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
 A: 
I am having trouble establishing that $\frac{\sqrt{2}}{\sqrt{3}} $ is a lower bound. 

The following direct calculation works - 
$$\frac{m+n\sqrt{2}}{m+n\sqrt{3}} -\frac{\sqrt{2}}{\sqrt{3}} = \frac{\sqrt{3}m + \sqrt{6}n - \sqrt{2}m-\sqrt{6}n}{\sqrt{3}m + 3n} = \frac{(\sqrt3-\sqrt2)m}{\sqrt{3}m + 3n} \ge 0.$$
A: Well,  $\frac {\sqrt 2}{\sqrt 3} < \frac{m+n\sqrt  2}{m+n\sqrt3} \iff$
$m\sqrt 2 + n\sqrt 6 < m\sqrt 3 + n\sqrt 6 \iff$
$m\sqrt 2 < m\sqrt 3$ which is always the case if $m > 0$.
.... so $\frac {\sqrt 2}{\sqrt 3}$ is a lower bound of D....
And $\frac{m+n\sqrt  2}{m+n\sqrt3} < \frac {\sqrt 2}{\sqrt 3} + \epsilon\iff$
$m\sqrt 3 + n\sqrt 6 < m\sqrt 2 + n\sqrt 6 + \sqrt 3\epsilon(m+n\sqrt3)\iff$
$m(\sqrt 3-\sqrt 2)< \sqrt3 \epsilon(m+n\sqrt 3)\iff$
$m\frac {\sqrt 3-\sqrt 2}{\sqrt 6\epsilon}-m < n$
If we set $m=1$ and $n>\frac {\sqrt 3-\sqrt 2}{\sqrt 6\epsilon}-1$ we can find this for any $\frac {\sqrt 3-\sqrt 2}3 > \epsilon > 0$.
So now, $\frac {\sqrt 2}{\sqrt 3} + \epsilon > \frac {\sqrt 2}{\sqrt 3}$ is not a lower bound.
So... that was a mess but... $\inf D =\frac {\sqrt 2}{\sqrt 3}$
=====
In hindsight:  I should have just taken my advice and just done it.
$\frac {m+n\sqrt 2}{m+n\sqrt 3} -\frac {\sqrt 2}{\sqrt 3}=$
$\frac {\sqrt 3(m + n\sqrt 2) -\sqrt 2(m+n\sqrt 3)}{\sqrt 3(m+n\sqrt 2)} =$
$\frac {m(\sqrt 3-\sqrt 2)}{\sqrt 3(m+n\sqrt 2)}:= \Delta(m,n)$
So as $\sqrt 3 > \sqrt 2$ and all other terms are positive,  $\Delta(m,n) > 0$. And so $\frac {\sqrt 2}{\sqrt 3}$ is a lower bound.
For any $\epsilon > 0$ we can ensure 
$\Delta(m,n) =\frac {m(\sqrt 3-\sqrt 2)}{\sqrt 3(m+n\sqrt 2)}< \epsilon$
by fixing $m$ and letting $n> \frac m{\sqrt 2}(\frac {(\sqrt 3-\sqrt 2)}{\sqrt 3\epsilon}-1)$.
And so $\inf D = \frac {\sqrt 2}{\sqrt 3}$.
The does require that we select an $\epsilon$ so that $\frac {\sqrt 3-\sqrt 2}{\sqrt 3} > \epsilon > 0$ but we can, wolog, assume that.
A: 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.
A: You can work the following result into the argument.
Let
$$f(x) = \frac{1 + \sqrt 2 x}{1 + \sqrt 3 x}$$
Using L'Hôpital's rule,
$${\displaystyle \lim _{x\to +\infty}f(x)=\frac{\sqrt 2}{\sqrt 3}}$$
