Question Number 175470 by mnjuly1970 last updated on 31/Aug/22
Answered by mahdipoor last updated on 31/Aug/22
$${get}\:\:\:\frac{{n}}{\mathrm{2}}\leqslant{x}<\frac{{n}+\mathrm{1}}{\mathrm{2}}\:\:\:\:{n}\in{Z} \\ $$$$\Rightarrow{n}\leqslant{y}={x}+\frac{{n}}{\mathrm{2}}<{n}+\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\left[{y}\right]={n}= \\ $$$$\Rightarrow{y}−\frac{\mathrm{1}}{\mathrm{2}}\left[{y}\right]=\left({x}+\frac{{n}}{\mathrm{2}}\right)−\left(\frac{{n}}{\mathrm{2}}\right)={x} \\ $$$$\Rightarrow\Rightarrow{f}^{\:−\mathrm{1}} \left({x}\right)={x}−\frac{\left[{x}\right]}{\mathrm{2}} \\ $$
Commented by mnjuly1970 last updated on 31/Aug/22
$${zendeh}\:{bashid}\:{mamnoonam} \\ $$$${jenabe}\:{mahdipoor} \\ $$
Commented by mahdipoor last updated on 31/Aug/22
$$\heartsuit \\ $$