help-4

[onyx peak]

And chose [0,pi] just because they wanted to or is that something I shouldn’t be concerned about for calc I and calc II?

[tiny torrent]

just because they wanted to

it could be any interval that allows cos to be one to one, there are infinite amounts but if ppl kept using different ones that would be a hassle so that is the agreed upon domain

and yeah you shouldn’t be concerned as long as you remember the restricted domains, its very important since calc 1 and 2 deals with a lot of trig

you wont be tested on why but its good to know

[onyx peak]

Well I really appreciate your help and @lyric sundial

Sorry if some of what I said sounded stupid

I had a teacher that just came in to check a box and leave

Never answered questions

[tiny torrent]

np and your alright, I tutor for calc and that’s usually something a lot of students don’t remeber so its good your working on it

[onyx peak]

I’m glad that I’m not alone

.close

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[open niche]

Hello is my proof correct ?

[tiny yacht]

i dont think line 1 is always correct

[open niche]

Wdym

[tiny yacht]

one sec, i try to find a counterexample

[open niche]

Sorry forgot to mention

We assume limit f exists

[tiny yacht]

ahhh ok then it's fine

[open niche]

Its on the start sorry this is just the 5th theorem

[open niche]

Alr ty

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[tiny yacht]

@open niche i only meant that line 1 is fine then

what are you trying to show here?
that when lim|f| = |lim f| => lim f = k?

or the other way around?

[open niche]

no

if limf exists

then lim|f| = |limf|

this is the theorem

and im proving it

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[open niche]

,

[open niche]

.reopen

[open niche]

im asking if the proof is correct

[tiny yacht]

i think i know what you mean but i also think this is a bit unclear

what you could do is
-show that lim |f| exists
-show that lim |f| = |k|
-say that |lim f| = |k| is trivial

to show that lim |f| exists, you can use the epsilon delta proof
if you start with the one for lim f, you can use triangle inequality: | |f| - |k| | <= | f - k | < epsilon

and that way you can show not only that lim |f| exists but also that it equals |k|

and you used lim f = k in the process

[tiny yacht]

this does make sense, but i have met tutors that would not give all points for this, because they want everything to be 100% precise

[open niche]

near xo means in 0<|x-xo|<δ

from the theorem that says if limf >0 then then there is δ such that 0<|x-xo|< δ => f(x)>0

i have already proved this previously thats why i used it

[tiny yacht]

ok then it makes sense

[open niche]

also something random

do we define kth root of a with a negative?

[tiny yacht]

i think i would include the delta there though
if k>0 => there exists a delta such that f(x)>0 near x_0

because if say it like this, it sounds like it is true for all delta

[open niche]

no i say near xo

near means could be very near or whatever but there is a small interval no matter how small that it is positive including xo

[tiny yacht]

ah ok, then it's fine

[tiny yacht]

what did you mean with this?

root as in

[open niche]

is 3rd root of -1 something defined?

the counter example my teacher gave me

is

what if we have 3rd root -1 then -1 ^ 1/3 = -1 ^2/6 = 1^ 1/6 =1 so -1 = 1 ?

when i was in highschool

but now that i come back to it there is an issue

the identity a^m/n= (a^m)^1/n is only true for positive a

idk i was confused back then

an other teacher told me we just defined it with a>=0 so we cant write a negative number

[tiny yacht]

yeah i get that, it get's fed into us that that rule is natural and makes perfect sense and should always work

[valid star]

Yes it’s defined. I assume by 3rd root you mean $\sqrt[3]{x}$ right?

[rocky lotus]

Chungus Kahn

[open niche]

but i didnt really give it much attention at the time as i had to focus on studying for entrance exams in a big uni

but never cleared it up

so

can someone explain if i can write 3sqrt(a) where a is negative?

i was told i could not but teacher might have just said it cause its complicated so we dont get more confused thats why im asking

[tiny yacht]

(-1)^1/3 asks, what number to the power of 3 is -1
if we look at real numbers, only -1 makes sense
if we look at complex numbers, you can think of the like vectors with direction and length
if we multiply two such complex numbers, their length multiplies and their angle adds
so we want a number z with length 1 and an angle phi such that 3phi=pi (because an angle of pi is 180° and alines with the negative x axis so that z^3 can land on -1)

[open niche]

talking about real numbers

[tiny yacht]

then only -1

[open niche]

thats something different you said

[valid star]

The cube root of a negative number is negative. So, for example, the cube root of -8 would be -2

[open niche]

you said that -1 ^3 is -1

[tiny yacht]

what

[open niche]

that doesnt mean that 3rd root of -1 is -1

the question is

[valid star]

?

