Algebraic Structures | Question-5(b)
Algebraic Structures | Question-6
Determine whether the following function is well defined:
\[
f: \mathbb{Q} \to \mathbb{Q}, \quad \text{defined by} \quad f\left(\frac{a}{b}\right) = \frac{a^2}{b^2}
\]
where \(a, b \in \mathbb{Z}, b \neq 0\), and \(\frac{a}{b}\) represents a rational number.
Question: Is \(f\) well-defined?
Solution
Definition of Well-Defined Function
A function is well-defined if:
- Every input in the domain produces exactly one output
- If an input has multiple representations, all representations give the same output
For rational numbers \(\mathbb{Q}\), this means:
\[
\text{If } \frac{a}{b} = \frac{c}{d} \text{ in } \mathbb{Q} \text{ (i.e., } ad = bc\text{)}, \text{ then } f\left(\frac{a}{b}\right) = f\left(\frac{c}{d}\right).
\]
Step 1: Understanding the Function
The function \(f\) takes a rational number \(\frac{a}{b}\) and returns \(\frac{a^2}{b^2}\).
We need to check if different representations of the same rational number give the same output. For example:
Test Case
\[
\frac{1}{2} = \frac{2}{4} = \frac{3}{6}
\]
Step 2: Apply Function to Equivalent Representations
Applying the function to our test case:
\[
f\left(\frac{1}{2}\right) = \frac{1^2}{2^2} = \frac{1}{4}
\]
\[
f\left(\frac{2}{4}\right) = \frac{2^2}{4^2} = \frac{4}{16} = \frac{1}{4}
\]
\[
f\left(\frac{3}{6}\right) = \frac{3^2}{6^2} = \frac{9}{36} = \frac{1}{4}
\]
✓ Observation: All equivalent representations give the same output \(\frac{1}{4}\)!
Step 3: General Algebraic Proof
Let \(\frac{a}{b} = \frac{c}{d}\) in \(\mathbb{Q}\), which means \(ad = bc\).
We need to prove that \(f\left(\frac{a}{b}\right) = f\left(\frac{c}{d}\right)\):
\[
f\left(\frac{a}{b}\right) = \frac{a^2}{b^2}, \quad f\left(\frac{c}{d}\right) = \frac{c^2}{d^2}
\]
Starting from \(ad = bc\), we square both sides:
\[
(ad)^2 = (bc)^2
\]
\[
a^2 d^2 = b^2 c^2
\]
Divide both sides by \(b^2 d^2\) (both nonzero since \(b, d \neq 0\)):
\[
\frac{a^2}{b^2} = \frac{c^2}{d^2}
\]
Step 4: Verification of Well-Definedness
The equality \(\frac{a^2}{b^2} = \frac{c^2}{d^2}\) shows that:
\[
f\left(\frac{a}{b}\right) = f\left(\frac{c}{d}\right)
\]
This holds for all equivalent representations of rational numbers.
✓ Key Insight: The function \(f\left(\frac{a}{b}\right) = \frac{a^2}{b^2}\) is actually the squaring function \(f(x) = x^2\) on \(\mathbb{Q}\), which is a legitimate, single-valued function.
Conclusion
The function \(f\) is WELL-DEFINED because:
- For \(\frac{a}{b} = \frac{c}{d}\), we have \(\frac{a^2}{b^2} = \frac{c^2}{d^2}\)
- Different representations of the same rational number produce identical outputs
- The function is equivalent to \(f(x) = x^2\) on \(\mathbb{Q}\)
Therefore, the function satisfies all requirements for being well-defined.
\[
\boxed{\text{Yes}}
\]
