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: Definition of injectivity
A function $f$ is said to be injective (one-to-one) if for all $a, b \in \mathbb{Z}\times\mathbb{Z}$, whenever $f(a) = f(b)$, we must have $a = b$. In other words, $f(x_1, y_1) = f(x_2, y_2)$ implies $(x_1, y_1) = (x_2, y_2)$.

Step 2: Given function
The function is defined as $f(x, y) = (2y, 3x)$. So, for two points $(x_1, y_1)$ and $(x_2, y_2)$, we have: 
$f(x_1, y_1) = (2y_1, 3x_1)$ and $f(x_2, y_2) = (2y_2, 3x_2)$.

Step 3: Checking injectivity
Assume $f(x_1, y_1) = f(x_2, y_2)$. This gives: 
$(2y_1, 3x_1) = (2y_2, 3x_2)$. 
So, from the equality of the first components, we get: 
$2y_1 = 2y_2$ ⟹ $y_1 = y_2$. 
From the equality of the second components, we get: 
$3x_1 = 3x_2$ ⟹ $x_1 = x_2$.

Step 4: Conclusion
Since $x_1 = x_2$ and $y_1 = y_2$, we conclude that $(x_1, y_1) = (x_2, y_2)$. Hence, the function $f(x, y) = (2y, 3x)$ is injective.