Let $$\mathbb{Z}[\sqrt 2]=\{a+b\sqrt 2 \mid a, b \in \mathbb{Z}\}$$ of the real numbers $\mathbb{R}$. Since the set of real numbers $\mathbb{R}$ forms a ring, the subset $\mathbb{Z}[\sqrt 2]$ also forms a ring if it is closed under addition and multiplication.
Addition:
Let $x, y \in \mathbb{Z}[\sqrt 2]$ with $x=a+b\sqrt2, y=c+d\sqrt2$ where $a, b, c, d \in \mathbb{Z}$. Then
$x+y=(a+b\sqrt2)+(c+d\sqrt2)$ $x+y=(a+c)+(b+d)\sqrt2$
Since $a, c \in \mathbb{Z} \implies a+c \in \mathbb{Z}$
and
Since $b, d \in \mathbb{Z} \implies b+d \in \mathbb{Z}$
$\implies x+y \in \mathbb{Z}[\sqrt2]$
Therefore, $\mathbb{Z}[\sqrt2]$ is closed under addition.
Multiplication:
Let $x, y \in \mathbb{Z}[\sqrt 2]$ with $x=a+b\sqrt2, y=c+d\sqrt2$ where $a, b, c, d \in \mathbb{Z}$. Then
$xy=(a+b\sqrt2)(c+d\sqrt2)$
$xy=(ac+2bd)+(ad+bc)\sqrt2$
Since $a, b, c, d \in \mathbb{Z} \implies ac+2bd \in \mathbb{Z}$
and
Since $a, b, c, d \in \mathbb{Z} \implies ad+bc \in \mathbb{Z}$
$\implies xy \in \mathbb{Z}[\sqrt2]$
Therefore, $\mathbb{Z}[\sqrt2]$ is closed under multiplication.
It is also an integral domain since $\mathbb{R}$ as a field is an integral domain.
Alternatively: A non-trivial solution of
$(x+y\sqrt2)(a+b\sqrt2)=0$ gives for
$xb+ya \neq 0$ the equation
$\sqrt2=-\frac{xa+2yb}{xb+ya}$, a contradiction to the irrationality of $\sqrt2$. If $xb+ya=0$ it follows that $xa+2yb=0$. Since the determinant of this linear system seen as an equation for x and y is $2b^{2}-a^{2}$, the only solution is $x=y=0$ if $2b^{2}-a^{2}\neq 0$. If $2b^{2}-a^{2}=0$, the irrationality of $\sqrt2 \implies a=b=0$. Thus either $x+y\sqrt2=0$ or $a+b\sqrt2=0$.
How to apply two axioms of Euclidean algorithm on $\mathbb{Z}[\sqrt2]$.
Thanks.