help-39

[queen rock]

Yes

Now the denominator

[safe prairie]

im gonna make my previous whole denominator ^2

since itself is just Pr(a1) expanded by lotp

[queen rock]

No you must square inside each conditional not after summing

[safe prairie]

but theres this

[queen rock]

Your denominator is P(a1 cap a1) = P(K1 cap a1 cap a1) + P(complement(K1) cap a1 cap a1)

You already have the first term

[safe prairie]

if each result is independent what makes Pr{a1}^2 wrong?

[queen rock]

They are only independent given the key

Both prints depend on the same chosen key, right?

[safe prairie]

Pr{ a1 cap a1 } never specifies they had to be correct/from k1 tho

[queen rock]

Yes exactly, which is why you have to sum over both possibilities

Each print has the same probability, but the two prints are not independent overall since they both depend on the same key

[safe prairie]

is this the nature of a1 cap a1 or is this part of my situation where both did in fact come from k1

[queen rock]

Due to your situation yes

It's just about both prints sharing the same key in this case

[safe prairie]

what makes you say this

[queen rock]

Only independent once the key is fixed

[safe prairie]

reading the question again i didnt infer this

is there any way you can elaborate as of maybe how its dependent when the key is not fixed

[queen rock]

'one impulse is chosen, then it is used twice' so both prints come from the same underlying key

[safe prairie]

by independence only when given a key, do you mean that, when not given a key, id know that first a1's p is not second a1's p (itd be 1-p and sth)

[queen rock]

If the first output is a1 for example, it becomes more likely the key was k1. but the same key is used again, so the second output is now more likely to also be a1

Does that make more sense?

[safe prairie]

ig it does more now

so now i have Pr{ a1 cap a1 cap k1' }, itd be Pr{ a1 cap a1 | k1' }Pr{ k1' }, and the first on would be Pr{ a1 | k1' }^2?

why is reading comprehension in math so much more difficult now

[queen rock]

Yes that looks right

[queen rock]

Yeah in probability you really have to pay attention to small phrases in the problem

[pearl pond]

@safe prairie Has your question been resolved?

Channel closed

Closed by @safe prairie

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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

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

[sharp smelt]

I'm confused here

[sharp smelt]

How is a_n≥0

[dusty jungle]

if a_n < 0 say a_n = - epsilion with epsilon>0 then the series converges to something <= - epsilon

because every term in the series after that is <= -eps

[livid aurora]

you can clearly see an is tending towards zero so it's never gonna be exactly zero or smaller than zero

[sharp smelt]

thanks

.close

[pearl pond]

Channel closed

Closed by @sharp smelt

Use .reopen if this was a mistake.

[unborn charm]

@sharp smelt what book is that

[brisk scarab]

likely "Understanding Analysis" by abbott

[sharp smelt]

abbott

[pearl pond]

Available help channel!

Send your question here to claim the channel.

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

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

[sharp smelt]

I have a proof

one minute

To begin our hypothesis is
\
(i) $\abs{a_1}≥\abs{a_2} \dots \abs{a_n}$
\
(ii)$\left( \abs{a_n} \right) \to 0$

Just want tot be sure of this

[jolly parrot]

Wai

[autumn fossil]

you sure?

is that what your theorem states?

[cinder flower]

(i) is poorly written

[autumn fossil]

and its also wrong as stated

[cinder flower]

it doesn’t even parse. what is n

[sharp smelt]

it states this

[autumn fossil]

why did you add the absolute values then?

[sharp smelt]

I thought I had to here 😭

[autumn fossil]

You don't have to, and you shouldnt. The hypotheses are written in the theorem, so you can just copy them down word for word

or rather symbol for symbol

i dont get why they didnt just say "an is decreasing"

[sharp smelt]

okay, I think I messed up

oops

I'll continue late, it's near 1 am

sorry

.close

[pearl pond]

Channel closed

Closed by @sharp smelt

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

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

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

[celest yoke]

Hey guys, I’m confused how is their x intercepts at -5, -1

Jm in adv func and how do I know which one intersects the asymptop

Something like this!

?

[dense jasper]

$$\frac{a}{b}=0 \implies a=0 \text{ and } b \neq 0.$$

[jolly parrot]

Civil Service Pigeon

[dense jasper]

,w 2x^2+12x+10=0

[jolly parrot]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by civil_service_pigeon

[dense jasper]

(this is immediate from factoring)

[dense jasper]

you can check this yourself on desmos

[pearl pond]

@celest yoke Has your question been resolved?

[celest yoke]

Oh yea I forgot abt that lmao

Thx .close

.close

[pearl pond]

Channel closed

Closed by @celest yoke

Use .reopen if this was a mistake.

[dense jasper]

[pearl pond]

Available help channel!

Send your question here to claim the channel.

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

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

[north talon]

Find all integers $a,b$ and prime $p$ such that $a^p + b^p = p$

[jolly parrot]

Copter

[supple hill]

What ideas you have?

[north talon]

i have that if a,b neq 0 mod p then (a,b) = 1

and by fermat a= -b mod p but other than plugging that in i have no idea

[supple hill]

We know that: $a^p \equiv a \pmod{p}$ y $b^p \equiv b \pmod{p}$, right?

[jolly parrot]

Ga³¹Br³⁵I⁵³9000✞

[north talon]

yes

[grave mirage]

Maybe first look for bounds for a and b

[supple hill]

Then $a+b$ it's multiple of $p$, right?

[jolly parrot]

Ga³¹Br³⁵I⁵³9000✞

[north talon]

yes, thats what i did

[north talon]

well a and b can be negative tho

if p = 2 we have solutions so we can consider odd p

[grave mirage]

The difference in general will still be bigger than p

[supple hill]

And since $a^p + b^p = (a+b)(\frac{a^p+b^p}{a+b}) = p$, doesn't that mean that $a+b$ can only be $1, -1, p$ or $-p$?

[jolly parrot]

Ga³¹Br³⁵I⁵³9000✞

[north talon]

what the freak

but p | a+b so a +b = p,-p right

[supple hill]

Yeah!

So if $a+b = p$, the other factor $\frac{a^p+b^p}{a+b}$ has to be equal to 1, and if $a+b = -p$, it has to be -1. Do you see any solution for $p > $2 that meets that?

[jolly parrot]

Ga³¹Br³⁵I⁵³9000✞

[north talon]

no

ohh

thansk!

close

.close

[pearl pond]

Channel closed

Closed by @north talon

Use .reopen if this was a mistake.

[supple hill]

You are welcome!

Have a nice day.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

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

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

[strong schooner]

https://tenor.com/view/ishowspeed-speed-i-show-speed-tweaking-tweaker-gif-4717444372669792224

[supple hill]

State your question and what have you tried?

[strong schooner]

https://tenor.com/view/ishowspeed-desert-speed-idiot-mario-bros-wii-gif-4579521032062283911

[dense jasper]

<@&268886789983436800> troll, also left server

[merry carbon]

If they've left, they won't need this then

.close

[pearl pond]

Channel closed

Closed by @merry carbon

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

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

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

[terse patio]

Help rq

[ruby cargo]

with which one and for what?

