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How do I do part C of this question?
well what is the formula for speed
given the position functions
how would you find the velocity
distance over time
in calculus terms
oh actually they gave you v=(x,y)
yeah
ahh ok my fault i assumed x and y were functions of position
ohh
were you able to do parts a and b?
yes
right so how did you do part a then
hmm you don’t need to graph
and solved it out
plugged 4 into what
yes but for what equation
both x and y
then you did what
that would give you two numbers
that isn’t the speed
that’s the velocity in each direction
right but that’s not the speed
that’s the velocity vector
speed is a scalar quantity
it gives only magnitude
*?
well given two vectors how would you find the magnitude of the resultant?
you have two vectors
velocity in the x and y direction
how do you find the resultant
hint: you used it in geometry class
the formula is a meme
does pythagorean theorem ring any bells
in general speed is the magnitude of velocity which is just =sqrt(vx^2+vy^2)
yes
OH
generally you’ll be given the position in each direction as a function of time
and have to find x’ and y’
well im taking ap pre calc rn and they gave me this as my last unit
then that will be your vx and vy
ohh you’re not in ap calc
yeahhh
i had a feeling this was calc stuff
i remember checking the curriculum in the beginning of last year
there’s a good amount of calc still
yeah
well vectors themselves aren’t calculus really
but the x’ and y’ being the derivatives of the position functions is
but that’s not needed here
oh kk
since they give you the velocity component functions
so for part a you need to find the resultant of what you got
and then for part c you set speed=0 then solve for t
using
^
mhm so you have essentially two legs of a triangle right?
yes
you can treat the vectors as such
then find the resultant
or hypotenuse so to speak
@rich plaza Has your question been resolved?
if I'm supposed to write this to solve for 0, how do I write a system of equations?
because its asking for a system of equations
so if the speed is zero that must mean that both x and y are zero
the only way for the sum of two squares to be zero is if both terms are zero
since squared terms are always positive
you’re solving for the time t such that the speed is zero
so my system of equations would be 0= (x equation) and 0= (y equation)
i dont want to type it out
exactly
it never equals 0
so that means that it doesnt exist
yup
and your reason is because 3^t≠0
that makes more sense now
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Can I have this verified please?
S(a set) denotes all the subsets of that set
Ah k so including duplicates?
Nah it'd be the same
So S(...) yields the same as P(...) just as a list instead of set then
You can always first verify by checking if you have found 2^n subsets where the original set contains n elements
I have 8 elements, yeah.
if you mean are those all the ways to choose elements from {R, Q, N}, basic combination verifies that what you wrote is all of them
if you mean smn else i have no clue 😭
Is my answer right, is my question.
That's one too many parentheses
IS THE LHS A SET
I see, thanks!
Sorry
Not my field, but shouldn’t the rhs be a set?
S denotes list / sequence of all subsets
I don't think it's a good notation, but it's whatever it is
there is also P that denotes the set of all subsets
Also @twilit field , perhaps you forgot to close this?
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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.
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4. I got an answer and would like my work checked.
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7. None of the above
i was sending it man 😭
i worked out the line equation but idk where to go from there
Where did this come from
You’re not necessarily drawing a tangent at a point where y=0
so then what do i do
Write the equation of the tangent for an arbitrary angle and then substitute x=3, y=0
Basically find the tangent for any value theta rather than a specific value
b/c you already found that $\frac{dy}{dx}=-\sin \theta$, so the equation is simply $$y=(-\sin \theta)( x-4\sin \theta)+\cos 2\theta$$
Civil Service Pigeon
oh i see
i found the cartesian equation instead
i got y = 1 - x^2/8
but idk what to do from there
Eh you do the same thing
what is 'the same thing'
Find the tangent for any point on the curve rather than a specific point
@urban viper Has your question been resolved?
ive done that but it was wrong
nvm found it
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this is a fairly unfortunate integral
Ohh no😂
Can you use Laplace transforms for this?
