Prove $H=\{ a+\sqrt2 b \mid a,b \text{ are rational numbers}\}$ is closed under multiplication.
All I have to prove is that if we multiple two elements of $H,$ then again we can write it in $a+\sqrt2 b$ form for $a,b$ rational. Right?
But $(a+\sqrt2 b)\cdot (a-\sqrt2 b)= a^2-2b^2.$ Then does it belong to $H$ or not, since I can’t write it in $a+\sqrt2 b$ form?
Let $a+\sqrt{2}b, c+\sqrt{2}d \in H.$ You need to show $(a+\sqrt{2}b)(c+\sqrt{2}d) \in H.$ Consider $$(a+\sqrt{2}b)(c+\sqrt{2}d)=(ac+2bd)+\sqrt{2}(ad+bc).$$ Since $a,b,c,d \in \mathbb{Q},$ therefore $ac+2bd,ad+bc\in \mathbb{Q}.$ Hence $(a+\sqrt{2}b)(c+\sqrt{2}d) \in H.$
Edit: In your example, $a-2b^2=a-2b^2+\sqrt{2} \ (0)\in H.$