I wouldn't take the intervals
#help-23
viscid monolith
glass edge
Intervals Frequency
10-19 4
20-29 7
30-39 4
40-49 6
50-59 2
viscid monolith
If i were to represent it, i'd say they are "groups"
viscid monolith
yup
glass edge
30-39?
viscid monolith
uhhh
glass edge
but what if it were even
glass edge
what if the sum of the frequencies were 24
24+1/2
25/2
12.5
which..
once again lies in the 30-39 interval?
viscid monolith
Also 24 is even
you would just take 24/2
12th position
but it is not exactly the middle
And if the 12th and 13th position is within the same group you're fine
If they are not in the same group, you have an issue
Imagine the median lies between the group of students who take bus vs group of students who takes the train
How will your median exist between bus and train commutes?
This video should be better for your example if there are even cases
Cumulative frequency
glass edge
Cumulative frequency
yes im aware of that
viscid monolith
i think its more suited for ur case
glass edge
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novel iron
"There are 8 men and 4 women. A team of 4 must atleast have 2 in it. How many ways are there to do this?"The answer is 201
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"There are 8 men and 4 women. A team of 4 must atleast have 2 in it. How many wa...
novel iron
My attempt: there are C(4,2) ways to choose 2 women from the woman's group
Now we have atleast 2 women
We now have 10 total people that we can choose from, out of which, we want to pick 2
quasi bison
a team of 4 must have at least 2 who in it
novel iron
So, C(4,2)*C(10,2) = 270
quasi bison
brb but i have an explanation for why your logic's wrong
novel iron
a team of 4 must have at least 2 **who** in it
2 woman?
novel iron
brb but i have an explanation for why your logic's wrong
no stress : )
quasi bison
There are 8 men and 4 women. A team of 4 must atleast have 2 in it.
you accidentally the word "women" here.
anyway what you've got is the number of ways to do the following:
choose 2 women, give them gold stars, and then pick 2 more people to go on your team.
i'm going to name the women Alice, Barbara, Cheryl and Diana, and the men Eric, Fred, George, Harry, Ivan, Jonah, Kevin and Larry
with your counting method, you count the following two teams:
{Alice*, Barbara*, Diana, Eric}
{Alice*, Diana*, Barbara, Eric}
as two different teams, when in fact they're the exact same
after all, if someone else does this choice, and sends you the team consisting of Alice, Barbara, Diana and Eric, how can you tell which of the women were "selected first"?
(you can't.)
novel iron
i'm going to name the women Alice, Barbara, Cheryl and Diana, and the men Eric, ...
This is a really nice way to show where I went wrong
I tried thinking about how to modify my approach to avoid this problem but I do not think I can use the approach (multipication principle)
quasi bison
you need to count subtractively
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novel iron
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hard crest
<@&268886789983436800> this's weird, a similar message was sent today
there seem to be a lot of them
dry sierra
Indeed
toxic stratus
cryptots
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shell horizon
can someone explain this move (linear transformations)
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can someone explain this move (linear transformations)
obsidian oracle
the set {x+1,x-1,x^2+2x} is a basis of R_2[X]
So every polynomial in there must be a linear combination of those 3 vectors
The question is, which one?
So you take f(x) = a + bx + cx^2, and you try to express it as a linear combination of x+1, x-1, and x^2+2x
the trick lies in seeing that (x+1) - (x-1) = 2, so a = a * ((x+1)-(x-1))/2
and similar trick with (x+1) + (x-1) = 2x
Otherwise, you can always try to solve a system of equations
Suppose $a+bx + cx^2 = A(x-1) + B(x+1) + C(x^2+2x)$
flat frigateBOT
Rafilouyear2026
obsidian oracle
with a,b,c "known" parameters
obsidian oracle
can someone explain this move (linear transformations)
solve for A,B,C
shell horizon
the trick lies in seeing that (x+1) - (x-1) = 2, so a = a * ((x+1)-(x-1))/2
okkeey the expression here is like 1 and its transformation is known
so it is like a*1
thank you
obsidian oracle
okkeey the expression here is like 1 and its transformation is known
The transformations of (x+1) and (x-1) are known at the beginning, and with this equality we can find the transformation of 1
shell horizon
got it thankss
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earnest mirage
Hello, could someone check if this proof looks good please?
\begin{Theorem}
Suppose $A$ and $B$ are sets, with universal set $U$. Then, $A \setminus B = A \cap B^c$.
\end{Theorem}
\begin{proof}
First, we prove that $A \setminus B \subseteq A \cap B^c$.
Let $x \in A \setminus B$.
Then, by the definition of the substraction, we have that $x \in A$ and $x \notin B$.
So, by the definition of the complement of complement on $B$, $x \in B^c$.
Then $x \in A$ and $x \in B^c$, thus, by the definition of the intersection, $x \in A \cap B^c$.
Since $x \in A \setminus B$ implies $x \in A \cap B^c$, by definition of subset,
we have that $A \setminus B \subseteq A \cap B^c$.
Next, we prove that $A \cap B^c \subseteq A \setminus B$.
Let $x \in A \cap B^c$.
Then, by the definition of the intersection, we have that $x \in A$ and $x \in B^c$.
So, by the definition of the complement on $B$, $x \notin B$.
Then $x \in A$ and $x \notin B$, thus, by definition of the substraction, $x \in A \setminus B$.
Since $x \in A \cap B^c$ implies $x \in A \setminus B$, by the definition of subset,
we have that $A \cap B^c \subseteq x \in A \setminus B$.
We proved that $A \setminus B \subseteq A \cap B^c$ and $A \cap B^c \subseteq x \in A \setminus B$,
therefore $A \setminus B = A \cap B^c$.
\end{proof}
```latexMor Bras
earnest mirage
\begin{Definition}[Subset of a set]
Suppose $A$ and $B$ are sets. If every element in $A$ is also an element of $B$, then $A$ is a \emph{subset} of $B$, which is denoted $A \subseteq B$.
\end{Definition}
\begin{Definition}[Intersection of sets]
The \emph{intersection} of sets $A$ and $B$ is the set $A \cap B = \{ x : x \in A \text{ and } x \in B\}$.
\end{Definition}
\begin{Definition}[Substraction of sets]
The \emph{substraction} of $B$ from $A$ is the set $A \setminus B = \{ x : x \in A \text{ and } x \notin B\}$.
\end{Definition}
\begin{Definition}[Universal set and complement]
Suppose $A$ and $U$ are sets.
If $A \subseteq U$, then $U$ is called the \emph{universal set} of $A$.
The \emph{complement} of $A$ in $U$ is $A^c = U \ A$.
\end{Definition}
Mor Bras
obsidian oracle
Hello, could someone check if this proof looks good please?
```latex
\begin{The...
There are some weird moments where you say things like "$A \cap B^c \subseteq x \in A \setminus B$"
flat frigateBOT
Rafilouyear2026
obsidian oracle
Is this intentional or a typo?
quasi bison
```latex
\begin{Definition}[Subset of a set]
Suppose $A$ and $B$ are sets. I...
are you typing out these definitions for us or for yourself btw
earnest mirage
It's a typo, thanks for pointing it out
earnest mirage
are you typing out these definitions for us or for yourself btw
I write the definitions for those checking the proof
\begin{proof}
First, we prove that $A \setminus B \subseteq A \cap B^c$.
Let $x \in A \setminus B$.
Then, by the definition of the substraction, we have that $x \in A$ and $x \notin B$.
So, by the definition of the complement of complement on $B$, $x \in B^c$.
Then $x \in A$ and $x \in B^c$, thus, by the definition of the intersection, $x \in A \cap B^c$.
Since $x \in A \setminus B$ implies $x \in A \cap B^c$, by definition of subset,
we have that $A \setminus B \subseteq A \cap B^c$.
Next, we prove that $A \cap B^c \subseteq A \setminus B$.
Let $x \in A \cap B^c$.
Then, by the definition of the intersection, we have that $x \in A$ and $x \in B^c$.
So, by the definition of the complement on $B$, $x \notin B$.
Then $x \in A$ and $x \notin B$, thus, by definition of the substraction, $x \in A \setminus B$.
Since $x \in A \cap B^c$ implies $x \in A \setminus B$, by the definition of subset,
we have that $A \cap B^c \subseteq A \setminus B$.
We proved that $A \setminus B \subseteq A \cap B^c$ and $A \cap B^c \subseteq A \setminus B$,
therefore $A \setminus B = A \cap B^c$.
\end{proof}
flat frigateBOT
Mor Bras
obsidian oracle
I write the definitions for those checking the proof
If it's not for yourself, then you don't need to include them. I doubt people checking the proof are complete beginners regarding the definitions in set theory
quasi bison
I write the definitions for those checking the proof
ngl basic shit like set intersection, subset, complement etc are really not necessary
it is like if you were to include a times table with every single algebra question you wanted checked
earnest mirage
ngl basic shit like set intersection, subset, complement etc are really not nece...
I could just put the theorem, then if the reader asks for clarification, I'll provide the definitions. What do you think?
It happened to me before for the definition of divisibility
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earnest mirage
<@&286206848099549185> Hello, could someone check if the proof looks good?
safe radishBOT
@earnest mirage Has your question been resolved?
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@earnest mirage Has your question been resolved?
earnest mirage
Here's the proof:
\begin{Theorem}
Suppose $A$ and $B$ are sets, with universal set $U$. Then, $A \setminus B = A \cap B^c$.
