If $x \geq 5$ and $x$ is a prime number, then what number lies between
$$\frac{x + 2}{x}$$
and
$$\frac{x + 3}{x}?$$
Here is my attempt:
Let $\theta \in \mathbb{R}$ such that
$$\frac{x + 2}{x} < \theta < \frac{x + 3}{x}.$$
Then,
$$1 < \theta < \frac{8}{5}.$$
Alas, this is where I get stuck. I was thinking of getting the average
$$\dfrac{\frac{x + 2}{x} + \frac{x + 3}{x}}{2} = \dfrac{2x + 5}{2x} \leq \frac{3}{2}$$
but how would I know whether this lies to the left or right of the quantity $\theta$?
Thanks!
No fixed number independent of $x$ will be between $\dfrac{x + 2}{x}$ and $\dfrac{x+3}{x}$.
To see this, note that it's equivalent to finding a number that is between $\dfrac{2}{x}$ and $\dfrac{3}{x}$, and then adding $1$. Then for any number $\theta > 0$, choosing $x$ larger than $\dfrac{3}{\theta}$ will make $\dfrac{3}{x} < \theta$.
For instance, if $\theta = \frac{1}{2}$, we might choose $x > \dfrac{3}{\frac{1}{2}} = 6$. To choose a prime, we might choose $x = 7$. Then $1 + \frac{1}{2}$ is not between $\dfrac{7 + 2}{7}$ and $\dfrac{7 + 3}{7}$.