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Consider the identity function I_(N):N to N defined as I _(N) (x)=x AA x in N. Show that although I _(N) is onto but I _(N) + I _(N) : N to Ndefined as (I _(N) + I _(N)) (x) = I _(N) (x) + I _(N) (x) x +x = 2xis not onto. |
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