help-49

[midnight plank]

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[last slate]

I know how to show isomorphism between two groups

But how do I do it with one being the external direct product

<@&286206848099549185>

[uneven sandal]

do you know how elements in Z_2 + Z_3 look like

[last slate]

Yes I think

It's the cross product

In a pair

( a , b)

[uneven sandal]

if a is in Z_2 and b is in Z_3 then (a,b)

yeah

[uneven sandal]

you need to find an isomorphism

[last slate]

Don't I need a function to to prove that

[uneven sandal]

yes

[last slate]

But I got none here

[uneven sandal]

the question is that you define your own function that satisfies all the properties that an isomorphism (function) has

so you need to identify which elements correspond to which

[last slate]

I don't know what function to take or how to define it

[uneven sandal]

first of all it has to go from Z_2 + Z_3 to Z_6 or the other way around

[last slate]

Yup

Can u help me out with defining the function

[uneven sandal]

sure

First of all for any isomorphism the neutral element of one group is sent to the neutral element of the other group

[last slate]

I think I can prove this with two previous theorems but I don't think that's what the question wants cause it's 5 mark

[last slate]

You mean the identity

[uneven sandal]

or identity element

sometimes called neutral element as well

however you wanna name it

[last slate]

I see okay

[uneven sandal]

So we know how to add numbers in Z_6

now our task is to find what elements of Z_2 + Z_3 correspond to elements of Z_6

[last slate]

Yes it's the mod 6

[uneven sandal]

for this lets just take an element from Z_2 + Z_3 choose any

[last slate]

Hmm let's take (1,1)

[uneven sandal]

Okay now add (1,1) to (1,1)

[last slate]

That's (2,2)

[uneven sandal]

yes but notice that the first slot is for elements in Z_2

[last slate]

Ohh yeah

Then it will be

(0,2)

My bad

[uneven sandal]

no worries

now add (1,1) to (0,2) again

[last slate]

(1,0)

[uneven sandal]

yes and if we do that again we get (2,1) = (0,1)

and once more then we get (1,2)

okay one last time

add (1,1) to (1,2)

[last slate]

Yup

But we are getting elements in z 2 x z3

[uneven sandal]

(1,1) + (1,2) = ?

[last slate]

(0,0)

Identity

[uneven sandal]

So we notice that

(1,1) + (1,1) + (1,1) + (1,1) + (1,1) + (1,1) = (0,0)

[last slate]

We did 6 time

[uneven sandal]

ye my bad changed it

[last slate]

Yup so the order of (1,1) is 6

Meaning z2 x z3 is cyclic

Generated by (1,1)

[uneven sandal]

yeah thats true

But I think not relevant for this question

[last slate]

Okay

[uneven sandal]

Now apply an isomorphism to both sides of our equation

[last slate]

How

[uneven sandal]

Lets assume you have an isomorphism f: Z_2 + Z_3 -> Z_6

[last slate]

Ok

[uneven sandal]

then

f((0,0)) = f( (1,1) + (1,1) + (1,1) + (1,1) + (1,1) + (1,1) )

And since we assumed f is an isomorphism we can do the following

f((0,0)) = f( (1,1) ) + f( (1,1) ) + f( (1,1) ) + f( (1,1) ) + f( (1,1) ) + f( (1,1) )

[last slate]

Yup

[uneven sandal]

Now f((0,0)) = 0

since this holds for any isomorphism

[last slate]

Yup

[uneven sandal]

and f( (1,1) ) is an elemnt out of Z_6

so what element in Z_6 do you have to add 6 times to get 0

[last slate]

1

[uneven sandal]

So we may define f( (1,1) ) = 1

[last slate]

Got it what about other elements

[uneven sandal]

Maybe take (0,2)

and check how many times you have to add (0,2) to itself to get (0,0)

[last slate]

Hmm 3 time

[uneven sandal]

(0,2) + (0,2) + (0,2) = (0,1) + (0,2) = (0,0)

[last slate]

Yup

[uneven sandal]

and again you can apply f on both sides

[last slate]

We get 2

f(0,2)=2

[uneven sandal]

yeah and do the same for all other elements in Z_2 + Z_3

[last slate]

I see let's assume I do that after that what do we do

We already assumed that there is an isomorphism

[uneven sandal]

Okay now you have a candidate for the isomorphism you are looking for

[last slate]

I don't quite understand

[uneven sandal]

Okay lets say you wirte everything out, you will get the following:

f( (0,0) ) = 0
f( (0,1) ) = a
f( (0,2) ) = b
f( (1,0) ) = c
f( (1,1) ) = d
f( (1,2) ) = e

[last slate]

Yup

[uneven sandal]

Then this is your educated guess for what the isomorphism should be

So how you start wrting this down is

You state your "guessed" isomorphism

and show that all the properties and isomorphism needs to have are satisfied

[last slate]

I see wait lemme find all the a b c d

[uneven sandal]

Yeah you first need to find out what your isomorphism should be, but when writing this down you can just say I define this function f and then show that your function is indeed an isomorphism

[last slate]

We get f(0,1) = 2

[last slate]

We again have to prove normally after finding the function?

I thought as we assumed an isomorphism and got the function

[uneven sandal]

are you sure its 2?

[last slate]

Therefore the function must be isomorphism

[last slate]

Yeah i got confused cause I got 2 for (0,2) too

[uneven sandal]

yeah (0,1) + (0,1) = (0,2)

so f(0,1) =! 2

you need to check how many times you have to add the same element to get the identity

in this case 4 * (0,1) = (0,0)

[last slate]

Yup doing that I got

f(1,2) = 1

[last slate]

No it's z3

Not z4

[uneven sandal]

the 2nd slot is for elements in Z_3

[last slate]

Yup z3 = { 0 , 1 , 2 }

[uneven sandal]

my bad

[last slate]

[uneven sandal]

ah lets work with you argument

f( (1,1) ) = 1

and you found the group is cyclic

so you can find all other elements from (1,1) and what they should correspond to

I believe we miscalculated somewhere

So f( (1,1) ) = 1 => f( (1,1) ) + f( (1,1) ) = 1 + 1

f( (1,1) ) + f( (1,1) ) = f( (1,1) + (1,1) )= f( (0,2) ) = 2

this way you can figure out a,b,c,d,e

[uneven sandal]

No we only made this assumption to guess the right function

[last slate]

I see

[uneven sandal]

you can see the current calculation as something you would write on a scratch paper

and after you found the function you throw all the calculations away and only keep the function you found

[last slate]

I see okay

I am getting the a b c now

[uneven sandal]

and then you rigorously show that your guessed function satisfies all the assumptions for an isomorphism

and if all of that works our then you know that Z_2 + Z_3 is isomorphic to Z_6

[last slate]

Okay got them

[uneven sandal]

What did you get?