[terse patio]

Question 3, I don’t wanna read just please tell me what I’m supposed to do🙏

[crystal dew]

idw read is kinda a bad idea when asking for help in a text-based platform, fyi

[terse patio]

Dw I read better when it’s on the digital screen instead of a textbook

🙏

[summer imp]

What you're supposed to do is interchanging x and y in vertex form and solving for y.

i.e. what is written

[terse patio]

Alr

[terse patio]

huh

Wait

The 2 on the y^2 got to go somewhere

Correct?

[summer imp]

Yes, 2 under the root.

[terse patio]

And f^-1(x) at the start replacing y

Got that

.close

[pearl pond]

Channel closed

Closed by @terse patio

Use .reopen if this was a mistake.

[terse patio]

Bru how do I answer 4 a

[buoyant cypress]

2(2^2) is not equal to 2 +4,
bracket means multipliccation,
2 is multiplied with 2(2^2)

[buoyant cypress]

also if you do not reopen the channel it will be recycled

[terse patio]

This?

Where did bro go

@buoyant cypress

[pearl pond]

Available help channel!

Send your question here to claim the channel.

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

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

[terse patio]

Br

[terse patio]

4a

[plush bramble]

mistake in this step

[buoyant cypress]

do you know BODMAS or PEDMAS

[plush bramble]

you have to multiply 2 with (3-1)^2 first before the subtraction step

https://assets.ltkcontent.com/images/104775/What-Is-PEMDAS-Order-of-Operations-Rules-in-Simple-Terms_7abbbb2796.jpg

[terse patio]

Oh

Bedmas

.close

[pearl pond]

Channel closed

Closed by @terse patio

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

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

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

[modest tartan]

2 things

1. did i do the log part right?
2. how do i find range of the thing inside log

[modest tartan]

also can i just keep this channel open so i dont spam make new channels

[plush bramble]

that would make it harder for people to help you

[modest tartan]

ok ill just make new channels

[plush bramble]

which step is the "log part" all 4 of them have log

[modest tartan]

hope no one minds

[modest tartan]

the last step

and the thing i made with the arrow to explain

[plush bramble]

the last expression is incomplete

what base is that supposed to be

[modest tartan]

yeah i didnt really didnt do whole thing

just to explain the step which is joined by the arrow

the base for last equals to is 5

[plush bramble]

are you just asking if $\sqrt{5} = 5^\tfrac{1}{2}$?

[jolly parrot]

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

[modest tartan]

im not sure if the base's power come in denominator or numinator

im wondering about that part

[merry carbon]

(also to step back a little bit, are you implying that sin(x) - cos(x) is between -1 and 1?)

[modest tartan]

yes?

oh wait shoot

its root2

man maybe i should take a break im messing up simple stuff

then the 2nd part of what i asked is just dumb now

anyways about the log part

does base power come in deno or numerator

[buoyant cypress]

bit

[modest tartan]

um

say log 10 base 5^3

[merry carbon]

How do you mean "base power"? as in where does the 1/2 go?

[modest tartan]

would it become 1/3 log10 base5

[modest tartan]

yeah thats what i mean

english is 2nd lang so not really the best

[merry carbon]

Pretty much, if you use the change of base rule, you can write $\log_{b^k}(c)$ as $\frac{\log_b(c)}{\log_b(b^k)} = \frac 1k \log_b(c)$

[jolly parrot]

@merry carbon

[modest tartan]

thanks alot

i just hv last question left in set

which i dont know how to appraoch

[merry carbon]

Sure, which question?

[modest tartan]

range of x^3 -12x for x belongs [-3,1]

i belive if i take x common it would creat a quadratic

however i dont know if i would get min value like that

caus min is at x=-3

wait no its "-12x"

so idk

[merry carbon]

Hmmmm, what have you learned up to this point? Have you e.g. covered calculus yet?

[modest tartan]

we havnt done calc yet

[merry carbon]

Fairs

[modest tartan]

wehv done quadratic graphs and the basic funtions

and the GM>AM stuff but i dont think thats usefull here

[modest tartan]

well we have done it in phy

differentiate then =0

then differentiate again to see if it a maxima or minima

is that what im supposed to do here?

[merry carbon]

You could do it that way, and that was the way I was wondering whether they expected you to do it that way, but it can be done without it

[modest tartan]

can u explain the way to do it without calc?

also i just tried the differentian way however the minima comes at x=2

which is outside of our range of x

so do i just put in the end points of our range of x and see which value is min?

[merry carbon]

I may need to leave an explanation for someone else tbh I can't remember how to do it off the top of my head, so I'd need to do some digging

[modest tartan]

no worries ill just ask my teach when we have our next class

[merry carbon]

And cause we have the minima being "after" x = 1, if we made a roughish sketch, we have that the minimum is gonna be what we get when we plug in x = 1 into the cubic

(it's not too hard to find the roots of the cubic, hence why we can figure out that will be the minimum )

[modest tartan]

hmm

well we havnt really studied the cubic graph so i though that if i put end points of range it would give max or min

im not quite sure if this is correct but i belive the cubic graph is somewhat similar to the quadratic graph

bro had to cross his explination 🥀

nvm he uncrossed

[merry carbon]

Ah no I crossed it out cause I forgot we found the turning points with calculus, I was gonna say "that assumes the turning points"

[modest tartan]

oh lol

[merry carbon]

The explanation is fine, I've only slightly embarrassed myself there

[modest tartan]

nw bro

if u saw the shi i was asking u wouldnt feel bad

[merry carbon]

Anyways, depends on how you mean by "similar", odd degree polynomials will either go "up" or "down" (e.g. from bottom left to top right, or top left to bottom right)

[modest tartan]

yeah thats what i was saying similar

the max and min of the interval would either be at end points or the turning points

[merry carbon]

Yep, that's pretty much always the case

I'm gonna cheat a little bit and get a graph for us

,w plot y = x^3 - 12x

[jolly parrot]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by chartbit

[modest tartan]

https://tenor.com/view/ok-nice-grafic-gif-13206845596055018221

[modest tartan]

min in our interval is at x=1

letzzzz gooo

im finly done wit this assigment

https://tenor.com/view/cat-dance-dancing-cat-chinese-dancing-cat-funny-cat-meme-cat-gif-221745127320028166

[merry carbon]

Do we have to get the max as well?

[modest tartan]

max is at x=2

from calc method

um the funtion becomes 16

final range becomes [-11,16]

[merry carbon]

Yep, -3 is between the left hand root and the max so we're happy

[modest tartan]

https://tenor.com/view/happy-snoopy-dance-gif-6346944927236436223

well ima go sleep now

its like 1am where i live

[merry carbon]

Hope you sleep well you definitely should get some rest, it's quite late

[modest tartan]

goodnight, morning,afternoon,evening

i can finly sleep peacfully

.close

[pearl pond]

Channel closed

Closed by @modest tartan

Use .reopen if this was a mistake.

[modest tartan]

thanks a lot

[merry carbon]

Have a good one, rest well

[modest tartan]

Zzz

[pearl pond]

Available help channel!

