@skullpetrol anyways, is that the advice you meant?

Yup.

You'll never know if you don't try.

can anyone tell me the difference between reciprocal and oppsite

4/5 and 5/4 are reciprocal right...what about opposite..what do they mean in math

Yes, 4/5 and 5/4 are reciprocals because (4/5) * (5/4) = 1.

Opposites, on the other hand, have opposite signs.

Such as, 6 and -6 or -2.5 and 2.5.

6 + (-6) = 0

@skullpetrol Oh ok so reciprocals equal to one when multiplied and opposite when added to zero?

what about inverses?

Yes.

Reciprocals are also called multiplicative inverses

@MATH i have a question: what's the inverse of 0? :P

Opposites are also called additive inverses.

Oh

@ZachHauk is it sin of 0 degrees

??

where did you get sine?

because sin 0 = 0

mixed up sin and cos there lol

anyways, 0 * 0 = 1??

oh or is it cosine idk let me check the unit circle real quick @ZachHauk

unit circle has nothing to do with this

this is

what is the inverse of 0?

wait so the inverses also equal to 1

the inverses multiply to 1

what number, when multiplied by 0, is 1?

not any

yep :P

so is it undefined?

in any "field", everything has an inverse

EXCEPT

the additive identity

we call 0 an "Additive identity"

because 0 + x = x

for ANY x

oh is zero only an additive identity or are there any other

knowing that, what would you say is the Multiplicative identity?

@MATHASKER nope, that's the only one

the multiplicative idenitity would be multiplied and equal to one

The multiplicative identity is 1

because x * 1 = x, right?

Hey @Daminark

oh were finding something that when multiplied against becomes whats in the other side of the equal

like

1/2 * 2

oh ok....so reciprocals when multiplied become one, inverses when added is one and opposite when added is one?

its 2?

no, when added it's zero

Opposites add to 0, Reciprocals/Inverses multiply to 1

a + (-a) = 0

Ohh ok so how would the arcsin(sqr3/2) = 60 if they don't multiply to 1

Hi

what??

OH

LOLL

@skullpetrol HE MEANT INVERSE TRIGONOMETRIC FUNCTIONS

@MATHASKER sorry, we misunderstood you

Forget all of that

:(

Inverse functions

are where you take in the OUTPUT

and spit back the INPUT

no its fine...i needed what the general idea of reciprcal, inverse and opposites so

for example, we have that sin(60) =sqr(3)/2

but, what if we only have the output?

and we want to find the angle

well, put it through the inverse function and we'll get back what we wanted

e.g. arcsin(sqr(3)/2) = 60

make sense?

Hi @Akiva

ohhh ok so we work our way back

yep :]

So this is the difference between "multiplicative inverses" and "inverse functions"

However, watch out

Because, there are actually infinitely many "inputs"

which would yield sqr(3)/2

("Additive inverses" are also a thing, though they're usually called "opposites")

For example, add or subtract 360 any time

sin(60) = sin(60 + 360) = sin(60-360) and so on

Yeah. Sin(1)=sin(361)=sin(721)=…, since sine is periodic

@Akiva should i at least apply? @skullpetrol said i should

so we only take from like first or fourth quadrant or something like that

So the solution is to take just whatever's between -90 and 90 (not counting 90)

...90?

oh what if its sin(-sqr(3/2), do we go clockwise?

Yeah?

oh

@MATHASKER well

is -sqrt(3/2) an angle??

does that look like an angle to you? :P

what if it says something like tan^-1(sin(sqr(3)/2)) is it tan^-(60)?

remember, sine functions take in an angle and spit out a value

For arccosine you take between 0 and 180, for arcsine you take between -90 and 90, I think @MATHASKER

no I meant inverse

you mean arcsin?

sin^-1(-sqr(3/2)*

It's the same as choosing the right quadrant

ya same thing right lol

@AkivaWeinberger we choose between 90 and -90

its kinda confusing but I got it a bit

(Sorry, typos)

@Akiva wait... what?

Fixed

180 degrees wouldnt cover it all

For cosine, 0 to 180 would

arcsin(sin(271))?

Remember,

sin(x)=sin(180-x)

wait, really?

Yeah

i thought it was periodic over 360 degrees?

It is

I said minus, not plus

wait

It's symmetric about the x=90 line

Check a graph

youre right

but

Or the circle diagram

Just two quadrants to cover every y-coordinate

so wait, what would 271 be?

Subtract 360. It's -89.

269 is also -89; subtract from 180.

ok let me check my graph lol

that's right

sin(271)=sin(269)

you dont need to go over the repeated values

sorry, i had a moment of stupidity

:/

@ZachHauk I suppose why not, what do you have to lose

my dignity /s

except maybe precious time and energy

Say, have you ever read Ted's chapter on non-euclidean geometries in his algebra book?

