
Re: simplify radicals  regular 12gon
Posted:
Dec 22, 2006 12:39 PM


JEMebius wrote:
> I guess one can look at the equality > > sqrt (2  sqrt3) = (sqrt6  sqrt2)/2
[snip]
> And =why= does a similar equality for sqrt (2  sqrt2) > not exist (regular octagon)? I don't know.
One of the results I alluded to in my other post in this thread is the following, which can be found in Chapter XI, Section 9, pp. 207210 (esp. bottom of p. 208 to top p. 209) of George Chrystal, "Algebra: An Elementary TextBook", Part I, 7'th edition, AMS Chelsea.
THEOREM: Let p,q be rational numbers. Then sqrt(p +/ sqrt(q)) can be expressed as sqrt(a) +/ sqrt(b) for some rational numbers a,b if and only if p > 0 and p^2  q is a perfect square.
Of course, one could also ask whether sqrt(p +/ sqrt(q)) could be written as a sum of three or more square roots. I think the answer is "no", as the p,q radical expression has degree at most 4, and the sum of 3 or more "dissimilar" square roots has degree at least 6. (By "degree", I mean the degree of the minimal polynomial over the rationals for the expression.)
For more about these and more complicated types of manipulations, google (as separate searches) the phrases "denesting radicals", "nested radical", and "nested radicals".
Here's how you can show your specific example, sqrt(2  sqrt(2)), can't be rewritten in the way that sqrt(2  sqrt(3)) can.
Assume sqrt(2  sqrt(2)) = sqrt(a)  sqrt(b), where a and b are rational numbers with b < a.
Both sides are positive, so we can square without introducing extraneous roots.
2  sqrt(2) = a + b  2*sqrt(ab)
or
2*sqrt(ab) = (a + b  2) + sqrt(2)
Squaring again gives
4ab = (a+b2)^2 + 2(a+b2)*sqrt(2) + 2
If a + b is not equal to 2, then this last equation allows us to write sqrt(2) as a rational combination of a and b, which contradicts the irrationality of sqrt(2).
Hence, we must have a + b = 2.
Using this in the first equation (cancel 2 = a+b from both sides) gives us sqrt(2) = 2*sqrt(ab), or 2 = 4ab.
Hence, we have
a + b = 2 4ab = 2
Plugging b = 2  a (rewrite of first equation) into 2ab = 1 (rewrite of second equation) gives us
2a(2  a) = 1
or
2a^2  4a + 1 = 0.
However, this equation has no rational roots, so we again have a contradiction.
Of course, we could also try a representation of the form sqrt(a) + sqrt(b), but you'll find the same thing happens. More generally, you can take care of both at once by originally starting with sqrt(2  sqrt(2)) = r*sqrt(a) + s*sqrt(b), where r, s, a, and b are rational. Following the steps above, we get
2  sqrt(2) = ar^2 + bs^2 + 2rs*sqrt(ab)
or
2rs*sqrt(ab) = (ar^2 + bs^2  2) + sqrt(2)
Squaring again gives
4(r^2)(s^2)ab = (ar^2 + bs^2  2)^2 + 2(ar^2 + bs^2  2)*sqrt(2) + 2
If ar^2 + bs^2 is not equal to 2, then this last equation allows us to write sqrt(2) as a rational combination of r, s, a, and b, which contradicts the irrationality of sqrt(2).
Hence, we must have ar^2 + bs^2 = 2.
Using this in the first equation (cancel 2 = ar^2 + bs^2 from both sides) gives us sqrt(2) = 2rs*sqrt(ab), or 2 = 4(r^2)(s^2)ab.
Hence, we have
ar^2 + bs^2 = 2 4(r^2)(s^2)ab = 2
Plugging bs^2 = 2  ar^2 (rewrite of first equation) into 2ar^2(bs^2) = 1 (rewrite of second equation) gives us
2ar^2(2  ar^2) = 1
or
2(ar^2)^2  4(ar^2) + 1 = 0.
This is a quadratic equation for ar^2. Solving, we get only irrational values for ar^2, which contradicts our assumption that a and r are rational.
Incidentally, the same method that I used in each of the above proofs can be used to prove the theorem I stated earlier.
Dave L. Renfro

