help-13

[torpid heath]

yes

[half stratus]

we proved that too

[half stratus]

so how do you prove it

like sketch it

[torpid heath]

what did you prove?

I don't know if you know but

[half stratus]

these are the steps

[torpid heath]

$\phi(n)$ is the cardinality of the group with product and multiplicative inverse

[wraith dagger]

Amiso_

[torpid heath]

(It wasnt clear so like $(\mathbb{Z}/n\mathbb{Z}, \cdot )*$

[wraith dagger]

Amiso_

[half stratus]

the cardinalty one isnt that good cause i have trouble with it

i was able to do the exercises for CRT

but then the whole thing didnt make sense

[torpid heath]

that's why we go from there

the totent function tell how many integer are invertible modulo n

therefore the cardinality of the group $(\mathbb{Z}/n\mathbb{Z}, \cdot)^*$

[wraith dagger]

Amiso_

[half stratus]

what the helly

[torpid heath]

so here we have the number $m \cdot n$

[wraith dagger]

Amiso_

[half stratus]

this is euelers therorem right

i remember taking cases for this one

[torpid heath]

this is true yes

[half stratus]

cause you have case where its composite and then you split it up ok

lets not do crt

crt is kinda confusing

[torpid heath]

it's just that without crt I don't see a way to do it

with CRT it is like 10 lines maximum

[half stratus]

so how do you do it with crt

lowk gpt and memrozie might be the sol

[torpid heath]

noooooo

[half stratus]

ok no gpt and sol

[torpid heath]

you have the group $(\mathbb{Z} / nm \mathbb{Z})$

[wraith dagger]

Amiso_

[half stratus]

ya thats like resude modulo or some shit like that

[torpid heath]

by the Chinese remainder you know that there is an bijection between

[torpid heath]

this means modulo nm

[half stratus]

ohh ok

[torpid heath]

$(\mathbb{Z}/nm\mathbb{Z}) = $ all the number modulo $nm$

[wraith dagger]

Amiso_

[half stratus]

ya because you take all the numbers that are not coprime out right

[torpid heath]

$(\mathbb{Z}/nm\mathbb{Z}, \cdot *) = $ all the number that has an inverse modulo mn

[half stratus]

wait htis makes no sense

the latex box broke

[torpid heath]

for instance if it is $(\mathbb{Z}/6\mathbb{Z})$ you have

[wraith dagger]

Amiso_

[torpid heath]

0, 1, 2, 3, 4, 5

[torpid heath]

yes

[half stratus]

also i prove these if it means anything

[torpid heath]

but $(\mathbb{Z}/n\mathbb{Z}) \neq (\mathbb{Z}/n\mathbb{Z}, \cdot *)$

[wraith dagger]

Amiso_

[half stratus]

so its just 6 multiplied by all the numbers in z?

so it takes out 6, 12, 18, etc

[torpid heath]

no

there is only 6 element in this group

0, 1, 2, 3, 4, 5

[half stratus]

is that an arbritary y

[half stratus]

oh ok alr

ohhh ok this makes sense sorry

[torpid heath]

no worries

[half stratus]

i dont get the second bit with the star

its a cross product but like a cross product of what

[torpid heath]

secdond with star means that it is the same group but we get rid of the one who doesn't have an inverse modulo n

for instace if we take the group with nm = 6

[half stratus]

i mean all of them have an inverse module except

[torpid heath]

the ones that are not coprime

that's why if p is prime you have for the set p^k: p^k - p^{k-1}

but if nm = 6

you got rid of 0, 2, 3, 4

you only have 1 and 5

[half stratus]

if ax = 1 mod 6 u ax = -1 mod 6 is all primes

[torpid heath]

and now if you compute $1 \cdot 1 = 1$ so this holds and $5 \cdot 5 = 1$ so this is true here

[wraith dagger]

Amiso_

[half stratus]

ya thats true

[torpid heath]

and here you have $\phi(6) = \phi(3)\phi(2) = 2 \cdot 1 = 2$

[wraith dagger]

Amiso_

[torpid heath]

now the proof

[half stratus]

thats true

this gotta be the weirdest question

so if you were to draw it how would you draw it

[torpid heath]

we need to find the size of the group $(\mathbb{Z}/mn\mathbb{Z})$

[wraith dagger]

Amiso_

[torpid heath]

what a group is?

[half stratus]

if its prime then its like k-1

but if its not prime then its like all the fta primes multipleid together

[torpid heath]

because of the chinese remainder theorem

[half stratus]

you use this

[torpid heath]

because you can create a bijection between
$(\mathbb{Z}/mn\mathbb{Z}) \leftrightarrow (\mathbb{Z} /n\mathbb{Z}) \cross (\mathbb{Z}/m\mathbb{Z})$

[half stratus]

ya

[wraith dagger]

Amiso_

[torpid heath]

therefore there are the same size

[half stratus]

ok yea that makes sense

also

idrk what the missing step here is

this what the prof did but i didnt get it

[torpid heath]

I see what the prof did

[torpid heath]

here the inducctive part is a bit weird

I start directly at n + 1 which is

not very "pretty"

[torpid heath]

but here you can remember the rule with power and modulo

[half stratus]

what do you think is the optimal way to prove this then

[half stratus]

i dont rlly remember

is it this oen

https://media.discordapp.net/attachments/903430334681608232/1406395463560859808/image.png?ex=68a24f58&is=68a0fdd8&hm=0c9c33db666ebcbcdae4d1a7a1d5dddd9c4a44d91be637298b0b99bc86bc278d&=&format=webp&quality=lossless&width=1368&height=104

[torpid heath]

no no it's modular algebra

like this

[half stratus]

ths is horrid bro i have my exam in 2 days

[torpid heath]

$(a^2)^3 \mod n \equiv (a^2 \mod n)^3 \mod n$

[wraith dagger]

Amiso_

[torpid heath]

you can go insite the power and do the modulo there

and the end of you proof you have this fact

$(a^{\phi(p^l)})^p \mod p^l$

[wraith dagger]

