help-43

[wispy fossil]

do i just plug the values into the formula? or how do i do this?

[pallid rune]

do you mean geometric?

[vernal frigate]

okay lets take question number 33 for example
what is it increasing by
like whats the difference between each term

[wispy fossil]

4

[vernal frigate]

4 becomes the d value in the arithmetic sequence formula, since thats the difference between each term

what does the sequence in question 33 start with

[wispy fossil]

oh so 3 + 4(n-1)?

[vernal frigate]

yeah thats the formula for question 33

now for a geometric sequence lets try question 37 an easy one

what is the first term in the sequence

[wispy fossil]

1

[vernal frigate]

okay

now what multiplication factor causes 1 to become -1

[wispy fossil]

-1

[vernal frigate]

okay

now a = 1, and r = -1
the formula for a geometric sequence is tn = ar^(n - 1)

plug in your values

r is the ratio between one term and the term before that btw, in this case -1/1 = -1

[wispy fossil]

so 1(-1)^(n-1)

[vernal frigate]

exactly

these two formulas will apply to all of those questions, but you need to look carefully to see if the differences between terms are equal every time or not

[wispy fossil]

ohhh okay i see

just decide whether its arithmetic or geometric and plug in the values?

[vernal frigate]

like for example question 44 is geometric but question 34 is arithmetic

yeah thats how it is

just feel glad that it didnt specify you to write recursive formulas

[wispy fossil]

heh yeah

thanks so much tho

.close

[compact pewter]

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[stark crown]

Let P be a point inside the regular hexagon ABCDEF. Prove that
AP^2 + DP^2 = BP^2 + EP^2 = CP^2 + FP^2

I'm really confused on what to do

[keen rose]

you can try the pytahgorean thoeream

pythagorean theorem

[stark crown]

Yeah but i don't get how to apply it

I'm confused lmao

[keen rose]

ok one idea might be

try it for AP^2 + DP^2

treate both as hypotenuses
(double application of the theorem)

[stark crown]

My friend told me to use some apollonius theorem or something

[stark crown]

ok lemme try

[cerulean steeple]

draw a diagram for starters

[keen rose]

^

[stark crown]

i've draw one and labelled like abcdef and the point p

that's basically what i had

[keen rose]

perfect

[stark crown]

wait so did you say i treat both AP^2 and DP^2 as hypotenuses

[keen rose]

yes 😄

and i guess i'm imagining the legs are vertical and horizontal

[cerulean steeple]

actually is this really true

[keen rose]

i think it is yes

[stark crown]

but like if i add the line to create a triangle like the missing line then its not a right angle triangle

if i add AD it won't be a right angle triangle

[keen rose]

here's what i mean

https://excalidraw.com/#room=d5bf1663d915cd481158,kqI1XqOXnAB8jmqkdyJ_Jg

[stark crown]

okay lemme see

ohh alright thannks

[keen rose]

👍

you can do it in terms of a, b, c, and d in the figure

i'd say leave the third pair off until the end

[stark crown]

okok

[keen rose]

that should be good

[stark crown]

is it like

something about how

each side of the hexagon

is equal

[keen rose]

yeah

[stark crown]

hmm

[keen rose]

i might try to redraw it on the side quickly

[cerulean steeple]

btw apollonius works here

(honestly i was pleasantly surprised it worked aswell lol)

[stark crown]

i haven't learnt it yet :/

i'm in 9th grade

my friend wanted me to use apollonius though

[stark crown]

wait is it equal because the dotted lines are straight

so then

on the other side

its like opposite

for each pair

[keen rose]

ok i don't have to redo the whole drawing but yess exactly

i think you get it

we will basically end up getting like a^2 + b^2 + c^2 + d^2 or something for both

[stark crown]

wait

but it's not though

because

on the other side

the triangles are smaller

[cerulean steeple]

is this comp math?

[keen rose]

yeah the contributions swap around

but the final answer should be the same

[stark crown]

what is contributions

[stark crown]

its like gauss problems

its my homework it's due tomorrow 😭

[cerulean steeple]

have you studied trig?

[stark crown]

i did some during tutoring but i wasn't paying attention and didn't get it ig

i think im doing in school soon though

[keen rose]

ok sry i changed my letters around. so like in my diagram how AP = a^2 + d^2

PD = b^2 + c^2

[stark crown]

okok lemme check

[keen rose]

sry i could change my letters to match yours

[cerulean steeple]

stewarts theorem maybe?

[keen rose]

this does def look cool too

[stark crown]

no let me just read it though

[keen rose]

but it can def be done just with the pythagorean theorem

[keen rose]

[stark crown]

what 💀

huh

i might be cooked

or slowaf

[keen rose]

just try AP^2 first

[cerulean steeple]

oh i get your idea now

[keen rose]

yeah nice

now try PD^2

[stark crown]

wait so is it equal because

both pairs

are

consist of

this

[keen rose]

exactlyy

[stark crown]

ohhh

that makes more sense now

[stark crown]

wait what

didn't i do that

[keen rose]

yeah you finished

[stark crown]

is the third pair the same though

because its like

rotated

[keen rose]

ok yeah exactly

the symmetry

it'd be exatly the same if we did it for AP^2 + DP^2, and CP^2 + FP^2

it is rotated kinda, it's easier to consider just the two pars again i think

then you can rotate it and flip and it'll look the same

you got it i think

it's the same geometry as before already

[stark crown]

oh

[keen rose]

yeah, your drawing already looks the same basically

and the position of P will actually be slightly different for this rotated version

but it doesn't matter, that doesn't affect any of the logic/calculations that we did

[stark crown]

OH

WE GOT IT

[keen rose]

🎉

(one slight logic thing, our a, b, c, d in this new diagram might be different. so technically we just show with the new one that FP^2 + CP^2 = AP^2 + DP^2, the same way as before, even though our a, b, c, d, changed this time)

[stark crown]

alright

[keen rose]

all the sums end up equaling each other anyway

[stark crown]

i'll write down my explanation really quickly and can you check it for me?

then i'll close the channel

[keen rose]

sure

[stark crown]

btw tysm ur the goat bro

[keen rose]

npp you got it on your own mostly

(you can argue that the vertical/horizontal side lengths, like a, b, c, d that we labeled twice, are equal by saying they are opposite sides on a rectangle)

[stark crown]

wait can you help me rq again

my friend is arguing it wouldn't work if i place point P differently

i put a new diagram on the whiteboar

[keen rose]

it is weirder but i think we're good

we can try to work it out

[stark crown]

i think thats right

[keen rose]

i think so too

[stark crown]

thanks again

[keen rose]

np it was a fun question

was good to check that

[stark crown]

btw what is apollonius theorem

[keen rose]

i don't know tbh

[stark crown]

oh

then i'll read through my gauss book later

[keen rose]

@cerulean steeple knew i think

sure

[keen rose]

interesting

[stark crown]

@keen rose how would i do fp and cp

[keen rose]

it's symmetric again for any of the two pairs given

[stark crown]

so bd and dc must be equal for the thoerem to work?

