help-39

[cursive wraith]

well A and B are indeed matrices of the same dimension

now

you want to show that something is a fundamental solution set

meaning it's a basis of Ker(AB)

how do we show it's a basis? Linear independence and surjectivity

so

I think linear independence is easier

[agile ridge]

and prove c1,...,c_r1, d1,...,d_r2 are 0?

[cursive wraith]

so

name this vector x + y

(I'll let you guess which part is x and which part is y)

oh before we do that

did you check that all xis and etas are solutions of ABx = 0?

[agile ridge]

no...

[cursive wraith]

well I'll let you do that, should be easy

[agile ridge]

[\text{For } \xi_i: \quad AB\xi_i = B(A\xi_i) = B \cdot 0 = 0]
[\text{For } \eta_j: \quad AB\eta_j = A(B\eta_j) = A \cdot 0 = 0]

[jolly parrot]

nino

[cursive wraith]

you got it

so

[cursive wraith]

let's go back into proof of linear independence

so we have x+y = 0

what should be x and y in here?

?

x and y are both vectors

[cursive wraith]

and based on this

you can define x and y easily

what would make sense

[agile ridge]

x=[c1,...c_r1] y=[d1,...,d_r2]

[cursive wraith]

uh

no really it's easier

this is x

this is y

[agile ridge]

aah okk

[cursive wraith]

so

what can we say about x, what can we say about y

[jolly parrot]

nino

[cursive wraith]

yes sorry for not responding

so

x is in nullA

y is in null B

but x+y = 0

so what more do we get?

[agile ridge]

\xi....\eta are linear indep

[cursive wraith]

not now

we've cut a big line

just deduce some stuff about x and y for now

knowing these three things:
x is in nullA
y is in null B
and x+y = 0

[agile ridge]

hmmm

[cursive wraith]

(recall that nulls are subspaces)

(meaning they're closed under addition, subtraction...)

[agile ridge]

a sum relationship between null A and null B

[cursive wraith]

not exactly

but something belongs to null A for example

x is in nullA

x = -y

[agile ridge]

so?

y is also in null A?

[cursive wraith]

finally yes

both x and y are in both nullA and null B

so

Ax = Bx = 0 for example

[cursive wraith]

do you see how to continue now?

perhaps we can look back at the question statement for info

[agile ridge]

why do we need to prove this

[cursive wraith]

recall that our end goal is that all xis and etas are equal to 0

to prove linear independence

so it'll be easier if x = 0 and/or y = 0

[agile ridge]

so we just need to apply A or B on the both side then?

[cursive wraith]

no we would be losing info

Ax = 0
Bx = 0

but perhaps

we can apply some OTHER matrix

[agile ridge]

D or C

[cursive wraith]

yes

what do we know about D and C

[agile ridge]

n×n square matrix and r(CA + DB) = n

[cursive wraith]

yep

what does that tell us about the null of that matrix, CA+DB?

[agile ridge]

nullspace =0

[cursive wraith]

yep

so now, do you see what we're gonna do, knowing Ax = 0 and Bx = 0?

[agile ridge]

(CA+DB)x=0

[cursive wraith]

yeah exactly

what does this imply about x?

[agile ridge]

x=0

[cursive wraith]

so this is 0

[agile ridge]

similar with y?

[cursive wraith]

y = -x = 0

[agile ridge]

ooh

[cursive wraith]

so now what do we do?

so we go back to x+y = 0 and back to start?

no the entire purpose of this was to get x=0 and y=0 separately

remind me what xis and etas are again

[agile ridge]

now we have to prove c1,...,c_r1, d1,...,d_r2 are 0?

[cursive wraith]

yes

[agile ridge]

since

[cursive wraith]

why did you apply A and B?

we have everything without A and B applied

$c_1\xi_1 + ... + c_{r_1}\xi_{r_2} = 0\
d_1\eta_1 + ... + d_{r_2}\eta_{r_2} = 0$

[jolly parrot]

rafilou2003

[agile ridge]

it's bc Ax=0 and Bx=0, right?

[cursive wraith]

remind me what xis are again

[cursive wraith]

yeah but we already know x = 0

[agile ridge]

fundamental solutions of Ax=0

[cursive wraith]

so it's no use applying A and B

[cursive wraith]

more than that

that's right

Fundamental

meaning that (xi1,....,xir1) is WHAT type of family

[agile ridge]

so we can prove that they're 0

then done

[cursive wraith]

you're still missing the keyword

[cursive wraith]

.

[agile ridge]

what type of familay means

[cursive wraith]

like

what property does (xi1,....,xir1) have

[agile ridge]

linear indep

[cursive wraith]

exactly

this allows us to conclude

[cursive wraith]

because we have a linear combination of xis that is 0

so all c_k are 0

(same thing for d_k)

thus linear independence of the whole family (xi1,....,etar_2)

[agile ridge]

yes

omg what kind of books I need to read so that i can solve this tough question so easy like you

thank you so much

[cursive wraith]

no problem, though we're not done here

still surjectivity to go through

so it's the same kind of reasoning

you need to show that for every x such that ABx = 0, there exists y and z such that x = y+z, with Ay = 0 and Bz = 0

(and then you can write y = sum (c_kxi_k) and z = ...)

