mersenneforum.org  

Go Back   mersenneforum.org > Great Internet Mersenne Prime Search > Math

Reply
 
Thread Tools
Old 2019-11-08, 22:31   #1
MattcAnderson
 
MattcAnderson's Avatar
 
"Matthew Anderson"
Dec 2010
Oregon, USA

7×89 Posts
Smile Nested Radicals

Hi again,

Wrote down (again) a nifty bit of mathematical trivia.

Let
sqrt(b+sqrt(c)) = sqrt(f)+sqrt(g)
and f<g
then
f = (b-sqrt(b^2-c))/2
and
g = (b+sqrt(b^2-c))/2.

I worked it out and checked with an example.
link1
link2

cheers
MattcAnderson is offline   Reply With Quote
Old 2019-11-09, 09:44   #2
Nick
 
Nick's Avatar
 
Dec 2012
The Netherlands

2×751 Posts
Default

If \(u+\sqrt{v}=x+\sqrt{y}\) then it does not necessarily follow that u=x and v=y.
Nick is online now   Reply With Quote
Old 2019-11-09, 13:24   #3
Dr Sardonicus
 
Dr Sardonicus's Avatar
 
Feb 2017
Nowhere

37×103 Posts
Default

\sqrt{2 + \sqrt{49}} = \sqrt{1} + \sqrt{4}

Last fiddled with by Dr Sardonicus on 2019-11-09 at 13:28 Reason: use smaller numbers
Dr Sardonicus is offline   Reply With Quote
Old 2019-11-09, 20:59   #4
MattcAnderson
 
MattcAnderson's Avatar
 
"Matthew Anderson"
Dec 2010
Oregon, USA

62310 Posts
Default

Nick,
You raise a good point. I do not right away see any counterexample. I just set rational parts equal and radical parts equal.
MattcAnderson is offline   Reply With Quote
Old 2019-11-10, 02:06   #5
MattcAnderson
 
MattcAnderson's Avatar
 
"Matthew Anderson"
Dec 2010
Oregon, USA

7×89 Posts
Default

with b+sqrt(c) assume b and c are rational numbers.
Also, if c is a perfect square, it simplifies a bit.
MattcAnderson is offline   Reply With Quote
Old 2019-11-10, 02:08   #6
MattcAnderson
 
MattcAnderson's Avatar
 
"Matthew Anderson"
Dec 2010
Oregon, USA

7×89 Posts
Default

Nick,
you are correct.
for example
2+sqrt(4) = 1+sqrt(9)
MattcAnderson is offline   Reply With Quote
Old 2019-11-10, 04:15   #7
LaurV
Romulan Interpreter
 
LaurV's Avatar
 
Jun 2011
Thailand

22·7·11·29 Posts
Default

can you find a square-free counterexample?
LaurV is offline   Reply With Quote
Old 2019-11-10, 13:48   #8
Dr Sardonicus
 
Dr Sardonicus's Avatar
 
Feb 2017
Nowhere

381110 Posts
Default

This exercise provides a nice illustration of the "wrong square root problem."

If you assume that f and g are positive integers, f < g, and that

x^2 - f, x^2 - g, and x^2 - f*g are irreducible in Q[x],

squaring both sides of the given equation leads to the stated relations, apart from the sign in the square root of b^2 - c, which is decided by f < g. Things can be put in terms of f and g as follows:

c = 4*f*g, b = g + f, and sqrt(b^2 - c) = g - f.

For example f = 2, g = 3 gives c = 24, b = 5, and

(sqrt(2) + sqrt(3))^2 = 5 + sqrt(24).

However, if f < g < 0, a minus sign goes missing in action, because sqrt(f)*sqrt(g) = -sqrt(f*g) (at least, assuming sqrt(f) and sqrt(g) are the positive square roots of |f| and |g|, multiplied by the same square root of -1). In this case,

sqrt(b + sqrt(c)) = sqrt(f) - sqrt(g) or sqrt(g) - sqrt(f), depending on whether you want pure imaginary numbers with positive or negative imaginary part.

f = -2, g = -1 give c = 8, b = -3, and b^2 - c = 1. Thus

sqrt(-3 + sqrt(8)) = sqrt(-1) - sqrt(-2) or sqrt(-2) - sqrt(-1).

I leave it to the reader to deal with the case f < 0 < g.
Dr Sardonicus is offline   Reply With Quote
Old 2019-11-10, 15:33   #9
LaurV
Romulan Interpreter
 
LaurV's Avatar
 
Jun 2011
Thailand

893210 Posts
Default

My point was that (related to Nick's post) if you have u+\sqrt{v}=x+\sqrt{y} then you subtract from both sides the min(u,x), and you get two square roots that differ by and integer. Assuming some square free stuff there, this shouldn't be possible unless you have the same root, which implies then you have the same integers...

Edit: whodahack broke matjax again?

Last fiddled with by LaurV on 2019-11-10 at 15:39 Reason: reverted matjax to normal TeX
LaurV is offline   Reply With Quote
Old 2019-11-10, 16:59   #10
Nick
 
Nick's Avatar
 
Dec 2012
The Netherlands

2·751 Posts
Default

Quote:
Originally Posted by LaurV View Post
My point was ...
Absolutely (we were leaving it to the OP to respond to your challenge!)

For those interested, these ideas also lead to Vitali sets which are a 2-dimensional analogue of the Banach-Tarski theorem (or "paradox").
Nick is online now   Reply With Quote
Old 2019-11-10, 17:06   #11
MattcAnderson
 
MattcAnderson's Avatar
 
"Matthew Anderson"
Dec 2010
Oregon, USA

7×89 Posts
Default

Dr. Sardonicus,

Thank you for your input.
I do not see where the equation
{1} sqrt(b^2 - c) = g - f.
comes from.

We start with
sqrt(b+sqrt(c))=sqrt(f)+sqrt(g).
As you pointed out, it follows that
b=f+g
and
c=4*f*g.
But, I do not yet understand the above equation {1}.
MattcAnderson is offline   Reply With Quote
Reply

Thread Tools


All times are UTC. The time now is 23:09.

Sat Nov 28 23:09:38 UTC 2020 up 79 days, 20:20, 3 users, load averages: 1.73, 1.43, 1.28

Powered by vBulletin® Version 3.8.11
Copyright ©2000 - 2020, Jelsoft Enterprises Ltd.

This forum has received and complied with 0 (zero) government requests for information.

Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation.
A copy of the license is included in the FAQ.