[open niche]

when defining kth root of a

[tiny yacht]

sqrt(-8) is not -2

[open niche]

did we allow a to be negative when being in real numbers

[tiny yacht]

if we look at 3sqrt(a) with a being negative, then there is no real solution

and for real numbers, the sqrt() function is defined such that the result is always positive

[open niche]

3root i meant

[tiny yacht]

sqrt() of a negative number will always be complex

[open niche]

like 3rd root

idk how to write it

[tiny yacht]

ahhhh

my bad

[open niche]

but like im not sure whats true

i was told in real analysis we only defined it for a positive or 0

[valid star]

Yeah, I said cube root

[tiny yacht]

turns out i can't read 😭

[open niche]

like strictly of of definition

[valid star]

It’s fine lol

[tiny yacht]

this is true for square root

[open niche]

did we restrict a >= 0

[valid star]

Not for cube roots

[open niche]

kth root

[tiny yacht]

if a<0 and we talk about cubic root (i think qbrt) then it's fine

[open niche]

is it ?

[valid star]

Not for where k is an odd number

For even index roots, taking them of negative numbers will give you a imaginary number answer; for odd index roots it can just be a negative real number

[open niche]

we are in R i say again

otherwise it gives infinite results

[valid star]

Yes. So for even index roots, a is restricted to a>=0. For odd index roots, you don’t need that restriction

[open niche]

its like multivalued in C

[open niche]

index roots?

oh yes

[open niche]

its not about if you need

but like in definition

did we restrict it

[valid star]

If you’re only working in R, then yes, it’s restricted. If you’re working outside of R, then no

[open niche]

in R

for odd k

[valid star]

Sorry, let me reword

In R: roots with odd indexes (3rd, 5th, etc) are not restricted, roots with even indexes are

Does that make sense?

[open niche]

yea idk

idk if context matters

[valid star]

Wdym?

[lyric sundial]

Sure it does

[open niche]

as in R or C context

[lyric sundial]

Yes

[open niche]

but does k have to be positive?

[valid star]

k can be negative as well

[open niche]

how do i symbolise integers?

Z?

[lyric sundial]

Yes, Z is the set of integers

(from the German word Zahlen, which means numbers/digits)

[open niche]

how do i symbolise the upside down U

in words

idk how to write this

[lyric sundial]

$\cap$

[rocky lotus]

Alberto Z.

[open niche]

oh yes ik the word didnt realise its from that

[lyric sundial]

This?

[open niche]

no like

in words

how would i write

[valid star]

Intersection maybe

[open niche]

we write the curly E like a in R instead of a curly E R

[valid star]

I always wondered why it was Z

[lyric sundial]

Yeah but is this the symbol you're referring to?

[open niche]

yes

[lyric sundial]

Yes, it's called intersection

I don't know what you mean with formalize

[open niche]

so i would say like

k in R intersection Z

to imply k isnt an integger

[valid star]

Well I’m pretty sure that just gives you Z

Because Z is a subset of R

[open niche]

intersection

[open niche]

like this

as in R-Z

$R \cap Z$

[rocky lotus]

taebek

[valid star]

Yeah, so intersection gives you the elements common to both sets.

And Z is fully contained in R, so it’s just Z

[open niche]

oh wait yea no

[valid star]

If you want all of the elements its the other U “union”

[open niche]

no

i want R-Z

[valid star]

Oh R without Z

[open niche]

yes

whats the symbol

[valid star]

Usually that’s R\Z or just R-Z

[lyric sundial]

The backslash

[open niche]

ight

thank you both for helping me

have a great rest of your days

.close

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[charred burrow]

$\setminus$

[rocky lotus]

DaveyLovesSocks

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[signal chasm]

can someone

[signal chasm]

explain

these type of graphs

i dont get it

when like its sin^2

or cos^2

[jovial edge]

you mean odd and even function?

[turbid valve]

Or what this notation means?

[jovial edge]

i believe it's sin(x)^2

[signal chasm]

this

tbh

with this problem

its just easier to sketch it

thing is

i did sketch it before

on a previous problem

not recent

and it led me to the wrong

ans

so i kinda

left it

[glass kelp]

Odd: (symmetrical abt origin)
f(-x) = -f(x)
Even: (symmetrical abt y axis)
f(-x) = f(x)

[turbid valve]

Try writing out a full utterance before pressing ENTER

Doing

Something like

this

Is much harder

to read

[jovial edge]

also what even is this graph

[turbid valve]

(I'd add "rotationally" and "reflectively" where appropriate, to make this clearer)

[turbid valve]

An example of a trapezium being used to approximate the area of a generic function y = f(x)? I'm guessing

[jovial edge]

even function suppose to be symmetrical , no? or this isn't function of sin(x)^2?

oh okay so it's a random function, i get it

[glass kelp]

It’s the integral

I think

Wait nvm

What tf is that

[patent hedge]

I am in 7 I want to clear olympiad of math can any one help me?

[jovial edge]

!help

[vale dock]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[patent hedge]

?

[tawny moon]

the channel is occupied by another person

please get another one!

also, welcome to the server!

[signal chasm]

just

made up

.close

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[signal chasm]

bro

am i dumb

if B is true

why cant A be true

[jovial edge]

reopen the channel

[signal chasm]

i did

.reopen

[vale dock]

✅

[jovial edge]

!xy

[vale dock]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[signal chasm]

bro

im not joking

thats it

[jovial edge]

Is this relate to the previous one?

[signal chasm]

thats the problem

no

its just

alone

[jovial edge]

oh

mb then

[signal chasm]

like

so

if D was true

we know that a and B

would be wrong

but i dont get

this

if and only if thing

[jovial edge]

D cover C, B cover A and A cover C and D

So the answer is C, no?