[stark crown]

could you help me i can't find where to put the dotted lines

[keen rose]

like we showed AP^2 + DP^2 = BP^2 + EP^2, it's symmetric if we swap out BP and EP for CP and FP

[cerulean steeple]

tldr focus on APD, BPE, CPF. AD=BE=CF, apollonius on the triangles gets what we want since alot of stuff cancels out

[keen rose]

oh ok yeah, ig the idea is, to pick another pair. like just do AP^2 + DP^2 = CP^2 + FP^2

just like last time

then it'll look exactly the same as last time again

[stark crown]

hmm

wait what

[keen rose]

i was drawing it sort of quick lol sry

i'm just trying to show that it looks exactly the same as before

except the location of P is different, but that's ok

yeah you got it again

[stark crown]

yea

thanks

[keen rose]

np!

[stark crown]

yo can i have any tips

on writing the proof

should i just do like

draw the diagram

and then say because of pythag it proves

and add my equations

[keen rose]

i think if you drew a single hexagon with the a, b, c, and d parts labeled (each twice)

and then wrote like AP = a^2 + d^2, etc.

that'd def be enough

and yeah in words, just mention the pythagorean theorem

you could label right angles in the diagram too, since the pythagorean theorem is for right triangles

[stark crown]

should i write it all in one diagram or split it into two seperate ones like we did

[keen rose]

and then also in words, say the CP^2 + FP^2 part is the same because of symmetry

[stark crown]

i mean use one diagram

or

[keen rose]

so in the diagram, you only need two pairs of lines

[stark crown]

split into two

[stark crown]

okay so like two diagrams right

[keen rose]

i think one is enough to show the idea

i meant just one hexagon with two pairs of lines

[stark crown]

i'll do 2 hexagon diagrams each with two pairs so that it covers all three just to be safe

[keen rose]

sure 😄

but there might be a problem writing all of these at once:
a^2 + d^2 = AP^2
b^2 + c^2 = DP^2
a^2 + b^2 = BP^2
c^2 + d^2 = EP^2
c^2 + b^2 = CP^2
a^2 + d^2 = FP^2

since the a, b, c, and d changed when we used the second hexagon

the way i think about it is that we can just do it for two pairs of lines, and then, just use symmetry to say that it works for another pair. and then we get basically X = Y and Y = Z and so X = Y = Z (representing the final equations)

so i kinda like just doing a single hexagon

but anyway it's up to you

[compact pewter]

@stark crown Has your question been resolved?

[stark crown]

i just need to write up my thing

[stark crown]

i'll write it seperately cuz i have two diagrams

[keen rose]

ok sounds perfect then 👍

[compact pewter]

@stark crown Has your question been resolved?

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[strange vine]

Guys I needed help with this

[strange vine]

What I've done is that I have taken the equality given as a constant 'k'.

And I don't know what to do to move ahead.

[wraith nacelle]

(y+z)=k(pb+qc)

same with the other 3

they have the same value

the rhs have a cool trick

[strange vine]

I tried that. But p and q come in the middle

We have to eliminate p and q.

[wraith nacelle]

have you sum all of them ?

[strange vine]

sum?

[wraith nacelle]

yep

sum y+z part with z+x part and x+y part

[strange vine]

Yeah that thing okay

I reached till smth

I'll send it

sry

[wraith nacelle]

i cant open the file

[strange vine]

Now?

perfect

[wraith nacelle]

nice

the second equation is hard

[strange vine]

Umm I think something is wrong with k

[wraith nacelle]

but

[wraith nacelle]

why ?/

[strange vine]

Nah alright

I felt that if it was 3k, then it would be like all the ratios added

[wraith nacelle]

nahh

[strange vine]

but then the denominator would be crazy

[wraith nacelle]

the second equation have this cool trick

if a/b = c/d = e/f then (xa+xc+xe)/(xb+xd+xf) is equal to a/b

try summing them

[strange vine]

ok thanks

[wraith nacelle]

you should be able to prove it

[strange vine]

yea done

[compact pewter]

@strange vine Has your question been resolved?

[compact pewter]

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[hazy obsidian]

why is proving it for N enough? dont we need to show it for all countable sets

[hushed magnet]

every countable set is N

[scenic forge]

but all countable sets must be a subset of N

[hazy obsidian]

oh

completely forgot😭

[hushed magnet]

well ok my comment was a joke

[hushed magnet]

this is just false

[hazy obsidian]

i remember its defined as card X <= card N right

[hushed magnet]

the point is, if X and Y are countable, then there are bijections X->N and Y->N. so then there is a bijection X x Y -> N x N

so if you find a bijection NxN -> N, then you also get a bijection XxY -> N

[hazy obsidian]

oh okay

thanks!

.close

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[bold sail]

Hii.. anyone familiar with derivatives? Can we use em to predict the next point?

[bold sail]

For eg f(x) = x^2 and I take 2 points (I wish they were infinitesimally close)
2, 3 = 4, 9
And so we measure their difference then add that to the derivative of the f(x) and add that to the previous point and this gets us the next point?
But what about x^3 or other functions?

Difference + derivative + previous point
(4-9) + 2 + 9

[inland nacelle]

Please elaborate on the second step

"measure their difference then add..."

[bold sail]

This would gets 16

[inland nacelle]

Apologies, I do not understand

[bold sail]

So the difference/slope is
9-4 = 5 right?

[inland nacelle]

Indeed

There is no slope here, its just the difference

[bold sail]

And the derivative of x^2 is 2? I mean second derivative?

[sullen gorge]

What difference? Distance?

[inland nacelle]

Second derivative is 2, yes

I would appreciate if you could write your procedure slightly clearer

thanks

[bold sail]

Oh how come if f(x) = x^2 then
We take 2 points 2, 3 the slope of this would be 5 no?

[inland nacelle]

Slope?

[bold sail]

9/3 - 4/2 = 5

[sullen gorge]

Slope of tangent?

[bold sail]

Difference ig?

[inland nacelle]

This is f(3) - f(2)

There is a misconception

This is unrelated to slope

[inland nacelle]

Yes

[quiet sentinel]

That doesnt correlate with the actual derivative

[bold sail]

What do you mean by actual derivative? U mean first derivative?

[quiet sentinel]

Yes

I should probably ask what is being investigated here

[bold sail]

Ok so
difference is 5 (9/3 - 4/2)
Second derivative is 2
And the last point is 9
Right?

