset B \implies A-B=\varnothing$?

[obsidian monolith]

Element118:

[pale kettle]

wtf

that is not a proof

[proud sparrow]

looks incomplete

[willow bear]

you said some things and then out of the blue asserted the very thing you wanted to prove

[valid flint]

i can see what you wanted to do, but as it stands it is incomplete.

[shrewd urchin]

All the elemnts of A are contained in B this implies that there doesnt exist a x which is in A and not in B

[valid flint]

you want to show a-b≠ null is a contradiction, so do that fully before saying implies a-b=null

[willow bear]

well you needed to say

[shrewd urchin]

I have to show $A \subset B \implies A -B=\varnothing$

[obsidian monolith]

Krishna (An unexperienced PS):

[willow bear]

"let's assume A - B ≠ ∅, this implies there exists an x such that x ∈ A and x ∉ B. but since A is a subset of B, x ∈ A implies x ∈ B"

and then

being honest to yourself

point out the contradiction

[shrewd urchin]

SO i should write that this leads to contradiction as $A \subset B $

[obsidian monolith]

Krishna (An unexperienced PS):

[willow bear]

no

after this

you should write

"This is a contradiction, so the assumption A - B ≠ ∅ is false. Therefore A - B = ∅, as desired."

[proud sparrow]

Let's assume A - B ≠ ∅, this implies there exists an x such that x ∈ A and x ∉ B. Since x ∈ A and A is a subset of B, then x ∈ B. This is a contradiction. as we have x ∉ B and x ∈ B. Hence, A - B = ∅.

[shrewd urchin]

Ohk

thats same

But ok

[proud sparrow]

It's best to be specific on what the contradiction is

[shrewd urchin]

Tbh i have never learned proofs

[valid flint]

its like you had the words but didnt use coherent grammar to form a coherent sentence. you had the ideas but didnt make a coherent proof

[shrewd urchin]

except of geometry

[proud sparrow]

Same idea, structure your ideas into a coherent line.

[shrewd urchin]

Today will complete the chapter of mathematical reasoning

yes yours and anns proof are better worded

[proud sparrow]

There's still the reverse to prove

we are not done yet

[shrewd urchin]

Yes i can do it

[proud sparrow]

Start with A - B = ∅, and assume A is not a subset of B....

[shrewd urchin]

$A \subset B \iff A-B=\varnothing$

[obsidian monolith]

Krishna (An unexperienced PS):

[shrewd urchin]

yes done

[willow bear]

well we've done =>

now we need <=

[shrewd urchin]

I mean on paper

done it in my notebook

[valid flint]

show us your proof maybe, it will help to have us see if you did everything correctly

[willow bear]

ok wait

no

[shrewd urchin]

need help with this proof

[willow bear]

okay let's just decide what you're doing rn

so as not to mix two different problems

[shrewd urchin]

Ok first lemme show you people the reverse proof

Proof:Lets say that $A \not\subset B$ $\implies x\in A and x \notin B$ . Now since $A-B=\varnothing$ $\implies x \in A then x \in B$ . this is a contradiction my assumption was wrong hence $a \subset B $

[valid flint]

\not \subset

theres a few problems with this. A-B≠ null doesnt mean x in A and x in B, can you see what it should mean

[shrewd urchin]

$A-B=\varnothing$ means that there is no element x that is in A but not in B

[obsidian monolith]

Krishna (An unexperienced PS):

[valid flint]

thats one way to put it, yeah. use implications to write this

[shrewd urchin]

ok

[valid flint]

$( p \land q) \neq (p \implies q)$

[obsidian monolith]

JohnDoeSmith:

[shrewd urchin]

I have seen this notation in the next chapter but dont know it

[valid flint]

uh that thing between the first p and q just means and

[shrewd urchin]

lazy

lol

[valid flint]

its a commonly used symbol

[shrewd urchin]

What do you want to say ? witht that proof is incorrect ?😫

[valid flint]

but anyways. (x in A) and (x in B) doesnt mean anything because we havent defined x. (x in A) implies (x in B) makes sense, as we are saying x that if x is in A then x must be in B

[shrewd urchin]

alright done

[valid flint]

show me

[shrewd urchin]

What?

