This Floor Function Is Slippery!

Algebra Level 1

For all $x \ge 0$ , and $x \in \mathbb {\text{R}}$ , is it true that

$\left \lfloor \sqrt{\lfloor x \rfloor}\right \rfloor = \left \lfloor \sqrt{x} \right \rfloor$

True False

1 solution

Leonardo Vannini
Mar 4, 2016

If $x$ is a perfect square ( $x={ n }^{ 2 }$ ) we have that $\left\lfloor \sqrt { \left\lfloor { n }^{ 2 } \right\rfloor } \right\rfloor =\left\lfloor \sqrt { { n }^{ 2 } } \right\rfloor$ which it's obv true.
If $x$ is a real number such that ${ n }^{ 2 }<x<{ (n+1) }^{ 2 }$ we can say that $x={ n }^{ 2 }+a$ with $a<({ n+1) }^{ 2 }-{ n }^{ 2 }=2n+1$ ; in this case we have $\left\lfloor \sqrt { \left\lfloor { n }^{ 2 }+a \right\rfloor } \right\rfloor =\left\lfloor \sqrt { { n }^{ 2 } } \right\rfloor =n$ and $\left\lfloor \sqrt { { n }^{ 2 }+a } \right\rfloor =\sqrt { { n }^{ 2 } } =n$ because $\sqrt { { n }^{ 2 }+a } <n+1$