Send your question here to claim the channel.

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

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

[hearty quartz]

[slim vigil]

Do you have a question?

[pearl pond]

Asking the actual question right away is more likely to get responses.

Asking "Can I ask...?" or "Does anyone know about...?" doesn't give people enough information to decide whether they can help, and answering can feel like a promise to help with the actual question, which they might find themselves unable to.

[pearl pond]

If you are done with this channel, please mark your problem as solved by typing .close

[proper nova]

hello?

you have any questions?

[pearl pond]

@hearty quartz Has your question been resolved?

[pearl pond]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

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

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

[sharp smelt]

Wanted a quick proof of this checked

[sharp smelt]

As $\sum a_n$ converges absolutely, $(\abs{a_n})$ is bounded, by say $M$. Then $\sum a_n^2 = \sum \abs{a_n} \abs {a_n} ≤ M \sum \abs{a_n}$. As $\sum \abs{a_n}$ converges, by the series comparison test we're done.

[sharp vigil]

do you mean absolutely

[jolly parrot]

Wai

[pearl pond]

@sharp smelt Has your question been resolved?

Channel closed

Closed by @sharp smelt

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

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

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

[dire tapir]

Have you tried anything?

[ivory basin]

I genuinely can't tell what your answer is here

If it's 2^2025-1 then I'm gonna ask you to read the definition of period again

[shadow elm]

mb
$(2^{2025} - 1) - ( \sum_{d|2025} 2^d - 1)$

[ivory basin]

Sigma, not sigma

[tropic saddle]

\sum

[jolly parrot]

parthisjoking

[ivory basin]

But yea you have the right idea then

Find everything of "period" 2025, and remove elements that have period dividing 2025

[shadow elm]

is this correct answer ?
do i leave it like this or do i actually calculate everything ?

[ivory basin]

I don't believe this is correct though, since you'd have to do some inclusion exclusion

Cuz like, smth with period 405 would include periods 81 and 5 and what not

So you can't just naively subtract all factors

[shadow elm]

someone told me to do

mobius

how do i do that

[ivory basin]

Unfortunately not a number theory person myself

Hopefully someone else can help you with that

I'd personally do this by some dynamic programming argument but that's quite computationally heavy

[shadow elm]

oh

[pearl pond]

@shadow elm Has your question been resolved?

[shadow elm]

<@&286206848099549185> 💔
what is mobius inversion

[proper nova]

God damn i just scrolled past your channel

[slim jackal]

Lol

Helpers ping though

[brazen blaze]

try to count many secrets as possible

something

[worthy spindle]

💀

[shadow elm]

💔

am i cooked

[red hull]

A mathematical operation

[grave mirage]

PIE

Apply PIE

[grave mirage]

No this was arguably the hardest from previous year alonside 2^a*3^b problem

[pearl pond]

@shadow elm Has your question been resolved?

[brazen blaze]

hey chatgpt what is mobius inversion

@terse tartan

[pearl pond]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

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

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

[plush prism]

whats 20+1

[summer imp]

Do you have an actual question?

[smoky gull]

21?

[plush prism]

thx

[smoky gull]

Ur wlcm ig?

[frank violet]

legal drinking age

-# in usa at least

[plush prism]

damn bro

ur a genius

[smoky gull]

Please . close if that was all

[plush prism]

how

im kinda new

to ts

[smoky gull]

Just type in . close

[plush prism]

.cole

..

.close

[pearl pond]

Channel closed

Closed by @plush prism

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

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

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

[jolly parrot]

Pom

[worldly jewel]

yes but it is bad style to write in this way

[jolly parrot]

Pom

[pearl pond]

Channel closed

Closed by @opaque stone

Use .reopen if this was a mistake.

✅ Original question: #help-39 message

[worldly jewel]

Were you given a statement in that format

because nobody writes like that

Are you in like an intro to proofs class that specifically tests if you can parse logic symbols

because otherwise that is a pretty atrocious way to write mathematics and there's no excuse for it

I guess that makes sense

carry on

[pearl pond]

Channel closed

Closed by @opaque stone

Use .reopen if this was a mistake.

✅ Original question: #help-39 message

Channel closed

Closed by @opaque stone

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

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

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

[jolly parrot]

Renato

[ruby cargo]

!status

[pearl pond]

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

@stoic imp Has your question been resolved?

[stoic imp]

https://cms.dm.uba.ar/academico/materias/1ercuat2016/taller_de_calculo_avanzado/axiomas_reales.pdf

[stoic imp]

can I get some help proving these using this axioms?

[ivory basin]

Cute, lemme just look up the field axioms once cuz I keep forgetting which ones are axioms and which ones are theorems

,w field axioms

[jolly parrot]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by xavier_dragon

[ivory basin]

Okay so for the first one, you wanna prove a ⋅ 0 = 0

The first step would be to rewrite the 0 on the LHS to something more useful

Also I'd highly recommend the natural numbers game for this sorta stuff. It's a very good way to learn how to prove stuff starting from axioms

[pearl pond]

@stoic imp Has your question been resolved?

[stoic imp]

i need to use the axioms that i attached above

[ivory basin]

Isn't the stuff in the image what you need to prove

[stoic imp]

https://cms.dm.uba.ar/academico/materias/1ercuat2016/taller_de_calculo_avanzado/axiomas_reales.pdf

[ivory basin]

Oh right that mb

Hold on

Okay so the axioms in this are:
Commutativity of addition and multiplication
Associativity
Existence of additive identity and additive inverse
Existence of multiplicative identity and inverse (except for 0)
Distributivity

Cool same stuff as what wa said

[ivory basin]

Anyway, let's write the 0 as b + (-b) and see what we get from that

[stoic imp]

a.0 = a.(0+0) = a.0 + a.0

[ivory basin]

Yup that would work as well

[stoic imp]

a.0 = a.0 + a.0
a.0 - a.0 = a.0 + a.0 - a.0

0 = 0 + a.0
a.0 = 0

[ivory basin]

Minor nitpick cuz this is foundational stuff, you wanna write + (- a.0) instead

Cuz you haven't defined subtraction

But yes that proof is correct

[stoic imp]

a.0 = a.(0+0) = a.0 + a.0
a.0 = a.0 + a.0
a.0 + (-a.0) = a.0 + (a.0 + (-a.0))
0 = a.0 + 0
a.0 = 0

[ivory basin]

Yup

[stoic imp]

which axioms are i am using

[ivory basin]

Firstly you're using the axiom of additive identity to write 0 as 0+0

[stoic imp]

s3?

[ivory basin]

Then you're using distributivity to write a.(0+0) as a.0 + a.0

Yes

[ivory basin]

And then D

And then you're using some properties of equations which aren't really field axioms

Stuff like "you can add the same thing to both sides of an equation and that doesn't change it"

[stoic imp]

where

[ivory basin]

Writing a. (0+0) as a.0+a.0

[stoic imp]

oh. distributivity

[ivory basin]

Ye

[stoic imp]

which axioms we used

[ivory basin]

Well after distributivity, we used existence of additive inverse

Which is S4

[stoic imp]

S3
a.0 = a.(0+0)
D
a.0 = a.0 + a.0
S4
a.0 + (-a.0) = (a.0 + a.0) + (-a.0)
S1
0 = a.0 + (a.0 + (-a.0))
S4
0 = a.0 + 0
S3
0 = a.0
P2
0 = 0.a

[ivory basin]

Yup.

