b implies c?
#help-28
rapid rain
neon yoke
is the butler lying
zenith kernel
G whether T or F both gives you C F, thus B F
neon yoke
You said yourself B, C both F
but C was telling the truth in the first one'
rapid rain
You said yourself B, C both F
"C = F" was found when in the case that B is T
so no longer valid unless shown now
neon yoke
"C = F" was found when in the case that B is T
but if B is F then C must be F right?
cuz of the first one
upbeat knot
if you are familiar with "proof by contradiction",
you assumed B
ended up with C and not C which is false
thus assumption is false
not B, i.e. Butler lies
rapid rain
but if B is F then C must be F right?
false can imply true
neon yoke
false can imply true
oh
zenith kernel
G whether T or F both gives you C F, thus B F
(C, G) is not (T,T), (G, H) is not (F, F)
G: T-> C:F
G: F -> H:T ->C :F
rapid rain
(C, G) is not (T,T), (G, H) is not (F, F)
G: T-> C:F
G: F -> H:T ->C :F
give some time for sean 😭
rapid rain
I'm not, I'm following sean's pace
rapid rain
well we have to go supposing more truth values now
rapid rain
looking at C being involved in a lot of statements
mainly "h implies not C"
this would be more constraining if C was true
because then h would have to be false
neon yoke
can i do this using rules of inference
rapid rain
only a false thing can imply a false thing
neon yoke
not b is a premise
but i cant apply modus ponens
kay now we have
b implies c
not c or not g
g or h
h implies not c
alright
b is F
rapid rain
once b is F, the first statement is useless btw
rapid rain
i will assume C is also F
try C is T first
neon yoke
try C is T first
I want to go by instincts
rapid rain
ok
neon yoke
(you wont be there in the exam hall)
alright if not C is T then what can i say abt h
0 -> 1 and 1 -> 1 both are true
its useless
now what should be my next step
rapid rain
well statements 1, 2 and 4 are now irrelevant when C is F
neon yoke
i could assume h = T
rapid rain
so the only statement left is "H or G"
rapid rain
its useless
you just said it
neon yoke
got it
neon yoke
i could assume h = T
.
neon yoke
that means C is F
rapid rain
that means C is F
no???
neon yoke
when i plug h = T i get c = F
rapid rain
"When C is F, then everything works. That means C must be F"
neon yoke
ohk
rapid rain
you haven't seen what happens in the other case, you can't assume it doesn't work without seeing it
neon yoke
b = F, c = F
neon yoke
you haven't seen what happens in the other case, you can't assume it doesn't wor...
are you talking abt c
rapid rain
are you talking abt c
yes
neon yoke
what other case?
rapid rain
"When C is F, then everything works. That means C must be F"
This is what you reasoned
neon yoke
This is what you reasoned
yeah
rapid rain
I'm saying that's not a good reasoning
rapid rain
well, C is T
neon yoke
how
neon yoke
oh so i assume c is t
rapid rain
All I'm saying is you've shown C = F is possible
to show that it's the only way
you must show C = T is impossible
that, you haven't done yet
rapid rain
oh so i assume c is t
you started off well
neon yoke
couldnt conclude anything
rapid rain
couldnt conclude anything
wdym
neon yoke
ok if c is T
rapid rain
you haven't tried "c is T"
neon yoke
kay now we have
b implies c
not c or not g
g or h
h implies not c
0 implies 1
true
0 or not g is true only if g is false
g or h is true if h is true
h implies not c is true for h = false
is that a contradiction
rapid rain
yes
rapid rain
g or h is true if h is true
just specify "only if h is true" here
neon yoke
what abt the rest
rapid rain
and G or H, we don't know
neon yoke
kay now we have
b implies c
not c or not g
g or h
h implies not c
neon yoke
and G or H, we don't know
what
rapid rain
once you have that B, C are both false
rapid rain
kay now we have
b implies c
not c or not g
g or h
h implies not c
1, 2 and 4 become useless
we've discussed so before
so only "G or H" is left to determine anything about G and H
rapid rain
can i say that
no, "G or H" is still a constraint
it's just not enough of a constraint to properly find both
neon yoke
the propositions work for all values of g and h, provided that they satisfy 2 and 4
rapid rain
recall what's the truth table of G or H
neon yoke
recall what's the truth table of G or H
okay
rapid rain
the propositions work for all values of g and h, provided that they satisfy 2 a...
provided they satisfy 3...
2 and 4 are irrelevant now
neon yoke
provided the satisfy 3
rapid rain
give me the truth values of G and H that work
rapid rain
yes, those are the 3 possibilities
rapid rain
well there was some hiccup in transforming the written statements into logic statements
neon yoke
lmao
rapid rain
"they can't both ..." means not(... and ...)
rapid rain
and so don't bring up Xor unless explicitely stated "exactly one of them ..."
neon yoke
alright
well so butler and the cook are liars and the others one we cant determine
at the start of this problem i assumed b = T and found a contradiction
what if i dont find the contradiction
can i conclude that my assumption was correct?
rapid rain
your assumption is plausible, not "correct"
your assumption only becomes correct when you find out the alternative is impossible
so you have to look at what happens when you assume the opposite at some point
zenith kernel
why did this take so long
I think you can get faster. By not questioning yourself. And set up straightforward to do what you are determined to do. When you halt and question yourself “is this step correct” it might break your train of thought
zenith kernel
Have confidence. As long as you don’t assume things unnecessary, only follow the basic conditions you don’t have room to make error
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neon yoke
Can we have more than one knight in these questions?
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Can we have more than one knight in these questions?
neon yoke
Cuz if the three people must be unique then 31. has no solution
robust nacelle
Can we have more than one knight in these questions?
no we can't
"…You know that one of these people is a knight, one is a knave, and one is a spy."
robust nacelle
Cuz if the three people must be unique then 31. has no solution
what do you mean ?
neon yoke
what do you mean ?
a knave can never say i am not the spy
robust nacelle
"when there is no unique solution, list all possible solutions or state that there are no solutions."
robust nacelle
a knave can never say i am not the spy
that is true
robust nacelle
"when there is no unique solution, list all possible solutions or state that the...
so there aren't any 👍
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sturdy vine
$lim {x\to 0} \frac{\ln(x+e^{-x})}{x\cdot \ln(1+2x)}$
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$lim {x\to 0} \frac{\ln(x+e^{-x})}{x\cdot \ln(1+2x)}$
glossy valveBOT
Professor Edward C. Hawthorne
icy hemlock
l'hopitals?
sturdy vine
Taylor mac laurin?
icy hemlock
er
sturdy vine
$\ln(1+x+e^{-x}-1)$
glossy valveBOT
Professor Edward C. Hawthorne
sturdy vine
$\ln(1+t)$
glossy valveBOT
Professor Edward C. Hawthorne
sturdy vine
$t-t^2/2=x+e^{-x}-1-(x+e^{-x}-1)^{2}/2$
fast peak
you could just differentiate
glossy valveBOT
Professor Edward C. Hawthorne
glossy valveBOT
Professor Edward C. Hawthorne
fast peak
its gonna be a painful calculation anyway I think
sturdy vine
$f'_N=(1-e^{-x})/(x+e^{-x})$
glossy valveBOT
Professor Edward C. Hawthorne
gritty rose
Can try to use log(1+u)/u -> 1 as u -> 0 for both u = 2x and u = x + exp(-x) -1
glossy valveBOT
Professor Edward C. Hawthorne
sturdy vine
$1-e^{-x}/4x$
glossy valveBOT
Professor Edward C. Hawthorne
sturdy vine
$e^{-x}/4 lim = 1/4$
glossy valveBOT
Professor Edward C. Hawthorne
sturdy vine
,w $lim {x\to 0} \frac{\ln(x+e^{-x})}{x\cdot \ln(1+2x)}$
glossy valveBOT
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zenith osprey
Hi, i need a little help on how to solve this, taking a practice test and i just dont really have a clear idea on how to do this, thank you!
sturdy vine
Hi
zenith osprey
Hi
sturdy vine
Only one straight line passes through two points, right?
zenith osprey
yeah
zenith osprey
would it be -6, -3?
sturdy vine
Ok this Is x , and y?
zenith osprey
3, -3?
sturdy vine
Yes!
So you have to find a straight line that passes through the points (-6,3) and (-3,-3), do you know how to do it?
zenith osprey
not really sorry
sturdy vine
y=mx+c
zenith osprey
ah
sturdy vine
So what do you know from the graph and seeing y=mx+c? What are you missing?
zenith osprey
sorry if im wrong but would it be m and c?
sturdy vine
?
sturdy vine
Ok
You know that a straight line passes through the point (-6,3)
So y=mx+c -> 3=m(-6)+c
zenith osprey
mhm
sturdy vine
You know that the other straight line passes through the point (-3,-3)
So y=mx+c -> -3=m(-3)+c
Clear?
zenith osprey
yes
sturdy vine
Now placing these 2 in a linear system, you find m and c that simultaneously respect the 2 equations (because only one straight line passes between 2 points)
Clear ? Do you know how to do it?
zenith osprey
i think i kinda get it
sour torrent
Hi
zenith osprey
Hi
sour torrent
Need help?
sturdy vine
i think i kinda get it
If it is clear you will replace the unknowns m and c in the general equation y=mx+c, do the same for the other 2 remaining ones
zenith osprey
sure, sorry im pretty slow 😭
for the graph itself, what is the difference between the bold point and the one that isnt fully shaded in?
sturdy vine
That the filled blue dot also accepts as input the x on which it is located, the non-filled one does not
As you can see on the right where the intervals are written
zenith osprey
oh i see, thank you!
i believe i was able to get an idea on how to work this thank you so much for your help
sturdy vine
👍
zenith osprey
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frosty gyro
i was wondering if we had a function u(x) is this allowed?
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i was wondering if we had a function u(x) is this allowed?
frosty gyro
im assuming its differentiable as well
leaden ermine
Whats star
frosty gyro
complex conjugate
leaden ermine
this seems very wrong
z* is not differentiable for example
complex differentiable
frosty gyro
complex differentiable
ok
im looking at this so i just thought we could swap it around
vagrant leaf
What is f?
frosty gyro
What is f?
they defined it as a complex-valued function
fast spear
is x real
frosty gyro
yes
fast spear
z* is not differentiable for example
i think you should use the derivative wrt x because of this
vagrant leaf
i was wondering if we had a function u(x) is this allowed?
