If $(B_t)$ is a Brownian motion, we say formally that $dB_t=(dt)^{1/2}$. And then we use the formal notation $dt\cdot dB_t=dt\cdot dt=0$ and $dB_t\cdot dB_t=dt$.
Q1) Is there a rigorous reason for that or is it only a convention ? In general, if $\alpha >1$, why (dt)^\alpha =0$ ? Can we prove this ?
Q2) I know that the fact that $(B_t)$ has no bounded variation, but only quadratic variation gives us that $dB_t=(dt)^{1/2}$. But how can we justify this fact rigorously ?
This question is quite correlated to my other question here. But since the question is not the same, I prefered to open a new question.