Example: .28verify the spread law in F13.29 Rational trigonometry

a triangle through points (2, 8), (9, 9), , (10, 0) of finite field-plane f13 × f13.



(2, 8), (9, 9) , (10, 0).

using pythagoras s theorem arithmetic modulo 13, find these sides have quadrances of:

(9 − 2) + (9 − 8) = 50 ≡ 11 mod 13


(9 − 10) + (9 − 0) = 82 ≡ 4 mod 13


(10 − 2) + (0 − 8) = 128 ≡ 11 mod 13

rearranging cross law as

s

3


=
1
−



(

q

1


+

q

2


−

q

3



)

2




4

q

1



q

2







{\displaystyle s_{3}=1-{\frac {(q_{1}+q_{2}-q_{3})^{2}}{4q_{1}q_{2}}}}

gives separate expressions each spread, in terms of 3 quadrances:

1 − (4 + 11 − 11)/4 × 4 × 11 = 1 − 3/7 ≡ 8 mod 13


1 − (11 + 11 − 4)/4 × 11 × 11 = 1 − 12/3 ≡ 10 mod 13


1 − (4 + 11 − 11)/4 × 4 × 11 = 1 − 3/7 ≡ 8 mod 13

in turn note these ratios equal – per spread law (at least in mod 13):

8/11 : 10/4 : 8/11

since first , last ratios match (making triangle isosceles) cross multiply, , take differences, show equality middle ratio also:

11 × 10 − 8 × 4 = 78 ≡ 0 mod 13

otherwise, standard euclidean plane taken consist of rational points, ℚ × ℚ, omitting non-algebraic numbers solutions. properties incidence of objects, representing solutions or content of geometric theorems, therefore follow number theoretic approach differs , more restrictive 1 allowing real numbers. instance, not lines passing through circle s centre considered meet circle @ circumference. incident such lines must of form

a
x
+
b
y



=
0





a

2


+

b

2





=

c

2







a
,
b
,
c
∈

q



{\displaystyle {\begin{aligned}ax+by&=0\\a^{2}+b^{2}&=c^{2}\end{aligned}}\quad a,b,c\in \mathbb {q} }

and meet circle in rational point.