Prove/Disprove that the set of rational numbers is uncountable (Real Analysis) Cantor digonalization method is used to prove that an open interval for real numbers $(0,1)$ is not countable. Since rational numbers also have decimal expansions, we could try to use this same method to prove that the set of rational numbers in $(0,1)$ is not countable. Explain why we could not conclude that the set of rational numbers in $(0,1)$ is uncountable using this method.
What i tried
Cantor diagonalizaton works by writing all the real numbers in the interval $(0,1)$ as an array of matrix and then finding a real number that does not belong to this set of array of matrix by making the diagonal entries of the real number different from that of the set of array of numbers in the matrix thus a contradiction. While for the set of rational numbers it can be expressed in the form of $\frac{a}{b}$ where gcd(a,b)=1, We do the same thing as for the rational numbers but i think this set of rational numbers somehow always tend to appear in the array of matrix no matter how we try thus this method of proof dosent apply for the case of rational number. But im unsure how. Could someone explain this to me. Thanks
 A: Its not the process but the conclusion of the process that fails.  When you write real numbers in the interval (0,1) as an array of matrix and then find a  number  by making the diagonal entries of the real number different from that of the set of array of numbers in the matrix, the resulting number is again a real number. 
When you apply the same trick to the rational numbers the resulting number may not be rational. Remember being rational means the decimal expression either terminates or repeating. How do you ensure these properties in the resulting number? 
A: Proof that the set of decimal expansions from $(0,1=.99999.....]$ are uncountable:
If they were countable we could list them all.  Then we could manipulate them all to get a number not on the list.  That new number is a decimal expansion.  But that contradicts that all decimal expansions were on the list.
Proof that the set of rationals in decimal representation from $(0,1=.99999.....]$ are uncountable:
If they were countable we could list them all.  Then we could manipulate them all to get a number not on the list. $\color {red} {\text{ That new number is a rational in decimal representation.}}$  But that contradicts that all rationals in decimal representation were on the list.
!!!BUT!!! There is no reason to believe that bit in red is true!
So instead we'd have to write:
If they were countable we could list them all.  Then we could manipulate them all to get a number not on the list. If $\color {red} {\text{ that new number is a rational in decimal representation.}}$ $\color {blue} {\text{then that would}}$ contradict that all rational in decimal representation were on the list.
And from that we can conclude the true (but wimpy) result:
Either the rationals in decimal representation for $(0,1=.99999....]$ are uncountable OR if they are countable then there exist decimal representations in $(0,1]$ that are not rational.
A: The rationals are countable.
Here is a intuitive sketch.
Make a table.  Put natural numbers across the top and down the left column.
the body of the table is the ratio of the number at the top of the table over the number in the right column.
This is a representation of the full set of $\mathbb Q+$
Now you can run a serpentine line that runs through every entry in the table.  this maps the rationals to $\mathbb N$
More formally:
There exists an injection from $\mathbb Q\to \mathbb Z\times \mathbb N$
i.e. $\frac pq = (p,q)$
the Cartesian product of countable sets is a countable set.