It is allowed if u is complex valued function of real variable. And for any n-th ddrivative it also holds as long it is differentiable enough
frosty gyro
It is allowed if u is complex valued function of real variable. And for any n-th...
ok ty
ty nyx and universe as well
vagrant leaf
You should try showing why it holds
whole lily
frosty gyro
You should try showing why it holds
i just needed this result to solve something related to this
ok thanks all
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lime rover
How do I solve this cubic?
3x³+5x²+4x+5=5
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How do I solve this cubic?
3x³+5x²+4x+5=5
sacred yarrow
👀
lapis holly
q = 2(5)²-9(3)(5)(4)+27(3)²(5) / 27(3)³
thick hedge
Δ=(q/2)²+(p/3)³
NO
thick hedge
How do I solve this cubic?
3x³+5x²+4x+5=5
notice the 5(s) cancel out
thick hedge
sure, but there's a better method here
so you have to solve 3x^3+5x^2+4x=0
which you should be able to
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thick hedge
😭 , I had given an alternate method
stone matrix
Lol
gritty rose
😭 , I had given an alternate method
indeed. but they chose to ignore you. just move on
hybrid yacht
How do I solve this cubic?
3x³+5x²+4x+5=5
you have a 5 on both sides
subtract those off and you have the exact same cubic as in your other help channel
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quartz flare
Would this be a valid proof of correctness? I've never proven an algorithm's correctness before.
\begin{proof}[Proof of the Correctness of Algorithm 1]
We will go by induction on $n$. Naturally, if $n = 1$, then the maximum is $a_1$ and we are done.
Now suppose that $max$ is the maximal value in $\{a_1, \dots a_k\}$. If $a_{k+1} > max$, then $a_{k+1}$ is the maximal element in $\{a_1, \dots a_{k+1}\}$; otherwise, $a_{k+1} \leq max$, and $max$ is the maximal element in this set. In the first case, the code sets $max := a_{k+1}$, so $max$ is the maximal value in the set; otherwise, the code keeps $max$ as is, so $max$ remains the maximal value in the set.
We can continue this process up to $k = n$ and we have $max$ is the maximum value in the set $\{a_1, \dots, a_n\}$.
\end{proof}
glossy valveBOT
Coolempire2026
maiden vapor
It looks fine. You should probably state the induction hypothesis more precisely
hallow charm
yes, this seems fine. what you are effectively doing is proving an invariant about the algorithm: in every iteration of the algorithm, max stores the maximum value among the elements in the subset {a_1, ..., a_k}
then clearly the algorithm terminates so by virtue of the invariant, it returns the max
hallow charm
it's relatively clear here but it may not be so obvious for recursive algorithms, for example
quartz flare
It's obvious to me because I can see it but proving it I don't know how I would
hallow charm
well the algorithm is simply doing a single pass through the list and then returning the max
so it "clearly" terminates
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rancid helm
in ques d, should we take exclusive or?
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in ques d, should we take exclusive or?
twilit leaf
no, thats part e's job
brittle steeple
oh god not the either discourse
but yes, garlic is right, it's unlikely that d and e are asking the exact same question phrased slightly differently, so likely inclusive or is desired
rancid helm
in this ques we take exclusive right?
grave elm
in this ques we take exclusive right?
It'll be probably best to ask your teacher
One one hand, "either" mildly suggests exclusivity. On the other hand, inclusive is usally the default unless explicitly stated otherwise
If I had to guess, I'd choose inclusive, but it's just a poorly worded question
zenith kernel
Inclusion exclusion, |A union B|=|A|+|B|-|A intersects B|
warped frost
this is confusing though, because the wording in 48 is the same as the wording in 22. d), and the wording in 22. d) suggests an inclusive OR while e) suggests an exclusive OR
so there's some kind of wording precedent here
grave elm
That's exactly what makes me think that it's meant inclusively
onyx glen
Inclusion exclusion, |A union B|=|A|+|B|-|A intersects B|
not what was asked
zenith kernel
I meant, i don’t know we should say, take inclusion, or take exclusion. That’s result is called inclusive-exclusive principle, so together
spiral vigil
i would read that as inclusive or
grave elm
I meant, i don’t know we should say, take inclusion, or take exclusion. That’s r...
