Question #343351

Determine whether each of the following functions from Z to Z is one to one and onto.


a.f(n)=n-1


b.f(n)=[n/2]


C.f(n)=n²+1

1
Expert's answer
2022-05-24T23:18:40-0400

a)


f(n)=n1,nZf(n)=n-1, n\in \Z

Let f(n1)=f(n2),n1,n2Z.f(n_1)=f(n_2), n_1, n_2\in \Z. It means that


n11=n21n_1-1=n_2-1

n1=n2n_1=n_2

The function f(n)=n1f(n)=n-1 is one-to-one from Z\Z to Z.\Z.

Let y=n1,yZ.y=n-1,y\in \Z. Then


n=y+1n=y+1

We see that nZ yZ.\exist n\in \Z\ \forall y\in \Z.

The function f(n)=n1f(n)=n-1 is onto from Z\Z to Z.\Z.


The function f(n)=n1f(n)=n-1 is one-to-one and onto from Z\Z to Z.\Z.


(b)


f(1)=1/2=1f(1)=\lceil1/2\rceil=1

f(2)=2/2=1f(2)=\lceil2/2\rceil=1

12, but f(1)=1=f(2)1\not=2, \ but\ f(1)=1=f(2)

The function f(n)=n/2f(n)=\lceil n/2\rceil is not one-to-one from Z\Z to Z.\Z.


Let y=k,yZ.y=k,y\in \Z. Take n=2k,nZ.n=2k, n\in \Z. Then


n/2=2k/2=k=k\lceil n/2\rceil=\lceil 2k/2\rceil=\lceil k\rceil=k

We see that yZ nZ\forall y\in \Z\ \exist n\in\Z such that n/2=y.\lceil n/2\rceil=y.

The function f(n)=n/2f(n)=\lceil n/2\rceil is onto from Z\Z to Z.\Z.


The function f(n)=n/2f(n)=\lceil n/2\rceil is onto but is not one-to-one from Z\Z to Z.\Z.


(c)


f(1)=(1)2+1=2f(1)=(1)^2+1=2

f(1)=(1)2+1=2f(-1)=(-1)^2+1=2

11, but f(1)=2=f(1)-1\not=1, \ but\ f(-1)=2=f(1)

The function f(n)=n2+1f(n)=n^2+1 is not one-to-one from Z\Z to Z.\Z.


nZ\forall n\in \Z the value of f(n)=n2+1f(n)=n^2+1 is non negative integer value.

Let y=1,yZ.y=-1, y\in \Z.

We cannot find nZn\in \Z such that f(n)=n2+1=1.f(n)=n^2+1=-1.

The function f(n)=n2+1f(n)=n^2+1 is not onto from Z\Z to Z.\Z.


The function f(n)=n2+1f(n)=n^2+1 is not ontoand is not one-to-one from Z\Z to Z.\Z.

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