#help-23
safe radishBOT
torpid acorn
no 😭
timid ridge
basically it's just the ratio of side lengths on a right triangle and they are extremely useful
winter whale
I recommend self studying and watching videos on it and if you don't get something feel free to ask
thin bridge
what do you consider to be "basic"
hard crest
remember how in geometry you had like SAS and SSS and AAS theorems? trigonometry is an extension of those
safe radishBOT
@torpid acorn Has your question been resolved?
uneven furnace
hardy sky
I’m trying to find integral using partial fractions, but first have to do long division. I’m lost ngl, what’s next? I did the long division
safe radishBOT
I’m trying to find integral using partial fractions, but first have to do long d...
hardy sky
Calculus 2 7.4 partial fractions
covert wren
I’m trying to find integral using partial fractions, but first have to do long d...
first was the long division which u did properly
,tex $(x + 4) + \frac{16}{x - 4}$
flat frigateBOT
WeirdBoi
vague phoenix
no
covert wren
uhh and yeah turns out this is the final solution lmao
vague phoenix
take integration of that
hardy sky
Ok so I do them separately
covert wren
yeah
hardy sky
Ok perfect thanks
vague phoenix
!done
safe radishBOT
If you are done with this channel, please mark your problem as solved by typing .close
hardy sky
.close
safe radishBOT
safe radishBOT
Available help channel!
gusty flicker
can someone walk me through how to redraw the circuit so its simpler to solve
gusty flicker
ping me if u can help, ima skip to the next problem for now
yk what, ill figure this out in the morning
help would still be great but im going off to bed
normal moss
help would still be great but im going off to bed
just saying, the channel will be closed if you don't respond to the almighty bot when it decides to ponder if your question has been answered
safe radishBOT
@gusty flicker Has your question been resolved?
stoic torrent
can someone walk me through how to redraw the circuit so its simpler to solve
see, same wire lengths without any interruption in middle are said to have same potential
by that virtue,
stoic torrent
safe radishBOT
safe radishBOT
lone void
...?
stoic torrent
...?
i solved his qsn but then channel was closed by bot so it registerd my soln as a doubt
safe radishBOT
Available help channel!
lone void
i solved his qsn but then channel was closed by bot so it registerd my soln as ...
.okay
whole locust
can someone help me
safe radishBOT
can someone help me
whole locust
im a yr 8 so it will be easy
lean otter
Do u know what's is congruent means?
lean otter
Correct we just have to proved that for triangle BAD and triangle BDC. Can u see the triangle I am mentioning?
whole locust
so how could i answer the question
vague phoenix
mb ik im not the smarttest 😅
U smart
lean otter
When I was in grade we use create table for statement and reason. Then we write prove on it.
spiral bronze
Did you learn SAS (etc.)
whole locust
side angle side right ?
spiral bronze
AB = BC, angle DBC = DBA, BD=BD
Is this SAS
spiral bronze
What is the angle on a straight line?
whole locust
90 ?
spiral bronze
Not quite, that's the angle on a quarter turn
whole locust
What is the angle on a straight line?
180
spiral bronze
You're not meant to give the full answer, you're helping
whole locust
umm my apologies but i dont quite fully understand
whole locust
Good, now is ADC a straight line?
yes
spiral bronze
Do you agree that the angle ADB is the same as the ange CDB?
fiery olive
Let em explain them to you
spiral bronze
umm my apologies but i dont quite fully understand
How so?
whole locust
How so?
i dident understand what he ment by all the letters
i just got a little confused
whole locust
Do you agree that the angle ADB is the same as the ange CDB?
yes
spiral bronze
i dident understand what he ment by all the letters
Yeah don't worry
fiery olive
And their sum is 180? Right
whole locust
yes
spiral bronze
What is 180 divided by 2?
fiery olive
So one angle is 180:2
whole locust
90
fiery olive
Then angle CDB is right?
whole locust
umm im pretty sure when there is a line which has another line though it and its 90 degrees
fiery olive
Perfect
fiery olive
Perfect
spiral bronze
like when they go through each other a a 90 degreen angle
They don't need to go through, they only need to meet
whole locust
i see
but one thing i didn't understand was when they said [corresponding angles of congruent triangles]
spiral bronze
We've shown the triangles to be congruent right?
whole locust
What is meant by congruent?
?
whole locust
What is meant by congruent?
oh that they are the same shape
spiral bronze
Not quite, although they are a pair of corresponding angles
There are three pairs^
safe radishBOT
Available help channel!
safe radishBOT
hearty fox
Prove BC=6GM
waxen wraith
how to prove it tho?
open wedge
Prove BC=6GM
bọn chuyên ah hay j đây
open wedge
can someone give me the full question?
karmic merlin
Idk im just guessing it
open wedge
translation: ABC is a right triangle at A, draw ray AD at the opposite direction of ray AB such that AD = AB
a) triangle CAD and triangle CAB is congruent
b) let M be the midpoint of CD, draw a parallel line with BC that passes through D and intersects BM at E. prove that BC = ED
c) let G be the intersection of AE and DM, prove that BC = 6GM
@karmic merlin
karmic merlin
Oh dw i understood it untranslated
open wedge
If you can prove DG=1/3DC then youre done
that involves DC is in some triangle
open wedge
gửi full hình ra đc ko, t nghĩ là dùng cái trung tuyến đấy
hình m đang bị cắt phần j đấy mà bọn t cần nhìn
karmic merlin
No no the image is enough
open wedge
it's truncated
i wonder what's on the left of the diagram
might be a triangle that might help
@hearty fox
looking at the name i knew what his nationality is 
karmic merlin
Think they just extended the line by accident cause the construction doesnt need anything else
hearty fox
cái geo này nó k cho đánh 2 cái = nhau
karmic merlin
If you can prove DG=1/3DC then youre done
All you need for this is thales theorem
open wedge
m đang để thừa ra một đoạn nào kìa
severe star
Lớp 8 hay 7
hearty fox
Lớp 8 hay 7
lop 7
open wedge
adu
hearty fox
=))
open wedge
t thấy m gửi j r =))))
severe star
Bây giờ có song song r đk
safe radishBOT
@hearty fox Has your question been resolved?