What was asked is whether it wants you to count strings which beign with 00 or end with 111 or both, or whether it wants you to count strings which begin with 00 or end with 111, but not both
i think it didnt refer to IE principle
zenith kernel
Oh 48.
warped frost
I think with precedent it's very likely inclusive OR, OP.
zenith kernel
Is there a difference between what either means in math and in daily English?
rancid helm
in 48, we should take both also right?
warped frost
48 reads as inclusive OR to most of us, yes, because of 22. d) and e)
zenith kernel
So A-B union B-A and A union B. What is math version what is daily version…?
warped frost
note that the wording is similar to 22. d), while 22. e) is unmistakably exclusive OR
spiral vigil
So A-B union B-A and A union B. What is math version what is daily version…?
this may be a question for another channel, friend
warped frost
that reads to me as if the author is consciously separating inclusive and exclusive OR
zenith kernel
I think it’s A union B, right? Because his teacher in 22 asked both d and e. And e means A-B union B-A, so d better mean A union B
onyx glen
these can both be clarified as exclusive by saying "but not both", and as inclusive by saying "or both"
brittle steeple
A xor B xor both
onyx glen
in daily enlgish, "either-or" implies exclusivity, but in math usually the implication is less strong if present at all
rancid helm
btw these are the questions from rosen's discrete mathematics book
grave elm
I think it’s A union B, right? Because his teacher in 22 asked both d and e. And...
Yeah, pretty much all of us agreed that it's most likely meant inclusively (A union B)
zenith kernel
Oh
warped frost
btw these are the questions from rosen's discrete mathematics book
no wonder
brittle steeple
Yes in context it must be A U B
rancid helm
is there any good solution book that i can refer to?
warped frost
wdym by "solution book"
rancid helm
solution manual
warped frost
well, I suppose you could try finding one online
I'm not aware if one exists though
grave elm
Try searching with "filetype:pdf"
adding that to the search query will only find results which are pdfs (which is the most common format for such manuals)
be aware that pdfs can sometimes be dangerous and so make sure the sources are somewhat trustworthy
warped frost
or, try searching in, well... the good places
I won't elaborate further - I presume you know what I mean.
ancient zinc
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ancient zinc
for b), I could only come up with eigenvectors (1,1,1,0,0,...), ... or more generally the ones where 3 consecutive terms are the same e.g.
a_{3k} = a_{3k+1} = a_{3k+2}
which is orthogonal and can be normalized but not really sure how to determine other possible eignevectors (and then determine if theres a hilbert basis)
rapid rain
which is orthogonal and can be normalized but not really sure how to determine o...
If it can help, I'm sure you've well noticed that the transformation works on groups of 3 coordinates (3k, 3k+1 and 3k+2)
Think about what happens to a much simpler transformation:
The one that only takes in (a_0,a_1,a_2) and outputs (a_2,a_0,a_1)
So a linear transformation from C³ to C³
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ancient zinc
checking for eigenvals/vecs
ancient zinc
If it can help, I'm sure you've well noticed that the transformation works on gr...
yeah some kind of cyclic permutation
ancient zinc
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neon yoke
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neon yoke
what do the first three terms of the sequence look like
its a root in a root in a root....
icy hemlock
yeah
neon yoke
I can also say that a_n+1^2 > 2
neon yoke
I can obtain a_n = a_n+1 ^ 2 -2
does that help with anything here
We are subtracting 2 from it
slate violet
you can square-root both sides of an inequality if both sides are positive
icy hemlock
well think of it this way
keen spire
are u familiar with newton sums
icy hemlock
lets assume for contradiction, maybe
keen spire
for quadratic
icy hemlock
that a_n+1 > 2
neon yoke
okay
icy hemlock
right, because if that was true the problem explodes
neon yoke
the question is about a_n tho
neon yoke
a_n < 2
icy hemlock
lets assume a_n >= 2
its the same thing
it's asking you to prove something for general n
does this imply anything about a_n-1
icy hemlock
I can obtain a_n = a_n+1 ^ 2 -2
using this
neon yoke
does this imply anything about a_n-1
im not sure
well no?
a_n only has a_n+1
in its formula
icy hemlock
well
you can shift it over by one
since a_n = a_n+1 ^ 2 -2
then you know that a_n-1 = a_n ^2 -2
right
icy hemlock
since a_n = a_n+1 ^ 2 -2
because this relationship just relates any two consecutive elements of the sequence
the n and n+1 are sort of placeholder
you could replace them with n-1 and n
or 2n and 2n+1
or 49n + 338 and 49n + 339 (i don't know why you'd do either of these though)
neon yoke
ok
icy hemlock
so
icy hemlock
does this imply anything about a_n-1
?
icy hemlock
let's say at some point a_n did actually go over 2 and a_n >= 2
outer lotus
This is very easy to prove by induction
icy hemlock
given that a_{n-1} = a_n^2 - 2, what does this mean about a_{n-1}
this is true
but im not sure if they know induction
outer lotus
This is very easy to prove by induction
Base case n=1 is true
icy hemlock
yeah
neon yoke
so induction is thew way
icy hemlock
for "prove for all positive integers n" questions induction should be one of the first things you try
outer lotus
so induction is thew way
$a_k < 2$
$2 + a_k < 4$
icy hemlock
it works unreasonably well
glossy valveBOT
a handsome russian dude
neon yoke
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neon yoke
outer lotus
**a handsome russian dude**
What do you do now?