[signal chasm]

yh it is

but how

did u do it

wdym D cover C

i dont get that

whats the difference

between if

and if and only if

[jovial edge]

D can be $(2,+\infty) \bigcup ( \text{another interval} )$

[rocky lotus]

Alexis_Fx

[signal chasm]

so can be C

[jovial edge]

But C is strictly $(2,+\infty)$

[rocky lotus]

Alexis_Fx

[jovial edge]

if and only if

[signal chasm]

ok

i was thinking that

OH MY

yk thanks

i thought abt that

in my head

and just forgot

[jovial edge]

you're welcome

[signal chasm]

thanks man

[stark wedge]

\cup

[jovial edge]

oops

mb

I copied it from gg

[stark wedge]

yeah sorry i was gonna say that \bigcup is for indexed unions

$\bigcup_{i=1}^n A_i$ like this

[rocky lotus]

Ann

[jovial edge]

yeah i see, thanks

[vale dock]

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[orchid mango]

need help how to find domain and range guys. pls explainnn, I don't understand cuz our teacher haven't taught us yet

[orchid mango]

the num 1

[hearty belfry]

sketching the graph would help

[orchid mango]

howw

[hearty belfry]

do you know what a domain and range is?

[feral coral]

Let's say you have a function f(x)
The domain of that function is all the values that x can take on so that the function has a defined output
the range is the set of the possible outputs that you get over the domain

[orchid mango]

all i know domain is X and range is Y

[hearty belfry]

Read what dockson said

[orchid mango]

I see I see

[feral coral]

Taking on 3 - x^2
What can you plug in for x? Is there anything that you can't plug in?

[orchid mango]

Based from ur description earlier, I think u can plug in any numbers

[hearty belfry]

yeah so what would the domain be?

[orchid mango]

infinity?

[hearty belfry]

eh sort of, but no

[orchid mango]

then what

[hearty belfry]

the domain would be for all real numbers

[feral coral]

you said "any numbers"
well the type of number we're dealing with is the real numbers, the numbers on a number line

which is R

[hearty belfry]

yeah, generally you write the domain like this ${x \in \mathbb{R}}$

[feral coral]

or just $\mathbb{R}$

[rocky lotus]

Dockson

[feral coral]

since it's the same set

[feral coral]

so what about 2, what is its domain?

[rocky lotus]

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

[orchid mango]

still kinda clueless but I'm guessing it's still any real number R

output 4?

[feral coral]

you're right

yeah so if the output is always 4, that means the range is the set {4}

but it can take any input, so any real number R is the domain

does that make sense now?

[orchid mango]

yeah, I think Im getting the hang of it. thanks!

[feral coral]

👍

[feral coral]

feel free to close if you don't need anymore help

[orchid mango]

.close

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[keen lily]

is this correct?

[jovial edge]

the answer seems correct

[hearty belfry]

looks correct

[keen lily]

the question says I do not need to fully simplify the result, I am guessing I did not fully simplify?
how would I do that, fully simplify the result I have I mean

[hearty belfry]

I don't think you can simplify it further but I'm guessing it involves using the trigonometric addition formula

[keen lily]

one of these?

[hearty belfry]

yeah

but that just makes it a lot more messier tbh

[keen lily]

oh huh

[keen lily]

so i should js leave it like this

[hearty belfry]

Yeah

the question doesn't ask for a full simplification

[keen lily]

alr ty

[keen lily]

yea it said fully simplify I just don't know how far is fully

[jovial edge]

you can factor sin(x^2-x)

but that's the simplest form ig

[keen lily]

.solved

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[novel canyon]

messy ballus

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[sterile mauve]

how to solve equations like a^x = x
with lambert w or any other ways

[tame wolf]

Hi! Here is a detailed example (with lambert W)
https://en.wikipedia.org/wiki/Lambert_W_function#Solving_equations

[sterile mauve]

a^x = x
x . a^(-x) = 1
-x.a^(-x) = -1

got it

ty

W(-ln(a)) = -x . ln(a)
-W(-ln(a))/ln(a) = x

[tame wolf]

You also have to be careful with the branches

[sterile mauve]

wdym

[tame wolf]

Depending on a the equation might have two solutions (or none)

[sterile mauve]

second solution is a complex?

[tame wolf]

Actually there is an infinite amount of complex solutions

I meant real solutions

This comes from the fact that there is not a single W function

[sterile mauve]

yea lamberts domain is 1 to inf as i remember

[tame wolf]

There is W_n for each integer n

[sterile mauve]

ye yea i heard can i understand them with undergraduate math

im not even in uni

[tame wolf]

Alright I will try to explain it with a similar but better known situation
Do you know euler's identity e^(2 pi i) = 1?

It is relatively well known

[sterile mauve]

i know but just know

idk how to use it in something

[tame wolf]

Well, if I ask you about the solutions of e^x = e, one obvious solution is x=1, right?

