Real Analysis Inf and Sup question I am hung up on this question for real analysis ( intro to anaylsis ). 
Find  $\inf D$ and $\sup D$
$$\mathrm{D}=\left\{\frac{m+n\sqrt{2}}{m+n\sqrt{3}} :m,n\in\Bbb{N}\right\}$$
I have spent enough time staring at this thing that I know the $\sup D=1$ and $\inf D=\frac{\sqrt{2}}{\sqrt{3}}$.
for $\sup D$:
$$m+n\sqrt{2}<m+n\sqrt{3}\implies\frac{m+n\sqrt{2}}{m+n\sqrt{3}}<1$$
so $1$ is an upper bound for $D$, and then for the confirmation that 1 is the least upper bound I can prove by contradiction that $\sup D$ cannot be less than $1$, because I could always find a $d \in D$ such that  $$\sup D<d<1$$, which is the contradiction since no $d \in D$ can be greater than $\sup D$.(proof omited)
So my problem is with $\inf D$. I am having trouble establishing that $\frac{\sqrt{2}}{\sqrt{3}}$ is a lower bound. I am just not seeing it. The intuition is that if $m$ is small and $n$ is large than the fraction $\frac{\sqrt{2}}{\sqrt{3}}$ dominates the expression, however it will always be slightly greater than $\frac{\sqrt{2}}{\sqrt{3}}$. Analytically I am just not able to show it.
Any help would be greatly appreciated

In general if $\alpha$ is a lower bound of $X$ and $Y$ is a nonempty subset of $X$ with $\alpha = \text{inf}(Y)$, then $\alpha = \text{inf}(X)$.
By fleablood's opening argument, we know that $\alpha = \frac{\sqrt 2}{\sqrt 3}$ is a lower bound for $D$.
Let $\mathrm{E}=\left\{\frac{1+n\sqrt{2}}{1+n\sqrt{3}} :n\in\Bbb{N}\right\}$.
We are going to show that the infimum of $E$ is equal to $\alpha$.
We have
$\tag 1 \frac{1+n\sqrt{2}}{1+n\sqrt{3}}= \frac{1}{1+n\sqrt{3}} + \frac{\sqrt{2}}{\frac{1}{n}+\sqrt{3}}$
Let $\varepsilon \gt 0$.
Find an $n$ such that both
$\quad \big| \frac{1}{1+n\sqrt{3}} \big| \lt \frac{\varepsilon}{2}$
$\quad \big| \frac{\sqrt{2}}{\frac{1}{n}+\sqrt{3}} -\alpha   \big|\lt \frac{\varepsilon}{2}$
are true.
Then, using $\text{(1)}$ and substitution, associativity and the triangle inequality,
$\quad \big| \frac{1+n\sqrt{2}}{1+n\sqrt{3}} - \alpha \big| = \big|  (\frac{1}{1+n\sqrt{3}} + \frac{\sqrt{2}}{\frac{1}{n}+\sqrt{3}}) -\alpha \big| = \big|  \frac{1}{1+n\sqrt{3}} + (\frac{\sqrt{2}}{\frac{1}{n}+\sqrt{3}}) -\alpha) \big| \le$
$\quad \quad \big|  \frac{1}{1+n\sqrt{3}} \big| + \big|(\frac{\sqrt{2}}{\frac{1}{n}+\sqrt{3}}) -\alpha) \big| \lt \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon$
we conclude that there are numbers in $E$ that are arbitrarlly close to the lower bound $\alpha$ of $E$. But then $\text{inf}(E) = \alpha$, as was to be shown.
