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Number and Operations: Solutions
 
Session 2 Part A Part B Part C Homework
 
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Solutions for Session 2, Part B

See solutions for Problems: B1 | B2 | B3

Problem B1

a.  

The counting numbers are infinite because we have shown a one-to-one correspondence between the set of counting numbers (1, 2, 3, ...) and a proper subset. Here the proper subset is every other counting number (2, 4, 6, ...); the correspondence is that each number in the first set is doubled to find the corresponding number in the second, and every element in the second set is halved to find the corresponding number in the first set.

b.  

Surprisingly, they are the same size! They must be the same size, because of the one-to-one correspondence we have just found.

c.  

The even counting numbers are a proper subset, because some numbers of the original set are missing. Specifically, the missing elements are all the odd numbers (1, 3, 5, ...).

<< back to Problem B1

 

Problem B2

Yes, the set of even counting numbers is countably infinite since it can be put into one-to-one correspondence with the counting numbers (as seen in Problem B1).

<< back to Problem B2

 

Problem B3

a. 

To do this, you need to find a way to list the integers in some order that can be matched with the counting numbers (1, 2, 3, ...). The difficulty lies in the fact that the integers are infinite in each direction (..., -3, -2, -1, 0, 1, 2, 3, ...). To work around this, we can start "counting" the integers from 0, working outward in both directions. Here is the one-to-one correspondence:

Counting Numbers

1

2

3

4

5

6

7

8

9

Integers

0

1

-1

2

-2

3

-3

4

-4

This pattern continues infinitely (though "countably" infinitely!).

b. 

They are the same size, because we have established a one-to-one correspondence between them.

<< back to Problem B3

 

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