It's not an answer bruh
How do you get k+1 here
outer lotus
Right
neon yoke
after doing the k+1 step
can I square both sides to get 2
or do i have to take care of the sign as well
icy hemlock
square root probably works better
outer lotus
Yeah, take sqrt of both sides
icy hemlock
be a bit careful with square rooting an inequality
icy hemlock
I am getting a_k+1 ^ 2 < 4
what do you get when you sqrt both sides of this
neon yoke
what do you get when you sqrt both sides of this
a_k+1 < 2
neon yoke
a_k+1 < 2 and a_k+1 > -2?
neon yoke
well so what do i do with the -2
icy hemlock
tbh not much
like
i think you know pretty well that a_k+1 is never actually going to be < 1
but
neon yoke
Well I do have a_k+1 < 2 always, cuz it is an AND clause
outer lotus
$\sqrt{2 + a_k} < \sqrt{4}$
neon yoke
so a_k+1 < 2 always
glossy valveBOT
a handsome russian dude
icy hemlock
it being -1 or something doesn't break anything
neon yoke
so proved?
icy hemlock
Well I do have a_k+1 < 2 always, cuz it is an AND clause
yep
so like
yeah
proved
neon yoke
cool
now move on to c
I am getting the difference of squares = 2
using the definition
outer lotus
Same thing. Prove for n=1, assume true for k. Prove for k+1
neon yoke
Same thing. Prove for n=1, assume true for k. Prove for k+1
is induction the only way
icy hemlock
for most questions involving "prove for all integers"
induction works
unreasonably well
especially when you have a really easy way to relate one case with the one right before it
which is the case here, you can kind of turn off your brain a bit
alternatively
i think you can probably get away with not using induction here
icy hemlock
c is kind of a non-statement
you should expand the RHS
and see what you end up getting
when you simplify
neon yoke
when you simplify
I used the recursive definition to plug in a_n+1^2 = 2+a_n
With that I get -(a_n^2 - a_n -2)
I should prolly faftorize this
-(a_n-2)(a_n+1)
alright
solved
icy hemlock
yeah
neon yoke
ty guys
icy hemlock
the hint does a lot
neon yoke
the hint does a lot
Well I can say that for a_n+1 - a_n > 0,
a_n+1 ^ 2 - a_n ^ 2 > 0 is also true
Then I can use (c) to get (2-a_n)(1+a_n) > 0
Which leads to
a_n < 2 or a_n < -1
Now from (b) I get that a_n < 2 so this clause is always true, (cuz of the OR)
So the sequence is strictly inceasing
icy hemlock
you should be a bit careful with the last step
if a_n < -1 you actually end up getting that (2-a_n)(1+a_n) < 0
but this is an easy fix
because you know a_n is always > 0 because square roots are positive
neon yoke
I got a_n >-1 mb
cuz it is (1+a_n) > 0
@icy hemlock Now how can I prove convergence and find the limit using these results?
(e)
Well I have that the sequence is strictly increasing
torn jolt
i think i can help here but shall read hold on
icy hemlock
so i dont feel comfortable answering with something possibly wrong
i will just watch
neon yoke
Well I have to find an upper bound for this sequence
zenith kernel
You finished b
neon yoke
steel solar
well you gotta look at the whole picture, sometimes zooming out and taking a break helps out
zenith kernel
And you finished c, thus it’s monotonically increasing
zenith kernel
Any continuous f, f preserves limits
i.e. lim f(a_n)=f(lim a_n)
Now plug in
f(x)=sqrt(x+2)
Yeah
neon yoke
why does lim a_n = lim a_n+1
zenith kernel
Definition
neon yoke
Definition
of what
zenith kernel
limit
zenith kernel
Any ε>0 there exists Ν>0 such that any n>=N, |a_n -L|<ε
Is the definition of lim a_n =L
Any ε>0, let M=N+1, any n+1>=M, |a_n+1 -L|<ε. Thus lim a_n+1 =L
neon yoke
alright
torn jolt
may i ask.. what the issue here..
zenith kernel
Idk, he halted
torn jolt
limit or the prove thing?
zenith kernel
Any continuous f, f preserves limits
i.e. lim f(a_n)=f(lim a_n)
Now plug in
f(x)...
He is at this stage, combine with lim a_n+1=lim a_n
zenith kernel
Yeah
steel solar
use hint perhaps 
midnight plover
Well, you use AM-GM or calculus for a.
zenith kernel
Yeah
steel solar
minima of a one variable function!
neon yoke
Why is replacing a_n with x okay?
steel solar
a_n is a positive real number
neon yoke
could I have done that in this part too?
zenith kernel
a_n is in the domain
torn jolt
Why is replacing a_n with x okay?
a_n+1 = f (a_n)
steel solar
we're showing that one part of the recursion has a certain minima, so a_{n+1} would have it as well
zenith kernel
inf f on (0, infty)<=inf of f on {a_n: n}
icy hemlock
oh i did not know the criteria for proving convergence was this simple
interesting, i thought it was more complicated
anyway sorry
neon yoke
could I have done that in this part too?
can i also do root(2+x) in here?
and then prove that it is always < 2? (for part b)
steel solar
eh, not really, sqrt(2+x) on its own doesn't have a maxima
midnight plover
Well in that case, you don't need any calculus
steel solar
for x a positive real
midnight plover
There induction works well.
neon yoke
Well in that case, you don't need any calculus
yea i did it using induction but why cant i do the same here
midnight plover
yea i did it using induction but why cant i do the same here
You could try, but AM-GM or calculus here is rather a lot easier.
steel solar
its just convenience here, induction is usually just a one-size-fits-all route
neon yoke
You could try, but AM-GM or calculus here is rather a lot easier.
