Question

Prove that if the real-valued function f is strictly increasing, then f is oneto-one.

Answer 1

See proof below

Explanation:

Let x,y be arbitrary real numbers. We want to prove that if x≠y then f(x)≠f(y) (this is the definition of 1-1).

If x≠y, we can assume, without loss of generality that x<y using the trichotomy law of real numbers (without loss of generality means that the argument in this proof is the same if we assume y<x).

Because f is strictly increasing, x<y implies that f(x)<f(y). Therefore f(x)≠f(y) because of the trichotomy law, and hence f is one-to-one.