\end{Theorem}
\begin{proof}
First, we prove that $A \setminus B \subseteq A \cap B^c$.
Let $x \in A \setminus B$.
Then, by the definition of the substraction, we have that $x \in A$ and $x \notin B$.
So, by the definition of the complement of complement on $B$, $x \in B^c$.
Then $x \in A$ and $x \in B^c$, thus, by the definition of the intersection, $x \in A \cap B^c$.
Since $x \in A \setminus B$ implies $x \in A \cap B^c$, by definition of subset,
we have that $A \setminus B \subseteq A \cap B^c$.
Next, we prove that $A \cap B^c \subseteq A \setminus B$.
Let $x \in A \cap B^c$.
Then, by the definition of the intersection, we have that $x \in A$ and $x \in B^c$.
So, by the definition of the complement on $B$, $x \notin B$.
Then $x \in A$ and $x \notin B$, thus, by definition of the substraction, $x \in A \setminus B$.
Since $x \in A \cap B^c$ implies $x \in A \setminus B$, by the definition of subset,
we have that $A \cap B^c \subseteq A \setminus B$.
We proved that $A \setminus B \subseteq A \cap B^c$ and $A \cap B^c \subseteq A \setminus B$,
therefore $A \setminus B = A \cap B^c$.
\end{proof}
Mor Bras
uneven knoll
**Mor Bras**
looks good to me, tho its probably better if you use less words for this proof
if you are writing more words than necessary deliberately for now just to get the hang of it then its alright
otherwise you could've written something like:\$x\in A\setminus B\iff x\in A$ and $x\notin B\iff x\in A$ and $x\in B^c\iff x\in A\cap B^c$\ and you would be done like that
flat frigateBOT
ali yassine
uneven knoll
although note that I am not saying using words is bad or using only symbols is great
what I am saying is that sometimes you dont need much words, and instead you can write more symbols to make a neat short proof
but other times you cant just use symbols, you would have to also use words
it depends on what you are trying to prove
but anyways good work, your proof is perfectly fine
I just felt like it might be good to point this out at this stage
earnest mirage
what I am saying is that sometimes you dont need much words, and instead you can...
For sure
Thank you for the advice
uneven knoll
if you are done from this channel please type .close to close the channel
earnest mirage
As of now, I'm practicing writing wordy proofs, since that's the advice from the guide I'm following
.close
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uneven knoll
As of now, I'm practicing writing wordy proofs, since that's the advice from the...
yea thats a good way to start, just keep what i told you in mind for the future when you become more comfortable with proofs
earnest mirage
<@&268886789983436800>
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timber karma
Let $a,b,c$ be real numbers such that the parabola $y = ax^2 + bx + c$ passes through 4 points, $(23,2026),(45,506),(67,p)$, and $(89,507)$. Which is true? $\$ A. $p \geq 1/2 $, B. $ 0 < p < 1/2$, C. $-1/2 < p \leq 0$, D. $p \leq -1/2$
safe radishBOT
Let $a,b,c$ be real numbers such that the parabola $y = ax^2 + bx + c$ passes th...
flat frigateBOT
Copter
timber karma
i have a feeling this is something about second differences, right? but im not too sure how to use it
glass carbon
Notice that e.g. 45-23 = 67 - 45 = 89 - 67
so the difference in increments in x is constant
you can use this to write the condition for increments in y
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quick wraith
Show that the set of differentiable real-valued functions 𝑓 on the interval
(−4, 4) such that 𝑓′(−1) = 3 𝑓 (2) is a subspace of 𝐑^(−4, 4)
quick wraith
Proving closure under addition given that condition
light shoal
how far did you get?
light shoal
do you understand what you need to show?
quick wraith
do you understand what you need to show?
Not really
light shoal
(for closure under addition)
suppose f and g are two such functions
you need to show that f+g is also such a function, which means you need to show:
- f+g is differentiable
- (f+g)'(-1) = 3(f+g)(2)
showing part 2 basically just involves identifying what (f+g)'(-1) and (f+g)(2) mean
quick wraith
I don't understand how functions can map to any number anything
I can make f(x) map to a zero real number
light shoal
well f and g both take a number in (-4,4) and return a number, right?
light shoal
yea that's ok
if you like i can explain how to do a similar problem
then you can use the same ideas here
light shoal
ok, suppose we're working with differentiable real-valued functions on (-4,4) like in your problem
and specifically the ones that satisfy f'(1) = f'(2)
(changed to derivatives to make it more interesting)
suppose f and g are two such functions
that means f and g are differentiable, and f'(1) = f'(2), and g'(1) = g'(2)
you want to show that the sum function, f+g, also has this property
the fact that f+g is differentiable is a basic result from calculus
in particular, sum of two differentiable functions is differentiable, and (f+g)'(x) = f'(x) + g'(x)
so, in particular if we plug in x = 1, we get
(f+g)'(1) = f'(1) + g'(1)
so far this is just using calculus results
now we use the fact that f'(1) = f'(2) and g'(1) = g'(2)
so (f+g)'(1) = f'(2) + g'(2)
and now use the calculus result again on the right hand side:
f'(2) + g'(2) = (f+g)'(2)
so tying it all together, you conclude that
(f+g)'(1) = (f+g)'(2)
quick wraith
so tying it all together, you conclude that
(f+g)'(1) = (f+g)'(2)
Is f'(1) = g'(2) and f'(2) = g'(1)?
light shoal
no, not in general
quick wraith
no, not in general
I edited it
light shoal
no, that's also not true in general
all you know is that f'(1) = f'(2) and g'(1) = g'(2)
quick wraith
I mean
@light shoal, f can be anything that is differentiable and where f'(1) = f'(2). I should be able to get anything within R, as long as it satisfies what I just mentioned
Okay
Thank you
Can you verify my scalar multiplication proof for this example?
light shoal
<@761155030513352705>, f can be anything that is differentiable and where f'(1) ...
yea the value of f'(1) and f'(2) don't matter, all that matters is that they are equal
quick wraith
yea the value of f'(1) and f'(2) don't matter, all that matters is that they are...
Okay
Where can I see the proof on the sum of differentiability
is differentiable
light shoal
check a calculus book
in this context they probably don't expect you to prove that, just to know it as background knowledge
quick wraith
check a calculus book
I don't remember that being told in mine
let c be any real number and f be an element of the set just mentioned
light shoal
I don't remember that being told in mine
example (from stewart's calculus book):
quick wraith
we prove that c times f'(1) = c times f'(2)
quick wraith
we prove that c times f'(1) = c times f'(2)
since multiplication by a real number to a differentiable function gives a differentiable function, then we must have that cf'(1) is differentiable and equal to cf'(2)
Thanks for ending this horrendous week of horror with that problem
😮💨
👋 bye
.close
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weary eagle
I nedd to find the general solution but it says my second and third answers are correct but my first answer is incorrect.
safe radishBOT
I nedd to find the general solution but it says my second and third answers are ...
weary eagle
hard crest
x1 - 4x3 = 2
hard crest
no, you wrote x1 = -4x3 + 2
hard crest
thats what i wrote, it wants me to rearrange it for x_1
then it would be x1 = 4x3 + 2
agile trench
Uhm help please. How do I find the cosine and sine of the circle?
hard crest
it's largely memorisation and special triangles
are you familiar with 45-45-90 and 30-60-90 triangles?
agile trench
are you familiar with 45-45-90 and 30-60-90 triangles?
Yeah
hard crest
ok great, so they didn't mark the angles which is weird, but the marks should be at 0, 30º, 45º, 60º, and 90º; and then continuing like that (so the next one would be 120º, 135º, etc)
agile trench
ok great, so they didn't mark the angles which is weird, but the marks should be...
Ok
But would I be able to see the cosine sine in a t1 84 calculator?
hard crest
you don't need a calculator for this
since you can use the ratio of side lengths in a 30-60-90 triangle
remembering that sine = opp / hyp
agile trench
remembering that sine = opp / hyp
Is this correct?
hard crest
yep!
and your adjacent will always be on the horizontal axis, and your opposite will always be the vertical component
agile trench
and your adjacent will always be on the horizontal axis, and your opposite will ...
Ohh ok
hard crest
and yeah for your answers they should be in radical form, not like decimal or whatever
that is, $\f{\s2}2$ is a totally reasonable answer 😅
flat frigateBOT
hayliänus austrǎlis
hard crest
none of this 0.7073213402... nonsense
agile trench
Is the sine 30/60? Or 60/30?
agile trench
remember the 1-√3-2 stuff?
Wdym?
hard crest
you said you were familiar with 30-60-90 triangles. what do you know about them?
agile trench
you said you were familiar with 30-60-90 triangles. what do you know about them?
They're right triangles with 1:2:3 ratios
hard crest
can you label the sides with their lengths? same for 45-45-90
agile trench
Is this correct?????
hard crest
yeah if we take the 30-60-90 triangle and flip it on itself we get an equilateral triangle
agile trench
maybe find `h`
How do I find h? I'm not the smartest in trig and I don't really knoww
hard crest
pythagorean theorem
wind python
How do I find h? I'm not the smartest in trig and I don't really knoww
tell us pythagorean theorem again
agile trench
tell us pythagorean theorem again
The only thing I remember is a² + b² = c²
wind python
do you remember what triangles that works on
agile trench
do you remember what triangles that works on
Right? If I remember correctly
wind python
yeah so they always have this thing
do you remember how to label a b c in the diagram of a right triangle for a^2 + b^2 = c^2?
agile trench
do you remember how to label `a` `b` `c` in the diagram of a right triangle for ...
