Proving that if $p$, $q$ are rationals and $p < q$, then there is a rational $v$ such that $p < v < q$

The question itself looks pretty simple, but I'm a complete beginner and have no idea where to start. Any help would be greatly appreciated

Proving that if $p$, $q$ are rationals and $p < q$, then there is a rational $v$ such that $p < v < q$

-
Try to think of a specific example of such a $v$ and prove it satisfies the requirements. –  Ragib Zaman Jul 29 '12 at 11:15
$\frac{mp+nq}{m+n}$ where m,n are positive integers. –  lab bhattacharjee Jul 29 '12 at 11:27

Proposition: for rational numbers $a$ and $b$, such that $a<b$, we have $a<\frac{a+b}{2}<b$.

Proof: The inequality is equivalent to $2a < a+b < 2b$ and we easily confirm that $2a < a+b \Leftrightarrow a < b$ and similarly $a+b < 2b \Leftrightarrow a < b$.

Now it is straightforward to obtain the result that for any two rationals $a,b$, there is (at least) countably infinitely many rational number between; it can formally be obtained by induction, but one see that one can just repeat the process and succesively obtain rationals between given ones.

There is also a nice geometric interpretation of this method. If we imagine rationals on a number line, then arithmetic mean $\frac{a+b}{2}$ lands exactly in the center of the interval $[a,b]$ (that is, it halves it).

-

The arithmetic mean of two numbers lies between the two, and the arithmetic mean of two rationals is itself a rational (both claims left as exercices...).

-

Just take average of these two rational numbers.

-