help-39

[sonic patrol]

Do you see how these are the same as infinite sequences of binary digits?

[barren root]

Yes

[sonic patrol]

Cool

So, we can identify T with $2^{\mathbb{N} }$

[jolly parrot]

Pseudonium

[barren root]

What does "identify" mean?

[sonic patrol]

I.e. we can say $T = 2^{\mathbb{N} }$

[jolly parrot]

Pseudonium

[barren root]

Yes, I see. We have an infinite set of infinite binary sequences

[sonic patrol]

Mhm

The proof starts by assuming T is countable, right?

[barren root]

Yes

[sonic patrol]

So, we have some $f : \mathbb{N} \to 2^{\mathbb{N} }$

[jolly parrot]

Pseudonium

[sonic patrol]

Which is supposed to be a bijection

But what cantor does is find an $s \in 2^{\mathbb{N} }$ that is outside the image of f

[jolly parrot]

Pseudonium

[sonic patrol]

However, s is still an infinite binary sequence

So it’s still an element of $2^{\mathbb{N} }$

[jolly parrot]

Pseudonium

[barren root]

Yeah, it was created on the assumption that it differs from all numbers in 2^N but it is there by its definition. Russel's Paradox. But why does it prove that there are uncountable sets? Because the assumption that T is countable doesn't hold?

[sonic patrol]

We derive a contradiction from the assumption that T is countable

Hence, T cannot be countable

So T must be uncountable

[barren root]

That indeed sounds true, but how can it be uncountable...

[sonic patrol]

Well

If it were countable, we could derive a contradiction?

And we don’t want contradictions

[barren root]

So that it is uncountable means what?

[sonic patrol]

It means there is no bijection with $\mathbb{N}$ (and that it’s not finite, but that’s easy)

[jolly parrot]

Pseudonium

[barren root]

So it appears we cannot make an enumeration of T.
But we supposedly can make an enumeration of Z.
Couldn't we make the same argument by e.g. taking the biggest number from Z and adding +1 to it?

I suppose I should make an example of the function you provided. The bijective function wouldn't allow such an argument since there would always be a number which is bigger by 1

Seems like being countable is a specific case of being in bijection, but here being in bijection with N

[sonic patrol]

Yes, or being finite

Typically we say that finite sets are also countable

Cause, well, you can count all their elements

[pearl pond]

@barren root Has your question been resolved?

[pearl pond]

@barren root Has your question been resolved?

[sonic patrol]

Wait do you still have a Q

[pearl pond]

@barren root Has your question been resolved?

[barren root]

Yeah I'm still pondering it over on why it causes a contradiction

[sonic patrol]

Wdym

you can think of it as just showing that any function $f : \mathbb{N} \to 2^{\mathbb{N} }$ is non-surjective

[jolly parrot]

Pseudonium

[sonic patrol]

So assuming that $2^{\mathbb{N} }$ is countable yields a contradiction

[jolly parrot]

Pseudonium

[sonic patrol]

Because given any function $f : \mathbb{N} \to 2^{\mathbb{N} }$, one can find an $x \in 2^{\mathbb{N} }$ that is not in the image

[jolly parrot]

Pseudonium

[barren root]

I don't understand why we need it to be countable to use the diagonal argument considering that we already use it on infinite sets

[sonic patrol]

So

The following is true for any set X

Let $f : X \to 2^X$ be a function

[jolly parrot]

Pseudonium

[sonic patrol]

Then there exists a $y \in 2^X$ outside the image of f

[jolly parrot]

Pseudonium

[barren root]

Perhaps the technique just shouldn't be used on sets which are defined to contain all numbers of infinite length? Because it yields an error in this case

[sonic patrol]

I don’t understand what the error is

Do you understand why the claim i just stated is true?

[barren root]

What is y?

Because if X has only e.g. 3 elements in it then the second set has only 8 elements and that's it

[sonic patrol]

So!

We can define y as follows

$y = {x \in X | x \not \in f(x) }$

[jolly parrot]

Pseudonium

[sonic patrol]

This is a subset of X

So we do indeed have $y \in 2^X$

[jolly parrot]

Pseudonium

[sonic patrol]

But we can also show that y is not in the image of f

Do you see why?

[barren root]

Because |X| is smaller than |2^X|

[sonic patrol]

Well

That’s the whole thing we’re trying to prove…

So unfortunately we can’t use that, no

[barren root]

I mean it's obvious

[sonic patrol]

The point is that

It’s not obvious for infinite sets

That’s why Cantor had to do his argument

[sonic patrol]

This statement holds for all sets X

Both finite and infinite

[barren root]

Given there is a proof for the case of infinite sets

[sonic patrol]

I’m showing you the proof for infinite sets!

[barren root]

Carton's diagonal argument?

[sonic patrol]

Yeah

[sonic patrol]

.

[sonic patrol]

.

Try using the definition of image

[barren root]

image in relation to an input value of a function is the output

[sonic patrol]

I’m… not sure what you’re trying to say

[barren root]

Citing the definition

And trying to see how to use it

[sonic patrol]

Can you tell me the definition of the set $\text{Im}(f)$?

[jolly parrot]

Pseudonium

[barren root]

The function's codomain?

[sonic patrol]

Nope!

The image is different from the codomain

[barren root]

The subset of codomain to which we have values leading from the domain

[sonic patrol]

Right, that’s roughly the idea

In this case, $\text{Im}(f) = {w \in 2^X | \exists z \in X \text{ such that } f(z) = w }$

[jolly parrot]

Pseudonium

[sonic patrol]

Using this, can you show that $y \not \in \text{Im}(f)$?

[jolly parrot]

Pseudonium

[sonic patrol]

This is the definition of y

[barren root]

Does $x \not \in f(x)$ mean that $x$ is not contained in the set to which $f(x)$ leads?

[jolly parrot]

Telov

[sonic patrol]

Indeed!

