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Ch1Q16

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็ฌฌ1็ซ 
้—ฎ้ข˜ 16
๐‘˜
ๅ‡่ฎพ k ≥ 3 ๅ’Œ ๐‘ฅ, ๐‘ฆ ∈ ๐‘… ๏ผŒ |x – y| = d > 0๏ผŒ
ๅ’Œ r > 0ใ€‚ ่ฏๆ˜Ž:
(a) ่‹ฅ 2r > d, ๅญ˜ๅœจๆ— ็ฉท็š„ z ∈ ๐‘… ๐‘˜ ไฝฟๅพ—
|z – x| = |z – y| = r.
(b)่‹ฅ2r = d๏ผŒๅญ˜ๅœจๅ”ฏๅทฒ็š„ z.
(c)่‹ฅ 2r < d๏ผŒไธๅญ˜ๅœจไปปไฝ• z.
่‹ฅk ๆ˜ฏ 2 ๆˆ– 1๏ผŒๅฟ…้กปๅฆ‚ไฝ•ไฟฎๆ”น่ฟ™ไบ›่ฏญๅฅ?
Chapter 1
Question 16
Suppose k ≥ 3 and ๐‘ฅ, ๐‘ฆ ∈ ๐‘…๐‘˜ , |x – y| = d > 0,
and r > 0. Prove:
(a) If 2r > 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 < d, there is no such z.
How must these statements be modified if k
is 2 or 1?
ๆ€่ทฏ
็”จๅฎš็† 1.37 (e),ๅฎš็† 1.35
็š„่ฏๆ˜Žๅ’Œๅฎš็† (d), (e) ไธŽ
(f)็š„่ฏๆ˜Žใ€‚
Thought
Use Theorem 1.37 (e), proof of
Theorem 1.35 and proof of Theorem
1.37 (d), (e) and (f).
่ฏๆ˜Ž
็”ฑไบŽ่งฃๅทฒ็ป™ๅ‡บ๏ผŒๆˆ‘ไปฌๅช้œ€้ชŒ่ฏ
๐‘ฅ−๐‘Ž =2 ๐‘ฅ−๐‘
ๅฝ“
ๅฝ“ไธ”ไป…ๅฝ“ ๐‘ฅ − ๐‘ = ๐‘Ÿ ไฝฟๅพ—
3c = 4b – a, 3r = 2 ๐‘ − ๐‘Ž .
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 โˆŠ 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 < d contradicts the above,
and thus no such z exists.
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