An easy proof that rational numbers are countable.

[Jehanne]

[polymath257]

[brewer]

I don't think I'll be getting there with 1 potato, 2...........

[Kernel Sohcahtoa]

Here's another way to approach the concept that the set of rational numbers is countable.  First, here are a few theorems:

Theorem 1: Let A_1, A_2,... be a countable family of countable sets (the index n of this countable family of countable sets is an element of the positive integers; the set of positive integers is the index set).  Then the union of all countable sets A_n is a countable set (Johnsonbaugh & Pfaffenberger, 30-31).

Theorem 2: There exists a bijection from the set of positive integers to the set of integers.  Therefore, the set of positive integers has the same cardinality as the set of integers, and so, the set of integers is countable (Hammack, 219).

Result to establish: The set Q of rational numbers is countable

Here's a proof from Foundations of Mathematical Analysis by Johnsonbaugh & Pfaffenberger

For each positive integer n, let A_n={...,-2/n,-1/n,0/n,1/n,2/n,...}.  Then A_n is countable for each positive integer n. (1)  By Theorem 1, the union of all sets A_n, which is equal to Q, is countable.(2) (Johnsonbaugh & Pfaffenberger, 31)

Notes

(1). In our constructed set A_n, notice that if we ignore the n in the denominator of each element, then we have the entire set of integers, and so, putting an n in the denominator of each element will not change the fact that each A_n has the same cardinality as the set of integers.  Since the set of integers is countable and each set A_n has the same cardinality as the set of integers, then it follows that each A_n is countable. 

(2). Notice that if we took the intersection of any two sets A_n, then the result would be the empty set.  Thus, each A_n is a distinct subset of the rational numbers. Hence, taking the union of all sets A_n produces the set of rational numbers. 



 References


 Hammack, Richard (2013). Book of Proof, 2nd ed. Virginia: Richard Hammack.

Johnsonbaugh, R and Pfaffenberger,W.E. (2002). Foundations Of Mathematical Analysis.  New York: Dover Publications, INC.

[Jehanne]

The proof/technique in my OP is also in Rosen, Discrete Mathematics, 7th edition.

Point of my thread is that there are things that are counterintuitive, yet provably true! These are sometimes referred to as "veridical paradoxes".

[polymath257]

[polymath257]

Here's another way to approach the concept that the set of rational numbers is countable.  First, here are a few theorems:

Theorem 1: Let A_1, A_2,... be a countable family of countable sets (the index n of this countable family of countable sets is an element of the positive integers; the set of positive integers is the index set).  Then the union of all countable sets A_n is a countable set (Johnsonbaugh & Pfaffenberger, 30-31).

Theorem 2: There exists a bijection from the set of positive integers to the set of integers.  Therefore, the set of positive integers has the same cardinality as the set of integers, and so, the set of integers is countable (Hammack, 219).

Result to establish: The set Q of rational numbers is countable

Here's a proof from Foundations of Mathematical Analysis by Johnsonbaugh & Pfaffenberger

For each positive integer n, let A_n={...,-2/n,-1/n,0/n,1/n,2/n,...}.  Then A_n is countable for each positive integer n. (1)  By Theorem 1, the union of all sets A_n, which is equal to Q, is countable.(2) (Johnsonbaugh & Pfaffenberger, 31)

Notes

(1). In our constructed set A_n, notice that if we ignore the n in the denominator of each element, then we have the entire set of integers, and so, putting an n in the denominator of each element will not change the fact that each A_n has the same cardinality as the set of integers.  Since the set of integers is countable and each set A_n has the same cardinality as the set of integers, then it follows that each A_n is countable. 

(2). Notice that if we took the intersection of any two sets A_n, then the result would be the empty set.  Thus, each A_n is a distinct subset of the rational numbers. Hence, taking the union of all sets A_n produces the set of rational numbers. 



 References


 Hammack, Richard (2013). Book of Proof, 2nd ed. Virginia: Richard Hammack.

Johnsonbaugh, R and Pfaffenberger,W.E. (2002). Foundations Of Mathematical Analysis.  New York: Dover Publications, INC.

The biggest problem here is that Theorem 1 is proven by essentially the same techniques as in the OP. A better way is to go as follows:

1. If there is an onto function f:A->B, then there is a one-to-one function g:B->A. This can use the axiom of choice, but if A is the set of naturals, we can use well-ordering of the naturals to get it.

2. If A is a subset of the naturals, then A is countable. This is proved by trivially if A is finite. But if A is infinite, we can send 1 to the smallest element of A, 2 to the next smallest, 3 to the next smallest, etc and this will give a one-to-one and onto function from N to A, showing A is countable.

3. Use the OP method to give an onto function from N to the rationals. Then 1 gives that the rationals are equivalent to a subset of N and thereby are countable.

[Jehanne]