[agile ridge]

omg where does z come from lemme digest a bit

[cursive wraith]

x and y are now completely different

and x is some random dude in null(AB)

take your time

[agile ridge]

y = sum (c_kxi_k) and z = sum(d_k\eta_k)

[cursive wraith]

you need to show that there EXISTS y and z such that this

you only have ABx = 0 for now

[agile ridge]

hmmm

how

[agile ridge]

isn't y and z here the x and y there

[cursive wraith]

no

y+z = x

not 0

ok I'll give you a hint

so

sometimes in order to show that those y and z exist

we suppose that they do and we find what they should be equal to

[cursive wraith]

so suppose ABx = 0 and x = ...

and deduce stuff about y and z

it's VERY similar to what we did with linear independence

[cursive wraith]

yep

perhaps you can now find the following quantities, in terms of x:
Ay, By, Az, Bz

[agile ridge]

they re all equal to 0

[pearl pond]

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[pearl pond]

@agile ridge Has your question been resolved?

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[shut frigate]

anyone

pls help me

and this

@pearl pond

[flat python]

rationalize denominator

[shut frigate]

ok

[flat python]

rationalize numerator and denominator

[shut frigate]

ok

[pearl pond]

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[magic ginkgo]

I get confused with problems like this:

Given tan(x) = -11/13, find the other trig functions (main 6, including tan, obviously).

I know that tan(x) = sin(x)/cos(x), so my brain wants to jump to just saying sin(x) = -11, cos(x) = 13.

This feels wrong.

It mostly feels wrong because we're on a section in which we use the main trig identities to solve problems. e.g., tan(x) + 1 = sec^2(x).

[inland ivy]

it is wrong

$\frac{a}{b}=\frac{c}{d}$ does NOT imply $a=c, b=d$

[jolly parrot]

kheerii

[inland ivy]

are you sure that's all the information you're given?

[magic ginkgo]

The Q is this:

[worthy lance]

Also, what values can sinx and cosx can take?

[magic ginkgo]

Also, thanks, seeing the fraction example made me chuckle because it's obvious lol.

Huh? sinx can take all real number, no?

Unless you meant the range.

[inland ivy]

ah, the information "alpha is in QIV" i.e the 4th quadrant is crucial information here

you will see why later

[inland ivy]

yes, they did mean the range

[magic ginkgo]

I agree it's crucial (it tells us where the signs go)

[inland ivy]

yeah

[magic ginkgo]

E.g., sin(x) will be neg

[inland ivy]

what do you think the range of sinx is?

[inland ivy]

indeed

[magic ginkgo]

-1 to 1.

[worthy lance]

Then sinx can never be -11

[magic ginkgo]

Oh

Well that's pretty clear, yeah.

[inland ivy]

yeah, which explains why your answer is wrong

I could have just as easily written $\tan(\alpha)=\frac{-22}{32}$ which, according to your assumption would imply $\sin(\alpha)=-22$ and $\cos(\alpha)=32$

[jolly parrot]

kheerii

[inland ivy]

what you actually need to do here is use the Pythagorean Identities given to you to relate tan(alpha) to the other trigonometric functions

[magic ginkgo]

No, I see now. It's clear the answer is wrong given the range of sinx.

But what am I doing in my head that's letting me think that since tanx=sinx/cosx, that I can just sub it in?

I agree that I need to use the identities, now that I've seen why my answer is silly.

[worthy lance]

13 * 2 is 26 btw

[magic ginkgo]

It's a major, silly mistake that I make often. tanx = sinx/cosx, therefore whatever is in the numerator is sinx. It's obviously not true, but I somehow keep doing it.

[inland ivy]

the original question has 16

[inland ivy]

your issue is stemming from assuming that this is true

which it isn't

what you CAN say is that the two numerators and the two denominators have to be in the same ratio

[magic ginkgo]

Because tanx is gives us a ratio of two numbers. Okay, that makes infinitely more sense.

I just have to stop doing silly stuff. I'll double check my range every time I catch myself doing something goofy.

Thank you, folks.

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[pearl nova]

Can someone give me the formula for Matrix for O level igcse?

[tropic saddle]

"the" formula?

[pearl nova]

Ye

Formulas

Anyone?

No one?

[rustic nacelle]

What formula?

[pearl nova]

Yk

The

Formulas needed to work with matrix

[silent salmon]

Liek determinant? Matrix multiplication?

[pearl nova]

Yeah

[warm current]

Here you go.
https://www.khanacademy.org/math/linear-algebra

[pearl nova]

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[craggy cedar]

been stuck on this question for ages

last one on my hwk

[rugged jolt]

useful hints: 1. the pink shape and the triangle are always on the top. 2: pay attention to the direction of the arrow

also make sure each combination of side shapes are actually possible when folded.

[craggy cedar]

okk

thank you

[craggy cedar]

i did it 🤣 only took me 35 minutes

geometry is not my strong point

thank you

[rugged jolt]

no problem

[craggy cedar]

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[dusty mango]

l

[pearl pond]

Please don't occupy multiple help channels.

[dusty mango]

.close

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[foggy parrot]

I'm working on finding the slope of the curve at the given point p(4,2) y=sqrt x. I have gotten this far I don't know where to go from here

[cosmic raptor]

you could multiply by the conjugate

[foggy parrot]

So multiply by ((sqrt 4+h)+(sqrt 4) / (sqrt 4+h)+(sqrt 4) ?

[cosmic raptor]

yep

should work

[foggy parrot]

Cool thanks.

