1.5 Estimation, Common Sense, and Intuition

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Learning Objectives

By the end of this section, you will be able to:

estimate computations without performing them, including using special numbers $\pi$ and $e$.
compare numbers and tell which is bigger or smaller
get the gist of scientific notation
assess possible answers to questions to see which are reasonable
recognize that intuition can fail us

One of the major complaints we hear from science departments about their students' math skill gaps is that students can't estimate numbers or tell when their answers are unreasonable. Maybe this is due to the national infatuation with calculators, but for whatever reason, students are arriving at college without a fluent sense of how numbers relate to each other. A physics student is tasked with calculating the amount of time it takes for a ball to fall to Earth, and after typing a few numbers into their calculator, they write down, "$-304,000$" and move on. They don't stop to realize that a) without checking their units, it's impossible to know for sure if that number is a reasonable amount of time (but I doubt it!!) and b) they shouldn't have gotten a negative answer in the first place! Unless this ball is a time machine. In which case, okay fine, I guess.

In this section, I want to start curing you of some of those blind spots. Hi, meet the real numbers. Get to know them, get a sense of their sizes, and how they relate to each other. Get in the habit of stopping after every problem or even every step and asking yourself, "Is that what I expected to get?" In fact, that's maybe the most important habit you can get into while studying math and science.

While working problems, adopt this routine:

Ask yourself, "What do I expect to see coming next, based on context or my goal for this problem? How do I expect the answer to look?"
Proceed with your attempt.
Pause again and ask yourself, "Was that what I expected to happen? Is my answer even feasible? Does the physical context give me any clues as to what would be reasonable?"

We need to start developing our estimation and common sense skills in order to answer these questions, so let's practice.

Estimation

I want you to read through this material, thinking about each problem on your own as much as possible, before moving on to the answer. It's a no-stakes situation, and you need to start developing the "guts" to take a stab at a problem without a clue. Let's go.

Example $\PageIndex{1}$

If I'm hosting a banquet for 212 guests and the dinner costs $37 a plate, about how much do I need to budget for the catering?

Solution

To calculate the exact cost, I would need to multiply 212 and 37. For a rough estimate, I say 212 is pretty close to 200, and 37 is pretty close to 40. Those numbers are much easier to multiply in my head. I estimate I need about $8000 in my food budget. I could also round 37 down to 35, since I can multiply 2 and 35 in my head pretty easily. My new estimate would be $7000, but I think that's low because I've now rounded down both my numbers! At least I've got a range that gives me a sense of the cost. The actual cost is $7844, so I'm not far off!

Exercise $\PageIndex{1}$

1. I'm about to type $ 21,300 \times 11 $ into my phone calculator, but my screen is broken. My phone says the answer is $256,300$, but I'm not sure the touchscreen was working right. About how much do I think the answer should be, so I'll know if I fatfingered it?

2. I'm buying a car for $4,667 in cash, and also buying an aftermarket stereo from my buddy to put in it for $140. I'm withdrawing cash at the bank right now and they're about to close! About how much should I ask for so I'll be sure to have enough?

Answer

1. Again, I say 11 is close to 10, and I can do $21300 \cdot 10 $ in my head knowing that I just need to tack on another zero to get $213,000 $ as my estimate. This gives me a clue that I must have typed something in wrong when my calculator gives me an answer of $ 256,300 $.

2. I round up to $4700 and $150 and ask the teller for $4850, as that will certainly cover my bases. When the pressure is off, I realize the exact total is $4807, but now I have some cash to hide in my glove box for emergencies. :)

It's also important to me that you get a sense of the most common irrational numbers that come up, namely $\pi$ (the Greek letter "pi") and $ e$ ("Euler's number," a special lil guy you will learn about in logarithms). You don't have to memorize a hundred digits of $\pi$ unless you have a really nerdy date you're trying to impress, but you should know:

\[ \pi \approx 3.14\text{ish} \quad \text{and} \quad e \approx 2.72\text{ish}. \notag \]

It's enough sometimes for our purposes just to think, "$\pi$ is a bit more than 3," or "$e $ is between 2 and 3." The next example might actually prove useful to you if you ever play the game Worldle and struggle to interpret its clues...

Example $\PageIndex{2}$

Fun fact: the radius of the Earth is about 3,959 miles. Using the fact that the circumference of a circle is $2 \pi r$, where $r$ is the radius of the circle, estimate the circumference of the Earth. (About how many miles you would travel if you followed the equator all the way around.)

Here's another fun fact: The point that is exactly on the opposite side of the globe from a location (you've heard of "digging a hole to China"?) is called its antipode. The antipode of New York City is in Australia. About how far, in miles, is it from NYC to Australia?

