Floor function with Multiplier

Question

I understand this: $f(x) = \lfloor x-4 \rfloor$ but how do you handle this $f(x) = 4\cdot\lfloor x-4 \rfloor$ . Do you distribute the $4$ then floor?

and the brackets denote floor.

Thank you

Answer 1

Note that for every $x$ there is a unique $n$ so that $n \le x < n+1$.

And that means there is also a unique $r: 0 \le r < 1$ so that $x = n + r$.

$\lfloor x\rfloor = n$ and $\{x\} = x -\lfloor x\rfloor = r$.

And we always have $x = \lfloor x\rfloor + \{x\}$.

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Okay get that under our belt.

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For any integer $n$ we have

$\lfloor x- n\rfloor= \lfloor x \rfloor -n$ because......

$x-n = \lfloor x \rfloor +\{x\} - n= (\lfloor x \rfloor - n) + +\{x\}$ and $0\le \{x\} < 1$ and $(\lfloor x \rfloor - n)$ is an integer so

$(\lfloor x \rfloor - n)\le (\lfloor x \rfloor - n) +\{x\} =x-n < (\lfloor x \rfloor - n)+1$ so

$\lfloor x -n\rfloor =\lfloor x \rfloor - n$.

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So $4\lfloor x -4\rfloor= 4(\lfloor x \rfloor - 4)=4\lfloor x \rfloor -16$