Why $1/(1+e^{-x}) = e^x/(1+e^x)$ I am currently learning the sigmoid/logistic function and have completely forgotten how the maths behind this equivalence works:
$$
\dfrac{1}{1+ e^{-x}} = \dfrac{e^{x}}{1+e^{x}}
$$
By this I mean how the left side equates with the right side. I know that the left side is simply multiplied by (1/exp(x)) but what are the rules governing this?
Any help would be much appreciated, especially the rules so that I can revisit them.

$$\begin{align} \dfrac{1}{1+ e^{-x}} &=\dfrac{1}{1+ e^{-x}} \\
 &= \dfrac{1}{1+ e^{-x}} \cdot \frac{e^x}{ e^x} \\
 &= \dfrac{e^{x}}{e^x+1} \end{align} $$

*

*Multiplying a number by 1 doesn't change its value, and $\frac{e^x}{ e^x}$ is just one way to write 1.


*Multiplication adds exponents of a common base: $a^b \cdot a^c = a^{b+c}$. We use this in the final line to get $e^x \cdot e^{-x}=e^0 = 1$.