[obsidian monolith]

Krishna (An unexperienced PS):

[valid flint]

ehh i guess then works. also we need to define x in the first line

use a $\exists$

[obsidian monolith]

JohnDoeSmith:

[shrewd urchin]

Will learn reasoning in the next chapter

anyways

[valid flint]

no your proof is still broken

[shrewd urchin]

why?

[valid flint]

$A \not\subset B$ $\implies x\in A and x \notin B$

[obsidian monolith]

JohnDoeSmith:

[valid flint]

what is x? we havent defined it

[shrewd urchin]

x is an element in A

[valid flint]

so the thing is (x in A) and (x not in B) is a logical test, it doesnt have defination built into it

we need to state somewhere that such an x exists, i.e

[shrewd urchin]

I dont have to write such a rigorous proof

[valid flint]

$A \not\subset B$ $ \implies \exists x : x\in A $ and $x \notin B$

[shrewd urchin]

I think

[obsidian monolith]

JohnDoeSmith:

[shrewd urchin]

yes correct

[valid flint]

if you arent writing rigorous proofs why write proofs at all though

but anyways, do you understand this

[shrewd urchin]

lol

continue

yes

[valid flint]

hmm ok i will expect you to do this on the next problem then

[shrewd urchin]

except that reflection of E

[valid flint]

that means there exists

[shrewd urchin]

OK

[valid flint]

we are saying that there exists some x such that (x in A) and (x not in B)

[shrewd urchin]

yes i know basic set therotic language

[valid flint]

cool. now we can move on to your other problem

[shrewd urchin]

Yes

[valid flint]

what have you tried

[shrewd urchin]

how to write therfore using latex

[valid flint]

\therefore

[shrewd urchin]

$x \in B\ \therefore x \in A \cup B \x \in A \cup C (A \cup B=A \cup C)\ x \in A or x \in C $

[valid flint]

or

[obsidian monolith]

Krishna (An unexperienced PS):

[shrewd urchin]

stook neow

[valid flint]

so far so good

[shrewd urchin]

What to do next

[valid flint]

well you have a second eqn with the intersects, try using that

[shrewd urchin]

$ x \in B \text{also} x \in A \ x \in A \cap B \ x \in A \cap C \ x \in A \text{and} x \in C $

[obsidian monolith]

Krishna (An unexperienced PS):

[shrewd urchin]

$B \subset C$

[obsidian monolith]

Krishna (An unexperienced PS):

[shrewd urchin]

similarly can show $C \subset B$

[obsidian monolith]

Krishna (An unexperienced PS):

[shrewd urchin]

and finally B=C

[valid flint]

uh did i miss the part where you showed any of those?

[shrewd urchin]

Yes i told simlarly i can show that

[valid flint]

you havent shown B subset C

[shrewd urchin]

x \in B alos x \in C hence B subset C

also*

[valid flint]

no you didnt even come close to that, you had x in A in both of your things remaining

[shrewd urchin]

I will be back my mother is calling me.

[valid flint]

ok

for when you come back, you have shown two things:

$ (x \in B) \implies (x \in A) \vee (x \in C)\ (x \in A) \land (x \in B) \implies (x \in A) \land (x \in C)$

[obsidian monolith]

JohnDoeSmith:

[valid flint]

we will try taking it from here

before we do though, i would like you to fully write out how you got these with good notation and such

[shrewd urchin]

Yes @valid flint

[shrewd urchin]

what nxt ?

