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.
This kind of "new wave" math teaching will probably make it even harder to learn algebra. I hope not.
I had never heard of a bijection before. So I looked up this website:
http://en.wikipedia.org/wiki/Bijection
It seems obvious that for every x there is a unique f(x), and for every f(x) there is a unique x. That fulfills the requirement for a bijection.
To prove that the function f(x) = mx + b is a bijection, we need to show two things: (1) that f is injective (or one-to-one) and (2) that f is surjective (or onto).
1. Injectivity: To prove that f is injective, we need to show that if f(x1) = f(x2), then x1 = x2. Let's assume that f(x1) = f(x2). Then, we have:
mx1 + b = mx2 + b
Subtracting b from both sides and dividing by m (since m is assumed to be nonzero), we get:
x1 = x2
Therefore, f is injective.
2. Surjectivity: To prove that f is surjective, we need to show that for every y in the codomain of f, there exists an x in the domain such that f(x) = y. The codomain of f is R (the set of real numbers), so let's take any y in R. We want to find an x such that f(x) = y.
We have f(x) = mx + b. Let's solve for x:
f(x) = y
mx + b = y
mx = y - b
x = (y - b)/m
Since m is assumed to be nonzero, (y - b)/m is defined for any y. Therefore, for any y in R, we can find an x in R such that f(x) = y.
Therefore, f is surjective.
Since f is both injective and surjective, we can conclude that f is a bijection.