Mathematics - 2017-06-12 (page 3 of 9)

Faust7

03:01

uh k i guess my first question would be why would one think that a function acting on a set would give outputs that wouldnt fit in a set?

Leaky Nun

because it isn't a function acting on a set.

Faust7

ok

what is it then?

Leaky Nun

this may seem circular... trust me it isn't.

the intuitive way of thinking about it would be circular

Dodsy

can somebody help me

with a trivial problem

Leaky Nun

@Faust7 you can view it as a function from the set that we have to the set whose existence we are trying to prove

this is the circular way of thinking about it.

@Dodsy just ask; don't ask to ask.

Faust7

03:03

ok with you so far...

Dodsy

How do you find the exact value of a trigonomic ratio without looking at the unit circle?

Leaky Nun

@Faust7 do you have any question about it?

Semiclassical

Not sure exactly what you mean by that.

Leaky Nun

@Dodsy for example?

Dodsy

$sin\frac{5\pi}{4}$

Obviously, if you have the unit circle it's easy

Leaky Nun

03:04

@Dodsy $\sin \frac{5\pi}4 = -\sin \frac\pi4 = -\frac{\sqrt2}2$

Dodsy

right.

Semiclassical

The manipulation way to it, imo, is $\sin(5\pi/4)=\sin(\pi+\pi/4)=-\sin(\pi/4)=-\sqrt{2}/2$.

Dodsy

but I don't see how you go to that.

Leaky Nun

I keep making mistakes.

Semiclassical

Well, in general $\sin(\pi+x)=-\sin(x)$.

Dodsy

03:06

@LeakyNun That's alright, no problem

Faust7

suprisingly i don't think so...

Semiclassical

Same with $\cos$, for that matter.

Dodsy

yes.

What about the last part

Leaky Nun

@Faust7 you already understand it?

Semiclassical

Ah, $\sin(\pi/4)=\sqrt{2}/2$?

Dodsy

03:06

right

Leaky Nun

you construct an isosceles right triangle

Faust7

Well theres a proof here on the wikipedia page

Semiclassical

tbh, the geometric way is the simplest one. If $\theta=\pi/4$, then that's a 45 degree angle.

Faust7

theres one part that im not sure what its defining

Leaky Nun

and argue that the opposite side divided by the hypotenuse is $\sqrt2/2$

Semiclassical

03:07

so the triangle you'd draw on the unit circle would be a 45-45-90 right triangle.

Leaky Nun

@Faust7 quote?

Dodsy

right because pi is 180

Faust7

one sec

Semiclassical

pi radians = 180 degrees, yeah.

Dodsy

this is how they explain it in my work...

Semiclassical

03:08

Another way to see it if you don't want to do geometry at all is via the double-angle identity

namely, $\cos(2x)=2\cos^2 x-1$

Dodsy

evaluate $sin\frac{5\pi}{6}$

$SA = \pi - RA$
$sin\frac{5\pi}{6} = \pi - RA$

Semiclassical

If you take $x=\pi/4$, then that's $\cos(\pi/2)=2\cos^2(\pi/4)-1$.

But $\cos(\pi/2)=0$, so $\cos(\pi/4)=1/\sqrt{2}$...actually, that only gives plus/minus.

Faust7

Formally, let \phi be any formula in the language of ZFC whose free variables are among x,y,A,w_{1},\dotsc ,w_{n} , so that in particular B is not free in \phi .

Semiclassical

Meh. Geometry > algebra :P

Faust7

whatdoes it mean for B to note be a free variable

Semiclassical

03:09

@dodsy | should be }

Faust7

whats the restriction being placed on?

Dodsy

Sorry

was editing it

Semiclassical

np

Dodsy

didn't even mean to send it

Then $RA = \pi - \frac{5\pi}{6}$

Leaky Nun

@Faust7 should I use an example to illustrate?

Dodsy

03:11

(related angle)

Faust7

sure

Dodsy

Then they jump to this

Leaky Nun

@Faust7 let's say we have the set {0,1}, and we're trying to prove the existence of {3,4}.

Faust7

ok

Leaky Nun

here, A={0,1} and B={3,4}.

