Prathmesh
#help-36
soft zealotBOT
dreamy warren
you used $dt/dx = -(1/v^2)(dv/dx)$ right
soft zealotBOT
anflo
dreamy warren
i think you have an extra v^2 on the left
final saddleBOT
@boreal smelt Has your question been resolved?
boreal smelt
you used $dt/dx = -(1/v^2)(dv/dx)$ right
yea
boreal smelt
i think you have an extra v^2 on the left
yea i did a mistake in differentiating
final saddleBOT
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warm python
I would like to fnd the fixed point(s) of $f(x)$. Just want to make sure the way I'm doing it is right.
Let $t \in R$ be a fixed point. Then $f(t)=t$. SO $t= \sqrt[3]{2x^3-x^2-x}$. We thus must solve $x^3=2x^3-x^2-x$ to find the fixed point. Then $x^3-x^2-x=0$ . The fixed points are the roots of this cubic.
final saddleBOT
I would like to fnd the fixed point(s) of $f(x)$. Just want to make sure the way...
soft zealotBOT
waimas
dreamy warren
yes the method looks correct
desert mantle
except for the typos
robust mulch
match variables
final saddleBOT
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torn heart
\int_{-\infty}^{\infty} e^{-t^2/2} dt = \sqrt{2\pi}
why
final saddleBOT
\int_{-\infty}^{\infty} e^{-t^2/2} dt = \sqrt{2\pi}
why
torn heart
$\int_{-\infty}^{\infty} e^{-t^2/2} dt = \sqrt{2\pi}$
soft zealotBOT
Chilll
ionic venture
What have you tried so far?
signal vector
Use the result of this integral
Oh, your screenshot appears to be from the same page
ionic venture
Gauss hiding behind 1.414
torn heart
Oh, your screenshot appears to be from the same page
yeah
opal plinth
That's not the same expression, but you can just make a change of variable
ionic venture
After working out the integral, you would need to figure out how you could turn..
torn heart
That's not the same expression, but you can just make a change of variable
yeah but same
ionic venture
$e^{-\frac{t^2}{2}}$ to $e^{-x^2}$
torn heart
it gives sqrt(2pi)
soft zealotBOT
Erebus
signal vector
Not much to figure out there
ionic venture
Not much to figure out there
@signal vector gives
final saddleBOT
@torn heart Has your question been resolved?
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tacit dust
Let circle (O) have diameter AB = 2R, and chord CD be perpendicular to OA at the midpoint M of OA. The tangent at C of (O) intersects OA at N. Draw a circle with center D and radius DM intersecting (O) at E and F. Draw diameter DP of (O), DP intersects BC at I and EF at H. Let K be the intersection of EF and CD. Calculate DK in terms of R.
final saddleBOT
Let circle (O) have diameter AB = 2R, and chord CD be perpendicular to OA at the...
soft zealotBOT
tacit dust
hmm, vietnamese i see
any progress?
leaden moon
any progress?
i read through your proof and ill just ask a question: How did you get $DM = \frac{R\sqrt{3}}{2}$?
soft zealotBOT
1 divided by 0 equals Infinity
final saddleBOT
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tacit dust
giờ thì đi kết luận là xong =)))
cx ng việt à 🙂
tacit dust
thêm 1 ng việt trên đây 🙂
final saddleBOT
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tacit dust
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sharp wing
how do u solve part b
final saddleBOT
how do u solve part b
final saddleBOT
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bleak notch
Might be a strange question, but why does observing that we have drawn a white ball increase the probability that the ball that was transferred was white? I get that if we do the math with conditionals, we see that there is a dependence, but I don't really get intuitively how it "affects" the outcome. There's nothing saying that x% of the time, a white ball is drawn, which would tell us something substantial, but in my mind this is a single observation telling us a lot.
final saddleBOT
Might be a strange question, but why does observing that we have drawn a white b...
whole halo
Might be a strange question, but why does observing that we have drawn a white b...
the bayes part of this really screws it up
what Im thinking is theres ways to explain the bayes formula or to make it plain obvious in other examples
maybe you look up a few videos, then try and use their words on this problem
Im just thinking if the ball was black, theres no way transferring a white ball is going to help the odds
tired walrus
or maybe you could draw a tree diagram for it idk
whole halo
but if the ball is white...
bleak notch
I can visualise the tree diagram and I (hope) understand Baye's formula, but I can't seem to accept that simply observing something once "increases" the odds
whole halo
...why is my drawing program drawing different thicknesses for differenc olors
bleak notch
there would be flying cars in 2025 they said
whole halo
bleak notch
I'm not really sure if I'm getting my question across well
bleak notch
this is basically how I solved the question
but something that bothers me is that observing something afterwards "affects" the probability in retrospect
if that makes sense
whole halo
oh thats all?
if it didnt, itd be 50%
if it did, itd not be 50%
thats it
you can see theres more white balls available in one of the options
so when you choose a white ball, its more likely its the top one
and so its more likely that a white ball was transferred than a black ball
indirectly because a black ball leads to fewer of the outcomes
and also indirectly because a white ball leads to more likely outcomes than a black ball
bleak notch
if it didnt, itd be 50%
at the point of transferring, shouldn't P(transferred ball is white) be 2/3, since whether a white or black ball is observed in the future doesn't affect the present?
whole halo
thats not true
if you throw a box off a cliff then you find out there wasnt a cat in the box at the time of throwing, that tells you there wasnt a cat when it was thrown off the cliff
that's shrodinger's cat experiment (the two dots above the o fell off the cliff which is the reason for the variant youre reading rn)
similarly you just found out the ball is white
youre saying that tells you nothing about the state of transferring the ball?
bleak notch
so in some sense, it's correct to say that the future affects the present?
whole halo
inuka use that common sense
the present lets us know about the past
thats called history
you did history pulling a white ball and then thinking about what mustve happened to let that happen
if youre pulling balls out, thats not the future either, thats the present
whole halo
if you throw a box off a cliff then you find out there wasnt a cat in the box at...
you can also just shift all the verbs to be future and present instead of present and past
same argument works there too
bleak notch
youre saying that tells you _nothing_ about the state of transferring the ball?
in my mind, even this doesn't feel true when it's just one observation
whole halo
..as opposed to how many observations?
