Why rational numbers are dense?

Question

So the books says that rational numbers are dense, meaning that for every two rational numbers there is another rational number in between them. Is it actually true? Why? It feels to me that there exists two rational numbers that do not have any other rational number in between...maybe I wrong but I can't understand why. I also cannot come up with counterexample...

Answer 1

given any two rational numbers $\frac{p_1}{q_1},\frac{p_2}{q_2}$, we can take the average of the two, $m$ = $\frac{\frac{p_1}{q_1}+\frac{p_2}{q_2}}{2} = \frac{p_1q_2 + p_2q_1}{2q_1 q_2}$ which is rational too. Thus proving there are always a rational number between any two.

Answer 2

You'll search for a long time for a counterexample.

If $\dfrac p q, \dfrac r s$ are distinct rationals, then their average: $$ \frac {ps + qr} {2qs} $$ is a rational between the two.

Answer 3

The other answers address why, between every two rationals, there's a third. However, this isn't actually the same as density (which your book may be using in an unusual way): density says that every real number has rational numbers "arbitrarily close" to it. Formally, "the rationals are dense in the reals" is the statement "For every real $r$, and every positive real $\epsilon$, there is a rational $q$ such that $\vert q-r\vert<\epsilon$ - that is, $q$ is within $\epsilon$ of $r$."

Here's why this is true:

• Fix a real $r$ and a distance (=positive real) $\epsilon$. There's some natural number $n$ such that $10^{-n}<\epsilon$. (Why?)

• So look at $r$. Since $r$ is a real, it has a decimal expansion $r= z.a_1a_2a_3 . . .$ (where "$z$" represents the part before the decimal point, and $a_i$ are digits). Now, this decimal expansion may not be unique - e.g. $0.99999 . . . =1.0000 . . .$ - but there is some decimal expansion.

• Now look at the number $q=z.a_1a_2a_3 . . . a_n$, that is, the number you get by cutting of $r$ at the $n$th decimal place. $q$ is rational (why?), and $\vert q-r\vert<10^n$ (why?), so - since $10^n<\epsilon$ - we are done.