The Real And Complex Number Systems

Proof: Choose a 0 = [x], and thus consider
[kx − ka 0 ] := a 1
then
0 ≤ k (x − a 0 ) < k ⇒ 0 ≤ a 1 ≤ k − 1
and
a 0 + a 1
k ≤ x ≤ a 0 + a 1
k + 1 k .
Continue the process, we then have
0 ≤ a i ≤ k − 1 for each i = 1, 2, ...
and
a 0 + a 1
k + a 2
k 2 + ... + a n
k n ≤ x < a 0 + a 1
k + a 2
k 2 + ... + a n
k n + 1
k n . (*)
(b) Let r n = a 0 + a 1 k −1 + a 2 k −2 + ... + a n k −n and show that x is the sup
of the set of rational numbers r 1 , r 2 , ...
Proof: It is clear by (a)-(*).
Inequality
1.23 Prove Lagrange’s identity for real numbers:
( n∑
) 2 ( n∑
) ( n∑
)
a k b k = a 2 k b 2 k −
∑
(a k b j − a j b k ) 2 .
k=1
k=1 k=1 1≤k