็ฌฌ1็ซ ้ฎ้ข 16 ๐ ๅ่ฎพ k ≥ 3 ๅ ๐ฅ, ๐ฆ ∈ ๐ ๏ผ |x – y| = d > 0๏ผ ๅ r > 0ใ ่ฏๆ: (a) ่ฅ 2r > d, ๅญๅจๆ ็ฉท็ z ∈ ๐ ๐ ไฝฟๅพ |z – x| = |z – y| = r. (b)่ฅ2r = d๏ผๅญๅจๅฏๅทฒ็ z. (c)่ฅ 2r < d๏ผไธๅญๅจไปปไฝ z. ่ฅk ๆฏ 2 ๆ 1๏ผๅฟ ้กปๅฆไฝไฟฎๆน่ฟไบ่ฏญๅฅ? Chapter 1 Question 16 Suppose k ≥ 3 and ๐ฅ, ๐ฆ ∈ 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? ๆ่ทฏ ็จๅฎ็ 1.37 (e),ๅฎ็ 1.35 ็่ฏๆๅๅฎ็ (d), (e) ไธ (f)็่ฏๆใ Thought Use Theorem 1.37 (e), proof of Theorem 1.35 and proof of Theorem 1.37 (d), (e) and (f). ่ฏๆ ็ฑไบ่งฃๅทฒ็ปๅบ๏ผๆไปฌๅช้้ช่ฏ ๐ฅ−๐ =2 ๐ฅ−๐ ๅฝ ๅฝไธไป ๅฝ ๐ฅ − 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)| < |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 โ R. However, by hypothesis |z – x| = |z – y| = r. Thus t = 1, and we have ๐ฅ+๐ฆ ๐ง= . 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.