bleak notch
that's interesting yeah
whole halo
any observeration is usable information
bleak notch
if we took 10 observations with replacement then it seems like it tells us something
but then if you just reduce the amount
surely 1 tells you something
that's very strange to me
whole halo
then what was even the point here
we cant get much out of 1 ball
but its bold to say you get nothing
not even a little bit of information? we saw the ball
bleak notch
ok that's fair
whole halo
you can split up the cases like this
bleak notch
since we're factoring that information into our calculation, surely that means we can quantify how much information a single observation gives us
whole halo
oh thats not good in dark mode
I dont know entropy so Im not answering that
as for this,
if upon seeing a white ball you guess that a white transfer mustve happened, you have a 4/5 chance of getting it right
this is because in two of the three scenarios, you were given more white balls to infer such
a white transfer directly boosts the chances to have picked a white ball
and so the chance changes, and moreover gets higher
bleak notch
oh interesting, so it's something like the sub-state-space that we select from is larger
whole halo
if on the other hand any white transfers never ended up in the second urn,
history is repeating itself, put that background back
you can see youd still have the usual 2/3 chances
in all of the scenarios, the white balls in the 2nd urn are the same before and after, so you gain no useful information had you pulled out a white ball
bleak notch
ok that's kinda interesting, I'll play around with this for a bit
thanks for the help
whole halo
np
bleak notch
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final saddleBOT
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barren pebble
H and P_A are the orthocenter and A-Humpty point of ∆ABC then H,P_A,B,C are concyclic. Why?
final saddleBOT
H and P_A are the orthocenter and A-Humpty point of ∆ABC then H,P_A,B,C are conc...
tranquil pine
H and P_A are the orthocenter and A-Humpty point of ∆ABC then H,P_A,B,C are conc...
what definition of humpty do u use
(or what equivalent facts do you know about it)
barren pebble
what definition of humpty do u use
∠P_AAB=∠P_ABC, ∠P_AAC=∠P_ACB
barren pebble
(or what equivalent facts do you know about it)
It's the intersections of circles that go through A and touch BC at B,C
barren pebble
I know the definition of that
Through some weird (kinda famous) proof of Ptolemy's theorem
But i'll prefer not to use it
tranquil pine
hmm ok well yeah actually we don't need that
lol
because you can just angle chase
Do you know <BHC in terms of the angles of triangle ABC?
barren pebble
∠BHC = 180°- ∠HBC - ∠HCB=180°-(90°- ∠B)-(90°- ∠C) = ∠B + ∠C
final saddleBOT
tranquil pine
lol yw
the reason I mentioned inversion was because a lot of properties about the humpty (and Dumpty) point are very easy to prove with inversion
its not needed in this one but it makes other properties much easier to prove
so when you learn inversion you can try to come back to this config and prove other properties with it
barren pebble
Thanks
Today is the first day i learn about those points though, i guess i'll stick to the basics
final saddleBOT
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lunar cairn
"Hi everyone! I'm from the Philippines and I'm aiming to win the Division Math Celebration Quiz Bee for senior high school. I'm in Grade 11. I think the competition will focus on statistics what i would do
final saddleBOT
"Hi everyone! I'm from the Philippines and I'm aiming to win the Division Math C...
chilly pollen
"Hi everyone! I'm from the Philippines and I'm aiming to win the Division Math C...
What do you need help with
leaden moon
you got any revision or smth like that?
final saddleBOT
@lunar cairn Has your question been resolved?
final saddleBOT
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lunar cairn
<@1377151609573740594> Has your question been resolved?
Not yet
light belfry
final saddleBOT
light belfry
Ya I am pretty sure this problem is nonsense
Cuz AEC is essentially AEB
currently trynna see this problem assuming BE is not a straight line
(But yea assumign that how would I know which one is straight and which one is curved
next thorn
Cuz AEC is essentially AEB
are you this sure this is true?
light belfry
~~are you this sure this is true?~~
Hmmm what do you want me to do find solid angle?
hearty tide
I see typo
next thorn
yeah I dont think any of the lines are curved here
next thorn
I see typo
it's just this
hearty tide
Let's try for ACB and check if any option matches, ig.
light belfry
Let's try for ACB and check if any option matches, ig.
Now what does that mean
hearty tide
Angle ACB
torn ore
How to find this limit ?
final saddleBOT
How to find this limit ?
torn ore
I know that the product of the dominant terms on both nominator and denominator[(2n^3)(3n^5)/(2n^6)(3n^2)] will decide the result .
blazing nymph
Well as n tends to inf, the +1 and -1 will become insignificatn so how else can u write the limit?
yep
final tangle
careful with ()
torn ore
How do I factor out the (2n) if there is an exponent, lets say top left
full jungle
How do I factor out the (2n) if there is an exponent, lets say top left
i think factoring just n will do the trick
torn ore
if I had like (2n-1) I would just "do" n(2-1/n), but how do I "get it out" when there's the exponent?
robust mulch
(2n-1)^3=(n(2-1/n))^3
full jungle
$\left(2n-1\right)^{3}\ =\ n^{3}\left(2-\frac{1}{n}\right)^{3}$
torn ore
oh, i jst keep it like that 
soft zealotBOT
Xerxes
torn ore
I see. thanks
full jungle
why dont we divide by n^8
its the same thing if we take n common from all the terms
next thorn
its the same thing if we take n common from all the terms
I see.
torn ore
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final saddleBOT
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warm python
I don't really get this
final saddleBOT
I don't really get this
warm python
The justification that is
warm python
The value of the parameter that maximises the joint dit
scarlet sequoia
so the value that maximizes L(theta)
L(theta) only has two values, 0 and (1/2)^n
since (1/2)^n is the max
any value of theta that gives you L(theta) = 1/2^n is an MLE
left echo
Also, @scarlet sequoia before the channel shut in our faces, you mentioned including an indicator function, but wouldn't that change nothing since the indicator function is the same every time?
warm python
I didn't mean to close the channel rudly last time , sorry
scarlet sequoia
Also, <@246787399991033858> before the channel shut in our faces, you mentioned ...
the indicator function is {2theta <= x_i <= 3theta}
so intersection is {2theta <= min(x_i) <= max(x_i) <= 3theta}
blazing kindle
In an equilateral triangle ([ABC]) with side length (\ell), the value of ((\vec{AB} + \vec{AC}) \cdot \vec{BC}) is:
(A) (\ell^2)
(B) (\sqrt{3},\ell^2)
(C) (\dfrac{\sqrt{3}}{2},\ell^2)
(D) 0The exercise does not include any image. I was looking at the solution, but I don’t understand why (\cos 60^\circ) and (\cos 120^\circ) appear, and why we are multiplying by them.
