<s>[INST] $\sqrt{2} + \sqrt{3}$ is not a rational number. I am trying to prove that $\sqrt{2} + \sqrt{3}$ is not a rational number.
I have used the following Theorems to prove that $\sqrt{2} + \sqrt{3}$ is not a rational number.
Theorem 3.2) Let $x$ be an integer. $x^2$ is odd if and only if $x$ is odd.
Theorem 5.6) If $x$ is a rational number and $y$ is an irrational number, then $x+y$ is irrational.
Theorem 5.9) Let $n$ be an integer. If $n^2$ is divisible by $3$, then $n$ is divisible by $3$.
The following is my proof.

The number $\sqrt{2} + \sqrt{3}$ is not a rational number.
Proof. Suppose that $\sqrt{2} + \sqrt{3}$ is a rational number. It follows that there exist an integer $p$ and a nonzero integer $q$ such that
\begin{align}
\sqrt{2} + \sqrt{3} = \frac{p}{q} \text{ for some $p,q \in \mathbb{Z}$ and $q \ne 0$}
\end{align}
Because the number $\sqrt{2} + \sqrt{3}$ is a nonzero number, the number $\frac{p}{q}$ is nonzero.
Multiplying both sides of the above equation by $(\sqrt{2} - \sqrt{3})$, one obtains
\begin{align}
(\sqrt{2} + \sqrt{3})(\sqrt{2} - \sqrt{3}) = \frac{p}{q}(\sqrt{2} - \sqrt{3}) \text{ for some $p,q \in \mathbb{Z}$ and $q \ne 0$}
\end{align}
By multiplying $-1$ on both sides of the above equation and making $\frac{p}{q}\sqrt{3} - \frac{p}{q}\sqrt{2}$ as the subject one obtains
\begin{align}
\frac{p}{q}\sqrt{3} - \frac{p}{q}\sqrt{2} = 1 \text{ for some $p,q \in \mathbb{Z}$ and $q \ne 0$}
\end{align}
Squaring both sides of the above equation, one obtains
\begin{align}
5\frac{p^2}{q^2} - 2\sqrt{2}\sqrt{3}\frac{p^2}{q^2} = 1 \text{ for some $p,q \in \mathbb{Z}$ and $q \ne 0$}
\end{align}
By multiplying both sides of the above equation by $\frac{q^2}{2p^2}$, one obtains
\begin{align}
\frac{5}{2} - \sqrt{2}\sqrt{3} = \frac{q^2}{2p^2} \text{ for some $p,q \in \mathbb{Z}$ and $q \ne 0$}
\end{align}
To show that $\sqrt{2}\sqrt{3}$ is irrational, suppose that $\sqrt{2}\sqrt{3}$ is on the contrary rational.
It follows that there exist an integer $r$ and a nonzero integer $s$ such that
\begin{align}
\sqrt{2}\sqrt{3} = \frac{r}{s} \text{ for some $r,s \in \mathbb{Z}$ and $s \ne 0$}
\end{align}
Assuming that $r$ and $s$ has no common factor. Squaring both sides of the above equation one obtains
\begin{align}
2 \cdot 3 = \frac{r^2}{s^2} \text{ for some $r,s \in \mathbb{Z}$ and $s \ne 0$}
\end{align}
Making $\frac{r^2}{3}$ as the subject, one obtains
\begin{align}
\frac{r^2}{3} = 2s^2 \text{ for some $r,s \in \mathbb{Z}$ and $s \ne 0$}
\end{align}
Because the number on the $RHS$ of the above equation is an integer, $r^2$ is divisible by $3$. It follows from Theorem 5.9) that $r$ is divisible by $3$.
Thus,
\begin{align}
r = 3i \text{ for some $i \in \mathbb{Z}$}
\end{align}
Substituting the above equation into the equation $\frac{r^2}{3} = 2s^2$ one obtains
\begin{align}
3i^2 = 2s^2 \text{ for some $i,s \in \mathbb{Z}$ and $s \ne 0$}
\end{align}
Because the number on the $RHS$ of the above equation is even, $3i^2$ must be even. Since $3$ is odd, the factor $2$ must be from the number $i^2$. Therefore, $i^2$ is even. It follows from Theorem 3.2) that $i$ is even.
Thus,
\begin{align}
i = 2u \text{ for some $i,u \in \mathbb{Z}$}
\end{align}
Substituting the above equation into the equation $r = 3i$, one obtains
\begin{align}
r = 2 \cdot 3u \text{ for some $u \in \mathbb{Z}$}
\end{align}
Substituting the above equation into the equation $2 \cdot 3 = \frac{r^2}{s^2}$ and making $s^2$ as the subject, one obtains
\begin{align}
s^2 = 2 \cdot 3 u^2 \text{ for some $s,u \in \mathbb{Z}$ and $s \ne 0$}
\end{align}
Because $s^2$ is even and divisible by $3$, it follows from Theorem 3.2) and Theorem 5.9) that $s$ is even and divisible by $3$.
Thus,
\begin{align}
s = 2 \cdot 3 v \text{ for some $v \in \mathbb{Z}$ and $v \ne 0$}
\end{align}
Substituting the equation $r = 2 \cdot 3u$ and the equation $s = 2 \cdot 3 v$ into the equation $\sqrt{2}\sqrt{3} = \frac{r}{s}$, one obtains
\begin{align}
\sqrt{2}\sqrt{3} = \frac{2 \cdot 3u}{2 \cdot 3 v} \text{ for some $u,v \in \mathbb{Z}$ and $v \ne 0$}
\end{align}
Because $r$ and $s$ have common factors, one obtains a contradiction with the assumption that $r$ and $s$ have no common factor.
Hence, the number $\sqrt{2}\sqrt{3}$ is irrational.
Since on the equation $\frac{5}{2} - \sqrt{2}\sqrt{3} = \frac{q^2}{2p^2}$ the number $\frac{5}{2}$ is rational and $\sqrt{2}\sqrt{3}$ is irrational, by Theorem 5.6), the number $\frac{5}{2} - \sqrt{2}\sqrt{3}$ is irrational. However, because $\frac{p}{q}$ is a nonzero rational number, the number $\frac{q^2}{2p^2}$ on the $RHS$ of the equation $\frac{5}{2} - \sqrt{2}\sqrt{3} = \frac{q^2}{2p^2}$ is a nonzero rational number.
Thus, one obtains a contradiction.
Therefore, the number $\sqrt{2} + \sqrt{3}$ is not a rational number.

Is the proof correct?

Reference
Daepp, U. and Gorkin, P., 2011. Reading, Writing, and Proving. 2nd ed. pp.29, 54, 55.
 [/INST] The proof in the original question is more work than is necessary to prove something so simple.
All you need to do is show that $(\sqrt{2} + \sqrt{3})^2$ is irrational${}^\dagger$, which amounts to showing that $\sqrt{6}$ is irrational, which you can prove in the same way you prove $\sqrt{2}$ is irrational.
$\dagger$ Note that $x^2$ irrational $\implies x\ $ is irrational. This is easy to see by considering the contrapositive.
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