Hi everyone.

Show that $\displaystyle \forall{x\in{\mathbb{Z^+}\cup{0}}}$ then $\displaystyle [\frac{n}{2}]+[-\frac{n}{2}]=n$

Thanks so much.

Printable View

- Apr 6th 2012, 04:55 PMFernandoInteger part
Hi everyone.

Show that $\displaystyle \forall{x\in{\mathbb{Z^+}\cup{0}}}$ then $\displaystyle [\frac{n}{2}]+[-\frac{n}{2}]=n$

Thanks so much. - Apr 6th 2012, 11:17 PMSylvia104Re: Integer part
You mean $\displaystyle \forall n\in\mathbb Z^+\cup\{0\},$ $\displaystyle \left\lfloor\frac n2\right\rfloor+\left|\left\lfloor-\frac n2\right\rfloor\right|=n?$

If $\displaystyle n$ is even, say $\displaystyle n=2k,$ $\displaystyle \left\lfloor\frac n2\right\rfloor+\left|\left\lfloor-\frac n2\right\rfloor\right|=k+\left|-k\right|=2k=n.$

If $\displaystyle n$ is odd, say $\displaystyle n=2k+1,$ $\displaystyle \left\lfloor\frac n2\right\rfloor+\left|\left\lfloor-\frac n2\right\rfloor\right|=\lfloor k+\tfrac12\rfloor+\left|\lfloor-k-\tfrac12\rfloor\right|=k+|-k-1|=2k+1=n.$ - Apr 7th 2012, 12:54 PMFernandoRe: Integer part
- Apr 7th 2012, 01:03 PMSylvia104Re: Integer part
Without absolute value, $\displaystyle \left\lfloor\frac n2\right\rfloor+\left\lfloor-\frac n2\right\rfloor=0$ if $\displaystyle n$ is even and $\displaystyle -1$ if $\displaystyle n$ is odd.

- Apr 9th 2012, 11:34 AMFernandoRe: Integer part
Thanks, maybe the exercise has a little mistake.

- Apr 9th 2012, 11:53 AMPlatoRe: Integer part
- Apr 10th 2012, 02:50 AMFernandoRe: Integer part