The Real And Complex Number Systems

For case x = a 0 .a 1 a 2 · · · a n . Then
∑ n
k=0
x =
10n−k a k
=
10 n
∑ n
k=0 10n−k a k
2 n 5 n .
For case x = a 0 .a 1 a 2 · · · a n 999999 · · · . Then
∑ n
k=0
x =
10n−k a k
+ 9
9
+ ... + + ...
2 n 5 n 10n+1 10n+m ∑ n
k=0
=
10n−k a k
+ 9 ∑ ∞
10 −j
2 n 5 n 10 n+1 j=0
∑ n
k=0
=
10n−k a k
+ 1
2 n 5 n 10 n
= 1 + ∑ n
k=0 10n−k a k
.
2 n 5 n
So, in both case, we prove that x is a rational number whose denominator is
of the form 2 n 5 m , where m and n are nonnegative integers.
1.9 Prove that √ 2 + √ 3 is irrational.
Proof: If √ 2 + √ 3 is rational, then consider
(√
3 +
√
2
) (√
3 −
√
2
)
= 1
which implies that √ 3 − √ 2 is rational. Hence, √ 3 would be rational. It is
impossible. So, √ 2 + √ 3 is irrational.
Remark: (1) √ p is an irrational if p is a prime.
Proof: If √ p ∈ Q, write √ p = a , where g.c.d. (a, b) = 1. Then
b
Write a = pq. So,
By (*) and (*’), we get
b 2 p = a 2 ⇒ p ∣ ∣ a 2 ⇒ p |a (*)
b 2 p = p 2 q 2 ⇒ b 2 = pq 2 ⇒ p ∣ ∣ b 2 ⇒ p |b . (*’)
p |g.c.d. (a, b) = 1
which implies that p = 1, a contradiction. So, √ p is an irrational if p is a
prime.
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