[last slate]

$$ f(0,0) = 0 /
f(1,1)=1 /
f(0,2)=2 /
f(1,0)=3 /
f(0,1) = 4 /
f(1,2)=5$$

[uneven sandal]

Nice :)

[grand pond]

Haruki

[last slate]

Lol i forgot latex I just used it once to write some paper for project

[uneven sandal]

nvm xd

[last slate]

Anyway what do I do now

I don't see a pattern to form a function

[uneven sandal]

Now you have a guess for what your isomorphism should be

Because we miscalculated somewhere

"Yup doing that I got
f(1,2) = 1"

but add (1,2) to itself

and see when you get zero

(0,0)

[last slate]

I got f(1,2)= 1 before but with generator I am getting 5

[uneven sandal]

I also get 5 this way

[last slate]

Can we just use theorem to prove this

We already got that z 2 x z3 is cyclic

And order 6

And z6 is cyclic

And order 6

Hence by another theorem it's isomorphism

Idk if this is applicable for 5 mark

[uneven sandal]

sure you can but you can also identify the isomorphism which you already have

[last slate]

I found all the mapping but

Hmm i have an idea

[uneven sandal]

Now you have to check if your function f is an isomorphism

[last slate]

Why not just put it like this

f is a function maps from Z2 X Z3 to Z 6 and be defined as then we put this

[last slate]

.

f(1,1)=1

And so on

[uneven sandal]

no f(1,1) =! 0

[last slate]

And check if f is isomorphism

Sorry

1

[uneven sandal]

Do you know what you have to check

[last slate]

Yes

We have to check if f(a o b) = f(a) o f(b)

Then bijection

[uneven sandal]

the first one is a bit annoying

the 2nd one is much easier

in your case

[last slate]

Yup

[uneven sandal]

for the 2nd one you can just define the inverse

or guess some other function and then see that its the inverse

that is the mathematically correct way to do it

[last slate]

Yup I have proved these few time so i think I can manage

Tricky part was getting the function

Defined

.close

[midnight plank]

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[midnight plank]

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[rare maple]

Given the vertex of the parabola, a point P on the parabola and the angle between line VP and axis, how do we construct the focus using straight edge and compass

[fervent minnow]

if you position the vertex at the origin, you could use the tangent of the angle to find the equation for the parabols

oh nevermind i forgot you have to use straight edge and compass

[shell wigeon]

and the angle between line VP and axis,
What exactly do you mean by that? Are you basically given the axis of symmetry of the parabola?

[rare maple]

Yes

[shell wigeon]

In that case all you need to do is draw the tangent at P

The tangent to x^2 at x=p goes through (p/2,0)

That should be all you need

[rare maple]

are there any messages after mine

[midnight plank]

@rare maple Has your question been resolved?

[midnight plank]

Channel closed

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[fickle sierra]

Hello, could someone check if this proof looks good please?

\begin{Theorem}
Assume $n \in \bZ$. If $n^2 -4n + 7$ is even, then $n$ is odd.
\end{Theorem}

\begin{proof}
We use the contrapositive.
Suppose $n$ is even. Then, $n = 2k$ for some integer $k$.
So,
\begin{align*}
n^2 -4n + 7 &= (2k)^2 -4(2k) + 7\\
&= 4k^2 - 8k + 4 + 3\\
&= 4(k - 1)^2 + 2 + 1\\
&= 2(2(k - 1)^2 + 1)+ 1\\
\end{align*}

Since $k$ is an integer, so is $2(k - 2)^2 - 1$.
And then, $n^2 -4n + 7 = 2l + 1$, where $l = 2(k - 2)^2 - 1$.
Thus, by the definition of odd integers, $n^2 -4n + 7$ is odd.
Therefore, by the contrapositive, if $n^2 -4n + 7$ is even, then $n$ is odd.
\end{proof}

[shell wigeon]

Not sure why you did the factorization but yes, the proof works

[low arrow]

this step looks off.

how did the (k-1)^2 become a (k-2)^2 and the +2 turn into a -1? I can understand if the +2 turned into a +1, but that doesn't explain the first issue.

[fickle sierra]

Typo, fixed 👌

[grand pond]

Mor Bras

[shell wigeon]

You need to fix it everywhere, not just on that line

[low arrow]

all mentions of (k-2) have to be (k-1).

I see two more spots.

[shell wigeon]

Though again, I question your reason for doing that factorization in the first place

[fickle sierra]

Yes, I see

😵‍💫

[last slate]

Guys anyone know references for number theories

[low arrow]

hi, I think #book-recommendations might be a better place to ask that question. this channel is unfortunately in use by another person.

[last slate]

Wdym by another person

[low arrow]

this help channel is being used by another user to ask their question.

[fickle sierra]

#❓how-to-get-help

[fickle sierra]

\begin{proof}
We use the contrapositive.
Suppose $n$ is even. Then, $n = 2k$ for some integer $k$.
So,
\begin{align*}
n^2 -4n + 7 &= (2k)^2 -4(2k) + 7\\
&= 4k^2 - 8k + 6 + 1\\
&= 2(2k^2 - 4k + 3)+ 1\\
\end{align*}

Since $k$ is an integer, so is $2k^2 - 4k + 3$.
And then, $n^2 -4n + 7 = 2l + 1$, where $l = 2k^2 - 4k + 3$.
Thus, by the definition of odd integers, $n^2 -4n + 7$ is odd.
Therefore, by the contrapositive, if $n^2 -4n + 7$ is even, then $n$ is odd.
\end{proof}

[grand pond]

Mor Bras

[shell wigeon]

Yes, that looks cleaner

[fickle sierra]

Does the proof looks good now?

[shell wigeon]

Looks like Nicole approved, so should be good

[fickle sierra]

Thank you for your comments!

.close

[midnight plank]

Channel closed

Closed by @fickle sierra

Use .reopen if this was a mistake.

[low arrow]

oh wait!

[low arrow]

minor mistake here!

and you'll have to correct every appearance of this.

[fickle sierra]

Thank you!

[midnight plank]

Available help channel!

