Question

Is g(n)=n-7 where g: R->R injective, surjective, or bijective

Answer 1

\(g(n) = n-7, g:R \to R\)

Let \(n_1 \ne n_2 \in R\)

Then 

\(n_1 - 7\ne n_2 - 7\)

⇒ \(g(n_1) \ne g(n_2) \,\forall\, n_1, n_2 \in R\)

\(\therefore g(n)\) is injective.

For every \(n \in R, \exists\, n-7\in R\) such that \(g(n) = n-7 \in R\)

\(\therefore g(n)\) is surjective also.

Hence, g(n) is a bijective function.