Infimum of $\left\{\left|\sqrt{m} - \sqrt{n}\right|\;: \;m,n \in\mathbb{N},\; m≠n\right\}$

Question

In my homework assignment I encountered this problem:

Find the infimum of the set $A=\left\{\left|\sqrt{m} - \sqrt{n}\right|\;: \;m,n \in\mathbb{N},\; m≠n\right\}$

How do i even start to find the infimum of this set??

Answer 1

First of all observe that $\left|\sqrt{m}-\sqrt{n}\right|\geq0$, So if we will prove that $0$ is the greatest lower bound we are done.

By the definition of the set $A$, we know that $\sqrt{n+1}-\sqrt{n}\in A\;\forall n\in\mathbb{N}$ (because $\sqrt{n+1}>\sqrt{n}$ we dont need the absolute value), and after a bit of simplification we would conclude that

$$ A\ni\sqrt{n+1}-\sqrt{n}=\frac{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{\sqrt{n+1}+\sqrt{n}}=\frac{1}{\sqrt{n+1}+\sqrt{n}} $$

Now let $\varepsilon>0$ we will take $n=\left\lceil\frac{1}{(2\varepsilon)^2}\right\rceil$. It happens to be the case that

$$ \frac{1}{\sqrt{n+1}+\sqrt{n}}\leq\frac{1}{2\sqrt{n}}\leq\frac{1}{2\cdot\frac{1}{\sqrt{(2\varepsilon)^2}}}=\varepsilon $$

Hence, by an equivalent definition of an Infimm of a set, we will conclude that $0$ is the infimum of $A$.

Answer 2

Hint:

We have for all $n,m \in \mathbb{N}$, that $\lvert{\sqrt{m}-\sqrt{n}}\rvert\geq 0$. So a lower bound is $0$. You now need to figure out if $0$ is the greatest lower bound (and prove if it is).

Answer 3

Write $m=n+k$, $k>0$, $m, n, k \in \mathbb{N}$.

0)$0$ is a lower bound

1)$\sqrt{n+k} - \sqrt{n} \ge \sqrt{n+1}-\sqrt{n}$,

since $\sqrt{}$ is strictly increasing.

2)$\sqrt{n+1} =\dfrac{n+1}{\sqrt{n+1}}$, and $\sqrt{n}=\dfrac{n}{\sqrt{n}}$;

3)$0<\sqrt{n+1}-\sqrt{n}<$

$\dfrac{n+1}{\sqrt{n+1}}- \dfrac{n}{\sqrt{n}}<$

$\dfrac{n+1}{\sqrt{n+1}}-\dfrac{n} {\sqrt{n+1}}=\dfrac{1}{\sqrt{n+1}} $.

4)Sandwich

5)Show that $0$ is least lower bound.

6)Assume $a>0$ is a lower bound.

7)Archimedean principle

There is a $n_0 \in \mathbb{N}$ s.t.

$n_0>\dfrac{1}{\sqrt{a}}-1$, or

$a>\dfrac{1}{\sqrt{n_0+1}}$, contradiction, and we are done.

Answer 4

As it was pointed out, since for $n \ne m$ we have that $|\sqrt{n}-\sqrt{m}|>0$, the infimum is surely greater or equal than zero. On the other hand, since $\lim_{n \to \infty}(\sqrt{n+1}-\sqrt{n})=0$, the infimum cannot be greater than zero.