# Ch1Q16

```็ฌฌ1็ซ
้ฎ้ข 16
๐
ๅ่ฎพ k ≥ 3 ๅ ๐ฅ, ๐ฆ ∈ ๐ ๏ผ |x – y| = d &gt; 0๏ผ
ๅ r &gt; 0ใ ่ฏๆ:
(a) ่ฅ 2r &gt; d, ๅญๅจๆ ็ฉท็ z ∈ ๐ ๐ ไฝฟๅพ
|z – x| = |z – y| = r.
(b)่ฅ2r = d๏ผๅญๅจๅฏๅทฒ็ z.
(c)่ฅ 2r &lt; d๏ผไธๅญๅจไปปไฝ z.
่ฅk ๆฏ 2 ๆ 1๏ผๅฟ้กปๅฆไฝไฟฎๆน่ฟไบ่ฏญๅฅ?
Chapter 1
Question 16
Suppose k ≥ 3 and ๐ฅ, ๐ฆ ∈ ๐๐ , |x – y| = d &gt; 0,
and r &gt; 0. Prove:
(a) If 2r &gt; d, there are infinitely many z ∈ ๐๐
such that
|z – x| = |z – y| = r.
(b) If 2r = d, there is exactly one such z.
(c) If 2r &lt; d, there is no such z.
How must these statements be modified if k
is 2 or 1?
ๆ่ทฏ
็จๅฎ็ 1.37 (e),ๅฎ็ 1.35
็่ฏๆๅๅฎ็ (d), (e) ไธ
(f)็่ฏๆใ
Thought
Use Theorem 1.37 (e), proof of
Theorem 1.35 and proof of Theorem
1.37 (d), (e) and (f).
่ฏๆ
็ฑไบ่งฃๅทฒ็ปๅบ๏ผๆไปฌๅช้้ช่ฏ
๐ฅ−๐ =2 ๐ฅ−๐
ๅฝ
ๅฝไธไปๅฝ ๐ฅ − ๐ = ๐ ไฝฟๅพ
3c = 4b – a, 3r = 2 ๐ − ๐ .
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)| &lt; |x – z| + |z – y|
⇒ |x – y| &lt; |z – x| + |z – y|.
Since by hypothesis |x – y| = d and
|z – x| = |z – y| = r, there exists
infinitely many such z if 2r &gt; 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)&middot;(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 โ R.
However, by hypothesis |z – x| = |z – y| = r. Thus
t = 1, and we have
๐ฅ+๐ฆ
๐ง=
.
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 &lt; d contradicts the above,
and thus no such z exists.
```