20170816, 20:40  #1 
Mar 2016
2^{2}×3×5×7 Posts 
pyth. trippel and rotation
A peaceful night for all,
I have the pythagoraic trippel 3, 4, 5 with 3^2+4^2=5^2, division by 5^2 is (3/5)^2+(4/5)^2=1, which means that the point is on the unit circle. tan a1 = 3/4 so that arctan (3/4) = 36.87 ° same thing for the pyth. trippel 5, 12, 13 (5/13)^2+(12/13)^2=1 tan a2 = 5/12 so that arctan (5/12) = 22.62 ° i know that there is a rotation matrix with discr (M)=1 with (1, 2, 2) (3,4,5) (2,1,2) = (5,12,13) (2,2,3) from http://mathworld.wolfram.com/PythagoreanTriple.html Is the degree of the rotation matrix = a2/a1 or a2  a1 and is the degree rational or irrational. Thanks in advance, if some one could give me an explanation. Greetings from Germany Bernhard 
20170816, 22:10  #2 
Dec 2012
The Netherlands
2^{4}·113 Posts 

20170816, 23:30  #3 
Mar 2016
2^{2}×3×5×7 Posts 

20170817, 09:49  #4 
Dec 2012
The Netherlands
2^{4}·113 Posts 
The matrix
\[\left(\begin{array}{rrr}1&2&2\\2&1&2\\2&2&3\end{array}\right)\] is not a rotation matrix (it is not orthogonal). If you want to rotate the plane so that the point \(\frac{1}{13}(12,5)\) on the unit circle is moved to the point \(\frac{1}{5}(4,3)\), use the matrix \[\left(\begin{array}{rr} \cos(\theta)&\sin(\theta)\\ \sin(\theta)&\cos(\theta) \end{array}\right)\] where \(\theta=a_1a_2\) and therefore \[ \tan(\theta)=\tan(a_1a_2)=\frac{\tan(a_1)\tan(a_2)}{1+\tan(a_1)\tan(a_2)} =\frac{16}{63}\]. In particular, this angle has a rational tangent. A classic book which starts with this sort of idea is the book "Rational points on elliptic curves" by Silverman and Tate, published by Springer. 
20170817, 13:40  #5  
Feb 2017
Nowhere
13·479 Posts 
Quote:
In any case, the angles Arctan(a/b), where a^2 + b^2 = p, p a prime congruent to 1 (mod 4) are all irrational multiples of pi radians. Furthermore, these angles are linearly independent over the rational numbers Q. For suppose that A and B are relatively prime and of opposite parity, and that p is a prime divisor of A^2 + B^2. If Arctan(A/B) were a rational multiple of pi, say m/n*pi, then we would have (A + B*i)^n = K, where K is an integer, since raising a complex number to the nth power multiplies its argument by n. Now in the ring R of Gaussian integers, gcd(p, K) is divisible by gcd(p, A + B*i). This is a prime divisor (a + b*i) of pR, so the gcd is not 1. But p and K are rational integers, so the gcd, being greater than 1, would have to be p. But it can't be, for the divisor a  b*i is relatively prime to a + b*i. Thus, no power of A + i*B can be a rational integer, so the angle in question is not a rational multiple of pi. Note that this argument does not apply to the prime p = 2; the argument of 1 + i is pi/4, and the 4th power of 1 + i is 4. Last fiddled with by Dr Sardonicus on 20170817 at 13:40 

20170817, 21:12  #6 
Mar 2016
2^{2}·3·5·7 Posts 
Thanks a lot for the helpful answers.
Greetings from the primes Bernhard 
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