$f(x) \in 2^X$

[jolly parrot]

Pseudonium

[sonic patrol]

So it’s a subset of X

And $x \in X$

[jolly parrot]

Pseudonium

[sonic patrol]

It’s an element of x

So, it makes sense to ask whether or not x is in f(x)

Sometimes it is, and sometimes it’s not

Whenever it’s not, we add it to y

[barren root]

How can a function belong to $2^X$? $2^X$ has sets, not functions

[sonic patrol]

Um

$f : X \to 2^X$ is our function, remember?

[jolly parrot]

Pseudonium

Telov

[sonic patrol]

So if we give it an input

Some $x \in X$

[jolly parrot]

Pseudonium

[sonic patrol]

Then the output is $f(x)$

[jolly parrot]

Pseudonium

[sonic patrol]

Which has $f(x) \in 2^X$

[jolly parrot]

Pseudonium

[sonic patrol]

So, f turns an element of X into a subset of X

Or if you prefer

Take some $a \in X$

[jolly parrot]

Pseudonium

[sonic patrol]

a is an element of X

So we can apply f to it

We get the output $f(a)$

[jolly parrot]

Pseudonium

[sonic patrol]

This is an element of the codomain of f

Which is 2^X

So $f(a) \in 2^X$

[jolly parrot]

Pseudonium

[sonic patrol]

In other words, f(a) is a subset of X

[barren root]

Oh yeah, sorry, not function belongs but one of its outputs, sorry

[sonic patrol]

It’s ok

I think maybe it was the f(x) notation that was confusing

Because often people say “let f(x) be a function…”

When really they should say “let f be a function…”

[barren root]

It was bound to happen given that I'm new to this notation honestly

[sonic patrol]

Mhm

That’s why i changed it to “a” and “f(a)”!

[barren root]

It helped, indeed

Back to our turtles

[barren root]

.

[barren root]

.

[sonic patrol]

Ping me if you’re stuck

[barren root]

Okay

[barren root]

Given that there are always more indicator functions than elements in the original set, there always will be out of the image indicator functions in the codomain which this y seems to contain. But that is for finite sets

[sonic patrol]

Indeed, your first statement is exactly what we’re trying to prove for infinite sets

That there are more indicator functions than elements

[barren root]

That there are more indicator functions for infinite sets than in the original infinite set...

[sonic patrol]

Indeed

[barren root]

So if $2^\mathbb{N}$ is uncountable then it is true for all infinite sets containing different whole numbers, at least

[sonic patrol]

Yes, so in particular what i stated is true for $X = \mathbb{N}$

[jolly parrot]

Pseudonium

Telov

[sonic patrol]

But it’s true for any set, finite or infinite

And essentially I’m trying to walk you through the proof

[barren root]

I'm afraid that the diagonal way of constructing such a number trips over any set which contains all numbers of given length

[sonic patrol]

So, X doesn’t necessarily have to be a set of numbers

It can just be any set

Id recommend going back to the definitions

To show that $y \not \in \text{Im}(f)$

[jolly parrot]

Pseudonium

[barren root]

y is the set of all subsets of the codomain of f which either are not in the image of f or are in the image of f but do not contain the element which leads to them

[sonic patrol]

Careful

y is not a set of subsets

y is just a subset of X

y is the set of elements x which are not in f(x)

Or if you prefer

[barren root]

y is a subset of X which contains all elements which when put in f do not lead to a set which contains them

[sonic patrol]

$y = { a \in X | a \not \in f(a) }$

[jolly parrot]

Pseudonium

[sonic patrol]

Yep

[barren root]

Actually is that leading to a set which does not contain them supposed to be important? Because I don't see its significance in the task

Not trying to be naughty or anything

[sonic patrol]

Yes!

Very important actually

You use this to show that $y \not \in \text{Im}(f)$

[jolly parrot]

Pseudonium

[sonic patrol]

If we instead defined something like

$w = { a \in X | a \in f(a) }$

[jolly parrot]

Pseudonium

[sonic patrol]

Then the argument wouldn’t work

It’s perfectly possible that $w \in \text{Im}(f)$

[jolly parrot]

Pseudonium

[sonic patrol]

It doesn’t have to be

It’s also possible that $w \not \in \text{Im}(f)$

[jolly parrot]

Pseudonium

[sonic patrol]

We can’t tell for sure

However, we know for certain that $y \not \in \text{Im}(f)$

[jolly parrot]

Pseudonium

[sonic patrol]

Because we can prove it

Which is what I’m trying to get you to do

[barren root]

To show that y is not in the image of f hmm

[sonic patrol]

Mhm

This is what cantor managed to do

[barren root]

Trying to understand what y not being in the image would mean

[sonic patrol]

That’s a good instinct

[barren root]

Thanks

So in regard to the function there are 3 kinds of elements in X:

1. those which do not lead to anything
2. those which lead to a set which contains them
3. those which lead to a set that does not contain them

y is the subset which contains all elements of types 1 and 3

[sonic patrol]

Well

There are no elements of type 1?

Because f is defined on all of X

[barren root]

Oh, it's the codomain what isn't used to the full

In finite ones at least

[sonic patrol]

Mhm

[barren root]

So all elements of X lead to an indicator function of X. \
y is a subset of X which contains all elements which lead to an indicator function not containing the element.\
We need to prove that $y \not \in \text{Im}(f)$

[sonic patrol]

y contains all elements leading to an indicator function that doesn’t contain the element

[barren root]

Yep, sorry

[jolly parrot]

Telov

[sonic patrol]

Mhm

[barren root]

Given that the elements lead to random elements in the codomain, it's difficult to say anything

[sonic patrol]

Would you like a hint?

[sonic patrol]

This is also a good instinct

[barren root]

Is it the hint yet?

[sonic patrol]

No lol

[barren root]

For instance they all can lead to sets which do not contain them

Or they all can lead to sets which do contain them

[sonic patrol]

This is true

[barren root]

In the first case y would contain all elements of X

In the latter case it wouldn't contain any

[sonic patrol]

Yep!

[barren root]

I think we should stop calling the elements of the 2^X set indicating functions because what they in fact are are all subsets of X

[sonic patrol]

Sure

[barren root]

We need to prove that all elements which lead to subsets of X not containing them are not together one of these sets?

So if each element is an assigned number then...

