# a2sol - MATH 242 Assignment 2 Solutions 1 Proof by...

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MATH 242 Assignment 2 Solutions 1. Proof by induction on n . In case n = 1 the statement reduces to 1 + a 1 1 + a 1 which is clearly true. So the induction starts. Now suppose that the statement is true for n we will establish it for n + 1 . We are assuming that 1 + n j =1 a j n j =1 (1 + a j ) . Then n +1 j =1 (1 + a j ) = (1 + a n +1 ) n j =1 (1 + a j ) , (1 + a n +1 ) 1 + n j =1 a j , since (1 + a n +1 ) 0 = 1 + n j =1 a j + a n +1 + a n +1 n j =1 a j , 1 + n +1 j =1 a j , since a n +1 0 and n j =1 a j 0 . This completes the induction step. 2. This is a straightforward induction. For n = 1 the assertion reads 1 2 2 1 4 , so the induction starts. For the induction step, we assume that the assertion is true for n and establish it for n + 1 . We find that 1 2 · 3 4 · 5 6 · . . . · 2 n - 1 2 n · 2 n + 1 2 n + 2 2 = 1 2 · 3 4 · 5 6 · . . . · 2 n - 1 2 n 2 · 2 n + 1 2 n + 2 2 1 3 n + 1 · 2 n + 1 2 n + 2 2 , so it will be enough to establish 1 3 n + 1 · 2 n + 1 2 n + 2 2 1 3 n + 4 (1) or equivalently (3 n + 4)(2 n + 1) 2 (3 n + 1)(2 n + 2) 2 . We find that (3 n + 4)(2 n + 1) 2 = 12 n 3 + 28 n 2 + 19 n + 4 and that (3 n + 1)(2 n + 2) 2 = 12 n 3 + 28 n 2 + 20 n + 4 so (1) follows since 0 n . This completes the induction step. 3. (i) Clearly for any fixed j N we have inf ( j,k ) N × N a j,k inf k N a j,k

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since the inf on the left ranges over a larger set than the one on the right. Since the inf on the left is a lower bound for the gadgets on the right as j varies, we have inf ( j,k ) N × N a j,k inf j N inf k N a j,k (2) because the right-hand side of (2) is the corresponding greatest lower bound. (ii) For any fixed j and k we will have a j,k inf k N a j,k and a fortiori a j,k inf j N inf k N a j,k . (3) So the right-hand side of (3) is a lower bound for the a j,k . Hence inf ( j,k ) N × N a j,k inf j
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