Check the injectivity and surjectivity of the following functions:(i) f:N→N given by f(x)=x2(ii) f:Z→Z given by f(x)=x2(iii) f:R→R given by f(x)=x2(iv)f:N→N given by f(x)=x3(v)f:z→Z given by f(x)=x3
(i) f:N→N is given by,f(x)=x2 It is seen that for x,y∈N,f(x)=f(y)⇒x2=y2⇒x=y∴f is injective.Now, 2∈N. But, there does not exist any x in N such that f(x)=x2=2∴f is not surjective.Hence, function f is injective but not surjective.(ii) f:Z→Z is given by,f(x)=x2It is seen that f(−1)=f(1)=1, but −1=1∴f is not injective.Now, −2∈Z. But, there does not exist any element x∈Z such that f(x)=x2=−2.∴f is not surjective.Hence, function f is neither injective nor surjective.(iii) f:R→R is given by,f(x)=x2It is seen that f(−1)=f(1)=1, but −1=1∴f is not injective.Now, −2∈R. But, there does not exist any element x∈R such that f(x)=x2=−2. ∴f is not surjective.Hence, function f is neither injective nor surjective.(iv) f:N→N given by,f(x)=x3It is seen that for x,y∈N,f(x)=f(y)⇒x3=y3⇒x=y∴f is injective.Now, 2∈N. But, there does not exist any element x in domain N such that f(x)=x3= 2. ∴f is not surjectiveHence, function f is injective but not surjective.(v) f:Z→Z is given by,f(x)=x3It is seen that for x,y∈Z,f(x)=f(y)⇒x3=y3⇒x=y∴f is injective.Now, 2∈Z. But, there does not exist any element x in domain Z such that f(x)=x3=2.∴f is not surjective.Hence, function f is injective but not surjective.