Semiclassical

03:11

Yeah, that's basically how I'd start. More intuitively: note that $5\pi/6$ rad = 150 degrees, so we're in the second quadrant. Hence we should take the supplement of that angle in order to land back in the first quadrant.

Dodsy

$RA = \frac{6\pi}{6} - \frac{5\pi}{6}$

right

Semiclassical

which amounts to writing $5\pi/6=\pi-\pi/6$.

And then I'd use $\sin(\pi-x)=\sin x$.

Leaky Nun

@Faust7 then, we define the formula $\phi$ as such: $\phi(x,y) := x+3 = y$.

Can you see that the first part of the implication is satisfied?

Dodsy

Okay

so then I could figure that out

Leaky Nun

Namely, $\forall x(x \in A \implies \exists!y \phi)$ @Faust7

Faust7

03:13

yes

Dodsy

but then you have let's say $\frac{\pi}{6}$

Semiclassical

that gets you down to $\sin(\pi/6)=\sin 30^\circ$.

Dodsy

how do you jump from that to -1/2

or whatever it is

1/2

Leaky Nun

@Faust7 then, the axiom says that there is a set $B$ such that $\forall x(x \in A \implies \exists y(y \in B \land \phi))$

Semiclassical

Here you have to note that this will be a 30-60-90 triangle.

Do you know the significance of that 60 degree angle?

Leaky Nun

03:14

@Faust7 can you list the elements in $B$ based on that?

Dodsy

Hm.

Faust7

not all of them

Dodsy

Is it a triple?

Faust7

but there are at least the elements 3,4 in B

Leaky Nun

@Faust7 yes

Semiclassical

03:15

Not sure what you mean by that.

Dodsy

oh a pythagorean triple

I might be wrong

Faust7

ok

Semiclassical

Ah. Well, all right triangles are pythagorean triples :)

But it's not an integer one.

Faust7

Thats intresting

Dodsy

oh right

so it's just a special triangle

Leaky Nun

03:15

@Faust7 your concept is really very clear.

I see a lot of potential in you.

Dodsy

special right

Leaky Nun

> The form stated here, in which $B$ may be larger than strictly necessary, is sometimes called the axiom schema of collection.

Semiclassical

Here's a hint: Draw the triangle for that in the first quadrant, and then draw its reflection across the x-axis.

(so that the two triangles share a common side)

Leaky Nun

@Faust7 that is, if you can solve the following problem:

Semiclassical

What do you notice about the angles of the combined triangle?

Faust7

03:16

Honestly the way the notation was im surprised i could even read it =p

Leaky Nun

1 hour ago, by Faust7

i find mathematics at least for me has been wrought with epiphanies that what was previously taken as fact; was at the very least a convenient misunderstanding or worst blatantly wrong.

Dodsy

hmmm

Faust7

lol

Dodsy

it makes an equilatteral?

Semiclassical

Right. All three angles are 60 degrees.

Leaky Nun

03:17

@Faust7 how many axioms there have you cleared?

Dodsy

right.

Just a little aside

Semiclassical

That means that the height of the right triangle, $\sin(30^\circ)$, is also half of one of the sides.

Dodsy

the penguins just won the stanley cup...

@Semiclassical right...

Semiclassical

But the sides are all the same, so they're all the same as the hypotenuse.

And what's the length of that?

(remember, unit circle)

Faust7

Well i don't know about "cleared" per se

Leaky Nun

03:19

@Faust7 I mean, how many axioms have you gone through?

Dodsy

uhhhh

well I'm not sure, we're in the cartesian plane.

Semiclassical

Remember how you drew the triangle. One coordinate is the origin, another is on the x-axis, and the last is on the unit circle.

Dodsy

but usually the hypoteneuse is equal to sqrt of a squared plus b squared.

Faust7

i guess technically 7 cause i looked up the axiom of infinity and ( i think extensionality) before i looked at this page

Semiclassical

So the hypotenuse goes from the origin to where?

Dodsy

03:21

right

Leaky Nun

@Faust7 can you list the ones you haven't gone through?

Faust7

axiom of power set and well ordering theorem

Dodsy

it connects to the y coordinate.

?

Semiclassical

Right, and that point is on the unit circle.