Amiso_

[torpid heath]

this it totally legal to do this;

$(a^{\phi(p^l)}\mod p^l)^p$

[wraith dagger]

Amiso_

[half stratus]

what the helly

oh wait

i do know this rule

[torpid heath]

for crypto

[half stratus]

i use this all the time

like 74^3234 mod 10

[torpid heath]

exactly

[half stratus]

then you basically keep spamming this rule

[torpid heath]

yes

but here what is $a^{\phi(p^l)}\mod p^l$

[wraith dagger]

Amiso_

[half stratus]

$(a^{\phi(p^l)}\mod p^l)^p \mod p^l$

[wraith dagger]

SushiMan

[half stratus]

like that no

[torpid heath]

yes

[half stratus]

then you get like 1

[torpid heath]

yes

[half stratus]

1 mod p^l is like 1

so you get 1

[torpid heath]

it is

[half stratus]

ohhh

[torpid heath]

however this is true

[half stratus]

wher is this applied tho

[torpid heath]

two to the left of "missing step"

[half stratus]

oh i see

ohh

FUCK

bro

youre a genius

i cant even lie i love you

[torpid heath]

lmao

I did the course a semester ago so I remember it well

[torpid heath]

the guy correcting you exam would prefer

the other way aroung

go grom

$

[half stratus]

wait

[torpid heath]

$P(n) \implies P(n+1)$

[wraith dagger]

Amiso_

[half stratus]

ya he wants it proven that way

[torpid heath]

because this is what inductive is

[half stratus]

wait

this shit makes no sense

a^theta pl mod pl + 1 = 1

[torpid heath]

hmmm

[half stratus]

cause i did this

its supposed to be to mod p^l + 1

[torpid heath]

yes

[half stratus]

i dont think what i have here makes sense then

[torpid heath]

but the equality $a^{\phi(p^l)}^p \mod p^{l+1} \equiv 1 \mod p^{l + 1}$ would be true?

[wraith dagger]

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

[half stratus]

what does this mean

[torpid heath]

the equality is true

but the way of proving it is wrong

[half stratus]

well ya but if you do it then youre assuming the conclusion

[torpid heath]

yes

so no $P(n+1) \implies P(n)$

[wraith dagger]

Amiso_

[half stratus]

im tryna prove it the other way

[torpid heath]

yes

[half stratus]

so what are you getting at

[torpid heath]

with the other way aroung?

[half stratus]

im not tryna prove this

[torpid heath]

yes

the good way around

[half stratus]

im tryna prove this

$P(n) \implies P(n+1)$

[wraith dagger]

SushiMan

[torpid heath]

yes

[half stratus]

yes

[torpid heath]

where are you atm

[half stratus]

im at home

i had to dip cause my dad needed something

but im back rn

[torpid heath]

no worries

[half stratus]

ok so what are we getting at now

[torpid heath]

we know that

$a^{\phi(p^k)} \equiv 1 \mod p^k$

[wraith dagger]

Amiso_

[half stratus]

yes

we know that

[torpid heath]

I can give you the hint

[half stratus]

sure

[torpid heath]

$a^{pb} - 1 = (a^b - 1)(a^{b(p-1)} + a^{b(p-2)} + \cdots + 1)$

[wraith dagger]

Amiso_

[half stratus]

but i didnt even get that step

[torpid heath]

yes

[torpid heath]

but it hints you into the proof

you have to have some sort of a^b - 1

[half stratus]

so you justsay that b = theta pl

so you justsay that b = theta pl

[torpid heath]

yes

[half stratus]

ok gimme one sec

[torpid heath]

you have to think about divisibility

[half stratus]

oh wait

wait

wait

a^b - 1

[torpid heath]

yes

[half stratus]

divided by that is equal to 1

wait

[torpid heath]

hm

[half stratus]

i dont think it divides $a^{b(p-1)} + a^{b(p-2)} + \cdots + 1)$

[wraith dagger]

SushiMan

[torpid heath]

no

[half stratus]

$(a^b - 1)$

[wraith dagger]

SushiMan

[half stratus]

that was kinda disapointing ngl

i thought i was goiig somewhere

[torpid heath]

we go from the fact that $p^k \mid (a^b - 1)$

[wraith dagger]

Amiso_

[half stratus]

ya thats true

thats our assumtion

[torpid heath]

then what happen if

we multiply by p

[half stratus]

then you get pK+1

[torpid heath]

I meant

yes

sorry it wasn't clear

[half stratus]

no problemo

but what this mean

[torpid heath]

what happend now if we took

$p^{k+1}$ and $a^{pb} - 1$

[half stratus]

it divides?

[wraith dagger]

Amiso_

[half stratus]

idrk

[torpid heath]

if $p^{k+1} | a^{pb} - 1$ then we finished the proof

[wraith dagger]

Amiso_

[half stratus]

simplywait

so lemme get this clear

[torpid heath]

yes

[half stratus]

wiat

are you free to hop on a call rn

@torpid heath

or vc or something

[torpid heath]

I guess I am

My english is not really good though

[half stratus]

ok no problem thats fine

[torpid heath]

you're so close

the only issue is the last step

[half stratus]

if you muliply both sides by p?

idrk

[torpid heath]

it not really mutliply by p

but doing $p(k+1)$ which means on the side of the $p^{k} \mapsto p^{k+1}$

[wraith dagger]

Amiso_

[torpid heath]

on the side of $a^{(\phi(p^k)} - 1 \mapsto a^{\phi(p^{k+1})} - 1$

[wraith dagger]

Amiso_

[half stratus]

yo

[torpid heath]

we are looking for $P(k+1)$ and in order to prove that this is true

[wraith dagger]

Amiso_

[torpid heath]

we needed the fact that $p^k \mid a^{\phi(p^k)} - 1$ and the relation I put earlier

[wraith dagger]

Amiso_

[half stratus]

yes

[torpid heath]

so if we take

$a^{\phi(p^{k+1})} = a^{p\phi(p^k)}$ (as we prove a long time ago when try the wrong way)

[wraith dagger]

Amiso_

[half stratus]

ya

[torpid heath]

therefore you need to prove

$p^{k+1} \mid a^{p\phi(p^k)}$

[wraith dagger]

Amiso_

[torpid heath]

then by the formula said before we have that

$a^{p\overbrace{\phi(p^k)}^{=b}} = (a^{\phi(p^k)} - 1) \left( a^{b(p-1)} + \cdots + 1\right)$

[wraith dagger]

Amiso_

[half stratus]

im so confused

[torpid heath]

from where?

