20191108, 22:31  #1 
"Matthew Anderson"
Dec 2010
Oregon, USA
7×89 Posts 
Nested Radicals

20191109, 09:44  #2 
Dec 2012
The Netherlands
1502_{10} Posts 
If \(u+\sqrt{v}=x+\sqrt{y}\) then it does not necessarily follow that u=x and v=y.

20191109, 13:24  #3 
Feb 2017
Nowhere
37×103 Posts 
Last fiddled with by Dr Sardonicus on 20191109 at 13:28 Reason: use smaller numbers 
20191109, 20:59  #4 
"Matthew Anderson"
Dec 2010
Oregon, USA
7·89 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. 
20191110, 02:06  #5 
"Matthew Anderson"
Dec 2010
Oregon, USA
7×89 Posts 
with b+sqrt(c) assume b and c are rational numbers.
Also, if c is a perfect square, it simplifies a bit. 
20191110, 02:08  #6 
"Matthew Anderson"
Dec 2010
Oregon, USA
7·89 Posts 
Nick,
you are correct. for example 2+sqrt(4) = 1+sqrt(9) 
20191110, 04:15  #7 
Romulan Interpreter
Jun 2011
Thailand
2^{2}·7·11·29 Posts 
can you find a squarefree counterexample?

20191110, 13:48  #8 
Feb 2017
Nowhere
37·103 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  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. 
20191110, 15:33  #9 
Romulan Interpreter
Jun 2011
Thailand
2^{2}·7·11·29 Posts 
My point was that (related to Nick's post) if you have 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 20191110 at 15:39 Reason: reverted matjax to normal TeX 
20191110, 16:59  #10 
Dec 2012
The Netherlands
2736_{8} Posts 
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 2dimensional analogue of the BanachTarski theorem (or "paradox"). 
20191110, 17:06  #11 
"Matthew Anderson"
Dec 2010
Oregon, USA
7·89 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=4*f*g. But, I do not yet understand the above equation {1}. 