help-39

[pearl pond]

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[sacred hollow]

.reopen

[pearl pond]

✅ Original question: #help-39 message

[sacred hollow]

i dont get it

ping me

[dense jasper]

Algebra bashing may work, but chances are that it'll be very tedious. Note that
$$P^2 \begin{bmatrix} 1 \ -1 \end{bmatrix}=P \left(P \begin{bmatrix} 1 \ -1 \end{bmatrix} \right)=P \begin{bmatrix} -1 \ 2 \end{bmatrix}$$
And since $P$ represents a linear transformation, you can find $P(e_1)$ and $P(e_2)$ (and thus $P$) quite easily.

[jolly parrot]

Civil Service Pigeon

[sacred hollow]

A linear transformation?

[west sapphire]

maybe you call it a linear map

[sacred hollow]

Nope, I was told I was supposed to use characteristic equation to solve this

[dense jasper]

having the matrix would be helpful to obtain the characteristic equation of the matrix

[sacred hollow]

Yea

[dense jasper]

hence why I said what I said above

[sacred hollow]

I don't know what linear transformation is though

[west sapphire]

do you know that P(au + bv) = aPu + bPv, where a,b are scalars and u,v are vectors?

that's what linearity means

[sacred hollow]

I don't think it was taught? Maybe there is another way to solve this?

[sacred hollow]

No

It sounds like something above highschool (mine)

[west sapphire]

so linearity is above high school but eigenvalues aren't...

i'm not sure what you're supposed to use tbh

[sacred hollow]

What's eigenvalues?

[west sapphire]

the roots of the characteristic equation, what do you call them?

[sacred hollow]

Like I am somehow supposed to use (A-lambda I) = 0

[sacred hollow]

Just roots of characteristic equation

[west sapphire]

well those are the eigenvalues

[dense jasper]

😐

[sacred hollow]

Oh

[dense jasper]

jee moment

[sacred hollow]

Yup

[west sapphire]

what a crazy exam it must be 😭

[sacred hollow]

So like? How do I solve it?

[west sapphire]

you know where P sends (1, -1) and (-1, 2)

use that to find out where it sends (1,0) and (0,1)

[sacred hollow]

Sends?

[west sapphire]

[read those as column vectors btw]

[west sapphire]

maps

[sacred hollow]

Maps?

[west sapphire]

um

the 'output' when you feed those vectors as 'input'?

i have no idea what terminology you are using

[sacred hollow]

Multiplication of P and the matrix (1,-1) and multiplication of P^2 and (1,-1)?

[west sapphire]

yes

view P as a function

[sacred hollow]

Ok

[west sapphire]

you feed it vectors and it outputs other vectors

that's what i mean by "send" and "map"

[sacred hollow]

Okok

Like p= f(x) and f(x) into matrix (1,-1) = (-1,2)?

[west sapphire]

you can view matrix multiplication y = Px as y = P(x) where P is behaving like a particular kind of function

one that satisfies P(ax + bz) = aP(x) + bP(z)

we call that a linear function (or linear map, or linear transformation)

it has that property because matrix multiplication has that property

you can exploit that here to find P(1,0) and P(0,1), and those are the two columns of the matrix P

(again i'm not typesetting here but those are column vectors)

[sacred hollow]

one that satisfies P(ax + bz) = aP(x) + bP(z) -> I don't get this

[west sapphire]

to give a numerical example:

$$P\left(2\begin{pmatrix}1 \ -1\end{pmatrix} + 4\begin{pmatrix}-1 \ 2\end{pmatrix}\right) = 2 P \begin{pmatrix}1 \ -1\end{pmatrix} + 4 P \begin{pmatrix}-1 \ 2\end{pmatrix}$$

sigh

i hate typsetting matrices

hold on while i fix

[sacred hollow]

Sorry 😓

[jolly parrot]

Bungo

[west sapphire]

had to go into actual latex on my computer to troubleshoot and fix haha

[sacred hollow]

I find that you can do this to be very cool

[west sapphire]

yea that's how you make it readable, but it's prone to making typos 🤣

[sacred hollow]

Eitherways, I don't understand how to use this to solve it. TwT

[sacred hollow]

I c i c

[west sapphire]

well i'll get you started

you have two "input" vectors to work with

namely (1, -1) and (-1, 2)

how can we form (1,0) and (0,1) from them

well if we add them together, we get:

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

maybe you can work out how to get (1,0)

you're allowed to multiply by scalars and add

[sacred hollow]

I was lost from here

It was a chocolate problem eitherways, I will ask my teach Ig

[west sapphire]

you know what P "does" to those vectors, in other words you know the values of P(1, -1) and P(-1, 2)

your goal is to find P

which is the same as finding P(1,0) and P(0,1)

what's chocolate problem, like bonus credit?

[sacred hollow]

Kinda like that.

[west sapphire]

ok

[sacred hollow]

Not exactly credits

[west sapphire]

ah but like a challenge problem anyway

[sacred hollow]

Yea

[west sapphire]

fair fair

well the hints i gave are enough to get you started but if you haven't done this kind of problem before then I can see why it's hard to understand

[sacred hollow]

shld i strt another help channel after closing this to ask another question?

[west sapphire]

yea best to do a new channel for a new problem

[sacred hollow]

Normally they give like values of P TwT

[sacred hollow]

kk

[sacred hollow]

thank you

[west sapphire]

yw

good luck!

[sacred hollow]

.close

[pearl pond]

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[stiff shell]

    \begin{problem}{Suppose $\dim V \ge 2$ and $T \in \End(V)$ such that $\ker T^{\dim V - 2} \ne \ker T^{\dim V - 1}$. Show that $T$ has, at most, $2$ distinct eigenvalues.}
        Let $n = \dim V$. Since the kernels do not equal eachtoher, there exists some $v \in V$ such that $T^{n - 2}v \ne 0$ but $T^{n - 1}v = 0$. Then the power sequence $\{v, Tv, \dots, T^{n - 2}v\}$ is linearly independent. Define $W = \Span\{v, Tv, \dots, T^{n - 2}v\}$. Its dimension is $\dim W = n - 1$ by linear independence of the vectors in the spanning set. Furthermore, suppose $w \in W$ is defined by $w = a_0v + a_1 Tv + \dots + a_{n - 2}T^{n - 2}v$, then note
            $$Tw = a_0 Tv + a_1 T^2 v + \dots + a_{n - 2}T^{n - 1}v.$$
        All terms except the last are in $W$ since they're scalar multiples of elements in the spanning set, and the last term is $0$ by construction of $v$, so $Tw \in W$ and thus $W$ is $T$-invariant.

