Nested Radicals

MattcAnderson · 2019-11-08, 22:31

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

Nick · 2019-11-09, 09:44

If $u+\sqrt{v}=x+\sqrt{y}$ then it does not necessarily follow that u=x and v=y.

Dr Sardonicus · 2019-11-09, 13:24

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

MattcAnderson · 2019-11-09, 20:59

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 · 2019-11-10, 02:06

with b+sqrt(c) assume b and c are rational numbers.
Also, if c is a perfect square, it simplifies a bit.

MattcAnderson · 2019-11-10, 02:08

Nick,
you are correct.
for example
2+sqrt(4) = 1+sqrt(9)

LaurV · 2019-11-10, 04:15

can you find a square-free counterexample?

Dr Sardonicus · 2019-11-10, 13:48

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.

LaurV · 2019-11-10, 15:33

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?

Nick · 2019-11-10, 16:59

Quote:

Originally Posted by LaurV

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").

MattcAnderson · 2019-11-10, 17:06

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}.

2019-11-08, 22:31	#1
MattcAnderson "Matthew Anderson" Dec 2010 Oregon, USA 2⁵·5² Posts	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

2019-11-09, 09:44	#2
Nick Dec 2012 The Netherlands 2×23×37 Posts	If \(u+\sqrt{v}=x+\sqrt{y}\) then it does not necessarily follow that u=x and v=y.

2019-11-09, 13:24	#3
Dr Sardonicus Feb 2017 Nowhere 4,643 Posts	$\sqrt{2 + \sqrt{49}} = \sqrt{1} + \sqrt{4}$ Last fiddled with by Dr Sardonicus on 2019-11-09 at 13:28 Reason: use smaller numbers

2019-11-10, 15:33	#9
LaurV Romulan Interpreter Jun 2011 Thailand 10010110001011₂ Posts	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

2019-11-09, 20:59	#4
MattcAnderson "Matthew Anderson" Dec 2010 Oregon, USA 800₁₀ Posts	Nick, You raise a good point. I do not right away see any counterexample. I just set rational parts equal and radical parts equal.

2019-11-10, 02:06	#5
MattcAnderson "Matthew Anderson" Dec 2010 Oregon, USA 2⁵×5² Posts	with b+sqrt(c) assume b and c are rational numbers. Also, if c is a perfect square, it simplifies a bit.

2019-11-10, 02:08	#6
MattcAnderson "Matthew Anderson" Dec 2010 Oregon, USA 1100100000₂ Posts	Nick, you are correct. for example 2+sqrt(4) = 1+sqrt(9)

2019-11-10, 04:15	#7
LaurV Romulan Interpreter Jun 2011 Thailand 7·1,373 Posts	can you find a square-free counterexample?

2019-11-10, 13:48	#8
Dr Sardonicus Feb 2017 Nowhere 4,643 Posts	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 - fg 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 = 4fg, 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(fg) (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.

2019-11-10, 17:06	#11
MattcAnderson "Matthew Anderson" Dec 2010 Oregon, USA 2⁵·5² Posts	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=4fg. But, I do not yet understand the above equation {1}.