Also you'd need P2 since the question asks for 0.a

And we have worked with a.0

[stoic imp]

we good?

[ivory basin]

Yup

[stoic imp]

now what

[ivory basin]

The second question I imagine?

[stoic imp]

ab = ac
P4
b = c

[ivory basin]

Basically yeah

[stoic imp]

jajaja

right?

[ivory basin]

Ye

[stoic imp]

ok

what about c)

[ivory basin]

Suppose a ≠ 0

Apply P4

[stoic imp]

but

we can use a)

[ivory basin]

You want ab = 0 => a = 0 or b = 0

a) would give you the other direction

[stoic imp]

ab = 0
Suppose a ≠ 0

P4
b = 0.a^-1

[worthy lotus]

??

[ivory basin]

You've also used P2 implicitly

But yes that works

[stoic imp]

where

no

[stoic imp]

p3 perhaps..

not p2

[ivory basin]

A full proof of what you've done would be

ab = 0
Suppose a ≠ 0. By P4, there exists a^-1
a^-1.(ab) = a^-1.0
By P3,
(a^-1.a)b = a^-1.0
By P2 and P4,
1.b = 0.a^-1
By P3 and (a),
b = 0

It is easy to forget just how much is done implicitly most of the time

[stoic imp]

yes

[stoic imp]

i shee

[stoic imp]

then this needs fix

correct?

[ivory basin]

Well not a fix, but an expansion

But yes, you should write it out explicitly

[stoic imp]

ab = ac
By P4, there exists a^-1
a^-1(ab) = a^-1(ac)
by p1
(a^-1a)b = (a^-1a)c
by p4
1.b = 1.c
by p3
b = c

[stoic imp]

what if a = 0

[ivory basin]

Well, yes what we have done is one half of it

We have shown that if a ≠ 0 then b = 0

If a = 0, then we already have a = 0

(we need a = 0 or b = 0)

(a = 0 already gives us that)

[ivory basin]

Yup

[stoic imp]

whats your point

[ivory basin]

Look at what we are trying to prove

Given ab = 0, either a = 0 or b = 0

[ivory basin]

We have two cases for this

Case 1, a = 0. In this case we already have what we want

Case 2, a ≠ 0. In this case, we showed that b = 0

So we have what we want in both cases

[stoic imp]

what if both a=b=0

[ivory basin]

That's fine too, it's not an exclusive or

[stoic imp]

you said two cases

[ivory basin]

Yes, a = 0 and b = 0 is a subcase of the firdt

The point is, once a = 0, we don't care about the value of b

Cuz we already have one part of the or

So we are done

The idea is that p or q is logically equivalent to "if not p, then q"

We don't care what happens if we have p

All we care about is having q if we don't have p

[stoic imp]

tautology?

[ivory basin]

Well not exactly tautology

But an axiom of logic

[stoic imp]

F -> ?

[ivory basin]

p → p ∨ q

We want p ∨ q, so assuming p already gives us that

[stoic imp]

what is your point

a = 0 and b = 0 are the only cases

[ivory basin]

Okay let me put this explicitly

We want to prove a = 0 or b = 0

Case 1: a = 0

In this case, we do not care what the value of b is

Because no matter what the value of b is, we already have a = 0 and therefore the or is already true

Case 2: a ≠ 0

In this case, we have proven that b = 0
Since b = 0 in this case, our or is true once again

These are the only possible cases because a is either zero or non zero

It cannot be anything else

[stoic imp]

ok

ab = 0
a = 0
0.b = 0 forall b in R by a)

[ivory basin]

We don't need to do anything if we assume a = 0

Assuming it makes the result we want true immediately

[stoic imp]

ok

help with d

[ivory basin]

This one's a bit tricky

Let's work by cases

If a = 0, we are done (why?)

[stoic imp]

(-1).a = (-ac)
-ac + ac = 0
-ac + ac + a = a

[ivory basin]

Make it explicit

[stoic imp]

fuck my life

[ivory basin]

Ahahaha

Yeah this stuff is annoying to do

[stoic imp]

(-1).a = a.(-1)
a.(-1) + a.(1) = a(-1+1) = a.0 = 0

a - a = 0

a.(-1) + a.(1) = 0 = a - a

[ivory basin]

This works if S4 mentions uniqueness

[stoic imp]

we can prove the inverse is uniK

suppose it isnt, then we

suppose x and y are both inverse additive of z

such that x ≠ y

then

[ivory basin]

That we can yeah, just makes your life easier if the axioms mention uniqueness

[stoic imp]

x + z = y + z = 0

basically we get to a contra

[ivory basin]

Yeye

If S4 doesn't mention it, then attach this proof

[stoic imp]

doesnt Mention uniqueness

[ivory basin]

Cool, then attach this as a lemma

[stoic imp]

x + z = y + z
x = y

[woven matrix]

the proof of uniqueness in a commutative field (I meant more that the group is commutative) is really short so you're ok

x+z = 0 if x is inverse, if y is another you can add y on both sides

x+(z+y) = y

x+0 = y

[ivory basin]

Commutative fields are really nice once you're done proving all this obvious bullshit

Unfortunately proving this obvious bullshit is extremely annoying

[stoic imp]

yep this better

d and e sucks

[stoic imp]

sorry

[ivory basin]

Why are you sorry not like you came up with the bullshit

If anything you're a victim

Yeah e is really really annoying, but you can use d to help somewhat

[stoic imp]

jejejehe

(-1).a = a.(-1)
a.(-1) + a.(1) = a(-1+1) = a.0 = 0

does this shit work or no

a - a = 0
a- a = a.(-1) + a.(1)
(a - a.(1)) -a = a.(-1)

[ivory basin]

Looks good to me but I'm also sleepy lol

[stoic imp]

fuck my life

[pearl pond]

@stoic imp Has your question been resolved?

[stoic imp]

can someone help me

for the love of god

[stoic imp]

this are the axioms I need to use

and I need to prove that

(-1).a = -a

[rough forge]

what's your progress on that

[stoic imp]

well

[stoic imp]

here

[stoic imp]

@rough forge

[rough forge]

the first line doesnt work, that's what you want to show

but the seocnd line is correct, because you apply P3

by adding both sides the additive inverse of -a

[stoic imp]

the first line doesnt work, that's what you want to show
care to elaborate?

