Mathematics

12:00 AM

@jdoe This was a contest question. I found the answer too, by the way, as it was given (sadly, only after the stupid thing was over)

jdoe

the nice thing about this is addition and multiplication give you back a number of that form: (a + b \sqrt 2)(c + d \sqrt 2) = (ac + 2 bd) + (ad + bc)\sqrt2

@Argon, so if I was to answer this quickly by hand I'd use the contined fraction to find that smallest solution (3,2) then perform something like (3+2 \sqrt{2})^5 by hand

by binary exponentiation

actualy that's ridiculous

Argon

With a calculator

jdoe

I'd just compute convergents of the continued fraction until I found x/y with x>100

that's much easier

Argon

@jdoe So can you explain this to me? Why continued fractions?

jdoe

@Argon, let x/y be a very good approximation of sqrt{2} (that's what continued fractions give you' the best possible rational approximation)

then x^2 - 2y^2 is going to be a very small integer but obviously it can't be zero

Argon

12:03 AM

@jdoe With Newtons method or something?

jdoe

hm?

(no, but what do you mean)

Argon

@jdoe Nevermind

jdoe

newtons method is not used in this

Argon

@jdoe But it will give a great rational approximation to $\sqrt{2}$

jdoe

quadratic irrationals have periodic continued fractions which are easy to find, e.g. \sqrt 2 = 1 + 1/(2 + 1/(2 + 1/(2 + 1/(2 + 1/( ...))))

so 1 + 1/(2 + 1/(2 + 1/(2 + 1/(2)))) = 41/29, for example, is a great approximation

and 41^2 - 2*29^2 = -1

Argon

12:06 AM

@jdoe Could I use any approximation that is close?

jdoe

if your approximation p/q isn't the convergent of a continued fraction it wont be 1/q^2 within the sqrt{2}, so you'll get p^2 - 2*q^2 = +/- something bigger than 1. This is legendre theorem from the theory of continued fractions.

Argon

P.S. 3 iterations w/ Newtons method gives $\frac{577}{408}$!

@jdoe Ok

jdoe

ok now I know what you meant about newtons method

Argon

Anyhow, so now I have $p$ and $q$ from a truncated continued fraction.

user19161

@Argon You like continued fractions? I know nothing about them, not even the definition!

Argon

12:11 AM

@WillHunting I'm not a huge fan, because they are generally not so useful. They are quite fascinating though!

jdoe

you might be surprised what you can compute with continued fractions

you can find primes of the form x^2 + y^2 and things

Argon

@jdoe Haha, $577^2 - 2\cdot408^2= 1$!!!!

HOW?

@jdoe Why does this happen?

jdoe

what?

Argon

@jdoe The numerator and denominators suffice the equation!!

jdoe

(p/q)^2 - 2 = 1/q^2

i.e. p/q is very close to sqrt{2}

user19161

12:19 AM

@Argon What? A and B suffice C? Is there such a sentence?

user19161

Oh, actually there is.

Argon

Hahahahahaha! Stumped the great Jasper!

@jdoe Ok :)

I like this

jdoe

does that make sense?

user19161

@Argon Rather, the little banana.

Argon

Not exactly

@WillHunting Nope, great Jasper

jdoe

12:20 AM

well look at it this way

$$p/q \sim \sqrt{2}$$

user19161

@Argon This reminds me of GC's "How long is your banana?"

jdoe

$$p^2/q^2 \sim 2$$

Argon

@WillHunting Hahha I remember that... and the flagging!

@jdoe Ok!

jdoe

$$p^2 \sim 2q^2$$

Argon

@jdoe Ok

jdoe

12:21 AM

$$p^2 - 2q^2 \sim 0$$

Argon

Right

user19161

@Argon Haha, I got flagged once too for saying "Between you and me, we have two bananas and four nuts"!

jdoe

but because $\sqrt{2}$ is irrational it can never actually equal 0, so it'll just be a very small integer

Argon

@WillHunting HAHAHAHAHA!

@jdoe Right

I got that part

But why $\frac{1}{q^2}$

jdoe

$$\left|\frac{p}{q} - \sqrt{2}\right| < \frac{1}{q^2}$$

Argon

12:23 AM

Ok

Oh @Will y, the blue boy No, he's not, he's Jasper Loy- M

@jdoe But we want the RHS to be 1!

user19161

I now have over 32k SE wide, yay!!!

Argon

YAY!

Link

Hi Again!

Argon

Hi, @Link!

user19161

@Argon My usual question: have you finished your HW?

Argon

12:26 AM

@WillHunting Sorta

user19161

@Link Hey Max!

Argon

}:)

Link

@WillHunting, congrats on 32K!

I'm almost to 3K >.>

Argon

I'm almost 6k

user19161

@Link Yup, though you don't really use many of the 101 accounts. =)

Link

12:27 AM

Who me?

:p

Yea, it just auto connected when I checked the site through googling :p

Argon

What kind of problems are solved w/ abstract algebra?

user19161

@Link I deleted 8 accounts that I didn't use very much, so now left with 3.

Link

@WillHunting The magical kind :p

user19161

@Argon Well, for example, insolvability of the quintic equation!

Link

Mine kept auto-connecting, so I thought, screw it.

Argon

12:29 AM

@WillHunting Cool

user19161

@Argon Also, some number theory results.

Argon

:)

Link

Any ideas for a senior quote for yearbook? I have 100 characters, preferably something witty :p

Argon

@Link Haha - I was actually thinking about what I would write for next year, this morning on my way to school - and came up with zilch.

Link

:p

I was thinking of a quote

But didn't know which

jdoe

12:36 AM

to show that p^2 - 2q^2 = 1 and no bigger you actually need a stronger result, that the approximations have quality $\frac{1}{2 \sqrt{2} q^2}$

Argon

Fancy stuff!

Link

I don't even understand the question :p

Argon

@Link Well, I want to find integer solutions to $x^2-2y^2=1$

jdoe

i might have that wrong

Link

Oh, okay

So just a hyperbola?

Argon

12:38 AM

@Link Yup. But with ints

Link

@Argon, I thought of this one: “People say nothing is impossible, but I do nothing every day.” -- Winne The Pooh

Argon

@Link Hahahhahahaha! That's great :)

Link

I know

its pretty good

Argon

The fact that it is by Winnie the Pooh is the icing on the cake :)

Link

I know right? I have this one too from calvin (Calvin and Hobbes) "(During a test.) As you can see, I have memorized this utterly useless piece of information long enough to pass a test question. I now intend to forget it forever. You've taught me nothing except how to cynically manipulate the system. Congratulations."

But it doesn't fit the 100 character limit

Argon

12:41 AM

I'm sure the teachers would love that one :) Hahahahaha

Link

I was thinking about using that on a test, but my teachers usually never ask for dates, just events on that date, instead of the date of the event. :C

@Argon, look at this math problem