Answer
Because \(f\) is both injective and surjective, it is a **bijection**. Consequently, \((0,1)\) and \((a,b)\) have the **same cardinality**.
Work Step by Step
To show that \(S = (0,1)\) and \(W = (a,b)\) (with \(a < b\)) have the same cardinality, it suffices to construct a **bijection** between them. A simple linear map works perfectly:
\[
f \colon (0,1) \;\to\; (a,b),
\quad
f(x) \;=\; a \;+\; (b - a)\,x.
\]
---
## 1. Injectivity
If \(f(x_1) = f(x_2)\), then
\[
a + (b - a)\,x_1 \;=\; a + (b - a)\,x_2
\;\;\Longrightarrow\;\;
(b - a)\,x_1 \;=\; (b - a)\,x_2
\;\;\Longrightarrow\;\;
x_1 \;=\; x_2,
\]
so \(f\) is **one-to-one**.
---
## 2. Surjectivity
Given any \(y \in (a,b)\), we want \(x \in (0,1)\) such that \(f(x) = y\). Solve for \(x\):
\[
y \;=\; a + (b-a)\,x
\;\;\Longrightarrow\;\;
x \;=\; \frac{y - a}{b - a}.
\]
Since \(y\) is strictly between \(a\) and \(b\), \(\frac{y - a}{b - a}\) lies strictly between 0 and 1. Thus **every** \(y \in (a,b)\) is obtained from **some** \(x \in (0,1)\), so \(f\) is **onto**.