Leaky Nun

@Faust7 the latter is equivalent to choice so you can skip it :p

Dodsy

03:21

Right.

Semiclassical

What's the distance of any point on the unit circle to its center?

Faust7

and the powerset axiom seems very close to the definition im used to

Dodsy

The distance matters!?

Leaky Nun

@Faust7 one must be very careful so as to distinguish between definition and axiom. In particular, axiom asserts that something exists.

Dodsy

oh right

Semiclassical

03:22

Sure, but it's a very simple distance.

Dodsy

the radius is 1

right?

Semiclassical

Right.

And every hypotenuse is just another radius.

Dodsy

so then the hyp is always 1

Semiclassical

Right.

Dodsy

so then since sin = o/h

it's just the y over 1

Semiclassical

03:23

Right.

But, going back to the equilateral triangle.

Dodsy

oh sorry

Semiclassical

If $\sin(30^\circ)$ is half the length of one of the sides of the equilateral triangle, and two of the sides are just radii of the circle, then what's $\sin(30^\circ)$?

Faust7

its seems the the axiom of power set is the definition combined with the axiom i was just working on

Dodsy

1/2

Semiclassical

Right-o.

Leaky Nun

03:24

@Faust7 what's the axiom you were just working on?

Dodsy

but how do you know that sin(30) is half the length of the hypoteneuse

Maybe I got lost along the way here

Semiclassical

Because the equilateral triangle is composed of two 30-60-90 triangles.

Dodsy

oh right.

TheGreatDuck

but sin(30) is the ratio

are we assuming that the hypotenuse is length 1?

Semiclassical

Yes.

Unit circle.

Dodsy

03:26

right.

TheGreatDuck

ah kk

carry on

watches intently

Faust7

it seems like a combination of this:

The form stated here, in which $B$ may be larger than strictly necessary, is sometimes called the axiom schema of collection.
and the definition of power set

Semiclassical

Once you've convinced yourself that $\sin(\pi/6)=\sin(30^\circ)=1/2$

Dodsy

uhhhhh

Semiclassical

The natural question is, what's $\cos(30^\circ)$?

Dodsy

03:27

hm.

Semiclassical

For that, examine the right triangle once more.

Dodsy

that's adjacent over hypoteneuse.

Semiclassical

Right.

Dodsy

but the adjacent is equal to 120

right?

Semiclassical

ummm

Dodsy

03:28

wait no.

Semiclassical

no. That's the supplement of 60 degrees.

Dodsy

let me think man

TheGreatDuck

probably a really stupid question but I feel like something just isn't quite right in my assumptions. I just want a bit of confirmation.

0

Q: If a parametric surface is continuous, is it reasonable to expect that an integer grid will also be continuous?

This is somewhat of a follow-up to a previous question which asked in a roundabout way about how to walk upon a parametric surface approximated by triangular faces. Now I want to know is whether or not my proposed construction of the model itself will actually work in the way I imagine. Suppos...

Dodsy

hmmm

so in this case the adjacent is the height of our triangle

that we've made

Leaky Nun

@Faust7 I don't get what you're trying to say.

Semiclassical

03:29

Dodsy

by reflecting in the x-axis

right

Semiclassical

This is the picture I've got.

Dodsy

right.

Faust7

@LeakyNun the Axiom schema of replacement

Semiclassical

Where $30^\circ$ is the angle between the x-axis and the two radii.

Dodsy

03:30

so cos(30) will be equal to $2\pi$

?

Faust7

doesnt it ensure the existence of the power set?

Semiclassical

you're not really approaching this right.

Dodsy

please help me

Semiclassical

By definition, what do we mean by "adjacent"?

Leaky Nun

@Faust7 are you familiar with the theorem "there is no bijection between a set and its power set"?

Dodsy

03:31

it means it's the line that connects the angle

Faust7

i guess if it did there wouldnt be a whole nother axiom for it

Dodsy

but isn't the hypoteneuse

Semiclassical

Eh, that's not how I"d say it.

Dodsy

Enlighten me.

Faust7

@LeakyNun theres is a one to one mapping from a set to its power set but no onto function

Semiclassical

03:32

What I'd say it as: It's the base length which is adjacent to the desired angle.

