Consider the function \( f: \mathbb{R} \to \mathbb{R} \) defined by \( f(x) = x^2 \). Determine whether \( f \) is One-to-one (injective) or Onto (surjective). Justify your answers. Here, \( \mathbb{R} \) represents the set of all real numbers.
Solution:
- One-to-one (Injective)?
A function is One-to-one (injective) if \( f(a) = f(b) \implies a = b \)
Take \( a = 1 \) and \( b = -1 \):
Then, \( f(1) = (1)^2 = 1 \quad \text{and} \quad f(-1) = (-1)^2 = 1 \).
Here, \( f(1) = f(-1) \) but \( 1 \neq -1 \).
Thus, \( f \) is not One-to-one (injective) because different inputs map to the same output.
- Onto (Surjective)?
A function is Onto (Surjective) if for every \( y \in \mathbb{R} \), there exists \( x \in \mathbb{R} \) such that: \( f(x) = y \).
Let \( y = -1 \). We check if there exists \( x \in \mathbb{R} \) satisfying \( x^2 = -1 \). But \( x^2 \geq 0 \) for all real \( x \), so no solution exists.
Thus, \( f \) is not surjective because negative numbers (e.g., \( y = -1 \)) have no corresponding element in the domain \( x \in \mathbb{R} \).
Therefore, the function \( f(x) = x^2 \) (with domain and codomain \( \mathbb{R} ) \) is neither injective nor surjective.
Please let me know in the comments if you find any errors in this solution.