I only remember c is the hypotenuse
wind python
right a and b are either leg (which doesn't matter just keep it organized)
agile trench
Ok
agile trench
2 is the hypotenuse right?
wind python
yes the matchup is a -> h, b -> 1, c -> 2
we can change a^2 + b^2 = c^2
into h^2 + 1^2 = 2^2
agile trench
I'm getting either h = 3 or square root 3
wind python
check your work and convince yourself it's root 3
agile trench
I convinced myself by rooting h² and 3 to cancel the ². And I got root 3
wind python
alright we have a 30-60-90 triangle
it has hypotenuse=2 which does not fit the unit circle
a similar triangle which has hypotenuse=1 tells us about the unit circle points
agile trench
a similar triangle which has hypotenuse=1 tells us about the unit circle points
Ohh ok
wind python
there's the similar triangle
just divide all edges in two
so now we can get sins and coss for 30 or 60 degrees from from (easy) soh cah on this trangle
agile trench
so now we can get `sin`s and `cos`s for `30` or `60` degrees from from (easy) so...
Ok
sin is opposite/hypotenuse soo root 3/1?
wind python
yes hypotenuse is conveniently 1 the way i set it so the legs just hand over the sins and coss
all you need to do is sanity checking at this point
hard crest
sin is opposite/hypotenuse soo root 3/1?
it would be √3 / 2
agile trench
it would be √3 / 2
Oh ok
wind python
regrettable that the assignment doesn't even label the angles
this is correct diagram tho
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earnest mirage
Hello, could someone check if this proof looks good please?
\begin{Theorem}
Suppose $A$, $B$ and $C$ are sets. Then, $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$.
\end{Theorem}
\begin{proof}
First, we prove that $A \cup (B \cap C) \subseteq (A \cup B) \cap (A \cup C)$.
Let $x \in A \cup (B \cap C)$.
By the definition of the union, we have that $x \in A$ or $x \in B \cap C$.
And, by the definition of the intersection, $$x \in A \text{ or } (x \in B \text{ and } x \in C).$$
Said differently,
$$(x \in A \text{ or } x \in B) \text{ and } (x \in A \text{ or } x \in C).$$
Then, by the definition of the union, $x \in A \cup B$ and $x \in A \cup C$.
So, by the definition of the intersection, $x \in (A \cup B) \cap (A \cup C)$.
Since $x \in A \cup (B \cap C)$ implies $x \in (A \cup B) \cap (A \cup C)$,
by the definition of the subset, we have that $A \cup (B \cap C) \subseteq (A \cup B) \cap (A \cup C)$.
A similar case can be made by letting $x \in (A \cup B) \cap (A \cup C)$ and following the definitions,
leading to $x \in A \cup (B \cap C)$, thus $(A \cup B) \cap (A \cup C) \subseteq A \cup (B \cap C)$.
We showed that $A \cup (B \cap C) \subseteq (A \cup B) \cap (A \cup C)$ and $(A \cup B) \cap (A \cup C) \subseteq A \cup (B \cap C)$,
therefore $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$.
\end{proof}
severe pond
why are you saying First, we prove that ... when you're just stating what you're supposed to prove
flat frigateBOT
Mor Bras
earnest mirage
why are you saying First, we prove that ... when you're just stating what you're...
Fixed 👍
severe pond
these kinds of proofs are always a bit strange because for example how do you justify the "said differently" part
i would just say, if x is in A then x is in A \cup B and x is in A \cup C. otherwise, if x is in B \cap C then x is in B and x is in C so x is in A \cup B and x is in A \cup C
relies less on this handwavy "said differently"
earnest mirage
i would just say, if x is in A then x is in A \cup B and x is in A \cup C. other...
Yes, this wording looks better.
\begin{proof}
First, we prove that $A \cup (B \cap C) \subseteq (A \cup B) \cap (A \cup C)$.
Let $x \in A \cup (B \cap C)$.
By the definition of the union, we have that $x \in A$ or $x \in B \cap C$.
And, by the definition of the intersection, $$x \in A \text{ or } (x \in B \text{ and } x \in C).$$
Now, if $x \in A$, then $x \in A \cup B$ and $x \in A \cup C$.
Else, if $x \in B \cap C$, then $x \in B$ and $x \in C$, thus $x \in A \cup B$ and $x \in A \cup C$.
So, by the definition of the intersection, $x \in (A \cup B) \cap (A \cup C)$.
Since $x \in A \cup (B \cap C)$ implies $x \in (A \cup B) \cap (A \cup C)$,
by the definition of the subset, we have that $A \cup (B \cap C) \subseteq (A \cup B) \cap (A \cup C)$.
A similar case can be made by letting $x \in (A \cup B) \cap (A \cup C)$ and following the definitions,
leading to $x \in A \cup (B \cap C)$, thus $(A \cup B) \cap (A \cup C) \subseteq A \cup (B \cap C)$.
We showed that $A \cup (B \cap C) \subseteq (A \cup B) \cap (A \cup C)$ and $(A \cup B) \cap (A \cup C) \subseteq A \cup (B \cap C)$,
therefore $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$.
\end{proof}
Mor Bras
severe pond
i would also say you don't have to spell out why this tells us its a subset
you can just say hence A \cup (B \cap C) \subseteq (A \cup B) \cap (A \cup C)
also the other direction isn't really the same thing
like its not a situation where its just a relabeling
might want to actually do the work
try wording it similar to how i did before with the union
earnest mirage
Ok, I'll work on that
safe radishBOT
@earnest mirage Has your question been resolved?
earnest mirage
\begin{proof}
First, we prove that $A \cup (B \cap C) \subseteq (A \cup B) \cap (A \cup C)$.
Let $x \in A \cup (B \cap C)$.
By the definition of the union, we have that $x \in A$ or $x \in B \cap C$.
And, by the definition of the intersection, $$x \in A \text{ or } (x \in B \text{ and } x \in C).$$
Now, if $x \in A$, then $x \in A \cup B$ and $x \in A \cup C$.
Else, if $x \in B \cap C$, then $x \in B$ and $x \in C$, thus $x \in A \cup B$ and $x \in A \cup C$.
So, by the definition of the intersection, $x \in (A \cup B) \cap (A \cup C)$.
Since $x \in A \cup (B \cap C)$ implies $x \in (A \cup B) \cap (A \cup C)$,
by the definition of the subset, we have that $A \cup (B \cap C) \subseteq (A \cup B) \cap (A \cup C)$.
Next, we prove that $(A \cup B) \cap (A \cup C) \subseteq A \cup (B \cap C)$.
Let $x \in (A \cup B) \cap (A \cup C)$.
Then, by the definition of the intersection, $$x \in (A \cup B) \text{ and } x \in (A \cup C).$$
Now, if $x \in A$, then $x \in (A \cup B)$ and $x \in (A \cup C)$.
Otherwise, if $x \in B \cap C$, then $x \in B$ and $x \in C$, so $x \in (A \cup B)$ and $x \in (A \cup C)$.
This means that $x \in A$ or $x \in B \cap C$.
Then, by the definition of the union, $x \in A \cup (B \cap C)$.
Since $x \in (A \cup B) \cap (A \cup C)$ implies $x \in A \cup (B \cap C)$,
by the definition of the subset, we have that $(A \cup B) \cap (A \cup C) \subseteq A \cup (B \cap C)$.
We showed that $A \cup (B \cap C) \subseteq (A \cup B) \cap (A \cup C)$ and $(A \cup B) \cap (A \cup C) \subseteq A \cup (B \cap C)$,
therefore $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$.
\end{proof}
Mor Bras
severe pond
hmm
not sure i follow from the otherwise bit
you're trying to show x is in A \cup (B \cap C)
you have x in (A \cup B) and x in (A \cup C)
so if x is in A then you're done
since x will be in A \cup (B \cap C)
the otherwise should be if x is not in A
but in (A \cup B) and (A \cup C)
do you understand
earnest mirage
I followed the same pattern as for the first subset: either x in A, or x in B \cap C, but now I understand what you pointed out. Let me rephrase it in the proof
severe pond
no but you just copied what i said without adjusting to the different direction
it is not the case now that we assume x is in A \cup (B \cap C)
we assume x is in (A \cup B) \cap (A \cup C)
and continue from there
its clear why if x is in A \cup (B \cap C) then i can just say otherwise after the x in A case
because of our definition of being in the union
severe pond
**Mor Bras**
you just jumped to what we wanted to conclude here though
earnest mirage
\begin{proof}
First, we prove that $A \cup (B \cap C) \subseteq (A \cup B) \cap (A \cup C)$.
Let $x \in A \cup (B \cap C)$.
By the definition of the union, we have that $x \in A$ or $x \in B \cap C$.
And, by the definition of the intersection, $$x \in A \text{ or } (x \in B \text{ and } x \in C).$$
Now, if $x \in A$, then $x \in A \cup B$ and $x \in A \cup C$.
Else, if $x \in B \cap C$, then $x \in B$ and $x \in C$, thus $x \in A \cup B$ and $x \in A \cup C$.
