Question

Prove that a strictly decreasing function is always one-one.

Answer 1

Suppose that f is not one-one. Then there exist $x_1\ne x_2$ such that $f(x_1)=f(x_2)$ .

If $x_1<x_2$ then $f(x_1)>f(x_2)$ , and the equality cannot hold. If $x_1>x_2$ then $f(x_1)<f(x_2)$ and the equality also cannot hold. We obtained a contradictory, which proves the statement.