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 = (bsqrt(b^2c))/2 and g = (b+sqrt(b^2c))/2. I worked it out and checked with an example. [URL="https://docs.google.com/viewer?a=v&pid=sites&srcid=ZGVmYXVsdGRvbWFpbnxtY2FzYW5kYm94fGd4OjVhNmZmZTUyMDE3ZDQ2YWI"]link1[/URL] [URL="https://docs.google.com/viewer?a=v&pid=sites&srcid=ZGVmYXVsdGRvbWFpbnxtY2FzYW5kYm94fGd4OjQ5ODY1Y2YyZDY4NTlmYmU"]link2[/URL] cheers 
If \(u+\sqrt{v}=x+\sqrt{y}\) then it does not necessarily follow that u=x and v=y.

[tex]\sqrt{2 + \sqrt{49}} = \sqrt{1} + \sqrt{4}[/tex]

Nick,
You raise a good point. I do not right away see any counterexample. I just set rational parts equal and radical parts equal. 
with b+sqrt(c) assume b and c are rational numbers.
Also, if c is a perfect square, it simplifies a bit. 
Nick,
you are correct. for example 2+sqrt(4) = 1+sqrt(9) 
can you find a [U]squarefree[/U] counterexample?

This exercise provides a nice illustration of the "wrong square root problem."
If you assume that f and g are [i]positive[/i] 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 [i]same[/i] 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. 
My point was that (related to Nick's post) if you have [TEX]u+\sqrt{v}=x+\sqrt{y}[/TEX] 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? :shock: 
[QUOTE=LaurV;530199]My point was ...[/QUOTE]
Absolutely (we were leaving it to the OP to respond to your challenge!) For those interested, these ideas also lead to [URL="https://en.wikipedia.org/wiki/Vitali_set"]Vitali sets[/URL] which are a 2dimensional analogue of the BanachTarski theorem (or "paradox"). 
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}. 
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