In this post, we will prove that the function \( f: \mathbb{Z} \to \mathbb{Z} \) defined by \( f(x) = 2x \) is injective (one-one) but not surjective (onto). Here \( \mathbb{Z} \) represents the set of all integers.
Proof:
- \( f \) is Injective (One-One):
A function is injective if: \( f(a) = f(b) \implies a = b \)
Let \( f(a) = f(b) \). Then: \( 2a = 2b \)
Divide both sides by 2: \(a = b \)
Thus, \( f \) is injective.
- \( f \) is Not Surjective (Onto):
A function is surjective if every element in the codomain has a corresponding element in the domain.
Consider \( y = 1 \in \mathbb{Z} \). Suppose there exists \( x \in \mathbb{Z} \) such that: \(f(x) = 2x = 1 \).
Solving for \( x \): \( x = \frac{1}{2} \)
But \( \frac{1}{2} \notin \mathbb{Z} \). Hence, there is no integer \( x \) that maps to \( 1 \).
Since not every element in \( \mathbb{Z} \) is mapped to by \( f \), the function is not surjective.
Therefore, the function \( f(x) = 2x \) is injective but not surjective when defined over the integers, \( \mathbb{Z} \). Hence proved.
Please let me know in the comments if you find any errors in this solution.