[sonic patrol]

We just need to prove $y \not \in \text{Im}(f)$

[jolly parrot]

Pseudonium

[sonic patrol]

I think - we’re mostly going in circles, so let me give you a hint

[barren root]

Yeah I translated it into words

[sonic patrol]

What it means for $y \in \text{Im}(f)$ is that $\exists a \in X \text{ such that } y = f(a)$

[jolly parrot]

Pseudonium

[sonic patrol]

So - what goes wrong if $y = f(a)$ for some $a \in X$?

[jolly parrot]

Pseudonium

[barren root]

Then an element of X leads to a subset of X which contains {all elements of X which lead to a subset of X not containing them} and that our element is one of them. Seems like a contradiction to me

[sonic patrol]

Hmm, I can’t exactly parse what you’re saying

[barren root]

Imagine how it is for me😂

[sonic patrol]

We don’t know whether $a \in f(a)$ or not

[jolly parrot]

Pseudonium

[sonic patrol]

But - suppose $a$ was in f(a)

[jolly parrot]

Pseudonium

[sonic patrol]

So $a \in f(a)$

[jolly parrot]

Pseudonium

[sonic patrol]

What would be the issue with $y = f(a)$?

[jolly parrot]

Pseudonium

[barren root]

It directly contradicts with the definition of y

[sonic patrol]

Because?

[barren root]

y contains only $a \not \in f(a)$

[jolly parrot]

Telov

[sonic patrol]

Indeed! So we couldn’t have $a \in y$

[jolly parrot]

Pseudonium

[sonic patrol]

So $a \in f(a)$ but $a \not \in y$, which contradicts $y = f(a)$

Ok

[jolly parrot]

Pseudonium

[sonic patrol]

So we know that $a \in f(a)$ is bad

[jolly parrot]

Pseudonium

[sonic patrol]

But we don’t know for sure that’s true

I mean, if we know $f(a) = y$, then we certainly can’t have $a \in f(a)$

[jolly parrot]

Pseudonium

[sonic patrol]

But there’s another possibility

Maybe $a \not \in f(a)$

[jolly parrot]

Pseudonium

[sonic patrol]

Would it be fine for $f(a)$ to equal y now?

[jolly parrot]

Pseudonium

[barren root]

Yes

[sonic patrol]

Actually, this situation is also bad!

This leads to a contradiction too

Can you see why?

[barren root]

Wait...

Ha!

y will be equal to f(a)

And a will be meant to belong to one of them but not the other one

But they are equal

[sonic patrol]

Exactly!

This is the issue

[barren root]

That's a subtle one

[sonic patrol]

f(a) and y are always guaranteed to be unequal

Because they differ in at least one place

Namely, a itself

Because either $a \in f(a)$, meaning $a \not \in y$

[jolly parrot]

Pseudonium

[sonic patrol]

Or $a \not \in f(a)$, meaning $a \in y$

[jolly parrot]

Pseudonium

[sonic patrol]

So they have to be different subsets

And so it’s absolutely impossible to have $y = f(a)$

[jolly parrot]

Pseudonium

[sonic patrol]

This is what cantor realised

[barren root]

Can we possibly continue tomorrow? I'm working on my subconsciousness now

[sonic patrol]

Yep sure though idk if I’ll be here tomorrow

But you can definitely open a help channel again and I’m sure someone will help

[barren root]

Can I ping you then?

[sonic patrol]

Not in a help channel I think but if you post in one of the other channels

[barren root]

Okay, good night

Thank you for your tremendous effort

[sonic patrol]

gn!

and no problem

cantor stuff is one of the most fun parts of set theory

[sonic patrol]

You should probably close the channel

[barren root]

Funny as in fun or as in difficult?

[sonic patrol]

Both

[barren root]

.close

[pearl pond]

Channel closed

Closed by @barren root

Use .reopen if this was a mistake.

[pearl pond]

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[pearl pond]

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[dusty stone]

Im a bit confused. I have triangle with given vertices and I have to calculate the double integral. Im what to make of the triangle, I mean how to get the lower and upper limits of integration. Let's say we integrate with respect to x first.

[frank goblet]

try to set up the limits

[dusty stone]

Does that mean that the lower limit of integration is the line that goes through the points (-1,0) (0,1) and the upper limit of integration the line that goes through the points (0,1) (1,0)?

[frank goblet]

it might be easier to split the integral in two

x goes from -1 to +1
y depends on x

or the other way around

[dusty stone]

So when we integrate with respect to x first, the limits are for x: lower limit of integration is the line that goes through the points (-1,0) , upper limit of integration the line that goes through the points (0,1) (1,0) and y goes from 0 to 1. When we integrate with respect to y first, x goes from -1 to 1 and y?

Im so confused lol

I thought when we integrate with respect to y, x goes from -1 to 1 and y from 0 to 1, but then we don't have the triangle

[frank goblet]

you are trying to cover the whole triangle
so let's say our x runs from -1 to 1, we get a line
now our y travels from that line at y=0 to y=x-1
then we get a triangle, but it will be wrong for the left part

so to get both parts, we either use the absolute of x, or we split it up before hand

now notice that the outer integration depends on x, that means our result will not be a number, it won't make a lot of sense
so we have to look if our order of integration makes sense
luckily there is a theorem saying that we can rearrange the order of integration
this theorem has some conditions tho, but these are met here iirc

[pearl pond]

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[dusty stone]

sorry for the writing, but I wanted to make sure I understand

[tulip ore]

try writing down the doubl eintegral with this information

[dusty stone]

like that?

[tulip ore]

keep in mind the diagonal edge is on the left in the left triangle, so the bounds y-1 and 0 should be swapped

y is from 0 to 1, then x is from y-1 to 0

other than that, its correct

[dusty stone]

you're right, my bad

[tulip ore]

another thing you can notice if you pay attention to the bounds,

both integrals can be combined to be $\int_0^1\int_{y-1}^{-y+1}(x^2+y^2);dx;dy$

[jolly parrot]

mtt

[tulip ore]

this would be one triangle with y from 0 to 1 and x from the left diagonal to the right diagonal

[dusty stone]

Oh I see

[tulip ore]

you can get this by reading the triangle but also from combining the integrals together

[dusty stone]

I see

I will try to calculate it now

[tulip ore]

alr gl

[frank goblet]

this would be what I would have done. However mtt's approach is much nicer and doesn't need to check for conditions which my approach does 😦

[tulip ore]

its a common rule to avoid absolute values anywhere in an integral

also you have a typo on the second line

[frank goblet]

oh really?

