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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

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  • $\begingroup$ Have you come across the sequence definition of $\inf$? $\endgroup$ – It'sNotALie. Sep 24 at 1:34
  • $\begingroup$ @It'sNotALie. Can't say I have, although I can't say that I haven't. I just checked and yes I have come across that in one of my books, however I have to say my professor has never used it in class so I feel there should some way to prove the inf without that. Although I'm definitely interested in learning how that applies. $\endgroup$ – jeffery_the_wind Sep 24 at 1:35
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    $\begingroup$ We have indeed $\forall m, n, \in \mathbb{N}$, $\frac{m + n\sqrt2}{m + n \sqrt3} > \frac{\sqrt2}{\sqrt3}$, otherwise there would be $M, N \in \mathbb{N}$ such that $\frac{M + N\sqrt2}{M + N\sqrt3} \leq \frac{\sqrt2}{\sqrt3}$, hence $M\sqrt3 + N\sqrt6 \leq M\sqrt2 + N\sqrt6$ which implies $M \sqrt3 \leq M\sqrt2$, an absurd! $\endgroup$ – Gabriel B. H. Lisboa Sep 24 at 1:46
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    $\begingroup$ @GabrielB.H.Lisboa $M\sqrt{3}\leq M\sqrt{2}$ is fulfilled if $M=0$. Which textbook does this come from, and is it one in which $0\in\mathbb{N}$? $\endgroup$ – probably_someone Sep 24 at 11:08
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    $\begingroup$ @probably_someone ok, you are right. It works fine for the $\geq$ inequality, though. The case when $m = 0$ becomes trivial (changing to $\geq$), showing $\frac{\sqrt2}{\sqrt3}$ is indeed an lower bound for the set. $\endgroup$ – Gabriel B. H. Lisboa Sep 24 at 18:02
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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... $\inf D =\frac {\sqrt 2}{\sqrt 3}$

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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.

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  • $\begingroup$ The first line there I seriously worked out several times i'm not sure why i missed it, but such is life. I had a similar kind of mess for the proof of the sup$D$. Thanks for the help. $\endgroup$ – jeffery_the_wind Sep 24 at 2:50
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I am having trouble establishing that $\frac{\sqrt{2}}{\sqrt{3}} $ is a lower bound.

The following direct calculation works - $$\frac{m+n\sqrt{2}}{m+n\sqrt{3}} -\frac{\sqrt{2}}{\sqrt{3}} = \frac{\sqrt{3}m + \sqrt{6}n - \sqrt{2}m-\sqrt{6}n}{\sqrt{3}m + 3n} = \frac{(\sqrt3-\sqrt2)m}{\sqrt{3}m + 3n} \ge 0.$$

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  • $\begingroup$ that doesn't guarantee it's a strict lower bound $\endgroup$ – It'sNotALie. Sep 24 at 1:41
  • $\begingroup$ @It'sNotALie. whats a "strict" lower bound? Its easy to show that $\inf D \le \frac{\sqrt{2}}{\sqrt{3}}$, so i jumped to the part that OP said was his/her problem. If you want "$> 0$" i.e. non-attainment, then this proof actually does do that $\endgroup$ – Calvin Khor Sep 24 at 1:42
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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.

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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,

$${\displaystyle \lim _{x\to +\infty}f(x)=\frac{\sqrt 2}{\sqrt 3}}$$

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  • $\begingroup$ -1 for using L'Hôpital for a completely elementary limit. $\endgroup$ – Martin Argerami Sep 24 at 11:19

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