Solution

Let's round 3,959 up to 4,000 and work with that. Knowing that $ \pi$ is not much more than 3, I can start with a very rough estimate of $2 \cdot 3 \cdot 4000 = 24,000$ miles. Maybe I even feel pretty good about my mental math doubling skills and approximate $ \pi$ as 3.1 instead, to get $ 2 \cdot 3.1 \cdot 4000 = 3.1 \times 8000 = (3 + 0.1)(8000) = 24000 + 800.0 = 24,800 $ miles. (The more you practice the decimal computations in Section 1.3, the more you will get comfortable with these ideas.) Anyway, the real value is around 24,901.55 miles.

If Australia is the antipode of NYC, it's halfway around the world! If the circumference of the Earth is around 25,000 miles, halfway would be about 12,500 miles.

Exercise $\PageIndex{2}$

My investment is expected to make 10% interest, compounded continuously, and I have a formula that computes the amount of money I'll have after 10 years. The formula gives an answer of $100e^{0.1 \cdot 10} = 100e$, but about how much money is that?

Answer: Knowing that $e \approx 2.72$, I need to multiply that by 100 using the trick of moving the decimal two places to the right. So I expect to have about $272. Since we round money to two decimal places, the real answer is $ 100(2.71828...) = $271.83 $

We should have some easy roots memorized, say at least the list below:

$\sqrt{4} = 2$	$ \sqrt{400} = 20 $
$\sqrt{9} = 3 $	$ \sqrt{3600} = 60 $, etc.
$\sqrt{16} = 4 $	$ \sqrt{10000} = 100 $, etc.
$ \sqrt{25} = 5 $	$ \sqrt[3]{8} = 2 $
$ \sqrt{36} = 6 $	$ \sqrt[3]{27} = 3 $
$ \sqrt{49} = 7 $	$ \sqrt[3]{64} = 4 $
$ \sqrt{64} = 8 $	$ \sqrt[3]{125}= 5 $
$ \sqrt{81} = 9 $	$ \sqrt[3]{1000} = 10 $
$\sqrt{100} = 10 $	$ \sqrt[4]{16} = 2 $
$ \sqrt{121} = 11 $	$ \sqrt[4]{81} = 3 $
$ \sqrt{144} = 12 $	$ \sqrt[4]{10000} = 10 $

Roots of other numbers, though, can at least be estimated between two integers, using our knowledge of the aforementioned perfect squares.

Example $\PageIndex{3}$

Estimate the square roots between two integers.

$ \sqrt{2} $
$ \sqrt{5} $
$ \sqrt{30} $

Solution

Well, 2 is between the perfect squares 1 and 4, so $ \sqrt{2}$ must be between their roots, 1 and 2. And since 2 is closer to 1 than it is to 4, I expect $ \sqrt{2}$ to be closer to 1 than to 2. In fact, $ \sqrt{2} = 1.41421456...$.
Again, 5 is between the perfect squares 4 and 9, and closer to 4. Then $ \sqrt{5} $ must be between 2 and 3, and closer to 2. In fact, $ \sqrt{5} = 2.2360679... $.
Since 30 is between the perfect squares 25 and 36, and pretty much in the middle, I expect $ \sqrt{30} $ to be something like 5.5, and in fact, $ \sqrt{30} = 5.477225575...$.

Exercise $\PageIndex{3}$

Estimate the square roots between two integers.

$ \sqrt{12} $
$ \sqrt{78} $
$ \sqrt{52} $
$ \sqrt{150}$

Answer

Between 3 and 4, maybe 3.5. Real value: 3.464101...
78 is almost 81, so I expect its square root to be a bit less than 9. Real value: 8.8317608...
Between 7 and 8 and closer to 7, maybe 7.2. Real value: 7.2111025...
150 is a bit more than 144, whose root is 12, so let's say 12.2ish. Real value: 12.247448...

Bigger or Smaller?

When you get to Calc I, there's a whole section in which you're looking for the maximum and minimum values a function takes on an interval. You'll learn the "Closed Interval Method," during which you'll calculate some values and have to compare them to see which are the max and min. But if those values are expressions with fractions or irrational numbers in them, you'll need to have a sense of what makes a number bigger or smaller than another!

Example $\PageIndex{4}$

Which is bigger?