[half stratus]

like everhwere

[torpid heath]

1. we took $P(n)$ we worked our way in and get the fact
   $P(n) \iff p^k \mid a^{\phi(p^k)} - 1$

2. we took $P(n+1)$ and by the assumption said before $p^k \mid a^{\phi(p^k)} - 1$

we know that we have to prove $p^{k+1} \mid a^{p\phi(p^k)}- 1$

[wraith dagger]

Amiso_

Amiso_

[torpid heath]

I forgot the - 1 everywhere srry

[wraith dagger]

Amiso_

[torpid heath]

by the hint I gave we can simplify the rhs of the expression to

this: $a^{p\overbrace{\phi(p^k)}^{=b}} - 1 = (a^{\phi(p^k)} - 1) \left( a^{b(p-1)} + \cdots + 1\right)$

[wraith dagger]

Amiso_

[half stratus]

can you hop on a call

[torpid heath]

yes

[half stratus]

ok

[torpid heath]

this one

$a^{pb} - 1 = (a^b - 1)(a^{b(p-1)} + a^{b(p-2)} + \cdots + 1)$

[wraith dagger]

Amiso_

[cedar kiln]

@half stratus Has your question been resolved?

[half stratus]

.close

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[simple ingot]

I’ve been trying to solve this engineering math problem for an exam thats coming up. I don’t know whether converting to spherical coordinates is helping me or not and I’m also not sure whether I can say that the “normal vector” is equal to the unit vector of r. Any ideas are welcome.

“A metal sphere of radius α and temperature equal to T0 is surrounded by a static fluid. Heat is transferred from the sphere to the fluid according to equation (1) (image). Calculate the directional derivative dT/dn perpendicular to the surface of the sphere as well as the rate of heat flow (equation 2) where k is the coefficient of heat conductivity of the fluid. “

[simple ingot]

I dont have any pictures of my work because I genuinely dont know where to start from.

Im not sure what vector to use for the derivative and Im also having trouble calculating it in spherical coordinates

[cedar kiln]

@simple ingot Has your question been resolved?

[simple ingot]

<@&286206848099549185>

[cedar kiln]

@simple ingot Has your question been resolved?

[cedar kiln]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[gleaming rover]

using combination notation how many rectangles can be made with an area >5cm^2. this was on my math test and i got it sooo wrong. had no clue where to start so i spent 20 minutes counting rectangles (ended up with an answer of like 405???)

[gleaming rover]

just dont even know how to approach an unfamiliar question like this one on the test? (it was only 2 marks as well 😭 )

[split ice]

so each square is 1mm x 1mm, and the total board is 6m x 1cm?

[gleaming rover]

well each of the small squares is 1mm x 1cm

and then the big middle bit is 6m x 1 cm (and then obviously 5 rows

i believe those mm measurements are so every rectangle has to include a 6m part therefore cutting the numbers

however i just dont know what to do

[gleaming rover]

.

[cedar kiln]

@gleaming rover Has your question been resolved?

[cedar kiln]

@gleaming rover Has your question been resolved?

[cedar kiln]

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• Show your work, and if possible, explain where you are stuck.
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[topaz elk]

I've been struggling with this a bit because I haven't been able to isolate y after inverting it. I can get to a point of isolating 1y but I'm not sure what to do with the rest. right now I have y=-x(5y+1)-2

[topaz elk]

although I was thinking instead of moving over the 2 to divide both sides by 5y+1 but after that I also am not sure what to do with it

[dreamy void]

You could also show f(f(x)) = x

[topaz elk]

what do you mean?

[dreamy void]

A function is self inverse if f(x) = f^(-1)(x)

that is equivalent to f(f(x)) = x

[topaz elk]

ohh I see ill try that

alright I got it thanks

.close

[cedar kiln]

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[spark dawn]

Hello, I would like to ask for some more opinions about the right way to answer this question:
"Apply De Morgan's Theorem to the term A' * B."

Someone kind enough to help me told me there's two ways to look at this:

1. Your teacher is asking you to find the equivalent for (A' n B)'
2. Your teacher is asking you to find the equivalent for (A' n B). If you're sure that this is what your teacher is asking you, remember that (A' n B) = ((A' n B)')' and apply the theorem to the inner parentheses.
   GENERALLY when you're asked to apply DMT it's the first case so that's what we did but if everyone is telling you to find 2. Then you should do that one.
   
   Showing my work for A' n B:
   A' n B
A' n C' // let B = C' and substitute in B = C'
(A u C)' // directly use De Morgan's Law which says that A' n B' is equivalent to (A u B)'
(A u B')' // substitute in C = B'
   
   is this correct or is it the general rule that i should negate A' n B first to solve for the negated form (A' n B)' to properly apply demorgans law?

Showing my work for (A' n B)':
(A' n B)'
(A n B')
(A u B') // switch operater

[dusk goblet]

i don’t know why you’re doing this substitution

it makes it confusing and hard to follow

also do you have a screenshot of the question

[dusk goblet]

because "apply de morgan’s" to me reads as take the complement of A’nB

[spark dawn]

the substitution is used to directly use De Morgan's Law which says that A' n B' is equivalent to (A u B)'

[dusk goblet]

B = (B’)’

[spark dawn]

can you explain the context on how to use this?

[dusk goblet]

(A’ n B) = (A’ n (B’)’) = (A u B’)’

[spark dawn]

oh so instead of let B = C', we do let B = (B')' ?