        \vspace{1em}

        Now we can define the induced map $\tilde{T}: V/W \mapsto V/W$ by $\tilde{T}(x + W) = Tx + W$. Also, note $\dim(V / W) = \dim(V) - \dim(W) = n - (n - 1) = 1$. Because this is an operator on a one-dimensional vector space, $\tilde{T} = \lambda I$ for some $\lambda \in \bF$. Thus $Tx + W = \lambda x + W$, and so $Tx - \lambda x \in W$. Let $w = Tx - \lambda x$ and apply $T^{n - 1}$ to both sides, yielding $0 = T^{n - 1}w = T^{n - 1}(Tx - \lambda x)$. This means $T^{n - 1}(T - \lambda I) = 0$ for the chosen $\lambda \in \bF$. Thus minimal polynomial divides $(z)^{n - 1}(z - \lambda)$, forcing only $0$ and $\lambda$ to be possible eigenvalues of $T$.
    \end{problem}

[jolly parrot]

Altanis

[stiff shell]

could someone verify my proof? note V/W is the quotient space

[pearl pond]

@stiff shell Has your question been resolved?

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[silent spade]

What's the best way to go about proving two partial orders are not isomorphic?

[silent spade]

I have the two partial orders
$$(L_2(\mathbb{N}), \leq_{lex})$$
and
$$(\mathbb{N}, \leq)$$

[jolly parrot]

jstN0body

[silent spade]

leq is just the usual ordering for natural numbers and leq_lex is lexicographic ordering for finite sequences of natural numbers

$$(0) \leq_{lex} (1) \leq_{lex} ... \leq_{lex} (0, 1) \leq_{lex} (0, 2)$$

basically just strings of numbers (or at least we can treat the sequences as strings) that should be sorted alphabetically using natural numbers as letters

L_2 refers to all finite sequences of length at most 2

[jolly parrot]

jstN0body

[silent spade]

I may have figured it out though it would be good to get a second look at my idea in case there are mistakes

assume that there is some function f : N -> L2(N) that models the isomorphism between L2(N) and N

$$\text{let}\ \sigma \in L_2(\mathbb{N})\ \text{with}\ \lg{\sigma} = 2$$

[jolly parrot]

jstN0body

[silent spade]

in this case lg(x) is the length of the finite sequence x

$$\text{let}\ n \in \mathbb{N}\ \text{s.t.}\ f(n) = \sigma$$

[jolly parrot]

jstN0body

[silent spade]

but f must preserve the usual order of natural numbers alongside the lexicographic order of L2(N)

so f(0) must be the empty sequence, f(1) must be (0), f(2) must be (1) etc.

so f(n) should be the singleton sequence (n - 1) which contradicts the fact that sigma has a length of 2

$$\therefore (L_2(\mathbb{N}), \leq_{lex})\ \text{is not isomorphic to}\ (\mathbb{N}, \leq)$$

[jolly parrot]

jstN0body

[west bluff]

I'd say it seems fine, however, you'd probably finish quicker just noting that the set of finite sequences of natural numbers is not well-ordered, while N is

This property is preserved by order isomorphisms

[silent spade]

why are the finite sequences not well-ordered? I actually noticed that in my textbook but I don't understand why it's not

[west bluff]

To see this, just note that 0<01<001<... Is an infinitely descending chain

[west bluff]

Has infinitely descending chains, which don't allow for well-ordering

If I remember correctly, you can define an order analogous to the lexicographical ordering which first compares lengths of sequences and THEN compares sequences lexicographically which IS a well-ordering

But since that's not the ordering you are considering, just noting one of the two is well-ordered and the other isn't should suffice

[silent spade]

ohh ok I see

thank you !

.close

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[silent spade]

.repoen

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[pearl pond]

✅ Original question: #help-39 message

[silent spade]

Actually it slipped my mind for a second but L2(N) only allows sequences of length up to and including 2 so my understanding is that there could not be an infinitely descending chain

in this case does it make more sense to proceed how I was before?

[west bluff]

If your sequences have length at most 2 and are compared lexicographically, I'm fairly sure that is well-ordered, so yeah

[silent spade]

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[green meteor]

For 4d, why reject -0.5

[iron basin]

the question probably mentioned a range for x like pi/2 < x < pi

this means x/2 also has some range and lies in a particular quadrant

[green meteor]

Ohh

The range is obtuse angle

Is there a way i can check without using arctan

And also

Instead of writing tanx=y

Can i simply use the quadratic formula on tanx = rathr than y=

[iron basin]

no i mean both your solutions are correct

but the question asks for the one within specified range

maybe you could show the original question

[iron basin]

its given that $90^{\circ} < x < 180^{\circ}$\
which means $45^{\circ} < x/2 < 90^{\circ}$

[iron basin]

x/2 would lie in the first quadrant

and from here you can eliminate the other value for tan x/2

[green meteor]

O

Another thing,

Can i divide both sides by a trig function for example, 2sin2x = cosx or should i put them on one side then factor

[thin mauve]

how does 13%=50%

[iron basin]

factoring is better yeah
pretty much the same logic as solving for x^2 = x
you can factor out terms, or divide both sides by x and consider for x = 0 later

[green meteor]

Oh ok

Thanks👍🙏

.close

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[thin mauve]

.reopen

[steep saddle]

if you need something just get your own channel

[thin mauve]

itll create too much channels

[plush bramble]

1 is not too many

[pearl pond]

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[zenith mural]

how would one do this?

i don't know where to begin, not even the formulas for length of coordinate

[solar ember]

Think of the line segment as the hypotenuse of two right-angled triangles sitting on the x-axis. Because the line is straight, these triangles are similar. This means the ratio of the side lengths (i.e. the segments of the line) must match the ratio of their vertical heights.

[zenith mural]

okok i got it

thanks

.close

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[shadow reef]

Im trying to use the shell method to find the volume, but i don't know how to find height and radius

[pearl pond]

@shadow reef Has your question been resolved?

[pearl pond]

@shadow reef Has your question been resolved?

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[pearl pond]

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[fading ledge]

Which of the following sets of vectors a=(iii a„)
in R» are subspaces o/R« ? (n ^ 3).
(/) all a such that a\ ^0;

[fading ledge]

How do I check it?

I meant quickly

[nocturne plover]

do you know when a subset is a subspace?

[random ermine]

@fading ledge

[worldly jewel]

abandoning their channel again

[random ermine]

his*

[fading ledge]

Yes

when alpha-beta belongs to subspace

And one more property which is related to scaler multiplication

[nocturne plover]

there's three properties

the zero vector has to be an element of the subset

the sum of any two elements of the subset has to be an element of the subset

[nocturne plover]

and any multiple of an element of the subset has to be an element of the subset

[fading ledge]

In my book two properties

Zero vector is included in first property

[nocturne plover]

or the second

but anyway

you can check if those properties hold for any of the cases

and if they do for a general element of the case then they do for the case

if you check the first one

i

then what happens if our $a_1 = 1$ and our constant $a = -1$

[jolly parrot]

Katharine

[nocturne plover]

the vector is an element of the subset

but is the multiple?