final saddleBOT
> **In an equilateral triangle ([ABC]) with side length (\ell), the value of ((\...
tired walrus
for vectors in general,
u • v = |u| * |v| * cos(theta)
where theta is the angle between u and v
@blazing kindle
blazing kindle
<@1158823031137906780>
i know we use cos but why 120 degrees and 60? in the exercise just saying equilateral triangle dont show any imagem or saying addictional info
tired walrus
in an equilateral triangle the interior angles are always each 60°
the angle between vectors AB and BC is 180 minus that
blazing kindle
in an **equilateral** triangle the interior angles are always each 60°
oh
i see
tysm
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final saddleBOT
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torn dome
prove that r is irrational if r^3 + r + 1 = 0
final saddleBOT
prove that r is irrational if r^3 + r + 1 = 0
soft zealotBOT
Sean [Ping On Reply Please!]
leaden moon
step 1: solve the cubic
torn dome
what now
leaden moon
step 1: solve the cubic
torn dome
i am a nub idk if that was sarcasm
leaden moon
**Sean [Ping On Reply Please!]**
anyways $\frac{p^3}{q^3} + \frac{p}{q} + 1 = 0$ right?
soft zealotBOT
1 divided by 0 equals Infinity
tired walrus
now $r^3+r+1 = \frac{p^3}{q^3} + \frac{p}{q} + 1$
p^3/q^3 + p/q + 1 = 0. how can you make this eq less scary
torn dome
yeah
torn dome
p^3/q^3 + p/q + 1 = 0. how can you make this eq less scary
put back r = p/q
leaden moon
put back r = p/q
congrats, you're back to square 1
tired walrus
why 1 on rhs
leaden moon
ok so $\frac{p^3q + pq^3 + q^4}{q^4} =0$
don't put spaces between $
tired walrus
also you shouldnt have put a space after the first dollar
soft zealotBOT
Sean [Ping On Reply Please!]
tired walrus
you could have used q^3 as your denom
leaden moon
having the denominator is scary
tired walrus
p^3 + pq^2 + q^3 = 0
torn dome
$p^3 + pq^2 + q^3 = 0$
soft zealotBOT
Sean [Ping On Reply Please!]
torn dome
now do i have to solve this
leaden moon
for strongness
idk what to call that
$p \in \mathbb Z$, $q \in \mathbb N$, $(p, q) = 1$
soft zealotBOT
1 divided by 0 equals Infinity
tired walrus
you don't have to "solve" this per se
torn dome
**Sean [Ping On Reply Please!]**
then do i have to prove that this is never 0
tired walrus
also yes it will be good to start off with r=p/q in simplest form
leaden moon
**1 divided by 0 equals Infinity**
which is this btw
torn dome
**1 divided by 0 equals Infinity**
is that the gcd (p,q) ?
tired walrus
this eq rearranges to q^3 = p(-p^2 - q^2)
leaden moon
is that the gcd (p,q) ?
yea
torn dome
this eq rearranges to `q^3 = p(-p^2 - q^2)`
$q^3 = p(-p^2 - q^2)$
soft zealotBOT
Sean [Ping On Reply Please!]
leaden moon
🔥
torn dome
contradiction?
soft zealotBOT
1 divided by 0 equals Infinity
torn dome
then what
leaden moon
but if you substitute back $p = q = 1$ then that does not satisfy the equation
soft zealotBOT
1 divided by 0 equals Infinity
leaden moon
$p = -1$, $q = 1$ also satisfies this, but try to plug back into the original equation
torn dome
**Sean [Ping On Reply Please!]**
this equation?
soft zealotBOT
1 divided by 0 equals Infinity
leaden moon
prove that r is irrational if r^3 + r + 1 = 0
like literally your question
torn dome
so are we done
leaden moon
yeah that completes the proof
torn dome
so if (p,q) = 1 then p not | q?
leaden moon
and except p = -1, q = 1
and except p = 1, q = -1
then it's true for all the remaining cases
torn dome
can we do this by breaking into cases
cases such as p is even and q is odd
and p is odd q is even
then both odd, etc
torn dome
yea
leaden moon
then it's true for all the remaining cases
apart from $\abs{p} = \abs{q} = 1$
soft zealotBOT
1 divided by 0 equals Infinity
leaden moon
so if (p,q) = 1 then p not | q?
then this statement is true
leaden moon
**1 divided by 0 equals Infinity**
so that's why we gotta consider the cases of $p$ and $q$ being $1$ or $-1$
soft zealotBOT
1 divided by 0 equals Infinity
leaden moon
which i believe if you try to substitute in your $r^3 + r + 1 = 0$ then it does not satisfy
soft zealotBOT
1 divided by 0 equals Infinity
bold turtle
You've essentially assumed that p/q is your answer, already simplified
Towards the end of the algebra stuff you've concluded that p and q are from {1, -1}
But then that makes your answer either 1 or minus 1; but you can easily check that these are not solutions
leaden moon
,w solve r^3 + r + 1 = 0
soft zealotBOT
leaden moon
proof done
torn dome
give me another proof question like this
leaden moon
ooh
bold turtle
prove that sqrt(2) is irrational
leaden moon
~~prove that sqrt(2) is irrational~~
nah that's too simple
torn dome
~~prove that sqrt(2) is irrational~~
what
thats the first prove they teach in a discrete math class
leaden moon
Prove that $\forall k \in \mathbb N$ that is not a square number then $\sqrt{k}$ is always irrational
soft zealotBOT
1 divided by 0 equals Infinity
leaden moon
how about now
torn dome
(idk how to do it)
leaden moon
~~prove that sqrt(2) is irrational~~
$\sqrt2$ too simple
soft zealotBOT
1 divided by 0 equals Infinity
leaden moon
i will use contraposition
contradiction you mean?