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[fickle sierra]

Hello, could someone check if this proof looks good please?

\begin{Theorem}
Assume $m,n \in \bZ$. If $mn$ is odd, then $m$ and $n$ are odd.
\end{Theorem}

\begin{proof}
We use the contrapositive.
Suppose either $m$ or $n$ is even. 
Without loss of generality, assume $m$ is even.
Then, $m = 2k$ for some integer $k$.
So, $mn = (2k)n = 2(kn)$.
Since $k$ and $n$ are integers, so is $kn$.
Then, $mn = 2l$, where $l = kn$.
Thus, by the definition of even integers, $mn$ is even.
Therefore, by the contrapositive, if $mn$ is odd, then $m$ and $n$ are odd.
\end{proof}

[low arrow]

minor nitpick: m and n are odd.

the last line too.

[grand pond]

Mor Bras

[low arrow]

mathematically I don't see anything wrong with the proof.

[shell wigeon]

(I'd probably write "suppose either m is even or n is even", but I would understand what you wrote)

[lusty python]

more generally, $\forall n \in \mathbb N, n > 1$, if $n \nmid ab \Leftrightarrow \begin{cases}
n \nmid a\
n \nmid b
\end{cases}$

[grand pond]

1 divided by 0 equals Infinity

[lusty python]

using the same way you proved

[midnight plank]

@fickle sierra Has your question been resolved?

[fickle sierra]

Thank you for your comments!

.close

[midnight plank]

Channel closed

Closed by @fickle sierra

Use .reopen if this was a mistake.

[midnight plank]

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[obsidian glen]

Let f : (a, b) → R be a monotone function. Show that f has at most countably many
point of discontinuity in (a, b), all of which are jump discontinuity.

[obsidian glen]

I start by considering f to be non decreasing since we can similarly show the other case as well

then [ \lim_{x \to d^{-}} f(x) \leq f(d) \leq \lim_{x \to d^{+}} f(x) ]

[grand pond]

ginny

[obsidian glen]

for d which is a point of discontinuity

but idk what I should do from here

(I am starting by showing it has only jump disc the countabloe discontinuity is for later)

[eternal pawn]

see if you can map each discontinuity to a something that is countable

[obsidian glen]

I mean idk how to

[obsidian glen]

I know I have to do this, with N or smth

[eternal pawn]

a bigger set would be easier

[obsidian glen]

bigger than N? Q?

[eternal pawn]

Q should do it

now, we need to map each discontinuity to a rational in an injective way
Hint ||use the fact that f is monotone||

[obsidian glen]

this right?

but no

[eternal pawn]

kinda

[obsidian glen]

that part assumes the limits exist

which would mean jump disc

but we have to show exactly that

[eternal pawn]

in this case, it will work

[obsidian glen]

I mean how so?

[eternal pawn]

since f is monotone

so it has a left/right limit at each point

[cosmic zodiac]

hi

[obsidian glen]

hello

[obsidian glen]

why

[visual tiger]

Monotone convergence

Non decreasing + upper bounded for left limit

Vice versa for right limit

[obsidian glen]

got it

okay so I can map d to m/M where m is the left limit and M is the right limit then? since every discontinuity has these two associated with it

and that should be that

[pulsar basalt]

am back online but soon

[eternal pawn]

it needs to be countable though

[obsidian glen]

ah yeah we need to have an injective map

not just a map to Q

so this map is no good

but since d is a point of discontinuity m != M, so I could map it to m/M + M/m

[nova yoke]

how will you use this to show that there are only countably many discontinuities?

[obsidian glen]

yeah this won't work mb, this isn't injective

because x+1/x isn't either so yeah

[nova yoke]

what set are you trying to inject into?

[obsidian glen]

Q

for countability

[nova yoke]

m/M isn't necessarily rational, right?

m and M can be any real numbers with m < M

[obsidian glen]

oh lol yeah lmao

m/M looks deceiving I forgot that doesn't make it rational

my bad

[nova yoke]

M could even be zero haha

to get you back on track... what can you say about the interval (m, M)?

[obsidian glen]

m < M

so an interval with measure M-m

with f(d) in it

too

[nova yoke]

well f(d) might not even be defined

oh wait it is

it might be m or M though

not necessarily in (m, M)

[obsidian glen]

it is a monotone function

it has to be

[nova yoke]

it's defined but not necessarily in the open interval (m, M)

do you need it to be?

[obsidian glen]

ah

no it doesn't

it's in [m,M]

yeah

[nova yoke]

let's say we consider two points of discontinuity

say d1 and d2

with corresponding points m1, M1, m2, M2

what can you say about the intervals (m1, M1) and (m2, M2)?

[obsidian glen]

disjoint is what I think it turns out to be but let me also think of an argument for it lol

[nova yoke]

yep that's what you want 😁

[obsidian glen]

yeah I cant argue why that is the case

if they weren't disjoint, so one of m_1 or M_1 will be in (m_2, M_2)

I mean the M_1 will be in the latter interval supposing M_2 >= M_1

[nova yoke]

it might help to draw a picture?

assume there's a point in both intervals

argue that it simultaneously has to be less than M_1 and greater than m_2 (assuming d1 < d2)

[obsidian glen]

and?

i mean once they are disjoint I can say there is ofc atleast 1 rational in them making the cardinality of number of these disjoint intervals less than that of Q so countable

[midnight plank]

@obsidian glen Has your question been resolved?

[midnight plank]

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[onyx jackal]

For d

[onyx jackal]

How do I simplify this as much as possible?

This is what I got so far

[fallow scarab]

next step would be to use 4 = 2^2

[onyx jackal]

Ok, I got 2^2n+1

It’s just that according to the textbook the answer should be 4?

[fallow scarab]

do you get something else ?

[onyx jackal]

I got 2^2n+1

Oh, that’s wrong

It should be 2^2n+2

Although I’m still unsure how I’m supposed to get 4 due to the unknown?

[fallow scarab]

too blurry to read

doesn't look like you distributed the 5n/3 power to the 2 here

[clever sedge]

= exists 😭

[onyx jackal]

Ahh I got it

[onyx jackal]

Yea that was my error

Easy peasy

[midnight plank]

@onyx jackal Has your question been resolved?

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[small warren]

Could I please get help understanding this formula? I would love an example of it if possible, and the name of it if it exists? Thanks!

[fossil knot]

It's like suppose you have a long stick that you want to divide into two pieces

and I said, I want one piece to be r times longer than the other piece

[small warren]

Right!

But how do you apply that?

