I will prove this claim using Zorn's Lemma. Note if $X$ is an empty set, then $X$ itself is trivially a maximal pairwise incomparable set, since there are no two distinct $a, b \in X$. So assume $X \neq \emptyset$.
#help-36
manic leaf
soft zealotBOT
toast
onyx peak
show that the poset of incomparable subsets of X ordered by subset relation is nonempty
onyx peak
lets just call the poset of incomparable subsets of X e.g. S btw to make it easier to write. So you'll be applying zorns lemma on S
you need to show:
- S is nonempty
- every chain of S has an upper bound in S
manic leaf
Kk that makes sense
manic leaf
you need to show:
1) S is nonempty
2) every chain of S has an upper bound in S
If S were to be empty wouldn’t that imply X is empty
onyx peak
If S were to be empty wouldn’t that imply X is empty
How?
you need to prove that
maybe X just has no incomparable subsets, who knows
manic leaf
If S is empty there are no incomparable elements of X
But if X has a single element, that single element itself is incomparable vacuously
So X must be empty
onyx peak
If S is empty there are no incomparable elements of X
But if X has a single ele...
okay yeah
singletons are incomparable, so if X is nonempty, S must be nonempty as well
manic leaf
ye, so could I instead perhaps make that a special case like “if X is empty then X itself is the maximal incomparable set”
Now consider sets where X is non empty
Then I can argue any chain of X is no empty
Non empty
onyx peak
Yeah, that works
manic leaf
Kk
onyx peak
Then I can argue any chain of X is no empty
wait wdym by chain? Why cahin?
onyx peak
you only need that S itself is nonempty, not that every chain is nonempty
But yeah, you can argue that S is nonempty as you did above
manic leaf
kk
onyx peak
so now you need to show that every chain of S has an upper bound in S
the general way we prove forall statements is this
so take an arbitrary chain in S and try to construct an upper bound for it
manic leaf
kk
onyx peak
hmmm
What does it even mean for some set U to be an upper bound of chain C in context of our S-order btw?
manic leaf
it contains all the possible incomparable elements for that chain
onyx peak
yeah, it has to contain all the elements of the incomparable subsets in C
what kind of set theoretic tool could allow you to create a set which contains all the elements of those many sets in C
onyx peak
that could work, but we have a better name for that set in this case
if C = {c1, c2, c3, ...} (this isnt very precise since it could be uncountable)
then you basically need a set that contains all the elements of c1 and all the elements of c2 and all the elements of c3, ...
manic leaf
union?
soft zealotBOT
MathIsAlwaysRight
manic leaf
From what I remember the Union of a set acts like the supremum
manic leaf
if C = {c1, c2, c3, ...} (this isnt very precise since it could be uncountable)
...
Oo this makes sense
ye
onyx peak
**MathIsAlwaysRight**
we still need to prove that this is an upper bound in S
manic leaf
Ah so I can say U C exists because of the union axiom and acts as an upper bound for a C, but we have not shown U C is in S
onyx peak
yep, the fact that its an upper bound is quite trivial
but you still need to prove that it belongs to S (i.e. that it's an incomparable subset of X)
manic leaf
could I pick any two terms in U C, show its imcomparable so we know its a incomparable set, then show any arbitrary element in U C is in x, so it’s a subset as well
Ok I need to review what the union axiom actually says
onyx peak
it basically asserts existence of this set
UC = {x | x in c for some c in C}
informally, you can think of it as
c1 u c2 u c3 u c4 u ...
where cn are the elements of C
<@&268886789983436800>
onyx peak
informally, you can think of it as
c1 u c2 u c3 u c4 u ...
where cn are the ele...
this is quite informal though, because C doesnt need to be countable
manic leaf
Take $c_1, c_2 \in \bigcup C$. By the Union Axiom, there exists sets $C_1, C_2 \in C$ where $c_1 \in C_1$ and $c_2 \in C_2$. WLOG assume $C_1 \subseteq C_2$. This implies $c_1 \in C_2$, hence $c_1$ and $c_2$ are incomparable.
soft zealotBOT
toast
manic leaf
Yep! Then just use Zorns lemma to finish it off
Tysm 😭😭
actually goated
I don’t think I would ever think of subset relation so i definitely need more practice
onyx peak
just to reflect on it a bit, this is the most typical ZL problem ever.
You just need to focus on the goal and not on the poset you're given. It's very tempting to apply ZL on the poset you're given, but you should instead think about what maximal element you need
you need to focus on the goal
the problem asks for a maximal pairwise incomparable subset
manic leaf
ye
onyx peak
aha, so our poset is gonna be the poset of pairwise incomparable subsets
and then it tells you how they are ordered
by subsets, that's the A' superset A thingy
and then its also nice to remember that union trick in the end, its a very common way to construct upper bounds on subset order relations
keeping those 2 things in mind, ZL proofs are gonna be much easier
manic leaf
keeping those 2 things in mind, ZL proofs are gonna be much easier
Ah okay
That makes a lot of sense
Tysm
onyx peak
np
manic leaf
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wise parrot
if right and left derivates of a function in a point are different does the function still have a tangent to the graph in that point?
final saddleBOT
if right and left derivates of a function in a point are different does the func...
gritty chasm
depends on the function
gritty chasm
mostly none.
night raft
well yeah it depends
gritty chasm
some may have more than one tangent.
night raft
for example |x| at (0,0)
gritty chasm
anything with too sharp a 'corner', intuitively.
royal gust
We're starting to get to a weird place of "how do you define tangent?"
wise parrot
so if its a sharp point does it have 1 tangent 2 tangent or none?
wise parrot
We're starting to get to a weird place of "how do you define tangent?"
the line made out of the x0 and that x that tends to it wether from right or the left side
i refer to this thing
gritty chasm
so if its a sharp point does it have 1 tangent 2 tangent or none?
more than one, atleast.
infinitely many, acc(bad terminology)
wise parrot
more than one, atleast.
so one like this whould have 3 tangents?
gritty chasm
so one like this whould have 3 tangents?
i dunno how thats gonna have any tangent at the point you indicated
royal gust
the line made out of the x0 and that x that tends to it wether from right or the...
