Prove that if $a$ is a real number with $a >2$, then there is an $n$ is an element of natural number such that $2+1/\sqrt{n}<a$

Question

Prove that if $a$ is a real number with $a >2$, then there is an $n$ is an element of natural number such that $2+1/\sqrt{n}<a$

---

The goal is to show $\inf\{2+1/\sqrt{n} : n\in\mathbb{N}\}=2$. I started off by proving that $\{2+1/\sqrt{n}\}$ is not an empty set and $2$ is a lower bound for it.

Answer 1

Simply solve for $n$: $$2+\dfrac{1}{\sqrt{n}} < a \Rightarrow \dfrac{1}{\sqrt{n}} < a-2\Rightarrow \sqrt{n} > \dfrac{1}{a-2}\Rightarrow n > ....$$

Answer 2

The square root is just there to confuse you. Any unbounded increasing function will do.

All that is needed is Archimede's axiom (AA) for the reals:

If x and y are positive reals then there is an integer m such that mx > y.

You want $\sqrt{n}(a-2) > 1$.

Applying AA, since $1$ and $a-2$ are positive reals, there is an integer $m$ such that $m(a-2) > 1$. Now, let $n = m^2$, and $\sqrt{n}(a-2) > 1$.