Number Less One is Greater than Square Root

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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 $n \ge 3$.

$\blacksquare$