[sterile mauve]

yea

but

if you divide e

x-1 = 2pii

oh

[tame wolf]

But if I choose x = 1 + 2 pi i, then e^x = e^(1 + 2 pi i) = e^1 * e^(2 pi i) = 1 * 1 = 1

Wait

So that x is not a solution

It is more complicated

But maybe you will believe me if I tell you that the equation has infinite solutions over the complex numbers

It is like the square root, which has two solutions

Or the cube root, which has three

[sterile mauve]

but it has no degree

so it makes sense it has inf sol

[tame wolf]

Like x^3 = 1 has three solutions over the complex numbers but only one of them is real

[sterile mauve]

yea

[tame wolf]

Degree?

[sterile mauve]

i mean x^1 x^2

degree as polinyomnal

[tame wolf]

Oh it's not really because of that

Like it is vaguely related because e^x is in some sense an infinite degree polynomial

But that is not relevant I think

What I mean is that the equation x^2 = a has two solutions and you get them using the two inverse functions (the square root sqrt(x) and -sqrt(x))

sqrt(x) makes sense for all complex numbers if you use the polar form

You divide the argument by two

And in a similar way there are also multiple inverse functions of xe^x

But an infinite number of them

And among all those, only W_0 and W_(-1) may take real values when x is real

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[signal chasm]

help

[signal chasm]

basically

this is a worked solution

but idk how i got it wrong

first set i out: 8,8,14

2nd set i put 9,9,18

so i got a range of 10

[hardy coral]

distinct integers

[signal chasm]

oh dead

cba

CBAAAAAA

ok

thanks

didnt see it

its

just trial and error right

[hardy coral]

It's not much trial

There's a pretty systematic way to do this

The medians basically give you one integer of each set of three. The smallest integer in each set should be as close as possible to the median because the mean is greater. So you'd like for the first set to be {7, 8, 15}, but then the second set has to have a 6 and that increases the overall range because it makes it {6, 9, 21}. So instead you put the 7 in the second set: {7, 9, 20} and that makes the first set {6, 8, 16}, with an overall range of [6, 20], aka 14

[signal chasm]

yh

makes sense

i kinda

uncounciosuly

di that

[vale dock]

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[midnight pier]

HI

[glass kelp]

hi vlad

[midnight pier]

can i solve this equation for x , y = i pi x - z ln(x)

using lambord function or something else i dont know ?

[glass kelp]

${y = i \pi x - z\ln x}$

[rocky lotus]

k

[midnight pier]

<@&286206848099549185>

[grim locust]

hm

first we can try to isolate the $\ln x$

[rocky lotus]

Amiso_

[grim locust]

$z \ln x = i\pi x - y \implies \ln x = \frac{i\pi x - y}{z}$

[rocky lotus]

Amiso_

[grim locust]

then $x = e^{\frac{i\pi x - y}{z}}$

[rocky lotus]

Amiso_

[midnight pier]

then multiplay both sides by e power to match lambord ?

[charred burrow]

just divide by $e^\frac{i\pi x}z$

[rocky lotus]

DaveyLovesSocks

[charred burrow]

on both sides

[grim locust]

yes

[charred burrow]

then you get an $xe^{\text{something}}$ and its pretty easy to use lambert from there

[rocky lotus]

DaveyLovesSocks

[midnight pier]

x/ e^((i pi x)/z) = e^((i pi x -y)/z - (i pi x)/z)

i would get this after dividing by e^ i pi x / x

[vale dock]

@midnight pier Has your question been resolved?

[midnight pier]

no i will try to find it my self

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[hushed crown]

Quick question this is the derivative of exp(a√x) / x (that the correction of an exercise) but for me there is an error and I'm not sure about it. See the derivative of exp(a√x) is ax / (2) * exp(a√x) but on the next step the x next to the a disappear. Do I miss something or is it an error ?

[hushed crown]

Ping me if you have the answer

[violet needle]

the first term looks kinda sus

because you have f' g

and g(x)=1/x

but you have x in the numerator instead of the denominator

the second term also looks sus

because g'(x) = -1/x^2

[strong raptor]

yeah, there shouldn't have a x in the first term

[violet needle]

but that seems to be missing

[strong raptor]

(in the numerator)

[violet needle]

oh shit my bad yeah none of this makes any sense

[strong raptor]

derivative of $\exp(a\sqrt{x}) is \frac{a\exp{a\sqrt{x}}{2\sqrt{x}}$

[rocky lotus]

