23b_hw6_solutions

23b_hw6_solutions - Math 23b Homework 6 Solutions 1 In base...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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....
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online