[pearl pond]

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[pearl pond]

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[pearl pond]

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[crisp hazel]

How did I know this was a translation

[nocturne grail]

hm?

[pearl pond]

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[crimson nebula]

What i understood:-

Since the sequence is decreasing , for any value of m , The set tm contains only a single value . That is the infimum.
While set um contains all the terms of the sequence.

Am i correct

[cinder flower]

“the set tm contains only a single value”

what do you mean by this?

[crimson nebula]

Since for each value of m , the set has infimum.

[glacial bluff]

are tm and um t_m and u_m respectively?

[crimson nebula]

Yes

I made a mistake

Sorry

tm and um are single elements

[cinder flower]

yes, for a fixed m, t_m and u_m are each just one number

[crimson nebula]

My mistake

[cinder flower]

are you still confused about anything?

[crimson nebula]

No

Thanks

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[crimson haven]

Is this right pls

[warped violet]

Yes

[crimson haven]

Erm how do I do it for the last k

[warped violet]

For the last k?

What is $\left(x^{\frac{1}{4}}\right)^4$

[crimson haven]

Yes

That

Oh

Ok

Ty

[jolly parrot]

USS-Enterprise

[warped violet]

What

no

I am asking how did you get that

[topaz terrace]

You gotta u sub, let x^1/4 = u and solve the quadratic

[crimson haven]

Trying to make x alone

[warped violet]

can we see the work

[crystal dew]

,rccw

[topaz terrace]

You can’t just raise both sides to the power of 4

What if one side is negative and the other isn’t

[crimson haven]

Oh

Thats how I was taught

[topaz terrace]

In this case if you haven’t learned about complex number yet there’s no solution

[topaz terrace]

You can keep doing that but maybe just check at the end if it makes sense

[crimson haven]

How pls

[topaz terrace]

Dyk functions and range

So for instance f(x)=x^0.25 has range y>=0

In words the fourth root of anything can’t be negative

[topaz terrace]

In your answer you found x=1, maybe just bring it back into your equation and check if 1^1/4 =-1

[crimson haven]

No

Is this correct pls

[topaz terrace]

Wdym no 😭

[crimson haven]

Like I haven’t learnt it

I didn’t mean to be rude

[topaz terrace]

Oh mbd 💀💀

[topaz terrace]

You have 16x^8 in the denominator right at the end

It got cut off

[crimson haven]

Yes

[topaz terrace]

It should be 8x^3 though in the numerator as well

[crimson haven]

Oh how

Pls

[topaz terrace]

The first line here you forgot the 2^2 on the first term

2^2 x^3/2 …

You took the square root from 2^4

[crimson haven]

Oh shi mb

my eyesit

Is this how this one’s done pls

[nocturne grail]

looks good to me

[crimson haven]

The book said no answer so Im scared

[nocturne grail]

no answer?

!xy

[pearl pond]

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

[crimson haven]

Also for the negative number that have powers how do you put then into the calculators

[nocturne grail]

parentheses

[crimson haven]

Do you put them in brackets

Oh ok

[nocturne grail]

yes

(-sadsja)^2 is gonna raise the whole -sadsja to the power of 2

that is lowkey hte point of a bracket.

[crimson haven]

Oh ok ok

Bc of the negative I front of the whole bracket I got confused

[pearl pond]

@crimson haven Has your question been resolved?

[pearl pond]

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[tacit bough]

Where am I going wrong here?

[spare lark]

First to second line

-3(x-1) = -3x + 3 not -3x -1

Else it seems good

[pearl pond]

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[late girder]

sorry but can i ask my question in this help? bot is down so there is no help available

[frank violet]

yes

[late girder]

what will be the maximum number of cylinder possible that could be fit inside the cuboidal box.

[unkempt yacht]

how many cans can be packed horizontally to span a length of 2p? @late girder

not asking you

and !nosols

!nosols

[tepid forge]

sorry

[unkempt yacht]

oh right, bot died

[tepid forge]

mb, will try not to directly give soln

[prisma vortex]

is it possible to do this without assumying it is symmetrical?

[jade rune]

32,Only two layers can be packed, so we can draw a top view. The container is very regular.

[late girder]

but i was thinking of any creative answer like if we can place in different order

[late girder]

it's a cube

[tepid forge]

u can just divide total volume of cube / volume of one cylinder

if u can cut the cylinder and fit it

[prisma vortex]

can some1 quickly check this pls

[clear shore]

assuming what

[prisma vortex]

the pattern is symmetrical

fully

[clear shore]

probably, how did you find the equations assuming it's symmetrical

[prisma vortex]

x^2-6x+13 =(x-3)^2+4 so vertex is (3,4) which has to be A. do the same for y=-x^2-6x-5 which is -(x+3)^2+4 so vertex is (-3,4) but to find the vertex of F or H how would we do it?

[clear shore]

Okay ig this's kinda ass but I think we have to assume that the distance between A's vertex and x-axis is equal to the distance that of D

[prisma vortex]

alr thats what i thought and i spent way too much time thinking 😭

[late girder]

i was thinking about any creative answer something like this but in 3d like we can tilt cans to fit more cans in container

[crimson haven]

The book says the answer is 1/2

[solar ember]

You added instead of multiplying

you should have done 1+3*2

[crimson haven]

Oh

Omds tysm

I was trying to get a channel bc none of them were pinging

[solar ember]

Yeah, the bot is down

[crimson haven]

Oh ok

Could y help me

[solar ember]

Not right now, i'm at work lol

[crimson haven]

H ok

Oh

Ok

[solar ember]

maybe someone else can

I can only occasionally drop in and chat something

[jade rune]

I have thought it, but container is very regular.

[solar ember]

@crimson haven I am free to help now if you still have doubts

[crimson haven]

oh it’s ok now, tysm for remembering me tho

[pearl pond]

@stoic imp Has your question been resolved?

[verbal whale]

<@&268886789983436800>

[solar ember]

No worries, i'm glad to help. Doing stuff like this also helps me de-stress

[ivory basin]

Weird channel, is bot up or nah

.close

.reopen

[pearl pond]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[sharp smelt]

Looks like it is now!

atleast somewhat

[pearl pond]

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[jolly parrot]

Renato

[stoic imp]

I need help because I struggle with strong induction

and I have a feeling I will need strong induction for this one

[toxic lichen]

formally yes

if you are ok with a lighter version of it that i call "2-steps-back induction" then you can also use that

but really it boils down to just... using the recurrence relation they give you lmfao

[stoic imp]

can we just use strong induction

I promise is not that obscure

is very similar to induction

[toxic lichen]

i know what strong induction is, renato.

how much do you have written down rn

[stoic imp]

nthg

because I don't know the inductive hypothesis for strong induction

[toxic lichen]

so really you dont know what strong induction is.

