# How do you use metric prefixes?

Jun 10, 2017

Hopefully correctly.........

#### Explanation:

$m L$ $=$ ${10}^{-} 3 \cdot L$;

$\mu s$ $=$ ${10}^{-} 6 \cdot s$;

$k m$ $=$ ${10}^{3} \cdot m$.

A good illustration of this is the strict use of ${\mathrm{dm}}^{3}$ in preference to $\text{litres}$. Now $1 \cdot {m}^{3}$ is an absurdly large volume, and if you ever mix and lay $5 - 6 \cdot {m}^{3}$ of concrete that is a good day's hard work.

Back to the problem, $1 \cdot L \equiv 1 \cdot {\mathrm{dm}}^{3}$; and $d = \text{deci} = {10}^{-} 1$. And thus $1 \cdot {\mathrm{dm}}^{3} \equiv {\left(1 \times {10}^{-} 1 \cdot m\right)}^{3} \equiv {10}^{-} 3 \cdot {m}^{3} \equiv \frac{1}{1000} \cdot {m}^{3}$ as required.

And so...............

$6.37 \times {10}^{-} 2 \cdot L = 6.37 \times {10}^{-} 2 \cdot \cancel{L} \times {10}^{3} \cdot m L \cdot \cancel{{L}^{-} 1} \equiv 63.7 \cdot m L$

Agreed?