I don't know actually about Laplace
If it works fine. I will surely use
Where did you see that question from?
Someone sent it to me @next bramble
i fail to see how laplace transform would help here
Yeah, it doesn't really help much. Laplace is my go-to when dealing with hard-looking calculus questions like that.
So is this not solvable?
it's solvable just not with a pretty process
Could you tell me the steps? I will try it
@dusky cradle Has your question been resolved?
@feral sedge
@dusky cradle Has your question been resolved?
If you have no tips, strategy to share then why say so..say NO like others said
Lol
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@bright shoal
yes
Yes
Sure
take t^3 common from denominator
Then?
Yeah
Sent
Okok
Use these sign
🤔
1/t^5=1/t*t^4
This one is correct
But
We have another term
Integration by parts
Obviously
So see
How?
It is
Now check on your tab
But i am doing
Yeah
,w int 1/[t^4*(1-1/t^3)^3]
Got it?
Oh k
Where did the 1 go
This image
Here
dv and u
Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online
wolfram alpha letdown
Lol
You are confusing me at these steps u and du
And dv
Let's back to it again
And i know how the integrating by parts
Okom
@dusky cradle
See imma tell simply
Integrate the 1/t^4 and stuff
And differentiate the 1/t
Integrate it
so i just went away for some time
Noo
also "NO" is wrong
U should type in a formal way
so i am not going to tell you "no" when the answer is "yes"
That sry for the inconvenience but I don't have sufficient info on this topic to help u
also you are listening to someone who does not know how to integrate 🤷♂️ so your choice here
U talking bout me?
yes
he is helping perfectly
We are just using ibp
Why you are saying such things sir?
Nevermind, we will continue
Ok
I guess that is why they are saying you wrong
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Given that subset $A={a_1, a_2, a_3, \dots, a_{2001}} \subset \mathbb{N}$, let m be the ammount of subsets of $A$ that has $3$ elements ${a_i, a_j, a_k}$ where $1\leq i < j < k \leq 2001$, such that $a_j=a_i+1$ and $a_k=a_j+1$, for every possibility of set $A$, what is the highest value of m?
im not sure if i translated this correctly, but i gave it my best shot
Skill_Issue
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1
Alr
So the first hint I would give you
Is that integers a,b,c such that b = a+1 and c=b+1 have a name...
(If you write them all in terms of a you will maybe see better)
Yeah
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The equivalent way of writing AF = BG is OF = OG
so it would be great if you could prove the second statement instead
try to create new triangles with one having OF as a side and other having OG as its side
and prove they are congruent
@feral barn Has your question been resolved?
Yep thanks
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1:2 isn't a number
it is simply an idea, however you may also treat it like a number
this is just a definition thing, not worth worrying about
no it's never true
let's say alice and bob both have apples in the ratio 1:2
that means for every apple alice has, bob has 2 apples
so bob has twice as many apples as alice
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|ax|+|by|=1 it will draw a square always?
Is there any rhombus possible?
Square is special case of rhombus when all amgles 90°
So only case?
didnt you post this question before
Yes I posted but i didn't get my desried answer properly
It will be always a square?
rhombus would be if if a != b I think
What is that?
a!?
a=/=b
a is not equal to
$a\neq b$
Skill_Issue
I think so
yes
by the way, you can use a graphing calculator and experiment with it
I have already done
That is why i am here there is no calculator who can give about angles
cough cough eyes cough cough
you can either arctan this, or use geogebra instead
you can calculate it yourself with trig
No issues blindly leave 😑
whar
Am I not doing?
huh
I don't use websites/calculators
I only know is slope formula
you said you used a graphing calculator 😭
And i used it and i got angles 90°@dreamy lichen
.
huh
let's try it here
(1/3/-1/2)=-2/3
so the slope is (1/3) / (1/2)
or this
Of course
2/3 and -2/3, depending on which line we take
Aren't m1m2=-1 here?
