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 2017-08-16, 20:40 #1 bhelmes     Mar 2016 22×3×5×7 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
2017-08-16, 22:10   #2
Nick

Dec 2012
The Netherlands

24·113 Posts

Quote:
 Originally Posted by bhelmes Is the degree of the rotation matrix = a2/a1 or a2 - a1 and is the degree rational or irrational.
Can you explain what you mean by the degree of a rotation matrix?

2017-08-16, 23:30   #3
bhelmes

Mar 2016

22×3×5×7 Posts

Quote:
 Originally Posted by Nick Can you explain what you mean by the degree of a rotation matrix? Perhaps then we can answer your question!
i mean the angle, degree was a bad translation.

Greetings from the primes,
Bernhard

 2017-08-17, 09:49 #4 Nick     Dec 2012 The Netherlands 24·113 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 $$\theta=a_1-a_2$$ and therefore $\tan(\theta)=\tan(a_1-a_2)=\frac{\tan(a_1)-\tan(a_2)}{1+\tan(a_1)\tan(a_2)} =\frac{16}{63}$. 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.
2017-08-17, 13:40   #5
Dr Sardonicus

Feb 2017
Nowhere

13·479 Posts

Quote:
 Originally Posted by bhelmes 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.
I'm not sure what "rotation matrix" you have in mind. I'm not sure what "discr" stands for. Did you mean "det" for determinant?

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 n-th 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 2017-08-17 at 13:40

 2017-08-17, 21:12 #6 bhelmes     Mar 2016 22·3·5·7 Posts Thanks a lot for the helpful answers. Greetings from the primes Bernhard

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