[rough forge]

You want to prove (-1)a=-a

but by writing -a as (-1)a you already assume that's true

(-1)a=-a is an assertion

The only things you know is P3: a=1a, D: distributivity and S4: that a has an additive inverse

[stoic imp]

I never did that

[rough forge]

(-1).a = a.(-1)
must have been the wind

[stoic imp]

but i am just doing xy = yx

[rough forge]

oh okay

[stoic imp]

correct me if im wrong tho

this are the axioms

[rough forge]

yeah that's fine

maybe you could state in each step which axiom you use

[stoic imp]

srry, my clipboard is dirty I pasted something else hehe

[stoic imp]

I am trying to figure out which stuff I did one sec

[rough forge]

[stoic imp]

uff sorry I went to a yogurt eating break for a bit

P2
(-1).a = a.(-1)
S4 and D
a.(-1) + a.(1) = a(-1+1)
no axioms
a(-1+1) = a.0
used exercise a)
a.0= 0

S4
a - a = 0
S3
a- a = a.(-1) + a.(1)
P3 ??
(a - a.(1)) -a = a.(-1)

@errant cedar @rough forge

[rough forge]

you basically want to prove that (-1)a is an additive inverse of a which means a+(-1)a=0 which you showed since a(-1+1)=a0=0 so this is where you finish

[stoic imp]

alrighty!!

[stoic imp]

can I get more help with e)

[rough forge]

what's e)

[woven matrix]

https://cdn.discordapp.com/attachments/908078029778083872/1487509800932741180/Screenshot_20260328_125753_Brave_-_Nightly.jpg?ex=69c966f5&is=69c81575&hm=9e2409a471136901123275fe6fe36a618fafe71c85d86372b24821826325c823&

[stoic imp]

this ^

used the proof from d)
(-a) . b = ((-1).a) . b

P1
((-1).a) . b = (-1).(ab)

P2

(-1).(ab) = (-1).(ba)

P1
(-1).(ba) = ((-1).b).a

used the proof from b)
((-1).b).a = (-b).a

P2
(-b).a = a.(-b)

[steel dagger]

pretty sure you can derive e) without using commutativity

[stoic imp]

care to elaborate?

[steel dagger]

consider (a-a)=0 for all a in the ring

[stoic imp]

where you going with this

dont get it

[steel dagger]

then use a)

[stoic imp]

how does that avoid P2

?????????

[steel dagger]

spoiler, try a bit before unspoilering: || start from using 0b= 0 and a0=0||

[stoic imp]

what

ab + (-ab) = 0

idk

i dont get it

[rough forge]

[stoic imp]

i am trying

[rough forge]

didn't you show -a=(-1)a

should make sense why that's 0

[stoic imp]

what i dont understand is how to avoid using p2

this lad is saying we can avoid commutativity, correct?

[steel dagger]

a hint if you havent unspoilered yet: prove that a(-b)= -(ab) and (-a)b= -ab instead

[stoic imp]

Mmm

[stoic imp]

i dont get it

wtff is going on

[steel dagger]

do you see the motivation of the hint?

[stoic imp]

nop

[steel dagger]

its using transitivity of =

if a=b and b=c then a=c

[stoic imp]

is this true?

[steel dagger]

its, i have done it on a call

[stoic imp]

call?

[rough forge]

is this proof-baiting

[stoic imp]

can I join

[rough forge]

sure

[stoic imp]

i was wondering because i dont see how

[rough forge]

well he's leading you there

if you see the proof you will learn nothing

[stoic imp]

not true

you can still learn a lot from others proof

IMO

[steel dagger]

im in vc

[rough forge]

yes but in your situation it's only instant gratification, you dont gain anything, you think you do, but when the paper is blank, your mind is blank

[stoic imp]

uff I already forgot

let me try

(-a).b + ab = (-a + a).b = 0.b = 0
(-a).b + ab = 0
(-a).b -ab + ab = -ab
(-a).b + (ab - ab) = -ab
(-a).b + 0 = -ab
(-a).b = -ab

[steel dagger]

do you remember the motivation of this?

[stoic imp]

not use commutativity?

or wdym

[steel dagger]

partially correct but not the actual intention

[stoic imp]

what is the real motivation

to use transitivity?

oh maybe that

[steel dagger]

how do you know if this is true?

[stoic imp]

if everything is equal to each other

I pressume

like, pairwise

if everything is pairwise equal to each other

[steel dagger]

?

[stoic imp]

[steel dagger]

i typed 2 equations not 3

how do you know initial two are correct

[stoic imp]

idk

[steel dagger]

you are missing implication signs

is it both ways or one way implication

[stoic imp]

should be iffs

[rough forge]

[pearl pond]

@stoic imp Has your question been resolved?

Channel closed

Closed by @stoic imp

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

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

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

[fading nexus]

I’m stuck on what to do next

[sweet junco]

keep going

[fading nexus]

I don’t know what else I can do tho

That’s the problem

[sterile python]

There are two -cos^2x

[sweet junco]

what if you multiplied the first one on the right by (1+cosx)

[sterile python]

You can try simplify numerator

[sweet junco]

both sides

[fading nexus]

The first one has a 1- but the second one doesn’t

[sweet junco]

by (1+cosx)/(1+cosx)

[fading nexus]

I’m confused

[sweet junco]

sinx/(1-cosx), multiplying it by (1+cosx)/(1+cosx)

[fading nexus]

This is how my teacher did it and I’m very confused

[sweet junco]

how did you randomly get 1-2cosx-cos^2x

I'm seeing it now

[fading nexus]

I factored it by putting it in the box

[sweet junco]

-cosx * -cosx = +cos^2x, u also had that one

[pearl pond]

@fading nexus Has your question been resolved?

[pearl pond]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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

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

[stark basin]

guyes

[stark basin]

in one is free for a call i need help

[plush bramble]

Just ask your question here

[pearl pond]

Asking the actual question right away is more likely to get responses.

Asking "Can I ask...?" or "Does anyone know about...?" doesn't give people enough information to decide whether they can help, and answering can feel like a promise to help with the actual question, which they might find themselves unable to.

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

@stark basin Has your question been resolved?

[pearl pond]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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

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

[wraith wraith]

explain inverse functions

[wraith wraith]

liek I wanyt you to start from the very basics

my teacher yapped abt random stuff but I got so confused by all of it

[sterile python]

google

[wraith wraith]

it does not make sense

[sterile python]

inverse function is instead y interm of x is now x interm of y

y=3x? then x = y/3

[vestal tapir]

it's when you flip the graph

[wraith wraith]

I know hwo to do the proecdure

I just do not understand why that works

or why we solve to make y the subject after

[sterile python]

wdym why that works

[wraith wraith]

I also understand this

[wraith wraith]

like WHY does changing the variables (x and y) GIVE the inverse

[sterile python]

reverse the role of input and output

[wicked kayak]

if f turns x's into y's, then the inverse turns y's into x's

[sterile python]

if input x gives output y, if you change, then the role also change

[wraith wraith]

ok but then why do we solve for y in that one after

[sterile python]

then tell me why you solve for x in normal functions

its the same but just inverse, thats it

[wraith wraith]

cuz thats what ur looling fo

[sterile python]

then we looking for y in the inverse

[wraith wraith]

ok but like if there is a restriction in an inverse but u rearrange it for the y

then u might lose a resitrction

[sterile python]

example?

