Register FAQ Search Today's Posts Mark Forums Read

 2019-11-08, 22:31 #1 MattcAnderson     "Matthew Anderson" Dec 2010 Oregon, USA 7·89 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
 2019-11-09, 09:44 #2 Nick     Dec 2012 The Netherlands 27408 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 1110111011012 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
 2019-11-09, 20:59 #4 MattcAnderson     "Matthew Anderson" Dec 2010 Oregon, USA 62310 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 11578 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 7·89 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 22·7·11·29 Posts can you find a square-free counterexample?
 2019-11-10, 13:48 #8 Dr Sardonicus     Feb 2017 Nowhere 3,821 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.
 2019-11-10, 15:33 #9 LaurV Romulan Interpreter     Jun 2011 Thailand 213448 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
2019-11-10, 16:59   #10
Nick

Dec 2012
The Netherlands

25×47 Posts

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").

 2019-11-10, 17:06 #11 MattcAnderson     "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}.