The Real And Complex Number Systems

x n1
x n
1 1 x n
x n
1 2 .
Remark: In (*), it is the derivative of 1 1 x at the point x 0. Of course, we can
prove (*) by L-Hospital Rule.
4.4 Two sequences of positive integers a n and b n are defined recursively by taking
a 1 b 1 1 and equating rational and irrational parts in the equation
a n b n 2 a n1 b n1 2 2 for n 2.
Prove that a n2 2b n2 1 for all n 2. Deduce that a n /b n 2 through values 2 ,and
that 2b n /a n 2 through values 2 .
Proof: Note a n b n 2 a n1 b n1 2 2 for n 2, we have
a n a2 n1 2b2 n1 for n 2, and
b n 2a n1 b n1 for n 2
since if A, B, C, andD N with A B 2 C D 2 , then A C, andB D.
Claim that a n2 2b n2 1 for all n 2. We prove the claim by Mathematical
Induction. As n 2, we have by (*)
a 22 2b 22 a 12 2b 12 2 22a 1 b 1 2 1 2 2 22 2 1. Suppose that as n k 2
holds, i.e., a k2 2b k2 1, then as n k 1, we have by (*)
a2 k1 2b2 k1 a k2 2b k2 2 22a k b k 2
a k4 4b k4 4a k2 b2
k
a k2 2b k2 2
1 by induction hypothesis.
Hence, by Mathematical Induction, we have proved the claim. Note that a n2 2b n2 1for
all n 2, we have
*
2 2
a n 1 2 2
b n bn
and
2
2b n
a n 2 2 2.
a2 n
a
Hence, lim n
1
n b n
2 by lim n b n
0 from (*) through values 2 ,and
2b
lim n
1 n a n
2 by lim n an
0 from (*) through values 2 .
Remark: From (*), we know that a n and b n is increasing since a n N and
b n N. That is, we have lim n a n , and lim n b n .
4.5 A real sequence x n satisfies 7x n1 x n3 6forn 1. If x 1 1 , prove that the
2
sequence increases and find its limit. What happens if x 1 3 or if x 2 1 5 2
Proof: Claim that if x 1 1 ,then0 x 2 n 1 for all n N. We prove the claim by
Mathematical Induction. Asn 1, 0 x 1 1 1. Suppose that n k holds, i.e.,
2
0 x k 1, then as n k 1, we have
0 x k1 x k 3 6
7
1 7
6 1 by induction hypothesis.
Hence, we have prove the claim by Mathematical Induction. Since
x 3 7x 6 x 3x 1x 2, then