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.
Best Answer
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.