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 f is a bijection, we need to show that f is both injective (one-to-one) and surjective (onto).
Injective (One-to-one):
To prove that f is injective, we need to show that if f(x) = f(y), then x = y.
Let f(x) = f(y).
This means mx + b = my + b.
Subtracting b from both sides, we have mx = my.
Since m is nonzero, we can divide both sides by m to get x = y.
Therefore, f is injective.
Surjective (Onto):
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 of f such that f(x) = y.
Let y be any real number in the codomain of f.
We need to find an x such that f(x) = mx + b = y.
Solving the equation for x, we have:
mx = y - b
Dividing both sides by m (since m is nonzero), we get:
x = (y - b)/m
Therefore, for any y in the codomain of f, we can find an x in the domain of f such that f(x) = y.
Thus, f is surjective.
Since f is both injective and surjective, f is a bijection.
To prove that function f(x) = mx + b is a bijection, we need to show that it is both injective (one-to-one) and surjective (onto).
1. Injective Proof:
To prove that f is injective, we need to show that if f(x₁) = f(x₂), then x₁ = x₂ for all real numbers x₁ and x₂.
Given f(x) = mx + b, let's assume f(x₁) = f(x₂).
So, mx₁ + b = mx₂ + b
We can subtract b from both sides to get mx₁ = mx₂.
Since m is nonzero, we can divide both sides by m to get x₁ = x₂.
This proves that f is injective, as for any pair of distinct real numbers x₁ and x₂, f(x₁) will not be equal to f(x₂).
2. Surjective Proof:
To prove that f is surjective, we need to show that for every y in the range of f, there exists at least one x in R such that f(x) = y.
Given f(x) = mx + b, let's take an arbitrary y in the range of f.
Since f(x) = mx + b, we have y = mx + b.
Rearranging this equation, we get mx = y - b.
Since m is nonzero, we can divide both sides by m to get x = (y - b)/m.
As x can take any value in R (the set of real numbers), we can find an x to satisfy the equation for any given y in the range of f.
This proves that f is surjective, as every element in the range has a corresponding pre-image in the domain.
Since f is both injective and surjective, it is a bijection.