[dusk goblet]

i mean i wouldn’t phrase it as "let" but sure

more like rewrite

substitute

etc

[spark dawn]

so what is the correct answer as a general rule? or are they both right
(A u B')' or (A u B')

[dusk goblet]

A’ n B ≠ A u B’

[dusk goblet]

i would interpret the question to mean this

if that’s what you’re asking

[spark dawn]

the complement of A'nB is (A'nB)' correct?
(A'nB)' = A u B’

[dusk goblet]

yes

[spark dawn]

so A u B’ is the correct answer to the question?

[dusk goblet]

that’s what i would say yes

[spark dawn]

what if the problem is (A' n B') instead?
do we apply demorgans rule to the complement for (A' n B') as a general rule?

(A' n B')' = (A u B)

[dusk goblet]

sure

[spark dawn]

why wouldnt we just apply demorgans rule to find the equivalent of (A' n B')? since demorgans rule says this:

ty for ur help so far

A' n B' = (A u B)'

[dusk goblet]

this is where the ambiguity comes in

and would be best just to email your professor for clarification

[spark dawn]

unfortunately they didnt respond to my emails

[dusk goblet]

how many days?

[spark dawn]

couple of weeks

[dusk goblet]

🤔

about this specific question?

[spark dawn]

yes

[dusk goblet]

or in general they don’t respond

[spark dawn]

they ignored it, they responded to other questions

[dusk goblet]

try asking again

[spark dawn]

i asked it again recently like 2 days ago, so we'll see

[cedar kiln]

@spark dawn Has your question been resolved?

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[swift wasp]

f(x) = 2/5x - 4

[swift wasp]

Zeros of functions smb help me understand this?? I suck w fractions

[fossil dawn]

so you want to find y = 0

let f(x) = 0 first

then before we continue

[swift wasp]

Mhm i understand that part

[fossil dawn]

is that $\frac{2x}{5}$ or $\frac{2}{5x}$?

[wraith dagger]

Hanako

[swift wasp]

Its like beside it

[fossil dawn]

ah ok so the first interpretation

[swift wasp]

Does that mean its in the numerator

Oh okay

[fossil dawn]

yeah you can consider it being in the num

in that case, the denom is 5

you can either find a common denominator between the two terms on the right, then cross-multiply

or you can clear the fraction by multiplying the entire equation by 5

[swift wasp]

How would u do it

[fossil dawn]

in this particular situation, the second way

but more generally the first way

[tropic oxide]

hold up. is the -4 meant to be inside the fraction or not

[swift wasp]

Its separate

0 = 2/5x (fraction) then - 4

[fossil dawn]

so $f(x) = \frac{2}{5}x - 4$

[wraith dagger]

Hanako

[swift wasp]

Yes

[tropic oxide]

it's best if you post a picture of the question in full so we don't have to interrogate you about what the question says btw

[swift wasp]

Okay ill do that next time ty

[fossil dawn]

but coming back to this, i would use the second way for this question simply because it's faster and the LHS is 0 anyway

there will be instances where you want to use the first way though

[swift wasp]

So i cancel out the 5 and multiply the 4 aswell?

[tropic oxide]

so what you're doing is solving the equation (2/5)x - 4 = 0, yes?

[swift wasp]

Yes

[tropic oxide]

this is a linear equation

[swift wasp]

Mhm

[tropic oxide]

do you know how to solve linear equations if all the numbers involved are integers and there's no fractions around?

e.g. if i gave you something like 4y - 21 = 15 and asked to solve for y, would you know what to do?

[swift wasp]

Yes

[tropic oxide]

ok can you tell me in your own words what you would do

in full detail with all steps

[swift wasp]

Add 21 to the other side then divide by 4, it would prob becomes 37/4 then simplify

[tropic oxide]

add 21 to both sides,

15+21 is not 37

[swift wasp]

36 oop

[tropic oxide]

you added two odd numbers, surely you can't expect another odd

[swift wasp]

Sorry i was tryna think quick

[tropic oxide]

don't.

math is not a race or a speedrun.

anyway yes, you add 21 to both sides and then divide by 4 on both sides, that's correct

now for your equation:
(2/5)x - 4 = 0

try to mimic at least one of these steps

though with the division there is going to be a slight difference in wording

but i would still like you to try it yourself

[swift wasp]

Add 4 to the other side so its 2/5x = 4

[tropic oxide]

ok so far so good

[swift wasp]

Idk what to do w fractions tho

[fossil dawn]

(note: probably prefer writing (2/5)x instead of 2/5x)

[swift wasp]

Okay ty imma rmb this for nxt time

[tropic oxide]

a fraction is a number like any other

when the number in front of y was 4 you knew to divide both sides by 4

[swift wasp]

I have a week n a half to do a 20 page algebra packet 🙂

[tropic oxide]

... ok that's great but i am trying to show you sth about this particular problem

trying to overcome your fractionphobia

[swift wasp]

Right im actually stuck here idk what to do

[swift wasp]

Right...

[tropic oxide]

the fraction is being multiplied by the x

2/5 is just a number, it's not that much of a different type of guy from 4 or -7 or 391

[swift wasp]

Should i turn it into a decimal then

Wait i think i'm getting somewhere...

Its 0.4 rite

[tropic oxide]

no need for that

what i was trying to get you to say is "divide both sides by 2/5"

ie to acknowledge that fractions are numbers in their own right

[swift wasp]

I was getting there, doing it as 0.4 sounds so much easier to my brain 😭

[tropic oxide]

you should definitely put effort into seeing fractions as numbers

even if decimals feel more "numbery"

[swift wasp]

So is x 10 guys....

[fossil dawn]

dividing a fraction is not too bad if you know what reciprocals are

[tropic oxide]

there's a bit of a difference in wording that you can make

[swift wasp]

I think i learned those but i forgot

[tropic oxide]

instead of "divide by 2/5" you can (and imo should) say multiply by 5/2.

