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In my text book I have following example to present computation of order of magnitude:

$$0.008 \space 6\space m \sim 10^{-2}m$$

I dont understand whether I am supposed to multiply $0.008$ with $6$, and then find order of magnitude or not. In that case it will be $0.048$ and then $4.8 \cdot 10^{-2}$, now according to the book, if the multiplier is greater than $3.162$ then order of magnitude is one larger than the power of 10 in the scientific notation. But that would mean the order of magnitude it $10^{-1}$ ?

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closed as unclear what you're asking by Jon Custer, John Rennie, mpv, user259412, ZeroTheHero Aug 21 '17 at 21:59

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  • $\begingroup$ That is 0.0086 - in analogy to using "," as a marker every three decades, sometimes books will use a slight space every three digits after a decimal point. That is all... $\endgroup$ – Jon Custer Aug 19 '17 at 18:48
  • $\begingroup$ @JonCuster yes, its now all clear - I have not read any physics nor math books for a long time. $\endgroup$ – mike Aug 19 '17 at 19:13
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It is fairly common to leave small gaps between groups of figures in a number with many decimal places, to make it easier to read the number or copy it (e.g. enter it into a calculator).

Look at the table "Some Physical Constants" at the start of the book for examples.

This is just the number "$0.0086$".

Separating last digit with a space is fairly pointless in this case, IMO, but I suppose the "style sheet" for the book was set up to format all numbers the same way.

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I don't understand the part with 3.162, but the line you quote

$$0.008 \space 6\space m \sim 10^{-2}m$$

I would understand to mean

$$0.0086\space m \sim 10^{-2}m$$

In words: 0.0086 is on the order of $10^{-2}$.

Decimals (and thousands) are often separated by spaces to make them more readable.

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  • $\begingroup$ 3.162 is the square root of 10, if the multiplier is greater than this then we add one to the power of 10 in the scientific notation to gain better aproximation of order of magnitude. $\endgroup$ – mike Aug 19 '17 at 19:12
  • $\begingroup$ @mike: Where/why do you square (or take the root)? $\endgroup$ – user1583209 Aug 19 '17 at 19:38
  • $\begingroup$ I dont, this is just an explanation how 3.162 is computed. You can find more on this here: physics.stackexchange.com/questions/107088/order-of-magnitude $\endgroup$ – mike Aug 19 '17 at 21:00

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