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 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.

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 [U]square-free[/U] 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 [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.

 LaurV 2019-11-10 15:33

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:

 Nick 2019-11-10 16:59

[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 2-dimensional analogue of the Banach-Tarski theorem (or "paradox").

 MattcAnderson 2019-11-10 17:06

Dr. Sardonicus,

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