(on both sides)

would you like to try seeing on paper how that works out

[swift wasp]

Yes

[tropic oxide]

well, do you have some paper with you

[swift wasp]

Give me a sec

OKAY so im here

[fossil dawn]

so far so good

i recommend writing 2/5 separately from x

[swift wasp]

Will do

How do u multiply 2/5 into 4

[tropic oxide]

so do what i said: multiply both sides by 5/2

[fossil dawn]

5/2, not 2/5

[swift wasp]

So i flip it first right?

[tropic oxide]

...

i told you one concrete thing to do. it would be nice if you did it instead of trying to look for some kind of justification prematurely!

in case you need to know why im having you do this, im gonna say that it'll become clear after you do it.

it is the kind of thing better shown than told.

[swift wasp]

Okay im here now

[tropic oxide]

10x/10, how can you simplify this?

[swift wasp]

1x/1 rite

[tropic oxide]

just x

[swift wasp]

Okayy found x

[tropic oxide]

note there wasn't actually any need to multiply the 5 and 2 together

gimme a minute to show you a somewhat better way to write this step

[swift wasp]

Okay tysm

[tropic oxide]

if you had something like 123456789/987654321 instead of 2/5

your way would demand that we work out the product 123456789*987654321 = 121932631112635269, twice. by all accounts, this would be rather unpleasant.

[swift wasp]

Right those numbers are hell

Tysm the visual helped some

[tropic oxide]

it's not a huge mistake and i wouldnt take points off for it but like

making your own life simpler whenever you can is a good philosophy to have

the upshot is that when there's a fraction (and not just then) being multiplied by the unknown, instead of dividing by it you can do the same thing phrased differently: multiplying (both sides) by the reciprocal of that fraction.

with the logic being to deliberately set up the cancellation that i showed you

and that's why you "flip".

[swift wasp]

Understood tysm

So i dont have to factor or anything for this?

[tropic oxide]

what exactly were you planning on factoring

as far as i can see there's nothing to factor...

[swift wasp]

Right okay im done here tysm 🙏 one step closer to finishing this

[tropic oxide]

ok

if you have nothing else to ask you can .close this channel

[swift wasp]

.close

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[serene depot]

how they get I2 as their final answer?

[fossil dawn]

if the question is what I_2 means, i think it's the 2x2 identity matrix?

[serene depot]

they asked to determine the inverse of A

and I dont seem to understand how they arrived at A^-1 = B = (-2,1 1.5, -0.5)

[fossil dawn]

because AB = identity matrix

so A and B must be inverses of each other

[serene depot]

Then why cant it be the matrix set from A?

[fossil dawn]

wdym matrix set from A

[serene depot]

why did they used (-2,1 1.5, -0.5) instead of (1,2 3,4)

[fossil dawn]

because the inverse of A can't be A itself unless A is the identity matrix

[void sand]

because it's B, not A, that is the inverse of A

[void sand]

(technically you need to verify that BA = I_2 as well, not just AB = I_2 )

[fossil dawn]

oh yeah

see another blunder

anyway, here the inverse of A is B, and the inverse of B is A (once both left and right inverses are proven, but that's a story for another time i suppose)

so the inverse of A cannot be A itself

[void sand]

given a matrix A, its inverse matrix A^-1 satisfies AA^-1 = A^-1A = I_2

you can verify for yourself that AA neq I_2, so A is not it's own inverse; on the other hand, y ou can also check that AB = BA = I_2 in this example, so B is A's inverse

[serene depot]

so that means inverse of B would be ( 1,2 3,4)?

[fossil dawn]

yep, and that would be A

you should verify this of course

[serene depot]

how do I do that?

[fossil dawn]

multiply BA

if they are inverses, this multiplication should also result in I_2

[serene depot]

thank you

I got another qn tho

[fossil dawn]

generally recommended you close this and open a new channel if you believe it's gonna be a long question

[serene depot]

What’s this formula used for

[fossil dawn]

quickly finding the inverse of a matrix, if it exists

[serene depot]

is there like an example in which you can provide?

[fossil dawn]

i can't think of an example off the top of my head, so i'm going to summon @void sand and hope he doesn't mind

[void sand]

sobbingcrying

[fossil dawn]

oh nvm i'll cook an example then

[void sand]

I don't have an example off the top of my head, but you can certainly find some examples online

[fossil dawn]

let's consider this matrix

the first thing you should always do when using the adjoint method is to evaluate the determinant of the matrix

can you do that?

[serene depot]

is it 14?

[fossil dawn]

i got 10

how did you get 14

[serene depot]

(3x4)+(2x1)

or is it supposed to be -?

[fossil dawn]

it's supposed to be ad - bc

as shown in the formula above

[serene depot]

oh then yeah its 10

[fossil dawn]

mhm

so it's not = 0, and we are safe

that means we can invert this matrix

[serene depot]

so I sub in 1/A as 1/10?

[fossil dawn]

hold on there

let's find the adjoint first

to find the adjoint matrix of A, there's a shortcut we can use here because it's a 2x2 matrix

[serene depot]

so I switch the numbers around right?

[fossil dawn]

this is the shortcut

swap d and a, then flip the signs of b and c

using this shortcut, find adj(A)

[serene depot]

why we change signs of b and c?

but not a and d?

[fossil dawn]

that is a good question. let me think about how i can put this because i am not overly sure myself

have you learnt about cofactors?

[serene depot]

I dont think so sorry

[fossil dawn]

ok i'm gonna try my best to explain this

the adjoint of a matrix is formed by replacing every element with its cofactor

and the formula and explanation for a cofactor is probably better explained by this page than i can hope to right now
https://www.cuemath.com/algebra/adjoint-of-a-matrix/

[serene depot]

ok thank you for helping

[fossil dawn]

quick correction: this method is NOT fast for finding the inverse of a matrix, especially if it's larger than 2x2

i am very sorry for this

but it remains feasible for a 2x2

the reason is because calculating all of those cofactors take a long while to do

[fossil dawn]

anyway, once you've understood, you can refer back to this formula and fill in the missing information

very sorry once again for having to dish you an article instead

[serene depot]

well what would be a better method to use then?

