For an integer $n\geq2$, we define the set $E=\left\{\frac{p}{q} : p,q \in \mathbb{N}, p+q=n\right\}$

Question

For an integer $n\geq2$, we define the set $E=\left\{\frac{p}{q} : p,q \in \mathbb{N}, p+q=n\right\}$

Show that this set is bounded, and determine $\sup E$ and $\inf E$.

By developing some elements of the set, I was able to notice that $\inf E = \frac{1}{n-1}$ and $\sup E=n-1$.

But I'm not even sure how the proof goes on....

Answer 1

$E$ is a nonempty finite subset of $\mathbb{R}$ and it is well known that such a subset has a smallest and a largest element.

Of course, this is not true for any infinite subset. Consider for example the interval $(0,1)$, which has no smallest and no largest element. It has nevertheless a "largest lower bound" = 0 and a "smallest upper bound" = 1.

Now, let's go back to :

$$E=\{\frac{n}{q}-1;\,1\le q\le n\}$$

The smallest element of $E$ is reached by choosing the largest value of $q$; and the largest element of $E$ is reached by choosing the smallest value of $q$. So we have :

$$\inf(E)=\min(E)=0\qquad\mathrm{and}\qquad\sup(E)=\max(E)=n-1$$

Note : I consider that $\mathbb{N}$ is the set of all natural numbers, including $0$. If you consider that $\mathbb{N}$ is the set of positive integers (1, 2, etc ...) then you have to adapt slightly what I wrote :

$$E=\{\frac{n}{q}-1;\,1\le q\le n-1\}$$

and the conclusion becomes :

$$\inf(E)=\min(E)=\frac{1}{n-1}\qquad\mathrm{and}\qquad\sup(E)=\max(E)=n-1$$