The normal distribution function $\Phi(z)$ has the definition $\Phi(z) \equiv \frac{1}{\sqrt{2 \pi}} \int_0^z e^{\frac{-x^2}{2}} \, dx$.
However the error function is conventionally defined such that $\frac{1}{2}\operatorname{erf}(\frac{z}{\sqrt{2}})\equiv\Phi(z)$.
I understand the $\frac{1}{2}$ scaling on the error function itself, since the $\Phi$ integral is symmetrical around $0$. But why scale $z$ in $\operatorname{erf}(z)$??