is it possible to count the number of rational numbers there are between any two integers?

Question

im having trouble in this can anyone help me?

is it possible to count the number of rational numbers there are between any two integers?

Answer 1

Depends what you mean by "count", and on how much you know about infinite cardinality.

There are infinitely many rational numbers between any two integers. That set of rational numbers, however, is countably infinite. This means that they can be put in a one-to-one correspondence with the set of counting numbers; informally, it means that it is possible to list them in a sequence with the property that any individual rational number eventually shows up after finitely many termns.

Such a sequence is easy to describe. First, list (in increasing order of the numerator) all of the rationals that lie between your two integers and have a denominator $2$. (If the integers are consecutive, there will be exactly one of them.) Then, list (again in increasing order of the numerator) all of the rationals that have a denominator $3$. Then do the $4$ths, and the $5$ths, and so on, omitting any that have already been listed in lower terms. The resulting sequence is a complete enumeration of all of the rationals in your desired interval.

For example, if your integers are $5$ and $6$, the enumeration would begin: $$\frac{11}{2}, \frac{16}{3}, \frac{17}{3}, \frac{21}{4}, \frac{23}{4}, \frac{26}{5}, \frac{27}{5}, \frac{28}{5}, \frac{29}{5}, \dots$$

Answer 2

$$ \frac 1 2,\quad \underbrace{\frac 1 3, \frac 2 3}_{\bullet/3}, \quad \underbrace{\frac 1 4, \frac 3 4}_{\bullet/4}, \qquad \underbrace{\frac 1 5, \frac 2 5, \frac 3 5, \frac 4 5}_{\bullet/5}, \quad \underbrace{\frac 1 6, \frac 5 6,}_{\bullet/6} \quad\underbrace{\frac 1 7, \frac 2 7, \frac 3 7, \frac 4 7, \frac 5 7, \frac 6 7,}_{\bullet/7} \quad \underbrace{\frac 1 8, \frac 3 8, \frac 5 8, \frac 7 8,}_{\bullet/8} \quad \ldots $$ This sequence "counts" all the rational numbers between $0$ and $1$.