Is B and A their change?

Or do you have to calculate P for both x and y

I don't get it

[fossil knot]

AB is what we're calling the line segment (i.e "long stick")

The left end is A, the right end is B

[small warren]

and if r is 1/3, then is it P=(1B+A)/3+1 or is it P=(1/3B+A)/1/3+1?

[fossil knot]

the second one since you're plugging in r=1/3

[small warren]

and each term has an x and y coordinate

[small warren]

Okay thanks!

[fossil knot]

in 2D yes

[small warren]

How does this look like in an example, if you don't mind?

Do i first solve for delta x for A and B and then delta y?

[fossil knot]

Sure let's say A = (2,3) and B = (10, 27) and we want r=1/3

[fossil knot]

You can just do both coordinates together, but if makes it more simple then yes you can also do this

[small warren]

Thanks!!!!!

[tepid yoke]

Can sm1 help me with dividing a fraction what has letters

[small warren]

How do you do that?

[tepid yoke]

Sm1 pls

[fossil knot]

go to an unoccupied channel

such as #help-12

#❓how-to-get-help

[tepid yoke]

Appreciate it

[fossil knot]

well let's do both coordinates together first since I think it's easier

Sure let's say A = (2,3) and B = (10, 27) and we want r=1/3

[small warren]

right

What's this formula called?

[fossil knot]

Then we have $P = \frac{\frac{1}{3}(2,3) + (10,27)}{\frac{4}{3}}$

[grand pond]

Ari

[small warren]

Yep

[fossil knot]

I'm just using the equation that you sent I don't think it has a conventional name

[small warren]

oh ok

[fossil knot]

$P = \frac{\frac{1}{3}(2,3) + (10,27)}{\frac{4}{3}}$

$= \frac{3}{4} \cdot ((\frac{2}{3}, 1) + (10, 27))$

[grand pond]

Ari

[fossil knot]

$= \frac{3}{4} \cdot (\frac{32}{3}, 28)$

$= (8, 21)$

[small warren]

What did you just do?

[grand pond]

Ari

[fossil knot]

I just simplified the expression

[small warren]

i know, but how?

I mean the 3/4 part

how do you do that to get rid of the denominator?

[fossil knot]

ok so step by step

1. dividing by 4/3 is the same as multiplying by 3/4. So I pulled that out.
2. I multiplied the 1/3 into (2,3). 1/32 = 2/3, 1/33=1. So it ends up being (2/3, 1)

[small warren]

Ohhhhhhhhh

Ty

Tysm

[small warren]

How about this? Why isn't this x+x = 2/3+10=60/3=30?

if you dont mind

[fossil knot]

2/3 + 10 = 2/3 + 30/3 = (2 + 30)/3 = 32/3

[small warren]

omg it's just 32

[fossil knot]

yes

[small warren]

Ty!

Tysmmmmm ❤️

And I suppose the other way is to just do A_x and B_x first and calculate for that for the coordinate of P and then the same for y?

Don't be rude @tepid yoke

[fossil knot]

yes

[small warren]

Ty!

[fossil knot]

but the way I did it just does both at the same time

[small warren]

Yea it's definitely what I'll do, I really get bored from tedious work, if i can half my work then that's what I'll do

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[near geyser]

I am still unsure how exactly one applies Rouche's Theorem to find the roots of polynomials. For example, to find how many roots z9+z5−8z3+2z+1
has in between the circles |z|=1
and |z|=2

[fossil knot]

it can tell you how many lie in D_2(0)

and how many lie in D_1(0)

then you can subtract them

[near geyser]

Yeah. But i am confused at how do we select the dominating polynomial@fossil knot

[fossil knot]

ideally you want to simplify the current polynomial if that makes sense

right now we have $f(z) = z^9 + z^5 + 8z^3 + 2z + 1$

[grand pond]

Ari

[fossil knot]

a good candidate to try could be $g(z) = z^9$

[grand pond]

Ari

[fossil knot]

since you know g has 9 zeros at the origin

then you can try to see if |f-g| < |f| everywhere in the disk

if not, you could also try h(z) = z^9 + 1 or something

[near geyser]

I see

Thanks for explaining

[fossil knot]

yes try to see a worked example of these kinds of problems

usually theyuse a pretty similar approach

[near geyser]

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[blissful trench]

What is the approximate ratio of AC resistance to DC resistance ($R_{AC}/R_{DC}$) for steady-state sinusoidal excitation at 60 Hz?

[grand pond]

# real car fan

[blissful trench]

answer says 1.6 but I'm not sure how they got it

[midnight plank]

@blissful trench Has your question been resolved?

[midnight plank]

@blissful trench Has your question been resolved?

[midnight plank]

@blissful trench Has your question been resolved?

[olive matrix]

kinda feel like very few people here are going to have the context for this

[modern sapphire]

What even is the difference in AC resistance and DC resistance tbh? Coz AC Resistance sounds like it just might be impedence

and for that youd need other components like inductors or capacitors in the circuit so it can be different from the resistance

[blissful trench]

I honestly dunno

is it just me, or does this question seem AI slop because the answer is provided without any solution

I'll close this before I explode lmao

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[twilit field]

why is commutativity even needed here

[lyric charm]

wait why are they asking you to prove R is an integral domain

did they not mean R/P

in any case what would a prime ideal be in a non comm ring

[twilit field]

oops

right

missed that part of the defn

[twilit field]

You think this isn't true?

I mean it's almost immediately true, which is weird for a problem so late into the chapter and exercise

( this is problem 10 for context, and while there are 41 in this exercises,yeah....)

[lyric charm]

well hm ok then it's more like saying there EXISTS a prime ideal of R without zero divisors

ig R itself will be integral then

[twilit field]

I mean if xy=0 then as P is prime , x or y in P and P has no zero divisors

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[dawn prism]

Case 1: f splits over k

$\implies$ f splits into linear factors over k

$\implies$ f $\in$ k[x] is not irreducible (since 2 $\leq$ deg(f) )

Need help with
Case 2: f doesn't split over k

[grand pond]

krumpler

[dawn prism]

I have to show that f is irreducible if it has no roots in any proper subfield of the splitting field

[midnight plank]

@dawn prism Has your question been resolved?