[stoic imp]

yeah exactly

[toxic lichen]

the inductive step for strong induction is

(∀k ∈ N)(k < n => P(k)) => P(n)

that left bit is your IH

can also be rephrased as P(1) & P(2) & ... & P(n-1)

[stoic imp]

so what do I need to prove

p(1) p(2) p(k)

[toxic lichen]

have you not overcome your bracket allergy yet

this is like the 3rd time you are mercilessly dropping every single () in sight

you need to prove that if your statement holds for all natural numbers <n then it holds for n itself as well

do you know "ordinary" induction? yes/no

[stoic imp]

yes, but strong induction is harder

[stoic imp]

what do I need to prove then p(1) p(2) p(k)

[toxic lichen]

no

p(1), p(2), ..., p(n-1) are your ASSUMPTIONS

they are the INDUCTIVE HYPOTHESIS, the very thing you ASKED FOR and which i GAVE

your GOAL is p(n)

[stoic imp]

I see

IH, there exists k < n such that P(k) => P(n)

[grave mirage]

They finally moved past the set theory pnc questions

[stoic imp]

Inductive Step, let k = 1

[toxic lichen]

incorrect

bro you either didnt read what i said at all or you did but then it went flying right out 😭

[toxic lichen]

right here

∀

this symbol. it means FOR ALL.

[stoic imp]

ok

[toxic lichen]

😭😭😭😭😭😭😭😭

[stoic imp]

IH, for all k <n, P(k) => P(n)

Inductive Step, take k = 1, then P(1) = 3^2 | a1

[stoic imp]

what is confusing in contrast with regular induction is that idk if i need to prove p(1)

[toxic lichen]

it doesnt HURT to prove p(1).

[stoic imp]

p(1), p(2), p(k=n)

[toxic lichen]

i think i have to go

[stoic imp]

?

[grave mirage]

Do you know why induction works renato?

[pearl pond]

@stoic imp Has your question been resolved?

[stoic imp]

i wouldn't say i know induction that well

[worldly jewel]

Yeah I wouldn't say that either

[stoic imp]

are you coming here just to roast me or are you planning to help

[worldly jewel]

I think you should go revise what induction is

You couldn't even state the definitions

You're going to have a hard time trying to apply induction if you don't even know what induction is

[stoic imp]

i see thank you for the help

[grave mirage]

Well I think if you know why it works, you'll be able to solve these problems pretty easily

Atleast you'd know what to do

[stoic imp]

why do you guys have to follow me around

im just asking my homework

[grave mirage]

I helped you gng 😭

I'm sorry if what I said felt mean

Never meant to discourage you or smth

[stoic imp]

did you

[grave mirage]

Ig ann did

[stoic imp]

then why you said you did

[grave mirage]

I helped

Tried to ig

[frosty sentinel]

if you're having trouble because you think strong induction requires ALL OF p(1), p(2), ..., p(n-1) to prove p(n)

for this question you just need p(n-1) and p(n-2) to prove p(n)

[stoic imp]

why just n -2 and n-1

an = 3an-1 + 5(an-2)^2

an+1 = 3an + 5(an-1)^2

[stoic imp]

if we prove p1 and p2, then we can take k+1 = n-1 where 2 < k < n and k = n -2

well what you are saying is that we dont need to prove p1 p2

but i still think it kinda is needed

[frosty sentinel]

(sigh) let me clarify

in any induction there are 3 steps

1. base case(s)
2. induction step using induction hypothesis
3. wrap up

for 1) base case(s) you still need to show that p(1) and p(2) are true, which is quite obvious

but im saying that for 2) induction step, you just need to assume p(n-2) and p(n-1) true to show that p(n) true

[stoic imp]

ok so how to do step 2

[frosty sentinel]

you first assume that p(n-1) and p(n-2) are true

then you need to show that as a result,

p(n) becomes true

[west bluff]

Wdym, set theory questions are the best ones

[frosty sentinel]

dont sub in a value of n, just generalise n from the getgo by treating n as a variable that represents any integer >= 3 (since we have p(n-2) starting from p1 )

[stoic imp]

base case
k = 1
3^2 | a1
3^2 | 3^2
k = 2
3^3 | 7 x 3^3
7 * 3^3 = 3^3 * c where c = 7

[stoic imp]

p(n-1): an = 3an-1 + 5(an-2)^2

[stoic imp]

p(n-2): an+1 = 3an + 5(an-1)^2

[frosty sentinel]

p(n) means 3^(n+1) divides an

[stoic imp]

yeah but do you mind i use 2 < k < n instead of reusing n

[stoic imp]

ok, sry

[frosty sentinel]

because we are trying to prove the assertion that

3^(n+1) divides an

so we want to make the statement p(n) the same as the assertion for induction

[stoic imp]

ok

[grave mirage]

Relation and functions part

[frosty sentinel]

ok can

now we are in the induction step

our goal is to show p(n) is true, meaning

3^(n+1) divides an

we know that an = 3an-1 + 5an-2^2

so for what values of k do you think we need to assume p(k)?

[stoic imp]

when k = n - 1
p(k) : 3^(k+2) | a_{k+1}
when k = n - 2
p(k) : 3^(k+3) | a_{k+2}

[stoic imp]

p(k) is 3^(k+1) | a_k
and a_k = 3a_{k-1} + 5a_{k-2}^2
so we need to know k-1 and k-2

[frosty sentinel]

but from p(k-1) and p(k-2),

you know that a_{k-1} has factor 3^k, and a{k-2} has factor 3^(k-1)

now you just need to use that information and the equation for a_{k} to show that a_k has a factor of 3^(k+1)

here's a hint, if a_{k-1} has factor 3^k, what factor does 3a_{k-1} have?

[stoic imp]

ah right we are assuming that p(n-1) and p(n-2) is true

[stoic imp]

3a{k-1) has a factor 3^(k+1)

what about 5a_{k-2}^2

[frosty sentinel]

that's your job as someone who's trying to write a proof

[stoic imp]

well if 3 | a_{k-2}

[frosty sentinel]

3^(k-1) | a_{k-2}

[stoic imp]

well

a_{k-2} has a factor 3^(k-1)

and 2 < k < n

thus, a_{k-2} has a factor at the very least 3^2

because k is integer

[frosty sentinel]

well, we're dealing with 5**(a_{k-2} )^2**

[stoic imp]

yes but

if we manage to show that a_{k-2} is divisible by 3^2

when 2< k < n

or is there any better idea

[frosty sentinel]

that is technically true

but since we want to prove in terms of k

it will be more useful to write out the largest power of 3 in terms of k

[stoic imp]

care to expand on that idea?

largest factor of 3 in terms of k?

