I want to prove that for $p,q,r$ different primes, $\sqrt{p}+\sqrt{q}+\sqrt{r}$ is irrational.
Is the following proof correct?
If $\sqrt{p}+\sqrt{q}+\sqrt{r}$ is rational, then $(\sqrt{p}+\sqrt{q}+\sqrt{r})^2$ is rational, thus $p+q+r+2\sqrt{pq}+2\sqrt{pr}+2\sqrt{qr}$ is rational, therefore $\sqrt{pq}+\sqrt{pr}+\sqrt{qr}$ is rational.
If $\sqrt{pq}+\sqrt{pr}+\sqrt{qr}$ is rational, then $(\sqrt{pq}+\sqrt{pr}+\sqrt{qr})^2$ is rational, therefore $pq+qr+pr+\sqrt{p^2qr}+\sqrt{pq^2r}+\sqrt{pqr^2}$ is rational, therefore $p\sqrt{qr}+q\sqrt{pr}+r\sqrt{pq}$ is rational.
Now suppose $p<q<r$. If $p\sqrt{qr}+q\sqrt{pr}+r\sqrt{pq}$ and $\sqrt{pq}+\sqrt{pr}+\sqrt{qr}$ are rational, then $$p\sqrt{qr}+q\sqrt{pr}+r\sqrt{pq}-p(\sqrt{pq}+\sqrt{pr}+\sqrt{qr})$$
is rational, therefore $(q-p)\sqrt{pr}+(r-p)\sqrt{pq}$ is rational.
If $(q-p)\sqrt{pr}+(r-p)\sqrt{pq}$ is rational, then $((q-p)\sqrt{pr}+(r-p)\sqrt{pq})^2$ is rational, thus $(q-p)^2pr+2(q-p)(r-p)\sqrt{p^2qr}+(r-p)^2pq$ is rational, thus $\sqrt{qr}$ is rational.
But $q,r$ are distinct primes, thus $qr$ can't be a square. Thus $\sqrt{qr}$ is irrational. Contradiction.
Also, is there an easier proof?
Best Answer
There is an simplier proof:
Suppose that $\sqrt{p}+\sqrt{q}+\sqrt{r}=w$, where $w$ is rational. Then$\sqrt{p}+\sqrt{q}=-\sqrt{r}+w$, so:
$$(\sqrt{p}+\sqrt{q})^2=(-\sqrt{r}+w)^2$$
Therefore $\sqrt{pq}+w\sqrt{r}$ is rational. But it isn't (the same argument). Suppose that $\sqrt{pq}-w\sqrt{r}=s$ where $s$ is rational, then:
$$\sqrt{pq}=s+w\sqrt{r}$$ $$(\sqrt{pq})^2=(s+w\sqrt{r})^2$$ So $sw\sqrt{r}$ is rational (we also should check case when $s=0$ or $w=0$, but it is simple).