# How many significant figures are appropriate when a new digit is created?

In my chem class the other day, we were working through a question about average atomic mass. We were given the following information about two isotopes of boron:

Boron-10: Atomic Mass = 10.013, % in nature: 19.8%

Boron-11: Atomic Mass = 11.009, % in nature: 80.2%

For the weighted average (before sig fig rounding), then, Boron 10 contributes 10.013*.198 = 1.982574 AMU, and Boron 11 contributes 11.009*.802 = 8.829218 AMU.

Now, if we were just finding one of these values, we would report 1.98 or 8.93 AMU (to 3 sig figs). Adding these together, we get 10.81 AMU. However, if we follow "convention" and do all rounding at the end, we get 10.8 AMU (3 sig figs).

My question is, which of these is correct? On the one hand, 10.8 is correct since it has the same number of sig figs as the least accurate given value (19.8% or 80.2%). On the other, why should we go from hundredths to tenths precision just because we got a new digit in the least precise position? One could argue that the "10" is essentially its own digit.

Personally, I agree with the response here (Significant Figures Interpretation) that says sig figs don't make a whole lot of sense. But for the purposes of this question, which is the proper way to express the answer?

The way significant figures is taught is a shortcut for a more exacting method since the exact precision of the original figures isn't typically given. Thus 0.198 is assumed to be +/-0.0005.

Many science papers will use something like 0.1976 +/-0.0014 to indicate the "true" standard deviation of the value. "True" meaning of course the best estimate since the true standard deviation cannot be known. The relative error is thus $\dfrac{0.0014}{0.1976} = 0.709 \%$. This is rather inconvient for calculating the error prorogation since it isn't +/- 1 digit in the least significant digit.

Obviously 10.013*0.198 = 1.982574 AMU is unjustifiable. Having seven significant figures is just wrong. The weighted value for the AMU can't have seven significant figures when the weight per cent is only known to three significant figures. However typically extra significant figures are carried in intermediate calculations and only the final answer is rounded.

Carrying such extra digits in intermediate calculations isn't a problem with a modern calculator. In the "olden days" when a slide rule was used three significant figures in intermediate calculations were the limit. Log tables allowed four significant figures.

So the gist is that since the percentages have three significant figures, then the answer should be rounded to three significant figures. Thus the answer is indeed 10.8 AMU.

Now using a modern computer all the values could be randomly sampled. So the value of 0.198 could for instance be assumed to have a standard deviation of 0.0005 and other variables treated the same way. Then the sampled values could be used for a series of calculations. A mean and a standard deviation of the various final answers could then be calculated. Such grinding is really beyond reasonable human effort but no problem for a computer.

Actually, you are correct - when adding significant figures together, you keep the largest last significant figure between the two numbers. For example, $15.31+2.2=17.5$. On the other hand, when you are multiplying numbers, you keep the same number of significant figures. Using the same numbers, $15.31*2.2=34$. While this decision might appear arbitrary, there is a good reason behind it.

Since we do not know anything past the tenths place about $2.2$, lets say this number is actually $2.2x$ where $x$ is a digit that would cause this number to round down (value between 0 and 4). $15.31+2.2x = 17.5[x+1]$ (where a number in brackets is a digit that is the result of a calculation. It can carry to the next place, but for reasons that I will explain later, we assume that it does not). Since we now know nothing about the hundredths place after adding these two numbers, we cannot include it in our result.

Likewise for multiplication, $15.31 * 2.2x = 33.[6+x][8+5x][2+3x][x].$ Since we know nothing about the value of $x$, we know nothing after the decimal point. Because our next digit is guaranteed to round up, regardless of the value of $x$, we round up to get $15.31*2.2x=34$. This is not a special case. In fact - the number of digits that will not depend on the value of $x$ is always equal to the number of significant figures in that number (this is - again - because we assume that $x$ will not cause a digit to carry), thus giving us our multiplication rule for significant figures.

Now for why it is okay to assume that $x$ will not cause a digit to carry. When you record an experimental value, you are supposed to record one more digit than your device can differentiate (ex. guessing the location of a meniscus between two numbers in a graduated cylinder). As such, the final digit that you use in your calculations is inaccurate anyways. Because of this, it is safe to assume that $x$ did not cause any additional overflow. If we were to assume that x overflowed, then it could cause your doubtful digit to also carry, which would result in numbers that disagree with the measured value.

I hope this helps (I know I went into a lot more detail than you were looking for).