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What is point of rounding off to significant figures in multiplication. e.g. $3\times123456789=370370367$, if I round off the answer to least sig fig it becomes $400000000$ which has a large error. So what is point of significant figures here?

I really don't understand the importance of sig fig.

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    $\begingroup$ No rounding in the example you give: the product of two integers is an integer. $\endgroup$
    – Ed V
    Jun 16 at 20:53
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    $\begingroup$ Significant figures are a waste of time. It is much easier and cleaner to simply report the absolute uncertainty and use error propagation to report the uncertainty of results of multiplication. $\endgroup$
    – Jagerber48
    Jun 16 at 22:41
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    $\begingroup$ @Jagerber48 Yep! And some physicists are not at all bashful about spending 100 pages or so pounding that home: av8n.com/physics/uncertainty.htm. I like where Denker says “People who care about their data don’t use sig figs.”. $\endgroup$
    – Ed V
    Jun 16 at 23:06
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    $\begingroup$ @Jagerber48 Soft disagree. The “significant figures” approach is a useful introduction to the idea of finite-precision data at the high-school level. Compare to the oversimplified model in intro biology courses where brown eyes and blue eyes are used to teach about dominant and recessive genes, even though eye color is a complex phenotype controlled by the interactions of many different genes. For people who haven’t yet internalized calculus, error propagation is a dark art. $\endgroup$
    – rob
    Jun 17 at 0:01
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    $\begingroup$ @Jagerber48 Have you actually done this though? I think you're forgetting what it's like to be a high school student. High school students have trouble remembering significant digits (one simple rule that you can justify to them at their level), let alone memorizing error propagation rules (several equations that seem to come out of nowhere and that they won't have a basis to understand for a few years). It's not going to stick. $\endgroup$
    – DKNguyen
    Jun 17 at 2:00

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I like to teach my students to trust their hunches about significant figures using their emotional response to the rounded number. I call this the “anger management method of error estimation.”

When I go hiking, I like to scrounge up a walking stick from the deadwood in the forest. The right length for a walking stick is pretty close to my height. I am (to one significant figure) about two meters tall. One meter is too short for a walking stick (that’s more like a cane, which serves a different purpose). If you imagine coming across a hiker with a three-meter walking stick, you will probably set yourself to laughing — an emotional response.

If you found two candidate walking sticks whose lengths were different by 20 cm, you might stand them up next to yourself and decide that one is a better length than the other. You might even break off a handspan (10 cm) or two to get the length right. But imagine coming across a hiker who has heard somewhere that a walking stick should be as long as he is tall, and he was 181.5 cm tall at his last doctor visit, so he is sitting on the trail with a measuring tape carefully sanding down the knob at the top of his piece of found deadwood. You’d probably have an emotional response to him, too.

So for the length of a walking stick, you really care about the first digit. You probably also care about the second digit. That last centimeter of length, through, is insignificant. The length of a walking stick has about two significant digits.

The trousers I wore yesterday are 110 cm long. (Two meters: not trousers. Half a meter: also not trousers.) If you buy trousers from a fancy shop, the tailor may fuss about the last centimeter of their length, which controls how the cuff folds where it lands at your shoe. If you have a tailor fixing the hem of your trousers, he’ll mark the correct length with a piece of chalk and then pin the fabric in place with sub-centimeter precision. But if you had a tailor who was fussing over sub-millimeter alterations in the length of your trousers (“the pin is between the wrong fibers!”), you would have an emotional reaction to him and leave the shop. If your trousers came back from the tailor 10 cm too long or 10 cm too short, you would likewise have an emotional reaction and leave the shop.

The millimeter digit is the first insignificant digit in the length of a pair of trousers: you would possibly be annoyed if the centimeter digit were wrong, but probably not notice if the millimeter digit were wrong. So for my trousers, 110 cm has three significant digits.

Of course, you might be shorter than I am, and your trousers might only be 95 cm long. Do your trousers only have two significant figures? If I decree “trousers have three significant figures,” have I doomed short people to a world filled with insane submillimeter tailors? No. Trust your intuition, keep an extra digit if you feel squashy about rounding too much, and be aware that there are more robust ways than counting significant figures for people who really need to be sure about their uncertainty analysis.

