Question #175381

Let f: Z®Z be such that f(x) = x +1.

Is f invertible? and if it is, what is its inverse?


1
Expert's answer
2021-03-29T08:58:42-0400

1) The map ff is invertible if and only if ff is bijective if and only if ff is surjective and injective.

ff is surjective because yZxZ:f(x)=x+1=y(x=y1)\forall y \in \mathbb{Z} \exists x \in \mathbb{Z}: f(x) = x + 1 = y (x = y - 1)

ff is injective because x1Z,x2Z,x1x2\forall x_1 \in \mathbb{Z}, \forall x_2 \in \mathbb{Z}, x_1 \neq x_2f(x1)=x1+1x2+1=f(x2)f(x_1) = x_1 + 1 \neq x_2 + 1 = f(x_2)

Thus, ff is invertible.

2) The map f1:ZZf ^{-1}: \mathbb{Z} \rightarrow \mathbb{Z} is inverse to ff if f1f=ff1=eZf^{-1} f = f f^{-1} = e_{\mathbb{Z}} , where eZe_{\mathbb{Z}} is the identical map ZZ\mathbb{Z} \rightarrow \mathbb{Z} .

f1(y)=y1f^{-1}(y) = y - 1 because f1f(x)=(x+1)1=x==(x1)+1=ff1(x)f^{-1}f(x) = (x + 1) - 1 = x = \\ = (x - 1) + 1 = ff^{-1}(x)


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