If we sum em all up we get the next point 16

And this is true for all!?

2 + 9 + 5 = 16

[quiet sentinel]

This kinda ties in with sequences and series

[bold sail]

Now take 3 and 4 you'd get the 25

[quiet sentinel]

The second difference of the sequence x^2 is 2

[bold sail]

And so on

[bold sail]

But what about x^3 this procedure is invalid for that!?

But true for x^2

[quiet sentinel]

Because theres a third difference there

[bold sail]

I think I'm taking the derivative until I end up with a real number?

Which in case of x^3 is 6?

[bold sail]

So it's only possible for x^2?

[quiet sentinel]

Yes

Every power has an nth difference in the seauence

[bold sail]

Btw how do you predict where the graph is gonna go next?

[bold sail]

Am I just adding the slope of secant line to the slope of tangent line here and last point?

Avg rate+ derivative+ last point

[bold sail]

Like the average rate
Should I explain it to you again??

[quiet sentinel]

So in a sequence

[agile onyx]

Actually you are just on your way of discovering Euler's method

[quiet sentinel]

Each difference is affected by the next one

So in a cubic sequence

The third difference is how much the second difference goes up for each term

And it would be constant

[compact pewter]

@bold sail Has your question been resolved?

[compact pewter]

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[dusky quartz]

I'm pretty sure there's only 1 possible solution

add the two expressions to get 15

[forest token]

yeah..is that answer wrong

[dusky quartz]

um

no solution provided

[forest token]

oh ok so u want to cross check?

[dusky quartz]

yeh

it's a 4 variable system given only 2 equations so

[forest token]

yeah

i think its 1 only

[boreal nimbus]

yes

[dusky quartz]

it's not some dark magic devil in the details thing right

[strange bolt]

no, its allowed to add the equations up and so there is just one solution. Could you close the channel when your done?

[dusky quartz]

.close

[compact pewter]

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[compact pewter]

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[kind viper]

there must be at least one what?

sure, if |G|=870 then there's at least one subgroup of order 29

[safe oyster]

Slight problem in this step

You are overcounting

You are undercounting here and also in the final inequality but still its enough to give you a contradiction

But this may not happen in other cases

Yes

And in the inequality you can do a +1 for the identity which isnt counted here

Yes

I cant recall off the top of my head but there are orders where you cant afford to undercount

[safe oyster]

Can you give an example? Also its normal to be momentarily confused when learning, so dont think that

[compact pewter]

@copper loom Has your question been resolved?

[safe oyster]

@copper loom if all your questions are answered you can type .close

how 1 on LHS? LHS needs to be a function and rhs will be the 0 function btw since in Z/(2), 1 + 1 = 0

[compact pewter]

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[fallen frost]

Lets say we have a vector [x,y]_B in V, V = R^2 that's expressed in terms of basis B={e1,e2}. When we carry out the linear combination of the vector components with basis B, like x*e1 + y*e2 = (x',y'), then does this mean that we turned the coordinate vector [x,b]_B in V into the abstract vector (x', y') in R^2 ? And, thus, this abstract vector is no longer expressed in terms of a basis -- the components x' and y' are simply elements of R^2?

[compact pewter]

@fallen frost Has your question been resolved?

[worn drum]

the tuple x' and y' is a basis dependent representation of the abstract vector

for example, let

(B = \left{ \begin{pmatrix} 1 \ 1 \end{pmatrix}, \begin{pmatrix} 1 \ -1 \end{pmatrix} \right}) and ([x, y]_B = [2, 3]_B)
[
\vec{v} = 2 \begin{pmatrix} 1 \ 1 \end{pmatrix} + 3 \begin{pmatrix} 1 \ -1 \end{pmatrix} = \begin{pmatrix} 5 \ -1 \end{pmatrix}

[boreal girder]

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

[worn drum]

oh 😦

$$ B = \left{ \begin{pmatrix} 1 \ 1 \end{pmatrix}, \begin{pmatrix} 1 \ -1 \end{pmatrix} \right} $$ and $$ [x, y]_B = [2, 3]_B $$

 $$
 \vec{v} = 2 \begin{pmatrix} 1 \\ 1 \end{pmatrix} + 3 \begin{pmatrix} 1 \\ -1 \end{pmatrix} = \begin{pmatrix} 5 \\ -1 \end{pmatrix}
 $$

[boreal girder]

narinelepen

[fallen frost]

So, this (5, -1) is a member of R^2, and its not a coordinate vector. Instead, it's an abstract vector which is not expressed in respect to a basis. Thus, there is no further linear combination to perform on (5, -1). Is this correct?

[devout latch]

Yeah, i mostly agree but i’m not 100% sure what you mean. (5,-1) as an element of the vector space R^2 is expressed here with respect to the basis B as (2,3)_B a coordinate vector. However with respect to basis B’ = {(1,0),(0,1)} you can also express (5,-1) as the coordinate vector (5,-1)_B’ because 5(1,0)-1(0,1)=(5,-1)

[devout latch]

so i would agree your interpretation is correct

[compact pewter]

@fallen frost Has your question been resolved?

[fallen frost]

I think we're in agreement, so I want to make sure we don't start disagreeing because of ambiguous language.
I'm basically looking for a confirmation of what I think it's going on.
There's 2 types of vectors:

1. Abstract ones, which are just ordered tuples that are members of a vector space, say R^2. These vectors are not described with respect to a basis, and thus their components are their literal value -- no further calculation needs to happen (such as a linear combination) to those component.
2. Coordinate vectors are also ordered tuples, but these vectors are expressed with respect to a basis. Thus, their coordinates (components) are not their final literal value; instead, coordinate vectors represent a linear combination of those coordinates with the basis on which the vector is expressed. If you carry out that linear combination, such as in the example above where the result is (5, -1), then that doesn't result in another coordinate vector. Instead, it results in an abstract vector.

It just so happens to be the case that abstract vectors, such as (5, -1), can be "expressed" in the standard basis, which just happens to have the same coordinates as the abstract version.

[devout latch]

I agree

[fallen frost]

Cool. I'm going to leave the question open for a minute, in case some one else reads it and has dissagreements with it.

