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In Velleman's "Calculus: a Rigorous Course," Example 9 from Section 1.3 tasks us with computing $ h(h(x)) $, where $ h(x) = \lfloor 5x - 2 \rfloor $.

My initial solution:

\begin{align*} h(\lfloor 5x - 2 \rfloor) &= \lfloor 5(\lfloor 5x \rfloor - 2) - 2 \rfloor \\ &= \lfloor 5\lfloor 5x \rfloor - 10 \rfloor - 2 \\ &= \lfloor 5\lfloor 5x \rfloor \rfloor - 12 \end{align*}

However, the provided solution from the book is:

$ 5\lfloor 5x \rfloor - 12 $

How can I get the correct solution ?

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    $\begingroup$ The output of the ceiling function is an integer; an integer multiplied by $5$ is again an integer. $\endgroup$
    – Mauro ALLEGRANZA
    Commented Mar 26 at 13:35
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    $\begingroup$ Thank you, Mauro ! $\endgroup$
    – F. Zer
    Commented Mar 26 at 13:43
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    $\begingroup$ @MauroALLEGRANZA That's the floor function, not the ceiling function. Not that it matters here. $\endgroup$
    – jjagmath
    Commented Mar 26 at 16:38
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    $\begingroup$ If $M$ is an integer then $[M]$. And as $[5x]$ is an integer so is $5[5x]$. $\endgroup$
    – fleablood
    Commented Mar 26 at 16:48

1 Answer 1

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You almost got it!

$\begin{align*} h(\lfloor 5x - 2 \rfloor) &= \lfloor 5(\lfloor 5x \rfloor - 2) - 2 \rfloor \\ &= \lfloor 5\lfloor 5x \rfloor - 10 \rfloor - 2 \\ &= \lfloor 5\lfloor 5x \rfloor \rfloor - 12 \end{align*}$

Then, since $\lfloor 5x \rfloor $ is always an integer we get that $5 \times \lfloor 5x \rfloor $ is also an integer so the outer floor is not needed indeed

So we get $h(h(x))= 5\lfloor 5x \rfloor - 12$

So we can say $\lfloor a\lfloor b \rfloor \rfloor = a \times \lfloor b \rfloor $ , when $a$ is an integer .

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