$ \frac{5}{2} $ vs $ \frac{7}{2} $
$ \frac{7}{4} $ vs $\frac{7}{5} $
$ \frac{5}{4} $ vs $ \frac{13}{12} $

Solution

If two fractions have the same denominator, look at the numerators. A bigger numerator makes a bigger number as a whole. The answer is $ \frac{7}{2} > \frac{5}{2} $.
If two fractions have the same numerator, look at the denominators. A bigger denominator makes a smaller number as a whole. Say I have a pizza that I'm cutting into pieces. Certainly a half of a pizza is bigger than a quarter of a pizza, so $ \frac{1}{2} > \frac{1}{4}$. Great. Now I'm hungry. No more pizzas! The answer is $ \frac{7}{4} > \frac{7}{5}$.
In the second fraction, both the numerator and the denominator are larger, so I can't tell if the number as a whole is bigger or smaller... But if I get a common denominator by multiplying the first fraction top and bottom by 3, I can compare instead $ \frac{15}{12} $ and $\frac{13}{12} $, at which point it's clear the first fraction is a bigger number.

Exercise $\PageIndex{4}$

Which is bigger?

$ \frac{37}{100} $ vs $ \frac{41}{100} $
$ \frac{9}{11} $ vs $ \frac{9}{13} $
$ \frac{7}{6} $ vs $ \frac{5}{4} $

Answer

$ \frac{41}{100} $
$ \frac{9}{11} $
Hint: change both fractions so they have a common denominator and then compare. It's $ \frac{5}{4} $.

Reviewing the section on converting between fractions and decimals will help you with the next exercise.

Exercise $\PageIndex{5}$

Which is bigger?

$2 $ vs $\frac{3}{2} $
$ \frac{5}{4} $ vs $ 1.5 $
$ \frac{105}{50} $ vs $2 $

Answer

1. 2 is bigger since the other is 1.5

2. 1.5 is bigger since $\frac{5}{4}$ is a whole $ \frac{4}{4} = 1 $ plus an extra fourth, so 1.25

3. Hint: Compare $ \frac{105}{50} $ to $\frac{100}{50} $. It's bigger.

Let's go back to $ \pi $ for a challenge round that came up last time I taught Calc I...

Example $\PageIndex{5}$

Which is bigger, $ 2\pi - 1 $ or $ \frac{3}{2} \pi$?

Solution

Let's estimate the value of each option and then compare. If $ \pi $ is 3.14ish, then $2\pi $ is 6.28ish, and subtracting 1 puts me around 5.28ish. Meanwhile, $ \frac{3}{2}\pi = \frac{3\pi}{2} $, and I know that $3\pi$ is bigger than 9 but less than 10... Well, then half of $3\pi$ is certainly less than 5.28ish! The true values are $ 2\pi - 1 = 5.283185... $ and $ \frac{3}{2} \pi = 4.712338... $.

Now you try a combo of ideas...

Exercise $\PageIndex{6}$

Which is bigger, $ 3\sqrt{5} $ or $ 5$?

Answer: 5 is a tad more than 4, so we know $ \sqrt{5} $ is a bit more than 2, which means $ 3\sqrt{5} $ is a bit more than 6. Certainly bigger than 5.

Scientific Notation

If you're doing anything in the sciences, you are going to run into scientific notation. This is a technique we use for writing gigantic numbers or super tiny numbers without getting a hand cramp from writing zeros, like the approximate number of oxygen atoms in a cubic centimeter of air,

\[ 5,300,000,000,000,000,000 \text{ (yikes)} \notag \]

It all comes down to the multiplying/dividing by powers of 10 trick. Observe:

\[ 1.2345 \cdot 10^5 = 1.2345 \cdot 100,000 = 123,450. \notag \]

This means I could write 123,450 in terms of a number with a ones digit and decimal places, multiplying by however many tens I need to be equivalent. This makes my life way easier when dealing with those oxygen atoms, because I can just count up the times I would move the decimal point to get it next to the 5 (it's 18 steps), and write $ 5.3 \cdot 10^{18} $ instead. Finally, notice that as the powers of ten go up, the numbers are increasing fast. $1\cdot 10^7 = 10,000,000$ is a lot bigger than $1\cdot 10^5 = 100,000 $.

In the US, the likelihood that a person who goes to beaches will be attacked by a shark is 1 in 11.5 million (according to Wikipedia), or

\[ \frac{1}{11,500,000} = 0.0000000869565 \notag \]

but that can also be written as $ 8.69565 \cdot 10^{-8} $, by counting how many times I would need to move the decimal to the right to get it to the right of 8. Notice that the more negative the power, the smaller the number, like how $1\cdot 10^{-6} = 0.000001 $ is smaller than $1 \cdot 10^{-3} = 0.001$. You will work with scientific notation in your science classes, but what I want you take away for now is this:

Something times 10 to a positive power is a large number, like the first example. When the power increases, the number increases rapidly!
Something times 10 to a negative power is a small number, like the second example. The more negative the power, the tinier the number!

Common Sense

Always, always, always, ALWAYS, ALWAYS, ALWAYS, I'M BEGGING YOU, always!!! ask yourself if your answer makes common sense at the end of a problem.