[upbeat gorge]

imo, you have the right idea, but an abuse of language:

1. a vector simply written (1,2) in R^2 is the "standard" vector, representing one step right and 2 steps up on the Cartesian plane. It is a linear combination, but of the standard basis S = {i, j}, so we don't need to convert the coords. these are described in terms of a basis, just the standard one.
2. a vector [1,2]B is the "standard" vector (1,2), but represented in terms of a basis B. you would need to "carry out that linear combination," as you said. note that the co-ordinates [1,2] are based on the representation in the standard basis S.
   2.5. a vector going from a basis C to a basis B would be written C[1,2]_B, and its coordinates (1,2) are the representation in C. like others said, better to get used to this idea as well as the standard basis.

note that these are all coordinate vectors. an abstract vector will be something like v, which is not guaranteed to have coordinates. this is the representation that doesn't rely on a basis, because there are no "co-ordinates." so when you say "abstract vector," it's understood to be v, and not (1,2) in S. This is because it allows us to describe a generic vector space, or describe a vector space without having to explicitly state what the vectors are.

[compact pewter]

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[fallen frost]

.reopen

[compact pewter]

✅

[fallen frost]

@upbeat gorge So, the vector (1,2) in R^2 can exist in R^2 without a reference to a basis, correct? So, then it would just be a tuple of numbers that are members of R^2. That's what I mean when I say abstract vector.

[upbeat gorge]

it can exist without a basis, but in that case you can't write a tuple of numbers

those numbers come from some kind of basis - usually the standard

the tldr would be: abstract vector is v, no tuple, and coordinate vector is with tuple

regardless of basis being standard or not

hopefully that helps?

[fallen frost]

This is the first time I heard of that.

I'm not saying you're wrong, but I just never encountered that info before.

[hushed magnet]

of course I can just write a tuple of numbers

(1,2)

here I just did it

I dont need to know the concept of a basis for this

the name abstract vector is not good tho

with that I agree

[fallen frost]

What nomenclature would you use?

[hushed magnet]

its just an element of R^2

calling it a vector is fine

[fallen frost]

The problem is that "vector" is then ambiguous.

[hushed magnet]

yes

and thats fine

[fallen frost]

Coordinate vector -- ok, I get what that is. But, the "other" vector.

[hushed magnet]

vector just means element of a vector space

[kind viper]

there is no distinction at all between vector qua list of numbers and vector qua "abstract" point in a vector space

they are not different beasts

[fallen frost]

There is some distinction between (1,2) in R^2 and that same vector [x,y]_B in R^2.

I understand they "represent" the same vector, but they carry inherent properties that make them somewhat different.

[upbeat gorge]

i mean to say, you require a basis of some kind to use the tuple representation. i.e. this tuple relies on the standard basis

[hushed magnet]

no it doesnt

its just a tuple of numbers

[upbeat gorge]

if (1,2) is a vector, then (1,2) = 1i + 2j, thus it depends on the basis {i, j}, right?

whether or not it's helpful to think about it that way remains to be seen ig

[hushed magnet]

just because I can write it as a linear combination doesnt mean that it depends on that basis

(1,2) is just a tuple of real numbers

its just an element of RxR

which is just a cartesian product of sets

[upbeat gorge]

this discussion should probably not continue here, so i'll just say: abstract vector is a not-so-good name, but your understanding is good

[cursive drift]

vector is my name

😁 😁

[fallen frost]

I don't particularly care of any specific nomenclature, but it should be named something. "Coordinate vector" has semantic implications; namely, there's a buncha numbers that have an associated basis. Then, the other kind of vector that doesn't have a basis can also benefit from a name. Whatever that name is, it just needs to communicate the idea that it isn't a coordinate vector. "Abstract vector" is what I've seen, but could be "geometric vector", or whatever. Not having a name is not useful.

But, anyway, my original question was answered, so thanks all.

🙂

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[timid dune]

I don't quite rememebr hoew to use stars and bars

you have 7 cookies, and you want to create 4 groups (group can have 0) so how many bars do you need

is it 3?

[arctic portal]

yea, 3 bars

[timid dune]

This is equivalent to a+b+c+d=7 for 0,1,2,3 .. right?

If each kid had at least one, would it be then equivalent to a+b+c+d=7-4-=3?

[arctic portal]

yes, that's right

[timid dune]

Just conceptually though

What if the kids we are talking about were not distinct

So if we said 4 groups instead

Then certain arrangements would be the same

Would we divide over by 4!?

So if it was arranging 7 cookies into 4 boxes, then it would be the ans from that q all over 4!?

Eg 1,3,2,1 is the same as 1,2,3,1

For the q above, I believe it takes 1321 and 1231 as distinct cases

[shadow terrace]

I dont think that would be it because then in this case it would be 120/4! = 5, whereas there are 11 cases.

you gotta manually count the cases for identical objects in identical boxes

[mortal nimbus]

This could just be done with division and distribution

But are all the cookies different?

[shadow terrace]

identical I believe

[mortal nimbus]

Then

[timid dune]

Hmm

Yes identical

That’s odd

I would assume some underlying symmetry but I guess not

[mortal nimbus]

Hmm

I think 120 should be it

Might be wrong

[timid dune]

Do you know why?

[shadow terrace]

it is

[timid dune]

I can’t think of a reason

Oh right

Cuz each case doesn’t occur an equal no of times I think that’s why

[mortal nimbus]

Just division problem

[timid dune]

eg 7000 occurs only 4!/3! times whilst 1331 occurs 4!/2!2!

So it’s not exactly symmetrical

Interesting

Alright ty

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[winged lion]

I tried proving the 3rd inequality from left

Can somebody check in

[hushed magnet]

oh index battles. fun

isnt it the same as first inequality?

upper bound on the sums should be m

[winged lion]

you mean the idea?

oh bruh

typo

[hushed magnet]

dont see a mistake rn

[winged lion]

yea i think it's okay too

ty for checking

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[fallen frost]

In linear algebra,
I guess it's technically possible for there to exist a basis B={e1,...,eN}, of some vector space R^n, where the components of the basis vectors e1,...,eN are NOT expressed in terms of the standard basis. What is an example of where such a thing would be useful?

[pseudo rapids]

diagonalization of a linear transformation

[fallen frost]

hmm, interesting. I just started on the eigenvector/eigenvalue stuff on the book I'm reading, and it made a passing mention of diagonalization. Though, I haven't studied diagonalization, specifically, yet. So, I'll keep a look out for those kinds of basis vectors, and see if I spot them.

Thank you for the example

🙂

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[agile moth]

question for all my statics engineering students, can somebody explain why Joint A - ABx force is 54.17kN?

[compact pewter]

@agile moth Has your question been resolved?

[agile moth]

negative

wheres all my eng students?

[royal fiber]

well the Poincaré Conjecture states that every simply connected, closed 3-dimensional manifold is homeomorphic to a 3-sphere, so you can deduce that joint A = ABx force is 54.17kN. understand?

