The most widely accepted definitions for an increasing function are the following:
- A (monotonically) increasing function, which says that $x_1 \leq x_2 \implies f(x_1) \leq f(x_2)$.
- A strictly (monotonically) increasing function, which says that $x_1 < x_2 \implies f(x_1) < f(x_2)$.
These two definitions are not the same, as any constant functions are increasing but not strictly increasing. Your definition would fit into the second one.
As for your second claim, indeed it holds. The contrapositive of $x_1 < x_2 \implies f(x_1) < f(x_2)$ asserts that $f(x_1) \geq f(x_2) \implies x_1 \geq x_2$. If $f(x_1) < f(x_2)$ is known, then we must have $x_1 \leq x_2$ by the contraposition, i.e. $x_1 = x_2$ or $x_1 < x_2$. But $x_1 = x_2$ can't be true, for otherwise we have that $f(x_1) = f(x_2)$, so we indeed have $x_1 < x_2$.
EDIT: OP asked a follow-up question in the comment on why in this case we are able to prove the bi-implication from just the implication alone here, but not in general. To illustrate this, we first note that more precisely the definition of a strictly increasing function should say:
$$
\forall x_1 \in \mathbb{R}, \; \forall x_2 \in \mathbb{R}, \; x_1 < x_2 \implies f(x_1) < f(x_2)
$$
The reason we can prove the reverse implication stems from the following two facts:
- For any real numbers $\mathbb{R}$, either $x_1 < x_2$, $x_2 < x_1$ or $x_1 = x_2$. That is, $<$ is a connex relation.
- There are two arbitrary variables involved in the definition above. Thus, $x_1$ and $x_2$ are just dummy variables that can be replaced by any other symbols.
With the contraposition, we see that we obtain:
$$
\forall x_1 \in \mathbb{R}, \; \forall x_2 \in \mathbb{R}, \; f(x_2) \leq f(x_1) \implies x_2 \leq x_1
$$
Replacing the dummy variables give:
$$
\forall x_1 \in \mathbb{R}, \; \forall x_2 \in \mathbb{R}, \; f(x_1) \leq f(x_2) \implies x_1 \leq x_2
$$
The two steps done above do not work for any general statements but are rather very special property of the relation $<$.
Best Answer
$$ f(b)= \frac{2^a-2^b}{\ln a-\ln b}$$
Then, $$f'(b)=\frac{-2^b\ln 2(\ln a-\ln b)-(2^a-2^b)\cdot (-1/b)}{(\ln a-\ln b)^2}$$ $$=\frac{2^b}{b(\ln a-\ln b)^2}(-b\ln 2(\ln a-\ln b)+2^{a-b}-1)$$
Let $g(b)=-b\ln 2(\ln a-\ln b)+2^{a-b}-1$.
Then, $$g'(b)=-\ln 2\ln a+\ln 2(\ln b+1)-2^{a-b}\ln 2=\ln 2(-\ln a+\ln b+1-2^{a-b})\lt 0$$ since $-\ln a+\ln b\lt 0$ and $1-2^{a-b}\lt 0$.
So, $g(b)$ is decreasing with $g(a)=0$, and so $g(b)\gt 0$ for $0\lt b\lt a$.
Therefore, since we have $f'(b)\gt 0$, we know that $f(b)$ is increasing.