a)
f(n)=n−1,n∈Z Let f(n1)=f(n2),n1,n2∈Z. It means that
n1−1=n2−1
n1=n2 The function f(n)=n−1 is one-to-one from Z to Z.
Let y=n−1,y∈Z. Then
n=y+1We see that ∃n∈Z ∀y∈Z.
The function f(n)=n−1 is onto from Z to Z.
The function f(n)=n−1 is one-to-one and onto from Z to Z.
(b)
f(1)=⌈1/2⌉=1
f(2)=⌈2/2⌉=1
1=2, but f(1)=1=f(2) The function f(n)=⌈n/2⌉ is not one-to-one from Z to Z.
Let y=k,y∈Z. Take n=2k,n∈Z. Then
⌈n/2⌉=⌈2k/2⌉=⌈k⌉=kWe see that ∀y∈Z ∃n∈Z such that ⌈n/2⌉=y.
The function f(n)=⌈n/2⌉ is onto from Z to Z.
The function f(n)=⌈n/2⌉ is onto but is not one-to-one from Z to Z.
(c)
f(1)=(1)2+1=2
f(−1)=(−1)2+1=2
−1=1, but f(−1)=2=f(1) The function f(n)=n2+1 is not one-to-one from Z to Z.
∀n∈Z the value of f(n)=n2+1 is non negative integer value.
Let y=−1,y∈Z.
We cannot find n∈Z such that f(n)=n2+1=−1.
The function f(n)=n2+1 is not onto from Z to Z.
The function f(n)=n2+1 is not ontoand is not one-to-one from Z to Z.
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