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[foggy parrot]

.reopen

[pearl pond]

✅

[foggy parrot]

Well now I'm stuck again and I'm pretty sure I messed up somewhere.

Ignore the circled answer I'm aware it's incorrect.

After multiplying by the conjugate I get a wierd irrational number.

I'm pretty sure that's not the right answer

Oh wait I see what I did wrong. Whoops I can math duh.

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[foggy parrot]

If anyone was curious it was 1/4th so the equation would be y=1/4x + 1

[pearl pond]

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[fossil drum]

At 36:20 he subistutes delta x and y for what they equal on the right,but what happened to the h?

https://youtu.be/tDPp5uWSIiU?si=-raLxUeVMJqhOovu

[fossil drum]

Since h represents the scalar multiple do we just omit it because it has nothing to do with the direction?

[wet osprey]

Love this guy let me have a look

Ok so he first wrote $f_x, dx + f_y,dy$

[jolly parrot]

frosst

[wet osprey]

That’s sort of “more correct” because it’s when you’ve already taken the limit

Writing $f_x\Delta x + f_y \Delta y$ is sort of he hasn’t taken the limit yet

[jolly parrot]

frosst

[wet osprey]

Ah

And lastly, the directional derivative cares only about the derivative wrt a unit vector in some direction

$PQ$ is some arbitrarily long vector, he’s used $h\hat u$ to represent that $PQ$ is going in the $\hat u$ direction with length $||PQ|| = h$, and $\hat u$ is a unit vector, ie $||\hat u|| = 1$

[jolly parrot]

frosst

[pearl pond]

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[midnight haven]

express (n - 1)(n - 2)(n - 3) as a factorial expression

[midnight haven]

i need help on how to do this

[pearl pond]

@midnight haven Has your question been resolved?

[summer imp]

Well those first three terms already should tell you you should find something like (n-1)! in there.

But that's too many factors, you'd have to divide by something ...

[pearl pond]

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[willow mauve]

Determine a rule for calculating , the number of toothpicks needed to
create an grid of squares. Explain your reasoning

[willow mauve]

by rule it means general term

[tulip ore]

@willow mauve you here?

[willow mauve]

yeah

[tulip ore]

have you done problems like this before

[willow mauve]

hmm

3a and b are pretty easy

ive only done problems with arithmetic sequences and geometric sequences

im not sure what question c is

[tulip ore]

question c does deal with a different kind of sequence

[willow mauve]

can it be found using common difference?

[tulip ore]

you dont need a common difference to be able to figure this out - you can get the sequence directly from the picture's patterns

but you can use a common difference to figure out question c

as a hint, its a quadratic sequence

[willow mauve]

I guess

but how do I get it?

[dapper kraken]

maybe try to find a pattern?

[willow mauve]

can it be found using second difference?

[dapper kraken]

a quadratic sequence is a second diffrence ye

[willow mauve]

how do I find the pattern

[dapper kraken]

just try to

[willow mauve]

4, 12, 24

hmm

[dapper kraken]

uh it should be enough

whsts the diffrence between them

[willow mauve]

hmmm

well

the difference is

uh

8, 12

second difference is 4

@dapper kraken

[dapper kraken]

yes

find the rule for it then

[willow mauve]

howwwwwwwwwwwwwwwww

@skill_us

[willow mauve]

howww

I wasnt tauight how to find the rule just from the second difference

@dapper kraken cmon

@tulip ore

[dapper kraken]

uh

[tulip ore]

do you want to continue this or have me continue it skill_issue

[dapper kraken]

you can continue it

im helping soneone else tn

rn

[willow mauve]

help 😭

[tulip ore]

lets try a more direct way of finding the quadratic sequence instead

lets say youre looking at the nth term in the sequence

that should mean an n x n square of toothpicks, right

[willow mauve]

ye

[tulip ore]

now if you look at each square, you can assign two toothpicks to each square

can you see how thats done?

[willow mauve]

wdym

[tulip ore]

we can count how many squares there are

but we need a number for toothpicks

can you connect the toothpicks to squares?

[willow mauve]

wdym

[tulip ore]

I thought you knew how to do these

this is the same idea

[willow mauve]

welllllllll

[tulip ore]

[willow mauve]

Idk if im stupid but the way your describing it

Idk

[tulip ore]

you did them all with first differences didnt you

[willow mauve]

yeah

they have a constant first difference

therefore they are linear

[tulip ore]

lets do a non-difference related way

where you look at the toothpicks and find the sequence based on them

[willow mauve]

the first term minus constatn second difference plus second difference times n?

[tulip ore]

youre just repeating formulas at me

[willow mauve]

yeahhh

[tulip ore]

what is that supposed to do to me

[willow mauve]

idk

wdytm non-difference related way

[willow mauve]

http://mrkennedy.pbworks.com/w/file/fetch/91539015/Nelson Functions 11 Textbook.pdf

page 449

[tulip ore]

please

stop repeating formulas at me

[willow mauve]

how do I do it through the textbook

[tulip ore]

do you not want to just see the sequence

[willow mauve]

ah

[tulip ore]

its right there

[willow mauve]

it will change by uh

[tulip ore]

are you listening to my questions

[willow mauve]

it was tn = 2n + 1

now its tn = 3n + 1

[tulip ore]

hey

[willow mauve]

?

[tulip ore]

are you listening to my questions

[willow mauve]

yeah\

[tulip ore]

do you just want to do it a difference-related way

or do you want to just see the sequence since its right there?

