I have a basic exponential growth model given by $N'(t)=N(t)\times r$ where $N(t)$ is the current population and $r>0$. My problem is to show if the initial population $N(0)=N_0>0$, then the population level is always positive. Solving for $N(t)$ gave $|N(t)|=N_0\times e^{rt}$.
I was trying to show this through contradiction assuming that there exists a value $t_1$ such that $N(t_1)<0$. Since we started at a positive initial population and the function is continuous, there must be a $t_2 \in (0,t_1)$ for which $N'(t_2)<0$. This implies that $N(t_2)<0$ and again there must exist $t_3 \in (0,t_2)$ such that $N'(t_3)<0$. Eventually, I wanted to show that this yields to $N_0<0$ which gives the contradiction, but I'm worried about the fact that I'm trying to use induction for a an uncountable subset of real numbers $(0,t_1)$.
Note: The solution to $N'(t) = N(t) \cdot r$ is $N(t) = N_0 e^{rt}$ (without the absolute value sign).
Assuming $N(0) = N_0 > 0$ then $N(t) > 0$ for all $t > 0$ since it's always true that $e^{x} > 0$ for real $x$.