Suppose that R is the set of real numbers. Let f: R--> R be defined by f(x)= mx+b, where m and b are real numbers and m is nonzero. Prove that f is a bijection.
A bijective function is a function that is both one-to-one (injective) and onto (surjective).
To prove that f(x)=mx+b is bijective requires two parts, f(x) is one-to-one and f(x) is onto.
Here are some example references that can help you prove that f(x) is one-to-one (injective), and f(x) is onto (surjective). If you can prove these two properties, then f(x) is bijective.
Definitions of injection (one-to-one) and surjection (onto):
http://en.wikipedia.org/wiki/Injective_function
http://en.wikipedia.org/wiki/Surjective_function
Example proofs:
http://www.math.csusb.edu/notes/proofs/bpf/node4.html
http://www.math.csusb.edu/notes/proofs/bpf/node5.html
You're welcome to post your proof for review.
To prove that a function is a bijection, we need to show that it is both injective (one-to-one) and surjective (onto).
1. Injectivity: To show that f is injective, we need to prove that for any two distinct real numbers a and b in the domain of f, their images under f are also distinct. In other words, if f(a) = f(b), then a = b.
Let's assume f(a) = f(b), i.e., ma + b = mb + b. Subtracting b from both sides, we get ma = mb. Since m is nonzero, we can divide both sides by m, resulting in a = b. This proves that f is injective.
2. Surjectivity: To show f is surjective, we need to prove that for any real number y in the codomain of f, there exists at least one real number x in the domain of f such that f(x) = y.
Let y be any real number. We want to find an x such that f(x) = y. Since f(x) = mx + b, we have mx + b = y. Rearranging the equation, we get mx = y - b. Since m is nonzero, we can divide both sides of the equation by m, resulting in x = (y - b)/m. Therefore, for any real number y, we can find an x such that f(x) = y. This proves that f is surjective.
Since f is both injective and surjective, it is a bijection.