[frosty sentinel]

oops i meant power mbmb

like the largest power of 3 that you can say for sure is a factor of a_{k-2}

and as a result the largest power of 3 that you can say for sure is a factor of 5 (a_{k-2} ) ^2

[stoic imp]

which is?

[frosty sentinel]

have you forgotten what you assumed?

we assumed p(k-2) way from the beginning

[stoic imp]

we assume 3^(k-1) | a_{k-2}

[frosty sentinel]

yes

so instead of starting from 3^2 | a_{k-2}

[frosty sentinel]

its more useful to just start from this

so what can we deduce about

5 (a_{k-2} )^2

[stoic imp]

we assume 3^(k-1) | a_{k-2}
which means there exists a C in Z such that
C . a_{k-2} = 3^(k-1)
then we square both sides
(C . a_{k-2})^2 = 3^(2k-2)
then we multiply by 5
5(C . a_{k-2})^2 = 5.3^(2k-2)

[frosty sentinel]

mmm mmm

so what is the highest power of 3 that we conclude is a factor of 5(a_{k-2})^2?

[stoic imp]

3^(2k-2)

[frosty sentinel]

and what do we need to show about 5(a_{k-2})^2 to finish the proof?

[stoic imp]

that 2 < 2k - 2 < n

?

[frosty sentinel]

here's the things we did so far:

want to show p(k), i.e.
a_k has factor 3^(k+1)

we assumed p(k-1), p(k-2)

we know
a_k = 3 a_k-1 + 5 (a_k-2)^2

we already showed
3 a_k-1 has factor 3^(k+1)

so what missing link do we need to show about 5(a_k-2)^2?

[stoic imp]

we need to show 5(ak-2) has a factor 3^k

?

[frosty sentinel]

are you sure its 3^k?

[stoic imp]

3^(k+1)

[frosty sentinel]

yes

[stoic imp]

well we know that 5(a_k-2)^2 has a factor 3^(2k-2)

right?

[frosty sentinel]

yes

on the right track

[stoic imp]

how does 3^(2k-2) | 5(a_{k-2})^2 => 3^(k+1) | 5(a_{k-2})^2

[frosty sentinel]

what did we define k as?

[stoic imp]

3^(2k-2) = 3^(-2(k+1))

[stoic imp]

an integer between (2,n)

[stoic imp]

i feel like I am missing something

yet i also think we are very close to finishing the proof

[frosty sentinel]

you have all the info you need:

[stoic imp]

how does 3^(2k-2) | 5(a_{k-2})^2 => 3^(k+1) | 5(a_{k-2})^2?

[frosty sentinel]

we know:

5ak-2^2 has factor 3^(2k-2)

integer k, 2 < k < n

want to show that:

5ak-2^2 has factor 3^(k+1)

[stoic imp]

yeah

3^(2k-2) = 3^(2(k+1) - 4) = [3^(k+1)]^2 * 3^(-4)

idk

[frosty sentinel]

ok but since you're stuck on this

lets try to play with small values of k

and see if you can spot a pattern

k = 3:

... has factor 3^(2*3-2) = 3^4

want to show ... has factir 3^(3+1) = 3^4

k = 4:

... has factor 3^(2*4-2) = 3^6

want to show ... has factor 3^(4+1) = 3^5

k = 5:

... has factor 3^(2*5-2) = 3^8

want to show ... has factoe 3^(5+1) = 3^6

what do you notice about the known powers 3^4, 3^6, 3^8, ... compared to the wanted powers 3^4, 3^5, 3^6, ...?

[stoic imp]

well i fixed what i was doing earlier 3^(2k-2) = 3^(2(k+1) - 4) = [3^(k+1)]^2 * 3^(-4)

is this good enough?

[frosty sentinel]

having a negative exponent 3^(-4) wont help much

maybe try to prove it for smaller values of k first

[frosty sentinel]

i have kindly typed out the thing we're trying to prove for k=3, k=4, k=5

how would you prove it for k=3, k=4, k=5?

can you spot a pattern on how you proved it?

[stoic imp]

well

maybe another induction

but thats a bad way thinking about it

[frosty sentinel]

no need for that

[hazy bear]

this is good
So it has factor of
3^(k+1) * [3^(k+1) * 3^(-4)]
= 3^(k+1) * 3^(k-3)

If 3^(k-3) is a positive integer then we can say it has factor 3^(k+1).
Given k>=3, we have 3^(k-3) positive integer.

I haven't read above chats but just wanted to point this out, next let Tom see if this is right or not.

[stoic imp]

well we can argue that 3^(2k-2) is strictly increasing

[frosty sentinel]

this works

[stoic imp]

ok so

we finished low key

if both terms in the sum have a factor 3^(k+1)

then the whole sum has a factor 3^(k+1)

[frosty sentinel]

yeah

so an has factor 3^(k+1) which is literally the end goal of p(n)

THE STRUCTURE OF THE PROOF

base cases: p(1), p(2) true

induction step:
for integer 2<k,
p(k-1) and p(k-2) true
=>
p(k) true

wrap up:
by semi-strong induction,

p(1) and p(2) => p(3)
p(2) and p(3) => p(4)
p(3) and p(4) => p(5)
...
=> p(k) true for all integer k = 1,2,3,4,5,...

[frosty sentinel]

man i would have loved for renato to discover for theyself that the powers of the known factors are higher than the powers of the factors we want to find as k>=3 instead of just reading spoilers but oh well

[frosty sentinel]

are you now able to organise / write the proof by induction?

if so you can close this channel

[stoic imp]

i wasnt that far away from doing it but was really stuck for quite some time though

[frosty sentinel]

here's a tip for next time

[stoic imp]

i had this question on my exam and got stuck in that very spot aswell

[frosty sentinel]

if the theorem isnt universally true for all values of the variable k or n or i,

then the trick is probably because of the range of values of of k or n or i or whatever variable you use

so try to use that definition

e.g. integer 2<k means k>=3

if you need a direction, then try completing the proof with small values of k to spot a pattern

[stoic imp]

yeah

[frosty sentinel]

very often in inductions,

they only work for a specific range of k or n or i

so its important to take note of that when doing inductions

[stoic imp]

ok

[frosty sentinel]

if you didnt use that fact but are stuck near the end of the proof, then you might probably need that

thats all

[stoic imp]

ok

thanks for the help guys, this was quite a long problem

maybe it was long because of my lack of knowledge

[frosty sentinel]

it probably felt long because you arent used to the rigour of number theory and factors

but then again

number theory and factors are just possible contexts to solve induction

you could end up with other contexts like geometry, algebra, e.t.c.