So, by the definition of the intersection, $x \in (A \cup B) \cap (A \cup C)$.
Since $x \in A \cup (B \cap C)$ implies $x \in (A \cup B) \cap (A \cup C)$,
by the definition of the subset, we have that $A \cup (B \cap C) \subseteq (A \cup B) \cap (A \cup C)$.
Next, we prove that $(A \cup B) \cap (A \cup C) \subseteq A \cup (B \cap C)$.
Let $x \in (A \cup B) \cap (A \cup C)$.
Then, by the definition of the intersection, $$x \in (A \cup B) \text{ and } x \in (A \cup C).$$
Now, if $x \in A$, then $x \in (A \cup B)$ and $x \in (A \cup C)$.
Otherwise, if $x \notin A$, then it must be $x \in B$ and $x \in C$,
so $x \in (A \cup B)$ and $x \in (A \cup C)$.
This means that $x \in A$ or $x \in B \cap C$.
Then, by the definition of the union, $x \in A \cup (B \cap C)$.
Since $x \in (A \cup B) \cap (A \cup C)$ implies $x \in A \cup (B \cap C)$,
by the definition of the subset, we have that $(A \cup B) \cap (A \cup C) \subseteq A \cup (B \cap C)$.
We showed that $A \cup (B \cap C) \subseteq (A \cup B) \cap (A \cup C)$ and $(A \cup B) \cap (A \cup C) \subseteq A \cup (B \cap C)$,
therefore $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$.
\end{proof}
Mor Bras
severe pond
since x will be in A \cup (B \cap C)
if its also in A then we are done because of this
if its not in A then it must be in B since x is in A \cup B but it also must be in C because x is in A \cup C
hence x must be in B and C
so x is in B \cap C
and thus x is in A \cup (B \cap C)
earnest mirage
\begin{proof}
First, we prove that $A \cup (B \cap C) \subseteq (A \cup B) \cap (A \cup C)$.
Let $x \in A \cup (B \cap C)$.
By the definition of the union, we have that $x \in A$ or $x \in B \cap C$.
And, by the definition of the intersection, $$x \in A \text{ or } (x \in B \text{ and } x \in C).$$
Now, if $x \in A$, then $x \in A \cup B$ and $x \in A \cup C$.
Else, if $x \in B \cap C$, then $x \in B$ and $x \in C$, thus $x \in A \cup B$ and $x \in A \cup C$.
So, by the definition of the intersection, $x \in (A \cup B) \cap (A \cup C)$.
Since $x \in A \cup (B \cap C)$ implies $x \in (A \cup B) \cap (A \cup C)$,
by the definition of the subset, we have that $A \cup (B \cap C) \subseteq (A \cup B) \cap (A \cup C)$.
Next, we prove that $(A \cup B) \cap (A \cup C) \subseteq A \cup (B \cap C)$.
Let $x \in (A \cup B) \cap (A \cup C)$.
Then, by the definition of the intersection, $$x \in (A \cup B) \text{ and } x \in (A \cup C).$$
Now, if $x \in A$, then $x \in A \cup (B \cap C)$.
Otherwise, if $x \notin A$, then it must be $x \in B$ since $x \in A \cup B$,
and $x \in C$ since $x \in A \cup C$.
So $x \in B \cap C$.
This means that $x \in A$ or $x \in B \cap C$.
Then, by the definition of the union, $x \in A \cup (B \cap C)$.
Since $x \in (A \cup B) \cap (A \cup C)$ implies $x \in A \cup (B \cap C)$,
by the definition of the subset, we have that $(A \cup B) \cap (A \cup C) \subseteq A \cup (B \cap C)$.
We showed that $A \cup (B \cap C) \subseteq (A \cup B) \cap (A \cup C)$ and $(A \cup B) \cap (A \cup C) \subseteq A \cup (B \cap C)$,
therefore $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$.
\end{proof}
🤔
good except the first sentence you say now if x is in A then x is in (A \cup B) and x is in (A \cup C)
you want to show x is in A \cup (B \cap C)
so it should be if x is also in A then x is in A \cup (B \cap C)
flat frigateBOT
Mor Bras
earnest mirage
It's quite late, maybe that's why I'm making such mistakes
severe pond
you're welcome
earnest mirage
<@&286206848099549185> Could someone else check the proof please?
cloud hound
This proof series has been nice to observe
cloud hound
i guess you don't trust me
No they've done that the whole series 😆 multiple opinions always
severe pond
i mean it would make sense
they did just copy a lot of what i said
and seemed shaky on the understanding
but didn't want to feel annoying by asking so many seemingly obvious questions
i get it
plucky elk
Much more annoying to ask some new helper catch up on the entire context from the beginning
I get it
earnest mirage
i guess you don't trust me
I appreciate the time you took to read the proof, pointing out the errors, correct them and reread the proof. I want a second reader to "approve" the proof in order to be sure that it looks good for more people. If there's no one else interested in reading the proof, I can understand.
severe pond
so you don't trust that it's correct
earnest mirage
AFIK, the writing of the proof is subjective, so it might look good me and you, but not for others, which is why second opinions are very appreciated.
astral glacier
AFIK, the writing of the proof is subjective, so it might look good me and you, ...
Learn lean and write the proof there
severe pond
only thing that can change is approach or style
astral glacier
Learn lean and write the proof there
Completely objective
solar hazel
Completely objective
proof?
earnest mirage
only thing that can change is approach or style
Yes, this is the part that differs from only using formulas
severe pond
are you asking for another way to prove this?
solar hazel
i have not read it but i don’t like your proof because it looks 3x longer than it needs to be
severe pond
it seems like you're just not confident with what i said
astral glacier
i have not read it but i don’t like your proof because it looks 3x longer than i...
Bestie it's a set theory proof
That's the appropriate length
severe pond
yea he adds too many words
solar hazel
if i was not on my phone i would write out a short af proof
astral glacier
Ehhh that comes with experience
severe pond
things like "Since x \in A \cup (B \cap C) implies x \in (A \cup B) \cap (A \cup C),
by the definition of the subset, we have that A \cup (B \cap C) \subseteq (A \cup B) \cap (A \cup C)."
cloud hound
The short version is the 1st order logic version
astral glacier
Let's not shame people for making their proofs explicit when they're starting out
solar hazel
the short version is
proof: trivial
warm ocean
proof: trust me bro
safe radishBOT
gray mulch
The short version is the 1st order logic version
Ye, first order logic makes it more concise
safe radishBOT
Available help channel!
nova parcel
In the line :
" In particular, we have .."
Can we write instead :-.
for any m≥k , we have xm<x+ ε=x+(-x)=0
glacial cairn
x + (-x) < 0 ?
If you meant = 0, then sure
The point is you don't need to bother with a "for any m >= K", you can just use one of them, in this case K
safe radishBOT
@nova parcel Has your question been resolved?
safe radishBOT
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Available help channel!
west grail
can someone explain how you draw these functions in the coordinate system like this
safe radishBOT
can someone explain how you draw these functions in the coordinate system like t...
west grail
basically in this case for the area all 3 must be fullfilled
hard crest
you know how to draw a line like 2x - 5y - 10 = 0 right
west grail
no i haven't done in ages, today first day of relearning calculus, well i started like 2 days ago tbh
ok i just looked it up
peak estuary
one option is to rearrange it to y=(something)
west grail
need to bring it into y =
peak estuary
another is to find two points on the line
west grail
another is to find two points on the line
oh you put in numbers into the equasion?
peak estuary
depending on what you mean by that, yes
west grail
well nvm that method sounds more complicated than the other one
y >= 2x/5 - 2
y > -x -3
ok i got the first one
dunno how to draw g: y > -x -3
or f: x-2 < 0
hard crest
west grail
ok i get it when there is a y now but what if there is no y. like x-2 < 0
hard crest
oh yeah that's a special case, it's just a vertical line
so solve for x instead
in that case it would be x < 2
so vertical (dashed) line at 2 and shade to the left
west grail
so if its -2 you go to the right side and for + 2 you go to the left side?
hard crest
no, it's to do with the direction of the < >
oh
hold on what exactly is your question?
west grail
can someone explain how you draw these functions in the coordinate system like t...
in this case they drew it on the right side
i assume cuz of the -2
they drew it through 2
hard crest
yes, they started by solving the inequality for x
to get x < 2
and then shaded all of the points where x is less than (to the left of) 2
west grail
ohhh
well i'll do another example and see if i can do it this time
yup this was another predone example but i did it on my own in my journal and checked and everything was correct
the only issue in this case is that h: -3x -y +6 <=0 doesn't even enclose an area
so only 2 of the 3 functions are used
@peak estuarywhat do you think about this. is the task still solved that way here at 2b)
hard crest
yup this was another predone example but i did it on my own in my journal and ch...
yes this happens sometimes
imagine if you had a system like { y > 0 ; y > 1 }
you also sometimes get systems where there are no solutions
west grail
ok wow i didn't expect the circle to give me so much trouble
safe radishBOT
ok wow i didn't expect the circle to give me so much trouble
west grail
1a i solved pretty fast
but B, C and D of this circle are quite nightmarish
what i calculated for B
B= (-3i -1)/4
ok for D = i get 1/4 i
i jsut did *(-2i) top and bottom for that
B should theoretically be right and with C i have no idea what to do
oh nvm i calculated B wrong cuz they switched up the imaginary part of B with the real one
.close
safe radishBOT
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languid monolith
Hello. Suppose that in a party, the guests clinked their glasses in pairs, and 21 clinks were heard. How many guests are there?
