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The mapping f : N to N given by f(n)=1+n...

The mapping f : `N to N` given by `f(n)=1+n^(2), n in N`, where N is the set of natural numbers, is

A

one-one and onto

B

onto but not one-one

C

one-one but not onto

D

Neither one-one nor onto

Text Solution

Verified by Experts

The correct Answer is:
C
Since, `" "f(n)=1+n^(2)`
For one-one, `1+n_(1)^(2)=1+n_(2)^(2)`
`implies" "n_(1)^(2)-n_(2)^(2)=0`
`implies" "n_(1)=n_(2)" "[becausen_(1)+n_(2)cancelimplies0]`
So, f(n) is one-one. But f(n) is not onto as some element of codomain is not the image of any element of domains.
Hence, f(n) is one-one but not onto.
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