rational numbers

It is clearly true that rational numbers are dense, because for rational numbers $a$ and $b$ with $a<b$, there is a rational number $(a+b)/2$. Furthermore this dense property of rational numbers is satisfied in real numbers.

If $a$ and $b$ are real numbers with $a<b$, then there is a rational number $p/q$ such that $a<p/q<b$.

It is said that rational numbers are dense in real numbers. That is, between any different two real numbers, there is a rational number.

[wrong answer]
For any rational number $q>0$, $qa<qb$. As real numbers are dense, there is a $p$ such that
\[ qa<p<qb \]
Therefore,
\[ a<\frac{p}{q}<b \]
But, we are not able to state assertively that $p$ and $p/q$ are rational numbers.

[almost sure]
On Archimedean property " $na>b$ ", let $a$ be $b-a$, $b$ be $1$, and $n$ be $q$. Then,
\[ q(b-a)>1 \]
In this form $a$ and $b$ are real numbers and $q$ is a integer number. Next, if we choose $p$ be a smallest integer number such that $p>qa$, then $p-1\leq qa$. Hence, as
\[ qa<p\leq qa+1<qa+q(b-a)=qb, \]
if we divide above all sides by $q$, then we get the equation which would be hoped. Obviously, $p/q$ is a rational number.