[serene depot]

np

[fossil dawn]

i'm sure others would have better methods, but even simple Gaussian elimination should be faster than this

in case you're wondering how it would be done using Gaussian elim, after you've determined that the matrix is invertible, construct an augmented matrix by placing the identity matrix next to your original matrix, like so

then, the goal is to use elementary row operations to transform the original matrix into the identity matrix while also modifying the identity matrix

when the left matrix is the identity matrix, whatever is on the right is the inverse of your original matrix

there might be some earlier stopping points you can make for higher row and column counts, but i'm out of my depth on those

[serene depot]

do I need to change anything in my original matrix?

[fossil dawn]

of course! the idea is to turn your original matrix into the identity matrix

[serene depot]

I meant like do I have to swap numbers before transforming it or?

[fossil dawn]

no, just put the matrix as it is

you're not using its adjoint

[serene depot]

I think I got a better idea now thanks for helping

[fossil dawn]

nps, and sorry for any confusion i've caused

!done

[cedar kiln]

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[serene depot]

np

I might need your help for further qns in the future, with your permission can I send a friend request?

[fossil dawn]

i suggest opening a new channel instead if you have any further questions

thanks for asking for permission first though

[serene depot]

ok can thank you

.close

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[eager hemlock]

The question is to find the value of the limit

[opaque harbor]

What have you tried?

[cedar kiln]

@eager hemlock Has your question been resolved?

[upper laurel]

it does not approach root 2

[latent bloom]

Is the third term $\frac{1}{n-\frac{1}{n}}$?

[wraith dagger]

VulcanOne

[boreal grove]

Yes my bad

[upper laurel]

limit here appears to be 1

adding 1/n isnt enough to counteract the sqrt

[cedar kiln]

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[upper laurel]

so youre essentially just square rooting a number an infinite number of times which leads to 1

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[proven summit]

<@&268886789983436800>, advertisement

[crude brook]

<@&268886789983436800>

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[eager hemlock]

Yes that's also my guess, I tried to use squeeze theorem but failed. I don't know how to properly prove that

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[quiet plover]

$\int \sqrt{5+3\cos(x)}$ is intriguing

[quiet plover]

did not start solving

no idea where to start

or $\int \sqrt{\sin^6 x + \cos^6 x} dx$

[proven summit]

I don't think either of them have an closed-form antiderivative

[nova snow]

,w integrate sqrt(5+3cosx)

[wraith dagger]

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[proven summit]

also make sure to include the dx part in your integrals

[nova snow]

,w integrate sqrt((sinx)^6 + (cosx)^6))

[wraith dagger]

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[proven summit]

like this $\int f(x) ,dx$

[wraith dagger]

឵឵MxRgD

[quiet plover]

uh oh

[quiet plover]

sorry forgor

[tropic oxide]

where did you get these integrals from?

given that WA spits out an answer in terms of elliptic integral functions, i would imagine that there's nothing better it could find!

[quiet plover]

this is the question

i tried it to integrate it with element as r dtheta

[tropic oxide]

this?

[tropic oxide]

can you show all of your work & confirm/deny that i displayed the region correctly

[quiet plover]

yeah that is it

[quiet plover]

between the limits 0 and 3pi/4

[tropic oxide]

oh so it turns out you actually wanted a definite integral but didn't tell us that before now

[quiet plover]

i guess that makes a lot of difference now that i think about it

but im still struggling to work with it as a definite integral

[tropic oxide]

hold on let me try to wrangle it

ok so i got it down to $\frac{3}{8\sqrt{2}} \int_0^{\pi} \sqrt{5 + 3 \cos(t)} \dd{t}$ with a bunch of trig shit

[wraith dagger]

Ann

[tropic oxide]

maybe even $\frac{3}{8 \sqrt{10}} \int_0^{\pi} \sqrt{1 + \frac35 \cos(t)} \dd{t}$

[wraith dagger]

Ann

[quiet plover]

exactly

[tropic oxide]

i think i might be cooking with a series expansion of that root tbh

ok i think i will stop cause that's definitely overkill and there is definitely a dirty trick.

[tranquil oracle]

it's one half-period of cos

[tropic oxide]

yeah and?

that doesnt seem like it helps us much at all

,w int[0,pi] sqrt(1 + 3/5 cos(t)) dt

[wraith dagger]

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[tropic oxide]

.........................

wow what.

[tranquil oracle]

okay that's still ugly

[tropic oxide]

sorry where is this question from again

what sadist gave you this crap

[quiet plover]

jee mains

[tropic oxide]

is there an answer key to this crap

[quiet plover]

only this is a troublesome one

other ones are pretty routine

[tropic oxide]

i think this is intentionally unsolvable

to crush as many hopes as possible

and cause misery

[tranquil oracle]

can you give the entire question

this could also be from a mistake in the intermediate working

[quiet plover]

umrmrmm

[tranquil oracle]

and i would like to ensure that possibility is eliminated

hmm, it's basically half the diamond thing

[quiet plover]

3/8th

i think

[tranquil oracle]

how is it 3/8

[quiet plover]

question

[tropic oxide]

notably, i did take for granted the integral that you wrote

so it is entirely possible that you screwed that part up somehow?

[tranquil oracle]

okay fine, it's 3/8, not the shaded region

[quiet plover]

oh

hm

[tranquil oracle]

but seriously, would do some symmetry stuff before formulating the integral

anyway, it would be 3/2 of a quarter of the diamond, so

$\frac{3}{2}\int_0^1(1-x^{2/3})^{3/2}dx$

[wraith dagger]

Element118

[tranquil oracle]

but still, x=sin^3 theta sub... sure you get cos^3 in the integrand but what

yeah it's quite solvable, but how do you really see that sub? maybe someone would try a (1-x^[2/3]) sub and then motivate further subs

$\frac{dt}{dx}=-2x^(-1/3)/3=-2\sqrt{1-t}/3$

and when the dust settles

$\frac{9}{4}\int_0^1\frac{t^{3/2}}{\sqrt{1-t}}dt$

[wraith dagger]

Element118

[tranquil oracle]

okay yeah maybe a u=sqrt(t) sub and then into trigo sub might be motivatable

probably intended path

[cedar kiln]

@quiet plover Has your question been resolved?