Alright, I got 0.5x + 3/2x decreasing from 0 to root 3 and increasing from root 3 to inf
how do I use this for the sequence?
neon yoke
(as explained in the hint)
Minimum is root 3
zenith kernel
Yeah
zenith kernel
inf f on (0, infty)<=inf of f on {a_n: n}
^
neon yoke
^
I proved that it is strictly decreasing. Also we have that a_n >= root 3
Now can we say that the limit of this sequence is root 3?
zenith kernel
By calculation it’s sqrt(3)
Though when doing these exercises, probably it requires you not to omit steps
neon yoke
By calculation it’s sqrt(3)
calculation?
I just used part a and b to infer that limit = sqrt(3)
zenith kernel
Okay, whatever works. As long as your steps are valid
midnight plover
There are some fancier theorems to conclude this as well in this case, for example sqrt(3) is a fixed point of f(x)=1/2(x+3/x).
zenith kernel
Banach fixed-point theorem seems to be an overkill. taking limits both sides like a normal student is fine already I think
neon yoke
Banach fixed-point theorem seems to be an overkill. taking limits both sides lik...
I did not take limit on both side
zenith kernel
You have to have some kind of steps anyway. If not taking limit both sides, what did you do
neon yoke
You have to have some kind of steps anyway. If not taking limit both sides, what...
well if lower bound = root 3 and function is decreasing
the limit = root 3
zenith kernel
It’s false
neon yoke
zenith kernel
sqrt(3) is a lower bound, not inf
sqrt(3)+1+1/n decreases, and have sqrt(3) as a lower bound
neon yoke
sqrt(3)+1+1/n decreases, and have sqrt(3) as a lower bound
well I got L = L/2 + 3/2L
After applying limits on both sides
i can get L = root 3 from this
zenith kernel
Yeah that’s what people usually do
brisk minnow
Hi, how can I prove this using contradiction?
full forumBOT
Hi, how can I prove this using contradiction?
brisk minnow
If x!=y then wither x>y or x<y right
But idk what to do w this info
Can I have a small hint
jovial junco
If x!=y then wither x>y or x<y right
what is the power ?
brisk minnow
I should state x,y>0
jovial junco
If x!=y then wither x>y or x<y right
you can do something like that
brisk minnow
Alr I’ll have a go
little prism
I suspect the strategy is to suppose x and y are different
and then use that x^n-y^n=0
factorise...use what you know...prove it's impossible. But need to know the formula of x^n-y^n as a product
brisk minnow
factorise...use what you know...prove it's impossible. But need to know the form...
No idea what the factorisation of this is icl
Is it common?
In real anal
zenith pilot
x^n = y^n and x,y > 0
The latter term cannot be zero since x,y > 0. Hence x - y = 0.
little prism
Yes, that one...
little prism
Is it common?
think so.
brisk minnow
Oh 😞 no way I was getting that lol
zenith pilot
Yes. The latter term is just a geometric progression.
little prism
There's one for sums too, but afaik it needs the exponents to be odd to work (or something like that).
brisk minnow
Then that leads us to a contradiction if x!=y
zenith pilot
There are other ways too. This isn't the only one.
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tame cradle
.close
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tame cradle
lets say we define N=p1.p2.p3.....ps-1 + ---- a sum of product of s-1 primes my friend wants to prove whether N-1 can be divisible by b^2 where b has atleast one prime p1,p2 ,p3 ... etc as its factor
grave elm
lets say we define N=p1.p2.p3.....ps-1 + ---- a sum of product of s-1 primes my...
if N-1 was divisible by b^2, what else would it be divisible by? (look at the ps)
grave elm
yeah, one of the primes
grave elm
N-1 is p1 * p2 * ... * ps - 2
so this has to be divisible by one of the primes
can you see sth from here?
Can e.g.
5 * 3 * 7 * 11 - 2 be divisible by 7?
(without calculating it)
tame cradle
its product of s-1 primes
i can understand this that sum is of s terms so s-1 terms would have p1 in denom and two would have 1/p1^@
tame cradle
Can e.g.
5 \* 3 \* 7 \* 11 - 2 be divisible by 7?
(without calculating it)
oh shit hello no used modulus
lmao
distribute
-2mod7
oh shi
grave elm
oh shit hello no used modulus
yeah, sure
tame cradle
but man hey lets not get excited here we have (1 term -1) mod p1 here
grave elm
p | kp + a iff p | a
tame cradle
yes
grave elm
but man hey lets not get excited here we have (1 term -1) mod p1 here
wait wdym?
tame cradle
look i understand but what about term which does not have p1
eg
p2.p3...ps-1mod p1
grave elm
oh wait, i mightve misunderstood the question
tame cradle
understand you are right s-1 terms would become 0 mod p1
would about sth term and -1 mod p1
:.
