0

I need some guidance in helping me solve this proof. Here is the question:

Give a proof by cases that $\lfloor 4x \rfloor = \lfloor x \rfloor + \lfloor x + 1/4 \rfloor + \lfloor x + 1/2 \rfloor + \lfloor x + 3/4 \rfloor$

Jake0991
  • 71
  • Let's write $x=\lfloor x\rfloor+{x}$, where ${x}$ denotes the fractional part of $x$. I'd suggest considering four cases depending on this fractional part: when it's in $[0,1/4)$ or in $[1/4,1/2)$ or in $[1/2,3/4)$ or in $[3/4,1)$. – zipirovich Nov 23 '16 at 00:57
  • This could also be closed for lack of context. – Jyrki Lahtonen Feb 28 '19 at 16:01

1 Answers1

1

HINT: Let $\alpha=x-\lfloor x\rfloor$, the fractional part of $x$. Your four cases are $0\le\alpha<\frac14$, $\frac14\le\alpha<\frac12$, $\frac12\le\alpha<\frac34$, and $\frac34\le\alpha<1$. Investigate $\lfloor 4x\rfloor$ separately for each case; it may be helpful to write $x$ as $\lfloor x\rfloor+\alpha$.

Brian M. Scott
  • 616,228