Can you say this a bit more carefully? I think the line passes through x0, which I can accept. But what does it mean for x to "tend to" a line?
gritty chasm
yh thats gonna have more than one tangent.
royal gust
I personally believe that does not have a single tangent
wise parrot
yh thats gonna have more than one tangent.
how do i know how many tangets it will have?
the number of function branches in a bracketed function it glues together?
royal gust
But, I can't refer to any definition of tangent that isn't just a picture idunno
gritty chasm
how do i know how many tangets it will have?
intuition? and if a func. is not differentiable at a certain point, then, ofc its either zero tangents or more than one.
wise parrot
in my language
we call this the tangent
idk the proper
terminology in english
srry
strange sparrow
I think for most elementary purposes it suffices to just say "No, there is no tangent at this point"
royal gust
We typically define tangent through the definition of the derivative itself, so to say "the derivative doesn't exist" is just saying "the tangent doesn't exist".
We could reasonably extend this notion to a "left tangent" and "right tangent" and you could say |x| at x = 0 has a "left tangent" and "right tangent". This isn't typically done, the usefulness of it doesn't go far.
wise parrot
We could reasonably extend this notion to a "left tangent" and "right tangent" a...
is this just for right and left or do u keep splitting the tangent if needed?
gritty chasm
is this just for right and left or do u keep splitting the tangent if needed?
we don't require it much.
tangent is tangent
to split it into left and right is uncanny
wise parrot
what i mean is
if it tends to that dot
with 3 diff formulas
theres a tangent for each formula?
gritty chasm
if the function is differentiable, yes.
differentiable at a point = unique tangent exists at that point
wise parrot
if the function is differentiable, yes.
why does it have to be differentiable, cant it have a derivate without being differentiable?
royal gust
In R there's only two directions, so you could only generate the two tangents.
In R^2 or C or something, you could have infinitely many tangents as you can approach a point from infinitely many directions
(I shouldn't be using the word tangent here, and these are not really tangents, but "kind of tangents")
gritty chasm
why does it have to be differentiable, cant it have a derivate without being dif...
to have a derivative is the definition of differentiable.
idk what is called in your language
wise parrot
in my language
or at least how ive been taught
pls tell me what it refers to
if we can calculate the derivate its said that the function has derivate in it
and if that derivate isnt equal to infinity or minus ifninity
we call that the function is derivable in it
which i think is what differentiable means
wise parrot
*idk what is called in your language*
also do you happen to know why maths differ from country to country?
if there have been other ppl confused like me
gritty chasm
also do you happen to know why maths differ from country to country?
its just the terms. happens here too :D>
gritty chasm
which i think is what differentiable means
you're correct.
final saddleBOT
@wise parrot Has your question been resolved?
wise parrot
i dunno how thats gonna have *any* tangent at the point you indicated
like this ?
fervent plover
help
south dirge
Have you tried anything?
Or are you stuck?
south dirge
Okay, so did you divide by y^2 first?
fervent plover
so y' - Ay = -By^2
y' = - By^2 + Ay
a = 2
u = y^1-a
u = y^-1
south dirge
okay so far so good
fervent plover
like i think homogeneaous and non homogeneous then combine it i
south dirge
find IF and solve?
fervent plover
its first ODE
fervent plover
so i check the answer and i says this
and idk how to get to the y = somthing
figure 4
south dirge
Yeah sorry
fervent plover
ouhhhh okayy
fervent plover
i dont really understand how to use it like i think itsn
u' + Au = 0
and u' + Au = B
yes
south dirge
You dont even need IF her
fervent plover
a and b are constant
south dirge
u' can be rewritten as du/dy correct?
fervent plover
yes
south dirge
Now isolate all terms of u on one side and prepare for intragting both sides
so you have
south dirge
nope whered your B go
fervent plover
so b does not equal 0 then
south dirge
so b does not equal 0 then
yeah
south dirge
is this correct
.
fervent plover
umm wait
south dirge
so b does not equal 0 then
i mean more like we dont know that yet
fervent plover
yea ??
south dirge
right, well now integrate
fervent plover
ok and then whats next
south dirge
is it ln|B-Au| = y + C
when you integrate this
there would be a negative sign outside
i think thats what you may have missed
fervent plover
wait where is the negative sign tho
south dirge
Right now you can just take to the power of e^ on both sides
south dirge
so i check the answer and i says this
hold on js realised what is t
fervent plover
ummm sooo what do i do next
candid hull
sure
fervent plover
its okk
south dirge
js let him know that his integration isnt right
candid hull
well you did it correct
south dirge
hes forgetting / A
fervent plover
okay so do i restart ?
fervent plover
what do i do know
candid hull
u' can be rewritten as du/dy correct?
but this should be du/dt in the first place not du/dy @fervent plover
candid hull
yes
fervent plover
okay soo -1/A ln|B-Au| = y + C
candid hull
take exponential on both sides
candid hull
nope
why is that 1/A still there untouched
exp(-1/A * ln(B-Au)) is what you get on the LHS
candid hull
nope again
candid hull
exp(a*b) = exp(a)^b = exp(b)^a, it works like powers
so exp(-1/A * ln(B-Au)) is exp(ln(B-Au))^(-1/A)
and that exp(ln(..)) you can simplify
candid hull
and how would you do that
fervent plover
so ln|B-Au| = -A(t+c)
candid hull
yeah sure you can do that too
candid hull
here's the subtle part now
let's use a simpler example
say you're interested in all functions y(t) such that |y(t)| = e^t
what can these functions look like ?
fervent plover
umm wdym by that
candid hull
tell me all the y(t)'s you think satisfy this equation
candid hull
why so?
fervent plover
because both sides will equal to 0
candid hull
we're not finished at all with the example
candid hull
say you're interested in all functions y(t) such that |y(t)| = e^t
also I wasn't clear on that, but I want the equation to be true for all possible t
fervent plover
ouhh okay umm so y(t) = e^t or -e^t
candid hull
yes indeed
candid hull
that's not all though
fervent plover
huh?? how
candid hull
what about something like this
it starts off as e^t
then it becomes -e^t
you'd also have |y(t)| = e^t everywhere
this does work
candid hull
well the thing is in differential equations, we're looking for functions that are differentiable
and this funky example I gave you is not, it's not even continuous
fervent plover
ohh so my equation is the same ??