torn dome
no contraposition
leaden moon
ahhhhhhhhh
bold turtle
$\sqrt2$ too simple
I mean in the sense that this is the same type of question
leaden moon
I mean in the sense that this is the same type of question
yea
bold turtle
Assume the opposite i.e. that your answer is rational; arrive at a contradiction
torn dome
root k is always rational for all k belong to N that are squares
bold turtle
Starting with that, and then showing sqrt(p) is irrational for any p prime, is good practice
torn dome
if root k is rational then k is a square
leaden moon
if root k is rational then k is a square
hm
torn dome
root k = p/q
leaden moon
so you arrived at a contradiction that fast
torn dome
k = p^2 / q^2
torn dome
k = (p/q)^2
leaden moon
so $k = p^2$
soft zealotBOT
1 divided by 0 equals Infinity
leaden moon
aight i see
torn dome
is this valid
torn dome
wat
bold turtle
so $k = p^2$
that's too fast
leaden moon
that's too fast <:KEK:586240877358350341>
torn dome
that's too fast <:KEK:586240877358350341>
is my proof valid?
leaden moon
yeah enough of the roots
torn dome
proof by contraposition
bold turtle
so k is a square
No, because of this line
"k is a square" is shorthand for "k is the square of some INTEGER"
But all you've said is that "k is the square of some RATIONAL p/q"
You've not shown that q must be 1
bold turtle
proof by contraposition
It is also, crucially, NOT a contraposition
proper dagger
there is a rather nifty argument using parity
leaden moon
that's why the $\gcd(p, q) = 1$ assumption is so important
soft zealotBOT
1 divided by 0 equals Infinity
proper dagger
but I would rather not spring that on OP atm
leaden moon
It is also, crucially, NOT a contraposition
it's called ||contradiction||
bold turtle
Contraposition is when you attempt to prove
[ Statement A is true ] implies [ Statement B is true ]
by demonstrating that
[ Statement B is false ] implies [ Statement A is false ]
torn dome
how is it called contradiction, i took the contrapositive of the original conditional
then i proved the contrapositive
leaden moon
Contradiction is when you want to prove the opposite is wrong
torn dome
i proved that the contrapositive is true
bold turtle
then i proved the contrapositive
Because that's not what contrapositive is.
leaden moon
i proved that the contrapositive is true
Contradiction is that you prove from what you have to prove statement B is impossible to happen
what you did here is contradiction
torn dome
**Because that's not what contrapositive is.**
the question:
is k belong to N is not a square then root k is irrational
contrapositive:
if root k is rational then k is a square
leaden moon
the question:
is k belong to N is not a square then root k is irrational
contrap...
but k is not a square
it's a contradiction
bold turtle
the question:
is k belong to N is not a square then root k is irrational
contrap...
ContraDICTION is a type of contrapositive which goes as follows:
[ the universe makes sense ] implies that [ A is true ]
Taking the contrapositive of THIS is what contradiction is:
[ A is false ] implies [ there is something wrong in the universe ]
torn dome
so k is a square
bold turtle
the question:
is k belong to N is not a square then root k is irrational
contrap...
If root k is rational... Okay, now you have to show that this fraction is over ONE
You HAVE NOT done that yet.
torn dome
If root k is rational... Okay, now you have to show that this fraction is over O...
what is over one
proper dagger
i proved it to you by k = (p/q)^2
you did not show that p/q is an integer.
leaden moon
what is over one
$\frac{A}{1}$
soft zealotBOT
1 divided by 0 equals Infinity
torn dome
you did not show that p/q is an integer.
p and q are ints, p/q is not
leaden moon
p and q are ints, p/q is not
yes
torn dome
(p,q) = 1
proper dagger
if p/q is not an integer, then how can k be a perfect square?
leaden moon
p/q is only a rational number
bold turtle
p and q are ints, p/q is not
Yes, but p OVER q IS NOT AN INTEGER, IT IS A FRACTION.
leaden moon
ragebait ahhhhh
bold turtle
am I using non-standard terminology or sth wtf
leaden moon
~~am I using non-standard terminology or sth wtf~~
i think you got ragebaited
proper dagger
the whole idea of the proof is that if k is a perfect square, it should be the square of some integer
not a rational
torn dome
~~am I using non-standard terminology or sth wtf~~
im jus slow sorry
proper dagger
unless the rational happens to be an integer, but not the case here.
at least you've not shown that to be the case
leaden moon
im jus slow sorry
ill have an easier question for you
bold turtle
Yeah, so, you CAN solve this by proving
root(k) is rational => k is a square (of an integer)
but this isn't done
leaden moon
prove that an irrational number + a rational number always results to an irrational number
torn dome
how can k be a square
bold turtle
prove that an irrational number + a rational number always results to an irratio...
Start with THIS one first
torn dome
its a fraction
bold turtle
You need practice at figuring out what it means to say something is rational or not
umbral steppe
the contrapositive method is really not the best method to attack that particular problem.
bold turtle
So start with 1/0=inf's question
umbral steppe
it is very hard to show that q = 1.
torn dome
prove that an irrational number + a rational number always results to an irratio...
ok i assume that rational + irrational is rational
then i give a counter example of root 2 and 1
leaden moon
So start with 1/0=inf's question
||why does everyone call me that lol||
leaden moon
||i go by Infinity||
torn dome
You can’t prove things by examples
i am disproving a statement using counter example
ok i assume that rational + irrational is rational
i am disproving my assumption
bold turtle
then i give a counter example of root 2 and 1
Okay, even if you tried that (and this doesn't really work here, because you're trying to prove this FOR ALL config.s); how do you KNOW that root(2) + 1 is irrational?
torn dome
Okay, even if you tried that (and this doesn't really work here, because you're ...
oh
right
umbral steppe
Sean, have you learnt about quantifiers?
bold turtle
You disprove it by a counterexample
Nah, this could work tho actually; he's disproving the negative of the desired statement, so...
torn dome
Sean, have you learnt about **quantifiers**?
yea
how do i express an irrational number?
umbral steppe
be aware of the quantifiers of the statement you want to prove.
torn dome
rational i can do with p/q
bold turtle
Those aren't quantifiers...
silver dew
You are showing any rational x, any irrational y, x+y is irrational. If you want to disprove that, logically you can try to find a rational x and a irrational y such that x+y is rational (though it will be impossible) and you didn’t achieve that
umbral steppe
you can only use an arbitrary counterexample if your quantifier is $\exists$, not $\forall$. $\forall$ requires a stronger argument.
soft zealotBOT
Lute
torn dome
alright im stuck
leaden moon
well, since one of them is rational
torn dome
where to go from a + b = p/q
leaden moon
assume $a$ is rational
soft zealotBOT
1 divided by 0 equals Infinity
bold turtle
Contradiction because...?
umbral steppe
may I know if you're doing discrete math, number theory, analysis, or something else?
torn dome
Contradiction because...?
i assumed that b is irrational
torn dome
may I know if you're doing discrete math, number theory, analysis, or something ...
discrete mathematics freshman year CS class
torn dome
This reads like you're assuming _a_ is irrational, not b
i assume a is rational and b is irrational
umbral steppe
i assumed that b is irrational
so what about b lets you express it as m/n? what is m and n? you'll have to state them clearly in your proof.
silver dew
At least you are warmer now.
