"The number $\sqrt{2}$ is not a rational number."
Proof. Suppose that $\sqrt{2}$ is a rational number.
Let $\mathbb{P}$ denotes the set of all prime numbers.
It follows that,
\begin{equation} \sqrt{2} = \frac{p}{q} \text{ where p,q $\in > \mathbb{Z}$ and $q \ne 0$} \end{equation}
By squaring both sides of the equation, one obtains:
\begin{equation} 2 = \frac{p^2}{q^2} \end{equation}
From
\begin{equation} 2q^2 = p^2, \end{equation}
it follows that,
\begin{equation} p^2 = 2r \text{ where r $\in \mathbb{Z^+}$}. > \end{equation}
From
\begin{equation} q^2 = \frac{p^2}{2}, \end{equation}
it follows that,
\begin{equation} q^2 = 2s \text{ where s $\in \mathbb{Z^+}$}. > \end{equation}
Because, for every positive even integers 2$\phi$ where $\phi \in > \mathbb{Z^+}$,
\begin{equation} 2\phi = uv \text{ where $u \in \mathbb{Z^+}$ and $v > \in \mathbb{P}$ ,} \end{equation}
it follows that,
\begin{equation} 2 = \frac{p^2}{q^2} = \frac{2r}{2s} = \frac{ax}{by} > \text{ where a,b $\in \mathbb{Z^+}$, $b \ne 0$, and $x,y \in > \mathbb{P}$}. \end{equation}
Consider the equation:
\begin{equation} x = 2\left(\frac{b}{a}\right)y \end{equation}
Because the $LHS$ of the above equation is a prime number and the $RHS$ is not a prime number, one obtains a contradiction.
Hence, the number $\sqrt{2}$ is not a rational number.
I have realized that the proof I had demonstrated before is incorrect, so I have put the incorrect proof in the above blockquote. The new proof is in the texts below.
I am trying to prove that $\sqrt{2}$ is not a rational number on my own using proof by contradiction. However, I am not sure whether my proof is correct or not.
The following is my proof.
"The number $\sqrt{2}$ is not a rational number."
Proof. Suppose that $\sqrt{2}$ is a rational number.
It follows that there must be an integer $p$ and a nonzero integer $q$ such that:
\begin{align} \sqrt{2} &= \frac{p}{q} \end{align}
By squaring both sides of the above equation, one obtains:
\begin{align} 2 &= \frac{p^2}{q^2} \end{align}
Rearranging the above equation so that $p^2$ is the subject, one obtains:
\begin{align} p^2 &= 2q^2 \end{align}
Thus,
\begin{align} p^2 = 2r \text{ where r $\in$ $\mathbb{Z^+}$} \end{align}
Also, from the equation $2 = \frac{p^2}{q^2}$, by rearranging the equation so that $q^2$ is the subject, one obtains:
\begin{align} q^2 = \frac{p^2}{2} \end{align}
Thus,
\begin{align} q^2 = 2s \text{ where s $\in$ $\mathbb{Z^+}$} \end{align}
Substituting $2r$ and $2s$ into the equation $2 = \frac{p^2}{q^2}$, one obtains:
\begin{align} 2 = \frac{p^2}{q^2} = \frac{2r}{2s} \end{align}
Since $p^2 = 2q^2$, it follows that $p^2 > q^2$.
Because both $p^2$ and $q^2$ are positive even numbers and $q^2 \geq 4$, it follows that $p^2 \geq 16$.
Thus, the equation $2 = \frac{p^2}{q^2} = \frac{2r}{2s}$ is expressed as:
\begin{align} 2 = \frac{p^2}{q^2} = \frac{(4+2m)^2}{(2+2n)^2} \text{ where m,n $\in$ $\mathbb{Z^*}$ } \end{align}
Because the term $\frac{(4+2m)^2}{(2+2n)^2} = \frac{(m+2)^2}{(n+1)^2}$ and $n \leq m+1$ for $\frac{(m+2)^2}{(n+1)^2}$ to be an integer, it follows that:
\begin{align} 2 = \frac{p^2}{q^2} = \frac{(4+2m)^2}{(2+2n)^2} = \frac{(m+2)^2}{(n+1)^2} \text{ where m,n $\in$ $\mathbb{Z^*}$ and $n \leq m+1$} \end{align}
Expanding $(m+2)^2$ as $m^2 +4m + 4$ and factor $2$ from the term, one obtians:
\begin{align} \frac{p^2}{q^2} = \frac{(m+2)^2}{(n+1)^2} = \frac{(m^2 +4m +4)}{(n^2 +2n +1)} = \frac{2(\frac{m^2}{2}+ 2m + 2)}{(n^2 + 2n +1)} = 2 \text{ where $m,n \in \mathbb{Z^*}$ and $n \leq m+1$} \end{align}
For the term $\frac{2(\frac{m^2}{2}+ 2m + 2)}{(n^2 + 2n +1)}$ to be equal to $2$, $\frac{(\frac{m^2}{2}+ 2m + 2)}{(n^2 + 2n +1)} =1$.
For $\frac{m^2}{2}+ 2m + 2$ to be an integer, $m=2k$ where $k \in \mathbb{Z^*}$.
Thus,
\begin{align} 1 = \frac{(\frac{m^2}{2}+ 2m + 2)}{(n^2 + 2n +1)} = \frac{(2k^2+ 4k + 2)}{(n^2 + 2n +1)} = \frac{2(k^2+ 2k + 1)}{(n^2 + 2n +1)} \text{ where $k,n \in \mathbb{Z^*}$ and $n \leq 2k +1$} \end{align}
Assuming there exists $k,n \in \mathbb{Z^*}$ such that $k^2 + 2k +1$ equals to the number $1$ and $n^2 + 2n +1$ equals to the prime number $2$.
Such $k,n$ make the term $\frac{2(k^2+ 2k + 1)}{(n^2 + 2n +1)} = 1$.
Because $n^2 +2n +1 \ne 2$ for all $n \in \mathbb{Z^*}$, one obtains a contradiction.
Hence, the number $\sqrt{2}$ is not rational.
