Is it correct to say that 9.0 is one order of magnitude smaller than 10.0?

Has anyone a link/source about confronting order of magnitudes, apart from wikipedia?

  • $\begingroup$ What do you mean by "confronting order of magnitudes"? That part of the question isn't clear, and the first part is trivially answered by a web search or by reading the Wikipedia article (even just the introductory paragraph). $\endgroup$ – David Z Oct 29 '11 at 9:40
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    $\begingroup$ If the first part is trivially answered, could you answer it please? By "confronting order of magnitudes" I refer to this goo.gl/T87sS $\endgroup$ – Mascarpone Oct 29 '11 at 9:45
  • $\begingroup$ So you mean finding order of magnitude comparisons, i.e. finding the order of magnitude of the ratio between two quantities? Is there some specific aspect of it as explained in the Wikipedia article that you are confused about? (The answer to the first part of the question is of course "no", though I have to wonder why you weren't able to figure that out yourself - perhaps an explanation would help clarify the question) $\endgroup$ – David Z Oct 29 '11 at 10:03
  • $\begingroup$ I thought the link explained it well enough. Confronting the order of magnitude is the ABS(p-q) where p and q are the exponent of the numbers written in scientific notation. By this definition the answer should be yes. $\endgroup$ – Mascarpone Oct 29 '11 at 10:06
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    $\begingroup$ Confront is not the same as compare! Someone speaking a roman language as mother tonge should know that. $\endgroup$ – Georg Oct 29 '11 at 11:05

I suppose the OP is looking for some general rule to be used when you want to say "A is N orders of magnitude bigger (smaller) than B".

In that case, consider $$N = || \log_b(A/B)) ||$$ (where $ || \dots || $ is taken to mean round to the nearest integer, and negative values just mean chose "A is smaller than B", but the magnitude retains the same significance.

Here $b$ the the base you are speaking in (10 generally, but it is sometimes useful in computer science circles to speak of binary orders of magnitude).

In this case $\log_{10}(10/9) = 0.045 \approx 0$ so 9 and 10 are of the same order of magnitude as one would naively expect.

You can manage this rule without having to extract logarithms by noting that $0.5 = \log_{10}(R)$ implies $R = \sqrt{10} \approx 3.16$. Just count the number of digits difference in the long-hand written form and add one if the ratio of the leading values is at least 3.2.

That is

  • 30 is the same order of magnitude as 10
  • 35 is one order of magnitude larger than 10
  • 300 is one order of magnitude larger than 10
  • 350 is two order of magnitude larger than 10
  • 3.5 is the same order of magnitude as ten
  • 3.0 is one order of magnitude smaller than 10

Final note: Don't obsess over this! Orders of magnitude are useful because they let you make quick and reasonably accurate guesses, and guesses are not subject to precise rules. For instance $\pi$ is close enough to $\sqrt{10}$ that it's OK to treat it as the same order of magnitude as either 1 or 10.

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  • $\begingroup$ Sensibly pragmatic. A factor of 3 is what I normally work on. $\endgroup$ – Rob Jeffries Dec 30 '14 at 21:32
  • $\begingroup$ I concur with not being obsessed. Nevertheless, this is a definitional issue to be aware of, especially when it comes to multiplying quantities. For example, on the one hand, $3 \times 3 = 9$ is the product of two numbers of order 1 that gives a number of order 2. On the other hand, $5 \times 5 = 25$ is a product of two numbers of order 2 that gives a number of order 2 again. This may be confusing. I guess it is good to state beforehand which definition of order of magnitude one adheres to. I had asked a related question in the Maths community in math.stackexchange.com/q/2379100/446004 $\endgroup$ – XavierStuvw Aug 7 '17 at 8:18
  • $\begingroup$ 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/107088/order-of-magnitude $\endgroup$ – XavierStuvw Aug 7 '17 at 8:27

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