safe radishBOT
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modest thistle
I get major confused by this long Set theory thing. where do i start?
modest thistle
$|D \cup E \cup F| = |D \cup (E \cup F)| = |D| + |E \cup F| - |D \cap (E \cup F)| = |D| + |E| + |F| - |E \cap F| - |D \cap (E \cup F)| = |D| + |E| + |F| - |E \cap F| - |(D \cap E) \cup (D \cap F)| = |D| + |E| + |F| - |E \cap F| - |D \cap E| - |D \cap F| + |D \cap E \cap F|$
flat frigateBOT
PainAndLight
astral glacier
Firstly by breaking it into lines
manic patrol
astral glacier
Hi bina
modest thistle
astral glacier
Good
astral glacier
Well are you familiar with this formula:
modest thistle
yes
astral glacier
$|A \cup B| = |A| + |B| - | A \cap B|$
flat frigateBOT
Xavier 🌺
astral glacier
...you didn't even see the formula
modest thistle
yes im familar wit that
its he cardinality of union
and I can swap around with it
astral glacier
Well this one is just a repeated application of that
manic patrol
I will leave this diagram here in hopes that it will help.
the numbers indicate how many times the area is included by just adding |A| + |B| + |C|.
modest thistle
solar hazel
modest thistle
Sadly I dont have to draw
astral glacier
Well the drawing is a good way to understand what's going on
modest thistle
Why is there so many more - in the solution
astral glacier
Well this one is just a repeated application of that
Do you understand what I'm saying here
astral glacier
What order
You can do it in any order, it doesn't matter
Every time you see a union, apply the formula
astral glacier
I imagine those are values that you've been given?
modest thistle
yses
astral glacier
...so you plug in the values
modest thistle
oh
it got moved
but shouldnt the | D n E n F | be negative
after moving it to the right
manic patrol
no.
modest thistle
but i swappped it
manic patrol
what did you swap it with?
modest thistle
manic patrol
and so their signs should not change.
you are moving the single-set cardinalities and the two-set intersections. the triple union and intersection do not move at all.
delicate shore
honestly you can just count on the venn diagram yourself
manic patrol
two-set inclusion-exclusion applied once on E U F.
delicate shore
the result will immediately pop out
modest thistle
honestly you can just count on the venn diagram yourself
I have to write it this way
manic patrol
you can use the diagram for intuition!
modest thistle
two-set inclusion-exclusion applied once on E U F.
mhhh
delicate shore
If you get how the formula arises then the proof comes more easily
one should never try to prove anything that is not obvious
solar hazel
wisdom from the great 0lante
delicate shore
well I didn't say it, 
did
manic patrol
if you need help understanding the whole thing, perhaps the Wikipedia article on inclusion-exclusion would help.
manic patrol
if that is from Wikipedia, right under that line:
This formula can be verified by counting how many times each region in the Venn diagram figure is included in the right-hand side of the formula. In this case, when removing the contributions of over-counted elements, the number of elements in the mutual intersection of the three sets has been subtracted too often, so must be added back in to get the correct total.
modest thistle
oof
manic patrol
I have already included that Venn diagram with annotations regarding the number of times each section is first included when adding |A| + |B| + |C|.
modest thistle
ill try
delicate shore
modest thistle
dafuq did I just watch. 
but made sense tbh
so to get the intersection of the all
Why again dont I just move the lhs stuff to the rhs and the |A n B C | thats on the rhs to the lhs?
and change their precedding operator symbols accordingly
modest thistle
you are moving the single-set cardinalities and the two-set intersections. the t...
i mean why D;
manic patrol
why overcomplicate things?
manic patrol
oh, you want to move the triple intersection to the left and the triple union to the right.
modest thistle
yes
manic patrol
well, that works, but then you'll get -(triple intersection) on the left, and to clear that negative sign you'll have to multiply both sides by -1, necessitating you to flip every single sign on the right. if you think that doing so is preferable, by all means use that method - the final answer is the same.
modest thistle
so but
how did the triple intersection get to left anyway
it was always on the right
ive were not movig
do you think just writing down this in a test would be enough?
manic patrol
I personally would derive the first line here through inclusion-exclusion, but I don't know how your examiner or teacher likes things.
for any question concerning tests, ask your teacher.
safe radishBOT
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Available help channel!
nimble helm
Is this correct and if yes, is there any way I could improve this?
solar hazel
open wedge
but kinda long for me and hard to keep track
open wedge
Is this correct and if yes, is there any way I could improve this?