[rough forge]

You are mirroring the function values across the y=x line

so in a sense it's a transformation

[wraith wraith]

y=2/x+3

change the x and y

x=2/y+3

yx=2+3y

yx-3y=2

y(x-3)=2

y=2/(x-3)

we have now lost the original restriction of y cannot equal -3

@sterile python I just don't understand the logic

[wraith wraith]

why

[rough forge]

blud

[wraith wraith]

wassup

[rough forge]

try to solve 3=2/x+3

you get 2/x=0 which is a contradiction

you will never attain y=3

you never lost something cause there was never something to begin with

[wicked kayak]

You don't have to swap the variables. Suppose you have a function y=f(x), and f is invertible. You can just write x = f^-1 (y). In other words just solve for x in the original equation.

[wraith wraith]

how can we find the inverse from just the original function

ok sorry I think I just am not getting it

are theer any videos you reccomend I watch?

[wraith wraith]

but I was referring to y=-3

[rough forge]

as mention ^ swapping y with x in other words y=x is only mirroring your original function along the first y=x line

[wicked kayak]

Try to pinpoint your question. It seems you are just getting caught up in notation. x and y are ultimately just symbols. in the notation y = f(x), x is the input of the function and y is the output. When we write x = f^-1 (y), we are saying y is the input of f^-1 and x is the output.

[rough forge]

why can't it be -3?

[wraith wraith]

loook at my word abnove

[wraith wraith]

^^

[rough forge]

i looked and it doesnt make sense

[wraith wraith]

I think I am overcompiating this

[wraith wraith]

why not

I want to work thru this step by step sorry im just very confused rn

like from the beginning

so f(x)=y

[rough forge]

let y=-3 then -3=2/x+3 => -6=2/x => x=-1/3 so f maps -1/3 to -3 and the invese -3 to -1/3

[wraith wraith]

inverse of f(x) = x

[wicked kayak]

It does seem you are overthinking a bit -- do you know how to draw a function-as-a-machine diagram?

[wraith wraith]

yea

[wicked kayak]

let's call our function f, you can put x's in it and it turns them into y's

[wraith wraith]

yes

[wicked kayak]

okay, so if f is invertible

[wraith wraith]

like f(x)=2x^2+6x+2 or whatevr

[wraith wraith]

it must pass the horizontal line test

[wicked kayak]

then there is another function f^-1, that takes in those same y's that came out of f, and turns them back into x's

[wraith wraith]

because if it does not, that means there are multiple x's for the same y

which altho is a function,

it means if u invert it, the function will not know which x to take

[wraith wraith]

yes

[wicked kayak]

Is any part of what i have said not clear atp

[wraith wraith]

what wud happen if u took the inverse of smth where it does not pass horizontal line test?

[wraith wraith]

nope! everything makes sense so far!

[wicked kayak]

you can't, because in that case the inverse does not exist

We say that a function f is invertible if it (1) is one-to-one (i.e. passes the horizontal line test) and (2) is "onto"

It seems that at your level of understanding you do not need to worry about "onto"-ness

[wraith wraith]

ok

[wicked kayak]

so if your function is not 1-1, then the question "what is the inverse?" makes no sense

does that help?

[wraith wraith]

yes

[wraith wraith]

yes

[wicked kayak]

what else you got

[wraith wraith]

but more specifically now I want to dig into why flipping the x and y does work for finding the inverse

sorry mb I mistyped

[wicked kayak]

okay so this is just notation, again you do not have to do it

we typically do it because we are interested in plotting f and its inverse on the same set of axes

[wraith wraith]

ok

wait I just realixed if u flip a horizontal line over the line y=x then you get a vertical line

does that have anything to do with inverse functions not working if it does not pas s horizontal line test?

[wicked kayak]

again, there is no "inverse function" if the function in question does not pass the horizontal line test, so there is nothing to either work or not work

[wicked kayak]

yes, but realize vertical lines do not pass the vertical line test and so are not functions, so this is irrelevant to the discussion of inverse functions

or functions in general

[wraith wraith]

I meant like

the fact that the above is true

is another reason the horizontal line test works right

[wicked kayak]

the "solve for y" part when finding the inverse is attempted in order to basically check whether the "inverse" is a function. If you can solve it for y then the inverse is a function. If you cannot, it is not.

[wicked kayak]

what do you mean reason the horizontal line test works?

what do you mean "works"?

[wraith wraith]

the reason the horizontal line test proves an inverse function exists

[wicked kayak]

the horizontal line test is a graphical characterization of the definition of injectivity (1-1)

Okay so tell me what a function is, precisely

we need to check that your understanding is correct

[wraith wraith]

every input gives one output

one output and exactly one

[verbal whale]

Very well

[wicked kayak]

correct. now a definition: we say a function is 1-1 if whenever f(x)=f(y), we have x=y.

graphically, a function is 1-1 if it passes the horizontal line test.

(1-1 is read "one-to-one")

a function needs to be 1-1 in order to be invertible, because if it is not, then the "inverse" could take in single numbers and output more than one thing, and thus won't be a function

make sense?

[wraith wraith]

ok

[wraith wraith]

yea

[wicked kayak]

okay good, feel better now?

[wraith wraith]

hm

I am still confused why after flipping the variabales

u need to solve for y = something

like for the inverse ^^

[wraith wraith]

I thought of it more like one to one means every y only corresponds to one x

how does this work

[verbal whale]

Because we usually write our functions with x as the input variable and y as the output variable

[wraith wraith]

I think all the variable changes is what ocnfuses me

[wicked kayak]

you don't "need" to necessarily. but your ability to do it gives you information about whether or not the function is invertible. Again, the notation y = f(x) means y is a function of x. We don't typically plot x = g(y) graphs on the same set of axes and call g a "function". More precisely, f is a function of x, and g is a function of y, but may not be a function of x

[wraith wraith]

ok

[wicked kayak]

if we have f(x) = x + 3, we know the inverse of "adding 3" is "subtracting 3". Hence f^-1(x) = x-3

[verbal whale]

For example, f(x) = x - 6

If we write y = x - 6 and, given a value of y we want to know the x that generated it, we isolate x

=> x = y + 6

So f^-1(y) = y + 6

But it isn't common to use y as independent variable, hence we change it to x and write:

f^-1(x) = x + 6

#help-39

[wraith wraith]

ok

[wicked kayak]

You are bogging yourself down in the roles that the symbols are playing. It does not matter that much. Like @verbal whale said, we typically use x as the independent variable, and y as the dependent variable, so writing f^-1(y) = x is a bit awkward

[wraith wraith]

ok

thanks for the help!

[wicked kayak]

for sure, good luck

[wraith wraith]

@wicked kayak @verbal whale !

[wraith wraith]

I have an exam on monday

[wicked kayak]

take some time to digest. draw some pictures.

[pearl pond]

@wraith wraith Has your question been resolved?

Channel closed

Closed by @wraith wraith

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

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

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

[prime stream]

Pinter, Abstract Algebra, Chapter 29, Ex. A7

Clearly, $\pi$ is algebraic over $\mathbb{Q}(\pi^3)$. Now, as I understand it, we need to prove that no polynomial of degree less than $3$ in $\mathbb{Q}(\pi^3)[x]$ has $\pi$ as a root.