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Actually, it becomes $4\times10^8$, not $400000000$. It seems pointless because you are thinking of exact numbers in your example, and if every number in there was an exact number it would be pointless. But in real life at least some of those numbers are going to come from measurements and measurements have uncertainty.

Basically, you do not write down digits that may not be true. Another more explicit method of expressing uncertainty is writing down the error bars like $3\pm0.1$ but even there the fact that $0.1$ was written instead of $0.10$ holds significance.

Significant digits is sort of an implicit way of expressing and interpreting uncertainty when there is nothing explicit to express uncertainty to keep things in check.

If your 3 is a factor of three (exact) then you don't need to round significant digits because it is exactly 3; In other words $3.\bar0$.

But if your 3 is actually a measurement like 3 meters then there is uncertainy in that number. Since you do not write digits you are not sure about and all you wrote down was $3m$ that means the actual number could be anywhere between $3.\bar0$m and $3.\bar9$. If you wrote $3.0$m then that means you were sure of more digits in your measurement and by doing that you are saying the real number is somewhere between $3.\bar0m$ and $3.0\bar9$m

So if we assume that 123456789 is an exact number in your example (or is a measurement accurate to that number of digits) if we multiply them together, then the actual answer could be anywhere between $370 370 367.\bar0$ and $493 827 156.\bar0$

Between 123456789 and 3, the 3 has the most uncertainty in it because it has the least significant digits (the digits you are sure about) therefore it is the weakest link holding back the your calculation. The result of the calculation can't be any more accurate than the 3 is so we use the same number of significant digits in the result.

That is why you cannot say the answer is 400000000, because if you did that would mean you are certain about every zero in that result because that's what you wrote down. Since you are not sure of all those zeroes, you can only say the answer is $4\times10^8$, which as written means the result could be anywhere between $4\times10^8$ and $4.\bar9\times10^8$

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    $\begingroup$ I'm used to the convention that something like "3m" is treated as the result of rounding so for the actual length x, 2.5m <= x < 3.5m. $\endgroup$
    – Jasper
    Jun 17 at 8:24
  • $\begingroup$ I believe that the least count is kept in mind while writing uncertain digit. E.g. 3m =2.9m or 3.1m. so why there is 3m= 3.0 to 3.9 in your answer? $\endgroup$
    – Level1
    Jun 17 at 9:00
  • $\begingroup$ @Jasper I think there are different schemes because as someone basically said in the comments: significant digits are kind of a sloppy way to handle things. Maybe they should be considered a lie to children to make them more aware of what they are writing down. $\endgroup$
    – DKNguyen
    Jun 17 at 13:18
  • $\begingroup$ @user359206 That could be true and would make a bit more sense. But it can also depends on the way your measuring instrument works too. There's a lot of different ways to go about things. $\endgroup$
    – DKNguyen
    Jun 17 at 13:24
  • $\begingroup$ Mathematically $4\times10^8 =400000000$. $\endgroup$
    – md2perpe
    Jun 18 at 6:34
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Sometimes the error one might have while conducting measurements is large. Then, you do not need to be performing calculations with accuracy larger than the one allowed by the experimental device. This is the point, if I am not mistaken.

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The concept of significant digits is used when working with measured numbers as opposed to exact numbers. It is a relatively simple way to determine how precisely you will know the result of a calculation involving measured numbers.

Say your numbers $3$ and $123456789$ come from measuring lengths by using a method that that has a precision of one unit. Perhaps you're pacing to measure length and you can measure to the nearest metre. The product of $3$ m and $123456789$ m might represent an area in square metres.

The thing is, if you are only measuring to the lengths to the nearest metre then the length you recorded as $3$ m is almost certainly not exactly three metres. The actual length might be anywhere between $2.5$ m and $3.5$ m. Similarly the other length could be up to half a metre more or less, though as percentage uncertainty this is much, much less important. So the actual area could be anywhere from $2.5 \times 123456788.5 = 3.1 \times 10^8$ to $3.5 \times 123456789.5 = 4.3 \times 10^8$. So, if you only know the shorter length to one significant digit, you would not be able to claim to know the area to the nearest square metre i.e. to nine significant digits. Rounding off to one significant digit, i.e. $400000000$, expresses what the calculated product is and also how precisely you know it. It may look like you are introducing a large error in the product, and you would be if you knew the two initial numbers exactly, but if they are measured values then $370370367$ is almost certainly not the correct product.

