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I have learned in class that to subtract decimal numbers and keep significant figures, one just lines up the decimal, then rounds the answer according to the operand with the fewest places after the decimal.

My question is how to handle an integer subtracted by a decimal.

112 - 12.0

If I add the decimal to 112.0, my concern comes with 100.0 being 4 significant figures, which has more than either of the original operands.

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  • $\begingroup$ If you only have 3 figures, then is like you don't really know if the input $112$ is $112$ or maybe $112.1$, or $112.4$. By writing $100.0$ you are claiming that the information given in the input tells you the digit after the period is $0$, but we don't know that. $\endgroup$ – user119256 Jan 10 '14 at 19:41
  • $\begingroup$ The rules are here en.wikipedia.org/wiki/… $\endgroup$ – user119256 Jan 10 '14 at 19:44
  • $\begingroup$ So even though 12.0 is more precise, the answer would be rounded at the integer level because I can only be as confident as the least confident operand? (being the integer) $\endgroup$ – Anduril_1251 Jan 10 '14 at 19:44
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When adding or subtracting, you don't count significant figures, you round to the least accurate. Your $112$ could be anywhere in $(111.5,112.5)$, so knowing that the $12.0$ is in $(11.95,12.05)$ is not much better than knowing it is in $(11.5,12.5)$. You should report your result as $100$, using whatever notation you have to show that the last zero (the one's place) is significant. What you really know is that the difference is in $(99.45,100.55)$, which could round away from $100$, but it is not likely.

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Of $112$ and $12.0,$ $112$ is less precise, since $12.0$ is significant to $10^{-1}$ while $112$ is significant only to $10^0.$ So the difference is significant only to the $10^0$ place.

So $112 - 12.0 = 100.$ where the decimal after the $100$ indicates that you know $100$ to three significant figures.

Writing the numbers in scientific notation removes any doubt. The expression with the correct number of significant figures is $1.00 \times 10^2.$ It's clear that you're dealing with three significant figures in this case.

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    $\begingroup$ Subtracting $1110.-1100.$ results in $10.$ with only two significant figures, despite the fact that the two numbers being subtracted have four each. It certainly should not give $10.00$ $\endgroup$ – Ross Millikan Jan 10 '14 at 19:53
  • $\begingroup$ Good point. So another restriction is that the absolute precision cannot be greater than that of the least precise term in the calculation? $\endgroup$ – John Jan 10 '14 at 20:02
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    $\begingroup$ For addition/subtraction you just look at the least precise number. The number of figures will take care of itself. For multiplication/division you look at the number of figures because what matters is fractional error, not absolute error. If I multiply a three figure number (with roughly $\pm 1\%$ error) by 12345, the resulting product still has roughly $\pm 1\%$ error, so should have three significant figures. $\endgroup$ – Ross Millikan Jan 10 '14 at 20:06
  • $\begingroup$ Edited my answer. Thanks again for spotting my error. $\endgroup$ – John Jan 10 '14 at 20:09

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