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An easy proof that rational numbers are countable.
#1
An easy proof that rational numbers are countable.
It's very counter intuitive that one could be able to "count" the set of rational numbers (those numbers that are quotient of two integers, say a/b), because such is an infinite set, but here's the proof:

[Image: rationals-countable.gif]



Quote:A set is countable if you can count its elements. Of course if the set is finite, you can easily count its elements. If the set is infinite, being countable means that you are able to put the elements of the set in order just like natural numbers are in order. Yet in other words, it means you are able to put the elements of the set into a "standing line" where each one has a "waiting number", but the "line" is allowed to continue to infinity.

In mathematical terms, a set is countable either if it s finite, or it is infinite and you can find a one-to-one correspondence between the elements of the set and the set of natural numbers. Notice, the infinite case is the same as giving the elements of the set a waiting number in an infinite line Smile.


https://www.homeschoolmath.net/teaching/...ntable.php
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#2
RE: An easy proof that rational numbers are countable.
(February 22, 2018 at 3:57 am)Jehanne Wrote: It's very counter intuitive that one could be able to "count" the set of rational numbers (those numbers that are quotient of two integers, say a/b), because such is an infinite set, but here's the proof:

[Image: rationals-countable.gif]



Quote:A set is countable if you can count its elements. Of course if the set is finite, you can easily count its elements. If the set is infinite, being countable means that you are able to put the elements of the set in order just like natural numbers are in order. Yet in other words, it means you are able to put the elements of the set into a "standing line" where each one has a "waiting number", but the "line" is allowed to continue to infinity.

In mathematical terms, a set is countable either if it s finite, or it is infinite and you can find a one-to-one correspondence between the elements of the set and the set of natural numbers. Notice, the infinite case is the same as giving the elements of the set a waiting number in an infinite line Smile.


https://www.homeschoolmath.net/teaching/...ntable.php

While this is a function that is onto the *positive* rational numbers, it is NOT a one-to-one function. So for example, 4/1=8/2=12/3. In fact each postitive rational number is hit infinitely often by this counting method.

Since the definition requires a function that is BOTH onto and one-to-one, this doesn't give the result (at least not directly).

There *is* a result that takes an onto function from the natural numbers to an infinite set and gives a one-to-one onto function. But the argument needs to be made for this.
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#3
RE: An easy proof that rational numbers are countable.
I don't think I'll be getting there with 1 potato, 2...........
God(s) and religions are man made and the bane of humanity. 

Of all the things I've lost, I miss my mind the most. Ozzy or Twain/take your pick
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#4
RE: An easy proof that rational numbers are countable.
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.











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#5
RE: An easy proof that rational numbers are countable.
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".
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#6
RE: An easy proof that rational numbers are countable.
(February 22, 2018 at 3:19 pm)Jehanne Wrote: 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".

Anyone who has studied math past the level of calculus has seen many of such.

One of my favorites is the Peano space-filling curve:
https://www.youtube.com/watch?v=MaoCp08hznM
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#7
RE: An easy proof that rational numbers are countable.
(February 22, 2018 at 1:30 pm)Kernel Sohcahtoa Wrote: 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.
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#8
RE: An easy proof that rational numbers are countable.
(February 22, 2018 at 4:17 pm)polymath257 Wrote:
(February 22, 2018 at 3:19 pm)Jehanne Wrote: 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".

Anyone who has studied math past the level of calculus has seen many of such.

One of my favorites is the Peano space-filling curve:
https://www.youtube.com/watch?v=MaoCp08hznM

WLC was a communications major (for actors, news anchors) at Wheaton College, and then he went to Bible school where people go to be evangelical Christian pastors.  I don't think that he took any STEM classes while in college.
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