[willow mauve]

uh

[dapper kraken]

you can maybe let the squares have 4 toothpicks and delete the doubled ones

[tulip ore]

couldnt do that earlier

asked me twice what that meant

[dapper kraken]

oh ok

[willow mauve]

bro 😭

[tulip ore]

do you just want me to do this for you man

you have a dirt poor method

[willow mauve]

its what the textbook taught

http://mrkennedy.pbworks.com/w/file/fetch/91539015/Nelson Functions 11 Textbook.pdf
page 449

[tulip ore]

and if the textbook teaches you to count things by moving them one toothpick at a time, would you follow it?

[willow mauve]

no

[tulip ore]

if you look here,

SH

stop typing

[willow mauve]

ok

[tulip ore]

please

you can begin with 1 toothpick facing vertically

then you can add on each square by placing 3 toothpicks to its right

to connect to form a square

[dapper kraken]

i dont want to shove formulas, but have you been teached on how to find a second diffrencd U_n before?

[willow mauve]

I do that

[tulip ore]

so 0 squares = 1 toothpick,

1 square = 4 toothpicks,

etc.

does that make sense

[willow mauve]

well

I did make the formula

[tulip ore]

so thats 3 toothpicks per square, right?

[willow mauve]

tn = 3n + 1

I found that

[tulip ore]

so thats

3 toothpicks

per square

[willow mauve]

that was easy part

[tulip ore]

right?

[willow mauve]

yeah

each term its + 3

begins wth 4

[tulip ore]

so you know what it means when I ask for "toothpicks per square"

right?

[willow mauve]

yeah

[tulip ore]

you assign particular toothpicks to each square

[willow mauve]

3 per square after the first term

[tulip ore]

blade please stop assuming you know whats going to happen

let me finish

[willow mauve]

ok

[tulip ore]

you assign particular toothpicks to each square, in this case you assigned 3 in a C shape

now you have to do the same thing for our n x n square

[tulip ore]

remember, you need to assign the same number of toothpicks to each square

now what can you do to assign them

just focus on assignment here

the formula comes later

[willow mauve]

ok

[tulip ore]

how many toothpicks per square

[willow mauve]

4, 12, 24

[tulip ore]

hey

no

I didnt ask for that

[willow mauve]

oh

ic

[dapper kraken]

bruh wtf 😭

[tulip ore]

here, we assigned 3 toothpicks per square

[tulip ore]

now lets try that again here

[tulip ore]

3 toothpicks in a C shape, to build each square up

[tulip ore]

now what about here

[willow mauve]

asigning 3 for each square wont work here tho?

[tulip ore]

yea what do you think Im asking you

Im asking you for toothpicks per square

how many toothpicks should you assign for each square?

3 wont work

1 is too little

what will?

[willow mauve]

uh

[tulip ore]

you dont have to assign every toothpick to a square

you remember earlier where we left 1 toothpick remaining

here, the same can happen

[willow mauve]

when did we leave one toothpick remaining

[tulip ore]

we didnt get to

because you interrupted in the middle of my explanation

by saying you already thought of it

it was easy you said

now let me finish

ok?

[willow mauve]

ok

[tulip ore]

we're going to use a different method of directly getting an equation from the pictures

first off, we assigned 3 toothpicks to each square

you noticed that this always worked

[willow mauve]

ye

[tulip ore]

now since there are n squares

and each square has 3 toothpicks

there are 3 * n toothpicks counted in red here

• 1 toothpick on the far left which was left out, but which we know the number of

and that gets you 3n + 1 toothpicks directly from the picture

[willow mauve]

ye

[tulip ore]

now for this

there are n x n squares

but this time you need to give toothpicks to each square

its ok if you leave some toothpicks out, we'll count them in later

[tulip ore]

see up here where I assigned toothpicks to each square?

[willow mauve]

yye

[tulip ore]

try to assign them here too

you just draw the same shape inside each square

[willow mauve]

do it 4 me

this one is a bit hard

please

[tulip ore]

blade

no

Ill give you a hint instead

[willow mauve]

ok

[tulip ore]

you need to assign 2 toothpicks to each square

now can you assign them

draw it on the picture or describe it

[willow mauve]

no

ok

@tulip ore

[tulip ore]

good job

now heres whats important

2 (red) toothpicks for each square

n x n squares

how many (red) toothpicks so far?

[willow mauve]

wdym

[willow mauve]

for the last square>

?

ok

[tulip ore]

now back to this

since there are n x n squares
and each square has 2 toothpicks
there are how many toothpicks counted in red here?

[willow mauve]

18

[tulip ore]

do you remember when you understood this earlier?

[willow mauve]

mhm

[tulip ore]

now we're doing it for this

n x n squares

times

2 toothpicks

is?

[willow mauve]

2n^2

[tulip ore]

since there are n x n squares
and each square has 2 toothpicks
there are 2n^2 toothpicks counted in red here

with the red toothpicks done,

youll notice whats left are Ls

can you figure out how many black toothpicks remain, based on n?

(for reference you know there are 2n^2 red toothpicks)

[willow mauve]

2n^2+2n

[tulip ore]

thats it, thats the sequence

[willow mauve]

is there an easier way

without drawing

like with second difference or smth

[tulip ore]

recapping first,

1. you can assign 2 toothpicks to each square
2. there are n^2 squares
3. so there are 2n^2 red toothpicks
4. theres 2n black toothpicks
5. so theres 2n^2 + 2n toothpicks total

you get all of these steps, right

[willow mauve]

ye

[tulip ore]

are those steps hard?

