Question

Check whether the function \( f : \mathbb{Z} \times \mathbb{Z} \rightarrow \mathbb{Z} \times \mathbb{Z} \) defined as \( f(x, y) = (2y, 3x) \) is injective (one-to-one) or not.

Hint:

Even though this function is injective, it is not surjective over integers! For instance, look at the output pair \((1, 1)\). To get \(2y = 1\), \(y\) would have to be \(0.5\), which is not an integer. So it's an injection but not a bijection!

Answer 1

Concept: A function \( f \) is injective (one-to-one) if map values of distinct elements are always distinct. Mathematically, for any two element coordinates \( (x_1, y_1) \) and \( (x_2, y_2) \) in the domain: \[ f(x_1, y_1) = f(x_2, y_2) \quad \Rightarrow \quad (x_1, y_1) = (x_2, y_2) \] If this implication holds true universally across the entire domain, the function is classified as injective.

Step 1: Apply the definition of injectivity.
Let us assume that there are two pairs \( (x_1, y_1) \) and \( (x_2, y_2) \) belonging to the domain \(\mathbb{Z} \times \mathbb{Z}\) such that their functional outputs are exactly identical: \[ f(x_1, y_1) = f(x_2, y_2) \]

Step 2: Substitute the functional rules into the equation.
Using the definition of the function \( f(x, y) = (2y, 3x) \), rewrite both sides: \[ (2y_1, 3x_1) = (2y_2, 3x_2) \]

Step 3: Equate corresponding components from both ordered pairs.
For two ordered pairs to be completely equal, their corresponding individual components must be equal to each other. This gives us two separate independent algebraic equations:
• Equating the first components: \[ 2y_1 = 2y_2 \] Dividing both sides by 2: \[ y_1 = y_2 \quad \cdots (1) \]
• Equating the second components: \[ 3x_1 = 3x_2 \] Dividing both sides by 3: \[ x_1 = x_2 \quad \cdots (2) \]

Step 4: Combine the component results.
From equations (1) and (2), we simultaneously find that \( x_1 = x_2 \) and \( y_1 = y_2 \). This implies: \[ (x_1, y_1) = (x_2, y_2) \] Since \( f(x_1, y_1) = f(x_2, y_2) \) cleanly leads to the conclusion that \( (x_1, y_1) = (x_2, y_2) \), the function is verified to be injective.