Given a unit $\epsilon = (a + b \sqrt{d} ) /2 > 1 $ of a real quadratic number field with discriminant $d$ and $a \geq 1$ and $b \geq 1$.
Why is there no unit $\epsilon$ in $\mathbb Q(\sqrt{2})$ with $1 < \epsilon < 1 + \sqrt{2}$ ?
We denote by $U_d = O_d^x$ the group of invertible elements in $O_d$. By definition, we know: $ \epsilon = \frac{1}{2} (a + b \sqrt{d}) \in U_d \iff \exists \epsilon^{'} \in O_d : \epsilon \epsilon^{'} = 1$
On taking the norm, one would get $1 = N(\epsilon \epsilon^{'}) = N(\epsilon) N(\epsilon^{'})$ which leads to $N(\epsilon) = N( \epsilon^{'}) = +/-1$. This implies $ \epsilon^{'} = \epsilon^{\sigma}$, equivalent to the Pell's equation: $a^2 -db^2 = +/- 4$.
You need $\epsilon$ to be an algebraic integer, so when $d=2$ that denominator $2$ cannot be there. Therefore $\epsilon=a+b\sqrt2$ for some integers $a,b$, and its conjugate is $\overline{\epsilon}=a-b\sqrt2$, both sharing the minimal polynomial $$ m(x)=x^2-2ax+(a^2-2b^2). $$ For them to be units, we need $a^2-2b^2=\pm1$. Therefore the quadratic formula tells us that $$ \epsilon=a\pm \sqrt{a^2\pm1}. $$ For $\epsilon$ to be $>1$ we need $a$ to be positive, and the first unknown sign to be $+$. If $a\ge2$, then $\epsilon>3>1+\sqrt2$. Therefore $1+\sqrt2$ is the smallest unit $>1$ in $\Bbb{Z}[\sqrt2]$.