Algebraic Structures | Solving the Matrix Commutator MX = XM | f:Q→Z, defined by f(a/b)=a

Algebraic Structures | Question-5(a)

Algebraic Structures | Question-5

Determine whether the following function is well defined:

\[ f: \mathbb{Q} \to \mathbb{Z}, \quad \text{defined by} \quad f\left(\frac{a}{b}\right) = a \]

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:

Table of Contents.

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: Test with Counterexample

Consider two equivalent fractions representing the same rational number:

Counterexample

\[ \frac{1}{2} = \frac{2}{4} \quad (\text{since } 1 \cdot 4 = 2 \cdot 2) \]

Step 2: Apply Function to Both Representations

Applying the function definition:

\[ f\left(\frac{1}{2}\right) = 1, \quad f\left(\frac{2}{4}\right) = 2. \]

⚠ Critical Observation: We get different outputs (\(1 \neq 2\)) for the same input in \(\mathbb{Q}\)!

Step 3: General Proof of Failure

Let \(\frac{a}{b} = \frac{c}{d}\) in \(\mathbb{Q}\), which means \(ad = bc\).

By the function definition:

\[ f\left(\frac{a}{b}\right) = a, \quad f\left(\frac{c}{d}\right) = c. \]

For the function to be well-defined, we would need \(a = c\) whenever \(ad = bc\).

However, the counterexample shows:

\[ 1 \cdot 4 = 2 \cdot 2 \quad \text{but} \quad 1 \neq 2. \]

Step 4: Fundamental Issue

The function \(f(a/b) = a\) depends on the specific representation of the rational number, not on its value.

Since rational numbers have infinitely many equivalent representations \(\frac{a}{b} = \frac{ka}{kb}\) for any integer \(k \neq 0\), the function gives different outputs:

\[ f\left(\frac{a}{b}\right) = a \quad \text{but} \quad f\left(\frac{ka}{kb}\right) = ka \]

Unless \(k = 1\), these are different integers.

Conclusion

The function \(f\) is NOT well-defined because different representations of the same rational number produce different outputs.

The function violates the fundamental requirement that each input in \(\mathbb{Q}\) must correspond to exactly one output in \(\mathbb{Z}\).

\[ \boxed{\text{No}} \]