Show that the sets R and (−1, 1) have the same cardinality by providing an explicit formula of abijection.

To show that the sets R and (-1, 1) have the same cardinality, we need to provide an explicit formula for a bijection between the two sets.

Let's define a function f: R -> (-1, 1) as follows:

f(x) = tan((π/2) * x)

To prove that f is a bijection, we need to show that it is both injective and surjective.

1. Injectivity: Let's assume that f(x1) = f(x2) for some x1, x2 in R. This means that tan((π/2) * x1) = tan((π/2) * x2). Since the tangent function is injective on the interval (-π/2, π/2), we can conclude that x1 = x2. Therefore, f is injective.

2. Surjectivity: For any y in (-1, 1), we can find an x in R such that f(x) = y. Let's solve for x in terms of y: tan((π/2) * x) = y (π/2) * x = arctan(y) x = (2/π) * arctan(y)

Since the arctan function maps (-1, 1) to (-π/2, π/2), we can see that x is in R. Therefore, f is surjective.

Since f is both injective and surjective, it is a bijection between R and (-1, 1). Thus, we have shown that the sets R and (-1, 1) have the same cardinality.