[turbid surge]

Unrelated, but tap on the ❌️ emote or it will time out

[agile moth]

no but ive got a follow up question

ABx is positive 54.17kN

but then when they calculate Fx its -54.17?

shouldnt it be positive

i have a feeling, nobody is gonna know this math since its so niche

[royal fiber]

ABx is positive 54.17 kN cause that’s just the size of the force in the x-direction (the magnitude). but when we do the ∑Fx equation, we care about direction. since AB is in compression, it pushes toward the joint, which means it’s pulling left. left is negative in the x-direction. so we plug in –54.17 in the equation. it’s the same force, just used with the correct direction

[silk agate]

If it is in equilibrium then it has to be

Lmao

[agile moth]

so a positive value means its compression?

[silk agate]

I mean no. You can equally use positive for extension and negative for compression

That will be equally valid

But the forces are opposite

[agile moth]

if you look at Joint A Fy, its -72.2 but that is also compression

[silk agate]

One min

[agile moth]

but then AB is +90.28 and that is also compression

so not sure how they are determining whether a force is compression or tension

[silk agate]

There they have taken this upward force as negative and the downward force as positive

[silk agate]

Just take one force as positive arbritrarily and its opposite counterpart should be negative

[agile moth]

this is how we were taught

we assume the force has tension

so that means force is pulling away from the join

t

if the calculation of the force ends up being negative, then we switch from tension to compression

agree?

so force AB is initially assumed to have tension (pulling away)

[silk agate]

Let me have a clearer look at the FBD

[agile moth]

but the calculation of force AB is +90.28

so shouldnt that mean our initial assumption is correct that force AB has tension?

so then why is AB showing as compression (C), and you can even see it in the free body diagram. they cross out the arrow at AB indiciating tension and then draw the arrow as compression (towards the joint)

[silk agate]

Yes. Here they have taken towards the joint as a positive force but the counterpart force which is balanced afterward is shown in the negative direction. Right?

[silk agate]

That might be the convention but that is not always true.

[agile moth]

??

for force ABx, how did they determine that it is compression

[silk agate]

I mean if i take compression to be positive and tension to be negative or vica versa the equations are gonna work the same

[agile moth]

for force ABx, how did they determine that it is compression

[silk agate]

In ABx the force is having a tendency to push through the system

This is compressing the wood plank

[agile moth]

ok but our initial assumption that it had tension (pulling away) fromthe joint

so then why did they change force AB to be compression even though the calculation for AB is positive value?

i thought we only change the direction if the value is negative

[silk agate]

I think the diagram is a little ambiguous in itself

[agile moth]

do you get what im saying though

cause it seems you're still confused as to what im trying to say

[silk agate]

You are saying that in the initial big diagram they seemed to have made two arrows and they have crossed one. And you wanna know why that thing is crossed

[agile moth]

look at AB

its crossed out when the arrow is pulling away (meaning tension)

and then they point a new arrow towards the join t(compression)

im asking...why are they doing this

force of AB is +54.17, so shouldnt the force remain as tension?

[silk agate]

Yeah

The diagram is alittle ambiguous

[agile moth]

in what way

and if AB force is actually tension (pulling away) then in Fx calculation, shouldn't it be AC - 100 + 54.17?

[silk agate]

And they have written -54.17 right? I think they have crossed that pulling away so it is tension.

And because it is tension the x component of AB will come exactly as pointed in the diagram.

Do you agree?

[agile moth]

just to be clear that -54.17 written in Fx is actually the value of ABx

agree?

but ABx is calculated to be +54.17

[silk agate]

Yeah but the direction of ABx is opposite to AC right?

[agile moth]

thats what im asking though...why is ABx direction going to the left..... ABx = +54.17....shouldnt it be pulling to the right???

[silk agate]

Well if they have calculated ABx positive in one line they should have taken 100kN to be negative but thats beyond me.

And if they have taken that to be positive so they have to take the second force acting in opposite sense negative

[silk agate]

Yes but in the next line they have flipped signs.

Now positive means compression and negative means tension.

Which is what i initially called ambiguous

[agile moth]

thats what im wondering also

how tf does this make any sense

[silk agate]

Yeah they should have taken that ABx positive

But now that they took that the equations will give the right value

Its just that if equation plops a negative value that will mean tension

Now

I hope ur question is resolved

[agile moth]

no...im still confused af

look at Joint B

[silk agate]

Yeah

[agile moth]

BEx is +54.17

and in the diagram, its showing that its pulling to the right

[silk agate]

But i hope that original doubt is resolved

[agile moth]

which makes sense

[silk agate]

Yeah

[agile moth]

but in Joint A ABx, ABx = +54.17 but its pulling to the left?

like how are they determining the direction?

BEx = +54 in joint B is going to the right..... ABx = +54 in joint A is going to the left

you see what im trying to get at here?

[silk agate]

One minute i will have to look at it again

[agile moth]

take your time lol

this is stressing me

if yuo couldnt already tell

[silk agate]

It is pulling to the right too bro look at the arrow clearly. The force ABx and BEx literally point to the same direction

Also why tf are we discussing truss bridges in a maths server

😭

[silk agate]

Also π²=g

[agile moth]

ABx is pointing to the left

what do you mean?

it was originally poitning to the right but then got crossed out for some reason and now points to the left

[silk agate]

Yeah this one is a good doubt and i will take some time to write my response.

[agile moth]

do you not see that?

and do you see here

all the positive values remain as tension cause that was our initial assumption

only Fx becomes copression because the force is a negative value which means our inital assumption was wrong

you know.....if the force is a negative value, im just going to assume the tension was wrong and it should be compression

fuck it

either im wrong or the question is wrong

thanks

[silk agate]

Consider figure on the roght side. I have marked the directions of force and tensions.

Now people have a doubt which is really common that the tension in the ball is acting only in the upward direction.

But in reality we can see that the wall is also applying a force on the upward direction (which is precisely the reason why it is not falling)

It is because tension is a bidirectional force

Just like spring force

[agile moth]

bro...i cant even understand your drawing

[silk agate]

Lemme draw it properly

[agile moth]

its ok....

[silk agate]

Ok

[agile moth]

im just ready to move on

[silk agate]

It is because they are internal forces

And not external forces

[silk agate]

Ok lmao

Ask ur teacher

They might explain u better

[agile moth]

what do you mean its because they are internal forces

[silk agate]

Have you been taught internal forces?

Like normal and tension?

[agile moth]

vaguely

[silk agate]

Ok can you tell me if you can make sense of the fact that tension is bidirectional?

[agile moth]

yea goes both ways

[silk agate]

Thats why the doubt is arising

[silk agate]

Yes thats precisely the reason of that too.

I mean im not sure how i can explain you that without taking a 1 hr lecture on internal forces

Lmao

So imma give up sorry

[agile moth]

no worries... namaste 🙏

.done

[silk agate]

Namaste

[agile moth]

,rotate

[agile moth]

@silk agate yo bro

i understand it now. want me to explain?

