Question

Check whether $f:\mathbb{Z}\times\mathbb{Z}\rightarrow\mathbb{Z}\times\mathbb{Z}$ defined by $f(x,y)=(2y,3x)$ is injective or not.

Hint:

To prove a function is injective, assume equal outputs and show that the inputs must also be equal.

Answer 1

Step 1: Recall the definition of injective function.
A function \(f:A\rightarrow B\) is called injective (one-one) if \[ f(x_1,y_1)=f(x_2,y_2) \] implies \[ (x_1,y_1)=(x_2,y_2) \] Thus, equal outputs must correspond to equal inputs.
Step 2: Assume equal images.
Suppose \[ f(x_1,y_1)=f(x_2,y_2) \] Then \[ (2y_1,3x_1)=(2y_2,3x_2) \] Step 3: Compare corresponding components.
From equality of ordered pairs, \[ 2y_1=2y_2 \] \[ 3x_1=3x_2 \] Divide the first equation by \(2\): \[ y_1=y_2 \] Divide the second equation by \(3\): \[ x_1=x_2 \] Step 4: Conclude equality of inputs.
Since \[ x_1=x_2 \] and \[ y_1=y_2 \] we get \[ (x_1,y_1)=(x_2,y_2) \] Step 5: Final conclusion.
Thus the function satisfies the condition of injectivity.
Final Answer:
\[ \boxed{\text{The function is injective}} \]