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Subtract 0.2 J from 7.26 J. Express your answer to the correct number of significant figures for the given data.

I think its 7, but the answer is 7.1. HOW?

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$7.26-0.2$ is surely equal to $7.06$ which rounds to $7.1$ right? –  Luboš Motl Jul 15 '12 at 7:25
Lubos answered your question, voting to close. This type of thing is only useful in elementary classes, and is sociology, not physics. The correct way to do error analysis is by standard deviations, and siginficant figures is just a poor approximation. –  Ron Maimon Jul 15 '12 at 8:11
I'm not sure we should get into the habit of closing questions just because they are elementary. We all had to start somewhere. Actually I think the OP is understandably confused because the question is poorly posed and 7.0 is a valid answer. –  John Rennie Jul 15 '12 at 10:20
This kind of analysis is regularly taught in introductory physics course at the university level in the US, and while it is a simplistic way of understanding the reliability of numbers it is still a useful one. –  dmckee Jul 15 '12 at 18:00
@JohnRennie this question is a homework problem, rather than a general question about why physicists work to so many significant figures or decimal points which would be a good question to ask. –  Physiks lover Jul 15 '12 at 18:28
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closed as off-topic by Chris White, akhmeteli, Dilaton, Waffle's Crazy Peanut, Manishearth Jul 16 '13 at 14:54

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3 Answers

When adding or subtracting quantities, you must consider the number of figures after (or before) the decimal point, not the total number of figures. In this case, the less precise quantity is 0.2 J, which has one decimal figure, so your final result must have one decimal figure after the point, and you do as Luboš Motl says above.

When multiplying or dividing, you count the total number of significant figures in each number. See the Wikipedia

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As Jellby mentioned additive operation work differently from multiplicative ones.

The thing to understand is that 7.26 means that the person reporting the number could not have reliably told the difference between 7.26X and 7.25Y for X and Y differing digits.1 Likewise 0.2 means that was uncertainty in the 0.01s column.

SO for the sake of addition (or subtraction) you should treat this as $$ \begin{array} 07.26? -\\ 0.2?? =\\ 7.0??\\ \end{array} $$ which tells us that the tenth-place will be the last significant one in the answer.

So why aren't we done? Because if we express 7.26 to the nearest tenth it would be 7.3.

Finally $7.3 - 0.2 = 7.1$ is the best answer that we can find under the limitation of our knowledge implied by the significance of the original figures.

As an aside, Ron's comment on the question to the effect that this kind of analysis is primitive is true. Learn it well enough to (1) pass the test and (2) use it sensibly when doing approximation and order of magnitude analysis and don't fret about it.

1 And here we run into the biggest weakness of this form of error analysis. What if they could tell the difference as long as X and Y differed by at least 3? They have no way of reporting that.

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The question is a bit confusing because although you're given one number, 7.26, to two significant figures you're given the other number, 0.2, to only one significant figure.

The way I'd do this is first to round everything to one significant figure first, so your numbers are 7.3 and 0.2. The subtraction then gives you 7.1.

However the number 0.2 could be as high as 0.24 and still be 0.2 to 1 significant figure. In that case the subtraction would give you 7.02, which is 7.0 to one significant figure. I think 7.0 and 7.1 are both valid answers.

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