Leaky Nun

@Faust7 yes. So, you can't replace things to generate the power set.

Dodsy

but you used the word in the definition

Semiclassical

Whereas the "opposite" is the base length which is opposite to the desired angle.

Faust7

ah

Semiclassical

...heh, so I did.

TheGreatDuck

03:32

@Dodsy an adjacent side of a triangle with regards to some angle is the line segment which shares the vertex of the angle.

Dodsy

:)

Semiclassical

Oh well. Do you see what I mean?

TheGreatDuck

wait no

Dodsy

Well, I've known what you meant by adjacent this whole time, Semi.

TheGreatDuck

i was thinking of a line segment adjacent to (some other line segment)

Semiclassical

03:33

Fair enough.

Dodsy

If that's what you were trying to get across

I don't know where my hang up is.

Faust7

but arent you still using the Axiom schema of replacement to ensure that powerset is itself actually a set?

Semiclassical

Point here is that it's that line segment that lies on the x-axis.

Dodsy

Right

Semiclassical

So we need to know the length of that.

TheGreatDuck

03:33

@Dodsy if you don't mind me asking, what are you having an issue with?

Dodsy

remember I said "it's the height of the triangle"

Semiclassical

Except the height is vertical :/

Dodsy

what.

Semiclassical

This would be the width.

Dodsy

no

Leaky Nun

03:34

@Faust7 no, that's what the axiom of powerset does. (Please ping me lol)

Dodsy

of our equallateral!

Semiclassical

Are we using different terminology?

Faust7

lol

Semiclassical

Ohh.

Dodsy

yes

TheGreatDuck

03:34

mind if I ask what the problem is?

Semiclassical

I thought you meant the height of the 30-60-90 triangle.

Dodsy

nono

the height of the equallateral

Semiclassical

Ahh. Then we're good.

Be that as it may, it's better to focus on the 30-60-90 triangle for this particular point.

Dodsy

I said "then it is the height of the triangle we've made by reflecting in the x-axis"

Semiclassical

Ahhhh

Dodsy

03:35

I am not very good at explaining myself :)

no worries, whatever you think.

TheGreatDuck

really confused what the actual problem we are trying to solve is

Semiclassical

So you've got a right triangle. You know that its height is y=op=1/2, and its hypotenuse is 1.

What relation do you know for the sides of a right triangle?

Faust7

@LeakyNun im an idiot its an axiom they dont have to prove it

Dodsy

well

Leaky Nun

@Faust7 lol

Dodsy

03:36

lets call adjacent "x"

and opposite "y"

Semiclassical

Right.

Avantgarde

@LeakyNun what a name you got there

TheGreatDuck

the image for the preview might prove useful

Polar coordinate system

In mathematics, the polar coordinate system is a two-dimensional coordinate system in which each point on a plane is determined by a distance from a reference point and an angle from a reference direction. The reference point (analogous to the origin of a Cartesian system) is called the pole, and the ray from the pole in the reference direction is the polar axis. The distance from the pole is called the radial coordinate or radius, and the angle is called the angular coordinate, polar angle, or azimuth. == History == The concepts of angle and radius were already used by ancient peoples of the...

Faust7

@LeakyNun the Axiom is stating that it exists

TheGreatDuck

nvmd

Leaky Nun

03:37

@Faust7 so you've gone through them all... should I give you questions?

@Faust7 yes

Dodsy

we can find the length of "x" by using the angle made

Leaky Nun

@Avantgarde anagram of "Kenny Lau"

Semiclassical

We could, but that'll be circular.

Avantgarde

haha nice

TheGreatDuck

Dodsy

03:37

cos30=(x/1)

Semiclassical

It'll give us $x=\cos(30^\circ)$.

TheGreatDuck

let r = 1

that might prove helpful

Semiclassical

But the point is to find $x$ in the first place.

Faust7

@LeakyNun sure (im still re-reading the infinity one though)

Dodsy

Sorry, duck. I'm sure you're just trying to help, but I'm finding it hard to follow Semi at the moment.

Semiclassical

03:38

@Dodsy Should that be cos(30) = x/1 ?

Dodsy

right

sorry I forgot the equals sign.

Semiclassical

Okay. So if we know x, we'll know cos(30).

