Mandatory Lecture on Significant Figures

I can't believe I have to do this, but I do. You are all here because one of your coworkers decided to cause a containment breach. Now we're all stuck in this lecture because some poor security guard stepped a bit too close to an SCP and died—all while in the designated safe zone, too. Now we're in the process of amending it, and part of that is refreshing everyone on how to use sigfigs.

I am Lead Researcher Tanner Wilkins. For most of you this would probably be the only time we're ever going to interact. To those unfamiliar with significant figures, don't worry—I've made this lecture as digestible as possible, and I'm very grateful that you are here. Good containment procedures and good research involves everyone, not just scientists, and this information can very well save your life or others one day.

Now, if you're planning to be a hotshot and ignore this entire scolding, just take note of this: Only round off when you're done calculating, not after each intermediate step. After that, please leave if you're not going to listen. I will take note and I will put you and your work towards a higher standard.

Now, without further ado.

What are Significant Figures?

Significant Figures, or sigfigs, signify how confident we are in a numerical value. We're dealing with a lot of measurements in our work, and not all of our instruments are perfectly precise. Just because they aren't precise, doesn't mean we don't know by how much.

Let's say I give you a measurement of some value. How do you know what's significant?

Non-zero digits are always significant.

Very easy to remember. $453.23$ has 5 significant digits, and $23$ has 2.

Zeros in between any non-zero significant digits are significant too.

This too. $10,001$ has 5 sigfigs, so does $32,012$.

Leading zeros, or zeros to the left of the first non-zero digit, are not significant.

I can give you $000,000,000,000,000,000,000,000,000,001$, and it will only have 1 significant figure. I can start it off with a million zeroes and it will still be 1. Same thing with decimals, $0.000000025$ only has 2 sigfigs.

Trailing zeroes might be significant, depending on how they're measured.

If I give a number of $124,000$ meters, how many sigfigs is that? If I just did a rough estimate and had a ridiculously long stick that was $1,000$ meters long, then the number of sigfigs is 3. If it was $100$ meters long, then it's 4. If I made some poor D-Class measure with a stick that's $0.1$ meters long and made him swear on his life that he was as precise as possible, how many sigfigs do we have? 7 sigfigs (and a terrified D-Class).

To be unambiguously clear with the significance of trailing zeros, we use standard scientific notation.

Where $m$ is a real number $0 < m < 10$, and $n$ is an integer.

$m \times 10^n$

Note: We could also write this is as $m\text{e}n$, with $\text{e}$ being shorthand for "10 to the power of". For example, $1.24 * 10^5$ can be abbreviated as $1.24\text{e}5$

How do we know the sigfigs? Just count the number of digits of $\text{m}$. $1.24 \times 10^5$ has 3 sigfigs, while $1.2400 \times 10^5$ has 5 sigfigs. Same number when you expand it, but here we won't have a containment breach.

A number that is exact has an infinite number of significant figures.

I can count exactly how many idiots caused the containment breach. It's $5$. Normally if I gave you that as a measurement, it would be 1 sigfig, but since I can name them exactly—as with things I can count exactly, the significant figures is infinite.

The most fundamental constants we use, like Planck's constant, Avogadro's number, and $\pi$, have infinite precision; in practice, however, we only use them out to a certain number of significant figures. For example, three sigfigs of Planck's constant ("$h = 6.63 \text{ J/Hz}$") are usually enough for simple calculations, while a physicist might go out to six or so ("$h = 6.62607 \text{ J/Hz}$").

Measurements and Estimations

Measuring

Figure 1

For this part, I've taken the liberty to do a sample measurement for all of you. If you look at Figure 1, I've measured a tiny tool with a ruler. Write down on your notepads how long you think it is.

If you answered "$3\text{ cm}$", you're on the right track, but that is incomplete. If you answered "$2.9\text{ cm}$", you're a bit closer, but not quite there.

In our line of work, we need to be as precise as possible, especially when dealing with dangerous and delicate scenarios. Setting the distance of a hypothetical Foundation personal teleporter into a building as $3.0\text{ m}$ may get you into the room you need to be, but $3.1\text{ m}$ might mean your head will be stuck in a wall.

So! Back to the ruler. The numbered marks show the length in whole centimeters. Very easy. Midway between that is a slightly shorter mark, which signifies $0.5$ centimeters. And each tiny mark signifies $0.1$ centimeters. So our ruler, when you line it up properly, has a resolution or precision of $0.1$ centimeters, or $1$ millimeter.

But wait! If you look closer, the tool lies between the lines! What do we do? This is where estimations come in.

Estimations

Since we can visually see that the length is between two millimeter lines, you can estimate what the length is to another digit, in this case, the hundredths. I myself would estimate the length to be $2.92\text{ cm}$. Any digit would be technically correct, but please use your best judgement—the kind that won't cause a containment breach and kill an innocent security guard.

