Answer
Because \(f\) is both injective and surjective, it is a **bijection**. Consequently, \((0,1)\) and \((2,5)\) have the **same cardinality**.
Work Step by Step
To show that the open intervals \(S = (0,1)\) and \(V = (2,5)\) have the same cardinality, you can construct a **bijection** between them. A convenient linear map is
\[
f : (0,1) \,\to\, (2,5),
\quad f(x) = 3x + 2.
\]
---
## 1. Injectivity
If \(f(x_1) = f(x_2)\), then
\[
3x_1 + 2 = 3x_2 + 2
\;\;\Longrightarrow\;\;
3x_1 = 3x_2
\;\;\Longrightarrow\;\;
x_1 = x_2.
\]
Thus, \(f\) is **one‐to‐one** (injective).
---
## 2. Surjectivity
For any \(y \in (2,5)\), we want an \(x \in (0,1)\) such that \(f(x) = y\). Solve for \(x\):
\[
y = 3x + 2
\quad\Longrightarrow\quad
x = \frac{y - 2}{3}.
\]
Since \(y\) is strictly between 2 and 5, \(\tfrac{y-2}{3}\) is strictly between 0 and 1. Therefore, every \(y \in (2,5)\) has a preimage in \((0,1)\). This shows \(f\) is **onto** (surjective).