Erk Gah

derivative of $\exp(a\sqrt{x}) is \frac{a\exp{a\sqrt{x}}{2\sqrt{x}}$
```Compilation error:```! File ended while scanning use of \frac .
<inserted text> 
                \par 
<*> 285089887798034436.tex
                          
I suspect you have forgotten a `}', causing me
to read past where you wanted me to stop.
I'll try to recover; but if the error is serious,
you'd better type `E' or `X' now and fix your file.```

[hushed crown]

this is the general form

[violet needle]

wait no sorry

it does almost make sense as I initally thought

the x just needs to be moved down to the denominator

[strong raptor]

that is correct, but the first term u'(x)v(x) is wrong

[hushed crown]

and this is f(x)

[strong raptor]

derivative of $\exp(a\sqrt{x}) \text{ is } \frac{a\exp(a\sqrt{x})}{2\sqrt{x}}$

[rocky lotus]

Erk Gah

[violet needle]

shit

I'm intellectually challenged sorry

nah I think the first term is fine

pretty sure the second term is missing an x though

[strong raptor]

it just has an extra x

in the numerator

[violet needle]

nah dude

[hushed crown]

Haha it's ok me too that why I'm here 😭

[strong raptor]

that shouldnt be there

[violet needle]

it comes from the v in your formula

[strong raptor]

oh yeh

[violet needle]

yeah

[strong raptor]

lol

[strong raptor]

aren't we all

[hushed crown]

yeah but then the next equality it disappears

[strong raptor]

i got too immersed in the derivative of exp(a*sqrt(x)), completely forgot about the other term lmao

[strong raptor]

i believe they factored exp(a\sqrt(x)) then divided somethin there

yea

no they just factored

[violet needle]

they either did something wrong or are missing steps

[hushed crown]

Oh I just saw it

There is x / √x

[violet needle]

they have the right answer but they definitely skipped a few steps

[hushed crown]

So if we make it rational it becomes x*√x / x so just √x

[violet needle]

oh what I'm really stupid

they didn't even skip any steps I just lack braincells

I haven't differentiated a quotient in years 😭

[strong raptor]

$\frac{\frac{axe^{a\sqrt{x}}}{2\sqrt{x}}-e^{a\sqrt{x}}}{x^2}= \frac{e^{a\sqrt{x}}\left(\frac{ax}{2\sqrt{x}}-1\right)}{x^2}$

huh

where di that 2 come from

[hushed crown]

Got the same reaction haha

[strong raptor]

ok

should be this

god damn

[violet needle]

the only nontrivial thing they do is they turn x / sqrt(x) into sqrt(x) but yeah joseph already noticed that

[rocky lotus]

Erk Gah

[hushed crown]

That it I think

[strong raptor]

oh

just multiply by sqrt(x)/sqrt(x) then it becomes sqrt(x)

[violet needle]

yeah or write it as x^1 / x^(1/2)

[strong raptor]

true...

[hushed crown]

Yeah I did not see it at first

[strong raptor]

yeha so

[hushed crown]

Well ty haha

[strong raptor]

everythings fine i think?

lol

[hushed crown]

Yup

[strong raptor]

incredible

[hushed crown]

cya

[strong raptor]

yw

[hushed crown]

.close

[vale dock]

Channel closed

Closed by @hushed crown

Use .reopen if this was a mistake.

[vale dock]

Available help channel!

Send your question here to claim the channel.

Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[night patio]

I am able to do it using venn diagrams

it is very intuitive

I am not able to write a proof for it tho

[hardy coral]

Can you write the distributive property?

[night patio]

wait i did use that

I wrote two proofs

for it

i just need help in

cleaning them a bit

sending wait

[night patio]

I used de morgan's law to get the 2nd step

is this proof correct?

In the last line, how do i prove that A ∪ (A ∩ B) = A

I wrote that because its kind of intuitive

how do i prove it

rest all of the line are correct ig?

oh wait i did something wrong in the start

sorry

[hardy coral]

It's called the absorption law
https://en.wikipedia.org/wiki/List_of_set_identities_and_relations#Other_identities_involving_two_sets

[night patio]

this is wrong

[hardy coral]

No that looks correct to me

Your proof is correct but it's way more complicated than it should be

[night patio]

wait yes it is correct

sorry

[night patio]

hmmm

how do i simplify it?

[hardy coral]

Just write down the distributive property for two sets X and Y

[night patio]

yes i did use the distributive property right away at the start

without de morgan's law

the steps were almost the same

[hardy coral]

Forget about the problem and your proof, consider two sets X and Y and write down what is called the distributive property

[night patio]

yes

wait

[hardy coral]

Ok that's correct, now note that these are equalities, they can be used both ways

[night patio]

ooooo

hmmm

oh wait

i get it

[turbid valve]

(cf how a(b+c) = ab + ac can be used both ways)

[night patio]

yes

SIMPLE

😭 😭 😭 😭 😭

how i am so dumb

[night patio]

crazy facts i should keep in mind

[night patio]

any proof for this?

[hardy coral]

Sure
https://proofwiki.org/wiki/Absorption_Laws_(Set_Theory)/Union_Absorbs_Intersection

[night patio]

how do i learn this art

this proof is so clean

