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In the problem $\frac{8.01-7.50}{3.002}$

Why would the answer be $0.17$ and not $0.170$? My least amount of sig figs is $3$ in the original equation. The only thing I can come up with is in the intermediate step.$8.01-7.50= 0.51$ exactly, which only has $2$ significant figures. Does the intermediate step really count in determining significant figures? Thank you. :)

PinkyWay
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kbeach
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  • I don't think the intermediate step matters as long as it isn't an approximation or a 'mistake', but an exact value. The approximation up to $3$ decimals in this case changes the $2^{\text{nd}};&;3^{\text{rd}}$ decimal, but it doesn't seem relevant. I've been told to write at least $4$ decimals and, frankly, this is the first time I heard of the expression (significant)$\text{figures}$, so I find this question rather interesting because there are many things our tutors don't pay attention to in my country. – PinkyWay Jan 23 '20 at 23:36
  • The rule I have seen is that the smallest significant digit in an addition or subtraction is given by the largest of the smallest significant digits in the numbers in the addition or subtraction, which in this situation means the hundredths place. A way of viewing what happened is to instead think about propagating uncertainty through interval arithmetic: if you know $x$ is between $8.005$ and $8.015$ and you know $y$ is between $7.495$ and $7.505$ then all you actually know about $x-y$ is that it is between $0.50$ and $0.52$--not even the hundredths place is totally certain. – Ian Jan 23 '20 at 23:42
  • You can perhaps see the point more transparently by considering a more extreme example. Consider something like $x-y$ where $x$ was measured to be $1.0000$ and $y$ was measured as $0.9999$. Would you report $x-y$ as $0.0001000$, or as $0.0001$? Surely it can't be the former? – Ian Jan 23 '20 at 23:45
  • In the last example you used, you would only stop at 0.0001 because of where the decimal lined up in the addition or subtraction. In the example I'm having issues with is that there is a division factor with it. When dividing or multiplying, the answer must have the same significant figures as the measurement with the least amount of digits (precision). – kbeach Jan 24 '20 at 00:47
  • Right, but then you need to compare the significant digits of the entire dividend vs. the significant digits of the entire divisor. So you wind up with $0.51/3.002$, calculate that in exact arithmetic, and then round to 2 digits, getting 0.17 instead of 0.170. – Ian Jan 24 '20 at 03:32

2 Answers2

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It is often regarded as good practice to give $1$ fewer sig. figs. than in the given numbers. So you are right, the least number of sig figs in the original equation is $3$ therefore give $2$ in the final answer.

(This has nothing to do with the intermediate step you mention.)

To see why this is regarded as best practice you could look at what the original numbers might mean. For example they might have been rounded from $8.014,7.495,3.0015$.

Your calculation with these numbers gives $0.173$ to $3$ sig figs . So $0.17$ to 2 sig figs gives a more honest degree of accuracy.

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Unless the 0 in $7.50$ is an exact figure, you can't deem it significant.

The significant figures (also known as the significant digits and decimal places) of a number are digits that carry meaning contributing to its measurement resolution. This includes all digits except:[1]

All leading zeros. For example, "013" has two significant figures: 1 and 3; Trailing zeros when they are merely placeholders to indicate the scale of the number (exact rules are explained at identifying significant figures); and Spurious digits introduced, for example, by calculations carried out to greater precision than that of the original data, or measurements reported to a greater precision than the equipment supports.

So unless the 0 is showing accuracy of a scale or an exact measurement ( like there being exactly 453592370 micrograms to a lb by definition) , there could be rounding inaccuracies ( like quoting a nutrition label 64 g of sugar is 21% of a 2000 Calorie diet, or 25 grams sugar to 8% for another, if they round from half percents the range on the first gives $128\over 41$ to $128\over 43$ the second gives $10\over 3$ to $50 \over 17$ (that's all assuming grams are completely accurate)).

So arguably, you could have as few as two significant figures, or the 0 could be from rounding.

Source wikipedia on significant digits.

Community
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  • Conventionally a non-leading zero to the right of the decimal point is significant; this is a bit more transparent when dealing with scientific notation, since then the "non-leading" distinction is irrelevant. – Ian Jan 24 '20 at 03:33