Measuring the significant figures is one of the most important things in experimental science. It is so because we never want to overestimate (or underestimate) the precision of our measurements.

Suppose, we are given to measure the circumference of a circle. We measure it to be $3.14150$ meters. From the knowledge of elementary math you might ask why I even bothered writing the $0$ in the end of that crazy number. The reason is simple. It indicates that I've measured the circumference to the 5 th digit after the decimal point. If I didn’t have the proper instruments to measure the $5$^{th} digit, I'd rather write it to the maximum precision.

I'm not going into a definition right now. But let's start with some examples. Notice the numbers given below:

**Which digits of these numbers tell me how precise my measurements are?**

In the first number ($0.00500$), the significant digits are $5$, $0$ and $0$ ($0.00500$). It actually has $3$ significant figures. Let's see why.

The 0’s after the decimal point (& before $5$) are important to define the number. But it isn’t necessary to tell me how precise the measurement is. Say, $0.00500$ is the measurement in kilograms. $0.00500$ is the same as $5.00$ grams. The zeros before $5$ come solely from the unit convention. There is no difference between $0.00500$ kilograms and $5.00$ grams. But the zeros after $5$ come from the precision of measurement. I didn’t have to write the two zeroes (since $5$ gm is the same thing as $5.00$ gm), but the two zeros indicate that I have the appropriate apparatus to measure two more digits after $5$. If the actual weight was $5.01$ gm, I could've measured it. If I had put only one zero after $5$ ($5.0$ gm), that would mean that I couldn't measure the weight of $5.01$ gm. So I'd rather round it up to the nearest digit.

Let’s talk about the second number ($0.021$). Again, the zeros before $2$ are not significant. Here, the significant figures are $2$ & $1$. The point is you don’t want to count the leading zeros before the first non-zero digit.

The $3$^{rd} number ($710.$) has $3$ significant figures. The decimal point in the end of the number tells you that you have the number of exact $710$, not more or less in your precision of measurement.

Accordingly, the number $7.0$ has two significant figures. The $5$^{th} number ($301.002$) has $6$ significant figures. Even though there are zeros in between, they are actually parts of the measurement.

Now, we can get a feeling about significant figures. It is actually the digits which give us information about the precision of measurements. Let’s move on the some formally written rules of significant digits.

Now let’s see how we round the significant digits when doing arithmetical operations.

If we multiply $2$ numbers with different significant digits,we round up the result to the smallest significant digits. For example:

$5.2$ x $2.34 = 12.168$. Here, $5.2$ has $2$ SFs and $2.34$ has $3$ SFs. So, we should round the result to $2$ SFs. $12.168$ should be written as 12 (after rounding up the 1 after the decimal point).

Another example:

$2.11$ x $3.5 = 7.385$. Rounding up to the $2$ SFs, the result should be written as $7.4$

Same rules apply for division.

For addition, let’s take the following example:

$2.25 + 3.1 = 5.35$, but the least precise measurement is $3.1$ .So, the answer should be written as $5.4$ (after rounding up)

Here is one important thing: For addition or subtraction, least precise measurement is the main issue, rather than the number of significant figures.

Let's add two numbers: $700.2 + 7.34 = 707.54$. Here, $700.2$ has $4$ significant figures and $7.34$ has $3$ significant figures. But the answer should be written with $4$ significant figures – $707.5$, because $700.2$ has least precise measure! Similarly $5000.56 + 5.305 = 5005.87$ (after rounding up)

One last point: We should never round to the significant figures until the end of all operations. Suppose, we have to add $500.56$ and $6.587$ and then divide the result by $3.41$. The steps are:

$(500.56 + 6.587) = 507.147$; we don’t round significant Figure at this point.

$\frac{507.147}{3.41} = 148.6803519$, which should actually be written as $149$, since the least significant figured number has 3 significant figures.

Once these rules are grasped, these can be used to further applications.

