Question Number 91588 by M±th+et+s last updated on 01/May/20
$${what}\:{is}\:{f}^{−\mathrm{1}} \:{for}\:{f}\left({x}\right)=\lfloor{x}\rfloor?? \\ $$
Commented by mr W last updated on 01/May/20
$${x}\rightarrow\lfloor{x}\rfloor\:{is}\:{a}\:{function},\:{but} \\ $$$$\lfloor{x}\rfloor\rightarrow{x}\:{is}\:{not}\:{a}\:{function},\:{because}\:{an} \\ $$$${element}\:{of}\:{set}\:\lfloor{x}\rfloor\:{is}\:{not}\:{associated} \\ $$$${with}\:{exactly}\:{one}\:{element}\:{of}\:{set}\:{x}. \\ $$
Commented by MJS last updated on 01/May/20
$${f}^{−\mathrm{1}} \:\mathrm{is}\:\mathrm{no}\:\mathrm{function} \\ $$$$\mathrm{same}\:\mathrm{problem}\:\mathrm{with}\:{g}\left({x}\right)={c};\:{c}\in\mathbb{R}.\:{g}^{−\mathrm{1}} \left({x}\right)=? \\ $$