thankyou

^^

[hardy coral]

By learning some proof theory / logic and also LaTeX I guess

[turbid valve]

Alternatively, think it through logically

I'm in S, or I'm in S and T

[night patio]

LaTeX

[turbid valve]

Well that second part is in S anyways

So it's just extra fluff

[night patio]

hmmmm

[turbid valve]

So I'm still just in S

[night patio]

will use this idea in other proofs too. i like it

[turbid valve]

So S u (S n T) is still just S

[night patio]

yupp

[turbid valve]

Trust me LaTeX isn't that bad

[hardy coral]

(you can click on those step descriptions:
https://proofwiki.org/wiki/Union_with_Superset_is_Superset)

[night patio]

no no i love it

hhahahaha

i have started using it a bit

while writing my notes in Obsidian

[night patio]

oh yes

its nice

thankyou everyone for your help

.close

[vale dock]

Channel closed

Closed by @night patio

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Send your question here to claim the channel.

Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

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[misty vessel]

Can anyone explain to me what permutations and combinations are? And how to solve basic questions such as

How many 3 digit even numbers can be formed from the digits 1, 2, 3, 4, 5 if the digits can be repeated?

[tawny moon]

combinations are any groupings of objects where order does not matter

eg: if i asked you to pick 3 coins from a pile of 4, there are 4 combinations here (ABC, ABD, ACD, BCD)

[misty vessel]

Ok

[tawny moon]

but if i now ask you to arrange what you picked, that becomes a permutation

because order matters

[misty vessel]

Which will be?

[night patio]

Try making arrangements of ABC

how many arrangements can you get?

[misty vessel]

Abc, acb, bca

[tawny moon]

there's more

[misty vessel]

Bac, cab

[tawny moon]

one more

[misty vessel]

Cba

[tawny moon]

there we go

[misty vessel]

Yeah

[tawny moon]

so you realize there are six ways to arrange three coins

[misty vessel]

Yes

[tawny moon]

but you also realize that there are four separate groups of three coins you could have picked

[night patio]

just so you don't miss any arrangements, you can try making them in a very symmetrical, ordered way. Like for ABC, you can do something like this:

ABC
ACB

BAC
BCA

CAB
CBA

[tawny moon]

so how many different arrangements of any three coins are there?

[violet needle]

If you're feeling adventurous and know what bijections are then you could think about how many bijections there are of the form f : {A, B, C} -> {A, B, C} and how they relate to permutations

[misty vessel]

This is where I get stuck

I don't understand the language

[tawny moon]

ok

so there are six ways to arrange ABC

[misty vessel]

Ok

[tawny moon]

how many ways are there to arrange ABD?

[misty vessel]

Yes

[tawny moon]

how many for ACD? BCD?

[misty vessel]

6?

[tawny moon]

mhm

[misty vessel]

6, 6?

[tawny moon]

good

so you realize that there are 6 ways to arrange any three coins and four different groups of three to consider

so first, we start from the fact that there are 4 different groups of three. then, from each group of three,, you can arrange the coins in six different ways

so the number of arrangements of three random coins from a pile of four is 6+6+6+6 = 6x4 = 24

[misty vessel]

So 4 groups of 3 then 3 groups of 3

Oh ok

Hmm ok

[tawny moon]

hold up. where did the 3 groups of three come from

[misty vessel]

Didn't you just edit that out?

Or my bad

[misty vessel]

What do you mean by "pile of four"?

[night patio]

there were 4 coins right

ABCD

[misty vessel]

Yes

[night patio]

so we have a pile of 4 coins

We can make 4 groups of 3 coins each
ABC
BCD
CDA
ABD

[misty vessel]

So basically there are 4 coins

[night patio]

yes

[misty vessel]

Ohhhh ok ok

[tawny moon]

apologies if my wording is unclear

[night patio]

Now

ABC
ACB
BAC
BCA
CAB
CBA

BCD
BDC
CBD
CDB
DBC
DCB
...

[misty vessel]

No problem, my understanding is the problem here

[misty vessel]

Yes

[night patio]

this is what they mean

total of 24 such arrangements for a pile of 4

[misty vessel]

Okk

Now I get it

[night patio]

Do it for 2 coins

AB

how many arrangements can you make?

[misty vessel]

2

[night patio]

How many can you make for 3 coins ABC?

[misty vessel]

9?

6

[night patio]

yes

How many for 4?

ABCD

[misty vessel]

12

[night patio]

if i tell you to make arrangements for 4 coins

we will get pairs like,

ABCD
ABDC
ACBD
ACDB
ADBC
ADCB
....

How many of these arrangements can we make?

[misty vessel]

I don't know how to do that

[misty vessel]

This

[night patio]

Okay lets go to 2 coins case,
AB
BA

[misty vessel]

Ok

[night patio]

if i give you a third coin C, how would you add C to these existing 2 pairs AB and BA

to make new pairs

[misty vessel]

Ok

[night patio]

think of a way of adding C to AB and BA

to make new pairs

[misty vessel]

Hm

Putting c in place of a and b?

Or like

[night patio]