[silk agate]

What do u understand?

The tension being internal force?

I mean it being bidirectional?

[agile moth]

the whole compression/tension, arrow direction thing

so basically

if you look at joint A, we first solved for Fy because there are two forces (AB and 72.2)

so that means we can solve for the one unknown variable AB

so in the solution for Fy you can see that ABy = -72kn

so now....this is where it all makes sense

because our inital asusmption that ABy was tension (pulling away), our calculation tells us that is it actually COMPRESSION because we have a negative value (-72kN)

so because ABy is compression, that means ABx and AB forces MUST ALSO BE COMPRESSION....why? because that is just how science works

so you can see for calculations for ABx and AC, even though we are getting POSITIVE values, its irrelevant because we already know ABx and AC forces is compression because we already solved that ABy is COMPRESSION

hope that makes sense

[silk agate]

Yeah

Sorry i cannot be with u for a longer time

I got a test tmrw

And i need to sleep

Sry

[agile moth]

yo bro i got 3 final exams tomorrow plus assignment

due

good luck...see you

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[rotund wave]

2^logx = ( e^log2 ) ^ logx = e ^ (log2 * logx) = (e^logx)^log2 = x^log2

where did i go wrong

[rotund wave]

@wintry condor

[wintry condor]

Alr bet

This is what you had written earlier, just making it look nice:

$2^{\log(x)} = (e^{\log(2)} )^{\log(x)} = e^{\log(2)\log(x)} = (e^{\log(x)})^{\log(2)} = x^{\log(2)}$

[boreal girder]

Sri

[rotund wave]

yes

[wintry condor]

You are correct that $2^{\log(x)} = x^{\log(2)}$, thats completely fine. But the steps to get there arent correct

[boreal girder]

Sri

[wintry condor]

So when you do the $e^{\log}$ idea, try to think of $(2^{\log(x)})$, as one complete term. Meaning that you arent taking it out of the pemdas equation and changing the ordering around

[boreal girder]

Sri

[wintry condor]

The difference between $e^{\log((2^{\log(x)}))$ is that you are doing $2\cdot 2\cdot 2\cdots$, log(x) times. However what you had written down was doing $(e^{\log(2)} ) \cdot (e^{\log(2)} ) \cdots$, log(x) times instead.

So I guess with that in mind, does that lead you to a beter understanding?

[boreal girder]

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

[rotund wave]

ru saying the first eqn is wrong

$e^{ \log(2^{\log{x}}) }$

[boreal girder]

69-category theorist

[wintry condor]

Yeah, this is what you should have written instead

Because of these log rules

[rotund wave]

ohh alr

but i was actually rewriting 2 as e^log2

[wintry condor]

Yeah exactly!

You can just try to memorize these log rules if youd like, but if you want to understand it intuitively. $\log(x^y)$ is the same as $\log(x\cdot x\cdot x\cdots) = \log(x) + \log(x) + \log(x) + \ldots = y\log(x)$

[boreal girder]

Sri

[wintry condor]

Parenthesis do matter

[rotund wave]

thanks

i don't think i'll forget log rules anytime soon tho 😂

[wintry condor]

nws

[worthy aspen]

wait why is this wrong

[wintry condor]

cc this w/o ping

[worthy aspen]

also sorry if i'm not following help channel rules, got here from math discussion

[wintry condor]

$e^{\log(2)} = 2$, which isnt our intention

[boreal girder]

Sri

[rotund wave]

i mean uhhh that was my intention

[worthy aspen]

this is true though

[rotund wave]

i think i just approached it differently than what ur thinking

[wintry condor]

Or I guess, better to say we didnt want that bc @rotund wave was trying to show $2^{\log(x)} = x^{\log(2)}$

[boreal girder]

Sri

[worthy aspen]

yeah that's true

at least for a valid domain or something

[wintry condor]

Yep

[worthy aspen]

but for positive reals it's true

[rotund wave]

x > 0 yes

[wintry condor]

!done

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[wintry condor]

.close

Oh I think you have to do it

[rotund wave]

lemme uhh let this mellow for a while if someone else wants to say smth

[wintry condor]

nws

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[royal tusk]

How to solve this real life application

[kind crane]

Do you know inverse sine or arcsine

[royal tusk]

I dont know arcsine

but i know inversine

would you just make it sin^-1 1/3=2pi/5

t

[kind crane]

Yea

[royal tusk]

now what would i do with that tho

[kind crane]

Find inverse sine of 1/3 and plug into a calculator to get a decimal value

Then use algebra and solve for t

[royal tusk]

would I use radians

or degree

when doing the inverse sine

[kind crane]

Depends on the units in the problem

[royal tusk]

in this question

what would you say

[kind crane]

π being there suggests radian

[royal tusk]

I use

ah

alr

uh

lowkey my alg is bad

i tried solving

but idk how

I did this

but i dont think im supposed to multiply 5

nvm i got this

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[livid spoke]

dumb question, but if i see something like x²+bx+c can i just add a =0 at the end so i can use the quadratic formula

[eager thunder]

well a=1 not 0 in the case you mentioned

so...

[livid spoke]

so i cant do that ever?

[upper bane]

i think OP meant to just stick an = 0 at the end

which is not valid in this case, because that would be an assumption

if you are told to assume that the expression = 0, though, then go ahead

[wide moss]

that only makes sense if you're trying to find the roots

[upper bane]

do you have an example question?

[livid spoke]

no

how do i do this ome

i think its not a or b right

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[heavy basalt]

how to do those questions

[kind viper]

do you know how to do volumes of revolution in general?

also is that your work or someone else's scribbled on

@heavy basalt

[heavy basalt]

that is my work but idk if it is correct or no

[heavy basalt]

yeah

im just not sure with something

[kind viper]

yknow like

the more specific you are with the things you're unsure about, the better

ok right

19 is correct

20 is incorrect: integrand should be (1 - y^2) not (1-y)^2.

washer method calls for pi * int(outer^2 - inner^2) dx

inner being whichever curve is closer to the rotation axis

[heavy basalt]

ohhhh i forgot that

[heavy basalt]

and you don’t need to square inside again?

[kind viper]

what like (1-y^2)^2 ?

[heavy basalt]

yeah

[kind viper]

no you dont.

[heavy basalt]

okay

[kind viper]

the washer method hinges on the area formula for an annulus

[heavy basalt]

i mixed with the other one ig

[kind viper]

which is pi(R^2 - r^2)

[heavy basalt]

i see

didn’t matter if it is a triangle or other shape ?

[heavy basalt]

and for 22 is like (1-x^2)?