You can assume there exists rational x, and irrational y, such that x+y is rational, then … … contradiction.
What you did originally was you assumed x=1 and y=sqrt(2), which doesn’t make sense
leaden moon
i miss the xmas icons
torn dome
a + b = p/q
r/s + b = p/q
b = p/q - r/s
bold turtle
I can't prove to you that I've bought apples without showing you the apples I've brought
leaden moon
a + b = p/q
r/s + b = p/q
b = p/q - r/s
now what
torn dome
take the lcm or smth
bold turtle
Show that p/q - r/s is rational
torn dome
b - ps - rq)/rs
torn dome
Show that p/q - r/s is **rational**
ps-rq is an integer
rs is also integer
rs != 0
so it is rational
umbral steppe
also, since 1/0=inf's question is no longer in contention, if you'd like, I'll show you an example proof for his statement of sqrt(k) being irrational iff k is not a perfect square.
leaden moon
also, since 1/0=inf's question is no longer in contention, if you'd like, I'll s...
i think i mentioned what do i go by in this channel...
torn dome
also, since 1/0=inf's question is no longer in contention, if you'd like, I'll s...
sure
umbral steppe
this does require you to recall what prime factorizations are and what a perfect square means in terms of its prime factorization.
leaden moon
what is $v_p$ here 😭
soft zealotBOT
1 divided by 0 equals Infinity
leaden moon
no need
worldly mesa
p-adic valuation
leaden moon
$nb^2 = a^2$, this means that $b^2 \mid a^2$
soft zealotBOT
1 divided by 0 equals Infinity
leaden moon
and $\gcd(a, b) = 1$
soft zealotBOT
1 divided by 0 equals Infinity
torn dome
contradiction
leaden moon
contradiction
not a contradiction yet
torn dome
not a contradiction yet
|a| and |b| = 1
leaden moon
|a| and |b| = 1
yep
umbral steppe
it was meant to be an example to get OP to see the kind of structure that is expected of a proof.
leaden moon
anywys
leaden moon
it was meant to be an **example** to get OP to see the kind of structure that is...
ah i see
bold turtle
it was meant to be an **example** to get OP to see the kind of structure that is...
And it surely is a marvellous one (unfortunately [margin joke here])
umbral steppe
but if you insist on it, I'll remove the proof, and I apologize for sending a complicated proof here.
leaden moon
anyways, this is just an example on how you can write your proof @torn dome
leaden moon
but if you insist on it, I'll remove the proof, and I apologize for sending a co...
no need to remove
oh mb im too late
gleaming anchor
in discrete math, or are you ok with other fields?
torn dome
in discrete math, or are you ok with other fields?
basic math
gleaming anchor
sure, let me get one from my stash
torn dome
how many proofs have you written @bold turtle
do you also write formal proofs (like not just for the sake of solving a homework question, but for something of real value)
bold turtle
"For something of real value" requires even MORE vigour, you know
The whole point of homework questions is SO THAT you learn how to write these in presentable manners in the first place
gleaming anchor
Prove that the square of a positive odd integer leaves a remainder of 1 when divided by 4.
(alternatively: prove that the square of any positive odd integer is always 1 mod 4.)
bold turtle
how many proofs have you written <@660973674870079488>
Also, I'm not sure this is answerable; I've completed a maths degree, so it's probably in the hundreds, if not thousands
gleaming anchor
hint: use the algebraic definition of an odd number.
torn dome
Also, I'm not sure this is answerable; I've completed a maths _degree_, so it's ...
are you a maths major
your role says undergrad
bold turtle
CS master's degree student; I've done a maths undergraduate degree
torn dome
CS master's degree student; I've done a maths undergraduate degree
alright
bold turtle
(I'm not American, so "major" is meaningless here)
leaden moon
who just pinged me in #graduate-applications 😠
bold turtle
Prove that the square of a positive **odd integer** leaves a remainder of 1 when...
Back to the question at hand... ^ @torn dome
leaden moon
who just pinged me in <#1013551332847784017> 😠
and in #math-pedagogy too 😭?
gleaming anchor
and in <#541382507656904704> too 😭?
!redir, please.
final saddleBOT
and in <#541382507656904704> too 😭?
This channel is only for on-topic discussion. Please take casual conversation to #discussion or #chill.
torn dome
Prove that the square of a positive **odd integer** leaves a remainder of 1 when...
i assumed any arbitrary n odd number = 2k+1 then i took the square
n^2 = 4k^2 + 4k +1
then i used long division
i am getting remainder = 1
gleaming anchor
ok not too bad of an attempt, and I presume you did your long div correctly
torn dome
yeah
gleaming anchor
but from 4k^2 + 4k + 1, you could have factored out a 4k from the first two terms to get 4k(k + 1) + 1
bold turtle
-# yh, tbh, I'm intrigued on what you did that you're calling long division
gleaming anchor
from here it's more directly obvious that you will have a remainder of 1 on division with 4, because k integer means k+1 integer, and 4k is obviously an integer multiple of 4
so (integer multiple of 4) times (integer) = (another integer multiple of 4), and + 1 makes the remainder 1
gleaming anchor
,rcw
soft zealotBOT
torn dome
but again, there's a faster and more direct method
which is better
gleaming anchor
I'd prefer the factoring method because it leaves the dangling + 1 as undeniable proof
torn dome
are you saying that i should write $n^2$ as $4k(k+1)+1$
soft zealotBOT
Sean [Ping On Reply Please!]
gleaming anchor
yes.
torn dome
how can i show that there will be remainder of 1
gleaming anchor
from here it's more directly obvious that you will have a remainder of 1 on divi...
^
bold turtle
bruh, the literal dangling 1 outside of that bracket IS the remainder
gleaming anchor
so (integer multiple of 4) times (integer) = (another integer multiple of 4), an...