Like creating new space for C

[misty vessel]

I can write them

Yes exactly

[night patio]

you can put C between A and B

to the left or to the right too

[misty vessel]

Yes

[night patio]

how many pairs will you get now

for C

[misty vessel]

Abc
Acb

Bca
Bac

Cab
Cba

[night patio]

yes for AB what you can do is you can place C to the left of A, in the middle, and to the right of B

[misty vessel]

Yes ok

[night patio]

for BA you can place C to the left of B, in the middle and to the right of A

so each of these 2 old pairs

[misty vessel]

Ok

[night patio]

give you 6 new pairs now

[misty vessel]

Yes

[night patio]

we just added 1 coin

and we got 3*2 pairs

[misty vessel]

Ok

[night patio]

what if we add a 4th coin D

[misty vessel]

So hence we got 6

[night patio]

we have 6 pairs now

in each pair

like ABC

where can you place D

[misty vessel]

Ohh yeah

[night patio]

how many places do you have?

[misty vessel]

So it should be 6*4

24

[night patio]

or we can say

we had

3*2 pairs already

on adding a 4th coin, each pair gives 4 new pairs

so we get 4*3*2 pairs

[misty vessel]

Oh h yeah

[night patio]

what if we add a 5th coin?

[misty vessel]

So that's factorial of 4?

[night patio]

yess

nicee

[misty vessel]

5 *4 32

[night patio]

yupp

[misty vessel]

Ok now I get it

Wow

[night patio]

so n! ways to arrange n distinct objects

[misty vessel]

Yes yes

And this we did arranging is it called combinations or permutations?

[night patio]

We made permutations

[misty vessel]

Okaye

[night patio]

n! is way of permuting things

making arrangements

combinations are different

[misty vessel]

Ok

[night patio]

for 5 distinct objects, we will get 5! = 120 ways of arranging

try making combinations

how many 3 object combinations can you make from ABCDE

[misty vessel]

What do you mean by objects?

[night patio]

like A B C D E

are the objects

how many groups of 3 can you make

using ABCDE

[misty vessel]

Yeah, wasn't this the same in permutations

[night patio]

grouping and arranging are two different things

[night patio]

one possible group would be ABC

group ABC and BAC are same

[misty vessel]

So you are asking me to * group * abcde? Right?

[night patio]

yes

group them

[misty vessel]

Ok

[night patio]

how many different groups can you get

[misty vessel]

So we can't repeat the letters?

Like we can't arrange it differently?

[night patio]

yes like if you go by the normal meaning of a group

ABC and BAC and CBA... are all the same

[misty vessel]

Yeah ok

So let me try once

Abc, BCD, cde, eab, eac, and so on?

[night patio]

yes

[misty vessel]

Okayyee

[night patio]

ABC
ABD
ACD
ADE
BCD
BDE
CDE
EAC
EBC
ABE

you will get these 10 groups right?

[misty vessel]

Ok

[night patio]

What if I tell you to make permutations of length 3 with the letter ABCDE

[misty vessel]

So 3 digits? Right?

[night patio]

yes

3 digits only

how many can you make

you know how to make 5 digit permuations

that would just be 5!

[misty vessel]

Oh ok

[night patio]

5*4*3*2

[misty vessel]

So 543

Not 2 right?

[night patio]

yess

think of it like

[misty vessel]

Okok hmhm

[night patio]

you have 3 spots

[misty vessel]

Yes

[night patio]

_ _ _

now in the first spot, you have 5 choices

A B C D E

if you pick any of them and place them, you will be left with only 4 letters for the next spot

[misty vessel]

Oh yeah

So there will be one letter less

[night patio]

if you place any of the 4 letters in the second spot, you will be left with only 3 letters for the last spot

[misty vessel]

Because that is already used?

[night patio]

yess

and see in this case

[misty vessel]

Ohh

Ok

[night patio]

since all the objects are distinct

[misty vessel]

So that's why we do factorials

[night patio]

there will be no duplicates

[misty vessel]

Ok

[night patio]

n! for n distinct object

[misty vessel]

Ok

[night patio]

in this 3-digit permuation case, we can say 5 * 4 * 3

you can think of it like a tree too

[misty vessel]

And if is a 4 digit number , we will include 2 as well?

[night patio]

yes

[night patio]

something like this

i didn't fully draw it

but I hope you get the point

[misty vessel]

Thanks for this, I did understand the tree

[night patio]

every path you take on this tree will give you a unique permutation

given that the objects are distinct

[misty vessel]

Ok

[night patio]

Also see

you can write this as 5!/2!

[misty vessel]

Yes

Of

Ok*

I think 2 will cancel out

[night patio]

yes and you will get 5 * 4 * 3

[misty vessel]

Ohh ok

[night patio]

lets say you have 7 distinct digits, and I tell you to make 4-length permutations

what would the answer be?

[misty vessel]

So 7x6x5x4

[night patio]

or you can write it as 7!/4!

[misty vessel]

Oh yeah

[night patio]

now lets say you have n distinct digits, and I tell you to make k-length permutations

[misty vessel]

But why 4!

Wont that be 3!

[night patio]

sorry 3

[misty vessel]

Ok