candid hull
ouhh okay umm so y(t) = e^t or -e^t
so really the only differentiable answers turn out to be only these two ones
fervent plover
ouhhhhh
fervent plover
okayyy
candid hull
so |B-Au| = e^-A(t+c)
so you end up with B-Au = +- e^(A(t+c))
candid hull
like if you were working with numbers
but they're not, they're whole functions
this step only works cause we want differentiable functions
fervent plover
okayyy
im sorry but i think i have to go . thanks a lot for ur help i really appreciate it .
candid hull
aight
final saddleBOT
@fervent plover Has your question been resolved?
wise parrot
what is an unghiular point for a derivate?
final saddleBOT
what is an unghiular point for a derivate?
wise parrot
this thing
that when you have a function that has 2 tangents in a point they have a 90 degree angle inbetween them?
<@&286206848099549185>
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Renato
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What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above
for a relation on A to happen, we need R subseteq AxA
so if A has n elements, AxA has n^2 elements
meaning, in total theres like n^2 number of possible relations right?
now, if R is reflexive, then R subseteq {(a,a) | a in A}
how we need to indentify of all the elements in AxA how many are part of the diagonal
thats exactly n elements
blissful meadow
meaning, in total theres like n^2 number of possible relations right?
There are more. n^2 is the number of elements in AxA. The number of relations is the number of subsets of AxA.
gentle zephyr
There are more. n^2 is the number of elements in AxA. The number of relations is...
well, its <= n^2 then
blissful meadow
No. {0,1} has 2 elements. The number of subsets of it is not <= 2...
gentle zephyr
There are more. n^2 is the number of elements in AxA. The number of relations is...
2^(n^2)
blissful meadow
Yes
blissful meadow
2^(n^2) can be interpreted as "for any one of the n^2 pairs in AxA, decide whether it's in the relation or not".
Now for reflexivity you're forced you have (a,a) in R for any a in A.
So you're expecting to make less choices.
How many pairs do you then have to make a choice whether they're in R or not?
gentle zephyr
How many pairs do you then have to make a choice whether they're in R or not?
2^n
I mean the answer to you question is "n" but 2^n is the answer for the number of reflexive relations
blissful meadow
How many pairs have the form (a,a) in AxA?
gentle zephyr
How many pairs have the form (a,a) in AxA?
n
blissful meadow
So for those n pairs, you're force to have them in your relation. There isn't a choice to make for them.
You started with n^2 choice
Now you have n less
gentle zephyr
what are you talking about
blissful meadow
2^(n^2) can be interpreted as "for any one of the n^2 pairs in AxA, decide wheth...
.
For a given relation, you have to choose for each of the n^2 pairs whether they're part of your relation.
For a reflexive relation, you have n less choices to make.
How many choices are left to make?
gentle zephyr
2^n
blissful meadow
You don't need to make a choice for those n pairs anymore.
You started with all relations for which you need to make a choice for n^2 pairs. Now you have n less choices to make.
gentle zephyr
You started with all relations for which you need to make a choice for n^2 pairs...
2^(n^2) + 2^n
help me for the love of god @blissful meadow
blissful meadow
2^(n^2) + 2^n
The fact that I'm probing should alert you that 2^n is wrong. I also said that you should expect less choices, since you have n pairs for which you don't need to make a choice.
If you had n^2 choices and now n less how many are left
gentle zephyr
help
gentle zephyr
If you had n^2 choices and now n less how many are left
2^(n^2 - n)
blissful meadow
Yes.
gentle zephyr
why
2^(n^2) is the number of relations of R
shouldnt it be 2^(n^2 + n)
@blissful meadow
gentle zephyr
2^(n^2 - n)
dude this is counting the number of relations on R which are NOT SYMMETRIC
correct? @blissful meadow
blissful meadow
No.
gentle zephyr
shit doesnt make any sense I dont get it
blissful meadow
Given a pair in (a,b) in AxA, you can choose whether it's in the relation or not. There are two outcomes, and n^2 such pairs, for a total of 2^(n^2) relations.
Now if you require that the choice for the pairs (a,a) in AxA is already made, you have n less choices to make. For each of the remaining n^2-n pairs left, you decide whether it's in the relation or not in succession, for a total of 2^(n^2-n) reflexive relations.
gentle zephyr
Given a pair in (a,b) in AxA, you can choose whether it's in the relation or not...
why 2^(n^2)
blissful meadow
Because you make n^2 choices between two things : in or out.
gentle zephyr
Because you make n^2 choices between two things : in or out.
no
the cardinality of AxA is n^2
blissful meadow
Yes, hence there are n^2 pairs in AxA, for each of which you decide whether it's in the relation or not in the relation.
gentle zephyr
blissful meadow
Giving you a total of 2^(n^2) relations, which fits with a relation defined as just a subset of AxA.
gentle zephyr
Yes, hence there are n^2 pairs in AxA, for each of which you decide whether it's...
what?
blissful meadow
What part is not clear?
gentle zephyr
there are 2^(n^2) possible relations
and out of each one of them since we don't know R is possible they are fulfilling the relation condition or not
so generally speaking at most there are 2^(n^2)
blissful meadow
A relation is just a collection of pairs in AxA. For any given pair, you can decide whether it's in the relation or not. There are n^2 pairs. Therefore there are 2^(n^2) relations.
gentle zephyr
A relation is just a collection of pairs in AxA. For any given pair, you can dec...
there are n^2 tuple elements
since the relation needs to be a subset of this set of n^2 tuple elements, we take the cardinality of the power set
gentle zephyr
say for example we have R = {}
blissful meadow
Yes, this is a relation.
blissful meadow
{} is not a pair
gentle zephyr
{} is not a pair
say for example we have a subset of the n^2 pairs
{} is a subset, and {(1,1), {}}
gentle zephyr
ok because I was wondering
if not (1,1) in R, meaning 1 R 1 is false
{} = {{}, (1,1)}
and we are double counting
blissful meadow
{} = {{}, (1,1)}
This doesn't make sense
There is no double counting. {} is a relation {(1,1)} is a relation.