^
consider that k is an integer. that means k+1 is an integer too.
torn dome
yeah
gleaming anchor
then obviously 4k is also an integer.
torn dome
yuh
gleaming anchor
more importantly, 4k is a multiple of 4.
and multiplying anything by a multiple of 4 makes the product a multiple of 4 too
so 4k(k + 1) is a multiple of 4
- 1 gives the remainder
end of story
bold turtle
cf. how if you write 194 = 14 x 13 + 12, then 12 is the remainder you get when you divide 194 by 13
torn dome
well i could have done $4(k^2+k)$
soft zealotBOT
Sean [Ping On Reply Please!]
bold turtle
Yh you can, and that's equally valid
The point being, you can clearly pull out a 4, and get a remainder of 1
gleaming anchor
a distinction without a difference
k^2 is clearly an integer from k being an integer
the sum of two integers is an integer, and multiplying by 4 gives a multiple of 4
- 1 still gives the remainder
end of story
.
torn dome
cool
gleaming anchor
I never said it was wrong.
gleaming anchor
but polylong div is prone to rather dumb errors from time to time
and thus I opt to not do it if I have better ways to show what I want to show
I'll leave you with one final question
take your time in doing this
given p, q are positive integers. show that p/q + q/p >= 2 always.
note: consider using contradiction.
bold turtle
That's pretty tame compared to the full AM GM tho wdym
bold turtle
In fact you're technically proving the AM-GM here for two numbers
leaden moon
yeah
torn dome
i arrived at p < q
gleaming anchor
or more insightful
torn dome
is that wrong
gleaming anchor
i arrived at p < q
ok so for one thing right
leaden moon
i arrived at p < q
idk how you get that
gleaming anchor
when doing proofs and wanting to get your answer checked
leaden moon
assumption?
gleaming anchor
it might help to show the whole proof
because arriving at a right conclusion from illogical steps renders the proof moot
for instance
bold turtle
"I have 3 apples because the spaceship man landed on the moon, and therefore I have 3 apples" is not a valid proof
gleaming anchor
sqrt(2) is irrational.
assume sqrt(2) is rational if I can be on this server.
I can be on this server, so sqrt(2) is rational (wtf?)
when writing proofs, the reasoning is as important as the conclusion
topaz tusk
indian mathematics questions solved here?
leaden moon
indian mathematics questions solved here?
idk
gleaming anchor
indian mathematics questions solved here?
you can ask for help in an available channel. please read #❓how-to-get-help
bold turtle
!occupied - jeez does no one read the damn info file
final saddleBOT
Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).
torn dome
ok well if p/q + q/p < 2
then p^2 + q^2 / pq < 2
p^2 + q^2 - 2pq < 2
(p-q)^2 < 0
p-q< 0
p< q
leaden moon
!occupied - jeez does no one read the damn info file
check his bio btw 😭
bold turtle
idk
-# They're asked, and seldom solved (because of the type of question and the nature of the answers sought 😭)
gleaming anchor
ok well if p/q + q/p < 2
then p^2 + q^2 / pq < 2
p^2 + q^2 - 2pq < 2
(p-q)^2 < 0...
your proof is correct up till you made two extraneous steps
p-q< 0
p< q
these two
torn dome
p < q and q < p
leaden moon
-# They're asked, and seldom solved (because of the type of question and the nat...
-# scam likely 😭
gleaming anchor
(p-q)^2 < 0
you should have stopped here
this is the contradiction you're looking for
bold turtle
Because that should've sent alarm bells ringing
leaden moon
these 2 steps don't make sense
torn dome
i didnt use parenthesis
gleaming anchor
also I suppose you meant < 0 on the second line there
bold turtle
i didnt use parenthesis
yh you're gonna wanna write them in esp. if you're typing these out
bold turtle
okay so (p-q)^2 < 0
alarm bells should be ringing - why?
gleaming anchor
what do you recall about squaring real numbers?
torn dome
what do you recall about squaring real numbers?
ok
got it
square is positive
ok so proved
bold turtle
or...?
bold turtle
ye
torn dome
yay
bold turtle
Try and write up a proof
gleaming anchor
Try and **write up** a proof
exactly.
torn dome
Try and **write up** a proof
ik how to write proofs trust me guys
bold turtle
...
torn dome
i did one in my MSE
bold turtle
I'm currently mashing the X-to-doubt button fwiw
torn dome
i got 2/5
leaden moon
yo @torn dome
since you learned your technique
why don't you go and try to help #help-2
gleaming anchor
that does not scream confidence, but if you do not want to have your proof writing reviewed, then sure, I won't force you to
leaden moon
why don't you go and try to help <#359052604149465088>
their question is to prove $\sqrt{3}$ is irrational
soft zealotBOT
1 divided by 0 equals Infinity
torn dome
i solved that one already
torn dome
that does not scream confidence, but if you do not want to have your proof writi...
ok i will send it here
leaden moon
i solved that one already
try and help him
bold turtle
It's beneficial to see if you've actually learnt it well if you can teach it to someone too
leaden moon
by yourself
torn dome
try and help him
i dont help, i am helpee
torn dome
It's beneficial to see if you've actually learnt it well if you can teach it to ...
i agree
but i have the discrete mathematics final tmr
leaden moon
i dont help, i am helpee
if you help him, you are actually trying to revise the stuffs you learned by yourself
just come and try to help
ill backup
gleaming anchor
to be fair, let him settle his finals first
plus, this is kinda advertising other help channels alrd
btw, I forgot to explain why your last two steps were invalid
from (p-q)^2 < 0, we know that no real number satisfies this inequality
therefore, taking a square root leads to complex numbers being introduced
and complex numbers are, well, not exactly well-ordered, or even ordered at all...
so the notion of p < q or q < p breaks
torn dome
@gleaming anchor
gleaming anchor
well written
torn dome
took you 10s
gleaming anchor
it's not hard to read font 24 writing
plus, I already caught the gist of your algebra
I just wanted to see your sentences
torn dome
thank you Celine!
gleaming anchor
rip the accent but nps
final saddleBOT
craggy plank
.solved
torn dome
final saddleBOT
torn dome
how are there 2^k terms (in last step)
young bobcat
wait who r u claiming is a troll
faint edge
@warm junco don't troll. It distracts people from getting help.