[stoic imp]

well it doesnt help that this is the very first class I am taking in my bachelor

[frosty sentinel]

so what you need to do is 2 things

[stoic imp]

number theory itself is very hard

[frosty sentinel]

1, work out your muscles on algebra / notation / language of proofs so that you can handle the rigour of different contexts more easily

2. understand and familiarise yourself with the 3 steps of induction

(base case, induction step, wrapup)

so that you can just worry about the context when doing induction

[stoic imp]

yeah Im working on that

but I'm just bad

[frosty sentinel]

once you truly know how induction works

you'll find the techniwue of induction much more intuitive

so keep it up

[stoic imp]

.close

[pearl pond]

Channel closed

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[pearl pond]

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• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
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[last hollow]

my problem: (x+y)(x+2y)^4=y+2
a=x+y
b=x+2y
=>ab^4=b-a+2
a(b^4+1)=b+2
so we know
b^4 % b+2==0
help dont know what next

[last hollow]

also know that x and y is intergers

<@&286206848099549185> is my problem written wrong?

[rigid lava]

chill out bro i am thinking

[maiden forum]

what grade math is this?

/year

[last hollow]

9

[rigid lava]

year 9 math??

[last hollow]

wait vietnamese math is that hard compare to other???

[rigid lava]

obviously

[maiden forum]

uh i'm g7 but i'll try to help

[rigid lava]

I think you should let x+2y=a

then x+y=a-y

[maiden forum]

man i am not in high school goodbye

[last hollow]

hmmm

[rigid lava]

it will help simplify LHS little bit

[last hollow]

let me try out

[clear shore]

not really? It's just kinda weird/different from others

[rigid lava]

you can obtain this

[clear shore]

what exactly is our task here

[last hollow]

get x and y

[rigid lava]

then just write 2 = 1x1 , -1x-2,2x1,1x2

[last hollow]

know that

[rigid lava]

yeah so do

lemme know if you get answer

[last hollow]

b^4 % b+2==0
oops mb this is wrong
(b^^4 +1)%(b+2)==0

up one is false

[main hinge]

What's the question

[last hollow]

sorry he nailed it

[last hollow]

find x and y

[last hollow]

all good now thanks man

[main hinge]

You got it alright then

[last hollow]

problem finish I just .close?

[main hinge]

Yes

[last hollow]

forever remember you man @rigid lava

.close

[pearl pond]

Channel closed

Closed by @last hollow

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[pearl pond]

Available help channel!

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Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
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• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[fathom sequoia]

don't need help, just need verification that this has
x = 11/2 as the answer

[somber adder]

type it into your calculator

[iron basin]

i guess bro

[somber adder]

see if both sides have the same value when x = 11/2

[heavy onyx]

,calc 5^(3(5.5)-2)

[jolly parrot]

Result:

1.3647875839232e+10

[fresh solar]

Yes it is

[heavy onyx]

,calc 5^(5.5+9)

[jolly parrot]

Result:

1.3647875839232e+10

[fathom sequoia]

answer key said 1/2 but i cant like verify it atm

no calc

[fathom sequoia]

thx

.close

[pearl pond]

Channel closed

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[pearl pond]

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• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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[blissful salmon]

Let P(n) be a nonconstant polynomial with rational coefficients. Is it possible to prove that there does not exist P(n) such that P(n) only generates prime numbers for natural n?

[toxic lichen]

you say function, do you mean polynomial?

[blissful salmon]

Oh yes, I meant polynomial

[glass meadow]

P(n) = 2 only generates primes

[blissful salmon]

Yes, but I was thinking about nonconstant polynomials

[glass meadow]

So, your question really is:

Is this statement provable: "there does not exist a non-constant polynomial P(n) with rational coefficients such that, for all natural number n, P(n) is prime" ?

[blissful salmon]

Yes, that’s my question

[glass meadow]

Well, if the polynomial has no constant term, then P(n) is divisible by n, so that doesn't work

If it has a constant term c, then P(c) is divisible by c, right? Now I suppose P(c) could equal c and be prime, but then I'm sure there's some other value c' such that P(c') would be divisible by c and not equal c

[tropic saddle]

rational coefficients

[glass meadow]

Oh yeah, I was thinking of integer coefficients, though I would think it still applies somehow

The no-constant-term case is still easy if I'm not mistaken: just let k equal the product of all denominators of the coefficients (or I guess the lcm, doesn't matter), then P(n*k) is divisible by n, right?

I can't think straight so I won't try to continue by my intuition says you can also use that k for the other case

[toxic lichen]

yeah i think you can

P(0) = p and consider P(kp) with k being the product of all coefficient denoms

[glass meadow]

Right so if you prove the statement for integer coefficients you also prove it for rational coefficients by scaling the input

[rancid depot]

dont forget about the constant term

[glass meadow]

<@&268886789983436800>

[pearl pond]

@blissful salmon Has your question been resolved?

[blissful salmon]

Maybe a slightly easier question. Does there exist a non constant polynomial P(n) with integer coefficients such that for all natural n, P(n) is prime?

[blissful salmon]

I already know that the constant term must be +1 or -1 but don’t know how to proceed further

[glass meadow]

What?

How is the constant term +1 or -1

[blissful salmon]

Oh wait, I’m not sure but it must be + or - 1 or a prime number

[glass meadow]

If you count 0 as a natural then P(0) must be prime...

[blissful salmon]

And I’m pretty sure it cannot be a prime number because say we called the constant term c which is a prime number. Then P(k * c) is divisible by c for integer k. That means P(k * c) = c. Now since the polynomial has a finite degree, if we plot P(n) as a function it must have a finite number of turning points. However there are infinitely many k so there exists no polynomial with constant term being prime I think

So now we are dealt with the constant term being +1 or -1

.close

[pearl pond]

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[jolly parrot]

Renato

[pearl pond]

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

[stoic imp]

1

[unique viper]

Do you know how induction works?

[stoic imp]

yes a little bit

[unique viper]

So if a statement about the integers is:
correct at n=1,
correctness at n implies correctness at n+1

then it is true for all integers>0, right?

[stoic imp]

ok what's your point

[unique viper]

It's asking you to show that the conditions are met

So it wants you to show:
1 >= 2/2

and that, assuming the statement is true at some integer, one can show it to be true at the next integer.

Do you think you know where to go from here?

[stoic imp]

no

[unique viper]

Is it true that:
1! >= 2^1/(1+1) ?

[stoic imp]

yes we already proved p(1)

that's the base case

[unique viper]

Have you written your inductive assumption?

