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.
To prove that the function f: R → R defined by f(x) = mx + b is a bijection, we need to show that it is both injective (one-to-one) and surjective (onto).
1. Injectivity (One-to-One):
To show that f is injective, we need to demonstrate that if f(x₁) = f(x₂), then x₁ = x₂ for any x₁, x₂ ∈ R.
Let's assume f(x₁) = f(x₂):
mx₁ + b = mx₂ + b
By subtracting b from both sides:
mx₁ = mx₂
Since m is non-zero, we can divide both sides by m:
x₁ = x₂
Thus, we can conclude that f is injective, as for any x₁ and x₂, if their images under f are equal, then the two input values must be equal.
2. Surjectivity (Onto):
To prove that f is surjective, we need to show that for every y ∈ R, there exists an x ∈ R such that f(x) = y.
Let y be an arbitrary real number. We need to find an x such that f(x) = y.
If we substitute f(x) = mx + b into the equation, we have:
mx + b = y
Now, we can solve for x:
mx = y - b
x = (y - b)/m
Since m is non-zero, we can always find an x that satisfies the equation for any y. Thus, f is surjective.
Therefore, since f is both injective and surjective, it is a bijection.