Question

let F be a field. Define f on F by f(x)=−x, and define g on F−{0} by g(x)=x－¹.Prove that f is a bijection from F to F and that g is a bijection from F−{0} to F−{0}

Answer 1

The function f(x) = -x is shown to be bijective on the field F by proving it is both injective and surjective. Similarly, the function g(x) = x⁻¹ is proven to be bijective on the set F - {0} by the same properties. Demonstrating that every element has a unique image and preimage in their respective domains and codomains establishes these bijections.

Step-by-step explanation:

To prove that f is a bijection from F to F, where f(x) = -x, we must show that f is both injective (one-to-one) and surjective (onto). For injectivity, assume f(a) = f(b) for some a, b in F. Then, -a = -b, and by applying the additive inverse, we have a = b, proving injectivity. To show surjectivity, for any y in F, let x = -y. Then f(x) = -(-y) = y, which means every element in F has a preimage, hence f is surjective.

Similarly, to prove that g is a bijection from F - {0} to F - {0}, where g(x) = x⁻¹, again we need to prove injectivity and surjectivity. For injectivity, assume g(a) = g(b) for some a, b in F - {0}. Then, a⁻¹ = b⁻¹. Multiplying both sides by ab (since neither a nor b is zero), we get b = a, proving injectivity. For surjectivity, for any y in F - {0}, let x = y⁻¹. Then g(x) = (y⁻¹)⁻¹ = y, showing every element in F - {0} has a preimage, so g is surjective.