Result:
-0.44444444444444
not perpendicular
Let me check another slope
in this pic there is no other slope
parallel lines have same slope
Wait let me check
the top right and down left have -2/3
the top left and down right have 2/3
Btw to calculate the exact angle, we can calculate angle of the bottom-right line relative to x axis
,w arctan(2/3)
,w arctan(2/3) to deg
that gave us the blue angle
note that blue = green
and yellow = 2*green
so ,calc 2*33.69
,calc 2*33.69
Result:
67.38
this is the yellow angle
orange + blue = 90
so orange = 90-blue
and 2*orange = purple
,calc 90 - 33.69
Result:
56.31
,calc 56.31*2
Result:
112.62
sure, if you dont undestand some step, just ask
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in this book it says the monotonic law holds for addition of integers
it says
monotonic law - from a < b it follows that a + b < b + c
is this a misprint
associative law : (a + b) + c = a + (b + c)
monotonic law : from a < b it follows that a + b < b + c
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Ok
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7. None of the above
I dont know where to begin
Ok
1.Commutative law
It says that if there are two numbers a and b
And if we add a and b keeping anyone as the first number
The addition is equal
like
a+b=b+a
yes
You know both the laws right?
yeah
You are mainly asking for the monotonic law
i dont know monotonic
okay i was like confused
Okay
is this a bad book then
Now is your doubt resolved
should i write a bad review on amazon
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Welcome
thanks a ton
Np
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i'm not stuck but somewhat confused as to what to do here, since r2,r3,r4 are all similar
@dense arch Has your question been resolved?
<@&286206848099549185>
after going through RREF, the 2 rows obtained do not correspond with the answer
nvm got it
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So they give us the 55%, would I put that in the bottom right box as .55
and then for the other two boxes I dont really know what to do
Like since its percentages, I was thinking that maybe u have to subtract 100-55 and that answer goes in the other 2 boxes or the top right one
or smth like that
idk tho
I've never done a matrix with percentages
or just a matrix like this
<@&286206848099549185>
<@&286206848099549185>
Assuming that there's a 100% chance a customer will buy some sort of item during their visit, and that you only buy either a food item or a non food item, then I believe you can think of the first column of the matrix as representing the "probability vector" of a person who purchased a food item originally
can you first tell what the NN entry should be?
so the top left would be 0.6 and the bottom left would be 0.4
i thought it would be .55
oh kk
yeah
so the other column would be .55 and .45
good
yes
now it depends on whether the rows represent the first time they bought or the columns
but if the rows do
the the rows need to sum to 1
so you can solve for the other 2
how do I know if the rows are the first time or the columns r
oh
so for the first row I would add .6 and .45
and the second one i would do .4 and .55
but they dont add up to 1
i think that's something you explicitly say
no no
I think maudran is doing it as columsn instead of rows
oh kk
yeah
since it's standard to multipy vectors to the right of matrices
so the columns would be the first time?
that's how I would interpret it
ohh
but maybe either way is acceptable
so like if I chose the row f and column n block
the top right one
I would say that they bought a non food item the first time and bought a food item the second time
and thats 45% of them
How would I do this>
*?
do I just plug in the same matrix twice, multiple them
and then thats my answer?
yeah but you'd need to specify which entry corresponds to the probability they're asking for
oh ok
You'd say that the probability a person chooses a food item the second time if they bought a non food item the first time is 45%
so would that be the top left box then?
ye
oh ok
tyyy
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np
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i was solving this problem where i have to prove RHS = LHS :
cos(2pi/7) + cos(4pi/7) + cos(8pi/7) = -1/2
but then i gave up on solving and checked the solution where they just multiplied and divided by 2sin(pi/7) and used product formula to expand.
my question is why would i even think of multiplying and dividing by 2sin(pi/7)??? how can someone think of approaching the problem this way? like what is the intuition behind multiplying and dividing by 2sin(pi/7)?
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Can some pls check
@visual comet Has your question been resolved?
@visual comet Has your question been resolved?