{{}, (1,1)} is not a relation as it is not a subset of AxA.
{} is not an element of AxA.
Again, for each of the n^2 pairs in AxA, you make a choice between whether it is in the relation or not in the relation.
You have n^2 choices between 2 disjoint options. There is no double counting.
gentle zephyr
There is no double counting. {} is a relation {(1,1)} is a relation.
how does the counting exactly work out?
blissful meadow
Again, for each of the n^2 pairs in AxA, you make a choice between whether it is...
.
This is the exact same argument detailing why the cardinality of the power set is 2^(n^2).
gentle zephyr
Again, for each of the n^2 pairs in AxA, you make a choice between whether it is...
why do you say YOU MAKE A CHOICEwhether it is in the relation or not
blissful meadow
Because this is essentially what you do to build an arbitrary relation.
You decide which pairs are in the relation
gentle zephyr
Because this is essentially what you do to build an arbitrary relation.
but the relation is not given
blissful meadow
Hence why you need to choose which pairs are in it.
blissful meadow
What?
gentle zephyr
then the number of relations are none
gentle zephyr
2^4 = 16
blissful meadow
and R iff x = y = 0
What is this
gentle zephyr
oh
gentle zephyr
you say R subseteq AxA
blissful meadow
I decide whether or not (1,1) is in R : 2 choices
I decide whether or not (1,2) is in R : 2 choices
I decide whether or not (2,1) is in R : 2 choices
I decide whether or not (2,2) is in R : 2 choices
Thus there's 2^4 possibilities.
gentle zephyr
and R iff x = y = 0
this is just when R = {}
gentle zephyr
I decide whether or not (1,1) is in R : 2 choices
I decide whether or not (1,2) ...
you are seeing it from another perspective
blissful meadow
It's the same perspective.
gentle zephyr
multiplicative principle of counting
blissful meadow
The cardinality of the power set is derived the same way...
It just makes it much easier seeing it as a succession of choices for the next parts because you won't be dealing with all subsets of AxA if you impose reflexivity or symmetry...
gentle zephyr
It just makes it much easier seeing it as a succession of choices for the next p...
can you help
blissful meadow
Given a pair in (a,b) in AxA, you can choose whether it's in the relation or not...
.
gentle zephyr
Given a pair in (a,b) in AxA, you can choose whether it's in the relation or not...
why are you subtracting
blissful meadow
Because there are n pairs you know have to be in R already.
gentle zephyr
you are discarding the reflexive relations
blissful meadow
I am not.
blissful meadow
Because there are n pairs you know have to be in R already.
.
If the n pairs are already in there, you don't have to make a choice for those.
There are n^2-n pairs that are "free"
gentle zephyr
There are n^2-n pairs that are "free"
how?
blissful meadow
how?
What part is confusing you? There are n^2 pairs in total. A choice has already been made for n of them. You still have n^2-n choices to make
final saddleBOT
@gentle zephyr Has your question been resolved?
gentle zephyr
What part is confusing you? There are n^2 pairs in total. A choice has already b...
wdym a choice has already been made for n of them
blissful meadow
They’re already in the relation because they’re forced to be otherwise it’s not reflexive
The n pairs (a,a)
gentle zephyr
n of the pairs of (AxA) are already in the relation
blissful meadow
Yes. There’s no choice involved in that
gentle zephyr
oh so thats why you subtract those
blissful meadow
Yes. You can think of it as the relation containing only the pairs (a,a) for each a in A being the “base” or any reflexive relation
And then you need to make your choice for each of the n^2-n pairs
Deciding whether to add them to it or not
You had a good hunch earlier in a way because this is the same thing as throwing away ALL reflexive pairs. So in fact there’s as many reflexive relations as there are relations with NO reflexive pairs.
(This isn’t the same as relations that aren’t reflexive, since those may contain some but not all reflexive pairs)
gentle zephyr
Yes. You can think of it as the relation containing only the pairs (a,a) for eac...
I see so the relations that are reflexive, are a subset of {(a,a) | a in A}?
@blissful meadow
blissful meadow
No the opposite
blissful meadow
Yes
gentle zephyr
I shee
however I still dont understand
why is it 2^(n^2 - n)
2^(n^2) is the total of relations AxA
to those we discard the relations that are surely reflexive, because they are a R a only
@blissful meadow
blissful meadow
You don't discard the relations that are reflexive.
You make the choice for the rest of the n^2-n pairs
That we won't know whether they're in R or not
gentle zephyr
isnt that discarding the relations that are only of the type {(a,a) with a in a}
at the end of the day, {(a,a) with a in A} subseteq R subseteq AxA
@blissful meadow
blissful meadow
When we say there's n^2-n choices left we're not discarding anything.
It's just the choice of (a,a) in R is already made for us from the get-go.
There's still n^2 - n pairs that may or may not be in R.
gentle zephyr
When we say there's n^2-n choices left we're not discarding anything.
It's just ...
It's just the choice of (a,a) in R is already made for us from the get-go.
meaning we ignore those
correct? @blissful meadow
blissful meadow
We ignore them in the sense that we don't make a choice for them.
gentle zephyr
there is this thing called complementary principle of counting
A + A^c = U
U - A^c = A
in this case, 2^(n^2) is our universe
@blissful meadow
blissful meadow
The cardinality of U is 2^(n^2), yes.
The thing is that now you need A^c, which is the set of relations that are not reflexive.
This is harder to count
blissful meadow
Yes, so it doesn't help.
gentle zephyr
I thought you was using that?
blissful meadow
No.
gentle zephyr
2^(N^2) - 2^N
that surely looks like it
maybe I am making a mistake tho
@blissful meadow which other way is it to see it
blissful meadow
Well neither A nor A^c has cardinality 2^n
blissful meadow
<@125307464647180288> which other way is it to see it
The choices is really the easiest way honestly.