[stoic imp]

p(n) => p(n+1)

where p(n) is $\sum_{k=1}^n k! \geq \frac{2^n}{n+1}$

[jolly parrot]

Renato

[stoic imp]

using p(n) we need to arrive to p(n+1)

[unique viper]

What can we say about $$\sum_{k=1}^{n+1}k!$$ in terms of our inductive assumption?

[jolly parrot]

Varixiuqlhfbgraijbzjnqghppxnqmvw

[brave sluice]

yeah replace n with n+1 and try to adjust the form so that it looks like p(n)

[stoic imp]

,, \sum_{k=1}^{n+1} k ! = (n+1)! + \sum_{k=1}^n k!

[jolly parrot]

Renato

[brave sluice]

yes, good

[stoic imp]

how to write the proof formally

with words and stuff?

suppose p(n) is true. thus \ $\sum_{k=1}^n k! \geq \frac{2^n}{n+1} \ (n+1)! + \sum_{k=1}^n k! \geq (n+1)! + \frac{2^n}{n+1} \ \sum_{k=1}^{n+1} k ! \geq (n+1)! + \frac{2^n}{n+1}$

[frosty sentinel]

formatting is ok so far

remember that your goal is to show p(n+1) is true, i.e.

SUM k! >= 2^(n+1) / [(n+1) +1]

[jolly parrot]

Renato

[stoic imp]

help

[brave sluice]

,align
\sum_{k=1}^{n+1} k ! &= (n+1)! + \sum_{k=1}^n k!\
&\geq (n+1)! + \frac{2^n}{n+1}

try something like this

[stoic imp]

now what

[jolly parrot]

Axe

[stoic imp]

yeah in my latex I just started from p(n) and added something to both sides

[stoic imp]

but got here aswell

[brave sluice]

oh i see, yeah what you wrote is fine too

[frosty sentinel]

mmm i think i know why you might be stuck

firstly it would be useful to try to fish out the wanted denominator

secondly, since equality is already met by p(1), you dont necessarily need ≥ for p(2) and above, you can use > for p(2) and above

[stoic imp]

care to elaborate

[frosty sentinel]

for p(1), SUMk! ≥ 2^1/(1+1) is technically true because

SUM k! = 2^1 / (1+1)

but notice that as n increases, for p(n),

SUM k! grows faster than 2^n/(n+1)

so for n>=2, in reality a stricter inequality is SUMk! > 2^n/(n+1)

but that means SUMk! ≥ 2^n/(n+1) is still technically true

so combine results for n=1 and n>=2

we can still get
SUMk! ≥ 2^n/(n+1)

[frosty sentinel]

so my point is that

in your induction step, you dont need to wreck your brain trying to ensure the equality in the ≥ holds when doing LHS ≥ RHS

even if you know that LHS > RHS, that is technically LHS ≥ RHS, so it is good enough to complete the proof

[brave sluice]

$(n+1)!+\frac{2^n}{n+1}=(n+1)!+\frac{n+2}{2(n+1)}\cdot\frac{2^{n+1}}{n+2}$

[jolly parrot]

Axe

[brave sluice]

this should help

hint: || each summand on the RHS is greater than or equal to half the desired amount ||

[pearl pond]

@stoic imp Has your question been resolved?

[pearl pond]

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[pearl pond]

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[royal galleon]

.close

[pearl pond]

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[robust lynx]

-# this is for you uss enterprise

[pearl pond]

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[blissful narwhal]

I have a really complex lattice I have to code in c, but I don't even have any experience with doing either. As a starting point what would be the most basic lattice I could code so I know how to scale it for what I'm copying?

[blissful narwhal]

If for whatever reason you want to know what I'm looking at, specifically this one from section 3.2 of this paper: https://link.springer.com/content/pdf/10.1007/978-3-662-48797-6_7.pdf for anyone that's allergic to links like me, this is: Computing Individual Discrete Logarithms Faster in GF(p^n) with the NFS-DL Algorithm by Aurore Guillevic

[pearl pond]

@blissful narwhal Has your question been resolved?

[pearl pond]

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[blissful narwhal]

.reopen

[pearl pond]

✅ Original question: #help-39 message

[ancient slate]

https://discord.gg/eF3Wjsd

[unborn abyss]

do you mean actual C or C++?

[blissful narwhal]

Actual C

[unborn abyss]

oh boy
maybe start by looking at GSL

but reading that paper it kinda looks like less numbers and more abstract algebra which is going to be rough

[blissful narwhal]

Yeah pretty much all of the values are representative and my eyes are seeing it but not really comprehending it. I'll try looking at GSL like you recommended

[blissful narwhal]

Thanks for this btw. I'll ask here as well

[ancient slate]

Np

There are more servers in #old-network , check that

[pearl pond]

@blissful narwhal Has your question been resolved?

[pearl pond]

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[terse patio]

Struggling to remember how exactly I answered these linear relation questions. What formula do I gotta use?

[brazen hare]

can u tell me the equation of a line

[torn flicker]

y = mx + c right?

[terse patio]

Y=2x-3?

[torn flicker]

yeah

so gradient is 2 right?

[brazen hare]

the general one

[brazen hare]

this one the person already said

so whats the m in this and c

[frail igloo]

in the form y=mx+c, c is the y intercept. to find the x intercept put y as 0 and solve.

the gradient tells us that one point is 2 blocks away from another

[brazen hare]

let gng derive that himself 😭🙏

giving him direct formula aint gonna help

[terse patio]

How exactly do I just make the y into 0?

Don’t I got to apply to both sides?

[frail igloo]

just say its 0

0=2x-3

3=2x

3/2=x

thus, the x intercept is at 3/2

which is..

1.5

and from ur graph, we confirm this

[terse patio]

Does this got to do with rise over run tho?

[frail igloo]

focus first

we know the gradient

2

we know the x intercept

3/2

now what should be the y intercept

.?

[terse patio]

-3?

[frail igloo]

correct

[terse patio]

Nice

So

Hold on

[frail igloo]

and look, from ur graph, the intercept is at -3

[terse patio]

What about B

[frail igloo]

okay

we have 3x+4y=12

isolate y

how can we do that?

[terse patio]

4y=12-3x

[frail igloo]

completely isolate it

[terse patio]

Divide everything by 4?

[frail igloo]

yea

[terse patio]

Y=3-0.75x

[frail igloo]

just write it in fraction

y=3-3/4x

right

?

[terse patio]

Yeah

[frail igloo]

so now

can we say that 2+3 is the same as 3+2

[terse patio]

Yeah

[frail igloo]

can we say that 2-3 is the same as -3+2

[terse patio]

Yeah

[frail igloo]

so we apply this here too

y=3-0.75x is the same as y=-0.75x+3

what does this tell us?

y=-0.75x+3

[terse patio]

Hold on do we have to always put numbers with exponents at the left?