[willow mauve]

no

[tulip ore]

now let me break the news to you

if your textbook doesnt tell you a formula for second difference

then I guess you just cant use a formula for second difference

[tulip ore]

this is the easiest method

[willow mauve]

ok

[tulip ore]

its the one method that works for 3a, 3b, and 3c

it directly counts the toothpicks that they drew for you

[willow mauve]

for 3a and 3b, I think that the easier method is seeing that there is a common difference

[tulip ore]

which is based on... seeing that you can add +3 or +2 more toothpicks

[willow mauve]

yes

[tulip ore]

which is literally identical to this method

3 toothpicks per square

[willow mauve]

but finding common difference is faster then drawing

[tulip ore]

= 3n

[willow mauve]

and testing

[tulip ore]

you dont have to draw 18 red lines

[willow mauve]

each time

[tulip ore]

you did not have to test anything either

[willow mauve]

u do?

u draw red lines

to see

[tulip ore]

I drew them for you because you couldnt see

you dont need to draw the red lines to do the method

[tulip ore]

you similarly had to count and draw the other terms in the sequence here to try your first difference method

of course you didnt actually draw them

you considered their shape and saw the +3

[tulip ore]

same thing here, I made you draw them to confirm that you saw what I saw

but really you can confirm 2 toothpicks per square and youre already halfway there

[willow mauve]

how old r u

[tulip ore]

why do you ask

[willow mauve]

ur smart

[tulip ore]

youre just finding excuses by this point

you can be just as smart

age is not a factor in smartness

[willow mauve]

Im just asking chill

[tulip ore]

really

[willow mauve]

yuh

[tulip ore]

after the 100 loaded questions you asked before

this one you say is chill?

[willow mauve]

idk

how old r u

im curious

U dont gotta answer

[tulip ore]

you gotta reveal your age first then 👀

[willow mauve]

15

[tulip ore]

19

[willow mauve]

dang

[tulip ore]

now the +4 year gap here doesnt mean anything

[willow mauve]

i know

[tulip ore]

I didnt study toothpicks for those 4 years

[willow mauve]

I know

[static narwhal]

hey can i get some help please i dont know if this is right or not:

[willow mauve]

ask in ur own place

[tulip ore]

yea it is, but you need to ask in a separate help channel next time

this is blade's channel

[static narwhal]

oh my bad idk how this server works

[willow mauve]

I dont mind

just that other people mught get anoyed

my question is alr answered

[tulip ore]

you can see the empty channels up there in the "Math Help (Available)" category

rn theres 4

[willow mauve]

#help-22

[static narwhal]

oh okay i just picked whatever sounded the coolest alr ill do that rn

[tulip ore]

lol alr

[last moth]

.close

[pearl pond]

Channel closed

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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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[sharp smelt]

if gcd(a,b)=1 and c|a+b then prove gcd(a,c)=gcd(b,c)=1

[sharp smelt]

I would like to do this using bezouts lemma

so I was thinking

$ax+by=1$

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[sharp smelt]

$\frac{a+b}{c}=k_1$

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[sharp smelt]

let x=y=1

so a+b=1

so 1=ck

so $c=k=1$

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[sharp smelt]

whicy proves gcd(a,c)=gcd(b,c)=1

[sharp smelt]

this step feels off

[tropic saddle]

yes because its wrong

bezout says there exist some x,y with ax+by=1

you cant choose them however you like

[sharp smelt]

right

let the gcd(a,c)=d

so $ta+eb=d$

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[sharp smelt]

and $ub+fc=d_2$

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[sharp smelt]

hmm

let $\frac{a+b}{c}=d$

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[sharp smelt]

can I have a hint

do I use bezout's lemma here?

just a hint please

so I think I should first find gcd(a,c)

that's the hard part

so that's the same as finding the $gcd(a, \frac{a+b}{c})$

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[sharp smelt]

hmm, and gcd(a,b)=1

[summer gorge]

i will probably do like this (a, b)=(a+b,b) which is the same as (c,b) similarly we can do for ( a,c)

[sharp smelt]

how's (a,b)=(a+b,b)?

[summer gorge]

d|a and d|b

[sharp smelt]

why, I could have a=2, b=6,d=4

[summer gorge]

Is d the gcd of a,b

[sharp smelt]

no

wiat

[summer gorge]

Do you get it now

[sharp smelt]

(a+b)/c=d

that's it

[sharp smelt]

why is it?

[summer gorge]

Here d is the (a,b)

[sharp smelt]

d is 1 though

right

[tropic saddle]

doesnt matter

[sharp smelt]

okay

[tropic saddle]

this is always true

[summer gorge]

There's a general statement that (a,b)=(a+kb,b)

[tropic saddle]

this is one of the first thing you prove about gcds

[sharp smelt]

ok, let me try to think of a proof in that case

[summer gorge]

Where k \in \N

[sharp smelt]

I think I'd use bezout's lemma to prove this

[tropic saddle]

overkill

[summer gorge]

It's simple if d|a and d|b then d|a+bk

[sharp smelt]

ok

[summer gorge]

Thus there's no difference in (a,b) and (a+bk,b)

[sharp smelt]

that's only if d|a, and d|b though

[summer gorge]

Yes

[sharp smelt]

what about if a=b=2 and d=4

oh right

[summer gorge]

That's not possible

[sharp smelt]

their gcd won't be 1

[summer gorge]

No like d=(a,b)

[sharp smelt]

d=gcd(a,b)

right?

