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).

This means that (x1y1, x1^3) = (x2y2, x2^3).

From this equation, we can see that x1y1 = x2y2 and x1^3 = x2^3.

Since x1y1 = x2y2, we can conclude that either x1 = x2 or y1 = y2.

If x1 = x2, then x1^3 = x2^3, which implies that y1 = y2.

Similarly, if y1 = y2, then x1y1 = x2y2, which implies that x1^3 = x2^3.

Therefore, in both cases, we have x1 = x2 and y1 = y2.

Hence, the function f is injective because different inputs always map to different outputs.

To determine if the function 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 as y = a/x.

Substituting this into the second equation, we get x^3 = b/a.

Taking the cube root of both sides, we find x = (b/a)^(1/3).

Substituting this value of x back into the first equation, we get y = a/[(b/a)^(1/3)].

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

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

Finally, since the function f is both injective and surjective, it is bijective.