[kind viper]

^squre
just write ^2 ...

yes number 22 will be $\pi \int_0^1 (1-x^2) \dd{x}$

[boreal girder]

Ann

[heavy basalt]

okay nice

is 21 also correct?

[heavy basalt]

I must declare my profound gratitude for thy unwavering support. Tysm Ms

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[plush portal]

How to prove menelau's theorem using Section formula?

[compact pewter]

@plush portal Has your question been resolved?

[obtuse tendon]

sorry what is section formula?

[plush portal]

when a line segment A(a,b) and B(c,d) is divided into ratio m:n by a point P, then P is $(\frac{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n})$

[boreal girder]

Double_mytrouble

[obtuse tendon]

ok skull.

<@&286206848099549185>

[plush portal]

when divided externally take sign of n negetive

[plush portal]

I got it nevermind

.close

[compact pewter]

Channel closed

Closed by @plush portal

Use .reopen if this was a mistake.

[compact pewter]

Available help channel!

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Remember:
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[nocturne kettle]

I can find a triangle like this

but I don't think there is enough info here to solve?

Well I'm sure there is

but I don't see it

I can find all these triangles

[civic quartz]

what the focus for this ellipse:
x^2/a^2 + y^2/b^2 = 1

[nocturne kettle]

I've never seen that before

can you explain what each letter represents?

[civic quartz]

general equation of an ellipse?

[nocturne kettle]

oh I see

a is the radius of the major axis?

[nocturne kettle]

So then a=5

I can do $\frac{5^2}{5^2}+\frac{0^2}{b^2}=1$

[boreal girder]

UCYT5040

[nocturne kettle]

$1+\frac{1}{b^2}=1\\frac{1}{b^2}=0$

[boreal girder]

UCYT5040

[nocturne kettle]

nvm I can't do this

I end up with 1=0

ohh I think I can use coordinates of the foci

$4^2=5^2-b^2\16=25-b^2\-9=-b^2\b=3$

[boreal girder]

UCYT5040

[nocturne kettle]

.close

[compact pewter]

Channel closed

Closed by @nocturne kettle

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
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[cerulean steeple]

im suddenly going back, what does "uniquely defined" exactly mean?

i thought it means like you can just choose it arbitrarily

but that doesent make sense

question is find the amount of positive integer solutions

basically the argument here is that if p=2 mod 3 then x mapto x^3 is surjective on Z/pZ->Z/pZ (call this lemma *). if 2027|b then a=b=2027 is a solution, if its not then you can choose any x, then rhs is uniquely defined and from *, b is uniquely defined. moreso, a^3=bx and from *, a is uniquely defined, thus there exists 2027 solutions to this case

[compact pewter]

@cerulean steeple Has your question been resolved?

[hushed magnet]

if x->x^3 is surjective, then also bijective

aka you can take cube roots uniquely

something like that probably

you dont have other third roots of unity which can mess with you

[cerulean steeple]

yeah i get that

[cerulean steeple]

i still dont get this though

[cerulean bramble]

"defined" - we know it exists
"uniquely" - ... it's unique

[cerulean steeple]

so like its unique to what?

like a is defined independently to other variables?

[hushed magnet]

given any number x, then we always have a unique b so that so x=-b^3

uh wait, b also appears on the left...

uhm...

[cerulean bramble]

"ditakrifkan secara unik", assuming it's Malay

[hushed magnet]

dunno what they are doing

but then again, I cant read it

[cerulean bramble]

Chi's only defined using two integers, a and b

[cerulean steeple]

indonesian

[cerulean steeple]

let me try to translate the whole thing then

[cerulean bramble]

oh then that might make things easier... "didefinisikan secara unik"

[cerulean steeple]

im not gonna lie i understand english better than my native tongue...

[cerulean bramble]

Rough translation into English:

Find the number of pairs of natural numbers [...] that satisfy
[...]

Solution: We claim that the answer is 2028. We will use the following well-known lemma.
CLAIM: If [...] is prime, then the map [...] is surjective.
To solve the problem, consider that 2027 = 2 (mod 3) and 2027 is prime. We consider two cases: - If 2027 | b, then 2027 | ... and therefore ... . Consequently, a = b = 2027 and this works. - If the opposite is true, then we have
[...]
Choose any .... . Then, the RHS is uniquely defined and from the claim, b is uniquely defined. Moreover, ... and from the claim, a is uniquely defined. Therefore, there are 2027 solutions to this case. We need to be careful about the case where RHS = ... and leads to a contradiction, but this is impossible because it leads to 2027 | ... but ...., a contradiction.

[cerulean steeple]

4.Find the total amount of pairs of positive integer $1\geq a,b\geq 2027$ such that
$$2027\mid a^6+b^5+b^2$$
Solution: we claim that the answer is $2028$. we will use the following popular lemma
Claim. If $p\equiv 2 \mod 3$ is a prime, then the mapping $x\mapsto x^3$ is surjective (under $Z/pZ\to Z/pZ$)
to complete the solution, consider that $2027\equiv 2 \mod 3$ and it is prime. Consider two cases: If $2027\mid b$ then $2027\mid b^5+b^2$ which implies $2027|a$, this means $a=b=2027$ and this works. If not, then
$$\left(\frac{a^3}{b}\right)^2+1\equiv -b^3 \mod 2027$$
Choose any random $X=\frac{a^3}{b}\in{1,2,\dots,2027}$. Then, RHS is uniquely defined and from the claim, b is uniquely defined. Also, $a^3=bX$ and from the claim, $a$ is uniquely defined. Because of that, there exists $2027$ solutions for this case. We have to be careful when RHS$=0\mod 2027$ which will lead to a contradiction, but this is not possible since $2027\mid \left(\frac{a^3}{b}\right)^2+1$ but $2027\equiv 3\mod 4$, which is a contradiction

[cerulean bramble]

well shet

[boreal girder]

ihave<skissue>

[cerulean bramble]

[I think this confusion stems from the fact that this is a zero-verb sentence - so in English, we use a conjugation of "to be", so "b is uniquely defined" but Bahasa Indonesia (and many other languages) don't use a verb here at all]

[It's not defining that b be uniquely defined; it's concluding that it is therefore uniquely defined]

[cerulean bramble]

Do you understand up to that middle congruence relation?

[cerulean steeple]

yes

i am lost on the second paragraph

[cerulean bramble]

So the idea is, this relation does hold somehow

But we're dealing with integers

So that fraction on the left has to be an integer

[cerulean steeple]

does it?

[cerulean bramble]

Well if not, then the LHS isn't an integer at all

[cerulean steeple]

cant you use modular inverse or that sort, or is that what you meant?