i have a better way to do
we still reach until $p^2 = 12q^2$
flat frigateBOT
1 divided by 0 equals Infinity
open wedge
but this time, we'll do $p^2 = 3(2q)^2$
flat frigateBOT
1 divided by 0 equals Infinity
nimble helm
yes
flat frigateBOT
1 divided by 0 equals Infinity
solar hazel
this is morally incorrect
nimble helm
$p = \pm 3^{1/2}2q$
flat frigateBOT
gribble19
flat frigateBOT
1 divided by 0 equals Infinity
open wedge
which is really easy to do
open wedge
this is morally incorrect
-# shush at least it works
solar hazel
you’re violating all the moral codes omg
open wedge
you’re violating all the moral codes omg
-# at least it works
solar hazel
this is so impure
flat frigateBOT
gribble19
flat frigateBOT
1 divided by 0 equals Infinity
open wedge
and some more unrelated conditions that me myself wouldn't want to say
so since $\gcd(p, q) = 1$ then $\gcd(p^2, q^2) = 1$
flat frigateBOT
1 divided by 0 equals Infinity
solar hazel
this is like, assuming you have developed basic ‘integer square root theory’ before you have done the things that should go into ‘integer basic square root theory’
extremely immoral
nimble helm
so since $\gcd(p, q) = 1$ then $\gcd(p^2, q^2) = 1$
how do you know this
open wedge
extremely immoral
extremely immoral and inappropriate when you make fun of someone's work while they're working on this
nimble helm
open wedge
or if ts is some ring theory that wai had before
solar hazel
extremely immoral and inappropriate when you make fun of someone's work while th...
if the work is immoral it is moral for me to call it out
open wedge
if the work is immoral it is moral for me to call it out
immoral
if it's moral for you to call it out, but is immoral if it's interupting
contradiction
nimble helm
slayla are you saying its better to do it how i did it or rather there is a third option that is better than both
open wedge
p^2 = 12q^2
let's see your moral way
there's no defined rule of what a proof should be moral look like right?
solar hazel
let’s ask snow he knows everything
@toxic stratus what do you think? do you think invoking the sqrt symbol for this problem is immoral?
as for a moral way… @nimble helm do you know euclid’s lemma?
nimble helm
no
solar hazel
well we can probably still use it even if you haven’t learned it by name
open wedge
can't you use a better name for a proof?
solar hazel
can't you use a better name for a proof?
worn pollen
💀
solar hazel
yep. so 6 divides p
nimble helm
so 6, 4, 3, 2 divide p^2
open wedge
6(2q^2) 🥀
she is really trying to waste 2 steps just to say that 🥀
solar hazel
now try what you did, but with the fact that 6 divides p instead
nimble helm
oh
yes that is what i originally wanted to do bc then it just simplifies to 18k^2 = q^2 for some integer k
and q is also even
solar hazel
nimble helm
but i couldn't figure out how to show that if 6 divides p^2 then 6 also divides p
nimble helm
6(2q^2) 🥀
she is really trying to waste 2 steps just to say that 🥀
no i think i get it
solar hazel
6(2q^2) 🥀
she is really trying to waste 2 steps just to say that 🥀
blud…
i am appealing to euclid’s lemma
which is the morally correct way to do ts
nimble helm
6 dividing p^2 doesn't mean 6 divides p but both 3 and 2 dividing p^2 means both divide p bc euclids lemma (which only works for primes (so not 6)) and then because 3 and 2 divide p, 6 also divides p
solar hazel
you probably want to use some intuition you have with the sqrt symbol or something
but really it follows from 3 dividing p and 2 dividing p
solar hazel
6 dividing p^2 doesn't mean 6 divides p but both 3 and 2 dividing p^2 means both...
beautiful
solar hazel
yes that is what i originally wanted to do bc then it just simplifies to 18k^2 =...
wouldn’t it be more like (6k)^2 = 36k^2 = 12q^2, which implies 3k^2 = q^2
then we have that 3 divides q
and 3 also divides p
and we are done
solar hazel
also your profile is so beautiful
nimble helm
also your profile is so beautiful
thank you 
toxic stratus
let’s ask snow he knows everything
<@118663865264242688> what do you think? do ...
you indeed speak the truth, amen
nimble helm
okay i understand it now im just writing it up
solar hazel
6 dividing p^2 doesn't mean 6 divides p but both 3 and 2 dividing p^2 means both...
well 6 divides p^2 does imply 6 divides p. but like… in general for a composite number in place of 6, that statement does not hold. that’s why i appealed to euclid’s lemma to show it in this case
that’s more of a message for @open wedge though
solar hazel
you indeed speak the truth, amen
as is this
open wedge
as is this
elaborate for me why moral is important?
nimble helm
so we say that because r is a rational r = p/q for integers p,q with greatest common denominator 1? or is there a cleaner way to word that
solar hazel
so we say that because r is a rational r = p/q for integers p,q with greatest co...
you can say that
there exist integers p and q with gcd(p,q) = 1
glacial cairn
so we say that because r is a rational r = p/q for integers p,q with greatest co...
p and q coprime
solar hazel
my preferred principle to use here is descent but you don’t need to worry about that
the represent by coprime integers perspective is also morally correct
open wedge
that’s more of a message for <@927839054639267891> though
i can't just always follow as your style of writing proofs here
that you define such a set of rules that makes your proofs moral and try to make others follow that rule
solar hazel
‘not invoking the sqrt symbol’ is not even style
it’s just really bad form and demonstrates you don’t know how number theory is built up and is inappropriate because you don’t build it with square root theory before ts
open wedge
it’s just really bad form and demonstrates you don’t know how number theory is b...
that does not mean im not allowed to use square roots
toxic stratus
how do you know the sqrt exists?
solar hazel
that does not mean im not allowed to use square roots
it literally does
open wedge
it literally does
why then
solar hazel
that you define such a set of rules that makes your proofs **moral** and try to ...
these are not my personal rules. this is how number theory is done
flat frigateBOT
1 divided by 0 equals Infinity
solar hazel
that’s just shorthand for saying there is no rational number whose square is 2
open wedge
answer me
solar hazel
yes sure that is number theory. but i am responding to your next point in advance about how there is a sqrt symbol there
nimble helm
moral AND easier to follow?
open wedge
it literally does
well then your argument contradicts this
solar hazel
moral AND easier to follow?
