# Number Less One is Greater than Square Root

## Lemma

Let $n \in \N$ be an integer.

Then for $n > 2$:

$n - 1 > \sqrt n$

## Proof

Let $n \in \N$ satisfy:

$n - 1 > \sqrt n$

Then:

 $\displaystyle n - 1$ $>$ $\displaystyle \sqrt n$ $\displaystyle \leadstoandfrom \ \$ $\displaystyle \paren {n - 1}^2$ $>$ $\displaystyle n$ squaring both sides $\displaystyle \leadstoandfrom \ \$ $\displaystyle n^2 - 3 n + 1$ $>$ $\displaystyle 0$ moving all terms to the left hand side

By definition, $n^2 - 3 n + 1$ is strictly increasing when the following inequality holds:

 $\displaystyle n^2 - 3 n + 1$ $<$ $\displaystyle \paren {n + 1}^2 - 3 \paren {n + 1} + 1$ $\displaystyle$ $=$ $\displaystyle n^2 - n - 1$ $\displaystyle \leadsto \ \$ $\displaystyle 2 \paren {n - 1}$ $>$ $\displaystyle 0$

The solution to this is $n \ge 2$.

By inspection, $n = 3$ satisfies $n^2 - 3n + 1 > 0$.

As $n^2 - 3 n + 1$ is strictly increasing, the inequality holds for all $n \ge 3$.

$\blacksquare$