Yeah it’s fine
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how would i prove this
SAS
how would i prove oc and ob congruent
they are both radii
how do you know
they both go from O to a point on the circle
by definition, points on the circle are which distance away from O?
yeah but what if they just are really close to the circle but theyre not actually on it
if it was just that i wouldve already solved it
is there no other method to prove the triangles congruent
????
I thought the drawing was obvious
what id the point isnt actually on the circlew
"what if A isn't actually on the circle"
in class sometimes they say u cant assume its on the circle
then what's the point of the circle being here
idfk
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Why is this the graph for the parametric equation?
The graph is y with respect to x
e^t is always positive
Express y in terms of the x here
How do u say e^2t in terms of e^t
?
is that like eliminating the parameter
Yes you can say that
so y= e^2x?
But more than "eliminate" you can rather say you are expressing one variable in terms of another
No
What is x?
x is e^t
And y is e^2t
So u want to express y in terms of x
It's like you know a foot is how many cm, and yard is how many cm, so now you want to "eliminate" the cms to express how many feet is a yard
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let $(a,b) \in (\mathbb{R} - \mathbb{Z}) \times \mathbb{N}$. It then follows , from definition, that $a \in (\mathbb{R} - \mathbb{Z}) \implies a \in \mathbb{R}, b \in \mathbb{N}. $ so we have $(a,b) \in \mathbb{R - Z} \times \mathbb{N}.$ However $ a\notin \mathbb{Z}$
(why am i here )= idk
This is my working so far?
Is it fine until now?
So what we essentially want to do is remove all points with integral abscissa
Which is what the operation on the right hand side achieves
but is this possible using set theory too?
(Like can I construct a proof using set theory ?
Yeah, distribution works here too
I know
but don't I have to prove that?
or is that an Axiom in set theory ?
yes/no please
Well when you ask it like that, I remeber distributive law applied to only intersections and unions
I mean we maybe could derive it using propositional logic
do you think the statement is true or false?
well, if you think it's true you'll need to show equality of sets, which I think you've started to do but got stuck?
Yes, pretty much
Yes, I got that too
It Is NOT , but how do I show that
there is nothing to express wdym
I meant I told the exact same thing in words
"since a \in R \setminus Z, a \notin Z, so (a,b) \notin (Z x N)"
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How would I draw a graph that makes this proof easier to understand?
!nopdf
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@past pulsar Has your question been resolved?
@past pulsar Has your question been resolved?
@past pulsar Has your question been resolved?
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Can anyone take photo and send solution for integration of x sin inverse x?
it takes time
It takes time
It does take time indeed
searching will be faster than someone doing it and sending it
because they most likely won't
if however you want people to guide you, they'll be more than happy to help
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Yeah, you can send your attempt at a solution or something of that sort
lol
or less happy now
/can someone solve this question?
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I checked answers and the 2/5 being probabilty of B was instead replaced with probabilty of chem if physics and I dont get why'
do you remember the formula for conditional probability?
alright so in the second image you are given
P(A), P(B), P(B | A) respectively
P(B|A) = P(A n B)/P(A)
now solve for P(A n B)
and you will see why they plugged in P(B | A) instead of P(B)
wdym which one
like between p(B) and P(B|A)
well the third case in the second image is clearly a conditional prob case so you know that you are given P(B | A) and P(A) (case 1) so all you have to do is multiply those two
P(B | A)P(A) = P(A n B)
or
P(A | B)P(B) = P(A n B)
ohh
ye so if they instead gave u the prob of students studying chemistry given they chose to study physics (P(A | B)) then you would have multiplied P(B)
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The definition of union states that a given element belongs to the union if it belongs to any of the sets. On the other hand the definition of intersection states that an element belongs to the intersection if and only if it belongs to all the sets. The only way for this to be true is if all the sets have the same elements.
would this suffice as a proof?
as always you need a mathematical proof
Hm I'd at least write the last sentence formally
"The only way for this to be true is if all sets have the same elements." Prove that directly
you have intuited the result, you should formally prove it
OK.