[summer gorge]

Yes

[sharp smelt]

ok

[summer gorge]

The question is quite simple if you know that property

[sharp smelt]

mhm

I'm lost

sorry

[summer gorge]

It's ok we will try again

Where are you lost exactly

[sharp smelt]

wait

nvm

the question was wrong

oops

[summer gorge]

No the question is right

[sharp smelt]

right, I misread it

[summer gorge]

Any doubts?

[sharp smelt]

ok, can we start from square one

[summer gorge]

Ok

now the question states that (a,b)=1 and c|a+b

[sharp smelt]

yes

[summer gorge]

we need to prove (a,c)=(b,c)

[sharp smelt]

ye

[summer gorge]

so we can approach like this:

let d=(a,b) (*note i will use () to denote g.c.d)

[sharp smelt]

ok

[summer gorge]

this implies that d|a & d|b

[sharp smelt]

yes

[summer gorge]

now we can use that fact to state that d|a+kb where k \in \N

[sharp smelt]

ok

[summer gorge]

now we will let k=1

so what do we get?

[sharp smelt]

d|a+b

[summer gorge]

as c|a+b

we can also state that c\geq d

[sharp smelt]

yes

[summer gorge]

so we can also state that (a+b,b)=(c,b)

similarly (a,b+a)=(a,c)

[sharp smelt]

ok

oh, I see

[summer gorge]

those both are the same thus (c,b)=(a,c)

[sharp smelt]

ah

thanks

I think I'll try the othe part(proving it's 1) by myself

tysm!

can I close this for now?

[summer gorge]

wdym?

[sharp smelt]

The problem is proving (a,c)=(b,c)=1

wait, but (a+b,b)=1

right

[summer gorge]

thats a given a

so yeah the proof is quite easy if you know that

[sharp smelt]

Thanks

.close

[pearl pond]

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[summer gorge]

:)

[pearl pond]

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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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[sly dagger]

$\overline{A}BC+AB \overline{C}$

[jolly parrot]

Galaxy

[sly dagger]

just doing some basic boolean algebra, I dont know how to simplify this down

I know I can pull a B from both, but thats where I get stuck

[dapper kraken]

huh

nvm

[vivid grove]

B(A either or C )

[sly dagger]

right

i know its B(A XOR C)

but idk how to get there using boolean algebra

[tulip ore]

do you know the definition of XOR in boolean algebra

[sly dagger]

not really

[tulip ore]

youll need to derive it

do you know how XOR acts?

[sly dagger]

yeah its true when exaclty 1 input is true

[tulip ore]

lets say you have inputs X and Y

can you try to type up X XOR Y in terms of ands, ors, and nots?

[sly dagger]

hmm

(X+Y)(!X!Y) maybe?

aight nvm i derived it using a truth table

!XY+!YX which is the situation i have above

[pearl pond]

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.

[sly dagger]

oh the bot

.close

[pearl pond]

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[pearl pond]

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[indigo urchin]

can someone explain why method 1 is wrong and method 2 is correct? and how to know when to use method 1 or two

[sly dagger]

they are the same

[indigo urchin]

the answers r different tho?

[sterile tusk]

nope

[indigo urchin]

or did i count it wrong in method 1?

[sly dagger]

the 2nd solution is just the first solution but rationalised

[sterile tusk]

the second just has the denominator rationalized

[indigo urchin]

OOHH

[cedar seal]

[indigo urchin]

thankzz

.close

[pearl pond]

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[cedar seal]

[pearl pond]

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[tawny tiger]

how can i find the resiude of this

i did this expansion but i get residue of 0 here

the answer is -1/2

i foudn my mistake sorry for the spam

.close

[pearl pond]

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[sterile turtle]

Hi, im curious on how wolphram works

[sterile turtle]

Or how i can use it

[wary hearth]

need help?

[marble oracle]

need help?

[sterile turtle]

Uhh

From someone who knows how to use wolphram yea

[marble oracle]

ok

It's a computational knowledge engine: it generates output by doing computations from the Wolfram Knowledgebase, instead of searching the web and returning links.

[tropic saddle]

do you mean wolfram in the sense of mathematica or just the website wolframalpha where you can enter questions

[sterile turtle]

Wolframalpha?

[marble oracle]

bye

[sterile turtle]

Man idk if its even called wolframalpha

The uhh

Formatting thing

On here

[tropic saddle]

latex?

$a^2+b^2=c^2$?

[jolly parrot]

Denascite

[sterile turtle]

A

Ye

Sorry i got it mixed up

[tropic saddle]

google for a latex tutorial

[sterile turtle]

Fair enough

,help

[jolly parrot]

A brief description and guide on how to use me was sent to your DMs!
Please use ,list to see a list of all my commands, and ,help cmd to get detailed help on a command!

[sterile turtle]

.close

[pearl pond]

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[pearl pond]

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[midnight haven]

Is infinity real no. or natural no. ? And why?