[tranquil oracle]

@quiet plover anything else you want to ask?

[quiet plover]

uh no lemme just try again

nothing else other than this

oh shoot

nvm the direct method workked

thanks though

.close

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[crimson sedge]

Could anyone help me this?
I was able to deduce m and n are even

So the eqn becomes
2^2x+3^2y=a^2
2^2x=(a-3^y)(a+3^y)

And after this I tried smth like this
a-3^y=2^i , a+3^y=2^2x-i

Now
a+3^y- (a-3^y)= 2*3^y = (2^2x+i) - 2^i

2 * 3^y= 2^i(2^2x-2i -1)

Therefore i=1
And
3^y=(2^2x-2) +1

But where do I go from here?
And am I even on the right path?

[crimson sedge]

<@&286206848099549185>

[lusty osprey]

how did you show m and n are even?

[lusty osprey]

line 2 reminds me of Pythagorean triplets

[crimson sedge]

Using congruence
Since 2^m+3^n is a perfect square
Its either 0 or 1 mod 3
Since 3^m is already divisible 3 and 2^m will never be divisible by 3
2^m=1 (mod 3)
And only when m is even will 2^m is of the form 3k+1

Similarly when I tried with mod 4 I got n is even asw

[lusty osprey]

ah I see

[crimson sedge]

The answer is m=4 and n=2
So basically 3 4 5 triplet
But idk how to reach there

[lusty osprey]

only that?

[crimson sedge]

Yes

Only 1 solution according to my textbook

[lusty osprey]

there was a condition on which numbers can form a Pythagorean triplet lemme try and remember

ah no I think it was that you can find a triple for any prime number congruent to 1 mod 4 (5, 13, 17, ...), which doesn't seem immediately useful

[crimson sedge]

I see

[lusty osprey]

also there was a way to generate them by squaring lattice points on the complex plane

maybe that helps

[crimson sedge]

We don't even have complex plane in our syllabus
But ill check it out

[cedar kiln]

@crimson sedge Has your question been resolved?

[cedar kiln]

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[sturdy escarp]

what's the difference between these two?

[slender ginkgo]

remainder theorem

states that the remainder is some number between x and a when applied to the next term

the estimate of remainder tells u how to estimate it using boundary involving that remainder

the phrasing is very bad

can someone plz help me

[tropic oxide]

the first expresses the remainder term in an exact but opaque form

the second gives you a way to put a bound on how big that term is, which is imprecise but stated in terms of more directly accessible things

[sturdy escarp]

hmm

okay I guess I kinda get it

I'm gonna try to do some problems maybe it'll make more sense that way, thanks

.close

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[ember heath]

whaaat

which chapter

[quiet plover]

area

its not that bad of a qn

[ember heath]

ohk havent covered that yet

[quiet plover]

its just integration and geometry being thick friends

maybe a small amount of series

and conic sections

[cedar kiln]

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[round lava]

aren't option (B) and (C) the same thing?

[round lava]

my approach:
(A.B)’ + (B.C)’ + A’ + C’
=> A’ + B’ + B’ + C’ + A’ + C’
=> A’ + B’ + C’

[crimson sedge]

no

AB+BC=AC

so option c is ) AC+A+C

[round lava]

so what i did is wrong?

[crimson sedge]

idk thermodynamics

[round lava]

are you sure? i do not think this is correct

this is not thermodynamics

[proven summit]

it's logic gates

[round lava]

yeah

[crimson sedge]

ah

I see

[latent bloom]

Logic gates and boolean algebra I think

[wintry monolith]

Logic gates

[round lava]

yeah

[crimson sedge]

crazy

[wintry monolith]

Ur preparing for gate ?

[round lava]

anyway, can someone please tell me why is this not correct?

[round lava]

i am not. i was preparing for a quiz with similar level of questions

[crimson sedge]

i told you

AB+BC=AC

[round lava]

you were not right

[crimson sedge]

I am

i mean

if they are vectors

it is correct

in vectorial geometry AB+BC= AC

[round lava]

just told you this is boolean algebra, not vectors or thermodynamics or whatever

anyway

i think why not option (A) too?

aren't all of them correct?

.close

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[plain granite]

Do I really have to use all of these big numbers as the answers? like question 7. I don’t like that I had to write 1 million there. So would I really have to write the full answer here?

[crimson sedge]

no u dont need to

u can just use like the exponent form

[plain granite]

what if i get points marked off

[crimson sedge]

why would you?

its not like you are really going to write 1 million in there

what if there is like 10^12, are you going to write that?

[plain granite]

hell no!

[crimson sedge]

that's what I mean

[latent bloom]

It tells you that you can expand if you need to

[crimson sedge]

yes expand means like

if you have

(x+y)^2

expand it

x^2+2xy+y^2

[latent bloom]

You can keep the number as 2^5 instead of writing 32

[plain granite]

ohh ok

may i ask another question

[crimson sedge]

sure

[latent bloom]

Ask away

[plain granite]

how do i expand division

also what the hell is #4

[latent bloom]

You don't need to use division here

[crimson sedge]

2^x/2^y = 2^x-y

[latent bloom]

Are you aware of the exponent rules?

[crimson sedge]

,exprules

nah i suck

[plain granite]

yes i know of dividing equal minus exponents

[fossil dawn]

just use that rule. it's sufficient

[latent bloom]

Yeah you don't need to use actual division here

[crimson sedge]

good, that's all you need here.

[latent bloom]

All the numbers in the fractions are of the same base anyways

[fossil dawn]

if there's a bunch of stuff on the top and bottom, pair the bases and only apply the exponent rules there

[plain granite]

but what do i do with the multiplication in the division??

sorry i feel so dumb rn 😔

[latent bloom]

Keep it that way

[crimson sedge]

You don't need to multiply

you just need to simplify top with the bottom

[latent bloom]

In no. 4, keep the numbers multiplied as is

Just modify their exponents

[plain granite]

oh ok ok

[fossil dawn]

apply te exponent laws between the 2s, x's and y's

then leave them be

[latent bloom]

You can also write the $z$ in the bottom as $z^{-1}$

[wraith dagger]

VulcanOne

[plain granite]

so for q2 i could go like this?