[tulip ore]

x in the outer bound is not defined

[frank goblet]

oh yeah, good point

I didn't even think about that from a physics point of view

[tulip ore]

it just sounds like you miswrote the integral

so that would mean the second line isnt necessary and you really began with this instead

[frank goblet]

ah no, i set it up as in line 2, because that was more intuitive to me, then i swapped the order of integration
but yeah, one could just start with line 3 instead of line 2

[tulip ore]

oh as like notes, so the first integral is the bounds for x and the second for y?

[frank goblet]

i don't think I understand
the outer integral in line 2 is with respect to y
the fact that the x in the limit is not defined tells me to swap the integration order
I know that to a mathematician that sounds like a crime haha

[tulip ore]

that to me sounds like you had the bounds in mind but not enough to write them in order, so they are like notes instead of like an actual integral to be interpreted

[tulip ore]

what I said after "notes" doesnt make any sense

so its more of swapping them until they made sense

what works out though is that you can just write them in the order they are defined, so -1 to 1 being the first integral youd write down LTR is a convenience

[pearl pond]

@dusty stone Has your question been resolved?

[tulip ore]

stuck figuring the integral out?

[dusty stone]

yeah

it takes me so long

lol

[tulip ore]

fyi it shouldnt take this long, do you need help with that

thats already too many steps

[dusty stone]

I got this

[tulip ore]

thats not correct

[dusty stone]

I mean that's only the left triangle

[tulip ore]

yea its not correct

[dusty stone]

oh

[tulip ore]

do you want to recheck your work or do this in a shorter way

[dusty stone]

Im not sure, depending if the shorter way is harder or not

[tulip ore]

its not

[dusty stone]

so maybe that

[tulip ore]

do you remember this integral

[dusty stone]

yes

[tulip ore]

thats an oddly long amount of time to remember that integral

lets go back to the original two then

you definitely remember $\int_0^1\int_{y-1}^0(x^2+y^2);dx;dy+\int_0^1\int_0^{-y+1}(x^2+y^2);dx;dy$

[jolly parrot]

mtt

[tulip ore]

right?

[dusty stone]

yes

[tulip ore]

now you can combine the integrals like this:

,,\int_0^1\qty(\int_{y-1}^0(x^2+y^2);dx+\int_0^{-y+1}(x^2+y^2);dx);dy

[jolly parrot]

mtt

[tulip ore]

theyre both integrals from 0 to 1 dy

then the integrals can be added together to be from y-1 to -y+1

,,\int_0^1\qty(\int_{y-1}^{-y+1}(x^2+y^2);dx);dy

[jolly parrot]

mtt

[tulip ore]

so you can add the integrals together into one integral so you dont have to recalculate the same integral

any difficulties so far?

[dusty stone]

no, I understand

[tulip ore]

this first integral is dx

wait a sec

you did calculate this integral correctly here

,,\int_0^1\qty[\frac13x^3+xy^2]_{y-1}^{-y+1}dy

[jolly parrot]

mtt

[tulip ore]

which is then as usual:

,,\int_0^1\qty(\qty(\frac13(-y+1)^3+(-y+1)y^2)-\qty(\frac13(y-1)^3+(y-1)y^2))dy

[jolly parrot]

mtt

[tulip ore]

are you with me?

[dusty stone]

yes

[tulip ore]

now the second time save you can do here is reduce how much you would need to expand

first, we can try combining the terms together so that less would be expanded

take a look at these terms

do you see a way of adding them together into one term (without expanding)?

if you cant see it, it looks like this:

,,\frac13(-y+1)^3+-\frac13(y-1)^3
\\frac13(-y+1)^3+\frac13(-y+1)^3
\\frac23(-y+1)^3

[jolly parrot]

mtt

[dusty stone]

I see, I wasn't sure if you can factor out -1 before expanding

[tulip ore]

theres a general rule you can use to see if its possible

what youre thinking of, in general, would be if -f(x) = f(-x)

right

[dusty stone]

yes

[tulip ore]

a function f like this is called "odd"

in general, polynomials with odd exponents only are odd functions

1/3 x^3 is one of them

you can notice 1/3 (-x)^3 = 1/3 (-1)^3 x^3 = -1/3 x^3

[dusty stone]

I see

[tulip ore]

,,\int_0^1\qty(\frac23(-y+1)^3+(-y+1)y^2-(y-1)y^2)dy

[jolly parrot]

mtt

[tulip ore]

after that being combined, the last two terms above also can

do you see how these can be combined?
(as a hint, this isnt related to odd functions but more of distributive property)

@dusty stone

can you use distributive property on this - and this (y - 1)?

[dusty stone]

like that?

[tulip ore]

no

[tulip ore]

try this instead

[tulip ore]

I didnt say that either

[dusty stone]

oh sorry

[tulip ore]

distributive property on the - and the (y - 1)

you should see that its the shortest sequence of steps out of the others

-(y - 1) is?

[dusty stone]

-y+1

[tulip ore]

yep

,,(-y+1)y^2-(y-1)y^2=(-y+1)y^2+(-y+1)y^2=2(-y+1)y^2

[jolly parrot]

mtt

[tulip ore]

so you can see it gets there shorter in two steps instead of 3 or anything similar

[dusty stone]

yeah, you're right

[tulip ore]

did you notice that -(y-1) matched with -y+1?

