# Guessing a numerical middle bound

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!

• There's a lot missing from this question. Why not $(x + 2.5)/x$? That's between those two numbers. – davidlowryduda Feb 20 '15 at 2:31
• @mixedmath, by number I meant an actual number without variables. My apologies for not being clear earlier. – Jose Arnaldo Bebita-Dris Feb 20 '15 at 2:36
• In general, it will be a number between $2/x$ and $3/x$ larger than $1$. You must depend on $x$ at least, since no number is larger than $2/x$ and less than $3/x$ for all positive $x$. Indeed, as $x$ increases, both of these go to $0$. – davidlowryduda Feb 20 '15 at 2:38
• No. For instance, $3/2$ is not between $(11 + 2)/11$ and $(11 + 3)/11$ – davidlowryduda Feb 20 '15 at 2:44
• Okay, thanks @mixedmath. I get it now. Can you write out your comments into an answer so that I may be able to accept it? Muchos gracias! =) – Jose Arnaldo Bebita-Dris Feb 20 '15 at 2:48

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 $\dfrac{7 + 3}{7}$.