[crimson sedge]

no

u didn't simplify the 3's

[plain granite]

oh uuhhh

where was i supposed to simplify?

[crimson sedge]

3^2/3 = ?

[plain granite]

oh im a dummy

[crimson sedge]

5^5/5^2 = 5^3 which u did

but 3^2/3 u didnt do anything to it

[fossil dawn]

also when writing numbers in multiplied form

put a center dot between them

like this: 3 • 5^2

otherwise some ppl will read it as 35^2

[plain granite]

GAAAH IM SORRRYYYY FOR BEING SO HARD TO WORK WITH

[fossil dawn]

nono, it's fine!

it's just a reminder

i'm not criticizing

[plain granite]

i know i know its a logical thing that isnt subjective

[crimson sedge]

Don't worry, you are learning:)

there's no such thing to be sorry when learning

[plain granite]

oookayy

wait so is it 1^2 • 5^3

[crimson sedge]

where ?

[plain granite]

guh?

[crimson sedge]

I mean what question

still 2) or 3)

[plain granite]

number 2

[crimson sedge]

ok so

no

can you tell me what 3^2/3 = ?

[plain granite]

uuuh well theres nothing to subtract when it comes to ex

OH MY GOD I FORGOT

THE DAMN INVISIBLE EXPONENT

[crimson sedge]

$$\frac{3^{2}}{3^{1}}$$

[wraith dagger]

sempre solo

[plain granite]

so its just 3 or 3^1?

[crimson sedge]

3 and 3^1 are the same thing

but

you want to write it just as 3

[plain granite]

true

[crimson sedge]

the ^1 is irrelevant because it is already known when you are writing just 3

[frosty chasm]

it s basically 9/3 ?

[fossil dawn]

implied 1 btw

[crimson sedge]

yes

[plain granite]

huuuuh ok….

so its 3 • 5^3

[frosty chasm]

i think i need wolfram alpha to check 😂

[fossil dawn]

if we had to write every implied 1 it'd be super annoying

[crimson sedge]

yes!

[fossil dawn]

imagine 1x^1 / 1

[plain granite]

YAAAH 🎉

[crimson sedge]

Good job!!!

[fossil dawn]

well done!

[crimson sedge]

I haven't typed in latex in so long I feel like a newborn

[plain granite]

this my answer for #4

[crimson sedge]

z^-1

remember, your z wasn't simplified by anything so its still in the denominator

[plain granite]

wait so

[frosty chasm]

anyone know how to get familiar with LaTeX ? like ytb videos or smth

[plain granite]

is it this then

[crimson sedge]

yes but you want to write the z in exponent form

your answer is still correct, just not the way they want you to write it

exponent form = a^x

[plain granite]

wait why though

[crimson sedge]

a is base, x is exponent

[plain granite]

ok ok

so 1^z ??????

[crimson sedge]

looks cleaner, just exponents

[crimson sedge]

no

$$\frac{1}{x} = x^{-1}$$

[wraith dagger]

sempre solo

[plain granite]

theres only 1 as the expo

OHHHH

ok ok because the singular exponent was supposed to remove something else but couldnt right

[crimson sedge]

not really

[plain granite]

hm

[crimson sedge]

x^3 = x * x * x right?

[plain granite]

yeah

[crimson sedge]

x^1 = x^2/x right?

[plain granite]

i think

[crimson sedge]

$$x^{1} = \frac{x^{2}}{x^{1}}$$

[wraith dagger]

sempre solo

[crimson sedge]

?

[plain granite]

oh yeah!

[crimson sedge]

okayy

what is x^0 = ?

based on the previous rules

[plain granite]

well uh

^0 is always 1 but this is a variable

[crimson sedge]

soo

would it be

x/x?

[plain granite]

i think so

im not sure

[crimson sedge]

what is x/x = ?

give x any value and see

[proven summit]

think about plugging a number to x

like let x = 5

[plain granite]

yeah

[proven summit]

you get 5/5 right?

[plain granite]

it equals 1

[crimson sedge]

ok

so

x^0 = x/x right?

[plain granite]

yeah

[crimson sedge]

okay

now

the most important part

x^-1 = ?

[plain granite]

1/x^-1 because an exponent cant be negative

well atleast thats what theyre teaching us idk

[proven summit]

1/x^-1 would just be x

since you're taking a reciprocal of it

[plain granite]

okay? (head empty)

[crimson sedge]

1/x^-1 = 1/(1/x) = x since 1/(a/b) = 1 * b/a = b/a

[proven summit]

you're close though to getting what x^-1 =

[crimson sedge]

vale

look at the pattern

[proven summit]

just need to alter some things

[crimson sedge]

x^3=x^4/x

x^2 = x^3/x

x^1 = x^2/x

[proven summit]

actually let me sketch the pattern so it's easier to visualise

wait a sec

[plain granite]

yeah thats the main problem for me rn

i cant visualise it

[plain granite]

x^0 ????

the ones on the division side are 1 more than on the left

[proven summit]

notice how as we go fowards, we times by x? (the dot just represents multiplcation)

what would going backwards be

[plain granite]

divide by x

[crimson sedge]

plug k = -1 what do you get

[plain granite]

x^-2

that was my first guess but wasnt sure

[proven summit]

correct! so what can you deduce x^0 be?

[plain granite]

hmmmmm

ah i cant tell 💔

https://tenor.com/view/anime-spy-x-family-dizzy-anya-cute-gif-25812281

[plain granite]

but is it this????

[crimson sedge]

no

how did you get that ?

[plain granite]

oookayyyy

iunno i thought it would go the opposite way instead

hold on i need to change my pfp so i can feel smarter i need mental manifestation

i’ll change it back after all this

okay

i hope this works