Question

Prove that if x is a positive real number, then

$\begin{array}{r}\left(\mathrm{a}\right)\lfloor \sqrt{\lfloor \mathrm{x}\rfloor }\rfloor =\lfloor \sqrt{\mathrm{x}}\rfloor \\ \left(\mathrm{b}\right)\left[\sqrt{\lfloor \mathrm{x}\rfloor }\right]=\lceil \sqrt{\mathrm{x}}\rceil \end{array}$

Short answer

$\begin{array}{r}\left(\mathrm{a}\right)\lfloor \sqrt{\lfloor \mathrm{x}\rfloor }\rfloor =\lfloor \sqrt{\mathrm{x}}\rfloor \\ \left(\mathrm{b}\right)\left[\sqrt{\lfloor \mathrm{x}\rfloor }\right]=\lceil \sqrt{\mathrm{x}}\rceil \end{array}$

Step-by-step solution

Step: 1

1. If x is a positive integer, then the two sides are equal. So suppose that $x=n^2+m+\epsilon$, where is the largest perfect square less than x, m is a nonnegative integer, and $0<\epsilon \le 1$. then both $\sqrt{x}\text{and}\sqrt{\lfloor x\rfloor }=\sqrt{n^2+m}$ between n and n+1, so both sides equal n.

Step: 2

1. If x is a positive integer, then the two sides are equal. So suppose that$x=n^2-m-\epsilon ,$ , where$n^2$ is the smallest perfect square less than x, m is a nonnegative integer, and $0<\epsilon \le 1$. then both $\sqrt{x}\text{and}\sqrt{\lfloor x\rfloor }=\sqrt{n^2-m}$between n-1 and n-1, so both sides equal n