[frank wolf]

i think you might have better luck in #groups-rings-fields

[spiral rock]

if f = gh in k[x], look at the splitting field of g over k. Denote the splitting field of a polynomial q by k(q).
If g is non-trivial, then g splits over k(g), and hence by the assumption since f has a root in k(g), it splits over k(g). This implies that k(f) is a subfield of k(g) so k(f)=k(g).
Now p | [k(f):k] and hence p | [k(g):k], but since p is prime the only way for p to divide [k(g):k] is for g to be of degree at least p, and f=g.

[dawn prism]

I get the first part.
Why is p | [k(f):k]?

[spiral rock]

k(f) contains k(a) for some root a of f in some algebraic closure, and [k(a):k]=p.

[dawn prism]

[k(a):k]=p would mean f is a minimal polynomial of a over k

right?

[spiral rock]

yes

ok this is a bit circular

oh.... Look at k(a), f has a root in k(a) and hence f splits over k(a), so k(f)=k(a).

[dawn prism]

yep , it works

thank you very much

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[chilly cobalt]

Let $P(x)$ be a polynomial of degree $2012$ with real coefficients satisfying [ P(a)^3+P(b)^3+P(c)^3 \ge 3P(a)P(b)P(c) ] Whenever $a,b,c$ are reals satisfying $a+b+c=0$. Can $P(x)$ have exactly $2012$ distinct real roots?

[grand pond]

Copter

[chilly cobalt]

i dont really know how to begin on this question, i dont think picking some value for a,b,c to find something helps at all

what would be the motivation to solve this?

[umbral scroll]

now that's some complicated math olympiad question

[midnight plank]

@chilly cobalt Has your question been resolved?

[chilly cobalt]

<@&286206848099549185>

[solemn badge]

Take P(a)=x, P(b)=y and P(c)=z
then $ x^3 + y^3 + z^3 \ge 3xyz implies \frac{1}{2} (x+y+z)((x-y)^2+(y-z)^2+(z-x)^2) \ge 0 implying x+y+z \ge 0 whenever a+b+c=0. $
After that I'm puzzled too

[chilly cobalt]

i mean obviously we only have to consider the case where one of P(a) < 0 otherwise its am gm

[grand pond]

Hi

[midnight plank]

@chilly cobalt Has your question been resolved?

[midnight plank]

@chilly cobalt Has your question been resolved?

[graceful drum]

leading term is positive

2P(a) + P(-2a)>=0

[atomic lava]

hi

[stiff bison]

!occupied

[midnight plank]

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).

[stiff bison]

are you going to help with Copter's question or go elsewhere

best channel for posting random stuff is #chill

[midnight plank]

@chilly cobalt Has your question been resolved?

[gilded jolt]

<@&268886789983436800> has only posted anything in ocupied help forums even after being asked not to

[patent snow]

@atomic lava please bring this to #chill; if you're not gonna help, then don't put that in here

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[umbral scroll]

Is it true that if GCD(a,b) = 1 then GCD(a+b , ab ) = 1 ?

[gaunt jetty]

yes

[umbral scroll]

how to prove it

[gaunt jetty]

you can do a proof by contradiction

[umbral scroll]

hmmm

[shell wigeon]

Rigorously?

[umbral scroll]

I'm thinkin on how to do it by contradiction rn

[gaunt jetty]

u gotta use euclid's lemma

[umbral scroll]

i do not know that

well we haven't learnt it

[radiant roost]

if a prime p divides ab then either p divides a or p divides b

<@&268886789983436800>

[craggy stump]

you can assume a+b and ab have another factor other than 1 suppose x. then a+b = xc (some number c). then a = x (c-b/x) and b = x (c-a/x)

this shows a and b have x as their common factor

which contradicts the condition that GCD of a and b is 1

am i right

[radiant roost]

started off good, but when you wrote b/x you assumed x divides b

[craggy stump]

we are talking about factors here no? arent they supposed to be natural numbers?

but thats right i didnt prove that b/x will be natural

[umbral scroll]

hmm

[umbral scroll]

I like the beginning but I cant find the contradiction

[radiant roost]

i would try to prove the contrapositive, i.e.
if GCD(a+b, ab) is not 1, then GCD(a,b) is not 1

[umbral scroll]

,rotate

[umbral scroll]

is there already a contradiction ?

[radiant roost]

the negation of "P implies Q" is not "P implies not Q"

the negation of "P implies Q" is "P and not Q"

[umbral scroll]

yoo

hmm

P and not Q

so I should say et instead of alors

[radiant roost]

yeah

french, right?

[umbral scroll]

do you speak french ?

[radiant roost]

i studied it in school

[umbral scroll]

oh ok haha

[umbral scroll]

alright but how do I get the contradiction here

[radiant roost]

let p be a prime dividing k

[umbral scroll]

ok

[radiant roost]

p | ab

[umbral scroll]

ye

[radiant roost]

so p | a or p | b

take it by cases

[umbral scroll]

ok

if p I a then a = pt with t c Z

sigma

[radiant roost]

sigma?

[umbral scroll]

sigma = ts is hard and my brain is thinkin

[radiant roost]

hint || p | a+b ||

[umbral scroll]

p I ab + a

[umbral scroll]

are we doing a contradiction proof here ?

[radiant roost]

yeah

[umbral scroll]

so this ?

no

[umbral scroll]

this

[radiant roost]

that's slightly different

[radiant roost]

yeah

[umbral scroll]

ok lm think

[radiant roost]

|| why were we able to say p | ab ? ||

<@&268886789983436800>

[umbral scroll]

because ab = (a^b)^2 * t , t c Z

[radiant roost]

^ means gcd?

[umbral scroll]

yeah

mb

ill name it t cause we alr use k

[radiant roost]

i don't follow

[umbral scroll]

me neither

[radiant roost]

k = gcd(a+b, ab), so k | a + b and k | ab

[umbral scroll]

we start with gcd(a,b) = 1 and GCD (a+b, ab) = k

if p I k then p I ab alr

and p I a + b

ok

[radiant roost]

yeah

[umbral scroll]

lets say a + b = sp and ab = tp

then a+b -sp = ab - tp

then a(1-b) + b = sp - tp

ie a (1-b) + b = p(s-t)

and this is a contradiction i guess ?

[radiant roost]

you should use the cases

[umbral scroll]

nah this leads to nowhere

so p I ab and p I a+b

if p I a then p I a + b - a

meaning p I b

[radiant roost]

yeah

[umbral scroll]

so p I a and p I b

and p is prime so p!= 1

which is a contradiction ?