:>
(p2....ps-1) mod p1 - 1 mod p1
wait
i got it
grave elm
oh i understand the question
tame cradle
(p2...ps-1) is always odd
and - 1 would be eveb
so even cannot be divided by odd
wait
are we right bro
?
grave elm
would about sth term and -1 mod p1
so the s-1 terms are -(s-1) (mod p) in total, is that right?
because each one is -1 mod p
grave elm
oh god
grave elm
you meant p1.p2.p3.....p_(s-1)
tame cradle
yes
grave elm
okay now i get it
tame cradle
yes
grave elm
i thought it was (p1.p2....) - 1
tame cradle
oh
lmao
no no
look we would remain with (p2.....ps-1 - 1) mod p1
this is an even number
:>
:>
:>
say it
grave elm
so its not divisible by p
tame cradle
:>
grave elm
i still dont get the question tho
grave elm
sum of product of s-1 primes
what does this mean
tame cradle
it means p1.p2.....ps-2 + p2.......ps-1 + -----
sum of s terms where each term is a product of s-1 primes
:>
grave elm
oh so you always leave out 1 prime?
grave elm
so there are s-1 terms?
tame cradle
he wants to show this is not divisible by an odd number b if any prime is a factor
:>
grave elm
what if one of the primes is 2?
grave elm
this is an even number
it could still be provable, just maybe not so easily
it wouldnt be an even number
grave elm
he wants to show this is not divisible by an odd number b if any prime is a fact...
is it about b^2 or b itslef?
tame cradle
it could still be provable, just maybe not so easily
what
why
s-2 terms are 0 mod p1
last term - 1
grave elm
oh and even number mod p can also be 0
tame cradle
mod p1
grave elm
14 mod 7 is 0
tame cradle
oh and even number mod p can also be 0
hell fuck
i forgot
lmao i feel sorry for him
i laughed at my friend
after which he got angry and then now i realise
why he was so troubled
even can also be divided by odd
how can we prove any idea
grave elm
he wants to show this is not divisible by an odd number b if any prime is a fact...
and is it about odd number b or b^2?
grave elm
so if you choose p1 = 2, p2 = 3, p3 = 5, it doesnt work
grave elm
oh
grave elm
2*3 - 1 is divisible by 5 (choose p1 = 5, p2 = 2, p3 = 3)
grave elm
it just doesnt seem like it could work
glossy valveBOT
CØSMOS
grave elm
Take any product of any primes - 1. Imagine if we were able to prove that this isn't divisible by any other prime p1 (then it couldnt be divisible by any prime at all)
it literally has to be divisible by one of them
tame cradle
proof by contradiction?
okay
lets us say
if its divisible by b
then we can say the p1.p2.p3.... last - 1 term is 0 mod p1
or its kp1+1
now
can a product of odd terms be kp1 + 1?
aah yes if k is even
fuck
grave elm
then we can say the p1.p2.p3.... last - 1 term is 0 mod p1
i dont even understand what you're saying lol
grave elm
kp1+1
well yes
But the p1 is completely independent of the last term
and the last term has to be divisible by some prime
and that prime cant be p2, p3, ..., ps-1
so it has to be divisible by some other prime
so if you choose p1 to be that prime, you get a counterexample
tame cradle
look
grave elm
so i can literally choose any possible last term, and i'll be able to construct a counterexample by choosing appropriate p1
tame cradle
(p1.p2.... should be kp1+1 where k is even
tame cradle
so i can literally choose any possible last term, and i'll be able to construct ...
bruh
ok
then any conditions do we have lmao
when this would be not div
tame cradle
look we can say one thing
grave elm
any condition you can possibly make will involve p1
tame cradle
lets say each prime is
2k+x1 form
2^s-1(k+x1/2)(k+x2/2)... can we think of something like this maybe
we can distribute mod
grave elm
most of that stuff will be fractions
tame cradle
(2^s-1)mod p1 (k+x1) mod p1 .... and so on
grave elm
you cant mod fractions
grave elm
no
grave elm
i dont think that modular fractions combine with normal fractions this nicely
tame cradle
look
it would be
mostly odd/2
now we have gcd(2,p1)=1
we can maybe try of
some shit
wait the hell
wait wait
we can
one sec
tidal coral
grave elm
i dont understand what kind of condition do you expect to get... give me any primes p2, p3, ..., p_(s-1) and ill always find a counterexample p1
grave elm
ur friends conjecture is wrong
present him with few counterexample
for extra magic points, let him choose the primes p2, p3, p4...
ill show u how it works
give me some primes
any small primes you want, i wont do big prime stuff
okay, say the primes are 7, 11, 13
,calc 71113-1
glossy valveBOT
Result:
1000
look
(p2.p3......ps1) mod p1 - 1 mod p1
now
this means
we would get at last
kp1+1 mod p1 - 1 mod p1
so this needs checking by prime p1
:>
@grave elm
can we but prove for b^2
an odd square is off the form 4m+1
grave elm
can we but prove for b^2
prove that it cant be divisible by b^2?
tame cradle
prove that it cant be divisible by b^2?
yes
grave elm
nope, we cant
still false
3*7*11 - 1 = 1000, choose b = 5. Then b^2 = 25 and it is divisible by that
finding any pattern in this would probably get you fields medal
tame cradle
lmao
grave elm
it would be a pretty significant pattern in primes
tame cradle
lmao
the actual lmao
lmao
my friend lmao
lmao
lemme tell him this
one sec
lmao
he is sad man
bro went to sleep
grave elm
the conjecture is false, it works only if p2 p3 p4 ... ps-1 - 1 has no square factors
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quartz flare
Back with my book again for feedback
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quartz flare
\begin{proof}[Proof of Correctness of Algorithm 4]
We will use induction on $n$ and denote the input sequence $\{a_1, a_2, \dots, a_n\}$ as $\{a\}$.
