k 2 2 k 1 k 1 2 k k 2 2 k k 1 2 k 2 2 k 1 k 2 2 k 1 k 2 2 k 1 Therefore k 1 S

# K 2 2 k 1 k 1 2 k k 2 2 k k 1 2 k 2 2 k 1 k 2 2 k 1 k

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= k + 2 2( k + 1) k + 1 2 k · k 2 + 2 k ( k + 1) 2 ? = k + 2 2( k + 1) k + 2 2( k + 1) = k + 2 2( k + 1) Therefore, k + 1 S and S = { n N : n 2 } . 20. Let S be the set of positive integers for which the equation holds. Step 1. 1 S : (cos x + i sin x ) 1 = cos x + i sin x . Therefore, 1 S . 2 Step 2. Assume that the positive integer k S . That is, assume (cos x + i sin x ) k = cos kx + i sin kx Step 3. Prove that k + 1 S . (cos x + i sin x ) k +1 = (cos x + i sin x )(cos x + i sin x ) k = (cos x + i sin x )(cos kx + i sin kx ) = cos x cos kx - sin x sin kx + i (sin x cos kx + cos x sin kx ) = cos( k + 1) x + i sin( k + 1) x Therefore, k + 1 S and S = N . 21(b). The inequality does not hold for n = 1 , 2 , 3. Prove that n 2 2 n for all n 4. Let S be the set of positive integers for which the inequality holds. Step 1. 4 S : 4 2 = 16 2 4 . Therefore, 4 S . Step 2. Assume that the positive integer k S . That is, assume k 2 2 k , ( k 4). Step 3. Prove that k + 1 S . ( k + 1) 2 = k 2 + 2 k + 1 2 k + 2 k + 1 2 k + 2 k = 2 · 2 k = 2 k +1 (here we used 2 k + 1 2 k ) Therefore, k + 1 S and S = { n N : n 4 } . Section 11 3(c). x negationslash = 0. Suppose (1 /x ) = 0. Then, 1 = x parenleftbigg 1 x parenrightbigg = 1 · 0 = 0 which is false. Therefore, (1 /x ) negationslash = 0. x = x · 1 = x parenleftbigg 1 x · 1 1 /x parenrightbigg = parenleftbigg x · 1 x parenrightbigg 1 1 /x = 1 · 1 1 /x = 1 1 /x 4. Given x 0. x epsilon1 for every epsilon1 > 0  #### You've reached the end of your free preview.

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• Fall '08
• Staff
• Integers, Natural number, Prime number
• • •  