Example $\PageIndex{6}$

My washing machine manufacturing business (De Wolf's De Washing Machines) made $120,000 in profits last year. My total revenue was $700,000 and I'm using a spreadsheet to total up my monthly costs because they varied according to season and demand... It says the costs I paid for the year total $650,000. Is that reasonable?

Solution

No, I must have gone wrong somewhere... Revenue is money coming in, and cost is money going out, and profit is what's left over, so Profit = Revenue - Cost. If I had $700,000 in revenue and $650,000 in costs, I would have only made a profit of $700,000-$650,000 = $50,000. That's not what my books say! My spreadsheet must have an error in its entries or formulas.

Exercise $\PageIndex{7}$

Water drains out of a full 5-gallon drum at a rate of 0.16 gallons per minute. I ask you calculate the amount of water left in the tub after half an hour, and you try a first time and get an answer of 6.18 gallons, but you try a second time and get an answer of 4.8 gallons. Are either of these answers reasonable?

Answer

The first one can't be the right number, because we started with 5 gallons and we're losing water, so we sure as heck can't magically have more than we started with! The second answer is at least less than 5, but look at that leaking rate. I'm losing more than 0.1 gal/min, so more than a tenth of a gallon. That means every 10 minutes I'm losing more than a whole gallon. After 30 minutes, I should lose more than 3 gallons, so I must be lower than 4.8! Neither is reasonable.

(What mistake probably happened here? Maybe you said, hey $ 0.16 \frac{\text{gal}}{\text{min}} \cdot 30 \text{ min } = 4.8 $ so that's the answer! You forgot that's the total amount of water that has flowed out, so the amount left in the drum is actually $ 5- 4.8 = 0.2 $ gallons.)

Intuition (a cautionary tale)

You may have been reading through this section and saying to yourself, "Well dagnab hecky durn, De Wolf! You're doing so well with all this rounding and mental math, what do I even need a math class for if I can just learn to guesstimate everything in my head??" If intuition is such a handy tool when solving problems and going through life, what is rigorous detailed math reasoning for?

Great question. Sometimes, intuition is a lie, Steven. A LIE! Let me give you some examples.

You and I are in prison (for being too funny, we're in a dystopia) and there's nothing to do but flip a quarter we found and see if it's heads or tails. I flip it six times in a row and get tails each time. Six tails in a row! You feel in your bones we're "due" for a heads, like it's more likely that the next time I flip it, we'll get heads. In fact, the odds of getting heads are 50% every single time, regardless of previous flips. If I describe the results of six flips by "TTTTTT" for all tails, or "THHTHT" for tails-heads-heads-tails-heads-tails, it ~feels~ like it should be "rarer" to get the TTTTTT outcome. But actually, the odds of getting exactly the TTTTTT outcome are the same as the odds of getting exactly the THHTHT outcome. (My intuition is even worse: after six tails in a row, I start expecting it to be tails every time, assuming I'm sitting in some kind of magical magnetic field singularity or something where everything is tails.)

Or another scenario: I have 25,000 miles worth of rope lying around so I decide to wrap it all the way around the Earth, right on the surface of the Earth. Then I decide I want to raise the rope 1 meter above the ground everywhere, all the way around. How much more rope do I need? I feel like a lot! The Earth is big! But no... The radius of the circle of rope, when it was lying on the ground, was some number of meters, call it $r$. If I want to raise the rope 1 meter above the ground, the new radius would be $r + 1 $. Then the circumference of the circle changes from $2 \pi r$ to $ 2 \pi (r+1) = 2\pi r + 2\pi $, for a change of only about 6.28 meters!

Hopefully in this section you've gained some skills that help you check your answers and impress your friends. Check out the exercises section next!

\(\sqrt{4} = 2\)	\( \sqrt{400} = 20 \)
\(\sqrt{9} = 3 \)	\( \sqrt{3600} = 60 \), etc.
\(\sqrt{16} = 4 \)	\( \sqrt{10000} = 100 \), etc.
\( \sqrt{25} = 5 \)	\( \sqrt[3]{8} = 2 \)
\( \sqrt{36} = 6 \)	\( \sqrt[3]{27} = 3 \)
\( \sqrt{49} = 7 \)	\( \sqrt[3]{64} = 4 \)
\( \sqrt{64} = 8 \)	\( \sqrt[3]{125}= 5 \)
\( \sqrt{81} = 9 \)	\( \sqrt[3]{1000} = 10 \)
\(\sqrt{100} = 10 \)	\( \sqrt[4]{16} = 2 \)
\( \sqrt{121} = 11 \)	\( \sqrt[4]{81} = 3 \)
\( \sqrt{144} = 12 \)	\( \sqrt[4]{10000} = 10 \)

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