No, I don't think I have that book

I'm sure he'd send it to you if you asked

currently reading about spectral theorem and linear algebra stuff

I think I'm fine at the moment

Hyperbolic geometry has some nice pictures, though

(and I say that without hyperbole)

oh ok

So here's a question

I draw finitely many curves on an infinite piece of paper

They don't intersect themselves and they don't intersect each other

(They're not infinitely long)

Given any two points A and B not on the curves, must I always be able to connect A and B with a curve that doesn't intersect itself or any of the lines?

Also: Is there a simple proof of this?

looks pretty trivial..

I mean, it's intuitive enough. I don't actually know if you have enough topolog knowledge to formalize the statement, so you probably can't give a formal proof

looks homotopy related?

I don't know of any simple (formal) proof at all, though

also, where can i buy that infinite piece of paper?

at hilberts hotel?

Sure. It's available at the front desk

Part of the paper is, anyway

Also — what about infinitely many lines

lines as in

Curves

segments?

oh

Curvy line segments

infinitely many curves on the piece of paper?

still seems to be true..

Oh, I see one way: Make them all vertical, starting at the $x$-axis and going up one inch

yeah

that's what i was thinking

and then draw that for every possible x-coordinate. Then things above and below can't get connected.

What about just countably many, though?

countably many curves?

Yeah

That don't intersect themselves or each other

well

that should be true

do you have any other problems?

@Akiva you should note that I haven't studied much. I still need to study real analysis and stuff like that

@ZachHauk I think it's false with countably many

hmm, do you have a counter example?

Consider an arc of a circle with an angle of 270 degrees, with the 90 degree opening facing up. Make it radius 1/2.

Now consider another arc of a circle with an angle of 270 degrees, with the 90 degree opening facing down, of radius 2/3. (Concentric with the first one)

And a third, with the opening facing up, of radius 3/4. And a fourth, with the opening facing down, of radius 4/5. Etc.

These are my countably many curves.

oh

so it's kind of like

no matter what, they must travel an infinite length

Can you connect the center of the circles to anything outside of the limit circle of radius 1?

Yeah @ZachHauk

Alternatively: Does the path first exit the "limit circle" out the top or out the bottom

Hey everyone!

neither

that's kind of like asking

$1-1+1-1+\dots$

Yeah

Hi @Daminark

The idea here is called "path-connectedness." The complement of these curves is not path connected.

But, if we only had finitely many of the curves, it should be connected

Hi all..

any1?

@greedIsGoodAha what is $m$?

do you mean to ask how to find $a$?

that's called the discrete log problem

hi @arctic

hello

yes, i mean how to find $a$

i edited my question, thank you

is fermat's little theorem the correct direction to go in? @arctictern

there's no "good" algorithm for solving the discrete log problem in Zp*

@Akiva could a proper statement be "Given finitely many finite paths which do not intersect eachother or self-intersect, is the complement of these curves path connected?"

Hi. Am I right in here mathb.in/128348?key=31a5064c3d30f14a213f72b0e26db3ebb335f9a5 ? A question on $\sigma$-algebras

is it fair to say then that although we can find $x^{p-1} \equiv a^{{(p-1)}^e} \equiv 1 \mod p$, there is no way to get to $a^{({(p-1)}^e) + 1} \equiv a \mod p$?

we don't have to do anything to find 1 mod p, one is just one

i don't quite understand, 1 is just 1 but i am trying to use $x = a^{(p-1)^{e}}$ to discover $a$ when $p$ and $e$ are known. so if i raise $a$ to an exponent such that it is equivalent to $1 \mod p$, then if i raise it to one plus that exponent won't i be able to discover the value of $a$?

@greedIsGoodAha you can't raise a to any exponent without knowing a

don't know why you're write down things like (p-1)^e either

because $x = a^e$, and $x^{p-1} = a^{(p-1)^e}$

no, $x^{p-1}=(a^e)^{p-1}=a^{e(p-1)}$

they aren't equivalent?

they are congruent mod $p$, but only because $e(p-1)$ and $(p-1)^e$ are both divisible by $p-1$

either way, why would anybody write down something like $(p-1)^e$?

i'm not sure if that's rhetorical but why not? if it's congruent and i'm trying to find $a$, then i want to express $x$ in terms of $a$

> why not

um, why though?

it's completely random and unhelpful

@ZachHauk Yeah, I think so. Do you know how a path is defined?

why is it random? it's using fermat's little theorem

It's just the image of a continuous function from [0,1] to R^2

continuous mapping from $[0,1] \to \Bbb R^2$?

Yeah

@greedIsGoodAha but where did you get the idea to pick (p-1)^e from, other than accidentally using exponent rules incorrectly?

you could have just as easily written $x^{(e(p-1))^{e(p-1)}}$ or other equally random and unhelpful things

@ZachHauk And, for "not self-intersecting", you'd want it to be one-to-one

one-to-one would be a non-self-intersecting path, right?

hah

we're in sync