We are asked to prove that the mapping $f: R \to R$, where $f(x) = x$ for all $x \in R$, is one-to-one (injective) and onto (surjective). Here, $R$ is the set of all non-zero real numbers.

1. Proving that $f$ is one-to-one (injective):

A function $f$ is injective if and only if for any two elements $x_1, x_2 \in R$, whenever $f(x_1) = f(x_2)$, it must follow that $x_1 = x_2$.

Let us assume that: $f(x_1) = f(x_2)$ By the definition of $f$, we know that: $x_1 = x_2$

Thus, the assumption that $f(x_1) = f(x_2)$ leads directly to $x_1 = x_2$. Therefore, $f$ is injective.

2. Proving that $f$ is onto (surjective):

A function $f$ is surjective if for every element $y \in R$, there exists an element $x \in R$ such that $f(x) = y$.

Let $y \in R$ be an arbitrary non-zero real number. We need to find an $x \in R$ such that: $f(x) = y$ By the definition of $f$, we have: $f(x) = x$ So, to satisfy $f(x) = y$, we can simply choose $x = y$. Since $y \in R$ (i.e., $y$ is a non-zero real number), $x = y$ is also a valid choice in $R$.

Thus, for every $y \in R$, we have found an $x \in R$ (namely, $x = y$) such that $f(x) = y$. Therefore, $f$ is surjective.

Conclusion:

The function $f: R \to R$, where $f(x) = x$ for all non-zero real numbers $x$, is both injective and surjective. Therefore, $f$ is a bijective function (both one-to-one and onto).

Would you like further details or have any questions?

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Related questions:

1. What is the difference between injective, surjective, and bijective functions?
2. Can you explain why $f(x) = x^2$ on $R$ is not injective?
3. How do you prove the inverse of a bijective function?
4. What happens if $f(x) = \frac{1}{x}$ for $x \neq 0$? Is it bijective?
5. Can a function be surjective but not injective? Give an example.

Tip: A bijection has an inverse function, meaning for every $y$ in the target set, there is a unique $x$ in the domain that maps to it.