If so, I proceeded as follows, although there might be a simpler way. Let us prove that $\pi \notin \mathbb{Q}(\pi^3)$. By a previously proven theorem, $\mathbb{Q}(\pi^3)$ consists of rational expressions of the form
[
\frac{a_n (\pi^3)^n + \cdots + a_0}{b_m (\pi^3)^m + \cdots + b_0}
]
(with nonzero denominator). Suppose $\pi \in \mathbb{Q}(\pi^3)$. Then
[
\pi = \frac{a_n (\pi^3)^n + \cdots + a_0}{b_m (\pi^3)^m + \cdots + b_0}.
]

Hence,
[
a_n (\pi^3)^n + \cdots + a_0 - b_m \pi^{3m+1} - \cdots - b_0 \pi = 0.
]

Since the powers of $\pi$ in the first part do not coincide with those in the second part, the expression can equal zero only if all coefficients are zero, which contradicts the assumption. Therefore, $\pi \notin \mathbb{Q}(\pi^3)$.

Thus, $\pi$ cannot be a root of a linear polynomial in $\mathbb{Q}(\pi^3)[x]$. Suppose $\pi$ is a root of a quadratic polynomial $x^2 + bx + c \in \mathbb{Q}(\pi^3)[x]$. Then it is divisible by $x - \pi$ in $\mathbb{Q}(\pi^3)[x]$, which again is impossible. Therefore, the minimal polynomial of $\pi$ is $x^3 - \pi^3$.

[jolly parrot]

Dedekind

[prime stream]

Check, please, if the proof is correct. Or maybe there is a simpler way to do it?

[pearl pond]

@prime stream Has your question been resolved?

[autumn trellis]

Looks correct, maybe you should say something like x-pi dividing x^2+bx+c is not possible due to the same reason as before, instead of just saying that's it's not possible (although maybe you meant that, some may take it wrongly)

As for an easier method, just showing that x^3-pi^3 is irreducible in Q(pi^3)[x] is enough (think of eisensteins criteria + gauss lemma)

[prime stream]

Thank you! But isn't the eisensteins criteria applied only in Q[x]?

[prime stream]

ah, I see, didn't know about it. But as I understand, we still need to prove that pi is not in Q(pi^3) in order to use it?

[autumn trellis]

take D=Q[pi^3].
(pi^3) Is a prime ideal be ause quotient by it is Q which is an integral domain

Then the above theorem (Eisenstein) says x^3-pi^3 is irreducible in Q[pi^3][x]

And Gauss lemma says that the irreducibility carries over to Q(pi^3)[x]

[prime stream]

think I understood, thank you very much!

.close

[pearl pond]

Channel closed

Closed by @prime stream

Use .reopen if this was a mistake.

[autumn trellis]

No problem!

[pearl pond]

Available help channel!

Send your question here to claim the channel.

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

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

[onyx otter]

Task: Given a poset $(X,\leq)$, prove that

$$A \preceq B : \Leftrightarrow \inf(A) \leq \inf(B)$$

never defines a partial order on the power set of $X$.

Hint: Do the cases of $X = \varnothing$ and $X \neq \varnothing$ seperately.

My question: If $X = \varnothing$ I'd just start checking the properties of a partial order one by one. But like... What is $\inf(\varnothing)$?

[jolly parrot]

Ruby

[onyx otter]

.close

[pearl pond]

Channel closed

Closed by @onyx otter

Use .reopen if this was a mistake.

[ivory basin]

What set are you working in

[onyx otter]

I noticed that since X is the empty set, the infimum can't even exist

[ivory basin]

$\inf{\varnothing} = \sup{X}$ usually

[jolly parrot]

Xavier 🌺

[onyx otter]

Because it would have to be an element of X

[ivory basin]

.reopen so it doesn't close on us

[pearl pond]

✅ Original question: #help-39 message

[onyx otter]

So it's obviously no partial order, since it isn't even a well defined relation

I'm working on the other part now

(X nonempty)

[ivory basin]

Ah fair

[ivory basin]

In general remember this

And the other way around for sup

[onyx otter]

Thank you :)

[onyx otter]

Ok, got the other direction too:

We know, that $\abs{X} < \abs{P(X)}$ (Cantor type shi), so $inf: P(X) \to X$ (just assume that all infima exist, cus otherwise it's not well defined) cannot be injective, so there are $X, Y \in P(X)$ s.t. $X \neq Y$, but $\inf(X) = \inf(Y)$, so $X \preceq Y \preceq X$, which shows that this is never antisymmetric.

[jolly parrot]

Ruby

[onyx otter]

Doneeee

.close

[pearl pond]

Channel closed

Closed by @onyx otter

Use .reopen if this was a mistake.

[smoky tartan]

Wtf is this

I thought maths was just 2+2

[onyx otter]

What is "2"? Have you previously defined this symbol? (I'm in uni. We don't really do numbers...)

[pearl pond]

Available help channel!

Send your question here to claim the channel.

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

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

[smoky tartan]

Stfu. I can't even do factorization. And how tf u do maths in uni. I once saw it and I almost died

[smoky tartan]

I don't need help lmao 💔

.close

[pearl pond]

Channel closed

Closed by @smoky tartan

Use .reopen if this was a mistake.

[smoky tartan]

.close

[onyx otter]

It's... An aquired taste, ig

[sterile python]

Valid crashout😂

[pearl pond]

Available help channel!

Send your question here to claim the channel.

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

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

[pearl pond]

Channel closed

Channel closed due to the original message being deleted.
If you did not intend to do this, please open a new help channel,
as this action is irreversible, and this channel may abruptly lock.

Available help channel!

Send your question here to claim the channel.

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

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

[arctic hatch]

Just like the 1st image end behavior (marked green), for the 2nd image, can I say that the end behavior is as follows:

3rd image

would this be correct?

[humble tendon]

i don't see why not

[pearl pond]

@arctic hatch Has your question been resolved?

[buoyant cypress]

for f(x) = $\frac{3x^5 - x^2}{x^3 + 3}$ \

$x \to \pm \infty, f(x) \approx 3x^2$\

$\lim_{x \to \pm \infty}, f(x) \to \infty$

[jolly parrot]

professor paradox

[arctic hatch]

if we look at the graph, as x is approaching -ve infinity, f(x) is approaching 3x^2 right? so can I not say that?

[sweet junco]

it would be appropriate to say acts like 3x^2 or is similar to 3x^2 the higher x-values we have, it also goes both ways(- and +)

but yeah approaching is also the right wording I would suppose

[arctic hatch]

in this image, they said it

but it's a different question

[buoyant cypress]

you can say that

[arctic hatch]

so this is correct?

[buoyant cypress]

mhm.. i dont feel its incorrect it like saying f(x) is asymptotic to 3x^2 as x-> +- infinity

[sweet junco]

it goes both ways, it does approach 3x^2 both ways

[arctic hatch]

but for +ve infinity, the curve is outside 3x^2 then how is it approaching, it just acts like 3x^2

[arctic hatch]

I want to say it this way bcz in the 1st image, they said it like this

[sweet junco]

it is not exactly 3x^2, as x approaches infinity it will act closer and closer to it but it is NEVER exactly 3x^2, it doesn't matter if it's inside or outside, that's just cuz I think (3x^2-1/x)/(1+3/x^3) gets higher if x is negative whereas it gets lower if x is positive

[arctic hatch]

if it's outside, how is it going towards 3x^2

it's not, so we can't say that, can we?

