Question

Asked Nov 1, 2019

Step 1

**4(i)**Consider the given mapping F: N→N defined by

F(n)=2-3n , for all natural numbers n belongs to N

for, F to be a function, for each n there will be exactly one value of F(n),

for example: n=1 gives F(n)=-1, n=2 gives F(n)=-4 but -1 and -4 are not natural numbers,

so, for each n there does not exist value of F(n) on the defined codomain.

Hence, **F(n) is not a function**.

Step 2

**(ii)**Now, when the codomain if F is modified to be Z (the set of all integers) , for each n there will be exactly one value of F(n) on defined codomain.

Hence, **F(n) is a function**.

So, here F:N→Z defined by F(n)=2-3n

Now, check injectivity:

Let n_{1} and n_{2} belongs to N such that

F( n_{1})= F( n_{2})

2-3n_{1}=2-3n_{2}

n_{1}= n_{2}

Thus, **F is one-to-one function.**

Now, check surjectivity:

It is known that a function f is surjective if for every y there exist an x such that f(x)=y,

Now, here, let y=F(n)=0 from Z, then

2-3n=0

n=2/3, which is not a natural number.

So, for every y there does not exist an x such that f(x)=y,

Hence, **F is not surjective.**

Step 3

**(iii)**

Now, Consider the given mapping G: R→R defined by

G(x)=2-3x ,

Now, let y=G(x) from R,then

y=2-3x

3x=2-y

x=(2-y)/3

Now, G(x)=G((2-y)/3)

=2-3[(2-y)/3]

...

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