Prop. $\mathbb{Q} \left( \sqrt {2}\right)$ under usual properties, i.e., $\mathbb{Q} \left( \sqrt {2}\right)$ is a field.
Proof. (Step of multiplicative inverse of $\mathbb{Q} \left( \sqrt {2}\right)$)
Let $a+b\sqrt {2}$ $\in$ $\mathbb{Q} \left( \sqrt {2}\right)$. We will find a',b' $\in$ $\mathbb{Q}$ such that $\dfrac {1} {a+b\sqrt {2}}$= $a'+b'\sqrt {2}$.
It suffices to show that if a+b$\sqrt {2}$ $\neq$ then $a^{2}-2b^{2}\neq 0$. Suppose not, say $a+b\sqrt {2}\neq 0$ but a^{2}-2b^{2}=0. Then 0 = $a^{2}-2b^{2}$= $(a+b\sqrt {2})$ $(a-b\sqrt {2})$. As $a+b\sqrt {2}$ $\neg$, so $a-b\sqrt {2}$=0. But then $\sqrt {2}$=a/b $\in$ $\mathbb{Q}$ provided by b $\neg$ 0. So b=0. However, (final step) if b=0 then a=0 as $a-b\sqrt {2}$. And this yields $a+b\sqrt {2}$=0. Contradiction.
My question is that is ''if b=0 then a=0 as $a-b\sqrt {2}$ $\equiv$ if $a-b\sqrt {2}$ then a=0 and b=0''? Because my proof is that (final step) if $a-b\sqrt {2}$=0 then a=0 and b=0. And this yields $a+b\sqrt {2}$=0. Contradiction
Suppose $\;c+d\sqrt2\;$ is the inverse of $\;a+b\sqrt2\neq0\;,\;\;a,b,c,d\in\Bbb Q\;$ , then:
$$(a+b\sqrt2)(c+d\sqrt2)=ac+2bd+(ad+bc)\sqrt2=1\implies$$
$$\implies\begin{cases}ac+2bd=1\\{}\\bc+ad=0\end{cases}$$
With $\;a,b\;$ as known quantities, take the coefficient matrix of the above non-homogeneous system:
$$A:=\begin{pmatrix}a&2b\\b&a\end{pmatrix}\implies A^{-1}\binom10=\frac1{a^2-2b^2}\begin{pmatrix}a&-2b\\-b&a\end{pmatrix}\binom10=\frac1{a^2-2b^2}\binom{\;a}{\!\!-b}$$
Thus, the inverse of $\;a+b\sqrt2\;$ is $\;\large\frac a{a^2-2b^2}-\frac b{a^2-2b^2}\sqrt2\;$
Observe that
$$a^2-2b^2=0\implies 2=\frac{a^2}{b^2}\in\Bbb Q\;,\;\;\;\text{contradiction}$$