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I have been studying about significant digits and addition rules and I can't quite digest the rules of addition completely.

It states that in the answer number of decimal places will be equal to the least number of decimal places in the terms. $//$What my teacher has taught me and what my book says

This makes sense first :

$1000.1+1.15=1001.2$

If we had went for number of significant digits rule we would have retained only $3$ significant digits.

But consider this case :

$1. 10^3+1.0=1001$

The number of significant digits in both of initial terms were $1$ and $2$ respectively, but in the final answer they are $4$. There are more significant digits in the answer. Isn't it wrong as last three digits of $1.10^3$ are insignificant

Please clear my doubt or whether the rule has an extension.

EDIT $1000$ changed to $1.10^3$. I don't think anyone understands what am I asking. I $know$ what the rules are and how to apply them but I want to know that $1.10^6+1.0=1000001$ . Don't $you$ think it is wrong as we are not sure of second last digit of $1.10^6$ but we are of $1000001$

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  • $\begingroup$ See, e.g., Bevington's book on statistical analysis. $\endgroup$ – Carl Witthoft Mar 4 '14 at 20:13
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The last digits in 1000 are absolutely significant, they state that you have not 1200, not even 1001 but exactly 1000. In scientific notation, you would write this as $1.000 \times 10^3$. Compare this to $1\times10^3$ where you have just one significant digit.

Update: consider the example from the question $1\times10^3+1.0$. The first term could be anything between 500 and 1500, so the answer lies between 501 and 1501. The expected value of the answer is 1001, but writing it so gives a false sense of precision. One could write it as $1001\pm500$, but this is almost the same as $1\times10^3+1.0=1\times10^3$, which is the answer according to the rule of significant digits.

The rule of significant digits is a simplification of the Propagation of uncertainty principle. As such, it can lead to erroneous results in some cases: $1+0.49=1$ doesn't look good. Use propagation of uncertainty when you need precise calculations.

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  • $\begingroup$ Okay... then 10^3 + 1.0 =? $\endgroup$ – evil999man Mar 6 '14 at 4:43
  • $\begingroup$ 1*10^3+1.0=10^3. I think that you understand it right, and I will add this to my answer. By the way, 10^3 has NO significant digits, it is just an order of magnitude. Could be 3000. $\endgroup$ – gigacyan Mar 6 '14 at 7:15
  • $\begingroup$ Thanks... that's exactly what I was arguing with my teacher... perhaps you can provide some link where this type of example is explained? I'll gladly accept your answer once edited. $\endgroup$ – evil999man Mar 6 '14 at 7:17
  • $\begingroup$ And yeah...that was 1.10^3 I was referring to $\endgroup$ – evil999man Mar 6 '14 at 7:29
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Working with significant digits is very prone to error, because it can be misleading. It is much better to work with explicit errors.

So to rewrite your example with explicit errors:

$\left(1.0 \pm 0.5\right) \times 10^3 + 1.00\pm0.05$

We now add the errors quadratically (assuming they are uncorrelated):

$\sqrt{(0.5\times 10^3)^2+0.05^2}=500.0000025000\ldots=0.5\times 10^3$

So, obviously the result stays

$\left(1.0 \pm 0.5\right) \times 10^3$

In actual physics you might call the $1.00\pm0.05$ negligible compared to $\left(1.0 \pm 0.5\right) \times 10^3$.

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When adding and subtracting, you can only go to the lowest number of decimal places. That is to say, we are dealing with precision and not significant figures when adding/subtracting numbers. If you have two measuring devices and one is accurate to 0.1mm and the other to 1mm, then you cannot definitively state the combined measure to 0.1mm, you can only state your certainty to 1mm due to the lesser measuring device.

For Case 1, your numbers have 1 decimal place and 2 decimal places, so the lowest is one decimal place, hence the .2 in the result.

For Case 2, your numbers have 0 decimal places and 1 decimal places, so the lowest is zero decimal places, hence the lack in the result.

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  • $\begingroup$ You do note that 1000 contains 1 significant digit and 1.0 2 significant digits, while the sum has 4 significant digits? $\endgroup$ – evil999man Mar 5 '14 at 1:46
  • $\begingroup$ Yes, I note that. You can't well add 0.843 and 0.98 together to get 1.823 because 0.98 is not that precise a value. The best you can do is 1.82. As stated, you are dealing with precision, not significant figures when adding/subtracting values. $\endgroup$ – Kyle Kanos Mar 5 '14 at 1:48
  • $\begingroup$ so answer should be $1000$ $\endgroup$ – evil999man Mar 5 '14 at 8:51
  • $\begingroup$ No, it is 1001. You can't add them to get 1001.0 because 1000 isn't that precise a measure. $\endgroup$ – Kyle Kanos Mar 5 '14 at 11:24
  • $\begingroup$ But last 3 digits of $1000$ are insignificant but in $1001$ all significant digits are significant $\endgroup$ – evil999man Mar 5 '14 at 12:37
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1000 has 4 significant digits as mentioned before, stating that the measured value is between 999.5 and 1000.5. 1.0 has 2 significant digits, staying it is measured within an accuracy of 0.05. Adding the numbers gives you a result within an accuracy of 0.5, so noting the result with 1 decimal digit is nonsense. If your measured value of 1000 indeed has only 1 significant digit, you should note it as $1\cdot10^3$, stating the value is between 500 and 1500. Adding 1.0 will not change the number, as in powers of 10, 1.0 written as $0.0010\cdot10^3$, the result being $1\cdot10^3$. When adding numbers with different powers of ten, always convert them to the same power while you make sure to preserve the significant digits.

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$1000$ has an ambiguous number of significant digits. It could have 1, it could have 4. I think it is generally assumed that it has four unless otherwise stated. (For example, I've seen people put a line over the last significant digit, like this: $1\overline000$.) This is why scientific notation is useful. If you are saying that $1000$ has $1$ significant, then we can write that as $1\times10^3$ and you can write $1.0$ as $1.0\times10^0$. Added together you get $1.001\times10^3$. If you carry through your significant digits, you still end up with $1\overline000$.

However, if you say $1000$ has 4 significant digits (that is, $100\overline0$), then you write it as $1.000\times10^3$ and then when you add $1.0\times10^0$, you still get $1.001\times10^3$ but this time when you carry through your significant digits, you end up with $1.001\times10^3$, or $1001$.

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