young bobcat
the denominators go from 2^k + 1 to 2×2^k inclusive
@torn dome does this answer ur question
torn dome
what is the best method of proof for this
final saddleBOT
what is the best method of proof for this
torn dome
direct proof doesnt work, i get stuck at 4k^2 + 4k
obtuse hedge
Can you factor 4k^2 + 4k
torn dome
Can you factor 4k^2 + 4k
4k(k+1)
terse dagger
yes
obtuse hedge
We want it to be divisible by 8, so since it has a factor of 4, what are we missing?
torn dome
2
obtuse hedge
Indeed, to show k(k+1) is divisibe by 2 it is possible!
torn dome
Indeed, to show k(k+1) is divisibe by 2 it is possible!
how
torn dome
an integer
obtuse hedge
With reference to division by 2 then, k, as an integer must be either _ or _
obtuse hedge
Right, so in either case, k(k+1) has an _ factor
torn dome
2
obtuse hedge
Exactly!, so k(k+1) is always divisible by 2
torn dome
Exactly!, so k(k+1) is always divisible by 2
what to write in the proof
i am proving for 8
terse dagger
you're left with 4 k (k + 1)
torn dome
yeah
terse dagger
and you can say that for any k
obtuse hedge
i am proving for 8
k(k+1) is divisible by 2 and 4 is divisible by 4, so the whole thing _
terse dagger
either k is even
meaning there is a factor of 2
or k + 1 is even
which also means there is a factor of 2
so that means that you can write 4 k (k + 1) as 4 * 2 * l
torn dome
how about we use induction on this
terse dagger
l being another integer
torn dome
btw how can i prove that k(k+1) is always even
terse dagger
You basically did before
torn dome
i could make two cases
obtuse hedge
if k is odd k +1 is even
You just proved it
torn dome
like when k is even and when k is odd
then i can multiply
okay
btw what abt induction
terse dagger
You don't need induction because for all the steps you've gone through you haven't said anything about n or k
obtuse hedge
btw what abt induction
Seems overkill when you can prove directly but it does work very nicely
One of the easier inductions to write out
torn dome
torn dome
i used to do these typa problems using induction
is this one also possible using direct proof or some other method
obtuse hedge
$11^{n+1} + (11 + 1)^{2n - 1} = 11^{n+1} + 11^{2n - 1} + 11^{2n - 2} + \dots + 11 + 1$ and $11^2 + 11 + 1 = 133$ so it should be possible to prove directly
soft zealotBOT
Coolempire2026
obtuse hedge
Likely easier by induction though
torn dome
how did you expand that power
obtuse hedge
Binomial theorem
torn dome
Base case
i will write my so far progres, you tell me how to prceed
holon
So after the base case and inductive step process I reached at $11^{k+2}+12^{2n+1}$ with the assumption that $133 | 11^{k+1}+12^{2k-1}$
shiite
anyway can you help me with the next step
how do i proceed
terse dagger
So what is your base case?
soft zealotBOT
Sean [Ping On Reply Please!]
torn dome
So what is your base case?
n = 1
torn dome
yeah
torn dome
that it works
of course it works, thats why they gave a whole question on it
so i am straight at n = k + 1
terse dagger
right so you assume it works for k
torn dome
yeah
terse dagger
so now you go look at $11^{k + 2} + 12^{2(k + 1) - 1}$
soft zealotBOT
Katharine
terse dagger
You wanna transform this into some shape such that you can use your assumption
torn dome
**Katharine**
$11^{k+2}+12^{2k+1}$
soft zealotBOT
Sean [Ping On Reply Please!]
torn dome
how to use this
to 'transform'
ik that i gotta add and subtract smth
should i do $11.11^{k+1}+12.12^{2k-1}$
terse dagger
what you can do is use the binomial theorem
soft zealotBOT
Sean [Ping On Reply Please!]
obtuse hedge
Are you sure about the 12
obtuse hedge
Check the exponents
obtuse hedge
(When you take out 1 twelve, the power should decrease by 1)
soft zealotBOT
Sean [Ping On Reply Please!]
final saddleBOT
obtuse hedge
The way I solved it is in #latex-testing for comparison
final saddleBOT
Available help channel!
terse moat
Guys, for context, I am still new at math, and I've been learning all the basics, so I'm not good at it, and I am the type of person that can't move on without understanding the why's to everything, my question is:
how do we know that [ACD] is similar to [DBC] ?
I see the square angle at D, and I understand it is a similar triangle but why? sorry if my questions are too basic, I'm going step by step
final saddleBOT
Guys, for context, I am still new at math, and I've been learning all the basics...
next thorn
Guys, for context, I am still new at math, and I've been learning all the basics...
they're both right triangles that share a length
I think that sufficies similiarty.
terse moat
oh so you only need to share the length and them both being right triangles? :O
obtuse hedge
🤔 no you need one other angle to show it as well, otherwise a bunch of triangles are similar
(Which is easily shown since the big triangle is similar to them both)
terse moat
oh okay, because I was trying to prove it with AA - Angle angle, or maybe SAS - Side angle side but I'm not being able to
next thorn
🤔 no you need one other angle to show it as well, otherwise a bunch of triangle...
apologies
terse moat
oh and this exercise is from khan academy
obtuse hedge
oh okay, because I was trying to prove it with AA - Angle angle, or maybe SAS - ...
Yeah AAA is how I did it when I looked as well
ACD is similar to ABC because they share angle A and the right angle (angle C/D)
ABC is similar to CBD because they share angle B and the right angle (angle C/D)
ACD is similar to CBD because they are both similar to ABC
terse moat
Yeah AAA is how I did it when I looked as well
ACD is similar to ABC because the...
One question, sorry, how do I know that both angle [ACD] and [ABC] are the same? I'm a bit confused there, sorry
next thorn
you dont need to apologize by the way! nor are your questions "basic". we've all been there we've all done that and It's really cool how you're willing to understand instead of memorizing!
terse moat
you dont need to apologize by the way! nor are your questions "basic". we've all...
