Consider the identity function `I_N : N->N` defined as, `I_N(x)=x` for all `x in N` . Show that although `I_N` is onto but `I_N+I_N : N->N` defined as `(I_N+I_N)(x)=I_N(x)+I_N(x)=x+x=2x` is not onto.

Explanation

From given `I_N` is onto

Here we can find an element 3 in the codomain N

Such that there does not exist any x in the domain N with `(I_N+I_N)(x)`=`2x`=3

`:.` `(I_N+I_N)` is not onto.