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Found in: Page 155

Discrete Mathematics and its Applications

Book edition 7th
Author(s) Kenneth H. Rosen
Pages 808 pages
ISBN 9780073383095

Prove that if x is a positive real number, then $\begin{array}{r}\left(\mathrm{a}\right)⌊\sqrt{⌊\mathrm{x}⌋}⌋=⌊\sqrt{\mathrm{x}}⌋\\ \left(\mathrm{b}\right)\left[\sqrt{⌊\mathrm{x}⌋}\right]=⌈\sqrt{\mathrm{x}}⌉\end{array}$

$\begin{array}{r}\left(\mathrm{a}\right)⌊\sqrt{⌊\mathrm{x}⌋}⌋=⌊\sqrt{\mathrm{x}}⌋\\ \left(\mathrm{b}\right)\left[\sqrt{⌊\mathrm{x}⌋}\right]=⌈\sqrt{\mathrm{x}}⌉\end{array}$

See the 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 $\sqrt{x}\text{and}\sqrt{⌊x⌋}=\sqrt{{n}^{2}+m}$ 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 $\sqrt{x}\text{and}\sqrt{⌊x⌋}=\sqrt{{n}^{2}-m}$ between n-1 and n-1, so both sides equal n