[dusty stone]

yes

[tulip ore]

so when you notice that, you can think of combining them or writing things in terms of them

[tulip ore]

if you went with that, you can find it this way

,,\int_0^1\qty(\frac23(-y+1)^3+2(-y+1)y^2)dy

[jolly parrot]

mtt

[tulip ore]

now for the third timesave

youll notice we'd have to expand (-y+1)^3 by now

thats -y^3 + 3y^2 - 3y + 1 but itd be ideal if we didnt need to do this

now you can do u = -y + 1

(and u' is -1)

[tulip ore]

do you want to try out the u-sub here

[dusty stone]

yes

[tulip ore]

go give it a shot

depending on how comfortable you are with u-subs, this can save more time than just expanding

if youre having trouble, the u-sub doesnt save that much time and should be skipped instead, since you already memorized that (-y + 1)^3 = -y^3 + 3y^2 - 3y + 1

[dusty stone]

what do I do with y^2?

[tulip ore]

you know that u = -y + 1

that means y = -u + 1, right

[dusty stone]

ah I see

[tulip ore]

ye

if you do u = mx + b of some sort, you can reverse that to get an inverse that works

[dusty stone]

was I wrong, factoring out -1 before the integral?

[tulip ore]

@dusty stone back

[tulip ore]

iirc what you posted was correct

one thing though

in definite integrals,

since u = 1 - y,

the bounds also change

the bounds usually went from y=0 to y=1

now they go from u=1 to u=0

changing the bounds along with the y means that what you have afterwards is an integral entirely within u

they both have the same value

along with doing this,

[tulip ore]

this is how I got this integral

[dusty stone]

I thought when you calculate the Indefinite integral first, you don't have to worry about changing the bounds

[tulip ore]

the problem with calculating the indefinite integral first is that what you do afterwards is... change the bounds

youre calcating those bounds one way or the other, its mandatory to eventually calculate the 1 - 0 and 1 - 1 change of bounds

you might as well do it in a way which doesnt require using y anymore

[tulip ore]

both of these integrals have the same value

[dusty stone]

but I didn't have to change the bounds here, no?

[tulip ore]

calculating -0 + 1 and -1 + 1 is still changing the bounds

it just doesnt have u= and y= labelled

additionally writing it this way requires you writing out the same thing but with (-y+1) instead of u

only to calculate -0+1 and -1+1 anyways

[dusty stone]

I see

[tulip ore]

regardless youve got the correct answer of 1/3

[dusty stone]

can't thank you enough

.close

[pearl pond]

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[tulip ore]

np

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[pulsar thorn]

Very quick question, feel like i shouldnt be taking up a whole channel for this but i am going to be doing a Dynamics and Fluids uni exam, i just wanted to ask what mode my calculator should be in for this because i am not sure

[pearl pond]

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[chrome flame]

Klein group 4

[chrome flame]

I am struggling with proper definition of it

[west sapphire]

isn't it just Z2 x Z2? what do the numbers in your screenshot have to do with it?

[chrome flame]

Actually I do not know Z2×Z2 mean

[last summit]

dark side gf does not know algebra like this

[autumn trellis]

They are saying that the subgroup {1,5,7,11} of Z12 is isomporphic to the klein 4 group, because they possess the same multiplication table

[west sapphire]

state the original problem in the future

!original

[pearl pond]

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.

[autumn trellis]

This is an odd way to define it, yeah. Usually defined as Z2 x Z2 or with a permutation group or via matrices

[pearl pond]

@chrome flame Has your question been resolved?

[chrome flame]

Tq very much for explaining

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[sharp smelt]

I'm not sure I get where lcm(a,b) came from

[sharp smelt]

will post an iamge ina a minute

[cosmic niche]

lcm(a, b) is defined as the smallest integer such that (1) it is a multiple of both a and b and (2) it is smaller than any number which is a multiple of both a and b

[sharp smelt]

yes

[cosmic niche]

So first they construct an m which clearly satisfies condition (1) right (in their first paragraph)

[sharp smelt]

ok, yeah

so I have c/m=gcd(a,b)

[cosmic niche]

And next they let c be any number which is a multiple of both a and b and they then show that m must be a divisor of c

And m is a divisor of c implies that m <= c such that the second condition is satisfied and m is the lcm

[cosmic niche]

No, c is an arbitrary multiple of both a and b

[sharp smelt]

yeah

but (c/b)x+(c/a)y=gcd(a,b) by bézout's lemma

[cosmic niche]

Hmm I'm not quite sure about that

Bézout states that there exists numbers x, y such that ax + by = gdc(a,b)

[sharp smelt]

yes

[cosmic niche]

In your case this would mean that gcd(a, b) = gcd(c/b, c/a) which I don't think is true in general

[sharp smelt]

hmm, ok

[cosmic niche]

But maybe back to the original question what part did you not understand

[sharp smelt]

from (c/b)x+(c/a)y, how did they get lcm(a,b)=ab/d?

[cosmic niche]

Well, the core idea of this proof is to define m = ab / d ( = ab/gcd(a,b)) and then show that this m satisfies the 2 conditions for being the lcm

So when c/m = vx + uy, we know that c = m(vx + uy) where vx + uy is an integer such that m divides c which in turn implies that m <= c

[sharp smelt]

mhm

ok

thanks

[cosmic niche]

I'd say there is a more elegant proof of this fact if you know how the gcd and lcm relate to the prime factorizations of both numbers

[sharp smelt]

I think I'm only introduced to primes in the next chapter . Thanks alot for the help

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[obsidian grail]

What is dominant and how to calculate it and for what is it used compared to average, median,

[pearl pond]

@obsidian grail Has your question been resolved?

[pearl pond]

@obsidian grail Has your question been resolved?

[pearl pond]

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[sharp smelt]

Find the gcd(143,227) using euclid's algorithm

[sharp smelt]

so this is what I've tried

224=143+84

143=84+49

49=35+14

35=2*14+7

14=2*7+0

which is wrong

why is this wrong?

[worthy lance]

First step

[sharp smelt]

oh

[tulip mason]

Yeah you have done addition wrong on multiple lines

[sharp smelt]

addition is going to be the death of of me

[worthy lance]

Back to school

[sharp smelt]

thanks

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[sharp smelt]

prove that gcd(a+b,a-b)=1 or 2

[sharp smelt]

if gcd(a,b)=1

I'm guessing I have to use EDA here?

[tulip mason]

What have u tried?