[valid flint]

oh did you write everything up

also we are going to consider two cases, one in which x is in A and one in which x is not in A

[shrewd urchin]

$ (x \in B) \implies (x \in A) \vee (x \in C)\ (x \in A) \land (x \in B) \implies (x \in A) \land (x \in C)$

[obsidian monolith]

Krishna (An unexperienced PS):

[valid flint]

\ for line break

two \

[proud sparrow]

\\ you mean

[shrewd urchin]

so i have clarely showed that b subset c

[valid flint]

no, this is not sufficient to show b is a subset of c

[shrewd urchin]

Then what shouldi show

[valid flint]

ok first do we understand why this is not sufficient

[willow bear]

\\

[proud sparrow]

||1. x in B implies x in A or x in C
2. x in A and x in B implies x in A and x in C
Suppose x in B.
Two cases:
x in B and x in A, the second statement implies x in C
x in B and x not in A. Therefore x in C by the first statement.||

Basically, you need to show x in B implies x in C

that, we don't see anywhere

[shrewd urchin]

Good

[shrewd urchin]

Conisdering both conditon can i write B subset C

can I ?

Case1 that x in A and x in B implies x in A and x in C

Case 2 x in B implies x in C

b subset C

B*

ann is this correct?

[proud sparrow]

Well, be more specific how your cases are exhaustive

[shrewd urchin]

i found this on interent

[viscid thistle]

How do i solve this: sum ((-1)^i) * (k^i) from i=0 to n

alternating sum of powers

oh yeah thanks

[shrewd urchin]

@willow bear is the proof correct that i posted above i am a litlle doubtful.

[willow bear]

idk if it's correct bc it's written too badly to tell

[shrewd urchin]

[viscid thistle]

Anyone able to help with question 2

I dont know how to approach the question

[proud sparrow]

Okay, what can you do?

Any formulas that might be useful?

Maybe half life formula?

[willow bear]

,rotate -90

[willow bear]

@viscid thistle do you know what it means for the half-life of Pu239 to be 24000 years?

[willow bear]

...

[shrewd urchin]

@willow bear what does this mean

Maximum and minimum of sets

[willow bear]

uhhh

[shrewd urchin]

🤔

[willow bear]

this is poorly written

[shrewd urchin]

Why are they say that equal

[willow bear]

their use of the min and max functions is definitely inappropriate

i'd dock points if someone wrote that on a test tbh

here's what i would write

[shrewd urchin]

How can the authors be so irresponsible these books are referred by students all over Maharashtra state

[willow bear]

For any sets $A$ and $B$, the following inequalities are true:
$$\max{n(A), n(B)} \leq n(A \cup B) \leq n(A) + n(B)$$
$$0 \leq n(A \cap B) \leq \min{n(A), n(B)}$$

i honestly don't know!

[obsidian monolith]

Ann:

[shrewd urchin]

[willow bear]

yeah really idk

lack of responsibility on the editors' part

notational sloppiness like this needs to be dealt with

[shrewd urchin]

They are irresponsible

And our teacher don't discuss theory

They try to make us memroise the steps

[willow bear]

yeah that's

[shrewd urchin]

[willow bear]

not how you teach math

at all

[shrewd urchin]

Yes

That's why I don't go to school

They make money but no sense at all

[viscid thistle]

@willow bear indian cram schools have even worse notation

as you have undoubtedly seen

stuff like $\sum_{n=1}^{n} f(n)$

[echo plaza]

I've only seen cram schools in anime

[willow bear]

Y I K E S

[echo plaza]

lol

[viscid thistle]

or $\ell n$ for $\ln$

[shrewd urchin]

I don't go to cram school.

[echo plaza]

that euler guy

[willow bear]

Y I K E S ^2

[spring thunder]

edexcel grade sheet incoming

[viscid thistle]

lol edexcel

[bronze canyon]

Wait what’s wrong with that first thing

[echo plaza]

Edexcel spec

[shrewd urchin]

I go to school

[viscid thistle]

lower bound is n

[shrewd urchin]

School started 2 days ago

[viscid thistle]

upper bound is n

[bronze canyon]

Oh didn’t notice

Oops

[shrewd urchin]

I only attend bio lecture .