more moral than before for sure
open wedge
tbh i don't even care if it's moral or not
solar hazel
the contradiction is slightly immoral but that’s more for style rather than correctness
so it’s ok
open wedge
tbh i don't even care if it's moral or not
math isn't defined by morality so i don't even care
safe radishBOT
solar hazel
math isn't defined by morality so i don't even care
blud… i’m using moral a certain way
perhaps not how you would use it in plain english in a serious way
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Available help channel!
atomic parrot
safe radishBOT
atomic parrot
How do this
quiet plume
What does it mean for T to be inversely proportional to w
quiet plume
Yeah ok
atomic parrot
And w=1/4 × cube root of d
quiet plume
Cube root you mean
atomic parrot
I just dont get how i find the value of d when T=36
atomic parrot
Cube root you mean
Yeah
quiet plume
Ok well if T = 36, can you find w?
atomic parrot
No?
steep jungle
Set up both equations
T = k/w and solve for k by plugging in the numbers 10 = k/9
W = k* cube root d
1= k* cuberoot 64
1 = k* 4
I think your set up was wrong
Let me know if you would like the values of k
atomic parrot
You have T = 90/w
Oh i just substitute 1/4 ×cube root of d for w?
quiet plume
Well just from T = 90/w, you could solve for w in 36 = 90 / w
Then that will give you w and you can do a similar thing for d
atomic parrot
Oh i justs substitute 36 into T=90/w
Then use what w is and substitute it in w=1/4 × cube root of d
Thx
So w=2.5
Then d is 2.15443469
quiet plume
If w = 2.5, then you get
2.5 = 1/4 * cbrt(d)
You need to cube to solve for d, not take the cube root
kindred sentinel
can somebody take a quick look at my math essay (10 pages but little text) and see if my calculations make sense?
safe radishBOT
can somebody take a quick look at my math essay (10 pages but little text) and s...
plucky elk
can you just write the math question
kindred sentinel
its not a question, its an investigation
i essentially have to find the optimal forearm angle for a receive in volleyball using vectors and reflection geometry
plucky elk
then try #recreational-math
kindred sentinel
whats that
burnt notch
A channel in this server
plucky elk
kindred sentinel
aah ok thx
faint glade
Use slopes to show that the points (-1, -3), (6,1), and (2, -5) form a right-angled triangle
The answer is supposed to be m1 = 3/2 & m2 = -2/3 but how did they reach that?
safe radishBOT
```
Use slopes to show that the points (-1, -3), (6,1), and (2, -5) form a right...
faint glade
I'm trying the slope equation, but (1-(-3)/(6-(-1) gives me 4/7?
timid ridge
I'm trying the slope equation, but (1-(-3)/(6-(-1) gives me 4/7?
what is the slope equation?
faint glade
$\frac{y2-y1}{x2-x1}$
flat frigateBOT
Vortac
timid ridge
is that what you have
timid ridge
$\frac{y2-y1}{x2-x1}$
what are the two points in this
faint glade
(-1, -3), (6,1), and (2, -5)
flat frigateBOT
Vortac
timid ridge
$\frac{1-(-3)}{6-(-1)}$
so you get the equation and the points, but why do you use x1 - y1 / x2 - y2
faint glade
so you get the equation and the points, but why do you use x1 - y1 / x2 - y2
I'm not? (x1: -1, y1: -3), (x2: 6, y2: 1)
y2: 1 - (y1: -3)
timid ridge
ok i see you are right my bad
well that's only one slope
a triangle has 3 sides
have you found the other 2
faint glade
The answer is supposed to be $m1 = \frac{3}{2} & m2 = \frac{-2}{3}$ so that first slope can't be right?
flat frigateBOT
Vortac
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
timid ridge
The answer is supposed to be $m1 = \frac{3}{2} & m2 = \frac{-2}{3}$ so that firs...
a triangle has 3 sides man
faint glade
yeah, and the answer key shows two slopes
timid ridge
the other two slopes could be the right angle
have you tried ?
are you aware of the fact that if there are two slopes M and N
if N = -1/M
then M and N are perpendicular and make right angles
so you have found ONE SIDE of the triangle
you dont know if the other two sides make right angles
it's not enough to say anything right now
faint glade
they haven't gone over right angles yet, so yeah that makes sense
thank you
.close
safe radishBOT
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Available help channel!
silent basin
This was in my answer key: I don't get it. If you pull the negation out through the forall y, doesn't that change it to: "there exists y" ?
safe radishBOT
This was in my answer key: I don't get it. If you pull the negation out through ...
tardy mango
This was in my answer key: I don't get it. If you pull the negation out through ...
De Morgan's law for quantifiers applies when you're trying to push a single negation inside
but here, you're using the implication rule to create a double negation on the outside
slim lion
This was in my answer key: I don't get it. If you pull the negation out through ...
p implies q
is equivalent to
not p or q
they used this ^
glacial cairn
That negation is on the first term of the implication. That implication is getting rewritten, the negation doesn't go through the quantifier
silent basin
But then wouldn't we have something like this?
glacial cairn
That's not what is written
silent basin
Sorry I meant:
slim lion
But then wouldn't we have something like this?
they rewrote the implication as not p or q
imagine we have
not p implies q
that's logically equivalent to saying
p or q
glacial cairn
Let $A(x,y) \equiv \neg N(x,y) \lor \neg P(y)$, then you have $\forall x (\neg \textcolor{orange}{[\forall y(A(x,y))]} \implies P(x))$
flat frigateBOT
Nel
glacial cairn
That orange part is untouched
slim lion
Sorry I meant:
(not forall y...) implies P(x)
not (not forall y...) or P(x)
silent basin
**Nel**
Wouldn't the orange part with the negation in front turn into a Existential quantifier?
glacial cairn
Let me rephrase
Let $B \equiv \forall y(\neg N(x,y) \lor \neg P(y))$ and $C \equiv P(x)$, then you have $$\forall x (\neg B \implies C) \equiv \forall x (B \lor C)$$
flat frigateBOT
Nel
safe radishBOT
safe radishBOT
Available help channel!
cosmic peak
for sequences, high school level! for An = n² - 8 does A₄n = 4n ² - 8?
summer coral
cosmic peak
it seems really easy but im not really sure so i came here
magic junco
for sequences, high school level! for An = n² - 8 does A₄n = 4n ² - 8?
