Consider the identity function I_N: N N defined as, I_N (x) = x x… - Vidyadip

Question

Consider the identity function $I_N: N \rightarrow N$ defined as, $ I_N (x) = x \forall x \in N.$
Show that although $I_N$ is onto but $I_N + I_N : N \rightarrow N$ defined as
$(I_N + I_N) (x) = I_N (x) + I_N (x) = x + x = 2x$ is not onto.

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Answer

We know that the identity function on a given set is always a bijection.
Therefore,$ I_N: N \rightarrow N $ is onto. We have,
$(I_N + I_N) (x) = 2x$ for all $x \in N$
This means that under $I_N + I_N$, images of natural numbers are even natural numbers.
So, odd natural numbers in $N$ (co-domain) do not have their pre-images in domain $N.$
For example, $1, 3, 5$ etc. do not have their pre-images.
So, $ I_N + I_N : N \rightarrow$ $N$ is not onto.

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