Consider the function f : R2 → R2 defined by the formula f (x, y) = (xy, x3). Is f injective? Is itsurjective? Bijective? Explain

To determine if the function f : R2 → R2 defined by f (x, y) = (xy, x3) is injective, we need to check if different inputs map to different outputs.

Let's consider two different inputs, (x1, y1) and (x2, y2), such that f (x1, y1) = f (x2, y2).

Using the formula for f, we have (x1y1, x1^3) = (x2y2, x2^3).

From this equation, we can see that for f to be injective, it must be the case that x1y1 = x2y2 and x1^3 = x2^3.

From the first equation, we can divide both sides by y1 and y2 (assuming they are not zero) to get x1 = x2.

Substituting this into the second equation, we have x1^3 = x2^3, which implies x1 = x2.

Therefore, we can conclude that f is injective since different inputs always map to different outputs.

To determine if f is surjective, we need to check if every element in the codomain R2 has a preimage in the domain R2.

Let's consider an arbitrary element (a, b) in R2.

To find a preimage for (a, b), we need to solve the equation f (x, y) = (a, b), which gives us the system of equations xy = a and x^3 = b.

From the first equation, we can solve for y by dividing both sides by x (assuming x is not zero), giving us y = a/x.

Substituting this into the second equation, we have x^3 = b, which implies x = b^(1/3).

Therefore, we have found a preimage (b^(1/3), a/b) for the element (a, b) in R2.

Since we can find a preimage for every element in the codomain, we can conclude that f is surjective.

Finally, since f is both injective and surjective, we can conclude that f is bijective.