Leaky Nun

@Faust7 I won't ask you about infinite sets

Faust7

lol ok

Semiclassical

Trig itself isn't going to help here, though. We need something more basic.

What's the simplest relationship you know between the three sides of a right triangle?

TheGreatDuck

03:39

@Semiclassical I might be able to help explain but i don't know what we are explaining? General trigonometry or some particular algebra or geometry problem? I came in halfway so I have a lack of context.

Dodsy

@Semiclassical $a^2 + b^2 = c^2$

Semiclassical

@dodsy Right. What's a,b,c here?

Faust7

@LeakyNun Sure ^^

Dodsy

where c is equal to the hypoteneuse of a right triangle.

Semiclassical

If you want to know the context, read the transcript.

Right. What about a,b here---how have you labelled them for your problem?

Dodsy

03:41

It doesn't really matter, does it?

I suppose we will call "a" the opposite.

and "b" the adjacent

Semiclassical

Sure. So that'll be y, and b will be x.

Leaky Nun

@Faust7 Let $a$ be a non-empty set that exists. Remember to state which axioms you are using in each step.
1. Prove that $a^+$ exists, where $a^+ := a \cup \{a\}$.
2. Prove that there is a set $b$ such that $a \cap b = \varnothing$.
3. Prove that there does not exist a set $x$ such that $x = \{x\}$.
4. Prove that there does not exist a set $x$ such that $x = \{a,x\}$.

Semiclassical

What do we know about a and c in this problem, though? What did we conclude earlier?

We just got through belaboring the point re: what the hypotenuse was, for instance :)

Dodsy

c = 1

a = 1/2

Semiclassical

Right.

So you've got $(1/2)^2+b^2=1^2$.

And you can solve that for $b$.

Dodsy

03:46

root 3 over 2

Semiclassical

Correct! So therefore $\cos(\pi/6)=\cos(30^\circ)=x=b=\sqrt{3}/2$.

And believe it or not, that's enough to get you pretty much every other special value on the unit circle:

goo.gl/images/wAlU0l

Leaky Nun

@Semiclassical I think you may still want $\cos(36^\circ)$.

Semiclassical

yes, but you'll notice I used the weasel words "pretty much" :P

There are some other special values, but multiples of 45 degrees and multiples of 30 degrees are the big ones.

The derivations for any other special values are decidedly more esoteric and the values aren't anywhere near as relevant.

Dodsy

oh I think I've figured it out

So what I can do then

is convert to degrees

then figure out the quadrant

and subtract what I need to

if the angle is 30

then it's 1/2, root 3 over 2

and can be positive or negative depending on the quadrant

if it's 45, its root 2 over 2, root 2 over 2.

Semiclassical

right.

Dodsy

03:54

if it's 60 it's root 3 over 2, 1/2

Faust7

@LeakyNun Im kind working on all of them at once which is probably not a great idea but w.e
for question 1) can i cheat and use the power set axiom to show that the P(a) is a set and then show $a^+$ is contained in the power set of a thus $a^+$ is a set?

Leaky Nun

@Faust7 how would you show that $a^+$ is contained in the power set of $a$?

Semiclassical

To clarify: If it's 30 degrees, then the sine is 1/2 and and the cosine is sqrt(3)/2.

Dodsy

right

Semiclassical

But usually you'd write the coordinates for that as (sqrt(3)/2,1/2).

i.e. (x,y)

Dodsy

03:56

cosine is always "x" though isn't it?

if it's 30 degrees its cosine 1/2, sine sqrt(3)/2

Semiclassical

If you're measuring the angle counter-clockwise from the positive x-axis, then yes.

Dodsy

no wait

you're right.

it depends which side you're measuring from.

Leaky Nun

@Faust7 also, by "contained", do you mean $\in$ or $\subset$?

Faust7

@LeakyNun oh shit one sec

Semiclassical

Yeah. If you measured the angle from the y-axis, for instance, then the adjacent side would be on the y-axis as well.

This actually comes up when you do so-called spherical coordinates.

See for instance here: goo.gl/images/MVitHO

Dodsy

03:59

huh, so it might not be as easy as I thought

Transcript for

$Mathematics$ Mathematics