[wet osprey]

It’s neither

Why? Its just not included

[midnight haven]

.close

[pearl pond]

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[chilly slate]

hey, I'm revising set theory for the next semester, just wanted to make sure my reasoning is correct, the question goes as follows:

$given a set S$ $${{\frac{1}{x} | x \in \mathbb{N}}}$$
$\text{subset of } \mathbb{R}$
Find the supremum, infimum, maximum and minimum of this set.
considering the input value of 1 results in the largest output value which is also 1, (because 1/1 = 1)
and 1 being a part of our subset, both maxS and supS are 1
for the infimum, the lowest value we can get is technically 0, when we use infinity as input
meaning our minimum doesn't exist cosidering 0 isn't a part of the natural numbers but it is the infimum
to sum it up:
infS = 0
minS = none
supS = 1
maxS = 1

[jolly parrot]

Mephisto

[pearl pond]

@chilly slate Has your question been resolved?

[pine jay]

Let me just retex this for you if that's fine, you're talking about the set $S={1/x : x\in\mathbb{N}\subset \mathbb{R}}$?

[tropic saddle]

well plugging in infinity is maybe not exactly how you should argue. depends on how rigorous you are supposed to be

[jolly parrot]

Aslan

[tropic saddle]

but the results are correct

[pine jay]

In such case, youre correct however I think your reasoning can be a bit improved but since you're just revising youll get the chance to practice the subtleties by then

[tropic saddle]

$S={1/x: x\in\bN}\subset \bR$

[pine jay]

omg

[jolly parrot]

Denascite

[pine jay]

yeah sorry didnt notice my garabge tex

Well i mean its not false lmao

[pearl pond]

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[plush maple]

hey could anyone help me with any of these questions here?

[spiral pivot]

Related rates.

!status

[pearl pond]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[spiral pivot]

!show

[pearl pond]

Show your work, and if possible, explain where you are stuck.

[spiral pivot]

@plush maple ^

Also nice pfp, Bruce Lee is badass

[pearl pond]

@plush maple Has your question been resolved?

[plush maple]

thanks 🙂

aand i’m not really sure where to begin

[spiral pivot]

@plush maple so part 1 you don't understand how to get the total amount of fence?

[plush maple]

L=20-piW?

[spiral pivot]

So we have the total length of fence is 2L (the two long sides of the rectangle, each length L) plus πW (circumference of a circle with diameter W, from two circle halves)

This is equal to 20

@plush maple ^

[plush maple]

yes okay

how do we write that in an expression to express L in terms of W?

[pearl pond]

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[barren root]

Trying to understand the proof of the existence of uncountable sets based on Carton's diagonal argument. I understand that we can construct a number which differs from all other numbers in a set using this technique.
However the proof concludes like this:

    The set T is uncountable.

The proof starts by assuming that T is countable. Then all its elements can be written in an enumeration s1, s2, ... , sn, ... . Applying the previous lemma to this enumeration produces a sequence s that is a member of T, but is not in the enumeration. However, if T is enumerated, then every member of T, including this s, is in the enumeration. This contradiction implies that the original assumption is false. Therefore, T is uncountable. ```
I think the number `s` is then just *isn't* in `T` because otherwise it would be in the enumeration. I think it's rather a chronological problem: the enumeration is created before `s` is added to `T`. Or the problem of considering that `s` is not in the enumeration if it's in `T` even though the enumeration is supposed to have all elements of `T`.
So I don't understand that proof. Please help me in understanding it.

[sonic patrol]

All cantor’s diagonal argument does is the following

Take any function $f : \mathbb{N} \to 2^{\mathbb{N} }$

[jolly parrot]

Pseudonium

[sonic patrol]

Cantor then constructs an $x \in 2^{\mathbb{N} }$ outside the image of f

[jolly parrot]

Pseudonium

[sonic patrol]

This shows that f cannot be surjective, so in particular cannot be bijective

Hence, there exists no bijection between N and 2^N

Indeed this works if you replace N by any set X

[barren root]

But it's true for the simple reason of the codomain being bigger than the domain

[sonic patrol]

How do you know the codomain is bigger than the domain though

After all, there’s a bijection between $\mathbb{N}$ and $\mathbb{Z}$

[jolly parrot]

Pseudonium

[barren root]

You say there's a bijection between Z and N but what -1 from Z will lead to in N?

I'm very new to set theory, sorry

[sonic patrol]

That’s fine that’s fine

I haven’t given you an explicit bijection after all

[sonic patrol]

Ok first - do you prefer 0 to be in N or not to be in N

Either is fine

[barren root]

To be

[sonic patrol]

Ok cool

Let $\mathbb{N}_{\geq 1}$ denote the set of strictly positive naturals

[jolly parrot]

Pseudonium

[sonic patrol]

Do you agree that there’s a bijection between $\mathbb{N}$ and $\mathbb{N}_{\geq 1}$?

[jolly parrot]

Pseudonium

[barren root]

For me they are the same so yes

[sonic patrol]

I don’t think they are - the first contains 0, whereas the second does not

[barren root]

Then I don't

[sonic patrol]

Ok, so let me explain

It is true that $\mathbb{N} \neq \mathbb{N}_{\geq 1}$

[jolly parrot]

Pseudonium

[sonic patrol]

However, they are still what mathematicians would call “in bijection”, or “isomorphic”

Have you met either of these terms?

[barren root]

I'm familiar with injection, surjection and bijection, but not "isomorphic"

[sonic patrol]

That’s fine

So, let me show you

Let $f : \mathbb{N} \to \mathbb{N}_{\geq 1}$ be defined by $f(n) = n + 1$

[jolly parrot]

Pseudonium

[sonic patrol]

Do you agree that f is a function between the sets I’ve shown?