[misty vessel]

Ok,

[night patio]

can you try writing a general formula for this

[misty vessel]

N!/k!?

Nono wait

[night patio]

yes you just corrected me before hahaha

[misty vessel]

It just means you're a good teacher : )

[misty vessel]

Is it

n!/(k-1)!

[night patio]

yesss

nicee

[misty vessel]

Ok

Wow

Heehe

[night patio]

just another way of writing it right

you don't have to memorize it

[misty vessel]

Yeah I can make it myself if I forget

[night patio]

now one last step

[misty vessel]

Ok

[night patio]

you will get k-length permutations with this formula right?

[misty vessel]

Whaaa

I don't know

[night patio]

this was the original question right

[misty vessel]

Yeah ok

[night patio]

the formula you derived the formula for this right

[misty vessel]

Yesh

[night patio]

like if there are 5 objects, how many 2-length permuations can you make

[misty vessel]

20

[night patio]

nicee

[misty vessel]

Hehehe

[night patio]

now can you list them if possible?

[misty vessel]

Can we say ABCDE

Are the objects?

[night patio]

yess

take 5 distinct objects

you will get 20 pairs

list them

[misty vessel]

ABCDE
Acbde
Acbed
Abced

Bacde
Bcade
Bcaed
Baced

And so on

My autocorrect makes things worse

[night patio]

no you have to make 2 pairs

2 length pairs

[misty vessel]

Oh

[night patio]

like
AB
BA
BC
CB

and so on

[misty vessel]

Ahhh

2 digit

Ok

[night patio]

2 length or 2 digit

yes

[misty vessel]

Ab
Ac
Ad
Ae

Bc
Ba
Bd
Be

Cd
Ca
Cb
Ce

De
Da
Db
Dc

[night patio]

EA
EB
EC
ED

right?

total of 20 pairs

[misty vessel]

Right

[night patio]

now can you see

that each group

has a permutation of itself here

like AB BA
ED DE
BC CB

and so on

[misty vessel]

Yes

[night patio]

we can say that since each group consists of 2 digits

[misty vessel]

Yes we can

[night patio]

each group is permuted in 2! ways

right?

[misty vessel]

Yes

But how

Each group

2x1=2

There are 4

[night patio]

by group i mean, the sequences which have the same letters

AB BA is the same group

[misty vessel]

Oh ok

[night patio]

they are permuations each other right?

[misty vessel]

Yes

[night patio]

yes

so each group like AB

can be permuted in 2! ways

because each group has 2 digits

[misty vessel]

Ok

[night patio]

now what if I remove the duplicates?

we have 5!/3! (2-length) permutations of ABCDE right

[misty vessel]

Yes

[night patio]

if i divide 20 by 2!

I can remove the duplicates

[misty vessel]

Wait wait

[night patio]

or (5!/3!)/2!

[misty vessel]

How does that relate

[night patio]

If i remove the duplicates

I can get the groups

[misty vessel]

Hey, is it okay if I ask too many questions?

[night patio]

yes

ask as many as you want

[misty vessel]

Ok

Alr

[night patio]

Which part do you not get. We can go over it again

[misty vessel]

I meant that how does the formula and removing duplicates relate

[night patio]

because the formula 5!/3! gave us the total number of pairs right?

5*4

[misty vessel]

Ooh ok ok

Yes yes

[night patio]

yess so out of these 5!/3! = 5*4 = 20 pairs

[misty vessel]

We removed duplicates, now we got 10 pairs

[night patio]

yes

[misty vessel]

Ok

[night patio]

because each 2 length pair is permuted right?

like AB BA

[misty vessel]

Yes

[night patio]

CD DC
BC CB
EB BE

so I know that I have 2-length pairs, I know that I made permuations using 5 digits and I got 20 of these pairs

And while making these 20 pairs are all permutations

[misty vessel]

Yes

[night patio]

The order matters in these permutations

becase we are arranging

[misty vessel]

Yes,

[night patio]

but I can group these permuations too

[night patio]

like this

[misty vessel]

Ok

[night patio]

so (5!/3!)/2!

this will be the total number of 2-length groups right?

or combinations you can say

[misty vessel]

Yes

[night patio]

Now what if I tell you to make 3-length pairs?

using the 5 digits

ABCDE

how many 3-length pairs can you make?

[misty vessel]

They must be groups?

[night patio]

no no just permutations

normal permuatations

like we did for 2-length permutations

[misty vessel]

Ok

60?

[night patio]

yess

[misty vessel]

Yess

[night patio]

Group 1 (A, B, C):
ABC, ACB, BAC, BCA, CAB, CBA

Group 2 (A, B, D):
ABD, ADB, BAD, BDA, DAB, DBA

Group 3 (A, B, E):
ABE, AEB, BAE, BEA, EAB, EBA

Group 4 (A, C, D):
ACD, ADC, CAD, CDA, DAC, DCA

Group 5 (A, C, E):
ACE, AEC, CAE, CEA, EAC, ECA

Group 6 (A, D, E):
ADE, AED, DAE, DEA, EAD, EDA

Group 7 (B, C, D):
BCD, BDC, CBD, CDB, DBC, DCB

Group 8 (B, C, E):
BCE, BEC, CBE, CEB, EBC, ECB

Group 9 (B, D, E):
BDE, BED, DBE, DEB, EBD, EDB

Group 10 (C, D, E):
CDE, CED, DCE, DEC, ECD, EDC

these 60 right?

[misty vessel]

Yes

[night patio]

now can you see that since these are 3-length pairs

they are permuted in 3! ways

[misty vessel]

Yes

[fading sierra]

so if the sky is piano what great wall of Russia gurt-yo?

quick question

[misty vessel]

What

Wrong channel my friend

[night patio]

now we have 60 ways