You can imagine your relation is a bag and for reflexive relations you already have all pairs (a,a) in the bag.
There's n^2 - n pairs left to pick from
For each of them either put it in the bag or don't. In any case you get a reflexive relation.
So that makes n^2-n successive choices between "in the bag" or "not in the bag"
So 2^{n^2-n} possibilities.
gentle zephyr
the problem is that we need to count the number of reflective relations my dude
2^n are surely reflective
the rest 2^(n^2 - n) we dont know
blissful meadow
My dude, I've explained this 20 times
gentle zephyr
could be or could not be reflexive
gentle zephyr
And also why is 2^(n^2 - n) = 2^(n^2) - 2^n ?
what is your question? why exponentiation rules work?
blissful meadow
Well not like that, that's for sure.
blissful meadow
because there are 2^n subseteq of {(a,a) | a in A}
We're not interested in subsets of {(a,a) | a \in A}
We're interested in supersets of it if anything.
{(1,1)} is a subset of {(a,a) | a in A}
It's not reflexive unless the only element in A is 1
gentle zephyr
yes the cardinal of the superset of {(a,a) | a in A} is 2^n
gentle zephyr
It's not reflexive unless the only element in A is 1
ohhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh
blissful meadow
It has to be 2^{n^2-n}
gentle zephyr
you are right. . .
if A = {1,2} then that shit is not reflexive because 2 R 2 is not part of the set
blissful meadow
Again, you start with the "basic" reflexive relation with n pairs in it. Then you have n^2 - n pairs left for each of which you decide whether it is in or not in the relation.
gentle zephyr
Again, you start with the "basic" reflexive relation with n pairs in it. Then yo...
can you give an down to earth example
because I am not catching up
blissful meadow
A = {1,2,3}. n=3
Pairs are {(1,1), (1,2), (1,3), (2,1), (2,2), (2,3), (3,1), (3,2), (3,3)}.
Start with R = {(1,1), (2,2), (3,3)}. There's 6 pairs left.
For each of them, decide whether or not you add it to R.
6 choices between 2 options gives 64. There are 2^6 = 64 = 2^{3^2 - 3} reflexive relations.
gentle zephyr
the R you chose could have also been R = {(1,1),(2,2)}
blissful meadow
No because that's not reflexive.
The rest of the argument wouldn't work because in one of the possibilities I may choose not to put (3,3) in R
Or for a simpler example with n=2.
A = {1,2}
Pairs are {(1,1), (1,2), (2,1), (2,2)}.
Start with R = {(1,1), (2,2)}.
There are 2 pairs left.
For (1,2), decide whether it's in R : 2 choices
For (2,1), decide whether it's in R : 2 choices.
In total, 4 possible outcomes, so 4 possible reflexive relations. 2^2 = 2^{2^2 - 2} = 4.
gentle zephyr
I dont understand because
if you start with R = {(1,1),(2,2)} you never add anything
blissful meadow
Replace "decide whether it's in R" with "decide whether you add it to R or not"
The point being that you don't make a choice for the reflexive pairs, since we already assume they're in R.
gentle zephyr
noo I see now
say for example
R = {(1,1),(2,2), (1,2)}
this shit is also reflexive
blissful meadow
Yes.
It doesn't matter what you add to it after you have all the reflexive pairs.
gentle zephyr
It doesn't matter what you add to it after you have all the reflexive pairs.
yes... I understand almost everything now
the thing is why you say 2^(n^2) - 2^n
blissful meadow
I never wrote that.
gentle zephyr
I never wrote that.
2^(n^2 - n) = 2^n^2 - 2^n
gentle zephyr
what?
blissful meadow
Because you have n^2-n choices to make
Between yes or no
So 2^(n^2 - n) outcomes
It turns out this is 2^(n^2) / 2^n because the choice of whether the n reflexive pairs are in or not is not being made.
gentle zephyr
2^(n^2) is the cardinal of the pset of AxA
2^n is the cardinal of the pset of pairs (a,a)
blissful meadow
Because you have n^2-n choices to make
Yes, but this doesn't make it any more obvious than that why you would divide those cardinalities.
It's really not that hard : you have n^2-n choices to makes between in or out.
Think of it however you want, a series of choice, a bit string with n^2 - n characters or whatnot.
The total is 2^(n^2-n).
gentle zephyr
can u put it in terms of combinations
blissful meadow
(2 choose 1)^(n^2 -n )
gentle zephyr
(2 choose 1)^(n^2 -n )
jaja
floral vault
hi
gentle zephyr
It's really not that hard : you have n^2-n choices to makes between in or out.
why nn - n
blissful meadow
Because you have n^2 pairs and n of them are certain to be in the relatino.
Leaving you
with
n^2 - n pairs
To decide for
gentle zephyr
I see
so when n = 2 and A = {1,2}
AxA = {(1,1),(2,2),(1,2),(2,1)}
then I have 2^2 pairs from AxA
out of which I know (1,1), (2,2) are in R
now I need to decide if I add or not the other 2 pairs, and the relation will remain reflexive
in this case the possible cases of R are
R1 = {(1,1),(2,2)}
R2 = {(1,1),(2,2), (2,1)}
R3 = {(1,1),(2,2),(1,2)}
R3 = {(1,1),(2,2),(1,2),(2,1)}
@blissful meadow
2(2^2 - 2) = 2^2 = 4
gentle zephyr
It's really not that hard : you have n^2-n choices to makes between in or out.
combinatorics is way too hard for me
blissful meadow
You haven't seen the end of it.
blissful meadow
It hasn't even broken off from the ice cap yet
gentle zephyr
It hasn't even broken off from the ice cap yet
so it gets tougher?
blissful meadow
Of course it does this is like counting the number of sandwiches you can make at subway
gentle zephyr
dude it was hard
blissful meadow
It's not supposed to be 2-hour hard though
gentle zephyr
Of course it does this is like counting the number of sandwiches you can make at...
i appreciate the help azyra
gentle zephyr
It's not supposed to be 2-hour hard though
i am bad tho
blissful meadow
Wait till you do symmetry
gentle zephyr
Wait till you do symmetry
care to elaborate?
blissful meadow
The second part of you question is to count the number of symmetric relations
gentle zephyr
The second part of you question is to count the number of symmetric relations
2^(n^2) - 2^n we discard the reflexive only
oh no we cant
because of them are reflexive and symmetric
A = {1,2,3}
AxA = {(1,1),(1,2),(1,3),(2,1),(2,2),(2,3),(3,1),(3,2),(3,3)}
gentle zephyr
The second part of you question is to count the number of symmetric relations
if they are symmetric they need to be reflexive or something
blissful meadow
No
gentle zephyr
unless its an empty relation or something
blissful meadow
{(1,2), (2,1)} is symmetric
Clearly not reflexive.
