Mathematics

@Argon i don't have one favorite..i like math in general, like to study as much as i can , i wanna know as much as i can. it's frustrating, 'cause there's a lot of things...

Argon

@Melvin I ask many people this question and this is a common answer :)

Melvin

@Argon good to know!

a Dangerous Idea

@Argon What would your answer be?

Argon

12:54 AM

@aDangerousIdea I can't say much about most fields of math because I don't know anything about them! However, so far, I enjoy analysis a lot.

a Dangerous Idea

Anyone seen Charlie?

Argon

@aDangerousIdea Hmmm... not for a few days. Odd.

a Dangerous Idea

Maybe the troll ate her.

Argon

I hope not! :)

a Dangerous Idea

Me too...

If he did, I Will Hunt for him :-D

Argon

1:00 AM

Hahahaha!!

Gustavo Bandeira

I've watched a class in which the teacher said that a function is something that builds a correspondence between x and y ($f: x \rightarrow y$), and that for every x, there's one and only one y, is it possible to have functions where you can have more than 1 y for one x?

If they exist, how are they called?

Argon

Multivalued?

Gustavo Bandeira

Dunno - if you're asking me.

Argon

I think this is what you mean

Multivalued function

In mathematics, a multivalued function (shortly: multifunction, other names: many-valued function, set-valued function, set-valued map, multi-valued map, multimap, correspondence, carrier) is a left-total relation; that is, every input is associated with at least one output. Strictly speaking, a "well-defined" function associates one, and only one, output to any particular input. The term "multivalued function" is, therefore, a misnomer because functions are single-valued. Multivalued functions often arise from functions which are not injective. Such functions do not have an inverse func...

Gustavo Bandeira

@Argon Thanks.

Argon

1:13 AM

No problem

Gustavo Bandeira

odd/even are terms that make sense only when describing integers, right?

Argon

Generally, yes

peoplepower

There are odd/even permutations, odd/even functions.

Argon

This is true as well

jshin47

if gcd(a,2k) = 1, does it follow that gcd(a^2,b) = 1 since a has no prime factors in common with b?

peoplepower

1:24 AM

What's b?

jshin47

*k

Im trying to prove that 1 +sqrt(2) all over 2 is irrational lol

Just making sure im not going crazy

Peter Tamaroff

@jshin47 So?

Do you know $\sqrt 2$ is irrational?

jshin47

Yes that one is easy ... but Im trying to do it w/o invoking closure of rationals or anything like that

Im trying to do it directly by contradiction

(1 + sqrt(2)) / 2 = a/b => b^2 = 4(a^2 - ab)

so b is even

then we can write b = 2k

Peter Tamaroff

@jshin47 Stop right there.

Let $s=\sqrt 2$. If $$m=\frac{s+1}2$$ were rational, so would be $2m-1=s$. Impossibru.

Argon

Hahaaha!

jshin47

1:29 AM

Thats nice, but im trying to do it as if i dont know that sqrt 2 is irrational.

as naively as possible

and naiveity usually comes pretty naturally for me

Peter Tamaroff

Well, you need to make things swifter and swifter everyday.

jshin47

So b = 2k => a^2 = k^2 + 2ak

peoplepower

It looks like you're on the right track, then. Go two/three steps further.

jshin47

Then k | a

* k | a^2

But gcd(a,b) = gcd(a, 2k) = 1

So if i can show that gcd (a^2, k) should be 1, but is actually some multiple of k, im done

is actually k i mean

peoplepower

Well gcd(a,2k)=1 implies ar+k(2s)=1, so gcd(a,k)=1...

jshin47

1:33 AM

so is it obvious enough that gcd(a^any integer power, k) = 1 as well?

peoplepower

You can easily verify that by contrapositive.

jshin47

gotcha. I was thinking that it should be "about as easy" to show that such a number is irrational directly as showing sqrt2 itself is irrational

right.

peoplepower

You took a little longer in your proof than necessary. After establishing $k^2=a^2+ab$, we have $a|k^2$, so gcd(a,k)>1 so gcd(a,b)=gcd(a,2k)>1.

Which is what I meant by "Go two/three steps further."

jshin47

I see. Yes that makes sense

user19161

@aDangerousIdea Please don't use my photo as your gravatar, that makes me very upset.

jshin47

1:53 AM

whats that about imitation and flattery?

user19161

Wow, how does this site get so many questions?

Argon

Curious people!

user19161

@Argon Did I miss your integral of the day?

Argon

@WillHunting Didn't have one yet. I was saving it for you

Argon's Integral of the Day: $$\int_0^1 \frac{\text{artanh}\, x \,\, \log x}{x\, (1-x)\, (1+x)}\, dx=-\frac{1}{16}\Big(7\zeta(3)+2\pi^2\log 2\Big)$$

user19161

@Argon You must be good at series too then.

Argon

2:05 AM

@WillHunting I didn't come up with it :)

jshin47

Did you find this using complex analysis?

I mean taking integral over a contour that has the segment 0 to 1 in R

Argon

@jshin47 I didn't prove it, but by the strange constants, I'm not sure if contour integration would work well here.

peoplepower

@jshin47 Try generalizing the irrationality problem to $\frac{\sqrt{p}+a}{b}$ where $a$ and $b$ are relatively prime.

Argon

The Zeta series implies to me that I this is evaluated using some type of summation, as @WillHunting suggested

jshin47

@peoplepower ill do that now

user19161

2:07 AM

It is 29 Nov, this reminds me of Jer 29:11.

jshin47

btw is there a way to make chat display tex

user19161

@Argon Suggested? I didn't suggest anything!

peoplepower

@jshin47 math.ucla.edu/~robjohn/math/mathjax.html

Argon

@WillHunting "You must be good at series too then." How is this not an implication of series, Sir [CENSORED]?

@WillHunting Favourite verse?

Nov 2 at 21:01, by Jasper Loy

Wow, I have 29 bronze and 11 silver badges now. Reminds me of Jer 29:11.

user19161

@Argon Nope, just happen to remember since it's a popular one. I don't really have a favourite verse and I am not Christian.

Argon

2:10 AM

@WillHunting Popular? Hmmm...

user19161

@Argon Well, it's often quoted...

xiamx

i need some help with orthogonal compliment math.stackexchange.com/questions/246967/…

Argon

@WillHunting Hm... Didn't know that

jshin47

@peoplepower does that link work in google chrome?

peoplepower

@jshin47 Yes

jshin47

2:11 AM

Ahhh I get it

lol

oops.

peoplepower

@jshin47 $p$ is prime btw.

jshin47

I figured

Otherwise your statement is not true :)

well actually doesnt $p$ only need to be not a perfect square

peoplepower

True, but the proof is harder.

jshin47

I see, gotcha

peoplepower

Eventually, you just have give up and invoke the closure of rationals under addition and multiplication and their inverses. :)

jshin47

2:17 AM

lol

Trying to follow the exact same strategy as the previous case, but unfortunately only for 2 - 1 = 1 is this so straightforward

user19161

3:32 AM

@xiamx It's spelled complement and not compliment.

Transcript for

$Mathematics$ Mathematics