# The floor function's relationship with odd and even functions

Question

Suppose $f: \Re \to \Re$ is a real valued function defined on the whole real line. For each a) and b) determine if the statement is correct and justify your answer.

a) If $f(x)$ is even then $g(x)= \left \lfloor{f(x)}\right \rfloor$ even.

b) If $f(x)$ is odd then $g(x)= \left \lfloor{f(x)}\right \rfloor$ odd.

Things I know

I know that the floor function is the greatest integer less than the value x.

$f(x)=f(-x)$ if $f(x)$ is even and the graphs are symmetrical about y=0

$f(x)=-f(-x)$ if $f(x)$ is odd

I have tried to think of a way, but with the knowledge I dont know how to come to the conclusion if the statements are correct and to justify. But my knowledge is up to date with the school syllabus (what we have been taught already). How do I do this?

• No.how about $f(x)=\sin x$ – Booldy Aug 22 '15 at 13:11
• @Booldy If $f(x)=\sin x$ then $f(x)$ is odd, and $g(x)$ is odd? So in that example its yes for b part. – M.S.E Aug 22 '15 at 13:16
• In "Things I know" it should be $f(x)=-f(-x)$ if $f(x)$ is odd. Personally I think it's better to write $f(-x)=-f(x)$ if $f(x)$ is odd (and better right $f$ instead of $f(x)$ when talking about the function not the value of the image of $x$). – Scientifica Aug 22 '15 at 13:16
• @Scientifica oh yes my bad :) Corrected it – M.S.E Aug 22 '15 at 13:18
• M.S.E $g(x)$ i not odd for $f(x)=\sin x$ – Booldy Aug 22 '15 at 13:19

## 1 Answer

b) is incorrect. The easiest counter example is taking $f(x)=x$. Booldy's counter example is correct too because if $f(x)=\sin x$ then $g(x)=\lfloor\sin (-x)\rfloor=\lfloor -\sin x\rfloor$ but nothing ensures that $\lfloor -\sin x\rfloor=-\lfloor \sin x\rfloor$. Take $x=\dfrac{\pi}{4}$ to be convinced.

For a) just use the definition you have. Let $x\in\mathbb{R}$.$f$ is even so $f(-x)=f(x)$. Now use this while calculating $g(-x)$.

• b) $x=\pi/2$ is one of the few points where $\lfloor-\sin x\rfloor=-\lfloor\sin x\rfloor$ – Empy2 Aug 22 '15 at 13:23
• @Michael totally out of topic: oh your from Adelaide, me too :D – M.S.E Aug 22 '15 at 13:25
• Im reading your answer :) – M.S.E Aug 22 '15 at 13:25
• :o my bad! I mean $\frac{\pi}{4}$ not $\frac{\pi}{2}$. So take this in consideration M.S.E. +1 Micheal – Scientifica Aug 22 '15 at 13:26
• @M.S.E off topic: You're embarassing me ^^ xD I'm not :p but thank you! – Scientifica Aug 22 '15 at 13:57