Do you have the original question?
cosmic peak
Do you have the original question?
yes hold on ill have to translate it
magic junco
It’s best to provide the screenshot as well so that we can verify it
cosmic peak
Determine the terms a₄n for the sequence defined by the formula:
the formula being An = n² - 8
magic junco
Is it a$_{4\text{n}}$?
flat frigateBOT
雙目
magic junco
If this is the case, you’re supposed to replace n with 4n entirely
Meaning that 4n should be squared instead of squaring n and multiply it with 4
safe radishBOT
@cosmic peak Has your question been resolved?
magic junco
@cosmic peak Since you reacted ❌ to the bot, is there anything you wanna ask?
cosmic peak
<@1477404522166092047> Since you reacted ❌ to the bot, is there anything you wan...
so does A₄n = 4n ² - 8?
magic junco
(4n)^2-8
magic junco
dw
cosmic peak
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tropic hatch
I have a finite cover of closed intervals for the cantor set. I want to show that this cover covers some Ck level from the construction of the C. How can i show this?
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I have a finite cover of closed intervals for the cantor set. I want to show tha...
@tropic hatch Has your question been resolved?
uneven ice
by Ck level you would have 2^k intervals right?
seach of the same lenght, also dependent on k
maybe use this to show a contradiction, using the fact that each closed Ck has C, which is compact
(im airballing)
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knotty matrix
hello
quiet plume
Have you tried something?
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@knotty matrix Has your question been resolved?
knotty matrix
Have you tried something?
i tried proof by absurd but it didn't work
hollow geode
for the first one
$(1, \sqrt{2}, \sqrt[4]{2})$
we will consider $(a, b, c) \in \mathbb{Q}^3$ :
$$a + b\sqrt{2} + c\sqrt[4]{2} = 0$$
$$a + b\sqrt{2} = -c\sqrt[4]{2}$$
$$(a + b\sqrt{2})^2 = c^2\sqrt{2}$$
$$a^2 + 2b^2 + 2ab\sqrt{2} = c^2\sqrt{2}$$
$$a^2 + 2b^2 = (c^2 - 2ab)\sqrt{2}$$
if $c^2 - 2ab \neq 0$
then $$\sqrt{2} = \frac{a^2 + 2b^2}{c^2 - 2ab}$$
that means $\sqrt{2} \in \mathbb{Q}$ contradiction
$c^2 - 2ab = 0$
$a^2 + 2b^2 = 0$
flat frigateBOT
akeanti💕
hollow geode
the a^2 + 2b^2 = 0 => a=0 and b=0
if u replace in the first equation then c = 0 too
knotty matrix
i see
i didn't think of squaring it
i was trying to make an equality such as a rational=irrationnal but i got stuck
flat frigateBOT
akeanti💕
hollow geode
after some simplification
knotty matrix
and for the second?
hollow geode
**akeanti💕**
$$(2c^3 + a^3 + 6ab^2) + (3a^2b + 2b^3)\sqrt{2} = 0$$
flat frigateBOT
akeanti💕
hollow geode
and u already found on the first question that (1, sqrt(2)) is independent
so both are = to zero
knotty matrix
**akeanti💕**
that's for the second or third
knotty matrix
so both are = to zero
yup
hollow geode
that's for the second or third
third, second looks like it might need more simplification
flat frigateBOT
akeanti💕
hollow geode
$$(a + b\sqrt{2}) + \sqrt{3}(c + d\sqrt{2}) = 0$$
flat frigateBOT
akeanti💕
hollow geode
that will imply
knotty matrix
and square it?
hollow geode
$$\sqrt{3} = - \frac{a + b\sqrt{2}}{c + d\sqrt{2}}$$
flat frigateBOT
akeanti💕
knotty matrix
yup for sure
hollow geode
$$\sqrt{3} = \frac{2bd - ac}{c^2 - 2d^2} + \frac{ad - bc}{c^2 - 2d^2}\sqrt{2}$$
flat frigateBOT
akeanti💕
hollow geode
$u = \frac{2bd - ac}{c^2 - 2d^2}$ avec $u \in \mathbb{Q}$
$v = \frac{ad - bc}{c^2 - 2d^2}$ avec $v \in \mathbb{Q}$
flat frigateBOT
akeanti💕
hollow geode
that will leave us with $$\sqrt{3} = u + v\sqrt{2}$$
flat frigateBOT
akeanti💕
knotty matrix
$$\sqrt{3} = - \frac{(ac - 2bd)+(bc-ad)\sqrt{2}{c - 2d}$$
flat frigateBOT
mdrrai_410
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)
flat frigateBOT
akeanti💕
knotty matrix
ah yeah alr
hollow geode
and u already found on the first question that (1, sqrt(2)) is independent
and u also have this again
knotty matrix
so sqrt 2 is equal to a rationnal
hollow geode
so sqrt 2 is equal to a rationnal
uhhh nope ig u can use smt else
hollow geode
**akeanti💕**
each part here is equal to 0
hollow geode
and u already found on the first question that (1, sqrt(2)) is independent
since u have this one again
according to question 1
$2uv = 0 \implies u = 0$ ou $v = 0$
flat frigateBOT
akeanti💕
knotty matrix
uhhh nope ig u can use smt else
ok but it is it fakse to say that?
