On the set of integers Z. define f: Z → Z as f(n) = begincases n/2 textif n text is even 0 textif n text is odd endcases then 'f' is

Question

On the set of integers Z. define f: Z → Z as f(n) = begincases n/2 textif n text is even 0 textif n text is odd endcases then 'f' is (A) bijective (B) injective but not surjective (C) neither injective nor suijective (D) surjective but not injective

Tardigrade Admin · Accepted Answer

Given, f(n)= begincases(n/2), n text is even 0, n text is odd endcases Here, we see that for every odd values of z, it will give zero. It means that it is a many one function. For every even values of z, we will get a set of integers (-∞, ∞). So, it is onto. Hence, it is surjective but not injective.

Q. On the set of integers Z. define f:Z→Z as f(n)={n/20​if n is evenif n is odd​ then ′f′ is

bijective

injective but not surjective

neither injective nor suijective

surjective but not injective

Given, f(n)={2n​,0,​n is even n is odd ​ Here, we see that for every odd values of z, it will give zero. It means that it is a many one function. For every even values of z, we will get a set of integers (−∞,∞). So, it is onto. Hence, it is surjective but not injective.

Q. On the set of integers $Z$ . define $f : Z \to Z$ as $f (n) = {n /2 0 if n is even if n is odd$ then $^{'} f^{'}$ is