Prove that for all $n$ $ \{\sqrt{n^2-1}\}=\sqrt{n^2-1}-n+1$ , $ \{\sqrt{n^2-1}\}$ is the fractional part

Prove that for all $n$ $\{\sqrt{n^2-1}\}=\sqrt{n^2-1}-n+1$

we know that $\{x\}=x-\left\lfloor {x}\right \rfloor $ and we also know that $\left\lfloor {x}\right \rfloor=n$ $\iff$ $n \leq x < n+1$

so I believe that we need to get to something similar to that

we know that $n^2-1 \geq 0$ because of the domain so we can do $0 \leq n^2-1 < n $ $\Rightarrow$ non negative values so we can square them $\sqrt{n^2-1} < \sqrt{n} = |n| =n$ so we got $\sqrt{n^2-1} < n$ and I could not figure out how to continue it or even if I started right

the book does not show full answers but it has this and I cannot understand why

in the book they got to $n-1 \leq \sqrt{n^2-1} <n$ and then they wrote "according to floor and ceiling properties" $\left\lfloor{\sqrt{x^2-1}}\right \rfloor =x-1$ these are supposed to be the last steps and we need to prove before using but the book only shows the final answers

$\left\lfloor{\sqrt{x^2-1}}\right \rfloor =x-1$ why does that happen?

and why $n-1 \leq \sqrt{n^2-1} <n$

thanks for any tips and help!

1. $\sqrt{n^2-1}<\sqrt{n^2}=n$.
2. If $n\ge 1$, $\enspace n-1 \le\sqrt{n^2-1}\iff(n-1)^2=n^2-2n+1\le n^2-1\iff 2(n-1)\ge 0$.

Therefore, we have $\; n-1\le \sqrt{n^2-1} <n$, so that $\;\Bigl\lfloor\sqrt{n^2-1}\Bigr\rfloor=n-1$, and consequently $$\Bigl\{\sqrt{n^2-1} \Bigr\}=\sqrt{n^2-1} - n+1.$$

$n-1<n+1$ and $n-1\ge0$, so...

$$ (n-1)^2\le(n-1)(n+1)=n^2-1<n^2 \\ n-1\le\sqrt{n^2-1}<n, [\sqrt{n^2-1}]=n-1, \{\sqrt{n^2-1}\}=\sqrt{n^2-1}-n+1 $$