Question

Consider the function \(f: \mathbb{R} \times \mathbb{N} \rightarrow \mathbb{N} \times \mathbb{R}\) defined as \(f(x, y)=(y, 3 x y) .\) Check that this is bijective; find its inverse.

Short answer

The function \(f(x, y)=(y, 3 x y)\) is bijective and its inverse is given by \(f^{-1}(a, b) = (\frac{b}{3a}, a)\).

Step-by-step solution

Prove injectiveness

Let's take two pairs \((x_1, y_1)\) and \((x_2, y_2)\) from \(\mathbb{R} \times \mathbb{N}\). Assume that \(f(x_1, y_1) = f(x_2, y_2)\), thus \((y_1, 3x_1y_1) = (y_2, 3x_2y_2)\). For these two pairs to be equal, both the first and second elements from each pair must be equal. This gives us two equations: \(y_1 = y_2\) and \(3x_1y_1 = 3x_2y_2\). Since \(y_1 = y_2\), it means \(3x_1y_1 = 3x_1y_2\), and simplifying further, it can be concluded that \(x_1 = x_2\). Therefore the function is injective.

Prove surjectiveness

A function is surjective if for every pair \((a, b)\) in the codomain, that is, in \(\mathbb{N} \times \mathbb{R}\), there exists a pair \((x, y)\) in the domain, such that \(f(x, y) = (a, b)\). We can see that by choosing \(x = \frac{b}{3a}\) and \(y = a\), where \((a, b) \in \mathbb{N} \times \mathbb{R}\), it results in \(f(x, y) = (a, 3 * (\frac{b}{3a}) * a) = (a, b)\). Hence, the function is surjective.

Find the inverse function

The inverse of a function reverses the direction of the mapping. The inverse of \(f: \mathbb{R} \times \mathbb{N} \rightarrow \mathbb{N} \times \mathbb{R}\), denoted as \(f^{-1}\), will map from \(\mathbb{N} \times \mathbb{R}\) to \(\mathbb{R} \times \mathbb{N}\). From the previous steps, we know that if \(f(x, y) = (a, b)\), you can solve for \(x\) and \(y\) in terms of \(a\) and \(b\) as \(x = \frac{b}{3a}\) and \(y = a\). Therefore, the inverse function \(f^{-1}(a, b) = (\frac{b}{3a}, a)\).