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Consider the identity function IN : N...

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.

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It can be easily seen that for 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`
`therefore (I_N+I_N)` is not onto.
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