# 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 $$\sup D$$: $$m+n\sqrt{2} so $$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, 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

• Have you come across the sequence definition of $\inf$? – It'sNotALie. Sep 24 at 1:34
• @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. – jeffery_the_wind Sep 24 at 1:35
• 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! – Gabriel B. H. Lisboa Sep 24 at 1:46
• @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 $0\in\mathbb{N}$? – probably_someone Sep 24 at 11:08
• @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. – Gabriel B. H. Lisboa Sep 24 at 18:02

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.

• 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. – jeffery_the_wind Sep 24 at 2:50

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

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

• that doesn't guarantee it's a strict lower bound – It'sNotALie. Sep 24 at 1:41
• @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 – Calvin Khor Sep 24 at 1:42

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

# $$\quad \big| \frac{\sqrt{2}}{\frac{1}{n}+\sqrt{3}} -\alpha \big|\lt \frac{\varepsilon}{2}$$

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.

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,

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

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