I mean this doesn't have any reasoning in it
this is basically just a fancy way of saying "it's true because obviously"
So I basically have to prove that they are subsets of one another
I think if you write your statement more mathematically then it's better
let $S=\bigcup_{
By Definition $x\in \bigcap_{\alpha \in I} A_{\alpha} \iff x\in A_i \forall i\in I $
on the other hand $x\in \bigcup_{\alpha \in I} if x\in A_i \text{ for some } i\in I$.
So thus if $x\in \bigcap A_{\alpha}$ it's also in $\bigcup A_{\alpha}$.
\
so $\bigcap A_{\alpha} \subseteq \bigcup A_{\alpha}$
would be the proof in one direction
\iff
ah
(why am i here )= idk
again, you need to say that I isn't empty
we went through the proof earlier
I'm taking that to be obvious because that's mentioned in my book
nothing is obvious to the one that's going to correct you
only union
intersection would be "for every i in empty set..."
which is always true
so it would be the universe set in which you're in
what??
OK. Added that
that just broke my brain
I can talk about it more in DM
sure!
Let $S=\bigcup_{i\in I}A_i=\bigcap_{i\in I}A_i$\
$$\forall x\in U: x\in S\iff x\in A_i\ \forall i\in I$$
kheerii
something like this
(why am i here )= idk
I think I have another idea
$x\in \bigcap_{\alpha \in I} A_{\alpha} \implies x\in A_{\alpha} \forall i \in I$
(why am i here )= idk
Now $x \in \bigcup_{\alpha \in I} \implies x \in A_{\alpha} \text{ for some } i \in I$
(why am i here )= idk
yeah this is basically what I wrote
oops
sorry
now if they are equal
that is $\bigcap_{\alpha \in I} A{\alpha} = \bigcup_{\alpha \in I} A_{\alpha}$
(why am i here )= idk
wait what are you setting out to prove
aren't you trying to prove that A_alpha are all equal to each other?
That all $A_{\alpha}$ are equal
(why am i here )= idk
ok
yeah, going to argue that now
can you check dm's?
try showing that $\bigcap_{\alpha \in I} A_{\alpha} \subseteq A_{\beta}$ and $A_{\beta} \subseteq \bigcup_{\alpha \in I} A_{\alpha}$ for all indices $\beta$
rbit
now for a certain $x$ to belong to the union, it has to belong to atleast one of the sets. But if the same is true for the intersection, it must belong to all the sets. As this is true for all x, the sets are equal
(why am i here )= idk
is this fine?
I would recommend that you start with a specific statement you wanna prove, write it down in mathematical terms then prove it.
you want to prove $$x\in A_i\iff x\in A_j\ \forall\ i,j\in I$$
kheerii
Simplest way to state it in my opinion would look like this:
for any two indicies $\alpha$ and $\beta$ where $\alpha \neq \beta$, $A_{\alpha} = A_{\beta}$
Mohamed Mohsen
mhm
yup, and I've explained that in words, have I not
i mean if you just want an insight as to why this is true then your thing is good enough
I think that this is your recurrent problem
For you proving = explaining in words
Problem is that proofs follow a very particular "syntax"
Such that explaining in words or visualising something isn't gonna fly
I could try writing it mathematically, I guess?
write it in such a way a 2 year old baby could understand. if you read it and you feel any ambiguity, it's not rigorous.
Got it
do 2 year old babies even know how to read?
I did , If I remember right
do people ever hear of metaphors ?
do people ever hear of not taking things too seriously?
it clearly meant make it simple and clear
bro i don't want to but you seem too serious idk why
i am the most unserious person you'll meet
i guess that's my bad xD
read ya wrong
you're good lol
OK. So $x\in \bigcup A_{\alpha} \implies x\in A_{\alpha}$ for some $\alpha \in I$. However, we already know that $x \in \bigcap A_{\alpha}$ which means that $\bigcup \subseteq \bigcap $ in this case
[Fyi its not inherently the fact that your "proof" is written in words that makes it not a proof (see hlounge discussion). It's that your argument written in words is missing logical steps]
is this better?