The solution to this exercise, for x guests, is to solve C(x,2) = 21. Why is it C(x,2) ?
safe radishBOT
Hello. Suppose that in a party, the guests clinked their glasses in pairs, and 2...
languid monolith
i cant even picture how the guests clinking their glasses would look
muted sapphire
well lets look at 4 people
languid monolith
okay
muted sapphire
how many clinks will there be
languid monolith
does each person clink with all other 3?
muted sapphire
no, each person only has 1 drink
languid monolith
oh
languid monolith
then with 4 people its 2 clinks
opaque fern
Hello. Suppose that in a party, the guests clinked their glasses in pairs, and 2...
isnt it simple application of combinations
muted sapphire
yes
opaque fern
order doesnt matter
muted sapphire
which is the same as saying
opaque fern
u are choosing 2
muted sapphire
how many pairs can we choose from 4 people
now look back at the problem
we know there are 21 clinks
so how many pairs can we choose from x people?
that'd be C(x,2)
languid monolith
hmm
but we have one specific choice of pairs dont we?
i suck at this lol
like, yeah, you can make pairs of x people in many different ways, but in our problem we choose one way to pair them
toxic stratus
i think it's just bad wording and you're supposed to assume every pair of guests clinked
muted sapphire
okay well we now know that with 4 people, we have 2 clinks true?
languid monolith
yeah
onyx tiger
i think it's just bad wording and you're supposed to assume every pair of guests...
like if two people clink simultaneously it is counted as one
languid monolith
with 5 there would be one who didnt clink tho
pliant citrus
n(n-1)/2 shd be the solution? classic handshake problem isnt it
toxic stratus
no no, its every single possible pair
pliant citrus
diagonal claculation
toxic stratus
so with 3 people A, B, C, it would be
AB
AC
BC
languid monolith
huh
toxic stratus
its just the bad wording
languid monolith
whats a better way to word the problem?
toxic stratus
Suppose that in a party, the guests each clinked their glasses with every other guest, and 21 clinks were heard in total. How many guests are there?
muted sapphire
the question is really interpretable
toxic stratus
theres only one interpretation that leads to the given solution though
languid monolith
oh i see
okay ill think it through then it should be possible
thanks
.close
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Available help channel!
versed wave
is there anything special about a box plot whose box tends towards the upper whisker?
safe radishBOT
is there anything special about a box plot whose box tends towards the upper whi...
versed wave
for example the right box
i can understand that median tending towards the upper/lower part of the box would tell you about the skewness, but what about the box itself to the whisker's ends?
safe radishBOT
@versed wave Has your question been resolved?
safe radishBOT
@versed wave Has your question been resolved?
hazy wasp
The whiskers may allow you to spot outliers
assuming the boxplot was made with the absolute min or max if it is very far from the box it is likely an outlier which you can then calculate
this is why some boxplots are made with the whisker drawn as 3rd quartile + 1.5 IQR and outliers as dots
earnest mirage
Hello, could someone check if this proof looks good please?
\begin{Theorem}
Suppose $A$, $B$ and $C$ are sets. Then, $A \setminus (B \cap C) = (A \setminus B) \cup (A \setminus C)$.
\end{Theorem}
\begin{proof}
First, we prove that $A \setminus (B \cap C) \subseteq (A \setminus B) \cup (A \setminus C)$.
Let $x \in A \setminus (B \cap C)$.
By the definition of the subtraction, we have that $x \in A$ and $x \notin B \cap C$.
This means that $x \notin B$ or $x \notin C$.
If $x \in A$ and $x \notin B$, then $x \in A \setminus B$.
Otherwise, $x \in A$ and $x \notin C$, then $x \in A \setminus C$.
So, we have that $x \in A \setminus B$ or $x \in A \setminus C$.
Thus, $x \in (A \setminus B) \cup (A \setminus C)$.
$x \in A \setminus (B \cap C)$ implies $x \in (A \setminus B) \cup (A \setminus C)$,
hence $A \setminus (B \cap C) \subseteq (A \setminus B) \cup (A \setminus C)$.\\
Next, we prove that $(A \setminus B) \cup (A \setminus C) \subseteq A \setminus (B \cap C)$.
Let $x \in (A \setminus B) \cup (A \setminus C)$.
By the definition of the union, $x \in A \setminus B$ or $x \in A \setminus C$.
In both cases, $x \in A$ and either $x \notin B$ or $x \notin C$, so $x \notin B \cap C$.
Thus, by the definition of the subtraction, $x \in A \setminus (B \cap C)$.
$x \in (A \setminus B) \cup (A \setminus C)$ implies $x \in A \setminus (B \cap C)$,
hence $(A \setminus B) \cup (A \setminus C) \subseteq A \setminus (B \cap C)$.\\
We proved that $A \setminus (B \cap C) \subseteq (A \setminus B) \cup (A \setminus C)$ and
$(A \setminus B) \cup (A \setminus C) \subseteq A \setminus (B \cap C)$, therefore
$A \setminus (B \cap C) = (A \setminus B) \cup (A \setminus C)$.
\end{proof}
Mor Bras
leaden saddle
Looks good and highly rigorous!
If you want (but thats really just a personal style preference) you can work a little more with \implies and (when suited)
\extoverset[8pt]{long expression}{=} (Though I had to look up the exact commands myself, so)
Okay, it doesnt really work the way i thought it would, anyway
(Once you're done, you can type .close)
safe radishBOT
wild cape
.reopen
safe radishBOT
✅ Original question: #help-23 message
wild cape
@earnest mirage its a bit of a jump to go from "x not in B cap C" to "x not in B or x not in C"
yes its intuitive but formally why is it true?
this is the most important part of the whole proof
cloud hound
This is funny
Mor Bras gets admonished yesterday for pinging helpers for a second opinion after getting helped once
Today, a mod reopens Mor Bras' question with an extra point of note
Sad vida
languid monolith
earnest mirage
this is the most important part of the whole proof
Certainly, I'll work a little more in the wording
wild cape
in other words, can you very carefully prove that if $x\notin B\cap C$ then $x\notin B$ or $x\notin C$?
flat frigateBOT
ロケット・ジャンプ
earnest mirage
This is funny
Mor Bras gets admonished yesterday for pinging helpers for a seco...
I'll rather let someone point out the flaws in the proof than let those flaws slip through 
cloud hound
I agree, which is why I was in support of you being able to reping after the first opinion
earnest mirage
I feel that, since I'm not using a lot of logic (yet), the proofs seems to be more wording than what they should
onyx tiger
This is funny
Mor Bras gets admonished yesterday for pinging helpers for a seco...
Lmao
severe pond
This is funny
Mor Bras gets admonished yesterday for pinging helpers for a seco...
this isn't even close to the same thing
delicate shore
this is a really long proof, you can get a cleaner proof simply by expanding the definitions and applying basic logic laws
example for this one:
x ∈ A \ (B ∩ C)
<=> x ∈ A and not(x ∈ B and x ∈ C)
<=> x ∈ A and (x ∉ B or x ∉ C)
<=> (x ∈ A and x ∉ B) or (x ∈ A and x ∉ C)
<=> x ∈ A \ B or x ∈ A \ C
<=> x ∈ (A \ B) ∪ (A \ C)
cloud hound
this isn't even close to the same thing
I wasn't there for the whole thing yesterday so I can't disagree with you, I'll take my leave 👍
severe pond
I wasn't there for the whole thing yesterday so I can't disagree with you, I'll ...
you can literally read the conversation lmao
wild cape
example for this one:
x ∈ A \ (B ∩ C)
<=> x ∈ A and not(x ∈ B and x ∈ C)
<=> x ...
the issue is whether demorgan is considered basic enough that it was already covered in class and can be invoked here
especially since this problem is basically the set version of demorgan
solar hazel
example for this one:
x ∈ A \ (B ∩ C)
<=> x ∈ A and not(x ∈ B and x ∈ C)
<=> x ...
amen 0lante. that’s exactly how i write my proofs of these types of statements
cloud hound
Yes grandpa plante wrote it the way I was thinking when I said first order logic yesterday
solar hazel
#help-23 message it was going to be exactly like that
wild cape
@earnest mirage theres actually a stronger statement that is key to proving both inclusions
$x\notin B\cap C$ if and only if $x\notin B$ or $x\notin C$
flat frigateBOT
ロケット・ジャンプ
safe radishBOT
@earnest mirage Has your question been resolved?
earnest mirage
<@582867906472181760> its a bit of a jump to go from "x not in B cap C" to "x no...
It should be clearer now:
\begin{proof}
First, we prove that $A \setminus (B \cap C) \subseteq (A \setminus B) \cup (A \setminus C)$.
Let $x \in A \setminus (B \cap C)$.
By the definition of the subtraction, we have that $x \in A$ and $x \notin B \cap C$.
By the definition of the intersection, $x$ must be in $B$ and $C$ for $x \in B \cap C$
If $x \notin B$ or $x \notin C$, then $x \notin B \cap C$.
This means that $x \notin B$ or $x \notin C$.
If $x \in A$ and $x \notin B$, then $x \in A \setminus B$.