[barren root]

Yes

[sonic patrol]

Do you agree that f is injective?

[barren root]

Yes

[sonic patrol]

Do you agree that f is surjective?

[barren root]

Yes. Then it means there is a bijection between them, I see

[sonic patrol]

Indeed

In fact, they are also what we call isomorphic

This means that f has a two-sided inverse

We can define $g : \mathbb{N}_{\geq 1} \to \mathbb{N}$ by $g(n) = n - 1$

[jolly parrot]

Pseudonium

[sonic patrol]

Do you agree that g is a bijective function between the sets I’ve shown?

[barren root]

Yes, and f is its reverse

[sonic patrol]

Indeed

So $f \circ g : \mathbb{N}{\geq 1} \to \mathbb{N}{\geq 1}$ is the identity

[jolly parrot]

Pseudonium

[sonic patrol]

And $g \circ f : \mathbb{N} \to \mathbb{N}$ is also the identity

[jolly parrot]

Pseudonium

[sonic patrol]

This is what it means for two sets to be isomorphic

It means “same shape”

[barren root]

Googling what the symbol and "identity" mean

[sonic patrol]

Ah, so you haven’t met function composition..

[barren root]

Probably I have but only a couple times

[sonic patrol]

It’s the process of chaining functions together

[barren root]

I see, I remember

[sonic patrol]

It might be worth proving that f o g and g o f are identities

[barren root]

So for reverse functions it's basically going from yourself to yourself

[sonic patrol]

Indeed

[barren root]

What would that mean? That it is always true? It's quite obvious in this case

[sonic patrol]

Just in this case

If it’s obvious that’s good!

The point I’m making is that

$\mathbb{N} \neq \mathbb{N}_{\geq 1}$

[jolly parrot]

Pseudonium

[sonic patrol]

Indeed, the LHS is, in a sense, strictly bigger than the RHS

And yet, these sets are in bijection

They are isomorphic

[barren root]

I don't know how shapes relate to sets but now I see that these 2 sets are bijective

Or rather

In bijection

[sonic patrol]

Mhm

So, we can’t just say that “it’s obvious there’s no bijection, the codomain is bigger than the domain”

Because here’s an example where one set is bigger than the other, and yet there’s a bijection

[barren root]

True

[sonic patrol]

That’s why cantor had to do his argument

[barren root]

In case of $\mathbb{Z}$ and $\mathbb{N}$ a function would be exempli gratia $g : \mathbb{N} \to \mathbb{Z}$ by $g(n) = n - 1$

[jolly parrot]

Telov

[barren root]

Injective? Yes

[sonic patrol]

The function you’ve provided is not a bijection between N and Z

Because it’s not surjective

[barren root]

Yes, was just checking that

Hm that's because not all images are used

Can you give a bijective function for that so we can move on?

I'm trying but it's a bit difficult

[sonic patrol]

Well…

You can use $f : \mathbb{N} \to \mathbb{Z}$ defined by $f(n) = \begin{cases} \frac 12 n \text{ if } n \text{ even } \ -\frac{n + 1}{2} \text{ if } n \text{ odd } \end{cases}$

[jolly parrot]

Pseudonium

[barren root]

Yes that works

All cantor’s diagonal argument does is the following\
Take any function $f : \mathbb{N} \to 2^{\mathbb{N} }$\
Cantor then constructs an $x \in 2^{\mathbb{N} }$ outside the image of f\
This shows that f cannot be surjective, so in particular cannot be bijective\
Hence, there exists no bijection between $N$ and $2^N$\
Indeed this works if you replace N by any set X

[jolly parrot]

Telov

[barren root]

Why is x outside the image of f?

[sonic patrol]

Cantor’s argument shows why

[barren root]

Because it differs from all numbers in the image. But if it differs from all numbers in the image, then it is not in the image

And there is no need to have a preimage for it

[sonic patrol]

That’s the idea

[barren root]

Ho, that's the same contradiction as the one used in the proof. I was not understanding it

However what confuses me is this:
The proof starts by assuming that T is countable.
Then all its elements can be written in an enumeration s1, s2, ... , sn, ... .
Applying the previous lemma to this enumeration produces a sequence s that is a member of T, but is not in the enumeration.

Why would s be in T? If it was in T, we would already have it in the enumeration

[sonic patrol]

So what is T?

[barren root]

Just some set I think

[sonic patrol]

You’ll need to give me more context - which set?

[barren root]

The only restriction placed on T is that it is assumed to be countable

[sonic patrol]

No, it must be the case that T refers to some specific set

Because, uh, there do exist countable sets

Like say T = N itself

[barren root]

Oh, yes, sorry

Cantor considered the set T of all infinite sequences of binary digits

[sonic patrol]

Right

We can identify that with $2^{\mathbb{N} }$

[jolly parrot]

Pseudonium

[sonic patrol]

Can you see why?

[barren root]

Could be just N, no? Bases are just representations rather than what the numbers are equal to

So 111 = 7

[sonic patrol]

Well, if T = N then the diagonal argument fails

Because N is countable

So - do you know what the notation $2^{\mathbb{N} }$ means?

[jolly parrot]

Pseudonium

[barren root]

Oh, it's all indicator functions

No?

[sonic patrol]

Yep