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Unformatted text preview: Math 23b Homework 6 Solutions 1. In base 3, 12012 2221 = 2021 and 12012 × 2221 = 120011122. In base 10, 140 79 = 61 and 140 × 79 = 11060, which agrees with the above. 2. Note that 57 = 2 × 27 + 3 and 40 = 27 + 9 + 3 + 1. From this, we see that 17 = 57 40 = 27 9 1. So we should place 17 , 9, and 1 one one side and 27 on the other. 3. Since each set A n is countable, there exist bijections f n : N → A n . Consider the function f : N × N → A given by f ( m,n ) = f m ( n ). We claim this is a surjection. To see this, let x ∈ A . Then, since A = ∪ i ∈ N A i , there exists some m ∈ N such that x ∈ A m . Since f m is a surjection, there exists n ∈ N such that f m ( n ) = x , so f ( m,n ) = f m ( n ) = x . Hence, f is surjective. By a result proved in class, N × N is countable, so there exists a surjection g : N → N × N . Then f ◦ g is a surjection N → A , so A is countable....
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This note was uploaded on 10/15/2011 for the course MATH 23b taught by Professor Bonglian during the Spring '09 term at Brandeis.
 Spring '09
 BongLian
 Math

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