Otherwise, $x \in A$ and $x \notin C$, then $x \in A \setminus C$.
So, we have that $x \in A \setminus B$ or $x \in A \setminus C$.
Thus, $x \in (A \setminus B) \cup (A \setminus C)$.
$x \in A \setminus (B \cap C)$ implies $x \in (A \setminus B) \cup (A \setminus C)$,
hence $A \setminus (B \cap C) \subseteq (A \setminus B) \cup (A \setminus C)$.\\
Next, we prove that $(A \setminus B) \cup (A \setminus C) \subseteq A \setminus (B \cap C)$.
Let $x \in (A \setminus B) \cup (A \setminus C)$.
By the definition of the union, $x \in A \setminus B$ or $x \in A \setminus C$.
In both cases, $x \in A$ and either $x \notin B$ or $x \notin C$, which is $x \notin B \cap C$.
Thus, by the definition of the subtraction, $x \in A \setminus (B \cap C)$.
$x \in (A \setminus B) \cup (A \setminus C)$ implies $x \in A \setminus (B \cap C)$,
hence $(A \setminus B) \cup (A \setminus C) \subseteq A \setminus (B \cap C)$.\\
We proved that $A \setminus (B \cap C) \subseteq (A \setminus B) \cup (A \setminus C)$ and
$(A \setminus B) \cup (A \setminus C) \subseteq A \setminus (B \cap C)$, therefore
$A \setminus (B \cap C) = (A \setminus B) \cup (A \setminus C)$.
\end{proof}
Mor Bras
wild cape
this is the claim i'd like you to rigorously prove now so that you can invoke it later in the problem
$x\notin B\cap C$ if and only if $x\notin B$ or $x\notin C$
flat frigateBOT
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wild cape
its a very intuitive claim and you might be tempted to try proving it on intuitive grounds, but the formal way to prove it is not as obvious
earnest mirage
its a very intuitive claim and you might be tempted to try proving it on intuiti...
Here's the lemma, it should be clear.
\begin{Lemma}
Suppose $A$ and $B$ are sets. Then, $x \notin A \cap B$ if and only if $x \notin A$ or $x \notin B$.
\end{Lemma}
\begin{proof}
~
\paragraph{$\Rightarrow$} First, suppose that $x \notin A \cap B$.
This means that $x$ is not in both $A$ and $B$.
So, either $x \notin A$ or $x \notin B$.
\paragraph{$\Leftarrow$} Now, suppose that $x \notin A$ or $x \notin B$.
By the definition of the intersection, $x$ must be in $A$ and $B$ for $x \in A \cap B$.
If either $x \notin A$ or $x \notin B$, then $x \notin A \cap B$.
\end{proof}
Mor Bras
wild cape
Here's the lemma, it should be clear.
```latex
\begin{Lemma}
Suppose $A$ and $B...
this is once again an intuitive leap but not a formal one
hint: to prove => use a certain exhaustive set of cases
grave flint
split kayak
mb
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@earnest mirage Has your question been resolved?
earnest mirage
hint: to prove => use a certain exhaustive set of cases
I think I could just use De Morgan's law in the theorem in the first place, since the guide I'm following already uses it and proves it intuitively just for these proofs on sets. Regardless, here's the correction for the lemma.
\begin{proof}
~
\paragraph{$\Rightarrow$} First, suppose that $x \notin A \cap B$.
This means that $x$ is not in both $A$ and $B$.
If $x \in A$ and $x \notin B$, then $x \notin A \cap B$.
Also, if $x \in B$ and $x \notin A$, then also $x \notin A \cap B$.
Lastly, if $x \notin B$ and $x \notin A$, then $x \notin A \cap B$.
So we have that either $x \in A$ and $x \notin B$, or $x \in B$ and $x \notin A$, or $x \notin A$ and $x \notin B$.
For this to be true, either $x \notin A$ or $x \notin B$.
\paragraph{$\Leftarrow$} Now, suppose that $x \notin A$ or $x \notin B$.
By the definition of the intersection, $x$ must be in $A$ and $B$ for $x \in A \cap B$.
If either $x \notin A$ or $x \notin B$, then $x \notin A \cap B$.
\end{proof}
Mor Bras
wild cape
I think I could just use De Morgan's law in the theorem in the first place, sinc...
it works but theres an easier set of cases. ill write it out since youve put in good effort
ideally the proof of <= also uses the framework of cases, "consider the set of cases x notin A, x notin B", but its not nearly as important as =>
ah wait lil edit
assume $x\notin A\cap B$. consider the exhaustive set of cases $x\notin A,~x\in A$. in the case $x\notin A$ we're done (the "or" holds). in the case $x\in A$ we must have $x\notin B$ (so the "or" holds), otherwise if $x\in B$ then $x\in A\cap B$, a contradiction. we conclude $x\notin A$ or $x\notin B$ since we showed its true in all cases
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earnest mirage
Thank you for your comment on the lemma. Returning to the original theorem, here's the correction citing De Morgan's law:
\begin{Theorem}
Suppose $A$, $B$ and $C$ are sets. Then, $A \setminus (B \cap C) = (A \setminus B) \cup (A \setminus C)$.
\end{Theorem}
\begin{proof}
First, we prove that $A \setminus (B \cap C) \subseteq (A \setminus B) \cup (A \setminus C)$.
Let $x \in A \setminus (B \cap C)$.
By the definition of the subtraction, we have that $x \in A$ and $x \notin B \cap C$.
This means that $x \in (B \cap C)^c$.
By De Morgan's law, $x \in B^c \cup C^c$.
So $x \in B^c$ or $x \in C^c$.
This means that $x \notin B$ or $x \notin C$.
If $x \in A$ and $x \notin B$, then $x \in A \setminus B$.
Otherwise, $x \in A$ and $x \notin C$, then $x \in A \setminus C$.
So, we have that $x \in A \setminus B$ or $x \in A \setminus C$.
Thus, $x \in (A \setminus B) \cup (A \setminus C)$.
$x \in A \setminus (B \cap C)$ implies $x \in (A \setminus B) \cup (A \setminus C)$,
hence $A \setminus (B \cap C) \subseteq (A \setminus B) \cup (A \setminus C)$.\\
Next, we prove that $(A \setminus B) \cup (A \setminus C) \subseteq A \setminus (B \cap C)$.
Let $x \in (A \setminus B) \cup (A \setminus C)$.
By the definition of the union, $x \in A \setminus B$ or $x \in A \setminus C$.
In both cases, $x \in A$ and either $x \notin B$ or $x \notin C$.
By the definition of the complement, $x \in B^c $ or $x \notin C^c$.
Then, by the definition of th union, $x \in B^c \cup C^c$.
Lastly, by De Morgan's law, $x \in (B \cap C)^c$, which means that $x \notin B \cap C$.
Thus, by the definition of the subtraction, $x \in A \setminus (B \cap C)$.
$x \in (A \setminus B) \cup (A \setminus C)$ implies $x \in A \setminus (B \cap C)$,
hence $(A \setminus B) \cup (A \setminus C) \subseteq A \setminus (B \cap C)$.\\
We proved that $A \setminus (B \cap C) \subseteq (A \setminus B) \cup (A \setminus C)$ and
$(A \setminus B) \cup (A \setminus C) \subseteq A \setminus (B \cap C)$, therefore
$A \setminus (B \cap C) = (A \setminus B) \cup (A \setminus C)$.
\end{proof}
Mor Bras
wild cape
Thank you for your comment on the lemma. Returning to the original theorem, here...
the thing is the theorem is basically the set version of demorgan (there are two formulations of it for logic and sets), so its quite awkward if not outright circular to invoke it here
thats why we did so much work "from scratch"
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@earnest mirage Has your question been resolved?
earnest mirage
the thing is the theorem is basically the set version of demorgan (there are two...
The proof uses De Morgan's law in order to have equivalences and proceed. I forgot that the book I'm using already stated the law, I believe, in order to be used that way. It was my fault to not include it.
wild cape
The proof uses De Morgan's law in order to have equivalences and proceed. I forg...
there are two laws. can you show how the book states and proves them?
earnest mirage
there are two laws. can you show how the book states and proves them?
Yes, let me write it.
earnest mirage
there are two laws. can you show how the book states and proves them?
Here it is:
\begin{Theorem}
Suppose $A$ and $B$ are subsets of a universal set $U$. Then,
$$(A \cup B)^c = A^c \cap B^c \text { and } (A \cap B)^c = A^c \cup B^c.$$
\end{Theorem}
\begin{proof}
First, we will prove that $(A \cup B)^c \subseteq A^c \cap B^c$.
Let $x \in (A \cup B)^c$.
Then, by the definition of the complement (in $U$), $x \in U$ and $x \notin A \cup B$.
By the definition, $x$ can be in neither $A$ or $B$. Said differently, $x \notin A$ and $x \notin B$,
which by the definition of the complement means $x \in A^c$ and $x \in B^c$.
And hence, by the definition of the intersection, $x \in A^c \cap B^c$.
We have shown that $x \in (A \cup B)^c$ implies $x \in A^c \cap B^c$, which means $(A \cup B)^c \subseteq A^c \cap B^c$.\\
Now, we will prove that $A^c \cap B^c \subseteq (A \cup B)^c$.
Let $x \in A^c \cap B^c$.