knotty matrix
**akeanti💕**
wait you speak french?
hollow geode
yea
hollow geode
**akeanti💕**
@knotty matrix u can take each case if u=0 or v=0
and apply it to the second term
that way u will find the contradiction
ig
knotty matrix
idk man the first seems straight forward
hollow geode
if $u = 0$, $2v^2 = 3 \implies v^2 = \frac{3}{2}$
flat frigateBOT
akeanti💕
flat frigateBOT
akeanti💕
flat frigateBOT
akeanti💕
knotty matrix
ok ok merci
hollow geode
**akeanti💕**
and also $a + b\sqrt{2} = 0$
flat frigateBOT
akeanti💕
hollow geode
and again (1, sqrt(2)) is independant
same as this one, so a=0 and b=0
hollow geode
ok ok merci
did u get what i was trying to do xD
knotty matrix
yup thank you bro
hollow geode
happy i could help ^^
knotty matrix
can you come dm?
hollow geode
sure
knotty matrix
.close
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kind quartz
I dont understand why it is taking the derivative like that. isnt it supposed to be in terms of dp/dt? Why is it just saying the derivative of 50,000p is just 50,000 instead of 50,000 dp/dt?
safe radishBOT
I dont understand why it is taking the derivative like that. isnt it supposed to...
kind quartz
Specifically i set revenue as R = p(rice) * x (quantity) and put x in terms of p, as shown there. so r = 50000p / root(4p +1)
And also when it takes the derivative of 4p+1, it just says 4, but if its with respect to price change why wouldnt it be 4 * dp / dt
and if thats what the outside one is then 1) why is it outside, and 2) why wouldnt it do the same type, for 50000p
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kind quartz
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winged flare
makes no sense
hardy lion
damn i was just about to solve world hunger, but this seems more important
ah well both problems seem unsolvable to me
unfortunate
olive basalt
guys use trigonometry to solve this
winged flare
guys use trigonometry to solve this
where did you get this from
olive basalt
well this question was made by me
hardy lion
well this question has undefined parameters
olive basalt
no solve is easy
r = 3
answer
well if u cant solve this solve this new question it's easy
hardy lion
sorry im kinda in a bad mood tonight
winged flare
you can go to #math-discussion if you want to challenge others
if you dont need help type .close
olive basalt
bruh
hardy lion
the line connected to the square should go to the top of the 6sqrt3 leg
olive basalt
yes
hardy lion
as the image stand right now, you cant solve it because some distances are undefined
you should fix that
olive basalt
wait lemme check
magic junco
sorry im kinda in a bad mood tonight
It’s okay, I take your apology
hardy lion
also please stop testing my and the rest of our abilities, i personally find it annoying
magic junco
jokes aside, you’re supposed to add more details as :/ said
wicked minnow
@everyone willy
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dusky totem
@everyone willy
Bro really tried pinging 315,000 people
wild cape
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hardy lion
wild cape
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void barn
Just wondering, whats the fastest way to do multiplication on paper, i do like simple algorithm rn
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Just wondering, whats the fastest way to do multiplication on paper, i do like s...
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normal moss
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boi
astral glacier
Please don't clog up help channels by texting like this, thanks
lean otter
huh
lean otter
Please don't clog up help channels by texting like this, thanks
wdym
quick crater
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tardy mango
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blissful oyster
I have this Poisson equation in x of the form (u'' + \omega^{2}u = f) with Dirichlet boundary conditions (u(0) = u(a) = 0) where f is some arbitrary function of x. I found the solution to be the following expression (\rightarrow)
safe radishBOT
I have this Poisson equation in x of the form \(u'' + \omega^{2}u = f\) with Dir...
flat frigateBOT
Amoomgus
blissful oyster
In Maple, when I use the diff command, Maple is not recognizing that the integral expressions are functions of x and instead treat them as some constant:
please someone help this is my third time askng this question
worldly spear
is the one in the blue
blissful oyster
The blue are outputs that are directly underneath their corresponding Maple code/script (in black).
blissful oyster
I know that this is the solution -- this is what I arrived at when integrating the arbitrary function f with the Kernal G aka Green's function, which was found with some basic ODE methods. I am just trying to verify that, when you apply L to G where L is the Laplacian, you get back f
and I'm trying to verify this with Maple for obvious reasons -- it is much too tedious to keep doing all this verification/validation by hand
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@blissful oyster Has your question been resolved?
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@blissful oyster Has your question been resolved?
primal bone
made it a pic
If you took a pic - just send the pic
Your text conversion fuxked up massively
blissful oyster
If you took a pic - just send the *pic*
I mean I know what they are saying but it doesn't tell me anything I don't already know. They are literally just restating my answer in a way that sounds like chat GPT to my brain/ears. I'm probably going to let this help section close because it's been over a day with no success
primal bone
yh, I did think it felt like some AI got in the mix
@blissful oyster Consider asking your question in some place dedicated for Maple, perhaps?
(idk if Maple has a Discord server, but I'd try looking there, or potentially on Reddit)
blissful oyster
yeahhh I tried looking in the Discord server list and via Google, but nothing for Maple specifically showed up
Honestly, I think I'd have better results just soldiering through Maple documentation and testing more code myself haha
primal bone
This is a general thing I'd recommend for coding; it might work in this case
Check if you get the same issue in much smaller cases, e.g. with writing smaller functions and see if Maple throws up errors
And then build it back up to the function you're trying to write
If you find a hitch, that's what you should seek to address
If writing a small function still throws an error, then it's likely you're not writing the function in a recognisable way
blissful oyster
Yup!