If $n = 2$, then the outer \textbf{for} loop is executed once, executing the inner \textbf{for} loop once, which checks if $a_1 > a_2$ and interchanges them if so. Thus, if $a_1 > a_2$, the algorithm yields the sequence $\{a_2, a_1\}$, and if not (i.e., $a_1 \leq a_2$), the algorithm yields $\{a_1, a_2\}$. In both cases, these are in increasing order.
Now suppose that for $n = k$, Algorithm 4 yields $\{a\}$ in ascending order, and let $n = k+1$.
The difference between execution when $n = k+1$ and when $n=k$ is one extra iteration of the outer \textbf{for} loop, and one extra comparison in the inner \textbf{for} loop.
In the inner \textbf{for} loop, the extra comparison of $a_k$ and $a_{k+1}$ places the greater as the $k+1$th element of the sequence (as shown in the $n = 2$ base case above).
At each iteration, then, the last two elements remain correctly ordered, and the first $k$ elements remain ordered because the algorithm yields the correct answer in the $n = k$ case.
Finally, in the last iteration of the outer \textbf{for} loop, when $i = k$, the inner \textbf{for} loop runs from $j = 1$ to $n - i = (k+1) - k = 1$, so only elements $a_1$ and $a_2$ are compared. As shown in the base case, these end up in the correct order.
Therefore, after the final iteration, all elements end in the correct (increasing) order, and so Algorithm 4 always produces the correct output.
\end{proof}
Added a line
glossy valveBOT
Coolempire2026
quartz flare
that's an easy to understand proof im reading here
Thank you 🙂
glossy valveBOT
1 divided by 0 equals Infinity
quartz flare
Yes 😆 I probably would have a lot more trouble writing a proof for a more efficient algorithm, so I'm just doing the simple ones to get the hang of it
rare dock
coolemplud
quartz flare
layla 💃
quartz flare
**Coolempire2026**
delicate torrent
coolemplud
how's your channel going 
rare dock
too hard for me to follow at this point
delicate torrent
too hard for me to follow at this point
your channel basically can't die now
rare dock
it can be killed at any time by many people
delicate torrent
it can be killed at any time by many people
it can't be closed basically
rare dock
that's what you think
delicate torrent
that's what you think
?
rare dock
little do you know...
delicate torrent
?
rare dock
delicate torrent
?
rare dock
delicate torrent
Back with my book again for feedback
you might consider sleighla's status
rare dock
quartz flare
you might consider sleighla's status
👀 what does that have to do with anything
delicate torrent
**Coolempire2026**
anyways, i think this proof is good enough
rare dock
👀 what does that have to do with anything
you read my status and it did not help with verifying your proof?
shocking really
quartz flare
anyways, i think this proof is good enough
That's good 👍 half the battle is making something valid 😂 now I just need comments on techniques/choices
I assume there's some more standard way to approach a proof like this
delicate torrent
That's good 👍 half the battle is making something valid 😂 now I just need comm...
i don't mind if arr.sort() comes in
twilit leaf
arr.sort() just magically appeared on everyones computers, good thing it works
delicate torrent
arr.sort() just magically appeared on everyones computers, good thing it works
std::sort(arr.begin(), arr.end(), [](T a, T b) {
// some lambda function that returns a fucking boolean
});
oof your code has an std, better get that checked out
twilit leaf
**Coolempire2026**
i think youre over compkicating this because once you finish the first iteration of the outer for loop, it reduces to the induction hypothesis
delicate torrent
oof your code has an std, better get that checked out
💀
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@quartz flare Has your question been resolved?
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@quartz flare Has your question been resolved?
spiral vigil
I assume there's some more standard way to approach a proof like this
usually you look at invariants
i have a hard time following your proof because of the formatting, but the main thing to show here is that n-1 iterations are actually enough
i think with boba sort you have that the top part of the array is sorted after each iteration
maiden vapor
**Coolempire2026**
this is not quite right
the first k elements remain ordered because the algorithm yields the correct answer in the n = k case.
That statement does not follow. Bubble sort on k+1 elements is not equivalent to Bubble sort on the first k elements, and then do extra comparison with a_{k+1}..
glossy valveBOT
The current time for coolempire93 is 12:08 AM (EST) on Thu, 01/01/2026.
quartz flare
I knew you would have interesting things to say
maiden vapor
happy new year coolemplud
quartz flare
I didn't want to ping you but I did check your time 😂
maiden vapor
quartz flare
happy new year coolemplud
happy new youear!