Then, by the definition of the intersection, $x \in A^c$ and $x \in B^c$.
By the definition of the complement (in $U$), $x \in U$ and $x \notin A$ and $ x \notin B$.
which by definition of the union means $x \notin A \cup B$.
And hence by the definition of the complement, $x \in (A \cup B)^c$.
We have shown that $x \in A^c \cap B^c$ implies $x \in (A \cup B)^c$ , which means $A^c \cap B^c \subseteq (A \cup B)^c$.\\
We have shown that $(A \cup B)^c \subseteq A^c \cap B^c$ and $A^c \cap B^c \subseteq (A \cup B)^c$.
Together, this demonstrate that $(A \cup B)^c = A^c \cap B^c$.
\end{proof}
Mor Bras
wild cape
Here it is:
```latex
\begin{Theorem}
Suppose $A$ and $B$ are subsets of a univer...
i only see a proof of the first law. did the book leave the proof of the second as an exercise?
earnest mirage
i only see a proof of the first law. did the book leave the proof of the second ...
Yes, it was left as a exercise.
wild cape
moreover this proof is rather informal. similar to your earlier proof it lacks details in the step from "x notin A U B" to "x notin A and x notin B"
earnest mirage
moreover this proof is rather informal. similar to your earlier proof it lacks d...
I can't comment on that. The chapter of these proof exercises is about sets, the next chapter is about induction, and the following chapter about logic.
All while the only proof technique is direct proofs, no contradictions or contrapositive.
wild cape
All while the only proof technique is direct proofs, no contradictions or contra...
if this is strictly necessary in this chapter then perhaps once you finish the chapter on logic then you can supply a more rigorous proof of demorgan (rigorous in the sense of following the rules of propositional logic)
round mango
<modping here>
vivid stream
Please don't delete modpings!
wild cape
bot showed up while i was replying to mor bras. never had such a free ban in a while
wild cape
All while the only proof technique is direct proofs, no contradictions or contra...
again demorgans laws are quite intuitive to understand so it feels like they shouldnt require complicated proofs, but it is surprisingly hard if you wish to adhere to formal rules of deduction
this is an example of following those formal rules
wild cape
Then, how should I proceed with the proofs on this chapter?
maybe first check out the chapter on logic. if it contains weird terms like "disjunctive syllogism" or "modus ponens" then it can probably teach you the propositional logic required to write formal proofs back here
radiant canopy
what topic is this
earnest mirage
**Mor Bras**
Are these proofs not good enough because the logic is intuitive?
wild cape
because the exercise is demorgans law, the proof where you cite demorgan is circular
your first proof can be viewed as good depending on your standard of rigor
regardless of your standard, it does not appeal to the rules of propositional logic
earnest mirage
I see
Thank you for your comments
I'll pay more attention to the logic, even if it's not yet covered
.close
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wild cape
no prob, happy studying
radiant canopy
what topic is this
@wild cape
wild cape
set theory and logic
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deep zenith
i got y=209(0.798)^x
safe radishBOT
i got y=209(0.798)^x
astral glacier
,rccw
flat frigateBOT
astral glacier
i got y=209(0.798)^x
Show your work perhaps?
deep zenith
candid gust
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deep zenith
thank you
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versed isle
A relation R on the set of complex numbers is defined by z1 R z2, if and only if (z1 - z2)/(z1 + z2) is real. Show that R is an equivalence relation
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A relation R on the set of complex numbers is defined by z1 R z2, if and only if...
versed isle
i had to see the solution in which they considered z1 = x1 + iy1 and so on, and proved it to be equivalence, but this doesnt hold when we take z2 to be 0 + 0i
astral glacier
A relation R on the set of complex numbers is defined by z1 R z2, if and only if...
Are you sure this is the correct question
Something seems wrong
!original please
safe radishBOT
Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.
versed isle
Are you sure this is the correct question
i am sure
astral glacier
Well the question is incorrect then cuz 0 is not related to 0 under R
versed isle
!original please
astral glacier
Unless for some reason they exclude 0 from the set of complex numbers
versed isle
Well the question is incorrect then cuz 0 is not related to 0 under R
i didnt quite get wht you meant
astral glacier
Set z1 = z2 = 0
astral glacier
If R was an equivalence relation, you want 0R0
But that's not true
So my best guess is they're working with ℂ*?
versed isle
Look at the fraction
yeah that doesnt even stisfy the relation so why bothering with 0/0 tho, i was talking about the case, z1 bein 2, z2 0 and z3 3i
astral glacier
Dude if 0R0 is false then R is not an equivalence relation
You don't need to check further
And if you're excluding 0 from complex numbers, none of the numbers you consider are allowed to be zero
versed isle
Dude if 0R0 is false then R is not an equivalence relation
but when 0R0 is not even in the solution set they want, why would the relation even be defined from 0 to 0
astral glacier
Do you know what an equivalence relation is
It's reflexive
Everything is related to itself
vestal carbon
but when 0R0 is not even in the solution set they want, why would the relation e...
Wdym solution set 😭
astral glacier
Like I said, either the complex numbers include zero and R isn't an equivalence relation, or the complex numbers exclude zero and then you cannot consider zero as any of the numbers concerned
vestal carbon
If its an equivalence relation, consequentially its reflexive too
versed isle
Do you know what an equivalence relation is
ohh okay nv,, forgot about the reflexive part
astral glacier
...brother
versed isle
...brother
yeah.... i feel it
astral glacier
Ask your teacher for clarification
Or wherever you got this from
My guess is they meant ℂ*
versed isle
Ask your teacher for clarification
okay
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sonic spear
idk what i got wrong, the answer key says it’s 330cm^2
safe radishBOT
idk what i got wrong, the answer key says it’s 330cm^2
nova parcel
idk what i got wrong, the answer key says it’s 330cm^2
330 is volume though...
Are u sure, it's 330 cm² and not cm³
sonic spear
it’s cm3 mb
😭
i’m a terrible reader
OH WE’RE LOOKING FOR VOLUME????
bro i cannot read
ty gang 😭
.close
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twin fulcrum
Could anyone check my stuff pls?
safe radishBOT
Could anyone check my stuff pls?
twin fulcrum
<@&286206848099549185> pls
gritty sable
<@&286206848099549185> pls
It's correct
Js the graph work
Ur answer is -1 ,but it's 2 actually
twin fulcrum
Ur answer is -1 ,but it's 2 actually
dang it 😭 where did i mess up?
gritty sable
dang it 😭 where did i mess up?
Above graph ur calculation is correct
But near graph u got it wrong
timid ridge
But near graph u got it wrong
no this is a different problem
gritty sable
no this is a different problem
Nahh
timid ridge
-1 is correct
timid ridge
the graph is labeled f(x) and g(x)
gritty sable
Though questions are same
onyx tiger
Could anyone check my stuff pls?
for 1) it is correct ,
you found (f-g)(x) = -2x^2 - 3x + 4
you can substitute x = -2 instead of finding f(-2) and g(-2) then doing f(-2) - g(-2)
timid ridge
yeah
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robust pollen
wut
robust pollen
how does this even work
nimble vine
how does this even work
these are rules
naive dragon
,tex .exp rules
flat frigateBOT
give me your fav cartoon series
neon oracle
I've been struggling with rational functions rnw, i dunno what to do.
limit definition of a derivative
naive dragon
what is our task here
mighty mango
I've been struggling with rational functions rnw, i dunno what to do.
limit def...
what is the question
neon oracle
sorry i
forgot to say
its finding the derivative of the given function at a given point
mighty mango
and you're supposed to use the defination????
neon oracle
and you're supposed to use the defination????
sorry, i don't understand
mighty mango
can you use normal derivative rules?
austere goblet
do you have to use the limit defn?
neon oracle
do you *have* to use the limit defn?
yes
mighty mango
neon oracle
that
mighty mango
aw shucks.... good luck
delicate shore
limit of difference quotients
hidden gyro
what's that called by any chance
Limit definition of a derivative
austere goblet
what's that called by any chance
the limit defn of a derivative, exactly what I asked if you must use it
neon oracle
oh
for more context, im stuck at lcd part
i alr have lcd, and i don't know what to do next 😭
mighty mango
I have to suggest that you learn the main widely-used formulas for derivatives. It'll make the process a lot simpler and much much faster.
if you are actually forced to the limit definition, then best of luck (I wouldn't do it that way for a million dollars).
naive dragon
i alr have lcd, and i don't know what to do next 😭
which problem we are working on
neon oracle
I have to suggest that you learn the main widely-used formulas for derivatives. ...
ok thanks (we are forced to do the limit of definition)
naive dragon
ok thanks (we are forced to do the limit of definition)
It's a good way to understand them dw 
Most people don't even know where those formulas are from
neon oracle
this part
neon oracle
<:pandathink:524885344995508235> which problem we are working on
slr
I remembered that i cross multiply it
then thats the result
neon oracle
just plug x=1
can't do that
neon oracle
i need to find the first derivative (if that was called)
naive dragon
naive dragon
i need to find the first derivative (if that was called)
You have to ?
neon oracle
You have to ?
yeah...
onyx tiger
can't do that
why
onyx tiger
i need to find the first derivative (if that was called)
you need f'(x) at 1 so you can plug x=1
or you just need the derivative?
neon oracle
derivative and the plug-in
onyx tiger
ok
neon oracle
after i solve the derivative, that's where i plug the 1
naive dragon
istg I hate this definition so much
can we use the other one
$\lim_{x\to x_0} \frac{f(x)-f(x_0)}{x-x_0}$
flat frigateBOT
give me your fav cartoon series
naive dragon
this right so far you need to simplify
neon oracle
this right so far you need to simplify
naive dragon
well
naive dragon
let f(x)=x+2
do this
naive dragon
what's harder, this or Differentiation Rules?
obviously this one
Differentiation Rules is kinda no brainer
mighty mango
what's harder, this or Differentiation Rules?
differentiation rules are 823049823094823094823094823094802394802 times easier
I remember my first calc lecture when I was taught this alone without the other rules and I quite ltierally skipped 85% of the questions.
if this is not for a grade then just please learn the rules
instead of smth like derivative of x^2 + x taking you 3-4 minutes, It would take you quite ltierally 7 seconds (2 seconds after you get used to it)
onyx tiger
differentiation rules are 823049823094823094823094823094802394802 times easier
lmao why that
naive dragon
if this is not for a grade then just please learn the rules
twin
the only reason I know all this is because I don't do my hw and have to much spare time
mighty mango
the organic chemistry tutor has videos on each and every one of hte rules
there is power rule ---> substraction and addition ---> quotient and product rule
then there's chain rule, doubt you will actually need it any time soon.
