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.
(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 n1 and n2 belongs to N such that
F( n1)= F( n2)
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
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.
Now, Consider the given mapping G: R→R defined by
Now, let y=G(x) from R,then
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