O wait

[frail igloo]

no

[terse patio]

We are replicating a formula

[frail igloo]

theres no exponents here

YES

y=mx+c

[frail igloo]

we've replicated it

[terse patio]

This one right here

[frail igloo]

so now what does -0.75 tell us

what is -0.75 contextually?

[terse patio]

The X

[frail igloo]

no

[torn flicker]

nope

[frail igloo]

y=mx+c
y=-0.75x+3

[terse patio]

Does the m stand for something

[frail igloo]

the m means gradient

[torn flicker]

duh

[frail igloo]

the c means y intercept

remember this

[terse patio]

What’s that💀

[frail igloo]

do u know slope?

[terse patio]

Rise over run?

[frail igloo]

yes

[terse patio]

Ahh

[frail igloo]

another way to say rise over run is slope or gradient

its faster

[terse patio]

Is that why you turned it into a fraction?

[frail igloo]

so the m means rise over run

[terse patio]

Or smthing like that

[frail igloo]

yeah, to make it easier, but to make u understand i used decimals to communicate

anyways

so what is -0.75

[terse patio]

The slope

[frail igloo]

good

now what is +3

[terse patio]

Y intercept

[frail igloo]

yes

nice

so to find the x intercept we..

put y as 0

0=-0.75x+3

[terse patio]

K

[frail igloo]

-3=-0.75x

[torn flicker]

x = 4

[frail igloo]

-3/-0.75=x

so

we get x as 4

x intercept btw

[terse patio]

Hold on

[frail igloo]

so

x intercept is 4
y intercept is 3
slope or rise over run is -3/4

[terse patio]

Did you just straight up removed the number from the x?

[frail igloo]

confirmed from ur graph

[terse patio]

And put it on the other side

[frail igloo]

huh

no i solved for x

[frail igloo]

solve for x here @terse patio

[terse patio]

It went from -3=-0.75x to this:

[frail igloo]

what do u infer the / could mean here?

[frail igloo]

here

[terse patio]

I guessing that to separate the number it must also be divided right and the x remains as x

[frail igloo]

correct.

im basically saying

-3 is divided by -0.75

which gives us the value of x as 4

-3/-0.75= -3 divided by -0.75

i would use textit but i suck

[terse patio]

kk nice

Think I got it now

[frail igloo]

$\frac{-3}{-0.75}=x$

[jolly parrot]

Tan

[terse patio]

Lemme try doin it by myself

[frail igloo]

type .close if ur finished with your query

[terse patio]

k got that

@frail igloo think I got it. Also quick question, when separating the -0.75 from x, we don’t have to change whether it’s positive or not right? Just only divide.

[frail igloo]

sorry

i was afk

yeah

just divid

divide

[terse patio]

Nice

[frail igloo]

@terse patio

[terse patio]

Yeah yeah

I got it

Nice

🙏

[frail igloo]

if u changed the sign

then u change the entire question

thats like saying 2-3 is -2-3

completely different stuff

[terse patio]

Cool

[frail igloo]

only happens in inequalities

[terse patio]

Think I’m done now

[frail igloo]

aight

[terse patio]

.close

[pearl pond]

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[pearl pond]

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[north talon]

Let $f,g$ be two nonzero polynomials with integer coefficients and $\deg f > \deg g$. Suppose that for infinitely many primes $p$ we have that $pf + g$ has a rational root. Show that $f$ has a rational root

[jolly parrot]

Copter

[north talon]

this is all i have right now (i dont even know if this is correct)

bwaaaa

[vernal jasper]

I LOVE your handwriting its actly so pretty

[north talon]

yay..

[jade rune]

your handwriting is pretty

[ancient slate]

true

[ivory basin]

As a hint, try showing that there's a subsequence of primes such that p_k*f+g has a common root

[north talon]

errrr

[north talon]

well clearly di cant be all 1 since then ri would be infinite integers that are bounded

[north talon]

by this you mean all the polynomials share a root, right?

[ivory basin]

Not all of them

But an infinite number of them

[north talon]

i mean, in the sequence

[ivory basin]

Yes, you should get a subsequence which shares roots

[north talon]

bwa....

wait how would you even get that

<@&286206848099549185>

[pearl pond]

@north talon Has your question been resolved?

[north talon]

is it true that m = n-1

[pearl pond]

@north talon Has your question been resolved?

[pearl pond]

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[trim relic]

Can somone help me with z^2 + 4z + 1 +4i =0

[solemn fractal]

Use quadratic formula right

And then probably try to alter the roots in the way you want them to be

[trim relic]

I put z^2 +4z = -1 -4i then added 4 to both sides

Then changing it to (z+2)^2 = 3 -4i

Then took the square root z+2 = sqrt3 -2sqrti

The I put real z+2=3

Then minus 2 so z = 1

My mistake I meant a = 1

B =-2 sqrti

And I meant a sqrt 3 -2

Maybe show me how you would solve it, so I can see what went wrong

[light helm]

Then changing it to (z+2)^2 = 3 -4i
Then took the square root z+2 = sqrt3 -2sqrti
you didn't take square roots correctly

[trim relic]

How do I do it correctly, my algebra is weak

[light helm]

if you want to find the square root of 3-4i,
you'd want to consider another complex number
(p + qi)^2 = 3 - 4i
p^2 - q^2 + 2pqi = 3 - 4i
and compare real and imaginary components and solve a system
but with that idea, could just express z as a general complex number immediately
the main issue with what you did is that
sqrt(a-b) isn't sqrt(a) - sqrt(b)
nor is sqrt(a-bc) the same as sqrt(a) - sqrt(b)c
or anything similar where you try to apply the root to terms individually

personally by observing coefficients,
i would group the terms as
(z^2 + 1) + (4z + 4i) = 0
and it should be relatively easy to identify one of the solutions

[trim relic]

It feels almost impossible for me did you have something like this and how do I get over that thing I’ve been stuck on this for 2 days

[light helm]

wdym something like this

[trim relic]

Like an impossible subject

[ancient slate]

its the way to calculate imaginary roots

like $\sqrt{i}$ is calculated the same way

[jolly parrot]

Roy

[trim relic]

What do you mean?

[light helm]

wdym impossible subject

[trim relic]

Im struggling to understand this subject and I’m wondering if you have had the same and if so how to get over the feeling of it being impossible

[light helm]

work on the basics

you could always try the brute force approach
you'll pick up tricks and ideas the more you do/practice

start with simpler problems

more complicated problems are more or less a mix of multiple smaller simpler problems

[trim relic]

Ok thanks i will try this

.close

[pearl pond]

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[pearl pond]

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[low matrix]

how to represent -97 in bcd