Question

Check whether \( f : Z \times Z \rightarrow Z \times Z \) (where \( Z \) is the set of integers) defined as \( f(x, y) = (2y, 3x) \) is injective or not.

Hint:

Injective checks focus on ensuring different inputs never map to the same output.
Linear transformations like \( (ay, bx) \) where \( a, b \neq 0 \) are always injective over numerical sets.

Answer 1

Concept:
• A function \( f: A \rightarrow B \) is injective (one-to-one) if \( f(a_1) = f(a_2) \) implies that \( a_1 = a_2 \) for all \( a_1, a_2 \in A \).
• For a mapping of ordered pairs, we assume \( f(x_1, y_1) = f(x_2, y_2) \) and check if it forces \( x_1 = x_2 \) and \( y_1 = y_2 \).

Step 1: Assume two inputs yield the same output
Let \( (x_1, y_1) \) and \( (x_2, y_2) \) be two elements in the domain \( Z \times Z \) such that:
\[ f(x_1, y_1) = f(x_2, y_2) \]

Step 2: Equate the corresponding components of the output pairs
Using the definition \( f(x, y) = (2y, 3x) \):
\[ (2y_1, 3x_1) = (2y_2, 3x_2) \]
This vector equality implies two scalar equations:
1) \( 2y_1 = 2y_2 \)
2) \( 3x_1 = 3x_2 \)

Step 3: Solve the resulting equations
From the first equation:
\[ 2y_1 = 2y_2 \implies y_1 = y_2 \]
From the second equation:
\[ 3x_1 = 3x_2 \implies x_1 = x_2 \]

Step 4: Verify the injective property
Since \( x_1 = x_2 \) and \( y_1 = y_2 \), it follows that the ordered pairs are identical:
\[ (x_1, y_1) = (x_2, y_2) \]
As \( f(x_1, y_1) = f(x_2, y_2) \) leads directly to \( (x_1, y_1) = (x_2, y_2) \), the function \( f \) is injective.