Thursday, 20 February 2014

Is this function bijective, surjective and injective?

:QZ with x:= floor of x.



I know a function is injective by using f(x1)=f(x2)x1=x2
and a function is surjective if each element of the codomain, yY, is the image of some element in the domain xX,
and bijective if the function is both injective and surjective.



I don't know what floor of x is.

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