Now I have $2.92\text{ cm}$ as my measurement. What are the sigfigs? Following the rules, we have 3 significant figures. This will be very important when I explain how to calculate them. This is especially why I told you that $3\text{ cm}$ and $2.9\text{ cm}$ are incomplete. They are not precise enough.

As a sidenote, any digital measurement tools will have the correct number on the display if they're working properly and calibrated correctly. Just make sure you actually know how to read a number.

Calculations

This is where it gets tricky.

Addition and Subtraction

When you add or subtract values together, the rightmost significant digit should be the same place as the leftmost last digit of each of the values.

As an example, let's say I have you add $54.3$ and $2.91$. The right most significant digit places are $54.\bf{3}$ (tenths) and $2.9\bf{1}$ (hundredths). The leftmost one is the tenths digit. Add the two up to $57.21$. Then round it off to the tenths place, so the final calculation is $57.2$.

Multiplication and Division

This one's much easier. Take the amount of significant digits of each value, and the final result should have the same amount of sigfigs as the value with the lowest amount of sigfigs. $23.4$ times $6.$? That's $140.6$, rounded off to a single sigfig and you get $100$.

Other operations like logarithms and transcendental functions

Y'know what, this gets harder. Chances are you won't need to use these for most applications, so this is out of the scope for this lecture. If you need this, please contact anyone from the Math Department.

And yes, I can hear your silent cheers.

Problem Solving

Now, I won't let you leave this lecture until we solve some relatively simple problems taken from actual scenarios. I will show the final answers in scientific notation, but as long as your answers are equivalent and have the same amount of sigfigs, you're fine. Do not forget the units. I do not want to see anyone give me a length of "100". 100 what? Toes? Inches? The length of how much paperwork the containment breach gave me?

Safe

Safe Problem 1

Given one of our containment chambers with the dimensions $4.5\text{ m} \times 4.5\text{ m} \times 4.5\text{ m}$, what is the total volume of the chamber?

Safe Problem 1 solution

We can multiply all those together. $4.5\text{ m} \times 4.5\text{ m} \times 4.5\text{ m} = 91.125\text{ m}^3$. Since each measurement only has $2$ significant figures, our final answer should have only $2$ significant figures. We'll round off $91.125\text{ m}^3$ to $2$ sigfigs, which leads to $91\text{ m}^3$.

Now here's something very very important!

Notice how I only rounded it off at the end? This simple problem shows off how errors will happen. Let's see what happens as an example when we round off after each calculation.

We can first do $4.5\text{ m} \times 4.5\text{ m}$, then multiply that with $4.5\text{ m}$. This should give the same answer due to the associative property of multiplication. This is what we'll do without rounding off first. $4.5\text{ m} \times 4.5\text{ m} = 20.25 \text{ m}^2$. Then $20.25\text{ m}^2 \times 4.5\text{ m} = 91.125\text{ m}^3$, which is what we got earlier.

Now with rounding per step, we get $4.5\text{ m} \times 4.5\text{ m} = 20.25\text{ m}^2$. If we round this off we get $20\text{ m}^2$. Then $20\text{ m}^2 \times 4.5\text{ m} = 90\text{ m}^3$. Look, $90\text{ m}^3$.

That's a whole $1$ cubed meter difference between the two answers. This may seem small for now, but again, with more dangerous margins this can mean the difference between life or death.

Answer: $9.1 \times 10^1\text{ m}^3$

Safe Problem 2

The entire population of an office building has accidentally seen an anomaly, and we need to amnestize them. If we have a headcount of exactly $200$ people, and each person needs an average of $1.275\text{ mL}$ of amnestics, what is the total volume of amnestics we need?

Safe Problem 2 solution

Here, it's $200 \times 1.275\text{ mL} = 255.0\text{ mL}$.

If you answered $300\text{ mL}$, it's probably because you thought that $200$ has only 1 significant figure. However, since we know it's an exact number, it effectively has infinite sigfigs. The $1.275\text{ mL}$ has 4 sigfigs, so our answer is $255.0\text{ mL}$.

If our estimate was $200$ people and not an exact number, then you could assume that it's 1 sigfig and $300\text{ mL}$ could be correct. This ambiguity is why we should be careful with our language.

Answer: $2.550 \times 10^2\text{ mL}$

Euclid

Euclid Problem 1

Figure 2. Please attempt to solve this problem before sliding.

Let's say we lost a lot of the documentation for SCP-9140. You're assigned to the reclamation team and you need to measure SCP-9140-2-1 with the given measuring device. How tall is SCP-9140-2-1? Please use Figure 2.

Euclid Problem 1 solution

The measuring resolution is tricky, and I made this so you gotta think a bit more. Each section is 2 cm, and we can clearly see that SCP-9140-2-1 is between 6 cm and 8 cm. You can technically give any answer between 6 cm and 8 cm while still being correct, but if you're not trying to get a good approximation I will smack you on the head with the ruler in Figure 1.

Technical Answer: Between 6.0 cm and 8.0 cm inclusive.

