Thread: Nested Radicals
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Old 2019-11-10, 13:48   #8
Dr Sardonicus
 
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Feb 2017
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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.
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