To prove that $S$ has an upper bound in $L$, we will use Zorn’s lemma. Let $T$ be the set of all chains in $L$ that are contained in $S$, and let $C$ be a chain in $T$. We want to show that $C$ has an upper bound in $L$.

Since $C$ is a chain in $L$, we know that $(C, \leq)$ is a partially ordered set. Let $c$ be an upper bound for $C$ in $L$. We claim that $c$ is also an upper bound for $S$.

Suppose for the sake of contradiction that $c$ is not an upper bound for $S$. Then there exists some $s \in S$ such that $s \nleq c$. Since $S \subseteq C$, we know that $s \in C$. Since $c$ is an upper bound for $C$, we have $s \leq c$. This contradicts the assumption that $s \nleq c$, so our claim is proven.

Therefore, $c$ is an upper bound for $S$. To complete the proof, we need to show that $c$ is in fact an element of $L$. Since every countable chain in $L$ has an upper bound, we know that $L$ is a complete partial order. In particular, every chain in $L$ has a least upper bound (lub).

Let $D$ be the set of all upper bounds of $S$ in $L$. We want to show that $c = \operatorname{lub}(D)$, which will complete the proof. First, we show that $c$ is an upper bound of $S$. We have already shown this above.

Now, let $d$ be an upper bound of $S$ in $L$. We want to show that $c \leq d$. By definition of $D$, we have $c \in D$, so $c$ is an upper bound of $S$. Therefore, we have $c \leq d$, since $d$ is also an upper bound of $S$.

Finally, we need to show that $c$ is the least upper bound of $D$. Suppose for the sake of contradiction that there exists some $d’ \in L$ such that $d’ < c$ and $d’$ is an upper bound of $D$. Since $c$ is an upper bound of $D$, we have $d’ \leq c$. But this contradicts the assumption that $c$ is the least upper bound of $D$. Therefore, $c$ must be the least upper bound of $D$.

We have now shown that $S$ has an upper bound in $L$. Therefore, by definition of a complete partial order, $L$ has a least upper bound for $S$, which completes the proof.

First, let’s assume that $\sqrt[k]{n}$ is rational. This means that there exists some integers $a$ and $b$ such that $\sqrt[k]{n}=\frac{a}{b}$, where $b\neq 0$. Raising both sides to the $k$-th power gives us $n=\frac{a^k}{b^k}$, which can be written as $nb^k=a^k$. This shows that $a^k$ is a multiple of $b^k$, which implies that $a$ must be a multiple of $b$. Let $a=cb$ for some integer $c$. Then we have $n=cb^k$, which can be written as $n=(b^{k-1}c)^k$. Therefore, $n$ is of the form $d^k$ for some integer $d=b^{k-1}c$.

Now let’s assume that $n=d^k$ for some integer $d$. We can write $d$ as $d=be$, where $b$ is the largest integer such that $b^k$ divides $d^k$, and $e$ is some integer. Then $n=d^k=(be)^k=b^ke^k$. Since $b$ is the largest integer such that $b^k$ divides $d^k$, $e$ is not divisible by $b$. Therefore, $b$ and $e$ are relatively prime. This implies that $b^k$ and $e^k$ are also relatively prime. Thus, $n=b^ke^k$ is in its simplest form as a product of relatively prime integers, and its $k$-th root is $\sqrt[k]{n}=e\in \mathbb{Z}$. Therefore, $\sqrt[k]{n}$ is rational.

Combining the two directions of the proof, we have shown that for all nonnegative integers $n \geq 0$ and $k>1$, $\sqrt[k]{n}$ is rational if and only if $n=d^k$ for some integer $d$.

Let $s=\sqrt{p}+\sqrt{q}+\sqrt{r}$.

Squaring both sides, we get:

$$s^2=p+q+r+2\sqrt{pq}+2\sqrt{qr}+2\sqrt{rp}$$

Since $s$ is rational, $s^2$ is also rational. Thus, $\sqrt{pq}$, $\sqrt{qr}$, and $\sqrt{rp}$ must be rational.

Suppose, for contradiction, that $p$ is not a perfect square. Then, $\sqrt{p}$ is irrational, and $\sqrt{q}+\sqrt{r}=s-\sqrt{p}$ is also irrational.

But this implies that $\sqrt{pq}$ and $\sqrt{qr}$ are irrational, which contradicts the fact that they are rational. Thus, $p$ must be a perfect square.

Similarly, we can show that $q$ and $r$ are also perfect squares. Thus, $p$, $q$, and $r$ are all perfect squares, as required.