[sharp smelt]

alternatively

I was able to determine one is always odd

and one number is always even

wondering if there's any way to use that

[tulip mason]

That's not true?

Take b=1

Then both are odd or both are even

[timid spindle]

^ you can't make or arrive at any assumptions about a or b since you asked to prove it for arbitrary a and b

[sharp smelt]

hmm, ok

[timid spindle]

(Well you could make assumptions if you're splitting into cases that cover everything but that doesn't seem the right strat here)

[sharp smelt]

is this supposed to be an easy problem ? Like should I revise the theory once more, or would I understand the theorom better by solving this?

[sonic patrol]

mmm there’s some subtleties involved

I’d recommend using the universal property

[sharp smelt]

that d|a , d|b \implies d|ab?

[sonic patrol]

no

so, lemme state it again

Given integers $m, n$, $\text{gcd}(m, n)$ is the unique integer which satisfies $\forall a \in \mathbb{Z}, a | m \wedge a | n \iff a | \text{gcd}(m, n)$

[jolly parrot]

Pseudonium

[sharp smelt]

hmm,ok

thanks

let me try something

hmm, $1|a+b; 1|a-b \implies 1|gcd(a+b,a-b)$

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[tulip mason]

You don't need to consider 1

$gcd(a,b)\ge 1,$ for all $a,b$

[jolly parrot]

Max

[timid spindle]

^ youre trying to show that if its bigger than 1, it must be 2

[sharp smelt]

hmm, but gcd(a,b)=1

[tulip mason]

Ah my bad I repeated letters

I meant for a generic a, b not related to this problem

[sonic patrol]

yes, which means that $d | a, d |b \iff d | 1$

[jolly parrot]

Pseudonium

[sonic patrol]

that’s the universal property

[sharp smelt]

if d|1 that implies d=1 though

right

[sonic patrol]

indeed!

hm, so

could you state the universal property for $\text{gcd}(a - b, a + b)$?

[jolly parrot]

Pseudonium

[sharp smelt]

$d|a-b; d|a+b \iff d|1$

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[sonic patrol]

nope

just use the definition

[sharp smelt]

$d|a-b \land d|a+b \iff d|gcd(a+b,a-b)$

[sonic patrol]

it’s an if and only if

[sharp smelt]

ah,ok

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[sonic patrol]

yep!

now, see how you can manipulate the LHS

[sharp smelt]

I could find a and b in terms of d and other constants?

like a-b/d=k_1

a+b/d=k_2

[sharp smelt]

how do you mean?

[sonic patrol]

idk just try stuff!

see how much you can deduce

[limber oasis]

it can be rewritten in arguably more meaningful ways

[sharp smelt]

hmm,could I have a hint please

[sonic patrol]

uh

hmm

what’s the first thing you learned in math

[sharp smelt]

operators

[limber oasis]

Full proof:||wlog a is odd (they're not both even), let d=gcd(a+b, a-b) = gcd(2a, a-b)||
||so d|2a, meaning d | 2 (ok) or d|a||
||in that second case d | a+b||
||this implies d | b, meaning d | 1||
Line 1 is your hint

[sonic patrol]

wait really

not like

how to count?

[sharp smelt]

oh, right

that

[sonic patrol]

mhm

and after that?

[sharp smelt]

odd and even numbers

[sonic patrol]

hmm

I would’ve thought like

addition..

[limber oasis]

odd/even is relevant here

[sonic patrol]

anyway that’s my hint

addition

[sharp smelt]

one has to odd?

[limber oasis]

yes

[sharp smelt]

and the other either even or odd

like 17 and 19 are coprime

[tulip mason]

yea

gcd(a,b)=1 means they both can't be even

[sharp smelt]

so I solve case-wise?

[limber oasis]

no need

[sonic patrol]

I don’t quite understand this proof

[sharp smelt]

hmm, if one of the numbers is 2l+1

the other can just be m

[sonic patrol]

How are you justifying ||d | 2a means d | 2 or d | a||?

[limber oasis]

we deduce that ||d | 1 or d | 2|| which is what we want
Notice ||a being odd|| is used on line 2

[sharp smelt]

so I have gcd(2l+1, m)=1

[sonic patrol]

This is the place I get stuck

[limber oasis]

oh

unless ||d=2a, but then b is even||

so it's fine

[sonic patrol]

I still don’t get why

[limber oasis]

Euclid's lemma?

[sonic patrol]

That says ||if d | ab and gcd(d, a) = 1 then d | b||

how are you using that here exactly?

[sharp smelt]

hmm, I'm lost, is it a bad idea to look at the proof and learn from it?

[limber oasis]

||if d doesn't divide 2 then gcd(d, 2) = 1, then d | a||

[sonic patrol]

well atm im trying to understand the proof bezier posted

[sonic patrol]

||this is false, 4 doesn’t divide 2 but gcd(4, 2) = 2||

[limber oasis]

||4 ain't odd||

[sonic patrol]

||im taking issue with your implication “if d doesn’t divide 2 then gcd(d, 2) = 1”||

[limber oasis]

that is poorly phrased, lemme go back

[sonic patrol]

I have a different proof in mind anyway

have you tried using my hint?

[sharp smelt]

I will try now, your hint was to use addition, right?

[sonic patrol]

Yes

So you have $d | a - b$ and $d | a + b$

[jolly parrot]

Pseudonium

[sharp smelt]

hmm, I could add them

to determine if d|2a

and subtract them to determine if d+2b?

[limber oasis]

||a=9, d=6 is more convincing, I'm thinking about it||
||that's impossible as then 3|b hence not coprime||

[sonic patrol]

that’s a good idea

[sharp smelt]

hmm

if d|2a, d is either a muliple of a or 2

[sonic patrol]

so

[sharp smelt]

similarly if d|2b

[sonic patrol]

are you saying you’ve shown that d | 2a?

if so, how?