There are more sophisticated methods of error analysis.

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    $\begingroup$ I'm sure you know this but the way your text is written basically implies that writing $3$m means $3m\pm0.5m$ which is not true. It's an alternate and more explicit approach to expressing the uncertainty than significant digits. Using significant digits for uncertainty $3m$ is more kind of the default way when there is nothing else mentioned so things don't unknowingly go out of control. $\endgroup$
    – DKNguyen
    Jun 16 at 22:41
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Play with this Rounding Significant Figures Calculator, and you will understand. https://www.calculatorsoup.com/calculators/math/significant-figures-rounding.php

You will have to provide the number of significant figures you want to round to. In your example you rounded to one significant figure.

If you round to four significant figures for example, you will get 370400000 So you will get the first four most significant figures and the rest will be zero. The last significant figure will be rounded: ...37.... will be rounded to ...40....

The number of significant figures you need depends on the application.

If for example, that number is an amount in your bank account, then you want to know it to the last dollar, so you may not want to round at all.

If instead, you are looking at the speed of light trying to get a grasp on its magnitude, then two or three significant figures are enough. Thousands separators will help, as in 299,000,000 m/s. This particular example cries for even more rounding, to 300,000,000 m/s or better 300,000 km/s

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Your example can be useful as an exercise in rounding, but is not a good example of how significant figures are used.

A more typical example is a simple electrical circuit with a battery and a resistor. You find the voltage over the resistor be $U = 7.23\ \mathrm{V}$ and the current be $I = 1.37\ \mathrm{mA}$. You use Ohm's law to calculate the resistance and enter the values into your calculator getting $$ R = \frac{U}{I} = \frac{7.23\ \mathrm{V}}{1.37\ \mathrm{mA}} = 5.277372262\ \mathrm{k\Omega} $$ Then you realize that the calculator has rounded or truncated the value and there is in fact an infinite number of decimals. This makes you wonder how many decimals you should write down. So you start thinking...

After a while you realize that also the Volt and Ampere meters have rounded the values. The real voltage is something between $7.225$ and $7.235\ \mathrm{V},$ and the real current is something between $1.365$ and $1.375\ \mathrm{mA}.$ Therefore the true resistance is something between $$ R_\text{min} = \frac{U_\text{min}}{I_\text{max}} = \frac{7.225\ \mathrm{V}}{1.375\ \mathrm{mA}} = 5.254545454\ldots\ \mathrm{k\Omega} $$ and $$ R_\text{max} = \frac{U_\text{max}}{I_\text{min}} = \frac{7.235\ \mathrm{V}}{1.365\ \mathrm{mA}} = 5.300366300\ldots\ \mathrm{k\Omega}. $$ Obviously it's hardly meaningful to write more than one decimal: $5.3\ \mathrm{k\Omega}$ to which both $R_\text{min}$ and $R_\text{max}$ can be rounded. But the value is probably close to the mean value of $R_\text{min}$ and $R_\text{max},$ which is $5.27745587\ldots\ \mathrm{k\Omega}$ and which is quite close to what you calculated first. So perhaps $5.28\ \mathrm{k\Omega}$ is a good answer...? Or as an interval, e.g. a mean value and an error: $5.28\pm 0.023\ \mathrm{k\Omega}$?

It has become a practice to round the result to the minimum number of significant figures among the input values. It's a rule of thumb, not a mathematically exact rule. In real science a more rigorous analysis should be made.

In the given example, both input values have three significant figures so the rule says that we should use that also for the result, i.e. the answer should be given as $5.28\ \mathrm{k\Omega}.$

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  • $\begingroup$ What if I measure the volume of a cube of length=9cm. How sig fig is going to produce some good answer. I believe the answer would be 729cm³, while using sig fig it would be 700cm³. Are both acceptable? $\endgroup$
    – Level1
    Jun 20 at 16:38
  • $\begingroup$ @user359206. How big is the error in your measure of the cube side length? Do you think that it's $0.5\ \mathrm{cm}$? I don't think so. Probably it's rather $0.05\ \mathrm{cm}$ (half a millimeter). That means that you have 2 significant figures and should answer with $730\ \mathrm{cm}^3.$ $\endgroup$
    – md2perpe
    Jun 21 at 7:38

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