[cerulean bramble]

But the RHS is an integer

We're literally dividing by b there, not multiplying by the mod-2027 inverse of b

[cerulean steeple]

oh?

so we want b|a^3?

[cerulean bramble]

Essentially

We have to check that that is actually what's going on though, which is what the following paragraph is doing

They've skipped some steps there, but that middle congruence relation is making use of the original divisor-fact we were given

[cerulean steeple]

divided by b^2 right?

[cerulean bramble]

yes

I missed by like three whole rows there lol

a | b + c implies b + c = 0 (mod a)

So b = -c (mod a)

Hence the minus appearing

[hushed magnet]

ok but thats the problem I had. chi is not defined in terms of a and b. a and b are defined in terms of chi. chi is chosen first, then there is a unique b so that chi^2+1=-b^3. and then there is a unique a so that a^3=b chi

[cerulean bramble]

Not quite

Chi is defined in terms of a and b

They're using that as a stand-in to refer to the fraction

[hushed magnet]

you cant define chi in terms of a and b if you dont even know that a and b exist

[cerulean bramble]

We're assuming they do

And checking what they are

(this is generally what you do with number theory questions tbh)

Supposing that this fraction is equal to anything mod 2027, then the middle congruence (henceforth MCE) gives us a unique number mod 2027 for b^3

Further, since b is between 1 and 2027, and 2027 is prime, then b is unique

[hushed magnet]

its fucked up logic to talk about a^3/b as if that was fixed if you didnt already know that a and b are unique and exist

[cerulean bramble]

Well, okay

[hushed magnet]

its perfectly fine to start with chi and then find a and b with chi=a^3/b and chi^2+1=-b^3

[cerulean bramble]

We can claim that this is some integer at least

If we suppose this has some value, then the MCE gives us a unique b, and then there's a unique a to follow - so then this actually does have a valid value

[cerulean steeple]

-# whats mce

[cerulean bramble]

-# middle congruence equation - but I'm typing this out far too many times

Another way of going about this is:
Pick a value for b

Then using the MCE we can find a unique a such that the fraction is in fact an integer

[hushed magnet]

no

why

[cerulean bramble]

Why must the fraction be an integer?

[hushed magnet]

at the very least there is a problem with a and -a

[cerulean steeple]

so if the MCE has some unique value, then there will be b that is unique mod p (mod p is diffrent for any other MCE) and using this you also get the ssame result for a?

[candid field]

So like my brother wants to be advanced fifth grader so don't blame me he just want to learn math sooooo idk

[cerulean bramble]

So like

[cerulean bramble]

!occupied

[compact pewter]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[cerulean bramble]

I think we're going around in circles

[cerulean steeple]

maybe

[cerulean bramble]

Ignoring chi altogether because it's just in the way,

If we pick any b, then the MCE can demonstrate that there is a unique a for this b so that this works

[cerulean bramble]

How so?

Because of the even power of a?

[hushed magnet]

yes

[cerulean steeple]

ohhhh

[cerulean bramble]

Finnicky, but I want to claim that it's because a and b are restricted to being between 1 and 2027

[hushed magnet]

well I mean -a obviously mod 2027

[cerulean bramble]

That's an expression, not an equation

dym " -a = 2027 - a (mod 2027) "?

[hushed magnet]

if a works, then 2027-a also works

so given b, there is no unique a that works

[cerulean bramble]

Oh wait

[cerulean bramble]

This claim's being used here

[hushed magnet]

that is irrelevant for what I mean

[cerulean bramble]

Claim - if p = 2 (mod 3) is a prime, then the map x |-> x^3 is surjective

oh true there's still the 2 there

[hushed magnet]

the strategy of the proof is: a,b are solutions to a^6+b^5+b^2 = 0 mod 2027 iff they are solutions to x^2+1 = -b^3, x=a^3/b mod 2027 for some x in {1, ..., 2027}

and for any fixed x in {1,...,2027} there is a unique pair (a,b) solving that second system of equations

[cerulean bramble]

hmm

I'm tempted to ask for a third opinion on this

There's gotta be some congruence relation with squares that applies here somewhere

e.g. how square numbers are either 0 or 1 mod 3

And 0, 1 or 2 mod 5... OH

wait no, shit.

[cerulean steeple]

is the solution incorrect then? or is it just terribly written/structured so theres aton of confusion now

[cerulean bramble]

Probably the latter

I think they're missing another lemma or something

I was about to say "yh I'm tempted to suggest you may summon the bat signal" at this point

[hushed magnet]

do you guys agree with this iff

[cerulean steeple]

which bat 👀

[scenic viper]

how to prove 1+1

[cerulean bramble]

(the @helpers signal)

[cerulean steeple]

does that really do anything

[hushed magnet]

1+1=succ(1)=2. for more details feel free to ask <@&268886789983436800>

[cerulean steeple]

its like a calling to the void

[cerulean bramble]

!occupied You've been warned. Also -

[compact pewter]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[scenic viper]

are u guys graduated in pure math?

[cerulean steeple]

as in the solution would look like that and it is true, or like what the image i posted is implying?

[hushed magnet]

just the iff claim

(a,b) solutions to this iff to that

[chilly basalt]

hi, let's move to #discussion cuz they're solving a problem

[cerulean bramble]

(yes, but) Again, this channel is occupied. Help channels are not a place for casual discussion. One would think that the channels labelled #discussion would be bloody obviously the places for discussions ...

[zealous rune]

Please move channels to not occupy someone else's help channel, for example visit #discussion.

[chilly basalt]

Let's not disturb them

[scenic viper]

sry

[hushed magnet]

so lets say a,b are solutions to a^6+b^5+b^2 = 0 mod 2027

and a,b nonzero

so then (a^3/b)^2 +1 = -b^3 mod 2027

so far we agree

setting x=a^3/b, then x^2+1 = -b^3 mod 2027

so a,b are solutions to x^2+1=-b^3 , x=a^3/b mod 2027 for some value x

on the other hand, let x be some value and let a,b be solutions to x^2+1=-b^3, x=a^3/b mod 2027

plugging the second equation into the first we get (a^3/b)^2+1=-b^3 mod 2027, so a^6+b^5+b^2 = 0 mod 2027

this proves that (a,b) is a solution to a^6+b^5+b^2 = 0 mod 2027 iff (a,b) is a solution to x^2+1=-b^3, x=a^3/b for some value x

is that ok

[cerulean steeple]

personally it makes sense but admittedly i do not fully grasp the concept so take my opinion with a grain of salt

[hushed magnet]

what concept are you not grasping

so far this just normal modulo and equation stuff

[cerulean bramble]

-# [lol that sounds antagonistic ]

[cerulean steeple]

actually nvm what i was kinda iffy on just clicked