[sharp smelt]

no, I'm saying if that's true

[limber oasis]

||d|2a. Assume d doesn't divide 2 nor a. Then there's k | a st d = 2k. Then k > 1 is a common divisor of a and b, contradiction||
That details line 2, proving its conclusion

[sharp smelt]

d|2b

[sonic patrol]

I would recommend showing this first

it is true that if d | a - b and d | a + b then d | 2a

[sharp smelt]

hmm

[sonic patrol]

indeed, it’s true generally that if d | k and d | r then d | k + r

[sharp smelt]

$a+b/d=k_1$

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[sharp smelt]

$a-b/d= k_2$

[sonic patrol]

order of operations..

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[sonic patrol]

as written these are not correct

[sharp smelt]

$d|a-b \implies \frac{a-b}{d} = K_1

right?

[sonic patrol]

what’s K?

[sharp smelt]

I meant K_1

k. is s a constant

[sonic patrol]

What’s K_1?

[sharp smelt]

a constant

[sonic patrol]

Unfortunately you have to be more specific than just “a constant”

What’s the definition of divisibility?

[sharp smelt]

I think something like one number is a multiple of another

[sonic patrol]

and what does that mean?

so specifically, if d and k are integers, what does d | k mean?

when is it true?

you really have to make sure you’re familiar with these definitions

otherwise these problems are gonna be a lot harder

[sharp smelt]

so having an intutive understanding isn't enough?

[sharp smelt]

$k=nd$

[sonic patrol]

no of course not

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[sonic patrol]

you need the definition

[sharp smelt]

where n is a constant

[sonic patrol]

nope!

you have to be more specific than that

[sharp smelt]

I'll qoute my book

An integer b is said to be divisible by another integer $a\neq 0$ in symbols $a|b$, if there exists some integer c such that b=ac.

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[sonic patrol]

exactly

so n has to be an integer

not just any ol constant

[sharp smelt]

oh, ok

[sonic patrol]

this is the part you were missing

after all

3 = 1.5 * 2

and 1.5 is a constant

[sharp smelt]

I see

ok

[sonic patrol]

in general for math

[sharp smelt]

so if I mention $k_1, k_2 \in \Z$ this is right?

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[sonic patrol]

you need to be absolutely watertight with definitions

[limber oasis]

tbf this is arithmetic it's implicit that we're exclusively working with integers

[sonic patrol]

well you still haven’t fixed the order of operations

[sonic patrol]

implicit can often lead to confusion

[sharp smelt]

this implies (a-b)/d = k_1

no?

[sonic patrol]

so

I would write that as

$\exists k_1 \in \mathbb{Z}, a - b = k_1 d$

[jolly parrot]

Pseudonium

[sonic patrol]

that’s what the definition says

[sharp smelt]

I see

and $a+b=k_2d$

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[sharp smelt]

so addinf them I get $a = \frac{(k_1+k_2)d}{2}$

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[sharp smelt]

now this isn't always divisble by d though, right? $\frac{k_1+k_2}{2}$ could be a non -integer

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[sonic patrol]

Yes

We’re not trying to show d | a

We’re trying to show d | 2a

[sharp smelt]

oh right

yeah, that's always an integer

similarly ${k_1-k_2}$ is always an integer

[sonic patrol]

No

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[sharp smelt]

or 2b

[sonic patrol]

And what do you mean

[sharp smelt]

K_1 is an integer, k_2 is an integer, so their sum is an integer

as is their difference

[sonic patrol]

Yes

so?

[sharp smelt]

just a minute

this implies d|2a and d|2b

[sonic patrol]

Right

Could you write out a complete proof that d| 2a?

[sharp smelt]

ok

[limber oasis]

oh that's so much simpler
I just missed that

[sharp smelt]

$\frac{a+b}{d} = k_1
\
\frac{a-b}{d}=k_2
\
2a= d(k_1+k_2)
\implies d|2a$

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[sharp smelt]

similarly $d|2b$

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[sonic patrol]

Ok

Can I show you how I’d write this proof?

Because there’s a few issues I take with how you’ve laid it out

[sharp smelt]

ok, thanks for the help

[sonic patrol]

So

We assume that $d | a - b$ and $d | a + b$

[jolly parrot]

Pseudonium

[sonic patrol]

By definition, this means $\exists k_1, k_2 \in \mathbb{Z}$ such that $a - b = k_1 d$ and $a + b = k_2 d$

[sharp smelt]

yes

[jolly parrot]

Pseudonium

[sonic patrol]

Then, summing, we get that $2a = (k_1 + k_2) d$

[jolly parrot]

Pseudonium

[sonic patrol]

Moreover, since $k_1, k_2 \in \mathbb{Z}$, we have $k_1 + k_2 \in \mathbb{Z}$

[jolly parrot]

Pseudonium

[sonic patrol]

So, by definition, $d | 2a$

[jolly parrot]

Pseudonium

[sonic patrol]

The main thing you were missing was this step

[sonic patrol]

And this step

[sharp smelt]

ah

ok

[sonic patrol]

I won’t make you write out the proof for the other one

But I would recommend doing it at some point

It’s good to keep things clear and explicit as much as possible in a proof

Appeal to definitions wherever you can

Anyway, we’ve successfully shown that $d | a - b, d | a + b \implies d | 2a, d | 2b$

[jolly parrot]

Pseudonium

[sonic patrol]

So… now what?

[sharp smelt]

hmm, gcd(2a,2b)=2

[sonic patrol]

How do you know that?

[sharp smelt]

bézout's lemma

[sonic patrol]

Ooh, very nice!

Yes, that’s correct

[sonic patrol]

So now, what can we use this for?

[sonic patrol]

(I said this cause I didn’t see that myself and that’s a neater way than what I had in mind)

[sharp smelt]

hmm

bezout's again?

(a+b)u+(a-b)t = gcd(a+b,a-b)

[sonic patrol]

So, I had something else in mind

Why not just use the universal property?

So, what does the universal property say for gcd(2a, 2b)?

[sharp smelt]

this

[sonic patrol]

Right

So what if m = 2a and n = 2b?

[sharp smelt]

a|2

so a is either 1 or 2

[sonic patrol]

Wait

How’d you get that?

Do you mean d | 2?

[sharp smelt]

yes

[sonic patrol]

Then that’s correct!