**Written By**: Santanu Bosu Antu

Suppose, we are given to measure the circumference of a circle. We measure it to be $3.14150$ meters. From the knowledge of elementary math you might ask why I even bothered writing the $0$ in the end of that crazy number. The reason is simple. It indicates that I've measured the circumference to the 5 th digit after the decimal point. If I didn’t have the proper instruments to measure the $5$

I'm not going into a definition right now. But let's start with some examples. Notice the numbers given below:

$0.00500$ | $0.021$ | $710.$ | $7.0$ | $301.002$ |

In the first number ($0.00500$), the significant digits are $5$, $0$ and $0$ ($0.00500$). It actually has $3$ significant figures. Let's see why.

The 0’s after the decimal point (& before $5$) are important to define the number. But it isn’t necessary to tell me how precise the measurement is. Say, $0.00500$ is the measurement in kilograms. $0.00500$ is the same as $5.00$ grams. The zeros before $5$ come solely from the unit convention. There is no difference between $0.00500$ kilograms and $5.00$ grams. But the zeros after $5$ come from the precision of measurement. I didn’t have to write the two zeroes (since $5$ gm is the same thing as $5.00$ gm), but the two zeros indicate that I have the appropriate apparatus to measure two more digits after $5$. If the actual weight was $5.01$ gm, I could've measured it. If I had put only one zero after $5$ ($5.0$ gm), that would mean that I couldn't measure the weight of $5.01$ gm. So I'd rather round it up to the nearest digit.

Let’s talk about the second number ($0.021$). Again, the zeros before $2$ are not significant. Here, the significant figures are $2$ & $1$. The point is you don’t want to count the leading zeros before the first non-zero digit.

The $3$

Accordingly, the number $7.0$ has two significant figures. The $5$

Now, we can get a feeling about significant figures. It is actually the digits which give us information about the precision of measurements. Let’s move on the some formally written rules of significant digits.

- Any non zero digits and zeros in between are significant. Ex: $0.00502$
- Leading zeros are not significant. Ex: $0.00502$
- Trailing zeros are significant. Ex: $0.500$

Here is a catch. If a number is given with no decimal point and some trailing zeros, the precision is ambiguous. The number $71000$’s precision isn’t so obvious. The last zeros don’t represent precise measurement. It has $2$ significant figures ($7$ & $1$). If we had to make those three zeros significant, we had to write the number in scientific notation including those zeros after $71$, that is $7.1000$ x $10$^{$4$}.

Now let’s see how we round the significant digits when doing arithmetical operations.

If we multiply $2$ numbers with different significant digits,we round up the result to the smallest significant digits. For example:

$5.2$ x $2.34 = 12.168$. Here, $5.2$ has $2$ SFs and $2.34$ has $3$ SFs. So, we should round the result to $2$ SFs. $12.168$ should be written as 12 (after rounding up the 1 after the decimal point).

Another example:

$2.11$ x $3.5 = 7.385$. Rounding up to the $2$ SFs, the result should be written as $7.4$

Same rules apply for division.

For addition, let’s take the following example:

$2.25 + 3.1 = 5.35$, but the least precise measurement is $3.1$ .So, the answer should be written as $5.4$ (after rounding up)

Here is one important thing: For addition or subtraction, least precise measurement is the main issue, rather than the number of significant figures.

Let's add two numbers: $700.2 + 7.34 = 707.54$. Here, $700.2$ has $4$ significant figures and $7.34$ has $3$ significant figures. But the answer should be written with $4$ significant figures – $707.5$, because $700.2$ has least precise measure! Similarly $5000.56 + 5.305 = 5005.87$ (after rounding up)

One last point: We should never round to the significant figures until the end of all operations. Suppose, we have to add $500.56$ and $6.587$ and then divide the result by $3.41$. The steps are:

$(500.56 + 6.587) = 507.147$; we don’t round significant Figure at this point.

$\frac{507.147}{3.41} = 148.6803519$, which should actually be written as $149$, since the least significant figured number has 3 significant figures.

Once these rules are grasped, these can be used to further applications.