[sweet junco]

bc the distance shortens

think of what happens with -1/x as you go to negative infinity or something, it becomes increasingly positive, what about the bottom part, 3/x^3 becomes decreasingly negative so it tends to be higher than the graph of 3x^2(or in this case "inside")

now what if x approaches positive infinity, the values of -1/x decreases 3x^2 or rather gets more subtracted to it or smth while the value of 3/x^3 keeps being positive so it would be smaller than 3x^2, thus outside as you are referring to,

however, as we keep going along, the deficit gets shorter and shorter, it gets trimmed as we're going along until there isn't any

wait

I'm not done typing

[sweet junco]

I prolly didn't make sense here but the behavior at the start of the graph doesn't always define the behavior of a graph at infinity

[arctic hatch]

yh I don't get it

[sweet junco]

-1/x if x is negative is positive, so the numerator increases
+3/x^3 if x is negative is negative, so the denominator decreases, as the denominator decreases and the numerator increases, it would be greater than JUST 3x^2, thus appearing to be "inside" 3x^2

-1/x if x is positive is negative, so the numerator decreases
+3/x^3 if x is positive, is positive, so the denominator increaseS
as the numerator decreases and the denominator increases, it would be lower than JUST 3x^2, thus appearing to be "outside" 3x^2

vice versa also applies here where 3x^2 would be inside of f(x) and the other one or smth

however these values LOWER to almost 0 as they approach infinity thus being negligible

[buoyant cypress]

even if its it outside youcan still say that it is moving closer the parabola

[arctic hatch]

so if it were an asymptote we would've said that the curve is approaching 3x^2 (just like the 1st image), but since it's not , we are saying that it approaches +- infinity instead.... right?

[buoyant cypress]

https://en.wikipedia.org/wiki/Asymptote#Curvilinear_asymptotes

[arctic hatch]

but there is no H.A or S.A in this problem

[sweet junco]

asymptote has a few different meanings, the high school one is super generalized to just a line or something

well not a few different meanings

[arctic hatch]

so it's an asymptote? 3x^2 ?

[sweet junco]

yea ig

[buoyant cypress]

not a linear asymptote

altho i do not much about the terminology we might need to consider the definition of asymptote

[arctic hatch]

so can I say that as x approaches both +ve and -ve infinity, f(x) approaches 3x^2?

[sweet junco]

prolly

[arctic hatch]

k

@sweet junco @buoyant cypress thank you both

.close

[pearl pond]

Channel closed

Closed by @arctic hatch

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

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

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

[jolly meadow]

hey i need some help with f(x) = ((1-5a)/(a+1))^x, i have to find the values of a so that f is defined in all R. I know a+1 must not be 0 but is there another condition that is neccesary?

[light helm]

you don't want the base to be negative

[random ermine]

first note that what ur exponentiating is a constant

wrt x

let b = (1-5a)/(a+1)

if b < 0 then f(1/2) = sqrt(b) which it not defined

[jolly meadow]

ok thanks guys

.close

[pearl pond]

Channel closed

Closed by @jolly meadow

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

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

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

[pearl pond]

Please don't occupy multiple help channels.

Channel closed

Channel closed due to the original message being deleted.
If you did not intend to do this, please open a new help channel,
as this action is irreversible, and this channel may abruptly lock.

Available help channel!

Send your question here to claim the channel.

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

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

[midnight haven]

need help in linear algebra, 2 questions:

1. If matrix A is an nxn matrix and Ax=b has more than 1 solution for some b, then the transformation x->Ax is not 1-to-1. What else can you say about this transformation?

2. Suppose A is an nxn matrix with the property that the equation Ax=b has at least one solution for each b in R^n. Without using Theorems 5 or 8, explain why each equation Ax=b has exactly one solution.

Theorem 5 & 8 will be provided with photos.

[midnight haven]

I'm confused by question 1 because Theorem 8 says an invertible matrix must have "at least one solution" (g) and the transformation x->Ax is 1-to-1 (f). In the question, though, it says Ax=b has more than 1 solution for some b, and that the transformation is NOT 1-to-1. I understand that an invertible matrix can only have one solution (linearly independent), but I'm confused by the "at least". It's worded strangely. If that's the case, then I guess it makes sense why it wouldn't apply here. I'm not sure what else I could say about it. That it's not invertible? Singular matrix?

[tropic saddle]

yes not invertible

thats about it

the reason th 8 states it like that is that its often convenient to have slightly weaker sounding statements that you can prove

[midnight haven]

And I'm not sure how to answer question 2 without using the 2 theorems. I could apply logical reasoning and just say "if it has more than 1 solution it will have infinitely many solutions which means there would be infinitely many inverses"

but I'm not sure if I'm expected to answer more in a more in depth manner

[tropic saddle]

if th 8 had both the statements "Ax=b has at least one solution for all b" and "Ax=b has exactly one solution for each b" then the first obviously seems easier to prove in some cases

and the second seems stronger as a tool to apply

I dont know what other things you have covered in your course that they expect you to say

rank nullity maybe?

[midnight haven]

furthest we've gotten is theorem 8 and invertible linear transformations

[midnight haven]

idk that

[tropic saddle]

hmm

do you know the concept of rank

[midnight haven]

no

[tropic saddle]

hmm

[midnight haven]

what do u think of this reasoning

[tropic saddle]

well the conclusion would be that it has no inverse, not infinitely many. but even then, so what

the matrix wasnt stated to be invertible

so it doesnt get you closer to your goal

[midnight haven]

let me see if the instructions contain any information i may have excluded

[tropic saddle]

what are theorems 1,2,3,4,6,7?

[midnight haven]

Nope

Suppose A is an n x n matrix with the property that the
equation Ax = b has at least one solution for each b in JE. 11 •
Without using Theorems 5 or 8, explain why each equation
Ax = b has in fact exactly one solution.

One sec

oops i forgot i can do photos

one sec

FUdge

@tropic saddle

[tropic saddle]

ok so all useless

hmm

well I suppose you can use th 7

[midnight haven]

why

[tropic saddle]

what does "there exists at least one solution" mean for solving linear systems by gauss/row reduction

[midnight haven]

it is invertible

[tropic saddle]

without th8

[midnight haven]

it will at least equal 0

[tropic saddle]

what will equal 0

[midnight haven]

the matrix

[tropic saddle]

that makes no sense

[midnight haven]

im confused then

[tropic saddle]

you have solved linear systems by row reduction, yes?

[midnight haven]

yeah

[tropic saddle]

at which point can you see that there is at least one solution

[midnight haven]

when Ax=0

[tropic saddle]

at least one solution for Ax=b

Ax=0 always has at least one solution

[midnight haven]

whats the difference

[tropic saddle]

that the thing on the right can have nonzero entries