Section 1.3: Functions

Solution
\(f\) is one-one, as \(f(x_1) = f(x_2) \Rightarrow 2x_1 = 2x_2 \Rightarrow x_1 = x_2\).
Also, given any real number \(y\) in \(\mathbb{R}\), there exists \(\frac{y}{2}\) in \(\mathbb{R}\) such that
\[f\left( \frac{y}{2} \right) = 2 \cdot \left( \frac{y}{2} \right) = y\]
Hence, \(f\) is onto.