(why am i here )= idk
how does this help? we already know $\bigcup = \bigcap$
I think this is still missing a few steps
rafilou2003
I think you would gain to write out exactly what you wanna prove at the end of your proof
something like:
another little comment related to what ΣAC was saying, you wanna prove that if x is in one of the As it's in the rest, why doesn't the proof start with, let x be in A_alpha or something like that. if you are trying to prove a statement you should start with the first given and end on what was required.
We suppose (Conditions)....$\$
Then... $\$
But... $\$
$(...) \$
thus (Goal)
rafilou2003
I guess what I'm trying to say that as x is in the union, it must belong to atleast one of the sets.
sure, but is that what we're after?
the proof is simple, just apply aesop tactic. Just kidding
But as this same x is in the intersection, it must belong to all the sets
what are you trying to prove
if a given x( An arbitrary element) belongs to all the sets, the sets must be equal
yes.
and btw do we need the intersection to be non-empty for "the sets must be equal" to be true?
But all the elements in the intersection, are also in the union
sure
and for an element to be in the union it needs to belong to atleast one of the sets
but I still don't think you understand what you want
what was the "thing" you wanted to prove
the sets must be equal right?
yup
yes
so our goal
can you write it mathematically?
so at least we know where we're supposed to go in the mathematical proof
just write the goal for now
The goal is to prove that $A_i = A_j $ for all $i,j \in I$
for some?
for ALL
(why am i here )= idk
let $x$ be an arbitrary element in $A_i$
?
(why am i here )= idk
there's a step before "Ai = Aj"
what is A_i?
A_i is a set
any set?
a set where i \in J
an arbitrary set

(why am i here )= idk
I think it would be enough to state that i is arbitrary though, but that's just my opinion
before going in the Ai=Aj territory
technically you need A_i and A_j both to prove that they're equal
so same with j
A forall statement proof starts with "Let ... be ..."
"forall i,j in I ..."
so the proof start with
"let i,j \in I"
Let $i,j \in I $ where both are arbitrary
drop " A_i, A_j be arbitrary sets such that"
they come for free with your "i,j in I" package
Just this?
let A_i A_j where i,j in I
Let $i,j \in I$ where both are arbitrary
(why am i here )= idk
now we know that $x\in \bigcap$
(why am i here )= idk
who's x
so $x \in A_i, A_j$
(why am i here )= idk
where $x \in \bigcap$
(why am i here )= idk
how is x defined here?
x is an element of the intersection
As I've said before $x \in \bigcup$
(why am i here )= idk
so it belongs to either A_i or A_j
why?
oh right
couldn't it belong to some other rando A_k
it could yeah
yeah so
well, x is an arbitrary elemnt
never proved a set equality before?
Not in 2 years or so
ok well do you remember double inclusion at least?
to prove X = Y, you show every x is in Y and every y is in X
great
so let's do that
so show A_i \subseteq A_j
and then vice versa will be easy
But why is saying that $x\in A_i$ and $x\in A_j$ means that $A_i=A_j$ where both are arbitrary worng ?
(why am i here )= idk
so Ai and Aj have an element in common
so what?
we now repeat this for all their elements
$x_1 from A_i$
(why am i here )= idk
then $x_2$ from A_i
(why am i here )= idk
Compile Error! Click the
reaction for more information.
(You may edit your message to recompile.)
and so on until $x_n$ from $A_i$
(why am i here )= idk
where $i \in I$
(why am i here )= idk
so you take a finite amount of x, and only from Ai?
And we then repeat this process for $A_j$
(why am i here )= idk
ok, just something to tell you
once you took an ARBITRARY guy from the first set
and shown it's in the other set
it's done
there's no "repeating the process"
I see
showing it always works once is gonna have the same result as showing it always works multiple times
Okay, so now I prove that they are subsets of one another