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Relations And Functions

Question
CBSEENMA12032510

Check the injectivity and surjectivity of the following functions.

f : Z → Z given by f(x) = x2

Solution

f : Z → Z given  
by f(x) = x2
It seen that for x,  straight y element of straight Z, f(x) = f(y) rightwards double arrow space space straight x cubed equals straight y cubed space rightwards double arrow space straight x equals straight y

therefore  f is injective 
Now, 2 space space element of Z but there does not exist any element x in domain Z such that f(x) = x3 = 2.
therefore    f is not surjective 
Hence, function f is injective but not surjective.