0
$\begingroup$

When finding out the order of magnitude of quantities, as said in my textbook, we compare the numerical part with $3.2$ (approximately $\sqrt{10}$ or rounded off version of $3.162$)

Thus, $9.12 \times 10^5$ has an order of magnitude of $6$ since $9.12$ is $>=$ $3.2$

But what if we had $3.17 \times 10^5$. According to my textbook, its order of magnitude would be $5$, as $3.17$ is less than $3.2$. But, as we have to compare with $\sqrt{10}$, which is $3.162$, the order of magnitude should be $6$, as $3.17$ is greater than $3.162$. This is my question.

Or to rephrase, when we are rounding off $\sqrt{10}$ to $3.2$, why aren't we rounding off the numerical part $3.17$ ?

I hope I made myself clear. I have looked at this, this, and this, but that doesn't satisfy me.

$\endgroup$
2
$\begingroup$

I think you are attaching excessive significance to your order of magnitude estimate. If I say something is $10^5$ to within an order of magnitude I mean that the $\log_{10}$ of that quantity is nearer $5$ than $4$ or $6$. If some quantity has a $\log_{10}$ of about $5.5$ does it really matter whether our order of magnitude estimate is $10^5$ or $10^6$?

Incidentally, if you wondered why we use $\sqrt{10}$ as the dividing line it's because $\log_{10}(\sqrt{10}) = 0.5$. So for example the $\log_{10}$ of something slightly less than $\sqrt(10)$ is nearer $0$ than $1$, while the $\log_{10}$ of something slightly greater than $\sqrt(10)$ is nearer $1$ than $0$.

$\endgroup$
  • $\begingroup$ But then, as I know in physics, we have to be precise, and can't just basing our facts on assumptions, can we ? And round something, while not the other ? $\endgroup$ – Gaurang Tandon Apr 6 '14 at 6:53
  • 2
    $\begingroup$ @GaurangTandon: you say we have to be precise, but the whole point is that with an order of magnitude estimate we aren't being precise. We're saying our estimate could be ten times too big or ten times too small. It seems odd to be quibbling about the difference between $3.16$ and $3.17$ when the real value could be as low as $0.31$ or as high as $31$. $\endgroup$ – John Rennie Apr 6 '14 at 7:00
  • $\begingroup$ Upvote for the explanation of the choice of $\log_{10}\, 10^\frac{1}{2} = 0.5$. Seen in on a log-scaled axis, this means discriminating based on the mid line across the interval between two successive orders of magnitude. This also implies that 0.5 is of order of magnitude 1, while 0.3 is of order of magnitude 0.1, for better and worse. $\endgroup$ – XavierStuvw Aug 7 '17 at 7:54
  • $\begingroup$ Note also that en.wikipedia.org/wiki/Order_of_magnitude suggests a cut-off value of 0.5 (5) rather than $\sqrt 10$. Also, see further talks at physics.stackexchange.com/questions/16322/… $\endgroup$ – XavierStuvw Aug 7 '17 at 8:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.