- #1

radou

Homework Helper

- 3,134

- 8

Is this proof OK?.. Thanks in advance.

So, we have to proove that the set Q of all rational numbers is countable.

Let's first look at strictly positive rational numbers. We can form the sets

[tex]S_{i}=\left\{s_{i1}, s_{i2}, \dots \right\}=\left\{\frac{1}{i}, \frac{2}{i}, \dots \right\}[/tex]. The union of these sets equals the set of strictly positive rational numbers. Now, we automatically know that some rational number [tex]\frac{k}{l}[/tex] is an element of the set [tex]S_{l}[/tex], and, since there is a 1-1 mapping between every set [tex]S_{i}[/tex] and the set of natural numbers N, we conclude [tex]s_{lk}=\frac{k}{l} \in S_{l}[/tex]. Now, since [tex]f(k, l) = \frac{k}{l}[/tex] is a 1-1 mapping defined on [tex]N\times N[/tex], and since [tex]N\times N[/tex] is countable, we conclude that the set of all strictly positive rational numbers is countable. Further on, there obviously exists a 1-1 mapping between this set and the set of all strictly negative rational numbers. Hence, the set of all strictly negative rational numbers

is countable, too. So, the union of the set of strictly positive, strictly negativerational numbers, and the set containing only zero {0}, is countable, since a finite or countable union of a finite or countable family of finite or countablesets is finite or countable.

P.S. Sorry the thread appeared twice, but I had some posting problems (as well as editing problems, logging on problems, etc.. contacted admin. )

So, we have to proove that the set Q of all rational numbers is countable.

Let's first look at strictly positive rational numbers. We can form the sets

[tex]S_{i}=\left\{s_{i1}, s_{i2}, \dots \right\}=\left\{\frac{1}{i}, \frac{2}{i}, \dots \right\}[/tex]. The union of these sets equals the set of strictly positive rational numbers. Now, we automatically know that some rational number [tex]\frac{k}{l}[/tex] is an element of the set [tex]S_{l}[/tex], and, since there is a 1-1 mapping between every set [tex]S_{i}[/tex] and the set of natural numbers N, we conclude [tex]s_{lk}=\frac{k}{l} \in S_{l}[/tex]. Now, since [tex]f(k, l) = \frac{k}{l}[/tex] is a 1-1 mapping defined on [tex]N\times N[/tex], and since [tex]N\times N[/tex] is countable, we conclude that the set of all strictly positive rational numbers is countable. Further on, there obviously exists a 1-1 mapping between this set and the set of all strictly negative rational numbers. Hence, the set of all strictly negative rational numbers

is countable, too. So, the union of the set of strictly positive, strictly negativerational numbers, and the set containing only zero {0}, is countable, since a finite or countable union of a finite or countable family of finite or countablesets is finite or countable.

P.S. Sorry the thread appeared twice, but I had some posting problems (as well as editing problems, logging on problems, etc.. contacted admin. )

Last edited: