Prove that between two unequal rational numbers there is another rational I understand that I probably asked a question that the users of this site would view as elementary, but I have only just dipped my feet into the waters of proof solving. Can somebody please tell me if my proof is valid?
Let there be two rational numbers $m/n$ and $p/q$ where $m$,$n$,$p$, and $q$ are integers, and $nq$ is not equal to zero.
$(m/n)<(p/q)$
$(p/q)-(m/n)=((pn-mq)/nq))$
The difference between the two numbers is a rational number. Two multiplied rationals has a rational product, and so if we multiply the (rational) difference by a rational number $>0$ and $<1$ and add that product to $(m/n)$ we have a rational number in between the two rationals.
 A: Short answer: If $a$ and $b$ are rational so is $\frac{a+b}{2}$. 

Long answer:
As you said you can construct an intermediate rational like this: $c=\min(a,b) + q |a-b|$ where $0<q<1$. $\min(a,b)<c<\max(a,b)$. My short answer is for $q=\frac 12$. 
An example to make a proof more formal:


*

*Prove that the sum and the product of two rational numbers are rational numbers. 

*Use 1. to construct a rational number strictly between a and b. For instance: $\min(a,b) + q |a-b|$ where $0<q<1$ is a rational to be rational. 



Extension:
With $q=\frac 1n$ $n \geq 2$ integer you would actually prove that there are infinite many rational numbers between two distinct rational numbers. 
A: Your proof is correct, but here is a curious fact that gives another 'nice' intermediate value that is not the average:

$\frac{a}{b} < \frac{a+c}{b+d} < \frac{c}{d}$ for every integers $a,c$ and positive integers $b,d$ such that $\frac{a}{b} < \frac{c}{d}$.

You can easily prove this inequality. Note that this fact solves the original question, because every rational is equal to some integer divided by some positive integer.
This intermediate fraction has some nice properties. For example, its denominator is the smallest possible if $ad+1 = bc$.
A: Your proof is correct. A similar approach is just taking the average of the two rational numbers (which is again a rational number).
EDIT: As pointed out, the average is given by taking the rational number to be $\frac12$ to multiply the difference (using your method).