[radiant roost]

yes

[umbral scroll]

and same logic if p I b

[radiant roost]

yeah

[umbral scroll]

wow ok

I need a minute to tink about this again

thanks for taking your time with me, very nice of you 🙂

[radiant roost]

you're welcome

[umbral scroll]

wait

why is it true that if p I ab then p I a or p I b @radiant roost

[radiant roost]

it's euclid's lemma

[umbral scroll]

but

[radiant roost]

https://en.wikipedia.org/wiki/Euclid's_lemma#Proofs

[umbral scroll]

15 I 6* 5
but 15 doesnt divide 6 and 15 doesnt divide 5

[radiant roost]

15 isn't prime

[umbral scroll]

ohh

alright

that's the catch

[radiant roost]

yeah

[umbral scroll]

alright thanksss

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[grand dagger]

Hi

[grand dagger]

I’m having trouble interpreting the force vector

I’m setting the plane where its origin is point P

<@&286206848099549185>

[midnight plank]

@grand dagger Has your question been resolved?

[main current]

Setting the plane?

Force vector is 10 lbs directly down.

It's 12 in away from the bolt, where the bolt is 30° above the horizontal

[grand dagger]

Yeah but how am I supposed to write the force vector

If the origin is point P

[main current]

With an arrow? Maybe I'm not understanding the question

They put the vector in red on the picture

Don't forget to tell the bot ❌, otherwise it will close the channel

[grand dagger]

Okay look

[grand dagger]

See how I set up the coordinate system

[main current]

Right

Are you looking for something like (0,-10) lbs?

[grand dagger]

(0,-10,0) yes

But again that would be if you’re measuring from up there

[main current]

Then the displacement vector would be (12cos(30), 12sin(30))

[grand dagger]

Btw this is in ℝ^3 because the cross product can’t exist in ℝ^2

[main current]

Yes

Then yeah you'd do the cross product

[grand dagger]

But do you see that the origin is at point P?

I need to remain consistent

Right

I can’t drag the origin now at the initial point of the force vector to interpret it

[main current]

Vectors don't have initial/final points. They can generally be moved

[grand dagger]

I don’t understand tbh

[olive matrix]

vectors have a direction and a magnitude; they do not have a position

[grand dagger]

Yes but

Here we’re reading the vector

Oh

Okay then

[grand dagger]

Is this always a set in stone rule

At least for force and magnitude vectors in ℝ^n

[olive matrix]

yes

a vector might describe a position or rather a displacement, but it does not inherently have a position

e.g. "3 kilometres north" is a vector, but it doesn't tell you where i started from

[grand dagger]

So the position of the vector in some diagram or word problem doesn’t affect the physics being calculated ?

[olive matrix]

it doesn't; but for a case like this, torque, the force is applied at a certain spot. We describe this using two vectors: one for the force itself (with a magnitude measured in pounds) and one for the position the force is being applied (with a magnitude measured in feet)

the blue vectors are all the same but the red vectors are not

[midnight plank]

@grand dagger Has your question been resolved?

[grand dagger]

What’s the intuition behind taking the cross product

[grand dagger]

<@&286206848099549185>

[olive matrix]

remember how the dot product gave us more when the vectors were pointing in the same direction

the cross product gives us more when the vectors are more perpendiculr

[grand dagger]

But how is that resultant vector’s magnitude the torque

[olive matrix]

well have you ever used a wrench to tighten / loosen a bolt?

[grand dagger]

Of course

[olive matrix]

okay, and which one do you think would be easier -- holding the wrench right next to the bolt, or holding it far away near the tip?

torque is proportional to the distance away from the load

and you'll note that the operator is applying a force perpendicular to the displacement vector, or at least as close to perpendicular as possible

[grand dagger]

Right next to it if I understand you correctly

[olive matrix]

in that picture, would the operator have an easier time if they were holding the red part of the wrench?

[grand dagger]

Ohhh

[grand dagger]

Nope

[grand dagger]

When they say torque what really is torque to begin with

Like work is some energy carried to a certain distance

Torque is what

[olive matrix]

ummm well it has to do with angular momentum

but ultimately it's a twisting force

[grand dagger]

Wdym by that

It’s the force being done on that point ?

[olive matrix]

yes; you can think about it as a force * a distance

https://www.reddit.com/r/askscience/comments/kf92de/what_actually_is_a_torque/

[midnight plank]

@grand dagger Has your question been resolved?

[midnight plank]

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[supple sigil]

mvt

[twilit field]

As $1$ is asymptotic to $f(x)$. It follows $\forall \varepsilon>0 \exists N \in R$ such that $n>N \implies \abs{f(x)-1}< \varepsilon$. We'd then have for fixed epsilon $0≤ \frac{\abs{f(x)-1}}{x}≤\frac{\varepsilon}{x}$

[grand pond]

wai

[twilit field]

mhm

[twilit field]

limiting this as x-> infty would be wrong, I suppose

[nova yoke]

probably need to also choose N large enough that f'(x) is near c

for x > N

[twilit field]

I also suppose we run into issues with what does the quotient really mean now

[twilit field]

Could you elucidate on this a bit pleas

[twilit field]

oh, consider arbitrary intervals and then

okay, I see

[nova yoke]

yea i think you need to form an actual difference quotient and then estimate from there

something like $\displaystyle \frac{f(x) - f(N)}{x - N}$

[grand pond]

Bungo

[twilit field]

yea

mhm

I"ll work on this

[nova yoke]

i think i got it to work if my scribblings are right

[supple sigil]

If you pick an N such that 1 – ε < f(x) < 1 + ε for all x > N, then for any l > N consider f(l) and f(l+1) ...

[twilit field]

Got it

thanks!

[twilit field]

I think I got it too. We have $0<\frac{1}{x-N} <\frac{1}{x}$. Further we have $0< \frac{\abs{f(x)-f(N)}}{x-N}< \frac{f(x)-f(N)}{x}$

[grand pond]

wai

[twilit field]

.close

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[midnight plank]

Please don't occupy multiple help channels.

[lyric charm]

⁨<@&268886789983436800>⁩ scam

[midnight plank]

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[woeful turret]

i got

this

where r=k

[cedar mason]

holy shit

yoda is going places

[woeful turret]

🤔

[cedar mason]

genuinely question

nvm

im dumb

[woeful turret]

?!!

