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.