We have that $d | 2a, d | 2b$

[jolly parrot]

Pseudonium

[sonic patrol]

So by the universal property, $d | \text{gcd}(2a, 2b)$

[jolly parrot]

Pseudonium

[sonic patrol]

Which means $d | 2$

[jolly parrot]

Pseudonium

[sonic patrol]

So d is either 1 or 2

[sharp smelt]

oh

right

[sonic patrol]

Does that make sense?

[sharp smelt]

how do I even think of this?

yeah

it does

[sonic patrol]

Cool

[sharp smelt]

but how do I learn to think like this?

[sonic patrol]

So, can I summarise what we’ve done so far?

[sharp smelt]

ok

[sonic patrol]

Just get used to using universal properties more

[limber oasis]

yeah that's why I made my commit that it was so much simpler

[sonic patrol]

So, what we’ve done is

We want to find $\text{gcd}(a - b, a + b)$

[jolly parrot]

Pseudonium

[sonic patrol]

Instead of looking at it directly, we’ve looked at the universal property

It doesn’t tell us what the gcd “is”, but it tells us what the gcd “does”, how it relates to other integers

Specifically, it tells us that for $d \in \mathbb{Z}$, $d | \text{gcd}(a - b, a + b) \iff d | a - b, d | a + b$

[jolly parrot]

Pseudonium

[sonic patrol]

Then, we showed that $d | a - b, d | a + b \implies d | 2a, d | 2b$

[jolly parrot]

Pseudonium

[sonic patrol]

And by another universal property, we have $d | 2a, d | 2b \iff d | \text{gcd}(2a, 2b)$

[jolly parrot]

Pseudonium

[sonic patrol]

Moreover, using Bezout’s lemma and gcd(a, b) = 1, we can deduce that gcd(2a, 2b) = 2

So, overall, we have shown the following

$d | \text{gcd}(a -b, a + b) \implies d | 2$

[jolly parrot]

Pseudonium

[sonic patrol]

Does that make sense?

[sharp smelt]

yes

[sonic patrol]

Can you see how to finish?

We’re not quite done

[sharp smelt]

yeah, so this implies d is either one or 2

hmm

[sonic patrol]

What were we trying to show originally?

[sharp smelt]

that the gcd is either 1 or 2

[sonic patrol]

Mhm

[sonic patrol]

.

Any ideas?

[sharp smelt]

not really, no

[sonic patrol]

We can use the Yoneda Lemma!

[sharp smelt]

what now?

[sonic patrol]

Namely

[sonic patrol]

This is true for all integers d, right?

[sharp smelt]

yes

[sonic patrol]

Can you think of any particular integer we can substitute

That makes the LHS true?

[sharp smelt]

a =5, b=3

[sonic patrol]

Not what I mean

I’m looking for d = …

[sharp smelt]

d=1 is always true

[sonic patrol]

Mhm

Anything else?

[sharp smelt]

d can be 2, but I don't see how that helps

[sonic patrol]

How about

$d = \text{gcd}(a - b, a + b)$?

[jolly parrot]

Pseudonium

[sharp smelt]

yes

[sonic patrol]

This is essentially applying the yoneda lemma

[sharp smelt]

what is the yoneda lemma though, google talks about group theory when I look it up

[sonic patrol]

Oh dw about it for now

[sonic patrol]

Anyway, what happens if we substitute this?

[sharp smelt]

then d=1 or d=2

[sonic patrol]

And isn’t that what we were trying to show?

[sharp smelt]

yeah

thanks a lot!

[sonic patrol]

In general, what the yoneda lemma says (for this category) is that

If you have $\forall d \in \mathbb{Z}, d | m \implies d | n$

[jolly parrot]

Pseudonium

[sonic patrol]

Then $m | n$

[jolly parrot]

Pseudonium

[sharp smelt]

I see

thanks

had a slightly unrelated question

so I'm going to be starting my UG in maths in around a month and a half, and NT is only a 2nd year course at my uni, would doing this now, as my first introduction to college math be a bad idea?

again, thanks a lot for all the help!

[sonic patrol]

I mean my first course at uni for math was “Numbers and Sets”

And it worked out

[sharp smelt]

thanks a lot, again

.close

[pearl pond]

Channel closed

Closed by @sharp smelt

Use .reopen if this was a mistake.

[sonic patrol]

lol I can’t believe I got to use category theory for a number theory proof

[pearl pond]

Available help channel!

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[daring wedge]

,rotaw

[daring wedge]

,rotate

Someone please 🥺🥺🥺

Check my question

[crude echo]

, rotate

[daring wedge]

Lmao thx

[crude echo]

Why V=2e^2x

[daring wedge]

If I take the derivative I would get e^2x?

Or am I wrong?

[crude echo]

V is supposed to be the other function itself

U=x^2

And V=e^2x

,w product rule

[jolly parrot]

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Requested by axiomofnochoice

[crude echo]

,w product rule integration

[jolly parrot]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by axiomofnochoice

[crude echo]

Nahh

I'm sorry

[daring wedge]

Isn't this the rule

[crude echo]

Also you are supposed to integrate V in first part

[crude echo]

That isn't product rule

[crude echo]

What you have written jn first line is

[daring wedge]

But I'm supposed to do this by the integration by parts method

[crude echo]

That's the product rule

Same

[daring wedge]

Look

It's the part one

[crude echo]

Yeah

[daring wedge]

Integ by parts and product rule is same?

[crude echo]

Yeah idl

But anyway

[daring wedge]

How do I solfe it 😢

Im stuck on the first part

Tf

[crude echo]

Tell me U and V

Don't worry it's alright

[daring wedge]

U is X squared
V is 2e^2x

[crude echo]

V is e^2x

[daring wedge]

But here it says dv not just v

Look at the LHS

[crude echo]

That is a formula for different things

[daring wedge]

What 😭

[crude echo]

Look at what you have written in first line

[daring wedge]

How do I use the integ by parts formula then

[crude echo]

That is integration by parts

[daring wedge]

I'm my picture?

[crude echo]

Yeah

[daring wedge]

Where I solved the quesuin?

[crude echo]

Yeah

[daring wedge]

Ok

This??

[daring wedge]

Is integration by parts?

[crude echo]

No