[viscid thistle]

lol

@echo plaza c r a m

[willow bear]

@viscid thistle any more examples of notational cancer

[shrewd urchin]

Crap school*

[spring thunder]

bio

[viscid thistle]

hmm let me think

[echo plaza]

let me get the edexcel spec

[spring thunder]

cramp schools

[shrewd urchin]

Cramp giving school

[echo plaza]

the one and only

[viscid thistle]

lol

[willow bear]

yikes

[spring thunder]

amazing

[willow bear]

wow just

[echo plaza]

this came up on the exam this year

[willow bear]

ok this isn't a yikes

this is a thonk

[shrewd urchin]

Why don't people follow convention

[viscid thistle]

they either don't know
or they think that everything will be understood from context

[shrewd urchin]

The bad rule is 70% attendance is important in school

[willow bear]

yknow what we should have

[spring thunder]

it's not a matter of following conventions, it's just utterly shit

[willow bear]

a YIKES emoji

[viscid thistle]

lol

[echo plaza]

I have a lot of those

[viscid thistle]

pepe yikes

[spring thunder]

just use ann

math lewdness

[echo plaza]

[shrewd urchin]

[willow bear]

no no

lewdness isn't yikes material

[shrewd urchin]

[willow bear]

this will do ig

[shrewd urchin]

Anyone knows a good book to learn algebra

[spring thunder]

yeah this one's ok

[echo plaza]

dummit and foote

[spring thunder]

algebra or algebra

[viscid thistle]

ohhhhh ok now i remember a really bad notational thing

$a_1 + a_2 + a_3 + \dots \infty$

[viscid thistle]

they always do the infty thing

always

[echo plaza]

bruh

[spring thunder]

$a_1 + a_2 + ... + a_\infty$

[echo plaza]

they could've just not put that there

[obsidian monolith]

emeric75:

[viscid thistle]

well yeah

but they do

[echo plaza]

and it would've been ok

[spring thunder]

[echo plaza]

oh my god

[viscid thistle]

https://files.askiitians.com/cdn1/images/2014822-112828350-3105-lightbulb_walking.gif

[spring thunder]

oh my god

louis the lightbulb

[viscid thistle]

lol

[shrewd urchin]

I remember a bad notation [-inf,inf]

[viscid thistle]

@willow bear please scroll up

also lol krishna

[echo plaza]

$[-\infty,\infty]=\bR^*$

[obsidian monolith]

CaptainLightning:

[shrewd urchin]

Why?

[viscid thistle]

where $\bR^*$ is the group $(\bR, \cdot)$

[willow bear]

oh god, jesus fuck

[viscid thistle]

group

[shrewd urchin]

How did you make this

[spring thunder]

who uses R* for extended reals jesus

[viscid thistle]

make what

[willow bear]

> like this

like this

[spring thunder]

guten tag

[shrewd urchin]

Oxide is my son

[spring thunder]

you need the space

[shrewd urchin]

Doesn't work

[echo plaza]

Krishna

[shrewd urchin]

Oxide is my son

[echo plaza]

lol

[shrewd urchin]

Oxide kadak mat ban tu thokla sirf pateli marta hai tu andar se khokla.
Tu chalata cycle me chalata beta rokla.

😎

@viscid thistle

[spring thunder]

[viscid thistle]

lol

[shrewd urchin]

You understand that? @viscid thistle

[willow bear]

please translate

[shrewd urchin]

You don't understand mumbai accent hindi

[viscid thistle]

@echo plaza

[shrewd urchin]

Android ka zamana aur tu pager hai. @viscid thistle

[echo plaza]

is that the screenshot I took

[shrewd urchin]

[echo plaza]

lol

[shrewd urchin]

People who write book in hurry must be hanged

[spring thunder]

retract sounds like an insult

[echo plaza]

where $\pi$ is the unit circle centered on $0$ traversed once counterclockwise

[obsidian monolith]

CaptainLightning:

[viscid thistle]

i see nothing wrong

[shrewd urchin]

Where the author is jerk. Who wrote the book in hurry

[viscid thistle]

krishna, did you know? that screenshot is an undergraduate entrance exam

[shrewd urchin]

I am taking about my book

Not that babbyish think that you send

Lol

[echo plaza]

so it is that screenshot

[shrewd urchin]

Lol

[spring thunder]

undergrad entrance complex anal

oof

[viscid thistle]

undergrad entrance metric space topo

[shrewd urchin]

Math phd entrance to ug entrance

[shrewd urchin]

Math is so much heck. People write bad books confusing notation

[echo plaza]

w8 I have red on my avatar?