So is {(1,2), (1,1), (2,1)}
blissful meadow
Yes. This is not what you claimed though.
gentle zephyr
Yes. This is not what you claimed though.
I changed my argument midway
say n = 3, A = {1,2,3}
R = {(1,1)} is symmetric not reflexive
blissful meadow
It's easier to see with a matrix for this one to determine the number of choices you have to make.
You write down a matrix with 0s and 1s as entries to signify whether the corresponding pair is in or out of the relation.
$\begin{bmatrix} 1 & 0 & 0 \
0 & 1 & 0 \
0 & 0 & 1 \end{bmatrix}$
soft zealotBOT
Azyrashacorki
gentle zephyr
can you make a drawing
blissful meadow
represents the relation {(1,1), (2,2), (3,3)} for instance
blissful meadow
If you look at it from this point of view with matrices, what do you think symmetric relations look like when written this way?
gentle zephyr
**Azyrashacorki**
so basically we need the cardinality of the pset of {(a,a) | a in A}
gentle zephyr
If you look at it from this point of view with matrices, what do you think symme...
$\begin{bmatrix} 1 & 0 & 0 \
0 & 1 & 0 \
0 & 0 & 0 \end{bmatrix}$
soft zealotBOT
Renato
gentle zephyr
this shit is symmetric and not reflexive
blissful meadow
Yes
gentle zephyr
If you look at it from this point of view with matrices, what do you think symme...
I am trying to think of cases of symmetry apart from (a,a)
there is this term
blissful meadow
So reflexive means the diagonal is all 1s.
gentle zephyr
in matrices
blissful meadow
What term?
gentle zephyr
that!
gentle zephyr
They're called symmetric matrices.
at least I am not so rusty with linear algebra, so, what about them?
blissful meadow
What can you say about choices in this case?
gentle zephyr
What can you say about choices in this case?
who tf knows at this point
we need to count the number of symmetric nxn matrices
@blissful meadow
blissful meadow
You really just need to check how many choices you have to make for your pairs.
gentle zephyr
in the case of a 3x3 matrix, we choose the lower triangular numbers, thats 3 choices out of 1 to 9 digits
blissful meadow
That's 3+2+1 = 6 entries you have to choose to get in or not.
gentle zephyr
You really just need to check how many choices you have to make for your pairs.
care to elaborate?
gentle zephyr
That's 3+2+1 = 6 entries you have to choose to get in or not.
the diagonal can be anything
blissful meadow
Yes
The number of entries on the diagonal or below is 6
It's the triangular numbers.
gentle zephyr
The choice of those entries determines a symmetric relation
how?
blissful meadow
For the entries on the diagonal you have two choices each that's fine.
But then once you make a choice for each entry below it it forces the ones above the diagonal to match.
So that if (2,1) is in R, then (1,2) is in R.
gentle zephyr
For the entries on the diagonal you have two choices each that's fine.
But then ...
care to give an down to earth example?
blissful meadow
Say you start with the empty relation
$\begin{bmatrix} & & \
& & \
& & \end{bmatrix}$
gentle zephyr
Say you start with the empty relation
$\begin{bmatrix} & & \\
...
so 0 means is not in R?
blissful meadow
Yes
gentle zephyr
and the index (i,j) is the pair
blissful meadow
You may choose to change entries in the diagonal to 1s.
Or not, so we know those we can choose from.
blissful meadow
I think it's easier if we start with just an actual empty matrix
gentle zephyr
R = {}
soft zealotBOT
Azyrashacorki
blissful meadow
It's ugly but ok
gentle zephyr
**Azyrashacorki**
by vacuous truth this is a equiv rela
blissful meadow
In any case, for the diagonal we can choose without issues.
For the entries below the diagonal, we can make our choices as normal, but every time we make a choice we have to match the corresponding entry above the diagonal to keep the matrix symmetric
So really only the choices of entries on the diagonal and below matter.
The rest is predetermined
In this case, there are 1 + 2 + 3 entries on and below the diagonal .
And for each of them we must decide whether it's a 1 or a 0
That gives 2^6 = 64 outcomes, each of which correspond to a symmetric relation
So the challenge resides in counting the number of entries on and below the diagonal of an n by n matrix
And then taking 2^(that)
Anyways I have to go
Good luck
gentle zephyr
Anyways I have to go
goodbye, ty for the help
gentle zephyr
the sum from 1 to n?
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patent grove
Where did I go wrong here
final saddleBOT
Where did I go wrong here
patent grove
its a SSS triangle
I filled in the angles after but before there were no angles
56.1 is right according to the ai but its all wrong after that
late gazelle
what's the original question?
patent grove
its just the 3 sides a=14, b=12, c=4 find all angles
late gazelle
kk
patent grove
bottom right of the paper is just me finding the biggest number that fits into 112 and 68 top right is where it went wrong I think
late gazelle
bottom right of the paper is just me finding the biggest number that fits into 1...
why do you need to do that?
late gazelle
are you not allowed to use a calculator for this?
