Stuck on a very very basic concept.

I have $1750\text{ mm}^2$ and need to get into $\text{m}^2$.

I figured $1750\text{ mm}^2 = (1750\text{ mm}) \cdot (1750\text{ mm})$, but I know this isnt right.

  • $\begingroup$ It's $1750 (mm)^2$ not $(1750 mm)^2$ so you don't square the $1750$, just the units. $\endgroup$ – John Rennie Mar 7 '14 at 7:18

The superscript $^2$ in $1750\text{ mm}^2$ refers to a squaring of the units, not the number $1750$.

A more transparent way to write this is $1750\text{ mm}\cdot\text{mm}$. The idea is now to multiply by $1$, but $1$ written in a clever way:

$$1=\frac{1\text{ m}}{1000\text{ mm}}$$.

Can you see how that number is conceptually equal to $1$? The top and bottom are the same thing.

Try multiplying your quantity by this number. Cancel out units carefully. Then see what your status is. There's one more step after this.

  • 1
    $\begingroup$ It seems simpler instead of multiplying by some strangely-looking identity to just substitute $\text{mm}=10^{-3}\text{m}$ and then simplify. $\endgroup$ – Ruslan Mar 7 '14 at 7:14
  • 2
    $\begingroup$ Yes, that does seem simpler here. $\endgroup$ – BMS Mar 7 '14 at 7:37
  • $\begingroup$ So I multiplied it all by 1750mm $\endgroup$ – user88720 Mar 11 '14 at 3:35

What you need to do is realize how many $\text{mm}$ fit into one $\text{m}$. The answer is $1000=10^3$. Therefore: $1 \text{m}=10^3 \text{mm}$, or equivalently, $1 \text{mm}= 10^{-3} \text{m}$. $$1750 \text{mm}^2=1750 (10^{-3}\text{m})^2=1750 * 10^{-6} \text{m}^2$$


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