[woeful turret]

also the r+1 is outside the ln and rest is inside

[cedar mason]

i thought for some reason that they didnt have a function of x in the integral lol

[woeful turret]

oh

[cedar mason]

i really hope that that is $(\ln(r+1))^2$ cuz otherwise thats horrendous notation

[grand pond]

Περσυ

[woeful turret]

its ln((r+1)^2/(r(r+2)))

[cedar mason]

just put extra brackets if not

oh wow

[woeful turret]

into (r+1)

[cedar mason]

that def needs extra brackets

[woeful turret]

ye mb

when i write its only for me to understand

[cedar mason]

are we sure that the integral is right tho

im not checking

[woeful turret]

uh i can show my work one sec

but its pretty easy

[cedar mason]

yeah

im just lazy

[cedar mason]

alright great

uhhh tbh i cant think of a nice follow up

[woeful turret]

oh alr

ahh question bro

[twilit field]

oh, I asked this on MSE a while ago I think

lemme pull it up

https://math.stackexchange.com/questions/2291997/proving-that-ln-left-frac4950-right-sum98-k-1-intk1-k-frac

not me asking though

[slow thorn]

rule of thumb would be to use 1 < k+1 < n for inequality sums like this in general

okay wait lower limit doesnt seem to use that

oh well

guess from the options maybe

[woeful turret]

:(

guess in advanced

-1000 rank

[slow thorn]

thats not what i meant

[woeful turret]

o

[slow thorn]

i meant guessing the inequality

[woeful turret]

how do u guess the inequality

[woeful turret]

this is very tough for me to understand

[last slate]

Spend more time in a q -time

[slow thorn]

the MSE one seems to have ln(49/50) instead of 49/50 though 🤔

brb

[last slate]

Yh tf is bernoulli

[cedar mason]

i only know the DE

and the fluids

[woeful turret]

thats fine but how did they directly get this after the summation i dont understand does it telescope or smth

[slow thorn]

their expression doesnt seem to be right but yeah all three sums in the previous step seem to be written in telescoping form

to get this

[woeful turret]

oh alright alright

then i can probably manage it then

if it telescopes

very tough to see that for me 😔

tyty

[slow thorn]

i think its still kind of hard to compare logs though..

[last slate]

Hope shit like this does not come

🤞

[woeful turret]

it came already 👍

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[woeful turret]

i differentiated

and i got

[lyric charm]

hang on a minute

[verbal pumice]

lol

[woeful turret]

?

[lyric charm]

we're integrating a positive function over an interval that's always positively oriented (i.e. x tan^-1(x) ≥ 0 always), yes?

[woeful turret]

💀

oh...

so

the minimum value

is 0?

[lyric charm]

so it should be, yes

[woeful turret]

ahhh alr alr

im dumb

thanks

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[balmy quartz]

heyt

[balmy quartz]

i found the area of the whole sector: 12.5 * pi / 3

then the area of the triangle: [25 * sqrt(3)] / 4

area k = area c

then i found area k (or c): sector - triangle = (25pi/6) - ([25 * sqrt(3)] / 4)

nvm

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[hallow basin]

You asked this question before right?

Or am I being dejavu

[balmy quartz]

i dont think so

i see my mistake tho

[hallow basin]

you have to be kidding me

#help-25 message

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[leaden seal]

Can someone check this (part in red may be wrong)

The working is shown on the second picture

It's integration by substitution

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[fickle sierra]

Hello, could someone check if this proof looks good please?
⁨⁨```latex
\begin{Theorem}
Assume $a,b \in \bR$. If $a \in \bQ$ and $b \notin \bQ$, then $a + b \notin \bQ$.
\end{Theorem}

\begin{proof}
Assume for a contradiction that $a \in \bQ$, $b \notin \bQ$, but $a + b \in \bQ$.
Since $a$ and $a + b$ are rationals, we can express both as
$a = \frac{p}{q}$ and $a + b = \frac{m}{n}$
where $p,q,m$ and $n$ are integers.
Then,
\begin{align*}
a + b = \frac{p}{q} + b &= \frac{m}{n} \
b &= \frac{m}{n} - \frac{p}{q}
\end{align*}
Since $\frac{m}{n}$ and $\frac{p}{q}$ are rationals, so is $\frac{m}{n} - \frac{p}{q}$.
This is a contradiction, since we assumed that $b$ is irrational.
Therefore, if $a \in \bQ$ and $b \notin \bQ$, then $a + b \notin \bQ$.
\end{proof}

[low arrow]

hm....

this 'and' would be better as a 'but' to indicate the contradicting supposition.

[grand pond]

Mor Bras

[junior flower]

i think ‘and’ is fine

[low arrow]

I see. I apologize for the unnecessary correction if so.

I'm more used to indicating contradictions with 'but', but I guess there is a point to using 'and'.

I don't see anything algebraically wrong with the proof though you should get a second opinion.

[junior flower]

the equation typesetting is awkward but the idea is fine

[fickle sierra]

I see that 'but' and 'and' are equivalent in the sense of a logical 'and', however, 'but' implies that it is contrary to something said previously

[low arrow]

that's how I see it as well.

[fickle sierra]

How would you typeset it?

[low arrow]

hence why I use 'but'. your assumption is contrary to the theorem.

[junior flower]

i’m on phone so this will take a sec

if you want to keep it all together and not break it up, one option would be something like $$a + b = \frac{p}{q} + b = \frac{m}{n} \implies
b = \frac{m}{n} - \frac{p}{q}.$$ but aligning by $=$ when it’s not an equality chain that looks like
\begin{align*}
a &= b\
&= c\
&= d
\end{align*} is poor style

[grand pond]

generating function shill

[junior flower]

another option is to do something in the same spirit as what i wrote but connect the statements with ‘hence’ or ‘therefore’ or ‘which implies’ (or however you want to style it) instead of \implies

[fickle sierra]

⁨```latex
\begin{proof}
Assume for a contradiction that $a \in \bQ$, $b \notin \bQ$, and $a + b \in \bQ$.
Since $a$ and $a + b$ are rationals, we can express both as
$a = \frac{p}{q}$ and $a + b = \frac{m}{n}$
where $p,q,m$ and $n$ are integers.
Then, $a + b = \frac{p}{q} + b = \frac{m}{n}$,
which means that $b = \frac{m}{n} - \frac{p}{q}$.
Since $\frac{m}{n}$ and $\frac{p}{q}$ are rationals, so is $\frac{m}{n} - \frac{p}{q}$.
This is a contradiction, since we assumed that $b$ is irrational.
Therefore, if $a \in \bQ$ and $b \notin \bQ$, then $a + b \notin \bQ$.
\end{proof}

[grand pond]

Mor Bras

[low arrow]

I don't know if I prefer this over the first attempt, to be honest.