[shrewd urchin]

Why did you send that ss Ann?

Oxide I was joking don't take my words seriously

Anyways I am.wrting what I understand from that symbols

Correct me if I am wrong

$\text{min}{n(A \cup B))}=max{n(A),n(B)}$ means A U B is min when B is a subset of A and and hence A U B=A. min(A U B)=n(A)%

[obsidian monolith]

Krishna (An unexperienced PS):

[willow bear]

the notation is inappropriate

also bad tex

[echo plaza]

$\min\max$

[obsidian monolith]

CaptainLightning:

[shrewd urchin]

Yes U=union

[willow bear]

not that only

anyway

it is true that the minimum possible value of n(A ∪ B), taken over all possible sets A and B with fixed cardinality, is equal to min{n(A), n(B)}.

[shrewd urchin]

Cardinality

[willow bear]

don't want to use "size"

but it means number of elements

[shrewd urchin]

Ok

[willow bear]

which i guess is fine since right now you only seem to really work with finite sets

[shrewd urchin]

Google searched

[willow bear]

but uh yeah once you get to infinite sets

that shit gets wack

[shrewd urchin]

Then it's undefined right

[viscid thistle]

no

[shrewd urchin]

I mean we cannot count the cardinality

[viscid thistle]

you can

[shrewd urchin]

Calculate the cardinality of real.numbers for examlle

[spring thunder]

(and "size" is pretty vague anyway, i could classify my sets by the number of multiples of 10 there is in them if i wanted)

[shrewd urchin]

This type of ser

[viscid thistle]

$|\bR| = 2^{\aleph_0}$

[viscid thistle]

where \aleph_0 is the cardinality of the natural numbers

[willow bear]

okay so uh

yeah cardinality has a more precise definition

[shrewd urchin]

Wot

[willow bear]

but TL;DR there are different kinds of infinity

you'll learn it in due time

...hopefully

[viscid thistle]

lol

[willow bear]

i think it'd only do harm if you tried to do anything with it now

[viscid thistle]

or you'll watch that one vsauce video and have weird ideas

[shrewd urchin]

This problem makes me feel.amgry how about you @viscid thistle

[viscid thistle]

lmao

[shrewd urchin]

#new edition Maharashtra board textbook

[viscid thistle]

it should be passed instead of failed

no way 160 students are passing jee just like that

[shrewd urchin]

Lol

[willow bear]

pfwahahahah

wow

fgsdjlfgsdg

holy shit

[viscid thistle]

lol

what happened ann

[willow bear]

HOLY SHIT THIS IS MOTHERFUCKING GOLD SDFGHSDJKLG

[spring thunder]

inb4 "infinities are infinite but some are more infinite than others" déclaration des droits de l'Homme et du citoyen, 2192

[willow bear]

déclaration des droits de l'Infini

[viscid thistle]

the authors of this textbook are truly immersed in post-irony meme culture

[shrewd urchin]

Lol

30 people with Dr. Label wrote this book

[spring thunder]

"all people are born equal but some are more equal than others"

mega meme

[viscid thistle]

lol

[shrewd urchin]

A dedicated non Dr. Will write a good book

Reading the books preface any one will laugh

[spring thunder]

what does neet stand for in here ?

[shrewd urchin]

We have written this book so that our students can perform better at competitive exams.

[echo plaza]

NEETJII

[willow bear]

NEET

how about

YEET

[spring thunder]

how about

FEET

[viscid thistle]

NEET

[shrewd urchin]

Neet is a national entrance exam for admission in Indians biggest medical institutes

[viscid thistle]

national eligibility cum entrance test

[shrewd urchin]

Jee is national exam for admission in Indians best engineering college

[viscid thistle]

cum entrance test

[spring thunder]

yea ik jee

[shrewd urchin]

jee= pressure

On kid

If he is not a competitive kid

[spring thunder]

jackass entrance examination

[shrewd urchin]

Lol

Enough of this I am.goinh to.listens emiway > I been that