Show that the mapping 𝑓: 𝑅 → 𝑅, 𝑓(𝑥) =𝑥, 𝑥 ≠0 𝑎𝑛𝑑 𝑥 ∈ 𝑅 is one one onto , where R is the set of non zero real numbers

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.

Show that the mapping f: R → R, f(x) = x, x ≠ 0 and x ∈ R is one-to-one and onto, where R is the set of non-zero real numbers.

Explore how to prove that the function f(x) = x, where x belongs to the set of non-zero real numbers, is both injective (one-to-one) and surjective (onto). This solution covers core concepts in functions, suitable for students in Grades 10-12.

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