[junior flower]

that is totally fine but i would put $$a + b = \frac{p}{q} + b = \frac{m}{n}$$ and $$b = \frac{m}{n} - \frac{p}{q}$$ in double dollar signs

[grand pond]

generating function shill

[junior flower]

that will look prettier

[low arrow]

did I miss something?

[fickle sierra]

⁨```latex
\begin{proof}
Assume for a contradiction that $a \in \bQ$, $b \notin \bQ$, and $a + b \in \bQ$.
Since $a$ and $a + b$ are rationals, we can express both as
$a = \frac{p}{q}$ and $a + b = \frac{m}{n}$
where $p,q,m$ and $n$ are integers.
Then, $$a + b = \frac{p}{q} + b = \frac{m}{n},$$
which means that $$b = \frac{m}{n} - \frac{p}{q}.$$
Since $\frac{m}{n}$ and $\frac{p}{q}$ are rationals, so is $\frac{m}{n} - \frac{p}{q}$.
This is a contradiction, since we assumed that $b$ is irrational.
Therefore, if $a \in \bQ$ and $b \notin \bQ$, then $a + b \notin \bQ$.
\end{proof}

[grand pond]

Mor Bras

[low arrow]

much better!

[junior flower]

agreed

[low arrow]

though I am also a fan of sprinkling in line breaks sometimes, but minor nitpick. this is already good.

personally I would format it as:

⁨```tex
\begin{proof}
Assume for a contradiction that $a \in \bQ$, $b \notin \bQ$, and $a + b \in \bQ$. \
Since $a$ and $a + b$ are rationals, we can express both as
$a = \frac{p}{q}$ and $a + b = \frac{m}{n}$
where $p,q,m$ and $n$ are integers.
Then, $$a + b = \frac{p}{q} + b = \frac{m}{n},$$
which means that $$b = \frac{m}{n} - \frac{p}{q}.$$
Since $\frac{m}{n}$ and $\frac{p}{q}$ are rationals, so is $\frac{m}{n} - \frac{p}{q}$. \
This is a contradiction, since we assumed that $b$ is irrational. \
Therefore, if $a \in \bQ$ and $b \notin \bQ$, then $a + b \notin \bQ$.
\end{proof}

[grand pond]

Nicole

[junior flower]

i usually advise never using new lines 😭

outside of environments like align

[low arrow]

that is fair. I just do it every time the next sentence conveys a new idea.

[junior flower]

also more generally, if you have some equations you are using for the proof, i advise not just stacking them on top of each other with nothing between them. align environment aligned with \implies and \iff (whichever one represents what you’re claiming about the equations) can be used. or you can explain how they are related to each other with words

[fickle sierra]

Thank you all for your help and recommendations!

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[leaden seal]

how does knowing the unit circle help with this question

[main current]

The integral will involve trig

[midnight plank]

@leaden seal Has your question been resolved?

[leaden seal]

couldmyou show me here

[main current]

Nice alternative method! Seems like you don't need any help.

[halcyon forge]

Need help for 1

[junior flower]

your handwriting is mid. also this channel is occupied

[halcyon forge]

Thought the help was done in this one

[low arrow]

it's not.

[leaden seal]

i still wanna know how id use the unit circle for it

[sudden yacht]

Probably by substituting x = sec(t)

[leaden seal]

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[stiff bison]

<@&268886789983436800>

[rich yew]

😭

[stiff bison]

yet again

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[stiff bison]

thank you

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[proper swift]

I have am STRUGGLING on this, please help

[proper swift]

please ping me if you have an answer

[civic lynx]

are you able to get any of these blanks?

[sudden yacht]

!status

[midnight plank]

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

[proper swift]

its decay, asymptote is obvious, domain is all real #s and thats it

1

[civic lynx]

for double-checking purposes: horizontal asymptote at y=3

can you circle the part of the graph that is positive and the part that is negative?

[plucky lark]

dude this problem is pretty self explanatory?

[slate ferry]

irrelevant 😔.

[slate ferry]

@proper swift

[proper swift]

stop making fun of me

[proper swift]

how do i circle it

[sudden yacht]

Wdym?

[civic lynx]

are you able to annotate the screenshot and then send it again?

[plucky lark]

i wasn't trying to sound rude lol

mb

[proper swift]

ill try

[proper swift]

isint it just whats above and below zero

[sudden yacht]

Sure

Positive means y > 0
Negative means y < 0

[proper swift]

understood

[civic lynx]

well yes, but we need to find the intervals of x for which the graph is above and below. it's usually easier to see that when you consider "pieces" of the graph

but we can skip some steps here: where is the graph above the x-axis, in terms of x values?

[sudden yacht]

Perfect

[proper swift]

question not understood

[sudden yacht]

So for what values of x is your function f(x) positive (i.e. above x-axis)?

[slate ferry]

they want the interval form of the part of the graph thats above the x-axis.

[sudden yacht]

#help-49 message is this, just rephrased if you'd like@proper swift

[proper swift]

is a value like 0,1

or 1,0

[civic lynx]

for a point like (1,0), the x-value is 1

[proper swift]

ohhh

so 2,1

it positive

https://tenor.com/view/congrats-noice-noice-job-nice-job-good-job-gif-16436948

[sudden yacht]

A point can't be positive or negative

[proper swift]

what

why not

[sudden yacht]

Also do never drop the brackets!

(2, 1)

[proper swift]

ok

[civic lynx]

here's an x-interval, to give you an idea:

[grand pond]

ηασιβ ♥

[proper swift]

what is this tellin gme

[grand pond]

ηασιβ ♥

[proper swift]

what

why is there money signs

[civic lynx]

so i can write the math properly

you can just look at the text in the pictures :)

[proper swift]

ok

ok

so how is that helpin gme

hello

[slate ferry]

this wants your "fill in the blank" answers in the form of intervals. intervals are just math text that that represents values of x. liek for example, with x > 4 (a random representation that tells you that x HAS to be bigger than 4) we can rewrite it as x is a part of ]4 , +inf[

the ]4 , +infinity[ is what we call an interval

[proper swift]

i have no idea what you are saying

[slate ferry]

referring to what alberto said here,

[proper swift]

please just tell me how to find the answers

[slate ferry]

boy im trying 😭

[proper swift]

there is no increase?

correct

?

[slate ferry]

correct... for the given part of the graph

[proper swift]

lets goo