第1章
问题 16
𝑘
假设 k ≥ 3 和 𝑥, 𝑦 ∈ 𝑅 ， |x – y| = d > 0，
和 r > 0。 证明:
(a) 若 2r > d, 存在无穷的 z ∈ 𝑅 𝑘 使得
|z – x| = |z – y| = r.
(b)若2r = d，存在唯已的 z.
(c)若 2r < d，不存在任何 z.
若k 是 2 或 1，必须如何修改这些语句?
Chapter 1
Question 16
Suppose k ≥ 3 and 𝑥, 𝑦 ∈ 𝑅𝑘 , |x – y| = d > 0,
and r > 0. Prove:
(a) If 2r > d, there are infinitely many z ∈ 𝑅𝑘
such that
|z – x| = |z – y| = r.
(b) If 2r = d, there is exactly one such z.
(c) If 2r < d, there is no such z.
How must these statements be modified if k
is 2 or 1?
思路
用定理 1.37 (e),定理 1.35
的证明和定理 (d), (e) 与
(f)的证明。
Thought
Use Theorem 1.37 (e), proof of
Theorem 1.35 and proof of Theorem
1.37 (d), (e) and (f).
证明
由于解已给出，我们只需验证
𝑥−𝑎 =2 𝑥−𝑏
当
当且仅当 𝑥 − 𝑐 = 𝑟 使得
3c = 4b – a, 3r = 2 𝑏 − 𝑎 .
Proof
(a) From the strict inequality of part
(e) of Theorem 1.37, replace x
by x – z and y by z – y, we have
|(x – z) + (z – y)| < |x – z| + |z – y|
⇒ |x – y| < |z – x| + |z – y|.
Since by hypothesis |x – y| = d and
|z – x| = |z – y| = r, there exists
infinitely many such z if 2r > d.
Proof
(b) If 2r = d then the strict equality of part (e) of
Theorem 1.37 applies and from its proof this can
hold only if
(x – z)·(z – y) = |x – z||z – x|.
This is true only when the strict equality of
Theorem 1.35 holds and from its proof occurs if
(x – z) = t(z – y), t ≠ 0 where t ∊ R.
However, by hypothesis |z – x| = |z – y| = r. Thus
t = 1, and we have
𝑥+𝑦
𝑧=
.
2
Proof
(c) By hypothesis |x – y| = d and
|z – x| = |z – y| = r and Theorem
1.37 (e), we have
|x – y| ≤ |z – x| + |z – y|
⇒ d ≤ 2r.
If 2r < d contradicts the above,
and thus no such z exists.