I created a test function g(x) defined in terms of some arbitrary definite integral and got exactly the output I was hoping for in the simple case
I guess I'll slowly build up from there back to u and see where the disconnect happens... I'll come back here when that happens
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crisp hull
When we check whether a function is defined can't we check by simplifying it first?
safe radishBOT
When we check whether a function is defined can't we check by simplifying it fir...
crisp hull
Suppose f of x equals xsq - 1 / x-1
At x equals 1
It's undefined can't we simplify first?
quiet plume
Even if you simplify stuff, the original expression remains undefined.
crisp hull
Even if you simplify stuff, the original expression remains undefined.
Open the formula above
X - 1 cancels out
It becomes 2
Isn't it defined
quiet plume
What I mean is even if you simplify and it becomes "defined"
The original expression at that point has the form 0/0
That's not defined.
late thorn
you cant simplify if they are zeros so you need to mention that they are not zeros before simplify
worldly lantern
It becomes 2
have you been studying a lot of limits lately
mint plume
when you simplify stuff like that, u also add the condition x≠1 for example
crisp hull
have you been studying a lot of limits lately
Continuity
quiet plume
That's not defined.
So like $f(x) = \frac{x-2}{x-2}$, clearly whenever $x\ne 2$, $f(x) = 1$, but $f(2)$ isn't defined
flat frigateBOT
Azyrashacorki
crisp hull
The original expression at that point has the form 0/0
Alr
earnest nacelle
X - 1 cancels out
yes, but recall what you are doing by cancellation
when you cancel common factors, you are dividing through the fraction by the same common factor
but at x = 1 in your example, you are dividing through by 0
which is obviously not allowed, hence that restriction remains even after you simplify
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worldly lantern
can you explain well ordering principle to me like im a newborn
safe radishBOT
can you explain well ordering principle to me like im a newborn
main mural
goo goo gah gah
no but are you referring to the well ordering on natural numbers or in general the set theoretic one?
proud tree
goo goo gah gah
this was very simple to follow, thank you!
cedar badger
Basically, suppose you have some natural numbers(can be infinite)
then there exists smallest number in that set of numbers
worldly lantern
how does that help
i already knew that if i have a collection of things , there will be a thing that is the smalles
what implication does this principle have
main mural
for natural numbers this is pretty obvious yes
but being able to claim this for any set out there requires axiom of choice
i might ask for some context
safe radishBOT
Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.
quiet plume
what implication does this principle have
It doesn't mean much more than what you think it does for natural numbers.
It can be useful because if you know you have some set of natural numbers with some property, you're guaranteed there's a LEAST one with that property. That's not true in general for sets with any order, like Z, Q or R.
Also as they quickly mention, it's actually equivalent to mathematical induction, which itself is a nice consequence.
quiet plume
Hum
quiet plume
One can prove that they imply each other.
The main idea is that if you have a set S like in induction and it's not N, then N \ S is nonempty and would have a least element. You can show that this would contradict the fact that N\S has a least element, so N\S must be empty, i.e. S = N. This shows WOP => induction at least
quiet plume
It's not exactly simple
quiet plume
Yeah it might be good to just take it as a fact that you get induction for free with WOP
quiet plume
Eventually I think it makes sense intuitively that the WOP has at least something to do with induction and vice versa
worldly lantern
where he tells something about WOP
quiet plume
Prob on the axiom of choice
worldly lantern
yeahh
quiet plume
Right well in natural numbers it's sort of intuitive that it should be well-ordered.
In general, given any set, you might want to (1) give an order to that set, (2) see if you can choose that order such that it is a well-order (has the same property as the WOP says N has).
Sets like Z and R with the usual order we have on them aren't well ordered for instance
quiet plume
Yeah that's not really it
But essentially if you accept what is called the Axiom of Choice (which people have good reasons to accept or reject), then you get what's called (NOT Zorn's Lemma) Well Ordering Theorem, which says that for ANY set there is an order which is a well-order on that set.
quiet plume
Well it doesn't say that any order is a well-order
For instance, a well-ordering of R would definitely not look like the usual ordering we have of R
But it would still be a valid order and that theorem says that it should exist.
quiet plume
Yes that's another consequence of the axiom of choice
worldly lantern
so the only problem with well ordering is infinity
quiet plume
Well in general for any finite set it's quite easy to order it
So yes the "issue" is with ordering infinite sets
worldly lantern
so the crux is that , we can assume that an infinite set has a well order , we dont know what it is
quiet plume
Yeah
worldly lantern
but we cant apply induction to the set of real numbers right
quiet plume
The thing is the axiom of choice has some very odd consequences like the Banach Tarski Paradox, but the Well-ordering theorem is quite useful in that it has some natural consequences, like that every vector space has a basis, which seems lik it should be true.
Heck on its own choice seems like it's quite intuitive
quiet plume
but we cant apply induction to the set of real numbers right
I think that you can define something akin to induction on any well-ordered set. So technically you could but again, the order wouldn't be workable
worldly lantern
understandable
this is why i love number theory
no real numbers in sight
thanks again
.close
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graceful lichen
_que : _ Find the sum of the solution of the equation |x|^3 - 15x^2 - 8|x| - 11 = 0
safe radishBOT
_que : _ Find the sum of the solution of the equation |x|^3 - 15x^2 - 8|x| - 11 ...
graceful lichen
my method is to let
t = |x| => (t^3 - 15t^2 - 8t - 11 = 0)
further this cubic isn't solveable
graceful lichen
the hit and try doesn't work
worldly lantern
in different regions
winged flare
my method is to let
t = |x| => (t^3 - 15t^2 - 8t - 11 = 0)
if $t=\alpha$ then $x = \pm \alpha$
worldly lantern
solve once while taking domain of fxn as R+
flat frigateBOT
hasty wagon
Vieta's ?
