a) f: Z→Z,f(x)=x2+1
1) f(−1)=f(1)=2
f is not injective
2) Since f(x)=x2+1≥1 , there doesn’t exists any x∈Z such that f(x)=0
f is not surjective
b) g: N→N , g(x)=2x
1) If g(x)=g(y) , 2x=2y , then x=y
f is injective
2) g(x)=2x=1 only if x=0 , but 0∈N
f is not surjective
c) h: R→R , h(x)=5x−1
1) If h(x)=h(y) , 5x−1=5y−1 , 5x=5y , then x=y
f is injective
2) For all y∈R there exists x=5y+1 such that f(x)=5x−1=y+1−1=y
f is surjective
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