@neon oracle
hard crest
chain rule is the best
mighty mango
if you wish, you shall start with the organic chem tutor's videos.
start with power rule
mighty mango
chain rule is the best
amen 🙏
flat frigateBOT
give me your fav cartoon series
naive dragon
This's how you derive quotient rule
naive dragon
after this step you can find f'(x) and g'(x) manually using limit definition
Imma go get dinner
safe radishBOT
@neon oracle Has your question been resolved?
onyx tiger
w writing
neon oracle
after this step you can find f'(x) and g'(x) manually using limit definition
thanks for that
@onyx tiger i have expanded the top, should i expand the lcd?
neon oracle
my teacher just gonna drop a random equation without knowing it will fill the whole paper.
onyx tiger
why does your teacher want to do it this way
naive dragon
You don't have to expanding
Just do what I did and you'll get the answer, genuinely the best way lol
Using differentiation rules in a legit way
neon oracle
why does your teacher want to do it this way
man i dunno
after i subtract it, it turn like −x²Δx+4Δx−4xΔx−x(Δx)²−2(Δx)²
after that uh, that over lcd over ∆x?
onyx tiger
You don't have to expanding
bro has done effort of expanding
onyx tiger
after i subtract it, it turn like −x²Δx+4Δx−4xΔx−x(Δx)²−2(Δx)²
factor ∆x
it cancels with ∆x in denominator
naive dragon
bro has done effort of expanding
I hate algebra, tbh there's no way I do that in a billion years
onyx tiger
should i write it or no
i dont think anything cancels so just simplify the brackets ddon't multiply
or dont simplify
just sub ∆x = 0
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glad fern
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@nova parcel Has your question been resolved?
nova parcel
.close
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spark lynx
Does anyone have any advice on how to do this?
Would it literally just be making individual nodes for the people and connecting them to the task nodes?
compact plover
Would it literally just be making individual nodes for the people and connecting...
it should be just that lol
safe radishBOT
@spark lynx Has your question been resolved?
spark lynx
So it will maybe look something like this then I guess?
Now I just need to figure out if it is bipartite
I'm going to say its not bipartite
I'm not sure why though
stoic dune
Make sure to read the definition of bipartite
spark lynx
Make sure to read the definition of bipartite
So the graph is bipartite, because its vertex's can be divided into two, distinct sets, people and skills
Thats what I've gathered from reading the definition..?
maybe im silly though and I didnt understand it properly
delicate shore
because it can be divided into two sets and every edge connects an elements of one set with an element of the other set
the latter part is the more important part
spark lynx
because it can be divided into two sets *and* every edge connects an elements of...
Ohhh
delicate shore
or in other words you can divide it into two sets such that among each set there are no "internal" edges
spark lynx
For the second question, I'm looking for Bipartite matching resources is?
Maximum bipartite matching the same as Bipartite matching
Or are they different
spark lynx
or in other words you can divide it into two sets such that among each set there...
Thank you 
🫶
delicate shore
For the second question, I'm looking for **Bipartite matching resources** is?
Ma...
I am having trouble parsing your question but most likely the answer is just yes
spark lynx
I am having trouble parsing your question but most likely the answer is just yes
okok ty <3
spark lynx
a.) The graph is bipartite, because it can be divided into two sets and every edge connects an element of one set with an element of another set.
b.) In a Bipartite matching every task is not assigned a staff; but every staff is assigned a task, this is because in a bipartite match only one edge can exist between a node in one set and another.
Thoughts?
safe radishBOT
@spark lynx Has your question been resolved?
safe radishBOT
@spark lynx Has your question been resolved?
delicate shore
It looks OK
this is because in a bipartite match only one edge can exist between a node in one set and another.
this sentence seems logically disconnected from the rest of your answer though
spark lynx
> this is because in a bipartite match only one edge can exist between a node in...
Can you explain more what you mean sorry?
delicate shore
It is not really related to your answer
spark lynx
I made it into one now
spark lynx
It is not really related to your answer
Ohh
Hmm I dont know how to say what I want
In a Bipartite match, there can only be one connection from one node in one set and another, what I mean is like it can't connect to any other nodes in the other set
spark lynx
It is not really related to your answer
I'm not sure how to say it though
Yk what I mean?
delicate shore
you want to say that since there are more tasks than staffs, a matching cannot match all tasks?
the simple explanation is just that every edge in a matching saturates exactly one new vertex on each side
and since there cannot be more edges in a bipartite matching than vertices in either component, the result follows
spark lynx
the simple explanation is just that every edge in a matching saturates exactly o...
yes that! That every edge in a matching saturates one node on both sides!!
Thank you
.close
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weary latch
I'm pretty sure this isn't the correct way to solve the problem. Do I need to use the Law of Sines to find one of the angles before doing the Law of Cosines?
safe radishBOT
I'm pretty sure this isn't the correct way to solve the problem. Do I need to us...
empty gyro
I'm pretty sure this isn't the correct way to solve the problem. Do I need to us...
Looks fine
weary latch
Looks fine
... oh, I realized my mistake.
I forgot to get the square root of the answer.
I got a big number and I was confused.
Okay, I got roughly 18.8.
Thanks for helping me rubber duck this!
.close
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empty gyro
I forgot to get the square root of the answer.
Right. Mb i ignored that too
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hot stone
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hot stone
Original question is the top line
Instructions were to change the LHS to match the RHS using trig identities
So I did that
But why can’t 1-cos/1-cos cancel out into 1
glacial cairn
It does, that's kind of the point
glacial cairn
... yes?
hot stone
So it doesn’t work out? 😭
glacial cairn
The exercise here is to prove $\frac{\sin \theta}{1+\cos \theta} = \frac{1-\cos \theta}{\sin \theta}$
flat frigateBOT
Nel
opal fractal
But why can’t 1-cos/1-cos cancel out into 1
it can cancel out, but i think the exercise wants to direct you towards using 1 - cos² = sin²
hot stone
Yeah
glacial cairn
The first step is to multiply the LHS by $\frac{1-\cos \theta}{1-\cos \theta}$
flat frigateBOT
Nel
hot stone
it can cancel out, but i think the exercise wants to direct you towards using 1 ...
That’s what my teacher did
Im just wondering how it can be done in a different format
glacial cairn
Which is allowed precisely because it equals 1
glacial cairn
If you wrote this, the third line is just wrong
hot stone
Why
glacial cairn
Because there's no valid step that leads to it
opal fractal
That’s what my teacher did
Im just wondering how it can be done in a different ...
probably not a way which is not a roundabout multiplication by 1-cos / 1-cos
since you want to have the 1-cos in the numerator, you kinda have to multiply by it eventually to get the term
hot stone
Because there's no valid step that leads to it
Can’t you treat the denominator as a single factor
So you divide the 1-cos(theta) on top by the one on the bottom
Which gets you 1-cos^-1(theta) which then you can put in the denominator as 1-cos(theta)
And that leaves sin(theta) in the numerator
glacial cairn
You think $\frac{1-\cos \theta}{1 - \cos^2 \theta} = \frac{1}{1-\cos \theta}$?
flat frigateBOT
Nel
glacial cairn
Well, that's wrong
opal fractal
if you explicitly factor out 1 - cos², you get (1-cos)(1+cos)
hot stone
Oh
opal fractal
in the third line, you basically undo what you did in the second line anyway
glacial cairn
You got $1 - \cos^2 \theta$ from $(1 + \cos \theta)(1 - \cos \theta)$ in the previous step, why would you think it can then transform into $(1 - \cos \theta)(1 - \cos \theta)$
flat frigateBOT
Nel
opal fractal
and in the 4th line, you flipped the LHS
glacial cairn
Yes
safe radishBOT
safe radishBOT
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exotic anvil
. @naive osprey post your question here
naive osprey
thanks
naive osprey
quasi timber
naive osprey
it’s in German
naive osprey
no uh the whole area
edgy breach
ok can you think of a way to kind of jigsaw piece the area
naive osprey
my English is very bad I don’t understand a thing