Answer that will get you not smacked by me: Between 7.4 cm to 7.8 cm.

Euclid Problem 2

As a precaution, SCP-3471 has a possible protocol to apply a layer of waterproof and rustproof paint to preserve the deck. The deck is $11.2\text{ m}$ long, composed of several iron plates, each measuring approximately $3$ meters wide. If $80.0 \text{ mL}$ of paint can cover $15 \text{ m}^2$, how many $\text{ mL}$ of paint do we need to cover the deck with 2 coats?

Euclid Problem 2 solution

Let's first calculate the area of the deck:

$11.2 \text{ m} \times 3 \text{ m} = 33.6 \text{ m}^2$

Now, let's figure out how much paint we need for that:

$33.6 \text{ m}^2 \times {80.0 \text{ mL} \over 15 \text{ m}^2} = 179.2 \text{ mL}$

Since we need two coats:

$179.2 \text{ mL} \times 2 = 358.4 \text{ mL}$

And since the original measurement has 1 significant figure—"approximately 3 meters wide", if you were paying attention like you should have been.

Answer: $4 \times 10^2 \text{ mL}$

Feels off right? But that's what significant figures and proper measurements are for—they tell us how certain we are—without having to deal with formal error propagation. If we want to be more specific, then let's go to SCP-3471 with a tape measure.

Euclid Problem 3

As an indirect measure during an SCP-6050-A event, we prevent any unauthorized personnel from going within $5.00000 \times 10^0 \text{ km}$ from the center of Lake Telaga Warna National Park. What is the total area that we cover?

Euclid Problem 3 solution

The answer to this will depend on what kind of $\pi$ we use. If you caught that, good! If not, that's alright. That's what this problem is supposed to teach you.

If you remember, the area of a circle is $\pi \times r^2$, where $r$ is $5.00000 \times 10^0$. That $5.00000 \times 10^0$ gives us 6 sigfigs total.

Now $\pi$ can be anywhere¹ from 3.14, to 3.1416, to 3.141592653, and all are valid². Let's check all different values.

I will omit the $10^0$ so it's easier to read.

$3.14 \times 5.00000^2 \text{ km}^2 = 78.5 \text{ km}^2$. Here $\pi$ has 3 sigfigs, which matches the answer.

$3.1416 \times 5.00000^2 \text{ km}^2 = 78.54 \text{ km}^2$. Here $\pi$ has 5 sigfigs, so the answer should have that same amount.

$3.141592653 \times 5.00000^2 \text{ km}^2 = 78.539816325 \text{ km}^2$. Here $\pi$ has 10 sigfigs, but our original measurement only has 6, so the answer should have 6.

Answer ($\pi$ = 3.14): $7.85 \times 10^1 \text{ km}^2$

Answer ($\pi$ = 3.1416): $7.8540 \times 10^1 \text{ km}^2$

Answer ($\pi$ = 3.141592653): $7.85398 \times 10^1 \text{ km}^2$

Keter

_{What? I can't say it was SCP-, oh alright. I'll keep their identities hidden too.}

Keter Problem 1

You are now in charge of SCP-███(█), where we recently had a containment breach. In order to make sure preventable containment breaches stay prevented, what would the safe zone be?

The exact effective radius of SCP-███(█) is unknown, but was estimated to be about 4 meters during its last test. If we put a standard 1.50 meter buffer, how far should the safe zone border be from SCP-███(█)?

Keter Problem 1 solution

Answer: 6 meters away

Not fucking 5.5 meters—6 meters. The 4 meter measurement only has one sigfig! It's that easy! That's why we have the buffers! We're not capable of throwing away so many resources to finding the exact danger zone of every single anomaly we encounter. That's why we have safe zones just in case. How this got past everyone in Site-██ is beyond me.

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Did you really think the Keter one would be more difficult? Most of the mistakes that actually happen are exactly like this, not because of some convoluted calculation that we need to use an AIC for. Just make sure to double check your work, triple check your coworker's work, and ask anyone from the Math Department if you feel unsure. I'd rather see you scared asking me a basic question than pale as a ghost after finding out you caused a completely avoidable catastrophe.

Tanner Wilkins out.

Show Footnotes

Footnotes

1. If anyone of you from the Engineering Department says $\pi$ is 3, I will personally get you demoted.

2. Same thing to you astronomists saying $\pi$ is 10.

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‡ Licensing / Citation

Cite this page as:

"Mandatory Lecture on Significant Figures" by sasarpillsa, from the SCP Wiki. Source: https://scpwiki.com/mandatory-lecture-on-significant-figures. Licensed under CC BY-SA.

For information on how to use this component, see the License Box component. To read about licensing policy, see the Licensing Guide.

Figure 1 picture taken by me.

Figure 2 SCP-9140 and original image of SCP-9140 by KatyaStrangelove. Edited by me to only get a single object and measuring device in the image.

SCP-3471 by TallShrimp.

SCP-6050 by nahyesq and PhoenixOfHope.