worldly lantern
and once by R-
hasty wagon
due to the mod
ah
worldly lantern
there are extra solutions
winged flare
but now i wonder if the solution is supposed to be trivial like that
worldly lantern
but now i wonder if the solution is supposed to be trivial like that
no
acoustic timber
my method is to let
t = |x| => (t^3 - 15t^2 - 8t - 11 = 0)
sum of solutions of t will be = -b/a = 15, maybe you can get somewhere with that
worldly lantern
we cant use this
winged flare
from here you see that the sum of roots for every value of t is 0
graceful lichen
from here you see that the sum of roots for every value of t is 0
then answer should be zero
graceful lichen
checking....
worldly lantern
is it 15?
worldly lantern
oh right
i fumbled
ans should be 0
the sum of solutions in negative domain is coming +ve and vice versa , which is not possible
graceful lichen
i do this step every time |x| = t then x = +- t but didn't notice we can say the sum like this
worldly lantern
i do this step every time |x| = t then x = +- t but didn't notice we can say the...
well you cant do this everytime i suppose
glacial cairn
You certainly can do it for (i), and you can do a very similar thing for (ii) and (iv), though (iv) is very easy anyway
lean otter
this occurs when the function is syymmetric ig
graceful lichen
i solved (i) and (iv) th by procedure
winged flare
(ii) seems to use the same idea with a bit of thinking
graceful lichen
yes there is manipulation in (ii)
i got t = [-1 +- sqrt(45)]/2 in (ii)
after that looking at solution i got to know some basic manipulation
glacial cairn
i got t = [-1 +- sqrt(45)]/2 in (ii)
You don't need to do any of that
graceful lichen
they let t = |x-3|
now |x-3| = [-1 +- sqrt(45)]/2
and then let y = |x-3|
so y = +- (x-3)
y = x -3 and y = 3 -x
so x = y +3 and x = 3-y and sum is 6
graceful lichen
You don't need to do any of that
how would you do it then
worldly lantern
write ( x-3)² as |x-3|² since its positive and let |x-3| =t
it will give you 4 soln
burnt pine
we doin ts with or without calc
glacial cairn
f(x) = (x-3)^2 + |x-3| - 11, consider f(x+3)
burnt pine
js making sure
worldly lantern
without
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whole void
who will solve will get my autograph
safe radishBOT
who will solve will get my autograph
quiet plume
Not how this works. You can try to solve it and people will help you along the way
!status
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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
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west venture
how do I logically solve this problem: find for which integers (2024+n)/n is a perfect square
safe radishBOT
how do I logically solve this problem: find for which integers (2024+n)/n is a p...
glacial cairn
2024 only has 14 divisors (not 1 or 2024) so you might as well just check them all
solar hazel
put 2024/n = k^2 - 1. 2024 does not have very many divisors
glacial cairn
(also n = -2024 is a solution if n is not given positive)
west venture
put 2024/n = k^2 - 1. 2024 does not have very many divisors
why k^2-1
solar hazel
equivalent to (2024+n)/n = k^2
west venture
yes
but it’s an odd number in that way isn’t it
I also don’t get why should I consider the divisors
solar hazel
if you put 2024 = n(k^2-1), then n is a divisor of 2024
so this is a pretty strong necessary condition on n
glacial cairn
I also don’t get why should I consider the divisors
(2024+n)/n = 2024/n + 1, so if n does not divide 2024, you don't even get an integer
solar hazel
so this is a pretty strong necessary condition on n
it makes the problem very finite
you can just check the finitely many divisors of 2024
west venture
(2024+n)/n = 2024/n + 1, so if n does not divide 2024, you don't even get an int...
that is true
brave wolf
If for some reason you wanna avoid checking 14 divisors, you could also do it like this:
2024/n = k^2 - 1 = (k-1)(k+1)
So you need a divisor of 2024 = 2^3 * 11 * 23 which can be factored as (k-1)(k+1). Those have same parity though, so either both factors are even, or both odd. They cant be both odd, because then it would have to be 11 and 23, which clearly doesnt work. So they must be both even, so
k-1 = 2 * _
k+1 = 2 * _
Now you can add 2, 11, 23
If we add only the 2, it works. So that's one solution, k-1 = 2; k+1 = 2*2
Now if we add one of 11 or 23, we must add both. Because otherwise the numbers would get too big to balance out
So we do
k-1 = 2 * 11 * _
k+1 = 2 * 23 * _
Now only 2 remains, and if we add it to k-1, we get k-1 = 44, k+1 = 46 as our last sol.
Now to actually reconstruct n from those, remember that
2024/n = (k-1)(k+1)
west venture
if you put 2024 = n(k^2-1), then n is a divisor of 2024
but shouldn’t I deduce it rather than putting it as an hypothesis?
solar hazel
sure? i’m not sure what putting it as a hypothesis even means though
glacial cairn
"which integers (2024+n)/n is a perfect square" directly translates to (2024+n)/n = k^2 where n and k are integers
west venture
yes
glacial cairn
That's not really a hypothesis
glacial cairn
You're fine as long as you don't transmit yourself
solar hazel
west venture
Can I solve this without using any system or solving any forth grade equation?
burnt notch
I guess so
safe radishBOT
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brave wolf
Can I solve this without using any system or solving any forth grade equation?
probably with a lot of terrible algebra
let r+s = p, rs = q
(r+s)^2 = r^2 + s^2 + 2q
(r^2 + s^2)^2 = r^4 + s^4 + 2q^2
so uhh
((r+s)^2 - 2q)^2 = r^4 + s^4 + 2q^2
(1 - 2q)^2 = 7 + 2q^2
now solve for q I suppose, it should be only qudratic