patent grove
no you are or else you can't solve it cause you need to get the sin inverse and cosine inverse of some of the fractions
like I said its just a habit its purely semantic
late gazelle
mhm ok
so were given all side lengths already 4, 12, and 14 so we should be able to use cosine rule for each of them
$A=\arccos{\frac{b^2+c^2-a^2}{2bc}}$ \
$B=\arccos{\frac{a^2+c^2-b^2}{2ac}}$ \
$C=\arccos{\frac{a^2+b^2-c^2}{2ab}}$
soft zealotBOT
KB
late gazelle
or if you wanted to use sine rule you could also do that, but ill have to check the numbers rq
so is A = 112.0deg, B=52.6deg, and C = 15.4deg roughly?
final saddleBOT
@patent grove Has your question been resolved?
static oak
I don’t remeber how to find the growth rate I thought I would divide the growth rate that happens every two instead of 1 by two but it doesn’t work
final saddleBOT
I don’t remeber how to find the growth rate I thought I would divide the growth ...
fossil geyser
Instead of dividing by two, you need to think "what number would I need to multiply by twice, that would be the same as multiplying by 25"
Dividing by 2 instead answers the question "what would I need to add twice, in order to add 25"
clear cove
You have $2 \cdot r^2 = 2 \cdot 25$
soft zealotBOT
Earthwyvern
static oak
Instead of dividing by two, you need to think "what number would I need to *mult...
I don’t know how to do that without just guess and checking a million times
clear cove
So then what is r
static oak
like there used to be a way I could find it mathematically but I don’t remember
static oak
So then what is r
5?
clear cove
yeah
fossil geyser
Yes indeed, the square root of 25. Which is 5
static oak
ah kk if it was like 3 steps instead of 2 steps between would I find the cube root then?
clear cove
Yeah
fossil geyser
Yes, that's right
static oak
And would the function formula here be f(x)=2(5)^x
clear cove
yes
static oak
Sorry this is kinda a stupid quesifon but I just don’t know if I’m setting this up correctly and I don’t want to do it all and end up it being all wrong
fossil geyser
Not quite, multiplying by 1.49 represents a percentage growth rate of 49%
Do you know how to express 1.49% as a pure decimal, without the percent symbol?
static oak
Do you know how to express 1.49% as a pure decimal, without the percent symbol?
49%?
Or 4.9
fossil geyser
"%" just means "divided by 100"
or, in other words, move the decimal two places to the left
so 1.49% is 0.0149
static oak
Then how would I write 49% as a decimal or 4.9%?
fossil geyser
49% is 0.49
In a whole number like 49, the decimal is implied to be at the end
Move it two places to the left, you get 0.49
Or alternatively, divide by 100. 49/100 = 0.49
static oak
so 1.49% is 0.0149
So in the () do I write the decimal version?
fossil geyser
You need to do 1 + the decimal version
static oak
I just don’t know how I would not write an expressing exponent
fossil geyser
Because, if something increases by 1%, that means that you have 101% of what you had before
static oak
cus I thought 49% decreasinf would be 0.49 and increasing would be 1.49
static oak
You need to do 1 + the decimal version
1.00149?
fossil geyser
No, the general form is (1 + the decimal), for an increase
or (1 - the decimal), for a decrease
static oak
How would I write it if it was increasing by 49%? Cus I thought it would be 1.49 but now 1.49 is 1.0149
fossil geyser
How would I write it if it was increasing by 49%? Cus I thought it would be 1.49...
no, no, you're right.
An increase of 49% is 1.49
An increase of 1.49% is 1.0149
fossil geyser
Sorry this is kinda a stupid quesifon but I just don’t know if I’m setting this ...
that's why this isn't correct. In this formula, you're increasing by 49% every year
static oak
fossil geyser
,rotate
soft zealotBOT
fossil geyser
That looks good
Now you just need to solve for x, and that's the number of years after 2000 when the population would be 1 million
static oak
So would it be 136.36?
soft zealotBOT
static oak
I think I took the log twice by accident
I fixed it
For this one I’m not sure not to put the half life of 65 days in the equation
fossil geyser
Looks good
(a) is correct, all they're asking for is that formula
(b) is good so far, just need to solve for t, that is the number of days
final saddleBOT
@static oak Has your question been resolved?
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<@&268886789983436800>
gritty chasm
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night raft
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hard blaze
Is the upwards force P, reaction force from newtons third law?
final saddleBOT
Is the upwards force P, reaction force from newtons third law?
gaunt raft
No, there's a force from hinge A
hard blaze
No, there's a force from hinge A
wait what
but why is that force going upwards then if it from hinge A
gaunt raft
The force from the hinge is different from P
Of course not marked in the diagram, it would be naive to assume zero hinge force, otherwise the rod would just tip over
hard blaze
fair, but why does P have a upward force then I'm having trouble seeing where its coming from.
gaunt raft
So your correct equation should be - W=P + (F_(hinge))
gaunt raft
fair, but why does P have a upward force then I'm having trouble seeing where it...
For rotational equilibrium
If force at P were absent, the rod would start spinning
hard blaze
For rotational equilibrium
oh, anticlock wise moment = clockwise moment?
gaunt raft
Are you familiar with torque?
gaunt raft
Best put, for any rigid body in rotational equilibrium, the torque about any point on it must be 0
Choose A as your reference origin (convenient here since it is the pivot about which the rod rotates). Now can you see weight provides clock wise torque?
gaunt raft
oh, anticlock wise moment = clockwise moment?
Precisely
hard blaze
Choose A as your reference origin (convenient here since it is the pivot about w...
and the force P counteracts that clockwise force to keep the rod stable, is that correct?
gaunt raft
and the force P counteracts that clockwise force to keep the rod stable, is that...
Yeah
Clockwise torque*
hard blaze
oops yeah thanks for the correction
It makes so much more sense now, thank you for taking the time to explain it to me. Have a good day dude!
.close
gaunt raft
Cheers
final saddleBOT
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hard blaze
From what point did they calculate the moment in this question? Because there are two pivots
final saddleBOT
From what point did they calculate the moment in this question? Because there ar...
final saddleBOT
@hard blaze Has your question been resolved?
hard blaze
<@&286206848099549185>
I'm mostly confused about the direction of motion the forces are moving because they didn't really specify from which pivot they are caluclating the moments from
coral zenith
From what point did they calculate the moment in this question? Because there ar...