Thank you a lot for your* comment! c:
next thorn
One question, sorry, how do I know that both angle [ACD] and [ABC] are the same?...
also for this, notice how ACD is 90-BCD. and guess what! DBC is also 90-BCD.
terse moat
omg you are right! :O
loud sundial
you dont need to apologize by the way! nor are your questions "basic". we've all...
hi 
terse moat
ohhhhhh I see what you mean, based on that you can conclude that no only is angle C = to angle B, but that also angle A equals Angle C (on BCD)
next thorn
first conclusion is from angle ACB as aw hole being acute, and thus, angles BCD and ACD must sum up to 90. BCD + ACD = 90 and thus ACD= 90 - BCD
second conclusion is from the right triangle DBC! sum of all angles in any triangle is 180. you already know BDC is 90 and thus 90 + DBC + DCB = 180 and thus DBC + DCB = 90 and thus DBC = 90 - DCB
terse moat
makes a lot of sense, I think like I can understand and prove similarities based on angles a lot better now
next thorn
makes a lot of sense, I think like I can understand and prove similarities based...
of course! keep practicing.
terse moat
because I can solve this exercise which is to just calculate the area of the triangle, but I always stop whenever I see something I don't 100% understand or can prove ,-,
thank you, how do I mark this as solved?
next thorn
you type .close
final saddleBOT
next thorn
always!! I wanna go to mathematics in University
AMAZING!
BEST OF LUCK
YOU CAN DEF DO THIS!!!
terse moat
maybe someday I will be able to help people out here too
next thorn
maybe someday I will be able to help people out here too
yes! you should start helping postgrads like @hasty mist because they dont know their maths.
terse moat
yes! you should start helping postgrads like <@1421147725335891981> because they...
so you want me to be completely demolished I see ahah, I will resume my studies now, thanks and thanks coolempire too! c:
final saddleBOT
Available help channel!
terse moat
I have been doing more math exercises about this topic, I will do a few more and move on, but I did a visual representation of why they are similar with the AA criteria with this drawing, this sometimes helps me understand concepts, trying to formulate them myself, just wanted to put it here, if there's anything I'm missing please tell me! Thanks a lot everyone!
final saddleBOT
I have been doing more math exercises about this topic, I will do a few more and...
terse moat
.close
final saddleBOT
final saddleBOT
Available help channel!
signal cipher
final saddleBOT
steep nest
because the outside part of the shape has a higher rotational distance than the cone part
the part with a large area of the cone moves less distance than the large area of the cylinder
signal cipher
because the outside part of the shape has a higher rotational distance than the ...
can you put it into simple words i dont get it
steep nest
well just look at the 2d image you posted
the ratio between the areas is 1/2 right
half
signal cipher
yea
steep nest
now if we rotate it along the line, you see how the inside becomes a cone
and the outisde becomes a cylinder, without the cone?
well as the triangle part is closer to the center, it will gain less volume. While the outside part, as it is further away from the center, will gain omre volume.
Try to visualize this
signal cipher
what i was thinking
is that the ratio would stay consistent
in every rotation frame
so when the 3d shapes are form how does the overall ratio change
ohhh
i get it
overlap
ohh
is overlap right reasoning?
.close
final saddleBOT
final saddleBOT
Available help channel!
restive sinew
final saddleBOT
tired walrus
incomes are 2x and 3x, expenses are 16y and 21y, savings are 4z and 9z, and 4z+9z = 312000/12
thorny otter
incomes are 2x and 3x, expenses are 16y and 21y, savings are 4z and 9z, and 4z+9...
also savings is income minus expenses 😊
Nevermind
iron basalt
The answer is Rs 60000 per month
tired walrus
The answer is Rs 60000 per month
!nosols
final saddleBOT
The answer is Rs 60000 per month
As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.
tired walrus
~~also savings is income minus expenses 😊~~
Nevermind
this op already said
tired walrus
incomes are 2x and 3x, expenses are 16y and 21y, savings are 4z and 9z, and 4z+9...
fixed a typo btw
iron basalt
As a helper, please do not give out answers that could be copied as a homework s...
Oh got it
tired walrus
312K is the annual savings of the two people combined
iron basalt
don't give out answers
Ok btw I solved it in my notebook
tired walrus
ok sure. good for you.
let's wait for OP to show signs of life again so he can tell us if he's stuck or if he also got it
@restive sinew you here?
iron basalt
I think he's gone
final saddleBOT
@restive sinew Has your question been resolved?
final saddleBOT
final saddleBOT
Available help channel!
restive sinew
final saddleBOT
restive sinew
How do I start it?
leaden moon
If the speed of $A$ is increased by 14 km/h, he will be 4.2 minutes early and if the speed is decreased by 2 km/h, then he will be 0.8 minutes late
soft zealotBOT
1 divided by 0 equals Infinity
leaden moon
i think this requires a system of equation
let the time $A$ makes through the distance as $x$ and the speed $A$'s going is $y$
soft zealotBOT
1 divided by 0 equals Infinity
leaden moon
so the distance we need to find is $xy$
soft zealotBOT
1 divided by 0 equals Infinity
leaden moon
How do I start it?
.
desert mantle
generally speaking, a good first idea is always to give names to all values
and then try to find some equations connecting them
final saddleBOT
@restive sinew Has your question been resolved?
leaden moon
generally speaking, a good first idea is always to give names to all values
@restive sinew
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@restive sinew Has your question been resolved?
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@restive sinew Has your question been resolved?
tired walrus
op is not showing any signs of life
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old quarry
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old quarry
AB: 3y-2x-5 = 0
AB distance = sqrt(13)
i took a point P (x,y)
distance of P from line. =
d=|2x1-3y1+5|/sqrt(13)
area = 1/2 x (2x1-3y1+5)/sqrt(13) x sqrt(13) = 10
2x1-3y1+5 =20
2x1-3y1 = 15
a=2 b=-3
which gives me ans 4
which is in the options but is wrong
where did i go wrong
left echo
What the hell is that font size
old quarry
sorry its from some website
floral zenith
distance of P from line. =
d=|2x1-3y1+5|/sqrt(13)
I guess here
fossil kiln
area = 1/2 x (2x1-3y1+5)/sqrt(13) x sqrt(13) = 10
2x1-3y1+5 =20
2x1-3y1 = 15
Forgot the modulus
fossil kiln
Also u gotta use the fact tht P is above line AB
old quarry
Also u gotta use the fact tht P is above line AB
so like 3y-2x-5>0?
left echo
If P is above the line then putting it into the line's equation will be positive
floral zenith
Also u gotta use the fact tht P is above line AB
It could be below too right? Its locus
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fossil kiln
It could be below too right? Its locus
Mentioned in ques
old quarry
It could be below too right? Its locus
question specifies above