From a quick glance , I can see that they have mentioned they are taking moments about the pivot on the right side
hard blaze
From a quick glance , I can see that they have mentioned they are taking moments...
where?
did i misread the question
coral zenith
where?
You are required to find out F, so obviously you shouldn't take moment about the pivot through which F is acting otherwise the torque due to it would evaluate to zero.
And more so in the answer key, they have mentioned
"As we are trying to find out F take moments from Fr"
hard blaze
"As we are trying to find out F take moments from Fr"
ohhhh
one sec let me draw something and can u tell me if im correct or not?
coral zenith
one sec let me draw something and can u tell me if im correct or not?
Alright
coral zenith
Did i draw the pivot and the direction of motion correctly?
Nope
Check the direction of motion for the 1800N force again
Try to isolate all other forces and focus on only that
hard blaze
Check the direction of motion for the 1800N force again
is my pivot correct atleast or did i take the wrong pivot aswell?
coral zenith
is my pivot correct atleast or did i take the wrong pivot aswell?
It is correct
hard blaze
okay give me a minute to think ill try again
coral zenith
okay give me a minute to think ill try again
Sure
final saddleBOT
hard blaze
actually wait i have one more quick question if u have the time @coral zenith
.reopen
final saddleBOT
✅ Original question: #help-36 message
coral zenith
actually wait i have one more quick question if u have the time <@14674102862345...
Yeah okay
hard blaze
diff question but how is the total moment same if the answer to the 1st question is postiive and the answer to the 2nd question is negative?
coral zenith
diff question but how is the total moment same if the answer to the 1st question...
I mean , they must be talking about the magnitude of the torque
Without taking into account the direction
hard blaze
I mean , they must be talking about the magnitude of the torque
oh so they're saying the direction doesn't matter?
coral zenith
oh so they're saying the direction doesn't matter?
See in this case, it's rather futile to compare directions
cause one is the torque of the couple and other one is moment about the point O
hard blaze
cause one is the torque of the couple and other one is moment about the point O
this might be a stupid question to ask, but whats the difference?
coral zenith
this might be a stupid question to ask, but whats the difference?
A couple are two parallel forces which are equal in magnitude and seperated by some perpendicular distance
And the forces are opposite in direction
If you focus on only the couple of forces given, you can see that it tends to rotate the body in a clockwise motion
hard blaze
If you focus on only the couple of forces given, you can see that it tends to ro...
yeah thats true
coral zenith
yeah thats true
And for the second case, they have written "let's assume this system is in equilibrium"
Which I don't get the reasoning
But you can see that they have taken anticlockwise moments as positive and clockwise as negative , so overall the moment comes out as negative and hence, it rotates in clockwise motion about O
Do you get it?
hard blaze
Do you get it?
not really, why does the fact that the moment has to be negative matters for it to rotate clockwise? what if it was positive would it then rotate anticlockwise?
are they trying to say that the torque of the couple and moment about O rotate in the same clockwise direction? or am i missing the point?
coral zenith
not really, why does the fact that the moment has to be negative matters for it ...
Moment of forces are vector quantities and they have subtracted clockwise moment from anticlockwise moments signifying that they have taken anticlockwise motion as postive
coral zenith
are they trying to say that the torque of the couple and moment about O rotate i...
They do rotate in clockwise direction
hard blaze
Moment of forces are vector quantities and they have subtracted clockwise moment...
okay i get this part
hard blaze
They do rotate in clockwise direction
is that what they were trying to say in the highlighted text?
coral zenith
is that what they were trying to say in the highlighted text?
They are talking about both of them having the same magnitude
They are probably trying to show that fact that the moment of force for a couple is independent of the pivot chosen
final saddleBOT
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left trail
where did our b/2 go?
final saddleBOT
where did our b/2 go?
final tangle
they subbed the whole x+b/2 with u
left trail
they subbed the whole `x+b/2` with `u`
but we only have x in our original integrand?
final tangle
did you skip past that middle section
kindred mortar
it is just a constant
left trail
no. that is how they handled the denominator
final tangle
which b/2 are you referring to then
kindred mortar
document says that the problem is reduced to calculate those 2 integrals
b/2 is a multiplicative constant so it doesn't change anything
desert mantle
they do $\int\frac{x}{(x^2+bx+c)^m} = \int\frac{x+b/2}{(x^2+bx+c)^m} - \int\frac{b/2}{(x^2+bx+c)^m}$ and the second integral is just from the second type
left trail
which b/2 are you referring to then
we have u = x+b/2. One of our integrals has x in the numerator and it seems as though they subbed in u for it
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sonic crystal
Let $f:\mathbb{R}^2 \to \mathbb{R}$ be twice continuously differentiable. Suppose the constraint
$F(x,y)=0$
defines $y=g(x)$ near a critical point $(a,b)$.
Show that
[
\left.\frac{d^2}{dx^2}\bigl[f(x,g(x))\bigr]\right|_{x=a}
\begin{bmatrix} 1 & g'(a) \end{bmatrix}
D^2(f+\lambda F)(a,b)
\begin{bmatrix} 1 \ g'(a) \end{bmatrix}.
]
final saddleBOT
Let $f:\mathbb{R}^2 \to \mathbb{R}$ be twice continuously differentiable. Suppos...
soft zealotBOT
water beam
sonic crystal
so i found the formula for d^2 / dx^2 of f(x,g(x)) but i have no idea how to turn it into the right hand side
sonic crystal
blud
soft zealotBOT
water beam
sonic crystal
ok this is what i got
its not nice looking so idk what to do
i guess im supposed to evaluate at x=a but i dont see how that simplifies stuff
drowsy epoch
Well what did critical point mean
drowsy epoch
huh
strange pelican
beam
drowsy epoch
What do you get applying lagrange multiplier
sonic crystal
uhm
sonic crystal
Erm.. I would check this
idk thats just what thre question says
bold turtle
It's when the gradient is 0
sonic crystal
What do you get applying lagrange multiplier
grad f + lambda grad F = 0
bold turtle
i. e. when all partial derivs are 0
sonic crystal
yeah but what am i even using this on